Complete Solutions Manual to Accompany
A First Course in Differential
Equations
with Modeling Applications
ELEVENTH EDITION, METRIC VERSION
And
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Differential Equations
with Boundary-Value Problems
NINTH EDITION, METRIC VERSION
Dennis G. Zill
Loyola Marymount University,
Los Angeles, CA
Prepared by
Roberto Martinez
Loyola Marymount University, Los Angeles, CA
Metric Version Prepared by
Aly El-Iraki
Professor Emeritus, Alexandria University, Egypt
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Ninth Edition, Metric Version
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Chapter 1
Introduction to Differential Equations
1.1
Definitions and Terminology
1. Second order; linear
2. Third order; nonlinear because of (dy/dx)4
3. Fourth order; linear
4. Second order; nonlinear because of cos(r + u)
5. Second order; nonlinear because of (dy/dx)2 or
1 + (dy/dx)2
6. Second order; nonlinear because of R2
7. Third order; linear
8. Second order; nonlinear because of ẋ2
9. Writing the differential equation in the form x(dy/dx) + y 2 = 1, we see that it is nonlinear
in y because of y 2 . However, writing it in the form (y 2 − 1)(dx/dy) + x = 0, we see that it is
linear in x.
10. Writing the differential equation in the form u(dv/du) + (1 + u)v = ueu we see that it is linear
in v. However, writing it in the form (v + uv − ueu )(du/dv) + u = 0, we see that it is nonlinear
in u.
11. From y = e−x/2 we obtain y = − 12 e−x/2 . Then 2y + y = −e−x/2 + e−x/2 = 0.
12. From y =
6
5
− 65 e−20t we obtain dy/dt = 24e−20t , so that
dy
+ 20y = 24e−20t + 20
dt
6 6 −20t
− e
5 5
= 24.
13. From y = e3x cos 2x we obtain y = 3e3x cos 2x−2e3x sin 2x and y = 5e3x cos 2x−12e3x sin 2x,
so that y − 6y + 13y = 0.
1
2
CHAPTER 1
INTRODUCTION TO DIFFERENTIAL EQUATIONS
14. From y = − cos x ln(sec x + tan x) we obtain y = −1 + sin x ln(sec x + tan x) and
y = tan x + cos x ln(sec x + tan x). Then y + y = tan x.
15. The domain of the function, found by solving x+2 ≥ 0, is [−2, ∞). From y = 1+2(x+2)−1/2
we have
(y − x)y = (y − x)[1 + (2(x + 2)−1/2 ]
= y − x + 2(y − x)(x + 2)−1/2
= y − x + 2[x + 4(x + 2)1/2 − x](x + 2)−1/2
= y − x + 8(x + 2)1/2 (x + 2)−1/2 = y − x + 8.
An interval of definition for the solution of the differential equation is (−2, ∞) because y is
not defined at x = −2.
16. Since tan x is not defined for x = π/2 + nπ, n an integer, the domain of y = 5 tan 5x is
{x 5x = π/2 + nπ}
or {x x = π/10 + nπ/5}. From y = 25 sec2 5x we have
y = 25(1 + tan2 5x) = 25 + 25 tan2 5x = 25 + y 2 .
An interval of definition for the solution of the differential equation is (−π/10, π/10). Another
interval is (π/10, 3π/10), and so on.
17. The domain of the function is {x 4 − x2 = 0} or {x x = −2 and x = 2}. From
y = 2x/(4 − x2 )2 we have
2
1
= 2xy 2 .
y = 2x
4 − x2
An interval of definition for the solution of the differential equation is (−2, 2). Other intervals
are (−∞, −2) and (2, ∞).
√
18. The function is y = 1/ 1 − sin x , whose domain is obtained from 1 − sin x = 0 or sin x = 1.
Thus, the domain is {x x = π/2 + 2nπ}. From y = − 12 (1 − sin x)−3/2 (− cos x) we have
2y = (1 − sin x)−3/2 cos x = [(1 − sin x)−1/2 ]3 cos x = y 3 cos x.
An interval of definition for the solution of the differential equation is (π/2, 5π/2). Another
one is (5π/2, 9π/2), and so on.
1.1 Definitions and Terminology
3
x
19. Writing ln(2X − 1) − ln(X − 1) = t and differentiating
4
implicitly we obtain
dX
1 dX
2
−
=1
2X − 1 dt
X − 1 dt
2
1
dX
−
=1
2X − 1 X − 1 dt
2
–4
–2
2
4
t
–2
2X − 2 − 2X + 1 dX
=1
(2X − 1) (X − 1) dt
–4
dX
= −(2X − 1)(X − 1) = (X − 1)(1 − 2X).
dt
Exponentiating both sides of the implicit solution we obtain
2X − 1
= et
X −1
2X − 1 = Xet − et
(et − 1) = (et − 2)X
X=
et − 1
.
et − 2
Solving et − 2 = 0 we get t = ln 2. Thus, the solution is defined on (−∞, ln 2) or on (ln 2, ∞).
The graph of the solution defined on (−∞, ln 2) is dashed, and the graph of the solution
defined on (ln 2, ∞) is solid.
y
20. Implicitly differentiating the solution, we obtain
4
dy
dy
− 4xy + 2y
=0
−2x
dx
dx
2
2
−x dy − 2xy dx + y dy = 0
2
–4
2xy dx + (x2 − y)dy = 0.
–2
2
–2
−
−1 = 0
Using the quadratic formula to solve
√
2 √
2
4
for y, we get y = 2x ± 4x + 4 /2 = x ± x4 + 1 .
√
Thus, two explicit solutions are y1 = x2 + x4 + 1 and
√
y2 = x2 − x4 + 1 . Both solutions are defined on (−∞, ∞).
y2
2x2 y
The graph of y1 (x) is solid and the graph of y2 is dashed.
–4
4
x
4
CHAPTER 1
INTRODUCTION TO DIFFERENTIAL EQUATIONS
21. Differentiating P = c1 et / 1 + c1 et we obtain
1 + c1 et c1 et − c1 et · c1 et
1 + c1 et − c1 et
c1 et
dP
=
=
dt
1 + c1 et
1 + c1 et
(1 + c1 et )2
=
c1 et
1 + c1 et
1−
c1 et
= P (1 − P ).
1 + c1 et
22. Differentiating y = 2x2 − 1 + c1 e−2x we obtain
2
dy
2
= 4x − 4xc1 e−2x , so that
dx
dy
2
2
+ 4xy = 4x − 4xc1 e−2x + 8x3 − 4x + 4c1 xe−x = 8x3
dx
23. From y = c1 e2x + c2 xe2x we obtain
4c2 xe2x , so that
dy
d2 y
= (4c1 + 4c2 )e2x +
= (2c1 + c2 )e2x + 2c2 xe2x and
dx
dx2
dy
d2 y
+ 4y = (4c1 + 4c2 − 8c1 − 4c2 + 4c1 )e2x + (4c2 − 8c2 + 4c2 )xe2x = 0.
−4
2
dx
dx
24. From y = c1 x−1 + c2 x + c3 x ln x + 4x2 we obtain
dy
= −c1 x−2 + c2 + c3 + c3 ln x + 8x,
dx
d2 y
= 2c1 x−3 + c3 x−1 + 8,
dx2
and
d3 y
= −6c1 x−4 − c3 x−2 ,
dx3
so that
x3
2
dy
d3 y
2 d y
+
2x
−x
+ y = (−6c1 + 4c1 + c1 + c1 )x−1 + (−c3 + 2c3 − c2 − c3 + c2 )x
dx3
dx2
dx
+ (−c3 + c3 )x ln x + (16 − 8 + 4)x2
= 12x2
In Problems
25–28, we use the Product Rule and the derivative of an integral ((12) of this section):
ˆ x
d
g(t) dt = g(x).
dx a
ˆ x −3t
ˆ x −3t
dy
e−3x 3x
e
e
3x
3x
dt we obtain
= 3e
dt +
· e or
25. Differentiating y = e
t
dx
t
x
1
1
ˆ x −3t
1
e
dy
= 3e3x
dt + , so that
dx
t
x
1
ˆ x −3t
ˆ x −3t 1
e
e
dy
3x
3x
− 3xy = x 3e
dt +
− 3x e
dt
x
dx
t
x
t
1
1
ˆ x −3t
ˆ x −3t
e
e
dt + 1 − 3xe3x
dt = 1
= 3xe3x
t
t
1
1
1.1 Definitions and Terminology
√
ˆ
ˆ
x
cos t
cos x √
√ dt + √ · x or
26. Differentiating y = x
x
t
4
4
ˆ x
1
cos t
dy
√ dt + cos x, so that
= √
dx
2 x 4
t
ˆ x
ˆ x
√
1
cos t
cos t
dy
√ dt + cos x − x
√ dt
√
− y = 2x
2x
dx
2 x 4
t
t
4
ˆ x
ˆ x
√
√
cos t
cos t
√ dt + 2x cos x − x
√ dt = 2x cos x
= x
t
t
4
4
ˆ
ˆ x
dy
5
sin x 10
10 x sin t
sin t
5 10
dt we obtain
=− 2 − 2
dt +
·
or
27. Differentiating y = +
x 1
t
dx
x
x 1
t
x
x
ˆ xx
5
10
10 sin x
sin t
dy
=− 2 − 2
dt +
, so that
dx
x
x 1
t
x2
ˆ
ˆ
5
5 10 x sin t
10 x sin t
10 sin x
2 dy
2
+x
+ xy = x − 2 − 2
dt +
+
dt
x
dx
x
x 1
t
x2
x
x 1
t
ˆ x
ˆ x
sin t
sin t
dt + 10 sin x + 5 + 10
dt = 10 sin x
= −5 − 10
t
t
1
1
ˆ x
ˆ x
dy
2
2
2
−x2
−x2
t2
−x2
−x2
= −2xe −2xe
e dt we obtain
et dt+ex ·e−x
28. Differentiating y = e +e
dx
0
ˆ x0
dy
−x2
−x2
t2
= −2xe
or
− 2xe
e dt + 1, so that
dx
0
ˆ x
ˆ x
dy
−x2
−x2
t2
−x2
−x2
t2
+ 2xy = −2xe
− 2xe
e dt + 1 + 2x e
+e
e dt
dx
0
0
ˆ x
ˆ x
2
−x2
−x2
t2
−x2
−x2
− 2xe
e dt + 1 + 2xe
+ 2xe
et dt = 1
= −2xe
x
1
cos t
dy
√ dt we obtain
= √
dx
2 x
t
0
29. From
0
y=
−x2 , x < 0
x≥0
x2 ,
y =
−2x,
2x,
we obtain
x<0
x≥0
so that xy − 2y = 0.
30. The function y(x) is not continuous at x = 0 since lim y(x) = 5 and lim y(x) = −5. Thus,
y (x) does not exist at x = 0.
x→0−
x→0+
31. Substitute the function y = emx into the equation y + 2y = 0 to get
(emx ) + 2(emx ) = 0
memx + 2emx = 0
emx (m + 2) = 0
Now since emx > 0 for all values of x, we must have m = −2 and so y = e−2x is a solution.
5
6
CHAPTER 1
INTRODUCTION TO DIFFERENTIAL EQUATIONS
32. Substitute the function y = emx into the equation 5y − 2y = 0 to get
5(emx ) − 2(emx ) = 0
5memx − 2emx = 0
emx (5m − 2) = 0
Now since emx > 0 for all values of x, we must have m = 2/5 and so y = e2x/5 is a solution.
33. Substitute the function y = emx into the equation y − 5y + 6y = 0 to get
(emx ) − 5(emx ) + 6(emx ) = 0
m2 emx − 5memx + 6emx = 0
emx (m2 − 5m + 6) = 0
emx (m − 2)(m − 3) = 0
Now since emx > 0 for all values of x, we must have m = 2 or m = 3 therefore y = e2x and
y = e3x are solutions.
34. Substitute the function y = emx into the equation 2y + 7y − 4y = 0 to get
2(emx ) + 7(emx ) − 4(emx ) = 0
2m2 emx + 7memx − 4emx = 0
emx (2m2 + 7m − 4) = 0
emx (m + 4)(2m − 1) = 0
Now since emx > 0 for all values of x , we must have m = −4 or m = 1/2 therefore y = e−4x
and y = ex/2 are solutions.
35. Substitute the function y = xm into the equation xy + 2y = 0 to get
x · (xm ) + 2(xm ) = 0
x · m(m − 1)xm−2 + 2mxm−1 = 0
(m2 − m)xm−1 + 2mxm−1 = 0
xm−1 [m2 + m] = 0
xm−1 [m(m + 1)] = 0
The last line implies that m = 0 or m = −1 therefore y = x0 = 1 and y = x−1 are solutions.
1.1 Definitions and Terminology
36. Substitute the function y = xm into the equation x2 y − 7xy + 15y = 0 to get
x2 · (xm ) − 7x · (xm ) + 15(xm ) = 0
x2 · m(m − 1)xm−2 − 7x · mxm−1 + 15xm = 0
(m2 − m)xm − 7mxm + 15xm = 0
xm [m2 − 8m + 15] = 0
xm [(m − 3)(m − 5)] = 0
The last line implies that m = 3 or m = 5 therefore y = x3 and y = x5 are solutions.
In Problems 37–40, we substitute y = c into the differential equations and use y = 0 and y = 0
37. Solving 5c = 10 we see that y = 2 is a constant solution.
38. Solving c2 + 2c − 3 = (c + 3)(c − 1) = 0 we see that y = −3 and y = 1 are constant solutions.
39. Since 1/(c − 1) = 0 has no solutions, the differential equation has no constant solutions.
40. Solving 6c = 10 we see that y = 5/3 is a constant solution.
41. From x = e−2t + 3e6t and y = −e−2t + 5e6t we obtain
dx
= −2e−2t + 18e6t
dt
and
dy
= 2e−2t + 30e6t .
dt
Then
x + 3y = (e−2t + 3e6t ) + 3(−e−2t + 5e6t ) = −2e−2t + 18e6t =
dx
dt
5x + 3y = 5(e−2t + 3e6t ) + 3(−e−2t + 5e6t ) = 2e−2t + 30e6t =
dy
.
dt
and
42. From x = cos 2t + sin 2t + 15 et and y = − cos 2t − sin 2t − 15 et we obtain
and
1
dx
= −2 sin 2t + 2 cos 2t + et
dt
5
and
1
dy
= 2 sin 2t − 2 cos 2t − et
dt
5
d2 x
1
= −4 cos 2t − 4 sin 2t + et
dt2
5
and
d2 y
1
= 4 cos 2t + 4 sin 2t − et .
dt2
5
Then
1
1
d2 x
4y + et = 4(− cos 2t − sin 2t − et ) + et = −4 cos 2t − 4 sin 2t + et = 2
5
5
dt
and
1
1
d2 y
4x − et = 4(cos 2t + sin 2t + et ) − et = 4 cos 2t + 4 sin 2t − et = 2 .
5
5
dt
7
8
CHAPTER 1
INTRODUCTION TO DIFFERENTIAL EQUATIONS
43. (y )2 + 1 = 0 has no real solutions because (y )2 + 1 is positive for all differentiable functions
y = φ(x).
44. The only solution of (y )2 + y 2 = 0 is y = 0, since if y = 0, y 2 > 0 and (y )2 + y 2 ≥ y 2 > 0.
45. The first derivative of f (x) = ex is ex . The first derivative of f (x) = ekx is kekx . The
differential equations are y = y and y = ky, respectively.
46. Any function of the form y = cex or y = ce−x is its own second derivative. The corresponding
differential equation is y − y = 0. Functions of the form y = c sin x or y = c cos x have second
derivatives that are the negatives of themselves. The differential equation is y + y = 0.
√
47. We first note that 1 − y 2 = 1 − sin2 x = cos2 x = | cos x|. This prompts us to consider
values of x for which cos x < 0, such as x = π. In this case
dy dx x=π
d
(sin x)
=
dx
= cos xx=π = cos π = −1,
x=π
but
√
1 − y 2 |x=π = 1 − sin2 π = 1 = 1.
Thus, y = sin x will only be a solution of y = 1 − y 2 when cos x > 0. An interval of
definition is then (−π/2, π/2). Other intervals are (3π/2, 5π/2), (7π/2, 9π/2), and so on.
48. Since the first and second derivatives of sin t and cos t involve sin t and cos t, it is plausible that
a linear combination of these functions, A sin t + B cos t, could be a solution of the differential
equation. Using y = A cos t − B sin t and y = −A sin t − B cos t and substituting into the
differential equation we get
y + 2y + 4y = −A sin t − B cos t + 2A cos t − 2B sin t + 4A sin t + 4B cos t
= (3A − 2B) sin t + (2A + 3B) cos t = 5 sin t
Thus 3A − 2B = 5 and 2A + 3B = 0. Solving these simultaneous equations we find A =
15
10
and B = − 10
13 . A particular solution is y = 13 sin t − 13 cos t.
15
13
49. One solution is given by the upper portion of the graph with domain approximately (0, 2.6).
The other solution is given by the lower portion of the graph, also with domain approximately
(0, 2.6).
50. One solution, with domain approximately (−∞, 1.6) is the portion of the graph in the second
quadrant together with the lower part of the graph in the first quadrant. A second solution,
with domain approximately (0, 1.6) is the upper part of the graph in the first quadrant. The
third solution, with domain (0, ∞), is the part of the graph in the fourth quadrant.
1.1 Definitions and Terminology
51. Differentiating (x3 + y 3 )/xy = 3c we obtain
xy(3x2 + 3y 2 y ) − (x3 + y 3 )(xy + y)
=0
x2 y 2
3x3 y + 3xy 3 y − x4 y − x3 y − xy 3 y − y 4 = 0
(3xy 3 − x4 − xy 3 )y = −3x3 y + x3 y + y 4
y =
y 4 − 2x3 y
y(y 3 − 2x3 )
.
=
2xy 3 − x4
x(2y 3 − x3 )
52. A tangent line will be vertical where y is undefined, or in this case, where x(2y 3 − x3 ) = 0.
This gives x = 0 or 2y 3 = x3 . Substituting y 3 = x3 /2 into x3 + y 3 = 3xy we get
1
1
x
x3 + x3 = 3x
2
21/3
3
3 3
x = 1/3 x2
2
2
x3 = 22/3 x2
x2 (x − 22/3 ) = 0.
Thus, there are vertical tangent lines at x = 0 and x = 22/3 , or at (0, 0) and (22/3 , 21/3 ).
Since 22/3 ≈ 1.59, the estimates of the domains in Problem 50 were close.
√
√
53. The derivatives of the functions are φ1 (x) = −x/ 25 − x2 and φ2 (x) = x/ 25 − x2 , neither
of which is defined at x = ±5.
54. To determine if a solution curve passes through (0, 3) we let t = 0 and P = 3 in the equation
P = c1 et /(1 + c1 et ). This gives 3 = c1 /(1 + c1 ) or c1 = − 32 . Thus, the solution curve
P =
(−3/2)et
−3et
=
1 − (3/2)et
2 − 3et
passes through the point (0, 3). Similarly, letting t = 0 and P = 1 in the equation for the
one-parameter family of solutions gives 1 = c1 /(1 + c1 ) or c1 = 1 + c1 . Since this equation
has no solution, no solution curve passes through (0, 1).
55. For the first-order differential equation integrate f (x). For the second-order differential equa´ ´
tion integrate twice. In the latter case we get y = ( f (x)dx) dx + c1 x + c2 .
56. Solving for y using the quadratic formula we obtain the two differential equations
y =
1
2 + 2 1 + 3x6
x
and y =
1
2 − 2 1 + 3x6 ,
x
so the differential equation cannot be put in the form dy/dx = f (x, y).
9
10
CHAPTER 1
INTRODUCTION TO DIFFERENTIAL EQUATIONS
57. The differential equation yy − xy = 0 has normal form dy/dx = x. These are not equivalent
because y = 0 is a solution of the first differential equation but not a solution of the second.
58. Differentiating we get y = c1 + 2c2 x and y = 2c2 . Then c2 = y /2 and c1 = y − xy , so
y = y − xy x +
y 2
1
x2 = xy − x2 y 2
and the differential equation is x2 y − 2xy + 2y = 0.
59. (a) Since e−x is positive for all values of x, dy/dx > 0 for all x, and a solution, y(x), of the
differential equation must be increasing on any interval.
2
(b)
dy
dy
2
2
= lim e−x = 0 and lim
= lim e−x = 0. Since dy/dx approaches 0 as
x→∞ dx
x→∞
dx x→−∞
x approaches −∞ and ∞, the solution curve has horizontal asymptotes to the left and
to the right.
lim
x→−∞
(c) To test concavity we consider the second derivative
d
d2 y
=
2
dx
dx
dy
dx
=
d
2
2
e−x = −2xe−x .
dx
Since the second derivative is positive for x < 0 and negative for x > 0, the solution
curve is concave up on (−∞, 0) and concave down on (0, ∞).
y
x
(d)
60. (a) The derivative of a constant solution y = c is 0, so solving 5 − c = 0 we see that c = 5
and so y = 5 is a constant solution.
(b) A solution is increasing where dy/dx = 5 − y > 0 or y < 5. A solution is decreasing
where dy/dx = 5 − y < 0 or y > 5.
61. (a) The derivative of a constant solution is 0, so solving y(a − by) = 0 we see that y = 0 and
y = a/b are constant solutions.
(b) A solution is increasing where dy/dx = y(a − by) = by(a/b − y) > 0 or 0 < y < a/b. A
solution is decreasing where dy/dx = by(a/b − y) < 0 or y < 0 or y > a/b.
1.1
Definitions and Terminology
(c) Using implicit differentiation we compute
d2 y
= y(−by ) + y (a − by) = y (a − 2by).
dx2
Thus d2 y/dx2 = 0 when y = a/2b. Since d2 y/dx2 > 0 for 0 < y < a/2b and d2 y/dx2 < 0
for a/2b < y < a/b, the graph of y = φ(x) has a point of inflection at y = a/2b.
(d)
y
y = a/b
y=0
x
62. (a) If y = c is a constant solution then y = 0, but c2 + 4 is never 0 for any real value of c.
(b) Since y = y 2 + 4 > 0 for all x where a solution y = φ(x) is defined, any solution must
be increasing on any interval on which it is defined. Thus it cannot have any relative
extrema.
(c) Using implicit differentiation we compute d2 y/dx2 = 2yy = 2y(y 2 + 4). Setting
d2 y/dx2 = 0 we see that y = 0 corresponds to the only possible point of inflection.
Since d2 y/dx2 < 0 for y < 0 and d2 y/dx2 > 0 for y > 0, there is a point of inflection
where y = 0.
(d)
y
x
11
12
CHAPTER 1
INTRODUCTION TO DIFFERENTIAL EQUATIONS
63. In Mathematica use
Clear[y]
y[x ]:= x Exp[5x] Cos[2x]
y[x]
y''''[x] − 20y'''[x] + 158y''[x] − 580y'[x] + 841y[x]//Simplify
The output will show y(x) = e5x x cos 2x, which verifies that the correct function was entered,
and 0, which verifies that this function is a solution of the differential equation.
64. In Mathematica use
Clear[y]
y[x ]:= 20Cos[5Log[x]]/x − 3Sin[5Log[x]]/x
y[x]
xˆ3 y'''[x] + 2xˆ2 y''[x] + 20x y'[x] − 78y[x]//Simplify
The output will show y(x) = 20 cos(5 ln x)/x − 3 sin(5 ln x)/x, which verifies that the correct
function was entered, and 0, which verifies that this function is a solution of the differential
equation.
1.2
Initial-Value Problems
1. Solving −1/3 = 1/(1 + c1 ) we get c1 = −4. The solution is y = 1/(1 − 4e−x ).
2. Solving 2 = 1/(1 + c1 e) we get c1 = −(1/2)e−1 . The solution is y = 2/(2 − e−(x+1) ) .
3. Letting x = 2 and solving 1/3 = 1/(4 + c) we get c = −1. The solution is y = 1/(x2 − 1).
This solution is defined on the interval (1, ∞).
4. Letting x = −2 and solving 1/2 = 1/(4 + c) we get c = −2. The solution is y = 1/(x2 − 2).
√
This solution is defined on the interval (−∞, − 2 ).
5. Letting x = 0 and solving 1 = 1/c we get c = 1. The solution is y = 1/(x2 + 1). This solution
is defined on the interval (−∞, ∞).
6. Letting x = 1/2 and solving −4 = 1/(1/4 + c) we get c = −1/2. The solution is y =
√
√
1/(x2 − 1/2) = 2/(2x2 − 1). This solution is defined on the interval (−1/ 2 , 1/ 2 ).
In Problems 7–10, we use x = c1 cos t + c2 sin t and x = −c1 sin t + c2 cos t to obtain a system of
two equations in the two unknowns c1 and c2 .
1.2 Initial-Value Problems
7. From the initial conditions we obtain the system
c1 = −1c2 = 8
The solution of the initial-value problem is x = − cos t + 8 sin t.
8. From the initial conditions we obtain the system
c 2 = 0 − c1 = 1
The solution of the initial-value problem is x = − cos t.
9. From the initial conditions we obtain
√
√
1
1 1
3
3
c1 + c2 = − c2 +
=0
2
2
2 2
2
√
Solving, we find c1 = 3/4 and c2 = 1/4. The solution of the initial-value problem is
√
x = ( 3/4) cos t + (1/4) sin t.
10. From the initial conditions we obtain
√
√
√
2
2
c1 +
c2 = 2
√2
√2
√
2
2
c1 +
c2 = 2 2.
[6pt] −
2
2
Solving, we find c1 = −1 and c2 = 3. The solution of the initial-value problem is x =
− cos t + 3 sin t.
In Problems 11–14, we use y = c1 ex + c2 e−x and y = c1 ex − c2 e−x to obtain a system of two
equations in the two unknowns c1 and c2 .
11. From the initial conditions we obtain
c1 + c 2 = 1
c1 − c2 = 2.
Solving, we find c1 =
3 x
1 −x
2e − 2e .
3
2
and c2 = − 12 . The solution of the initial-value problem is y =
12. From the initial conditions we obtain
ec1 + e−1 c2 = 0
ec1 − e−1 c2 = e.
Solving, we find c1 =
1
2
and c2 = − 12 e2 . The solution of the initial-value problem is
1
1
1
1
y = ex − e2 e−x = ex − e2−x .
2
2
2
2
13
14
CHAPTER 1
INTRODUCTION TO DIFFERENTIAL EQUATIONS
13. From the initial conditions we obtain
e−1 c1 + ec2 = 5
e−1 c1 − ec2 = −5.
Solving, we find c1 = 0 and c2 = 5e−1 . The solution of the initial-value problem is y =
5e−1 e−x = 5e−1−x .
14. From the initial conditions we obtain
c1 + c2 = 0
c1 − c2 = 0.
Solving, we find c1 = c2 = 0. The solution of the initial-value problem is y = 0.
15. Two solutions are y = 0 and y = x3 .
16. Two solutions are y = 0 and y = x2 . (Also, any constant multiple of x2 is a solution.)
2
∂f
= y −1/3 . Thus, the differential equation will have a unique
∂y
3
solution in any rectangular region of the plane where y = 0.
17. For f (x, y) = y 2/3 we have
18. For f (x, y) =
√
xy we have ∂f /∂y =
1
2
x/y . Thus, the differential equation will have a
unique solution in any region where x > 0 and y > 0 or where x < 0 and y < 0.
∂f
1
y
we have
= . Thus, the differential equation will have a unique solution
x
∂y
x
in any region where x = 0.
19. For f (x, y) =
20. For f (x, y) = x + y we have
∂f
= 1. Thus, the differential equation will have a unique
∂y
solution in the entire plane.
21. For f (x, y) = x2 /(4 − y 2 ) we have ∂f /∂y = 2x2 y/(4 − y 2 )2 . Thus the differential equation
will have a unique solution in any region where y < −2, −2 < y < 2, or y > 2.
22. For f (x, y) =
∂f
−3x2 y 2
x2
we have
. Thus, the differential equation will have a
=
3
1+y
∂y
(1 + y 3 )2
unique solution in any region where y = −1.
23. For f (x, y) =
x2
y2
∂f
2x2 y
=
we have
. Thus, the differential equation will have a
2
+y
∂y
(x2 + y 2 )2
unique solution in any region not containing (0, 0).
1.2 Initial-Value Problems
24. For f (x, y) = (y + x)/(y − x) we have ∂f /∂y = −2x/(y − x)2 . Thus the differential equation
will have a unique solution in any region where y < x or where y > x.
In Problems 25–28, we identify f (x, y) = y 2 − 9 and ∂f /∂y = y/ y 2 − 9. We see that f and
∂f /∂y are both continuous in the regions of the plane determined by y < −3 and y > 3 with no
restrictions on x.
25. Since 4 > 3, (1, 4) is in the region defined by y > 3 and the differential equation has a unique
solution through (1, 4).
26. Since (5, 3) is not in either of the regions defined by y < −3 or y > 3, there is no guarantee
of a unique solution through (5, 3).
27. Since (2, −3) is not in either of the regions defined by y < −3 or y > 3, there is no guarantee
of a unique solution through (2, −3).
28. Since (−1, 1) is not in either of the regions defined by y < −3 or y > 3, there is no guarantee
of a unique solution through (−1, 1).
29. (a) A one-parameter family of solutions is y = cx. Since y = c, xy = xc = y and y(0) =
c · 0 = 0.
(b) Writing the equation in the form y = y/x, we see that R cannot contain any point on the
y-axis. Thus, any rectangular region disjoint from the y-axis and containing (x0 , y0 ) will
determine an interval around x0 and a unique solution through (x0 , y0 ). Since x0 = 0 in
part (a), we are not guaranteed a unique solution through (0, 0).
(c) The piecewise-defined function which satisfies y(0) = 0 is not a solution since it is not
differentiable at x = 0.
d
tan (x + c) = sec2 (x + c) = 1+tan2 (x + c), we see that y = tan (x + c) satisfies
30. (a) Since
dx
the differential equation.
(b) Solving y(0) = tan c = 0 we obtain c = 0 and y = tan x. Since tan x is discontinuous at
x = ±π/2, the solution is not defined on (−2, 2) because it contains ±π/2.
(c) The largest interval on which the solution can exist is (−π/2, π/2).
1
1 1
d
is a solution of the differ= y 2 , we see that y = −
−
=
31. (a) Since
dx
x+c
(x + c)2
x+c
ential equation.
(b) Solving y(0) = −1/c = 1 we obtain c = −1 and y = 1/(1−x). Solving y(0) = −1/c = −1
we obtain c = 1 and y = −1/(1+x). Being sure to include x = 0, we see that the interval
of existence of y = 1/(1 − x) is (−∞, 1), while the interval of existence of y = −1/(1 + x)
is (−1, ∞).
15
16
CHAPTER 1
INTRODUCTION TO DIFFERENTIAL EQUATIONS
(c) By inspection we see that y = 0 is a solution on (−∞, ∞).
32. (a) Applying y(1) = 1 to y = −1/ (x + c) gives
1=−
1
1+c
or
1 + c = −1
Thus c = −2 and
y=−
1
1
=
.
x−2
2−x
(b) Applying y(3) = −1 to y = −1/ (x + c) gives
−1 = −
1
3+c
or
3 + c = 1.
Thus c = −2 and
y=−
1
1
=
.
x−2
2−x
(c) No, they are not the same solution. The interval I of definition for the solution in part
(a) is (−∞, 2); whereas the interval I of definition for the solution in part (b) is (2, ∞).
See the figure.
33. (a) Differentiating 3x2 − y 2 = c we get 6x − 2yy = 0 or yy = 3x.
(b) Solving 3x2 − y 2 = 3 for y we get
y
y = φ1 (x) = 3(x2 − 1) ,
1 < x < ∞,
4
y = φ2 (x) = − 3(x2 − 1) ,
1 < x < ∞,
2
y = φ3 (x) =
3(x2 − 1) ,
y = φ4 (x) = − 3(x2 − 1) ,
−∞ < x < −1,
–4
–2
2
4
2
4
x
–2
−∞ < x < −1.
–4
(c) Only y = φ3 (x) satisfies y(−2) = 3.
34. (a) Setting x = 2 and y = −4 in 3x2 − y 2 = c we get
y
12 − 16 = −4 = c, so the explicit solution is
4
y = − 3x2 + 4 , −∞ < x < ∞.
2
–4
–2
–2
–4
x
1.2 Initial-Value Problems
(b) Setting c = 0 we have y =
√
√
3x and y = − 3x, both defined on (−∞, ∞).
In Problems 35–38, we consider the points on the graphs with x-coordinates x0 = −1, x0 = 0,
and x0 = 1. The slopes of the tangent lines at these points are compared with the slopes given by
y (x0 ) in (a) through (f).
35. The graph satisfies the conditions in (b) and (f).
36. The graph satisfies the conditions in (e).
37. The graph satisfies the conditions in (c) and (d).
38. The graph satisfies the conditions in (a).
In Problems 39–44 y = c1 cos 2x + c2 sin 2x is a two parameter family of solutions of the secondorder differential equation y + 4y = 0. In some of the problems we will use the fact that
y = −2c1 sin 2x + 2c2 cos 2x.
π
39. From the boundary conditions y(0) = 0 and y
= 3 we obtain
4
y(0) = c1 = 0
y
π
π
π
= c1 cos
+ c2 sin
= c2 = 3.
4
2
2
Thus, c1 = 0, c2 = 3, and the solution of the boundary-value problem is y = 3 sin 2x.
40. From the boundary conditions y(0) = 0 and y(π) = 0 we obtain
y(0) = c1 = 0
y(π) = c1 = 0.
Thus, c1 = 0, c2 is unrestricted, and the solution of the boundary-value problem is y =
c2 sin 2x, where c2 is any real number.
41. From the boundary conditions y (0) = 0 and y π6 = 0 we obtain
y (0) = 2c2 = 0
y
√
π
π
= −2c1 sin
= − 3 c1 = 0.
6
3
Thus, c2 = 0, c1 = 0, and the solution of the boundary-value problem is y = 0.
42. From the boundary conditions y(0) = 1 and y (π) = 5 we obtain
y(0) = c1 = 1
y (π) = 2c2 = 5.
5
5
Thus, c1 = 1, c2 = , and the solution of the boundary-value problem is y = cos 2x + sin 2x.
2
2
17
18
CHAPTER 1
INTRODUCTION TO DIFFERENTIAL EQUATIONS
43. From the boundary conditions y(0) = 0 and y(π) = 2 we obtain
y(0) = c1 = 0
y(π) = c1 = 2.
Since 0 = 2, this is not possible and there is no solution.
44. From the boundary conditions y π
2
= 1 and y (π) = 0 we obtain
y
π
= 2c2 = −1
2
y (π) = 2c2 = 0.
Since 0 = −1, this is not possible and there is no solution.
45. Integrating y = 8e2x + 6x we obtain
ˆ
y = (8e2x + 6x) dx = 4e2x + 3x2 + c.
Setting x = 0 and y = 9 we have 9 = 4 + c so c = 5 and y = 4e2x + 3x2 + 5.
46. Integrating y = 12x − 2 we obtain
ˆ
y = (12x − 2) dx = 6x2 − 2x + c1 .
Then, integrating y we obtain
ˆ
y = (6x2 − 2x + c1 ) dx = 2x3 − x2 + c1 x + c2 .
At x = 1 the y-coordinate of the point of tangency is y = −1 + 5 = 4. This gives the initial
condition y(1) = 4. The slope of the tangent line at x = 1 is y (1) = −1. From the initial
conditions we obtain
2 − 1 + c1 + c 2 = 4
or
c1 + c 2 = 3
6 − 2 + c1 = −1
or
c1 = −5.
and
Thus, c1 = −5 and c2 = 8, so y = 2x3 − x2 − 5x + 8.
47. When x = 0 and y = 12 , y = −1, so the only plausible solution curve is the one with negative
slope at (0, 12 ), or the red curve.
1.2 Initial-Value Problems
48. We note that the initial condition y(0) = 0,
ˆ y
1
√
dt
0=
t3 + 1
0
is satisfied only when y = 0. For any y > 0, necessarily
ˆ y
1
√
dt > 0
3
t +1
0
because the integrand is positive on the interval of integration. Then from (12) of Section 1.1
and the Chain Rule we have:
d
d
x=
dx
dx
ˆ
y
0
√
1
dt
+1
t3
and
dy
1= 3
dx
y +1
1
dy
= y3 + 1
dx
√
dy y (0) =
= (y(0))3 + 1 = 0 + 1 = 1.
dx x=0
Computing the second derivative, we see that:
dy
3y 2
d 3
3y 2
3
d2 y
=
·
=
y
+
1
=
y3 + 1 = y2
3
3
dx2
dx
dx
2
2 y +1
2 y +1
3
d2 y
= y2 .
2
dx
2
This is equivalent to 2
d2 y
− 3y 2 = 0.
dx2
49. If the solution is tangent to the x-axis at (x0 , 0), then y = 0 when x = x0 and y = 0.
Substituting these values into y + 2y = 3x − 6 we get 0 + 0 = 3x0 − 6 or x0 = 2.
50. The theorem guarantees a unique (meaning single) solution through any point. Thus, there
cannot be two distinct solutions through any point.
51. When y =
1 4
16 x ,
y = 14 x3 = x( 14 x2 ) = xy 1/2 , and y(2) =
⎧
⎪
x<0
⎨0,
y=
1
⎪
⎩ x4 , x ≥ 0
16
we have
y =
⎧
⎨0,
⎩ 1 x3 ,
4
x<0
x≥0
=x
1
16 (16)
⎧
⎨0,
x<0
⎩ 1 x2 ,
x≥0
4
= 1. When
= xy 1/2 ,
1
and y(2) = 16
(16) = 1. The two different solutions are the same on the interval (0, ∞), which
is all that is required by Theorem 1.2.1.
19
20
CHAPTER 1
1.3
1.
INTRODUCTION TO DIFFERENTIAL EQUATIONS
Differential Equations as Mathematical Models
dP
= kP + r;
dt
dP
= kP − r
dt
2. Let b be the rate of births and d the rate of deaths. Then b = k1 P and d = k2 P . Since
dP/dt = b − d, the differential equation is dP/dt = k1 P − k2 P .
3. Let b be the rate of births and d the rate of deaths. Then b = k1 P and d = k2 P 2 . Since
dP/dt = b − d, the differential equation is dP/dt = k1 P − k2 P 2 .
4.
dP
= k1 P − k2 P 2 − h, h > 0
dt
5. From the graph in the text we estimate T0 = 180◦ and Tm = 75◦ . We observe that when
T = 85, dT /dt ≈ −1. From the differential equation we then have
k=
−1
dT /dt
= −0.1.
=
T − Tm
85 − 75
6. By inspecting the graph in the text we take Tm to be Tm (t) = 80 − 30 cos (πt/12). Then the
temperature of the body at time t is determined by the differential equation
π dT
= k T − 80 − 30 cos
t
,
dt
12
t > 0.
7. The number of students with the flu is x and the number not infected is 1000 − x, so dx/dt =
kx(1000 − x).
8. By analogy, with the differential equation modeling the spread of a disease, we assume that
the rate at which the technological innovation is adopted is proportional to the number of
people who have adopted the innovation and also to the number of people, y(t), who have
not yet adopted it. If one person who has adopted the innovation is introduced into the
population, then x + y = n + 1 and
dx
= kx(n + 1 − x),
dt
x(0) = 1.
9. The rate at which salt is leaving the tank is
A
A
kg/L =
kg/min.
Rout (10 L/min) ·
1000
100
Thus dA/dt = A/100. The initial amount is A(0) = 25.
10. The rate at which salt is entering the tank is
Rin = (10 L/min) · (0.25 kg/L) = 2.5 kg/min.
1.3
Differential Equations as Mathematical Models
Since the solution is pumped out at a slower rate, it is accumulating at the rate of (10 −
7.5)L/min = 2.5 L/min. After t minutes there are 1000 + 2.5t gallons of brine in the tank.
The rate at which salt is leaving is
A
7.5A
kg/L =
kg/min.
Rout = (7.5 L/min) ·
1000 + 2.5t
1000 + 2.5t
The differential equation is
7.5A
dA
= 2.5 −
.
dt
1000 + 2.5t
11. The rate at which salt is entering the tank is
Rin = (10 L/min) · (0.25 kg/L) = 2.5 kg/min.
Since the tank loses liquid at the net rate of
10 L/min − 12 L/min = −2 L/min,
after t minutes the number of liters of brine in the tank is 1000 − 2t liters. Thus the rate at
which salt is leaving is
12A
6A
A
kg/L · (12 L/min) =
kg/min =
kg/min.
Rout =
1000 − 2t
1000 − 2t
500 − t
The differential equation is
6A
dA
= 2.5 −
dt
500 − t
or
dA
6
+
A = 2.5.
dt
500 − t
12. The rate at which salt is entering the tank is
Rin = (cin kg/L) · (rin L/min) = cin rin kg/min.
Now let A(t) denote the number of kilograms of salt and N (t) the number of liters of brine
in the tank at time t. The concentration of salt in the tank as well as in the outflow is
c(t) = x(t)/N (t). But the number of liters of brine in the tank remains steady, is increased,
or is decreased depending on whether rin = rout , rin > rout , or rin < rout . In any case, the
number of liters of brine in the tank at time t is N (t) = N0 + (rin − rout )t. The output rate
of salt is then
A
A
kg/L · (rout L/min) = rout
kg/min.
Rout =
N0 + (rin − rout )t
N0 + (rin − rout )t
The differential equation for the amount of salt, dA/dt = Rin − Rout , is
dA
A
= cin rin − rout
dt
N0 + (rin − rout )t
or
dA
rout
+
A = cin rin .
dt
N0 + (rin − rout )t
21
22
CHAPTER 1
INTRODUCTION TO DIFFERENTIAL EQUATIONS
13. The volume of water in the tank at time t is V = Aw h. The differential equation is then
1 dV
1
cAh dh
−cAh 2gh = −
=
=
2gh .
dt
Aw dt
Aw
Aw
Using Ah = π
50
1000
2
= 0.0025 π , Aw = 42 = 16, and g = 9.8, this becomes
√
dh
= −0.00217 c h .
dt
14. The volume of water in the tank at time t is V = 13 πr2 h where r is the radius of the tank
at height h. From the figure in the text we see that r/h = 3/6 so that r = 0.5h and
1
πh3 . Differentiating with respect to t we have dV /dt = 14 πh2 dh/dt or
V = 13 π (0.5h)2 h = 12
4 dV
dh
=
.
dt
πh2 dt
50 2
√
, and g = 9.8.
From Problem 13 we have dV /dt = −cAh 2gh where c = 0.6, Ah = π 1000
√
Thus dV /dt = −0.00664π h and
√ 4
0.0266
dh
=
−0.00664
h = − 3/2 .
2
dt
πh
h
15. Since i = dq/dt and L d2 q/dt2 + R dq/dt = E(t), we obtain L di/dt + Ri = E(t).
16. By Kirchhoff’s second law we obtain R
1
dq
+ q = E(t).
dt
C
17. From Newton’s second law we obtain m
dv
= −kv 2 + mg.
dt
18. Since the barrel in Figure 1.3.17(b) in the text is submerged an additional y feet below
its equilibrium position the number of cubic feet in the additional submerged portion is
the volume of the circular cylinder: π×(radius)2 ×height or π(s/2)2 y. Then we have from
Archimedes’ principle
upward force of water on barrel = weight of water displaced
= (9810) × (volume of water displaced)
= (9810)π(s/2)2 y = 2452.5πs2 y.
It then follows from Newton’s second law that
w d2 y
= −2452.5πs2 y
g dt2
or
d2 y 2452.5πs2 g
y = 0,
+
dt2
w
where g = 9.8 and w is the weight of the barrel in pounds.
1.3
Differential Equations as Mathematical Models
19. The net force acting on the mass is
F = ma = m
d2 x
= −k(s + x) + mg = −kx + mg − ks.
dt2
Since the condition of equilibrium is mg = ks, the differential equation is
m
d2 x
= −kx.
dt2
20. From Problem 19, without a damping force, the differential equation is m d2 x/dt2 = −kx.
With a damping force proportional to velocity, the differential equation becomes
m
dx
d2 x
= −kx − β
dt2
dt
or
m
d2 x
dx
+ kx = 0.
+β
dt2
dt
21. As the rocket climbs (in the positive direction), it spends its amount of fuel and therefore the
mass of the fuel changes with time. The air resistance acts in the opposite direction of the
motion and the upward thrust R works in the same direction. Using Newton’s second law we
get
d
(mv) = −mg − kv + R
dt
Now because the mass is variable, we must use the product rule to expand the left side of the
equation. Doing so gives us the following:
d
(mv) = −mg − kv + R
dt
v×
dv
dm
+m×
= −mg − kv + R
dt
dt
The last line is the differential equation we wanted to find.
22. (a) Since the mass of the rocket is m(t) = mp + mv + mf (t), take the time rate-of-change
and get, by straight-forward calculation,
d
d
d
m(t) = (mp + mv + mf (t)) = 0 + 0 + mf (t) = mf (t)
dt
dt
dt
Therefore the rate of change of the mass of the rocket is the same as the rate of change
of the mass of the fuel which is what we wanted to show.
(b) The fuel is decreasing at the constant rate of λ and so from part (a) we have
d
d
m(t) = mf (t) = −λ
dt
dt
m(t) = −λt + c
23
24
CHAPTER 1
INTRODUCTION TO DIFFERENTIAL EQUATIONS
Using the given condition to solve for c, m(0) = 0 + c = m0 and so m(t) = −λt + m0 .
The differential equation in Problem 21 now becomes
v
dv
dm
+ m + kv = −mg + R
dt
dt
−λv + (−λt + m0 )
(−λt + m0 )
dv
+ kv = −mg + R
dt
dv
+ (k − λ)v = −mg + R
dt
k−λ
dv
−mg
R
+
v=
+
dt
−λt + m0
−λt + m0 −λt + m0
k−λ
R
dv
+
v = −g +
dt
−λt + m0
−λt + m0
d
(c) From part (b) we have that dt
mf (t) = −λ and so by integrating this result we get
mf (t) = −λt + c. Now at time t = 0, mf (0) = 0 + c = c therefore mf (t) = −λt + mf (0) .
At some later time tb we then have mf (tb ) = −λtb + mf (0) = 0 and solving this equation
for that time we get tb = mf (0)/λ which is what we wanted to show.
23. From g = k/R2 we find k = gR2 . Using a = d2 r/dt2 and the fact that the positive direction
is upward we get
k
gR2
d2 r
=
−a
=
−
=
−
dt2
r2
r2
or
d2 r gR2
+ 2 = 0.
dt2
r
24. The gravitational force on m is F = −kMr m/r2 . Since Mr = 4πδr3 /3 and M = 4πδR3 /3 we
have Mr = r3 M/R3 and
F = −k
r3 M m/R3
mM
Mr m
=
−k
= −k 3 r.
2
2
r
r
R
Now from F = ma = d2 r/dt2 we have
m
mM
d2 r
= −k 3 r
dt2
R
or
d2 r
kM
= − 3 r.
dt2
R
25. The differential equation is
dA
= k(M − A).
dt
26. The differential equation is
dA
= k1 (M − A) − k2 A.
dt
27. The differential equation is x (t) = r − kx(t) where k > 0.
1.3
Differential Equations as Mathematical Models
28. Consider the right triangle fromed by the waterskier (P ),
the boat (B), and the point on the x-axis directly below
the waterskier. Using Pythagorean Theorem we have that
the base of the triangle on the x-axis has length a2 − y 2 .
Therefore the slope of the line tangent to curve C is
y = −
a2
y
− y2
Notice that the sign of the derivative is negative because as the boat proceeds along the
positive x-axis, the y-coordinate decreases.
29. We see from the figure that 2θ + α = π. Thus
y
y
2 tan θ
= tan α = tan(π − 2θ) = − tan 2θ = −
.
−x
1 − tan2 θ
Since the slope of the tangent line is y = tan θ we have
y/x = 2y [1−(y )2 ] or y−y(y )2 = 2xy , which is the quadratic
equation y(y )2 + 2xy − y = 0 in y . Using the quadratic
(x, y)
θ
θ α
formula, we get
y =
−2x ±
−x ± x2 + y 2
4x2 + 4y 2
=
.
2y
y
θ
x
α
y
φ
x
Since dy/dx > 0, the differential equation is
dy
−x + x2 + y 2
=
dx
y
or
y
dy 2
− x + y 2 + x = 0.
dx
30. The differential equation is dP/dt = kP , so from Problem 41 in Exercises 1.1, a one-parameter
family of solutions is P = cekt .
31. The differential equation in (3) is dT /dt = k(T − Tm ). When the body is cooling, T > Tm ,
so T − Tm > 0. Since T is decreasing, dT /dt < 0 and k < 0. When the body is warming,
T < Tm , so T − Tm < 0. Since T is increasing, dT /dt > 0 and k < 0.
32. The differential equation in (8) is dA/dt = 2.5 − A/100. If A(t) attains a maximum, then
dA/dt = 0 at this time and A = 250. If A(t) continues to increase without reaching a
maximum, then A (t) > 0 for t > 0 and A cannot exceed 250. In this case, if A (t) approaches
0 as t increases to infinity, we see that A(t) approaches 250 as t increases to infinity.
33. This differential equation could describe a population that undergoes periodic fluctuations.
25
26
CHAPTER 1
INTRODUCTION TO DIFFERENTIAL EQUATIONS
34. (a) As shown in Figure 1.3.23(a) in the text, the resultant of the reaction force of magnitude
F and the weight of magnitude mg of the particle is the centripetal force of magnitude
mω 2 x. The centripetal force points to the center of the circle of radius x on which the
particle rotates about the y-axis. Comparing parts of similar triangles gives
F cos θ = mg
and F sin θ = mω 2 x.
(b) Using the equations in part (a) we find
tan θ =
mω 2 x
ω2 x
F sin θ
=
=
F cos θ
mg
g
or
dy
ω2 x
=
.
dx
g
35. From Problem 23, d2 r/dt2 = −gR2 /r2 . Since R is a constant, if r = R + s, then d2 r/dt2 =
d2 s/dt2 and, using a Taylor series, we get
d2 s
R2
2gs
+ ··· .
=
−g
= −gR2 (R + s)−2 ≈ −gR2 [R−2 − 2sR−3 + · · · ] = −g +
dt2
(R + s)2
R
Thus, for R much larger than s, the differential equation is approximated by d2 s/dt2 = −g.
36. (a) If ρ is the mass density of the raindrop, then m = ρV and
dV
d 4 3
dm
dr dr
=ρ
=ρ
πr = ρ 4πr2
.
= ρS
dt
dt
dt 3
dt
dt
If dr/dt is a constant, then dm/dt = kS where ρ dr/dt = k or dr/dt = k/ρ. Since the
radius is decreasing, k < 0. Solving dr/dt = k/ρ we get r = (k/ρ)t + c0 . Since r(0) = r0 ,
c0 = r0 and r = kt/ρ + r0 .
d
[mv] = mg, where v is the velocity of the raindrop. Then
dt
dv
dm
4
dv
4
+v
= mg
or
ρ πr3
+ v(k4πr2 ) = ρ πr3 g.
m
dt
dt
3
dt
3
(b) From Newton’s second law,
Dividing by 4ρπr3 /3 we get
dv 3k
+ v=g
dt
ρr
or
dv
3k/ρ
+
v = g, k < 0.
dt
kt/ρ + r0
37. We assume that the plow clears snow at a constant rate of k cubic kilometers per hour. Let
t be the time in hours after noon, x(t) the depth in miles of the snow at time t, and y(t)
the distance the plow has moved in t hours. Then dy/dt is the velocity of the plow and the
assumption gives
dy
= k,
wx
dt
where w is the width of the plow. Each side of this equation simply represents the volume
of snow plowed in one hour. Now let t0 be the number of hours before noon when it started
1.3
Differential Equations as Mathematical Models
snowing and let s be the constant rate in kilometers per hour at which x increases. Then for
t > −t0 , x = s(t + t0 ). The differential equation then becomes
k
1
dy
=
.
dt
ws t + t0
Integrating, we obtain
k
[ ln(t + t0 ) + c ]
ws
where c is a constant. Now when t = 0, y = 0 so c = − ln t0 and
t
k
ln 1 +
.
y=
ws
t0
y=
Finally, from the fact that when t = 1, y = 2 and when t = 2, y = 3, we obtain
2 2
1 3
1+
= 1+
.
t0
t0
Expanding and simplifying gives t20 + t0 − 1 = 0. Since t0 > 0, we find t0 ≈ 0.618 hours ≈ 37
minutes. Thus it started snowing at about 11:23 in the morning.
38. At time t, when the population is 2 million cells, the differential equation P (t) = 0.15P (t)
gives the rate of increase at time t. Thus, when P (t) = 2 (million cells), the rate of increase
is P (t) = 0.15(2) = 0.3 million cells per hour or 300,000 cells per hour.
39. Setting A (t) = −0.002 and solving A (t) = −0.0004332A(t) for A(t), we obtain
A(t) =
−0.002
A (t)
=
≈ 4.6 grams.
−0.0004332
−0.0004332
dP
= kP is linear
dt
(2) :
dA
= kA is linear
dt
(3) :
dT
= k (T − Tm ) is linear
dt
(5) :
dx
= kx (n + 1 − x) is nonlinear
dt
(6) :
dX
dA
A
= k (α − X) (β − X) is nonlinear (8) :
=6−
is linear
dt
dt
100
40. (1) :
(10) :
dh
Ah =−
2gh is nonlinear
dt
Aw
(11) : L
(12) :
d2 s
= −g is linear
dt2
(14) : m
(15) : m
d2 s
ds
= mg is linear
+k
2
dt
dt
(16) :
1
d2 q
dq
+ R + q = E(t) is linear
2
dt
dt
C
dv
= mg − kv is linear
dt
d2 x 64
− x = 0 is linear
dt2
L
(17) : linearity or nonlinearity is determined by the manner in which W and T1 involve x.
27
28
CHAPTER 1
INTRODUCTION TO DIFFERENTIAL EQUATIONS
Chapter 1 in Review
y
d
c1 e10x = 10 c1 e10x ;
1.
dx
dy
= 10y
dx
y
d
2.
(5 + c1 e−2x ) = −2c1 e−2x = −2(5 + c1 e−2x −5);
dx
3.
dy
= −2(y − 5)
dx
or
dy
= −2y + 10
dx
d
(c1 cos kx + c2 sin kx) = −kc1 sin kx + kc2 cos kx;
dx
y
d2
2
2
2
(c
cos
kx
+
c
sin
kx)
=
−k
c
cos
kx
−
k
c
sin
kx
=
−k
(
cos
kx
+
c
sin
kx);
c
1
2
1
2
1
2
dx2
d2 y
= −k 2 y
dx2
4.
or
d2 y
+ k2 y = 0
dx2
d
(c1 cosh kx + c2 sinh kx) = kc1 sinh kx + kc2 cosh kx;
dx
y
d2
2
2
2
(c
cosh
kx
+
c
sinh
kx)
=
k
c
cosh
kx
+
k
c
sinh
kx
=
k
(
cosh
kx
+
c
sinh
kx);
c
1
2
1
2
1
2
dx2
d2 y
= k2 y
dx2
or
5. y = c1 ex + c2 xex ;
d2 y
− k2 y = 0
dx2
y = c1 ex + c2 xex + c2 ex ;
y = c1 ex + c2 xex + 2c2 ex ;
y + y = 2(c1 ex + c2 xex ) + 2c2 ex = 2(c1 ex + c2 xex + c2 ex ) = 2y ;
y − 2y + y = 0
6. y = −c1 ex sin x + c1 ex cos x + c2 ex cos x + c2 ex sin x;
y = −c1 ex cos x − c1 ex sin x − c1 ex sin x + c1 ex cos x − c2 ex sin x + c2 ex cos x + c2 ex cos x +
c2 ex sin x
= −2c1 ex sin x + 2c2 ex cos x;
y − 2y = −2c1 ex cos x − 2c2 ex sin x = −2y;
7. a, d
8. c
9. b
y − 2y + 2y = 0
10. a, c
11. b
12. a, b, d
13. A few solutions are y = 0, y = c, and y = ex .
14. Easy solutions to see are y = 0 and y = 3.
15. The slope of the tangent line at (x, y) is y , so the differential equation is y = x2 + y 2 .
16. The rate at which the slope changes is dy /dx = y , so the differential equation is y = −y or y + y = 0.
17. (a) The domain is all real numbers.
Chapter 1 in Review
(b) Since y = 2/3x1/3 , the solution y = x2/3 is undefined at x = 0. This function is a
solution of the differential equation on (−∞, 0) and also on (0, ∞).
18. (a) Differentiating y 2 − 2y = x2 − x + c we obtain 2yy − 2y = 2x − 1 or (2y − 2)y = 2x − 1.
(b) Setting x = 0 and y = 1 in the solution we have 1 − 2 = 0 − 0 + c or c = −1. Thus, a
solution of the initial-value problem is y 2 − 2y = x2 − x − 1.
(c) Solving the equation y 2 − 2y − (x2 − x − 1) = 0 by the quadratic formula we get
√
y = (2 ± 4 + 4(x2 − x − 1) )/2 = 1 ± x2 − x = 1 ± x(x − 1) . Since x(x − 1) ≥ 0
for x ≤ 0 or x ≥ 1, we see that neither y = 1 + x(x − 1) nor y = 1 − x(x − 1) is
differentiable at x = 0. Thus, both functions are solutions of the differential equation,
but neither is a solution of the initial-value problem.
19. Setting x = x0 and y = 1 in y = −2/x + x, we get
1=−
2
+ x0
x0
x20 − x0 − 2 = (x0 − 2)(x0 + 1) = 0.
or
Thus, x0 = 2 or x0 = −1. Since x = 0 in y = −2/x + x, we see that y = −2/x + x is a
solution of the initial-value problem xy + y = 2x, y(−1) = 1, on the interval (−∞, 0) because
−1 < 0, and y = −2/x + x is a solution of the initial-value problem xy + y = 2x, y(2) = 1,
on the interval (0, ∞) because 2 > 0.
20. From the differential equation, y (1) = 12 +[y(1)]2 = 1+(−1)2 = 2 > 0, so y(x) is increasing in
some neighborhood of x = 1. From y = 2x+2yy we have y (1) = 2(1)+2(−1)(2) = −2 < 0,
so y(x) is concave down in some neighborhood of x = 1.
21. (a)
y
y
3
3
2
2
1
1
–3 –2 –1
1
–1
–2
–3
y = x2 + c1
2
3
x
–3 –2 –1
–1
1
2
3
x
–2
–3
y = –x2 + c2
(b) When y = x2 + c1 , y = 2x and (y )2 = 4x2 . When y = −x2 + c2 , y = −2x and
(y )2 = 4x2 .
(c) Pasting together x2 , x ≥ 0, and −x2 , x ≤ 0, we get y =
⎧
⎨−x2 ,
x≤0
⎩x2 ,
x > 0.
√
22. The slope of the tangent line is y (−1,4) = 6 4 + 5(−1)3 = 7.
29
30
CHAPTER 1
INTRODUCTION TO DIFFERENTIAL EQUATIONS
23. Differentiating y = x sin x + x cos x we get
y = x cos x + sin x − x sin x + cos x
and
y = −x sin x + cos x + cos x − x cos x − sin x − sin x
= −x sin x − x cos x + 2 cos x − 2 sin x.
Thus
y + y = −x sin x − x cos x + 2 cos x − 2 sin x + x sin x + x cos x = 2 cos x − 2 sin x.
An interval of definition for the solution is (−∞, ∞).
24. Differentiating y = x sin x + (cos x) ln(cos x) we get
− sin x
− (sin x) ln (cos x)
y = x cos x + sin x + cos x
cos x
= x cos x + sin x − sin x − (sin x) ln (cos x)
= x cos x − (sin x) ln (cos x)
and
y = −x sin x + cos x − sin x
− sin x
cos x
− (cos x) ln (cos x)
= −x sin x + cos x +
sin2 x
− (cos x) ln (cos x)
cos x
= −x sin x + cos x +
1 − cos2 x
− (cos x) ln (cos x)
cos x
= −x sin x + cos x + sec x − cos x − (cos x) ln (cos x)
= −x sin x + sec x − (cos x) ln (cos x).
Thus
y + y = −x sin x + sec x − (cos x) ln(cos x) + x sin x + (cos x) ln (cos x) = sec x.
To obtain an interval of definition we note that the domain of ln x is (0, ∞), so we must have
cos x > 0. Thus, an interval of definition is (−π/2, π/2).
25. Differentiating y = sin (ln x) we obtain y = cos (ln x)/x and y = −[sin (ln x) + cos (ln x)]/x2 .
Then
sin (ln x) + cos (ln x)
cos (ln x)
2 2
+x
x y + xy + y = x −
+ sin (ln x) = 0.
2
x
x
An interval of definition for the solution is (0, ∞).
Chapter 1 in Review
26. Differentiating y = cos (ln x) ln (cos (ln x)) + (ln x) sin (ln x) we obtain
sin (ln x)
sin (ln x)
cos (ln x) sin (ln x)
1
−
+ ln (cos (ln x)) −
+ ln x
+
y = cos (ln x)
cos (ln x)
x
x
x
x
=−
ln (cos (ln x)) sin (ln x) (ln x) cos (ln x)
+
x
x
and
1
sin (ln x)
1
cos (ln x)
+ sin (ln x)
−
y = −x ln (cos (ln x))
x
cos (ln x)
x
x2
1
sin (ln x)
1
cos (ln x) 1
+ ln (cos (ln x)) sin (ln x) 2 + x (ln x) −
− (ln x) cos (ln x) 2
+
x
x
x
x2
x
=
sin2 (ln x)
1
+ ln (cos (ln x)) sin (ln x)
−
ln
(cos
(ln
x))
cos
(ln
x)
+
x2
cos (ln x)
− (ln x) sin (ln x) + cos (ln x) − (ln x) cos (ln x) .
Then
x2 y + xy + y = − ln (cos (ln x)) cos (ln x) +
sin2 (ln x)
+ ln (cos (ln x)) sin (ln x) − (ln x) sin (ln x)
cos (ln x)
+ cos (ln x) − (ln x) cos (ln x) − ln (cos (ln x)) sin (ln x)
+ (ln x) cos (ln x) + cos (ln x) ln (cos (ln x)) + (ln x) sin (ln x)
=
sin2 (ln x)
sin2 (ln x) + cos2 (ln x)
1
+ cos (ln x) =
=
= sec (ln x).
cos (ln x)
cos (ln x)
cos (ln x)
To obtain an interval of definition, we note that the domain of ln x is (0, ∞), so we must
have cos (ln x) > 0. Since cos x > 0 when −π/2 < x < π/2, we require −π/2 < ln x < π/2.
Since ex is an increasing function, this is equivalent to e−π/2 < x < eπ/2 . Thus, an interval
of definition is (e−π/2 , eπ/2 ). (Much of this problem is more easily done using a computer
algebra system such as Mathematica or Maple.)
In Problems 27 - 30 we use (12) of Section 1.1 and the Product Rule.
27.
ˆ
x
y = ecos x
te− cos t dt
0
ˆ x
− cos x dy
cos x
cos x
xe
− sin xe
=e
te− cos t dt
dx
0
ˆ x
ˆ x
dy
cos x − cos x
cos x
− cos t
cos x
− cos t
+ (sin x) y = e
xe
− sin xe
te
dt + sin x e
te
dt
dx
0
0
ˆ x
ˆ x
te− cos t dt + sin xecos x
te− cos t dt = x
= x − sin xecos x
0
0
31
32
CHAPTER 1
INTRODUCTION TO DIFFERENTIAL EQUATIONS
28.
ˆ
x2
x
2
et−t dt
y=e
0
dy
2
2
2
= ex ex−x + 2xex
dx
ˆ
dy
2
2
2
− 2xy = ex ex−x + 2xex
dx
29.
ˆ
x
t−t2
e
ˆ
x
2
et−t dt
0
ˆ
2
dt − 2x ex
0
x
t−t2
e
= ex
dt
0
e−t
dt
t
1
ˆ x −t
ˆ x −t
e−x
e
e
−x
y =x
+
dt = e +
dt
x
t
t
1
1
x
y=x
e−x
x
y = −e−x +
x2 y + x2 − x y + (1 − x) y = −x2 e−x + xe−x
ˆ
2 −x
2
+ x e +x
ˆ
+ x
x
1
30.
ˆ
x
y = sin x
ˆ
t2
e cos t dt − cos x
0
x
e−t
dt − xe−x − x
t
1
ˆ x −t e−t
e
2
dt − x
dt = 0
t
t
1
x
ˆ
y = sin x e
x
= cos x
cos x + cos x
ˆ
x
2
t2
x2
e cos t dt − cos x e
ˆ
sin x + sin x
ˆ
2
et cos t dt + sin x
2
et sin t dt
x
2
et sin t dt
0
ˆ
2
y = cos x ex cos x − sin x
x
ˆ
2
2
et cos t dt + sin x ex sin x + cos x
0
⎛
=e
x
0
0
et sin t dt
0
x2
1
e−t
dt
t
0
x2
ˆ
x
⎜
cos x + sin x − ⎜
⎝sin x
2
x
0
ˆ
2
y
x
ˆ
t2
e cos t dt − cos x
0
0
2
= ex − y
y + y = ex − y + y = ex
2
2
x
⎞
⎟
e sin t dt⎟
⎠
t2
2
et sin t dt
Chapter 1 in Review
31. Using implicit differentiation we get
x3 y 3 = x3 + 1
3x2 · y 3 + x3 · 3y 2
dy
= 3x2
dx
3x2
3x2 y 3 x3 3y 2 dy
=
+
3x2 y 2 3x2 y 2 dx
3x2 y 2
y+x
dy
1
= 2
dx
y
32. Using implicit differentiation we get
(x − 5)2 + y 2 = 1
2(x − 5) + 2y
2y
dy
=0
dx
dy
= −2(x − 5)
dx
dy
= −(x − 5)
dx
dy 2
y
= (x − 5)2
dx
y
dy
dx
2
=
(x − 5)2
y2
Now from the original equation, isolating the first term leads to (x − 5)2 = 1 − y 2 . Continuing
from the last line of our proof we now have
2
(x − 5)2
1 − y2
1
dy
=
=
= 2 −1
2
2
dx
y
y
y
Adding 1 to both sides leads to the desired result.
33. Using implicit differentiation we get
y 3 + 3y = 1 − 3x
3y 2 y + 3y = −3
y 2 y + y = −1
(y 2 + 1)y = −1
y =
−1
y2 + 1
33
34
CHAPTER 1
INTRODUCTION TO DIFFERENTIAL EQUATIONS
Differentiating the last line and remembering to use the quotient rule on the right side leads
to
2yy y = 2
(y + 1)2
Now since y = −1 y 2 + 1 we can write the last equation as
3
−1
2y
2y
−1
= 2y
y = 2
= 2y(y )3
y = 2
(y + 1)2
(y + 1)2 (y 2 + 1)
y2 + 1
which is what we wanted to show.
34. Using implicit differentiation we get
y = exy
y = exy (y + xy )
y = yexy + xexy y (1 − xexy )y = yexy
Now since y = exy , substitute this into the last line to get
(1 − xy)y = yy
or (1 − xy)y = y 2 which is what we wanted to show.
In Problem 35–38, y = c1 e3x + c2 e−x − 2x is given as a two-parameter family of solutions of the
second-order differential equation y − 2y − 3y = 6x + 4.
35. If y(0) = 0 and y (0) = 0, then
c1 + c2 = 0
3c1 − c2 − 2 = 0
so c1 =
1
2
and c2 = − 12 . Thus y =
1
2
e3x − half e−x − 2x.
36. If y(0) = 1 and y (0) = −3, then
c1 + c2 = 1
3c1 − c2 − 2 = −3
so c1 = 0 and c2 = 1. Thus y = e−x − 2x.
37. If y(1) = 4 and y (1) = −2, then
c1 e3 + c2 e−1 − 2 = 4
3c1 e3 − c2 e−1 − 2 = −2
so c1 =
3
2
e−3 and c2 =
9
2
e. Thus y =
3
2
e3x−3 + 92 e−x+1 − 2x.
Chapter 1 in Review
38. If y(−1) = 0 and y (−1) = 1, then
c1 e−3 + c2 e + 2 = 0
3c1 e−3 − c2 e − 2 = 1
so c1 =
1
4
e3 and c2 = − 94 , e−1 . Thus y =
1
4
e3x+3 − 94 e−x−1 − 2x.
39. From the graph we see that estimates for y0 and y1 are y0 = −3 and y1 = 0.
40. The differential equation is
cA0 dh
=−
2gh .
dt
Aw
Using A0 = π(0.6/0.0125)2 = 4.906−4 , Aw = π(0.6)2 = 1.134, and g = 9.8, this becomes
c4.906−4 √
c √
dh
=−
19.6h = −
h.
dt
1.134
522
35
Chapter 2
First-Order Differential Equations
2.1
Solution Curves Without a Solution
1. x
3
2. x
y
5
2
1
–3
–2
y
10
x
0
–1
1
2
3
x
–5
–1
–2
–10
–5
5
0
10
–3
3. x
y
4
y
4. x
4
2
2
0
x
0
x
–2
–2
–4
–4
5. x
–2
0
2
–4
4
y
4
6. x
2
0
2
4
y
4
2
0
x
0
x
–2
–4
–2
–2
–2
0
2
4
–4
36
–2
0
2
4
2.1
7. x
y
8. x
4
Solution Curves Without a Solution
37
y
4
2
2
x
0
x
0
–2
–2
–4
–4
9. x
–2
0
2
–4
4
y
4
10. x
2
x
2
4
y
4
x
0
–2
–2
–4
–2
0
2
–4
4
y
4
12. x
–2
0
2
4
y
4
2
2
x
0
x
0
–2
–2
–4
13. x
0
2
0
11. x
–2
–2
0
2
–4
4
y
–2
2
4
y
14. x
3
0
4
2
2
1
x
0
–1
0
x
–2
–2
–4
–3
–3 –2 –1
0
1
2
3
–4
–2
0
2
4
y
15. (a) The isoclines have the form y = −x + c, which are straight
lines with slope −1.
3
2
1
–3
–2 –1
1
–1
–2
–3
2
3
x
38
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
y
(b) The isoclines have the form x2 + y 2 = c, which are circles
centered at the origin.
2
1
–2
1
–1
x
2
–1
–2
16. (a) When x = 0 or y = 4, dy/dx = −2 so the lineal elements have slope −2. When y = 3 or
y = 5, dy/dx = x − 2, so the lineal elements at (x, 3) and (x, 5) have slopes x − 2.
(b) At (0, y0 ) the solution curve is headed down. If y → ∞ as x increases, the graph must
eventually turn around and head up, but while heading up it can never cross y = 4
where a tangent line to a solution curve must have slope −2. Thus, y cannot approach
∞ as x approaches ∞.
17. When y < 12 x2 , y = x2 − 2y is positive and the portions of
solution curves “outside” the nullcline parabola are increasing.
When y > 12 x2 , y = x2 − 2y is negative and the portions of the
solution curves “inside” the nullcline parabola are decreasing.
y
3
2
1
x
0
–1
–2
–3
–3
–2
–1
0
1
2
3
18. (a) Any horizontal lineal element should be at a point on a nullcline. In Problem 1 the
nullclines are x2 − y 2 = 0 or y = ±x. In Problem 3 the nullclines are 1 − xy = 0 or
y = 1/x. In Problem 4 the nullclines are (sin x) cos y = 0 or x = nπ and y = π/2 + nπ,
where n is an integer. The graphs on the next page show the nullclines for the equations
in Problems 1, 3, and 4 superimposed on the corresponding direction field.
y
y
y
4
3
4
2
2
2
1
x
0
x
0
–1
x
0
–2
–2
–2
–4
–3
–3
–2
–1
0
1
Problem 1
2
3
–4
–4
–2
0
2
Problem 3
4
–4
–2
0
2
Problem 4
4
2.1
Solution Curves Without a Solution
(b) An autonomous first-order differential equation has the form y = f (y). Nullclines have
the form y = c where f (c) = 0. These are the graphs of the equilibrium solutions of the
differential equation.
19. Writing the differential equation in the form dy/dx = y(1 − y)(1 + y) we see that
critical points are y = −1, y = 0, and y = 1. The phase portrait is shown at the
right.
1
(a) x
(b) x
y
y
5
0
1
4
3
–1
2
1
–2
1
(c) x
2
–1
(d) x
1
2
1
2
x
x
y
–2
–1
y
1
2
x
–1
x
–2
–3
–1
–4
–5
20. Writing the differential equation in the form dy/dx = y 2 (1 − y)(1 + y) we see that
critical points are y = −1, y = 0, and y = 1. The phase portrait is shown at the
right.
1
(a) x
(b) x
y
y
5
0
1
4
3
1
2
1
–2
1
(c) x
2
2
(d) x
1
2
x
x
–1
–1
–2
–3
–1
x
y
–2
–1
1
x
y
–2
–1
–4
–5
39
40
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
21. Solving y 2 − 3y = y(y − 3) = 0 we obtain the critical points 0 and 3. From the
phase portrait we see that 0 is asymptotically stable (attractor) and 3 is unstable
(repeller).
3
0
22. Solving y 2 − y 3 = y 2 (1 − y) = 0 we obtain the critical points 0 and 1. From the phase
portrait we see that 1 is asymptotically stable (attractor) and 0 is semi-stable.
1
0
23. Solving (y − 2)4 = 0 we obtain the critical point 2. From the phase portrait we see
that 2 is semi-stable.
2
2.1
Solution Curves Without a Solution
24. Solving 10 + 3y − y 2 = (5 − y)(2 + y) = 0 we obtain the critical points −2 and 5.
From the phase portrait we see that 5 is asymptotically stable (attractor) and −2 is
unstable (repeller).
5
–2
25. Solving y 2 (4 − y 2 ) = y 2 (2 − y)(2 + y) = 0 we obtain the critical points −2, 0, and
2. From the phase portrait we see that 2 is asymptotically stable (attractor), 0 is
semi-stable, and −2 is unstable (repeller).
2
0
–2
26. Solving y(2 − y)(4 − y) = 0 we obtain the critical points 0, 2, and 4. From the phase
portrait we see that 2 is asymptotically stable (attractor) and 0 and 4 are unstable
(repellers).
4
2
0
41
42
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
27. Solving y ln(y+2) = 0 we obtain the critical points −1 and 0. From the phase portrait
we see that −1 is asymptotically stable (attractor) and 0 is unstable (repeller).
0
–1
–2
28. Solving yey − 9y = y(ey − 9) = 0 (since ey is always positive) we obtain the
critical points 0 and ln 9. From the phase portrait we see that 0 is asymptotically
stable (attractor) and ln 9 is unstable (repeller).
1n 9
0
29. The critical points are 0 and c because the graph of f (y) is 0 at these points. Since f (y) > 0
for y < 0 and y > c, the graph of the solution is increasing on the y-intervals (−∞, 0) and
(c, ∞). Since f (y) < 0 for 0 < y < c, the graph of the solution is decreasing on the y-interval
(0, c).
y
c
0
c
x
2.1
Solution Curves Without a Solution
43
30. The critical points are approximately at −2, 2, 0.5, and 1.7. Since f (y) > 0 for y < −2.2
and 0.5 < y < 1.7, the graph of the solution is increasing on the y-intervals (−∞, −2.2) and
(0.5, 1.7). Since f (y) < 0 for −2.2 < y < 0.5 and y > 1.7, the graph is decreasing on the
y-interval (−2.2, 0.5) and (1.7, ∞).
y
2
1.7
1
0.5
–2
–1
1
2
x
–1
–2
–2.2
31. From the graphs of z = π/2 and z = sin y we see that
(2/π)y − sin y = 0 has only three solutions. By inspection
we see that the critical points are −π/2, 0, and π/2.
1
–π
–
π
2
π
2
π
y
–1
From the graph at the right we see that
2
y − sin y
π
<0
>0
>0
2
y − sin y
π
<0
for
for
y < −π/2
y > π/2
π
2
0
for
for
− π/2 < y < 0
0 < y < π/2
–
π
2
This enables us to construct the phase portrait shown at the right. From this portrait we see
that π/2 and −π/2 are unstable (repellers), and 0 is asymptotically stable (attractor).
32. For dy/dx = 0 every real number is a critical point, and hence all critical points are nonisolated.
33. Recall that for dy/dx = f (y) we are assuming that f and f are continuous functions of y
on some interval I. Now suppose that the graph of a nonconstant solution of the differential
equation crosses the line y = c. If the point of intersection is taken as an initial condition
we have two distinct solutions of the initial-value problem. This violates uniqueness, so the
44
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
graph of any nonconstant solution must lie entirely on one side of any equilibrium solution.
Since f is continuous it can only change signs at a point where it is 0. But this is a critical
point. Thus, f (y) is completely positive or completely negative in each region Ri . If y(x) is
oscillatory or has a relative extremum, then it must have a horizontal tangent line at some
point (x0 , y0 ). In this case y0 would be a critical point of the differential equation, but we saw
above that the graph of a nonconstant solution cannot intersect the graph of the equilibrium
solution y = y0 .
34. By Problem 33, a solution y(x) of dy/dx = f (y) cannot have relative extrema and hence must
be monotone. Since y (x) = f (y) > 0, y(x) is monotone increasing, and since y(x) is bounded
above by c2 , limx→∞ y(x) = L, where L ≤ c2 . We want to show that L = c2 . Since L is a
horizontal asymptote of y(x), limx→∞ y (x) = 0. Using the fact that f (y) is continuous we
have
f (L) = f
lim y(x) = lim f (y(x)) = lim y (x) = 0.
x→∞
x→∞
x→∞
But then L is a critical point of f . Since c1 < L ≤ c2 , and f has no critical points between
c1 and c2 , L = c2 .
35. Assuming the existence of the second derivative, points of inflection of y(x) occur where
y (x) = 0. From dy/dx = f (y) we have d2 y/dx2 = f (y) dy/dx. Thus, the y-coordinate of a
point of inflection can be located by solving f (y) = 0. (Points where dy/dx = 0 correspond
to constant solutions of the differential equation.)
36. Solving y 2 − y − 6 = (y − 3)(y + 2) = 0 we see that 3 and −2
are critical points. Now d2 y/dx2 = (2y − 1) dy/dx = (2y − 1)(y −
3)(y + 2), so the only possible point of inflection is at y = 12 ,
although the concavity of solutions can be different on either side
of y = −2 and y = 3. Since y (x) < 0 for y < −2 and 12 < y < 3,
and y (x) > 0 for −2 < y < 12 and y > 3, we see that solution
curves are concave down for y < −2 and 12 < y < 3 and concave
up for −2 < y < 12 and y > 3. Points of inflection of solutions of
autonomous differential equations will have the same y-coordinates
because between critical points they are horizontal translations of
each other.
y
5
–5
5
x
–5
37. If (1) in the text has no critical points it has no constant solutions. The solutions have
neither an upper nor lower bound. Since solutions are monotonic, every solution assumes all
real values.
2.1
Solution Curves Without a Solution
38. The critical points are 0 and b/a. From the phase portrait we see that 0 is an
attractor and b/a is a repeller. Thus, if an initial population satisfies P0 > b/a,
the population becomes unbounded as t increases, most probably in finite time,
i.e. P (t) → ∞ as t → T . If 0 < P0 < b/a, then the population eventually dies out,
that is, P (t) → 0 as t → ∞. Since population P > 0 we do not consider the case
P0 < 0.
b
a
0
39. From the equation dP/dt = k (P − h/k) we see that the only critical point of the autonomous
differential equationis the positive number h/k. A phase portrait shows that this point is
unstable, that is, h/k is a repeller. For any initial condition P (0) = P0 for which 0 < P0 < h/k,
dP/dt < 0 which means P (t) is monotonic decreasing and so the graph of P (t) must cross the
t-axis or the line P − 0 at some time t1 > 0. But P (t1 ) = 0 means the population is extinct
at time t1 .
40. Writing the differential equation in the form
k mg
dv
=
−v
dt
m k
mg
k
we see that a critical point is mg/k.
From the phase portrait we see that mg/k is an asymptotically stable critical point.
Thus, lim v = mg/k.
t→∞
41. Writing the differential equation in the form
k
k mg
mg
mg
dv
2
=
−v =
−v
+v
dt
m k
m
k
k
we see that the only physically meaningful critical point is mg/k.
mg/k is an asymptotically stable critical
From the phase portrait
we
see
that
point. Thus, lim v = mg/k.
√
mg
k
t→∞
42. (a) From the phase portrait we see that critical points are α and β. Let X(0) = X0 .
If X0 < α, we see that X → α as t → ∞. If α < X0 < β, we see that X → α
as t → ∞. If X0 > β, we see that X(t) increases in an unbounded manner,
but more specific behavior of X(t) as t → ∞ is not known.
β
α
45
46
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
(b) When α = β the phase portrait is as shown. If X0 < α, then X(t) → α
as t → ∞. If X0 > α, then X(t) increases in an unbounded manner. This
could happen in a finite amount of time. That is, the phase portrait does not
indicate that X becomes unbounded as t → ∞.
(c) When k = 1 and α = β the differential equation is dX/dt = (α − X)2 .
X(t) = α − 1/(t + c) we have dX/dt = 1/(t + c)2 and
1
(α − X) = α − α −
t+c
2
2
α
For
1
dX
=
.
2
(t + c)
dt
=
For X(0) = α/2 we obtain
X(t) = α −
1
.
t + 2/α
X(t) = α −
1
.
t − 1/α
For X(0) = 2α we obtain
x
x
2α
α
α
α/2
–2 / α
t
1/α
t
For X0 > α, X(t) increases without bound up to t = 1/α. For t > 1/α, X(t) increases
but X → α as t → ∞.
2.2
Separable Variables
In many of the following problems we will encounter an expression of the form ln |g(y)| = f (x)+c.
To solve for g(y) we exponentiate both sides of the equation. This yields |g(y)| = ef (x)+c = ec ef (x)
which implies g(y) = ±ec ef (x) . Letting c1 = ±ec we obtain g(y) = c1 ef (x) .
1. From dy = sin 5x dx we obtain y = − 15 cos 5x + c.
2. From dy = (x + 1)2 dx we obtain y = 13 (x + 1)3 + c.
2.2 Separable Variables
3. From dy = −e−3x dx we obtain y = 13 e−3x + c.
4. From
1
1
1
= x + c or y = 1 −
.
dy = dx we obtain −
2
(y − 1)
y−1
x+c
5. From
4
1
dy = dx we obtain ln |y| = 4 ln |x| + c or y = c1 x4 .
y
x
6. From
1
1
1
dy = −2x dx we obtain − = −x2 + c or y = 2
.
2
y
y
x + c1
7. From e−2y dy = e3x dx we obtain 3e−2y + 2e3x = c.
1
8. From yey dy = e−x + e−3x dx we obtain yey − ey + e−x + e−3x = c.
3
x3
1
1
y2
dy = x2 ln x dx we obtain
+ 2y + ln |y| =
ln |x| − x3 + c.
9. From y + 2 +
y
2
3
9
10. From
1
1
2
1
=
+ c.
dy =
dx we obtain
(2y + 3)2
(4x + 5)2
2y + 3
4x + 5
1
1
dy = − 2 dx or sin y dy = − cos2 x dx = − 12 (1 + cos 2x) dx we obtain
csc y
sec x
− cos y = − 12 x − 14 sin 2x + c or 4 cos y = 2x + sin 2x + c1 .
11. From
12. From 2y dy = −
13. From
14. From
15. From
sin 3x
dx or 2y dy = − tan 3x sec2 3x dx we obtain y 2 = − 16 sec2 3x + c.
cos3 3x
−ex
ey
dy
=
dx we obtain − (ey + 1)−1 =
(ey + 1)2
(ex + 1)3
y
(1 +
y 2 )1/2
dy =
x
(1 +
x2 )1/2
dx we obtain 1 + y 2
1/2
1
2
(ex + 1)−2 + c.
= 1 + x2
1/2
+ c.
1
dS = k dr we obtain S = cekr .
S
1
dQ = k dt we obtain ln |Q − 70| = kt + c or Q − 70 = c1 ekt .
Q − 70
1
1
1
dP
=
+
dP = dt we obtain ln |P | − ln |1 − P | = t + c so that
17. From
P − P2
P
1−P
c1 et
P
P
= c1 et . Solving for P we have P =
.
ln
= t + c or
1−P
1−P
1 + c1 et
16. From
1
t+2
t+2
dN = tet+2 − 1 dt we obtain ln |N | = tet+2 − et+2 − t + c or N = c1 ete −e −t .
N
x−1
5
5
y−2
dy =
dx or 1 −
dy = 1 −
dx we obtain
19. From
y+3
x+4
y+3
x+4
x+4 5
y − 5 ln |y + 3| = x − 5 ln |x + 4| + c or
= c1 ex−y .
y+3
18. From
47
48
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
2
5
dy = 1 +
dx we obtain
y−1
x−3
(y − 1)2
y + 2 ln |y − 1| = x + 5 ln |x − 3| + c or
= c1 ex−y .
(x − 3)5
x+2
y+1
dy =
dx or
20. From
y−1
x−3
1+
1
21. From x dx = dy we obtain 12 x2 = sin−1 y + c or y = sin
2
1−y
x2
+ c1 .
2
1
ex
1
1
dx we obtain − = tan−1 ex + c or
dy
=
dx
=
2
x
−x
x
2
y
e +e
(e ) + 1
y
1
.
y=−
tan−1 ex + c
22. From
1
dx = 4 dt we obtain tan−1 x = 4t + c. Using x(π/4) = 1 we find c = −3π/4. The
x2 + 1
3π
3π
−1
or x = tan 4t −
.
solution of the initial-value problem is tan x = 4t −
4
4
23. From
1
1
1
24. From 2
dy = 2
dx or
y −1
x −1
2
1
1
−
y−1 y+1
1
dy =
2
1
1
−
x−1 x+1
dx we obtain
c(x − 1)
y−1
=
. Using y(2) = 2 we
y+1
x+1
x−1
y−1
=
or y = x.
find c = 1. A solution of the initial-value problem is
y+1
x+1
ln |y − 1| − ln |y + 1| = ln |x − 1| − ln |x + 1| + ln c or
1
1−x
1
1
1
25. From dy =
dx we obtain ln |y| = − − ln |x| = c or xy = c1 e−1/x .
dx =
−
y
x2
x2 x
x
Using y(−1) = −1 we find c1 = e−1 . The solution of the initial-value problem is xy = e−1−1/x
or y = e−(1+1/x) /x.
1
dy = dt we obtain − 12 ln |1 − 2y| = t + c or 1 − 2y = c1 e−2t . Using y(0) = 5/2 we
1 − 2y
find c1 = −4. The solution of the initial-value problem is 1 − 2y = −4e−2t or y = 2e−2t + 12 .
26. From
27. Separating variables and integrating we obtain
√
dy
dx
−
= 0 and
2
1−x
1 − y2
sin−1 x − sin−1 y = c.
√
Setting x = 0 and y = 3/2 we obtain c = −π/3. Thus, an implicit solution of the initialvalue problem is sin−1 x − sin−1 y = π/3. Solving for y and using an addition formula from
trigonometry, we get
−1
y = sin sin
π
π π
x
= x cos + 1 − x2 sin = +
x+
3
3
3
2
√ √
3 1 − x2
.
2
2.2 Separable Variables
28. From
1
−x
dy =
dx we obtain
1 + (2y)2
1 + (x2 )2
1
1
tan−1 2y = − tan−1 x2 + c or
2
2
tan−1 2y + tan−1 x2 = c1 .
Using y(1) = 0 we find c1 = π/4. Thus, an implicit solution of the initial-value problem is
tan−1 2y + tan−1 x2 = π/4 . Solving for y and using a trigonometric identity we get
2y = tan
y=
π
− tan−1 x2
4
π
1
tan
− tan−1 x2
2
4
1 tan π4 − tan (tan−1 x2 )
=
2 1 + tan π4 tan (tan−1 x2 )
=
1 1 − x2
.
2 1 + x2
29. Separating variables and then proceeding as in Example 5 we get
dy
2
= ye−x
dx
ˆ
4
x
1 dy
2
= e−x
y dx
ˆ x
1 dy
2
dt =
e−t dt
y(t) dt
4
ˆ
x
x
2
e−t dt
ln y(t) =
4
4
ˆ
x
ln y(x) − ln y(4) =
e−t dt
2
4
ˆ
x
ln y(x) =
e−t dt
2
4
´x
y(x) = e
4
2
e−t dt
49
50
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
30. Separating variables and then proceeding as in Example 5 we get
dy
= y 2 sin (x2 )
dx
1 dy
= sin (x2 )
y 2 dx
ˆ x
ˆ x
1 dy
dt =
sin (t2 ) dt
2
−2 y (t) dt
−2
ˆ x
−1 x
=
sin (t2 ) dt
y(t) −2
−2
ˆ x
1
−1
+
=
sin (t2 ) dt
y(x) y(−2)
−2
ˆ x
−1
sin (t2 ) dt
+3=
y(x)
−2
ˆ
y(x) = 3 −
x
sin (t2 ) dt
−1
−2
31. Separating variables we get
2x + 1
dy
=
dx
2y
ˆ
2y dy = (2x + 1) dx
ˆ
2y dy = (2x + 1) dx
y 2 = x2 + x + c
√
The condition y(−2) = −1 implies c = −1. Thus y 2 = x2 + x − 1 and y = − x2 + x − 1 in
order for y to be negative.
an interval containing −2 for values of x such that
Moreover√for
1
5
.
x2 + x − 1 > 0 we get −∞, − −
2
2
2.2 Separable Variables
32. Separating variables we get
(2y − 2)
dy
= 3x2 + 4x + 2
dx
(2y − 2) dy = 3x2 + 4x + 2 dx
ˆ
ˆ
(2y − 2) dy =
3x2 + 4x + 2 dx
ˆ
ˆ
2 (y − 1) dy =
3x2 + 4x + 2 dx
(y − 1)2 = x3 + 2x2 + 2x + c
√
The condition y(1) = −2 implies c = 4. Thus y = 1 − x3 + 2x2 + 2x + 4 where the minus
sign is indicated by the initial condition. Now x3 + 2x2 + 2x+ 4 = (x + 2) x2 + 1 > 0 implies
x > −2, so the interval of definition is (−2, ∞).
33. Separating variables we get
ey dx − e−x dy = 0
ey dx = e−x dy
ex dx = e−y dy
ˆ
ˆ
x
e dx = e−y dy
ex = −e−y + c
The condition y(0) = 0 implies c = 2. Thus e−y = 2 − ex . Therefore y = − ln (2 − ex ). Now
we must have 2 − ex > 0 or ex < 2. Since ex is an increasing function this imples x < ln 2
and so the interval of definition is (−∞, ln 2).
34. Separating variables we get
sin x dx + y dy = 0
ˆ
ˆ
ˆ
sin x dx + y dy = 0 dx
1
− cos x + y 2 = c
2
The condition y(0) = 1 implies c = − 12 . Thus − cos x + 12 y 2 = − 12 or y 2 = 2 cos x − 1.
√
Therefore y = 2 cos x − 1 where the positive root is indicated by the initial condition. Now
we must have 2 cos x − 1 > 0 or cos x > 12 . This means −π/3 < x < π/3, so the the interval
of definition is (−π/3, π/3).
35. (a) The equilibrium solutions y(x) = 2 and y(x) = −2 satisfy the initial conditions y(0) = 2
51
52
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
and y(0) = −2, respectively. Setting x = 14 and y = 1 in y = 2(1 + ce4x )/(1 − ce4x ) we
obtain
1
1 + ce
, 1 − ce = 2 + 2ce, −1 = 3ce, and c = − .
1=2
1 − ce
3e
The solution of the corresponding initial-value problem is
y=2
1 − 13 e4x−1
3 − e4x−1
=
2
.
3 + e4x−1
1 + 13 e4x−1
(b) Separating variables and integrating yields
1
1
ln |y − 2| − ln |y + 2| + ln c1 = x
4
4
ln |y − 2| − ln |y + 2| + ln c = 4x
ln
c(y − 2)
= 4x
y+2
c
y−2
= e4x .
y+2
Solving for y we get y = 2(c + e4x )/(c − e4x ). The initial condition y(0) = −2 implies
2(c + 1)/(c − 1) = −2 which yields c = 0 and y(x) = −2. The initial condition y(0) = 2
does not correspond to a value of c, and it must simply be recognized that y(x) = 2 is a
solution of the initial-value problem. Setting x = 14 and y = 1 in y = 2(c + e4x )/(c − e4x )
leads to c = −3e. Thus, a solution of the initial-value problem is
y=2
3 − e4x−1
−3e + e4x
=
2
.
−3e − e4x
3 + e4x−1
36. Separating variables, we have
dx
dy
=
2
y −y
x
Using partial fractions, we obtain
ˆ ˆ
or
1
1
−
y−1 y
dy
= ln |x| + c.
y(y − 1)
dy = ln |x| + c
ln |y − 1| − ln |y| = ln |x| + c
ln
y−1
=c
xy
y−1
= ec = c1 .
xy
Solving for y we get y = 1/(1 − c1 x). We note by inspection that y = 0 is a singular solution
of the differential equation.
2.2 Separable Variables
53
(a) Setting x = 0 and y = 1 we have 1 = 1/(1 − 0), which is true for all values of c1 . Thus,
solutions passing through (0, 1) are y = 1/(1 − c1 x).
(b) Setting x = 0 and y = 0 in y = 1/(1 − c1 x) we get 0 = 1. Thus, the only solution passing
through (0, 0) is y = 0.
(c) Setting x =
1
2
and y =
1
2
we have
(d) Setting x = 2 and y = 14 we have
y = 1/(1 + 32 x) = 2/(2 + 3x).
1
2
1
4
= 1/(1 − 12 c1 ), so c1 = −2 and y = 1/(1 + 2x).
= 1/(1 − 2c1 ), so c1 = − 32 and
37. Singular solutions of dy/dx = x 1 − y 2 are y = −1 and y = 1. A singular solution of
(ex + e−x )dy/dx = y 2 is y = 0.
38. Differentiating ln (x2 + 10) + csc y = c we get
dy
2x
− csc y cot y
= 0,
x2 + 10
dx
x2
2x
1
cos y dy
−
·
= 0,
+ 10 sin y sin y dx
or
2x sin2 y dx − (x2 + 10) cos y dy = 0.
Writing the differential equation in the form
2x sin2 y
dy
= 2
dx
(x + 10) cos y
we see that singular solutions occur when sin2 y = 0, or y = kπ, where k is an integer.
39. The singular solution y = 1 satisfies the initial-value problem.
1.01
y
1
–0.004 –0.002
0.98
0.97
0.002 0.004
x
54
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
40. Separating variables we obtain
−
dy
= dx. Then
(y − 1)2
x+c−1
1
= x + c and y =
.
y−1
x+c
Setting x = 0 and y = 1.01 we obtain c = −100. The solution
is
x − 101
.
y=
x − 100
1.02
y
1.01
–0.004 –0.002
0.002 0.004
x
0.99
0.98
41. Separating variables we obtain
dy
= dx. Then
(y − 1)2 + 0.01
10 tan−1 10(y − 1) = x + c and y = 1 +
1
x+c
tan
.
10
10
Setting x = 0 and y = 1 we obtain c = 0. The solution is
y =1+
x
1
tan
.
10
10
y
1.0004
1.0002
0.002 0.004
–0.004 –0.002
x
0.9998
0.9996
dy
= dx. Then,
(y − 1)2 − 0.01
1
with u = y − 1 and a = 10
, we get
42. Separating variables we obtain
y
1.0004
1.0002
10y − 11
5 ln
= x + c.
10y − 9
–0.004 –0.002
Setting x = 0 and y = 1 we obtain c = 5 ln 1 = 0. The
solution is
10y − 11
= x.
5 ln
10y − 9
0.002 0.004
x
0.9998
0.9996
Solving for y we obtain
y=
11 + 9ex/5
.
10 + 10ex/5
Alternatively, we can use the fact that
ˆ
dy
1
y−1
=−
tanh−1
= −10 tanh−1 10(y − 1).
2
(y − 1) − 0.01
0.1
0.1
(We use the inverse hyperbolic tangent because |y − 1| < 0.1 or 0.9 < y < 1.1. This
follows from the initial condition y(0) = 1.) Solving the above equation for y we get y =
1 + 0.1 tanh (x/10).
2.2 Separable Variables
43. Separating variables, we have
dy
dy
=
=
3
y−y
y(1 − y)(1 + y)
1
1/2
1/2
+
−
y 1−y 1+y
dy = dx.
Integrating, we get
ln |y| −
1
1
ln |1 − y| − ln |1 + y| = x + c.
2
2
When y > 1, this becomes
ln y −
1
y
1
ln (y − 1) − ln (y + 1) = ln = x + c.
2
2
2
y −1
√
√
Letting x = 0 and y = 2 we find c = ln (2/ 3 ). Solving for y we get y1 (x) = 2ex / 4e2x − 3 ,
√
where x > ln ( 3/2).
When 0 < y < 1 we have
ln y −
Letting x = 0 and y =
where −∞ < x < ∞.
1
y
1
ln (1 − y) − ln (1 + y) = ln = x + c.
2
2
1 − y2
1
2
√
√
we find c = ln (1/ 3 ). Solving for y we get y2 (x) = ex / e2x + 3 ,
When −1 < y < 0 we have
ln (−y) −
1
1
−y
ln (1 − y) − ln (1 + y) = ln = x + c.
2
2
1 − y2
√
√
Letting x = 0 and y = − 12 we find c = ln (1/ 3 ). Solving for y we get y3 (x) = −ex / e2x + 3 ,
where −∞ < x < ∞.
When y < −1 we have
ln (−y) −
1
1
−y
ln (1 − y) − ln (−1 − y) = ln = x + c.
2
2
y2 − 1
√
Letting x = 0 and y = −2 we find c = ln (2/ 3 ). Solving for y we get
√
√
y4 (x) = −2ex / 4e2x − 3 , where x > ln ( 3/2).
y
y
y
y
4
4
4
4
2
2
2
2
1
2
3
4
5
x
–4 –2
2
4
x
–4 –2
2
4
x
1
–2
–2
–2
–2
–4
–4
–4
–4
2
3
4
5
x
55
56
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
y
44. (a) The second derivative of y is
8
d2 y
dy/dx
1/(y − 3)
1
=−
=−
=−
.
dx2
(y − 1)2
(y − 3)2
(y − 3)3
6
The solution curve is concave down when d2 y/dx2 < 0
or y > 3, and concave up when d2 y/dx2 > 0 or y < 3.
From the phase portrait we see that the solution curve
is decreasing when y < 3 and increasing when y > 3.
2
4
–4
3
–2
2
x
4
–2
y
(b) Separating variables and integrating we obtain
8
6
(y − 3) dy = dx
4
1 2
y − 3y = x + c
2
2
y 2 − 6y + 9 = 2x + c1
–1
1
2
3
4
5
x
–2
(y − 3)2 = 2x + c1
√
y = 3 ± 2x + c1 .
The initial condition dictates whether to use the plus or minus sign.
√
When y1 (0) = 4 we have c1 = 1 and y1 (x) = 3 + 2x + 1 where (−1/2, ∞).
√
When y2 (0) = 2 we have c1 = 1 and y2 (x) = 3 − 2x + 1 where (−1/2, ∞).
√
When y3 (1) = 2 we have c1 = −1 and y3 (x) = 3 − 2x − 1 where (1/2, ∞).
√
When y4 (−1) = 4 we have c1 = 3 and y4 (x) = 3 + 2x + 3 where (−3/2, ∞).
45. We separate variables and rationalize the denominator. Then
dy =
1 − sin x
1 − sin x
1 − sin x
1
·
dx =
dx
dx =
2
1 + sin x 1 − sin x
cos2 x
1 − sin x
= sec2 x − tan x sec x dx.
Integrating, we have y = tan x − sec x + C.
√
√
46. Separating variables we have y dy = sin x dx. Then
ˆ
ˆ
ˆ
√
√
2 3/2
√
y
y dy = sin x dx
= sin x dx.
and
3
To integrate sin
√
√
1
x. Then du = √ dx =
2 x
ˆ
ˆ
ˆ
√
sin x dx = (sin u) (2u) du = 2 u sin u du.
x we first make the substitution u =
1
2u
du and
2.2 Separable Variables
Using integration by parts we find
ˆ
√
√
√
u sin u du = −u cos u + sin u = − x cos x + sin x.
Thus
2
y=
3
ˆ
sin
√
√
√
√
x dx = −2 x cos x + 2 sin x + C
and
√
√
√
y = 32/3 − x cos x + sin x + C .
√
47. Separating variables we have dy/
√
y + y = dx/ ( x + x). To integrate
ˆ
dx/
we substitute u2 = x and get
ˆ
ˆ
√
2u
2
du
=
2
ln
|1
+
u|
+
c
=
2
ln
1
+
du
=
x + c.
u + u2
1+u
Integrating the separated differential equation we have
2 ln (1 +
√
y) = 2 ln 1 +
Solving for y we get y = [c1 (1 +
√
√
x + c or
ln (1 +
√
y) = ln 1 +
2
x) − 1] .
48. Separating variables and integrating we have
ˆ
ˆ
dy
= dx
y 2/3 1 − y 1/3
ˆ
y 2/3
dy = x + c1
1 − y 1/3
−3 ln 1 − y 1/3 = x + c1
ln 1 − y 1/3 = −
x
+ c2
3
1 − y 1/3 = c3 e−x/3
1 − y 1/3 = c4 e−x/3
y 1/3 = 1 + c5 e−x/3
3
y = 1 + c5 e−x/3 .
√
x + ln c1 .
√
x +x
57
58
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
√
√
49. ˆ
Separating variables
we have y dy = e x dx. If u = x , then u2 = x and 2u du = dx. Thus,
ˆ
√
e x dx = 2ueu du and, using integration by parts, we find
ˆ
ˆ
y dy =
√
e
x
dx
1 2
y =
2
so
ˆ
√
√ √
2ueu du = −2eu + C = 2 x e x − 2e x + C,
and
√
√ √x
y=2
xe − e x + C .
To find C we solve y(1) = 4.
√
√
√ √
1e 1 − e 1 + C = 2 C = 4
y(1) = 2
so
C = 4.
√
√ √x
xe − e x + 4.
and the solution of the intial-value problem is y = 2
50. Seperating variables we have y dy = x tan−1 x dx. Integrating both sides and using integration
by parts with u = tan−1 x and dv = x dx we have
ˆ
y dy = x tan−1 x dx
1
1
1 2 1 2
y = x tan−1 x − x + tan−1 x + C
2
2
2
2
y 2 = x2 tan−1 x − x + tan−1 x + C1
y = x2 tan−1 x − x + tan−1 x + C1
To find C1 we solve y(0) = 3.
y(0) =
02 tan−1 0 − 0 + tan−1 0 + C1 =
and the solution of the initial-value problem is y =
C1 = 3
√
so
C1 = 9,
x2 tan−1 x − x + tan−1 x + 9 .
√
51. (a) While y2 (x) = − 25 − x2 is defined at x = −5 and x = 5, y2 (x) is not defined at these
values, and so the interval of definition is the open interval (−5, 5).
(b) At any point on the x-axis the derivative of y(x) is undefined, so no solution curve can
cross the x-axis. Since −x/y is not defined when y = 0, the initial-value problem has no
solution.
2
52. The derivative of y = 14 x2 − 1 is dy/dx = x 14 x2 − 1 . We note that xy 1/2 = x 14 x2 − 1 .
We see from the graphs of y (black), dy/dx (red), and xy 1/2 (blue), below that dy/dx = xy 1/2
on (−∞, 2] and [2, ∞).
2.2 Separable Variables
Alternatively, because
xy 1/2
√
59
X 2 = |X| we can write
⎧
⎪
−∞ < x ≤ −2
x 14 x2 − 1 ,
⎪
⎪
2
⎨
1 2
1 2
√
x − 1 = x x − 1 = −x 14 x2 − 1 , −2 < x < 2
=x y=x
⎪
4
4
⎪
⎪
⎩ 1 2
2 ≤ x < ∞.
x 4x − 1 ,
From this we see that dy/dx = xy 1/2 on (−∞, −2] and on [2, ∞).
53. Separating variables we have dy/
1 + y 2 sin2 y = dx
which is not readily integrated (even by a CAS). We
note that dy/dx ≥ 0 for all values of x and y and that
dy/dx = 0 when y = 0 and y = π, which are equilibrium
solutions.
y
3.5
3
2.5
2
1.5
1
0.5
–6
–4
–2
2
4
6
8
x
54. (a) The solution of y = y, y(0) = 1, is y = ex . Using separation of variables we find that the
solution of y = y [1 + 1/ (x ln x)], y(e) = 1, is y = ex−e ln x. Solving the two solutions
simultaneously we obtain
ex = ex−e ln x,
ee = ln x
so
e
and
e
x = ee .
e
(b) Since y = e(e ) ≈ 2.33 × 101,656,520 , the y-coordinate of the point of intersection of the
two solution curves has over 1.65 million digits.
55. We are looking for a function y(x) such that
2
y +
dy
dx
2
= 1.
Using the positive square root gives
dy
= 1 − y2
dx
dy
= dx
1 − y2
sin−1 y = x + c.
60
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
Thus a solution is y = sin (x + c). If we use the negative square root we obtain
y = sin (c − x) = − sin (x − c) = − sin (x + c1 ).
Note that when c = c1 = 0 and when c = c1 = π/2 we obtain the well known particular
solutions y = sin x, y = − sin x, y = cos x, and y = − cos x. Note also that y = 1 and y = −1
are singular solutions.
56. (a) x
y
3
–3
3
x
–3
√
(b) For |x| > 1 and |y| > 1 the differential equation is dy/dx = y 2 − 1 / x2 − 1 . Separating variables and integrating, we obtain
dx
dy
=√
and cosh−1 y = cosh−1 x + c.
x2 − 1
y2 − 1
Setting x = 2 and y = 2 we find c = cosh−1 2 − cosh−1 2 = 0 and cosh−1 y = cosh−1 x.
An explicit solution is y = x.
57. Since the tension T1 (or magnitude T1 ) acts at the lowest point of the cable, we use symmetry
to solve the problem on the interval [0, L/2]. The assumption that the roadbed is uniform
(that is, weighs a constant ρ pounds per horizontal foot) implies W = ρx, where x is measured
in feet and 0 ≤ x ≤ L/2. Therefore (10) becomes dy/dx = (ρ/T1 )x. This last equation is a
separable equation of the form given in (1) of Section 2.2 in the text. Integrating and using the
initial condition y(0) = a shows that the shape of the cable is a parabola: y(x) = (ρ/2T1 )x2 +a.
In terms of the sag h of the cable and the span L, we see from Figure 2.2.5 in the text that
y(L/2) = h + a. By applying this last condition to y(x) = (ρ/2T1 )x2 + a enables us to express
ρ/2T1 in terms of h and L: y(x) = (4h/L2 )x2 + a. Since y(x) is an even function of x, the
solution is valid on −L/2 ≤ x ≤ L/2.
58. (a) Separating variables and integrating, we have
(3y 2 + 1) dy = −(8x + 5) dx and y 3 + y = −4x2 − 5x + c.
Using a CAS we show various contours of
f (x, y) = y 3 + y + 4x2 + 5x. The plots shown on
[−5, 5]×[−5, 5] correspond to c-values of 0, ±5, ±20, ±40,
±80, and ±125.
y
4
2
x
0
–2
–4
–4
–2
0
2
4
2.2 Separable Variables
61
y
(b) The value of c corresponding to y(0) = −1 is f (0, −1) = −2;
to y(0) = 2 is f (0, 2) = 10; to y(−1) = 4 is f (−1, 4) = 67;
and to y(−1) = −3 is −31.
4
2
x
0
–2
–4
–4
–2
0
2
4
59. (a) An implicit solution of the differential equation (2y + 2)dy − (4x3 + 6x) dx = 0 is
y 2 + 2y − x4 − 3x2 + c = 0.
The condition y(0) = −3 implies that c = −3. Therefore y 2 + 2y − x4 − 3x2 − 3 = 0.
(b) Using the quadratic formula we can solve for y in terms of x:
−2 ±
y=
4 + 4(x4 + 3x2 + 3)
.
2
The explicit solution that satisfies the initial condition is then
y = −1 −
x4 + 3x3 + 4 .
(c) From the graph of the function f (x) = x4 + 3x3 + 4 below we see that f (x) ≤ 0 on the
approximate interval −2.8 ≤ x ≤ −1.3. Thus the approximate domain of the function
y = −1 −
x4 + 3x3 + 4 = −1 −
f (x)
is x ≤ −2.8 or x ≥ −1.3. The graph of this function is shown below.
–1 – √f(x)
f(x)
–4
4
–2
2
–2
2
–4
–4
x
–2
–6
–2
–4
–8
–10
x
62
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
–1 – √f(x)
(d) Using the root finding capabilities of a CAS, the zeros of f are found
to be −2.82202 and −1.3409. The domain of definition of the solution
y(x) is then x > −1.3409. The equality has been removed since the
derivative dy/dx does not exist at the points where f (x) = 0. The
graph of the solution y = φ(x) is given on the right.
2
x
–2
–4
–6
–8
–10
60. (a) Separating variables and integrating, we have
(−2y + y 2 ) dy = (x − x2 ) dx
y
4
2
and
1
1
1
−y 2 + y 3 = x2 − x3 + c
3
2
3
Using a CAS we show some contours of
x
0
–2
–4
f (x, y) = 2y 3 − 6y 2 + 2x3 − 3x2 .
–6
–4
–2
0
2
4
6
The plots shown on [−7, 7] × [−5, 5] correspond to c-values of −450, −300, −200, −120,
−60, −20, −10, −8.1, −5, −0.8, 20, 60, and 120.
(b) The value of c corresponding to y(0) = 32 is
f 0, 32 = − 27
4 . The portion of the graph between the dots corresponds to the solution curve
satisfying the intial condition. To determine the
interval of definition we find dy/dx for
2y 3 − 6y 2 + 2x3 − 3x2 = −
27
.
4
y
4
2
x
0
–2
–4
–2
0
2
4
6
Using implicit differentiation we get y = (x − x2 )/(y 2 − 2y), which is infinite when y = 0
and y = 2. Letting y = 0 in 2y 3 − 6y 2 + 2x3 − 3x2 = − 27
4 and using a CAS to solve for x
we get x = −1.13232. Similarly, letting y = 2, we find x = 1.71299. The largest interval
of definition is approximately (−1.13232, 1.71299).
2.3 Linear Equations
(c) The value of c corresponding to y(0) = −2 is
f (0, −2) = −40. The portion of the graph to the
right of the dot corresponds to the solution curve
satisfying the initial condition. To determine the
interval of definition we find dy/dx for
63
y
4
2
0
x
–2
–4
–6
–8
2y 3 − 6y 2 + 2x3 − 3x2 = −40.
–4
y
–2
0
2
4
6
8
10
(x − x2 )/(y 2 − 2y),
Using implicit differentiation we get =
which is infinite when y = 0
3
2
3
2
and y = 2. Letting y = 0 in 2y − 6y + 2x − 3x = −40 and using a CAS to solve for x
we get x = −2.29551. The largest interval of definition is approximately (−2.29551, ∞).
2.3
Linear Equations
1. For y − 5y = 0 an integrating factor is e−
´
5 dx
= e−5x so that
d −5x e y = 0 and y = ce5x
dx
for −∞ < x < ∞.
´
2. For y + 2y = 0 an integrating factor is e
2 dx
= e2x so that
d 2x e y = 0 and y = ce−2x for
dx
−∞ < x < ∞. The transient term is ce−2x .
´
3. For y + y = e3x an integrating factor is e
dx
= ex so that
d x
[e y] = e4x and y = 14 e3x + ce−x
dx
for −∞ < x < ∞. The transient term is ce−x .
4. For y +4y =
4
3
´
an integrating factor is e
4 dx
= e4x so that
d 4x 4 4x
e y = 3 e and y = 13 +ce−4x
dx
for −∞ < x < ∞. The transient term is ce−4x .
´
5. For y + 3x2 y = x2 an integrating factor is e
y=
1
3
3x2 dx
3
= ex so that
d x3 3
e y = x2 ex and
dx
+ ce−x for −∞ < x < ∞. The transient term is ce−x .
3
3
´
6. For y + 2xy = x3 an integrating factor is e
y = 12 x2 −
1
2
2x dx
2
= ex so that
d x2 2
e y = x3 ex and
dx
+ ce−x for −∞ < x < ∞. The transient term is ce−x .
2
2
´
1
1
1
c
1
d
[xy] = and y = ln x+
7. For y + y = 2 an integrating factor is e (1/x) dx = x so that
x
x
dx
x
x
x
for 0 < x < ∞. The entire solution is transient.
64
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
8. For y −2y = x2 +5 an integrating factor is e−
and y = − 12 x2 − 12 x −
9. For y −
11
4
´
2 dx
= e−2x so that
d −2x e y = x2 e−2x +5e−2x
dx
+ ce2x for −∞ < x < ∞. There is no transient term.
´
d 1
1
1
y = x sin x an integrating factor is e− (1/x) dx = so that
y = sin x and
x
x
dx x
y = cx − x cos x for 0 < x < ∞. There is no transient term.
10. For y +
´
3
2
d 2 y = an integrating factor is e (2/x)dx = x2 so that
x y = 3x and y = 32 +cx−2
x
x
dx
for 0 < x < ∞. The trancient term is cx−2 .
11. For y +
´
4
d 4 y = x2 − 1 an integrating factor is e (4/x)dx = x4 so that
x y = x6 − x4 and
x
dx
y = 17 x3 − 15 x + cx−4 for 0 < x < ∞. The transient term is cx−4 .
´
x
y = x an integrating factor is e− [x/(1+x)]dx = (x + 1)e−x so that
(1 + x)
cex
2x + 3
d (x + 1)e−x y = x(x + 1)e−x and y = −x −
+
for −1 < x < ∞. There
dx
x+1
x+1
12. For y −
is no transient term.
´
2
ex
d 2 x
13. For y + 1 +
y = 2 an integrating factor is e [1+(2/x)]dx = x2 ex so that
[x e y] = e2x
x
x
dx
1 ex
ce−x
ce−x
and y =
+
for
0
<
x
<
∞.
The
transient
term
is
.
2 x2
x2
x2
´
1
1
14. For y + 1 +
y = e−x sin 2x an integrating factor is e [1+(1/x)]dx = xex so that
x
x
1
ce−x
d
[xex y] = sin 2x and y = − e−x cos 2x +
for 0 < x < ∞. The entire solution
dx
2x
x
is transient.
15. For
´
dx 4
d −4 −4
− x = 4y 5 an integrating factor is e− (4/y) dy = eln y = y −4 so that
y x = 4y
dy y
dy
and x = 2y 6 + cy 4 for 0 < y < ∞. There is no transient term.
´
2
d 2 dx
+ x = ey an integrating factor is e (2/y) dy = y 2 so that
y x = y 2 ey and
dy
y
dy
2
2
c
c
x = ey − ey + 2 ey + 2 for 0 < y < ∞. The transient term is 2 .
y
y
y
y
16. For
2.3 Linear Equations
´
17. For y + (tan x)y = sec x an integrating factor is e
tan x dx
= sec x so that
d
[(sec x)y] = sec2 x
dx
and y = sin x + c cos x for −π/2 < x < π/2. There is no transient term.
´
18. For y + (cot x)y = sec2 x csc x an integrating factor is e
cot x dx
= eln | sin x| = sin x so that
d
[(sin x) y] = sec2 x and y = sec x + c csc x for 0 < x < π/2. There is no transient term.
dx
´
2xe−x
x+2
y =
an integrating factor is e [(x+2)/(x+1)]dx = (x + 1)ex , so
x+1
x+1
x2 −x
c
d
[(x + 1)ex y] = 2x and y =
e +
e−x for −1 < x < ∞. The entire
dx
x+1
x+1
19. For y +
solution is transient.
´
5
4
[4/(x+2)] dx = (x + 2)4 so that
y =
an
integrating
factor
is
e
x+2
(x + 2)2
5
d (x + 2)4 y = 5(x + 2)2 and y = (x + 2)−1 + c(x + 2)−4 for −2 < x < ∞. The
dx
3
20. For y +
entire solution is transient.
´
dr
+ r sec θ = cos θ an integrating factor is e sec θ dθ = eln | sec x+tan x| = sec θ + tan θ so
dθ
d
[(sec θ + tan θ)r] = 1 + sin θ and (sec θ + tan θ)r = θ − cos θ + c for −π/2 < θ < π/2.
that
dθ
21. For
There is no transient term.
22. For
´
d t2 −t dP
2
+ (2t − 1)P = 4t − 2 an integrating factor is e (2t−1) dt = et −t so that
P =
e
dt
dt
2 −t
(4t − 2)et
23. For
y +
2
2
and P = 2 + cet−t for −∞ < t < ∞. The transient term is cet−t .
1
3+
x
and y = e−3x +
y=
´
e−3x
d 3x an integrating factor is e [3+(1/x)]dx = xe3x so that
xe y = 1
x
dx
ce−3x
for 0 < x < ∞. The transient term is ce−3x /x.
x
´
x+1
2
[2/(x2 −1)]dx = x − 1
y
=
an
integrating
factor
is
e
2
x −1
x−1
x+1
d x−1
y = 1 and (x − 1)y = x(x + 1) + c(x + 1) for −1 < x < 1. There is no
so that
dx x + 1
24. For y +
transient term.
65
66
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
´
25. For y − 5y = x an integrating factor is e
ˆ
y = e5x
−5 dx
d −5x e y = xe−5x and
dx
1
1
1 −5x
1
e
+ ce5x .
+c =− x−
xe−5x dx = e5x − xe−5x −
5
25
5
25
If y(0) = 3 then c =
1
1 76
1
and y = − x− + e5x . The solution is defined on I = (−∞, ∞).
25
5
25 25
´
26. For y + 3y = 2x and integrating factor is e
−3x
ˆ
y=e
If y(0) =
= e−5x so that
3x
2xe
−3x
dx = e
3 dx
= e3x so that
2 3x 2 3x
xe − e + c
3
9
d 3x e y = 2xe3x and
dx
=
2
2
x − + ce−3x .
3
9
5
2
2 5
1
then c = and y = x − + e−3x . THe solution is defined on I = (−∞, ∞).
3
9
3
9 9
´
1
1
d
1
c
y = ex an integrating factor is e (1/x)dx = x so that
[xy] = ex and y = ex +
x
x
dx
x
x
1 x 2−e
. The solution is defined on
for 0 < x < ∞. If y(1) = 2 then c = 2 − e and y = e +
x
x
27. For y +
I = (0, ∞).
´
1
1
dx
− x = 2y an integrating factor is e− (1/y)dy =
so that
dy
y
y
d 1
x = 2 and
dy y
49
x = 2y 2 + cy for 0 < y < ∞. If y(1) = 5 then c = −49/5 and x = 2y 2 − y. The solution is
5
28. For
defined on I = (0, ∞).
´
E
d Rt/L E
di R
+ i=
an integrating factor is e (R/L) dt = eRt/L so that
e
i = eRt/L
dt
L
L
dt
L
E
E
+ ce−Rt/L for −∞ < t < ∞. If i(0) = i0 then c = i0 − E/R and i =
+
and i =
R
R
E
i0 −
e−Rt/L . The solution is defined on I = (−∞, ∞)
R
29. For
30. For
´
d −kt
dT
− kT = −Tm k an integrating factor is e (−k) dt = e−kt so that
[e T ] = −Tm ke−kt
dt
dt
and T = Tm +cekt for −∞ < t < ∞. If T (0) = T0 then c = T0 −Tm and T = Tm +(T0 −Tm )ekt .
The solution is defined on I = (−∞, ∞)
31. For y +
´
1
1
d
y = 4 + an integrating factor is e (1/x) dx = x so that
[xy] = 4x + 1 and
x
x
dx
ˆ
c
1
1
y=
(4x + 1) dx =
2x2 + x + c = 2x + 1 + .
x
x
x
2.3 Linear Equations
If y(1) = 8 then c = 5 and y = 2x + 1 +
5
. The solution is defined on I = (0, ∞).
x
´
32. For y + 4xy = x3 ex an integrating factor is e
2
−2x2
ˆ
y=e
3 3x2
x e
−2x2
dx = e
If y(0) = −1 then c = −
2
= e2x so that
4x dx
1 2 3x2
1 3x2
x e −
e +c
6
18
=
d 2x2
2
[e y] = x3 e3x and
dx
1 2 x2
1 x2
2
x e −
e + ce−2x .
6
18
1
1 x2 17 −2x2
17
2
and y = x2 ex −
e −
e
. The solution is defined on
18
6
18
18
I = (−∞, ∞).
33. For y +
´
ln x
1
d
y=
an integrating factor is e [1/(x+1)] dx = x+1 so that
[(x+1)y] = ln x
x+1
x+1
dx
and
y=
x
c
x
ln x −
+
x+1
x+1 x+1
If y(1) = 10 then c = 21 and y =
for
0 < x < ∞.
x
21
x
ln x −
+
. The solution is defined on
x+1
x+1
x+1
I = (0, ∞).
´
1
1
y =
an integrating factor is e [1/(x+1)] dx = x + 1 so that
x+1
x (x + 1)
1
d
[(x + 1) y] = and
dx
x
34. For y +
1
y=
x+1
ˆ
If y(e) = 1 then c = e and y =
1
1
ln x
c
dx =
(ln x + c) =
+
.
x
x+1
x+1 x+1
e
ln x
+
. The solution is defined on I = (0, ∞).
x+1 x+1
´
35. For y − (sin x) y = 2 sin x an integrating factor is e
(− sin x) dx
= ecos x so that
d cos x
[e
y] =
dx
2 (sin x) ecos x and
− cos x
y=e
ˆ
2 (sin x) ecos x dx = e− cos x (−2ecos x + c) = −2 + ce− cos x .
If y(π/2) = 1 then c = 3 and y = −2 + 3e− cos x . The solution is defined on I = (−∞, ∞).
67
68
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
´
36. For y + (tan x)y = cos2 x an integrating factor is e
tan x dx
= eln | sec x| = sec x so that
d
[(sec x) y] = cos x and y = sin x cos x + c cos x for −π/2 < x < π/2. If y(0) = −1
dx
then c = −1 and y = sin x cos x − cos x. The solution is defined on I = (−π/2, π/2).
37. For y + 2y = f (x) an integrating factor is e2x so that
⎧
1
⎪
⎨ e2x + c1 , 0 ≤ x ≤ 3
ye2x = 2
⎪
⎩
x > 3.
c2 ,
y
1
y=
⎧
1
−2x
⎪
⎪
⎨ 2 (1 − e ),
0≤x≤3
⎪
⎪
⎩ 1 (e6 − 1)e−2x ,
2
x > 3.
38. For y + y = f (x) an integrating factor is ex so that
⎧ x
⎨e + c1 ,
0≤x≤1
yex =
⎩−ex + c , x > 1.
2
If y(0) = 1 then c1 = 0 and for continuity we must have
c2 = 2e so that
⎧
⎨1,
0≤x≤1
y=
⎩2e1−x − 1, x > 1.
39. For y + 2xy = f (x) an integrating factor is ex so that
⎧
1 2
⎪
⎨ ex + c1 , 0 ≤ x ≤ 1
2
yex = 2
⎪
⎩
c2 ,
x > 1.
x
5
If y(0) = 0 then c1 = −1/2 and for continuity we must
have c2 = 12 e6 − 12 so that
y
1
5 x
–1
2
If y(0) = 2 then c1 = 3/2 and for continuity we must have
c2 = 12 e + 32 so that
⎧
1 3 −x2
⎪
⎪
0≤x≤1
⎪
⎨2 + 2e ,
y= ⎪
3 −x2
1
⎪
⎪
e , x > 1.
⎩ e+
2
2
y
2
3
x
2.3 Linear Equations
40. For
y +
⎧ x
⎪
⎪
⎨ 1 + x2 , 0 ≤ x ≤ 1
2x
y=
⎪
1 + x2
−x
⎪
⎩
,
1 + x2
If
y
1
x > 1,
5
x2
an integrating factor is 1 +
so that
⎧
1 2
⎪
⎪
0≤x≤1
⎨ 2 x + c1 ,
2
1+x y =
⎪
⎪
⎩− 1 x2 + c2 , x > 1.
2
69
x
–1
y(0) = 0 then c1 = 0 and for continuity we must have c2 = 1 so that
⎧
1
1
⎪
⎪
−
, 0≤x≤1
⎪
⎨ 2 2 (1 + x2 )
y=
⎪
⎪
1
3
⎪
⎩
− , x > 1.
2 (1 + x2 ) 2
41. We first solve the initial-value problem y + 2y = 4x,
y(0) = 3 on the interval [0, 1]. The integrating factor is
´
e 2 dx = e2x , so
d 2x
[e y] = 4xe2x
dx
ˆ
e2x y = 4xe2x dx = 2xe2x − e2x + c1
y
20
15
10
5
y = 2x − 1 + c1 e−2x .
3
x
Using the initial condition, we find y(0) = −1 + c1 = 3, so c1 = 4 and y = 2x − 1 + 4e−2x ,
0 ≤ x ≤ 1. Now, since y(1) = 2 − 1 + 4e−2 = 1 + 4e−2 , we solve the initial-value problem
y − (2/x)y = 4x, y(1) = 1 + 4e−2 on the interval (1, ∞). The integrating factor is
´
e
(−2/x) dx
= e−2 ln x = x−2 , so
4
d −2
[x y] = 4xx−2 =
dx
x
ˆ
4
dx = 4 ln x + c2
x−2 y =
x
y = 4x2 ln x + c2 x2 .
(We use ln x instead of ln |x| because x > 1.) Using the initial condition we find
70
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
y(1) = c2 = 1 + 4e−2 , so y = 4x2 ln x + (1 + 4e−2 )x2 , x > 1. Thus,
⎧
2x − 1 + 4e−2x ,
⎪
⎨
y=
0≤x≤1
⎪
⎩ 2
4x ln x + 1 + 4e−2 x2 , x > 1.
42. We first solve the initial-value problem y + y = 0,
y(0) = 4 on the interval [0, 2]. The integrating factor
´
is e 1 dx = ex , so
y
1
5 x
d x
[e y] = 0
dx
ˆ
x
e y = 0 dx = c1
–1
y = c1 e−x .
Using the initial condition, we find y(0) = c1 = 4, so c1 = 4 and y = 4e−x , 0 ≤ x ≤ 2. Now,
since y(2) = 4e−2 , we solve the initial-value problem y + 5y = 0, y(1) = 4e−2 on the interval
´
(2, ∞). The integrating factor is e
5 dx
= e5x , so
d 5x e y =0
dx
ˆ
5x
e y = 0 dx = c2
y = c2 e−5x .
Using the initial condition we find y(2) = c2 e−10 = 4e−2 , so c2 = 4e8 and y = 4e8 e−5x =
4e8−5x , x > 2. Thus, the solution of the original initial-value problem is
⎧ −x
4e ,
⎪
⎨
y=
0≤x≤2
⎪
⎩ 8−5x
, x > 2.
4x
2.3 Linear Equations
43. An integrating factor for y − 2xy = 1 is e−x . Thus
2
d −x2
2
[e y] = e−x
dx
√
ˆ x
π
−x2
−t2
erf(x) + c
e y=
e dt =
2
0
√
π x2
2
e erf(x) + cex .
y=
2
√
From y(1) = ( π/2)e erf(1) + ce = 1 we get c = e−1 −
√
π
2
erf(1). The solution of the
initial-value problem is
√
π x2
π
2
−1
e erf(x) + e −
erf(1) ex
y=
2
2
√
π x2
2
e (erf(x) − erf(1)).
= ex −1 +
2
√
44. An integrating factor for y − 2xy = −1 is e−x . Thus
2
d −x2
2
[e y] = −e−x
dx
√
ˆ x
π
−x2
−t2
erf(x) + c.
e y=−
e dt = −
2
0
From y(0) =
√
π/2, and noting that erf(0) = 0, we get c =
y = ex
2
√
π/2. Thus
√
√ √
√
π
π
π x2
π x2
−
erf(x) +
=
e (1 − erf(x)) =
e erfc (x).
2
2
2
2
45. For y + ex y = 1 an integrating factor is ee . Thus
x
d ex x
e y = ee
dx
From y(0) = 1 we get c = e, so y = e−e
x
ex
ˆ
and e y =
x
t
ee dt + c.
0
´x
0
t
x
ee dt + e1−e .
1
y = x. An integrating factor is e1/x . Thus
x2
ˆ x
d 1/x 1/x
1/x
and e y =
te1/t dt + c.
e y = xe
dx
1
46. Dividing by x2 we have y −
From y(1) = 0 we get c = 0, so y = e−1/x
´x
1
te1/t dt.
71
72
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
47. An integrating factor for
y +
10 sin x
2
y=
x
x3
is x2 . Thus
sin x
d 2 x y = 10
dx
x
ˆ x
sin t
dt + c
x2 y = 10
t
0
y = 10x−2 Si (x) + cx−2 .
From y(1) = 0 we get c = −10 Si (1). Thus
y = 10x−2 Si (x) − 10x−2 Si (1) = 10x−2 (Si (x) − Si (1)) .
48. The integrating factor for y − sin x2 y = 0 is e−
´x
0
sin t2 dt
. Then
d − ´ x sin t2 dt y =0
e 0
dx
e−
´x
0
sin t2 dt
y = c1
´x
y = c1 e
Letting t =
π/2 u we have dt =
ˆ
x
0
0
sin t2 dt
π/2 du and
ˆ √2/π x
π π
π
2
u2 du =
S
x
sin t2 dt =
sin
2 0
2
2
π
√
√
√
√
π/2 S
2/π x
2/π x
π/2 S
. Using S(0) = 0 and y(0) = c1 = 5 we have y = 5e
.
so y = c1 e
49. We want 4 to be a critical point, so we use y = 4 − y.
50. (a) All solutions of the form y = x5 ex − x4 ex + cx4 satisfy the initial condition. In this case,
since 4/x is discontinuous at x = 0, the hypotheses of Theorem 1.2.1 are not satisfied
and the initial-value problem does not have a unique solution.
2.3 Linear Equations
(b) The differential equation has no solution satisfying y(0) = y0 , y0 > 0.
(c) In this case, since x0 > 0, Theorem 1.2.1 applies and the initial-value problem has a
unique solution given by y = x5 ex − x4 ex + cx4 where c = y0 /x40 − x0 ex0 + ex0 .
51. On the interval (−3, 3) the integrating factor is
´
e
x dx/(x2 −9)
= e−
´
x dx/(9−x2 )
1
= e 2 ln (9−x
2)
=
9 − x2
and so
d 9 − x2 y = 0
dx
and y = √
c
.
9 − x2
52. We want the general solution to be y = 3x − 5 + ce−x . (Rather than e−x , any function that
approaches 0 as x → ∞ could be used.) Differentiating we get
y = 3 − ce−x = 3 − (y − 3x + 5) = −y + 3x − 2,
so the differential equation y + y = 3x − 2 has solutions asymptotic to the line y = 3x − 5.
53. The left-hand derivative of the function at x = 1 is 1/e and the right-hand derivative at x = 1
is 1 − 1/e. Thus, y is not differentiable at x = 1.
54. (a) Differentiating yc = c/x3 we get
yc = −
3c
3 c
3
=−
= − yc
x4
x x3
x
so a differential equation with general solution yc = c/x3 is xy + 3y = 0. Now using
yp = x3
xyp + 3yp = x(3x2 ) + 3(x3 ) = 6x3
so a differential equation with general solution y = c/x3 + x3 is xy + 3y = 6x3 . This
will be a general solution on (0, ∞).
73
74
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
(b) Since y(1) = 13 −1/13 = 0, an initial condition is y(1) = 0.
Since y(1) = 13 +2/13 = 3, an initial condition is y(1) = 3.
In each case the interval of definition is (0, ∞). The initialvalue problem xy +3y = 6x3 , y(0) = 0 has solution y = x3
for −∞ < x < ∞. In the figure the lower curve is the
graph of y(x) = x3 − 1/x3 , while the upper curve is the
graph of y = x3 − 2/x3 .
y
3
5
x
–3
(c) The first two initial-value problems in part (b) are not unique. For example, setting
y(2) = 23 − 1/23 = 63/8, we see that y(2) = 63/8 is also an initial condition leading to
the solution y = x3 − 1/x3 .
´
55. Since e
´
c1 e
P (x) dx+c
´
= ec e
P (x) dx
ˆ
P (x) dx
y = c2 +
´
c1 e
´
= c1 e
P (x) dx
P (x) dx ,
we would have
´
f (x) dx
and e
ˆ
P (x) dx
y = c3 +
´
e
P (x) dx
f (x) dx,
which is the same as (4) in the text.
56. We see by inspection that y = 0 is a solution.
57. The solution of the first equation is x = c1 e−λ1 t . From x(0) = x0 we obtain c1 = x0 and so
x = x0 e−λ1 t . The second equation then becomes
dy
= x0 λ1 e−λ1 t − λ2 y
dt
or
dy
+ λ2 y = x0 λ1 e−λ1 t
dt
which is linear. An integrating factor is eλ2 t . Thus
d λ2 t e y = x0 λ1 e−λ1 t eλ2 t = x0 λ1 e(λ2 −λ1 )t
dt
eλ2 t y =
y=
x0 λ1 (λ2 −λ1 )t
e
+ c2
λ2 − λ1
x0 λ1 −λ1 t
e
+ c2 e−λ2 t .
λ2 − λ1
2.3 Linear Equations
From y(0) = y0 we obtain c2 = (y0 λ2 − y0 λ1 − x0 λ1 ) / (λ2 − λ1 ). The solution is
y=
x0 λ1 −λ1 t y0 λ2 − y0 λ1 − x0 λ1 −λ2 t
e
+
e
.
λ2 − λ1
λ2 − λ1
58. Writing the differential equation as
1
dE
+
E = 0 we see that an integrating factor is
dt
RC
et/RC . Then
d t/RC e
E =0
dt
et/RC E = c
E = ce−t/RC
From E(4) = ce−4/RC = E0 we find c = E0 e4/RC . Thus, the solution of the initial-value
problem is
E = E0 e4/RC e−t/RC = E0 e−(t−4)/RC .
59. (a) x
(b) Using a CAS we find y(2) ≈ 0.226339.
y
5
5 x
60. (a) x
2
y
1
1
–1
–2
–3
–4
–5
2
3
4
5
x
75
76
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
(b) From the graph in part (b) we see that the absolute maximum occurs around x = 1.7.
Using the root-finding capability of a CAS and solving y (x) = 0 for x we see that the
absolute maximum is (1.688, 1.742).
61. (a) x
y
10
5
–10
–5
5
10
x
(b) From the graph we see that as x → ∞, y(x) oscillates with decreasing amplitudes ap√
1
proaching 9.35672. Since lim S(x) = , we have lim y(x) = 5e π/8 ≈ 9.357, and
x→∞
x→∞
2
√
1
since lim S(x) = − , we have lim y(x) = 5e− π/8 ≈ 2.672.
x→−∞
x→−∞
2
(c) From the graph in part (b) we see that the absolute maximum occurs around x = 1.7
and the absolute minimum occurs around x = −1.8. Using the root-finding capability of
a CAS and solving y (x) = 0 for x, we see that the absolute maximum is (1.772, 12.235)
and the absolute minimum is (−1.772, 2.044).
2.4
Exact Equations
1. Let M = 2x − 1 and N = 3y + 7 so that My = 0 = Nx . From fx = 2x − 1 we obtain
f = x2 − x + h(y), h (y) = 3y + 7, and h(y) = 32 y 2 + 7y. A solution is x2 − x + 32 y 2 + 7y = c.
2. Let M = 2x + y and N = −x − 6y. Then My = 1 and Nx = −1, so the equation is not exact.
3. Let M = 5x + 4y and N = 4x − 8y 3 so that My = 4 = Nx . From fx = 5x + 4y we obtain
f = 52 x2 + 4xy + h(y), h (y) = −8y 3 , and h(y) = −2y 4 . A solution is 52 x2 + 4xy − 2y 4 = c.
4. Let M = sin y − y sin x and N = cos x + x cos y − y so that My = cos y − sin x = Nx . From
fx = sin y − y sin x we obtain f = x sin y + y cos x + h(y), h (y) = −y, and h(y) = − 12 y 2 . A
solution is x sin y + y cos x − 12 y 2 = c.
2.4 Exact Equations
5. Let M = 2y 2 x − 3 and N = 2yx2 + 4 so that My = 4xy = Nx . From fx = 2y 2 x − 3 we obtain
f = x2 y 2 − 3x + h(y), h (y) = 4, and h(y) = 4y. A solution is x2 y 2 − 3x + 4y = c.
6. Let M = 4x3 − 3y sin 3x − y/x2 and N = 2y − 1/x + cos 3x so that My = −3 sin 3x − 1/x2
and Nx = 1/x2 − 3 sin 3x. The equation is not exact.
7. Let M = x2 − y 2 and N = x2 − 2xy so that My = −2y and Nx = 2x − 2y. The equation is
not exact.
8. Let M = 1 + ln x + y/x and N = −1 + ln x so that My = 1/x = Nx . From fy = −1 + ln x we
obtain f = −y + y ln x + h(x), h (x) = 1 + ln x, and h(x) = x ln x. A solution is −y + y ln x +
x ln x = c.
9. Let M = y 3 − y 2 sin x − x and N = 3xy 2 + 2y cos x so that My = 3y 2 − 2y sin x = Nx . From
fx = y 3 − y 2 sin x − x we obtain f = xy 3 + y 2 cos x − 12 x2 + h(y), h (y) = 0, and h(y) = 0. A
solution is xy 3 + y 2 cos x − 12 x2 = c.
10. Let M = x3 + y 3 and N = 3xy 2 so that My = 3y 2 = Nx . From fx = x3 + y 3 we obtain
f = 14 x4 + xy 3 + h(y), h (y) = 0, and h(y) = 0. A solution is 14 x4 + xy 3 = c.
11. Let M = y ln y − e−xy and N = 1/y + x ln y so that My = 1 + ln y + xe−xy and Nx = ln y.
The equation is not exact.
12. Let M = 3x2 y + ey and N = x3 + xey − 2y so that My = 3x2 + ey = Nx . From fx =
3x2 y + ey we obtain f = x3 y + xey + h(y), h (y) = −2y, and h(y) = −y 2 . A solution is
x3 y + xey − y 2 = c.
13. Let M = y − 6x2 − 2xex and N = x so that My = 1 = Nx . From fx = y − 6x2 − 2xex
we obtain f = xy − 2x3 − 2xex + 2ex + h(y), h (y) = 0, and h(y) = 0. A solution is
xy − 2x3 − 2xex + 2ex = c.
14. Let M = 1 − 3/x + y and N = 1 − 3/y + x so that My = 1 = Nx . From fx = 1 − 3/x + y
3
we obtain f = x − 3 ln |x| + xy + h(y), h (y) = 1 − , and h(y) = y − 3 ln |y|. A solution is
y
x + y + xy − 3 ln |xy| = c.
15. Let M = x2 y 3 − 1/ 1 + 9x2
and N = x3 y 2 so that My = 3x2 y 2 = Nx . From
fx = x2 y 3 − 1/ 1 + 9x2 we obtain f = 13 x3 y 3 − 13 arctan (3x) + h(y), h (y) = 0, and
h(y) = 0. A solution is x3 y 3 − arctan (3x) = c.
16. Let M = −2y and N = 5y − 2x so that My = −2 = Nx . From fx = −2y we obtain
f = −2xy + h(y), h (y) = 5y, and h(y) = 52 y 2 . A solution is −2xy + 52 y 2 = c.
17. Let M = tan x − sin x sin y and N = cos x cos y so that My = − sin x cos y = Nx . From
fx = tan x − sin x sin y we obtain f = ln | sec x| + cos x sin y + h(y), h (y) = 0, and h(y) = 0.
A solution is ln | sec x| + cos x sin y = c.
2
2
18. Let M = 2y sin x cos x − y + 2y 2 exy and N = −x + sin2 x + 4xyexy so that
2
2
My = 2 sin x cos x − 1 + 4xy 3 exy + 4yexy = Nx .
77
78
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
From fx = 2y sin x cos x − y + 2y 2 exy we obtain f = y sin2 x − xy + 2exy + h(y), h (y) = 0,
2
and h(y) = 0. A solution is y sin2 x − xy + 2exy = c.
2
2
19. Let M = 4t3 y − 15t2 − y and N = t4 + 3y 2 − t so that My = 4t3 − 1 = Nt . From
ft = 4t3 y − 15t2 − y we obtain f = t4 y − 5t3 − ty + h(y), h (y) = 3y 2 , and h(y) = y 3 .
A solution is t4 y − 5t3 − ty + y 3 = c.
20. Let M = 1/t + 1/t2 − y/ t2 + y 2 and N = yey + t/ t2 + y 2 so that
2 2
2
2
2
we obtain
My = y 2 − t 2 / t 2 +
y = Nt . From ft = 1/t + 1/t − y/ t + y
1
t
f = ln |t| − − arctan
+ h(y), h (y) = yey , and h(y) = yey − ey . A solution is
t
y
t
1
+ yey − ey = c.
ln |t| − − arctan
t
y
21. Let M = x2 + 2xy + y 2 and N = 2xy + x2 − 1 so that My = 2(x + y) = Nx . From
fx = x2 + 2xy + y 2 we obtain f = 13 x3 + x2 y + xy 2 + h(y), h (y) = −1, and h(y) = −y. The
solution is 13 x3 +x2 y +xy 2 −y = c. If y(1) = 1 then c = 4/3 and a solution of the initial-value
problem is 13 x3 + x2 y + xy 2 − y = 43 .
22. Let M = ex + y and N = 2 + x + yey so that My = 1 = Nx . From fx = ex + y we
obtain f = ex + xy + h(y), h (y) = 2 + yey , and h(y) = 2y + yey − ey . The solution is
ex + xy + 2y + yey − ey = c. If y(0) = 1 then c = 3 and a solution of the initial-value problem
is ex + xy + 2y + yey − ey = 3.
23. Let M = 4y + 2t − 5 and N = 6y + 4t − 1 so that My = 4 = Nt . From ft = 4y + 2t − 5
we obtain f = 4ty + t2 − 5t + h(y), h (y) = 6y − 1, and h(y) = 3y 2 − y. The solution is
4ty + t2 − 5t + 3y 2 − y = c. If y(−1) = 2 then c = 8 and a solution of the initial-value problem
is 4ty + t2 − 5t + 3y 2 − y = 8.
24. Let M = t/2y 4 and N = 3y 2 − t2 /y 5 so that My = −2t/y 5 = Nt . From ft = t/2y 4 we
3
3
t2
3
t2
+
h(y),
h
(y)
=
,
and
h(y)
=
−
.
The
solution
is
− 2 = c. If
obtain f =
4
3
2
4
4y
y
2y
4y
2y
3
5
t2
y(1) = 1 then c = −5/4 and a solution of the initial-value problem is 4 − 2 = − .
4y
2y
4
25. Let M = y 2 cos x − 3x2 y − 2x and N = 2y sin x − x3 + ln y so that My = 2y cos x − 3x2 = Nx .
From fx = y 2 cos x − 3x2 y − 2x we obtain f = y 2 sin x − x3 y − x2 + h(y), h (y) = ln y, and
h(y) = y ln y − y. The solution is y 2 sin x − x3 y − x2 + y ln y − y = c. If y(0) = e then c = 0
and a solution of the initial-value problem is y 2 sin x − x3 y − x2 + y ln y − y = 0.
26. Let M = y 2 + y sin x and N = 2xy − cos x − 1/ 1 + y 2 so that My = 2y + sin x = Nx . From
−1
, and h(y) = − tan−1 y.
fx = y 2 + y sin x we obtain f = xy 2 − y cos x + h(y), h (y) =
1 + y2
The solution is xy 2 − y cos x − tan−1 y = c. If y(0) = 1 then c = −1 − π/4 and a solution of
π
the initial-value problem is xy 2 − y cos x − tan−1 y = −1 − .
4
2.4 Exact Equations
27. Equating My = 3y 2 + 4kxy 3 and Nx = 3y 2 + 40xy 3 we obtain k = 10.
28. Equating My = 18xy 2 − sin y and Nx = 4kxy 2 − sin y we obtain k = 9/2.
29. Let M = −x2 y 2 sin x + 2xy 2 cos x and N = 2x2 y cos x so that
My = −2x2 y sin x + 4xy cos x = Nx . From fy = 2x2 y cos x we obtain f = x2 y 2 cos x + h(y),
h (y) = 0, and h(y) = 0. A solution of the differential equation is x2 y 2 cos x = c.
30. Let M = (x2 + 2xy − y 2 )/(x2 + 2xy + y 2 ) and N = (y 2 + 2xy − x2 )/(y 2 + 2xy + x2 ) so
that My = −4xy/(x + y)3 = Nx . From fx = x2 + 2xy + y 2 − 2y 2 /(x + y)2 we obtain
2y 2
+ h(y), h (y) = −1, and h(y) = −y. A solution of the differential equation is
f = x+
x+y
x2 + y 2 = c(x + y).
´
31. We note that (My −Nx )/N = 1/x, so an integrating factor is e dx/x = x. Let M = 2xy 2 +3x2
and N = 2x2 y so that My = 4xy = Nx . From fx = 2xy 2 +3x2 we obtain f = x2 y 2 +x3 +h(y),
h (y) = 0, and h(y) = 0. A solution of the differential equation is x2 y 2 + x3 = c.
´
32. We note that (My − Nx )/N = 1, so an integrating factor is e dx = ex . Let
M = xyex + y 2 ex + yex and N = xex + 2yex so that My = xex + 2yex + ex = Nx . From
fy = xex + 2yex we obtain f = xyex + y 2 ex + h(x), h (x) = 0, and h(x) = 0. A solution of
the differential equation is xyex + y 2 ex = c.
´
33. We note that (Nx − My )/M = 2/y, so an integrating factor is e 2 dy/y = y 2 . Let M = 6xy 3
and N = 4y 3 + 9x2 y 2 so that My = 18xy 2 = Nx . From fx = 6xy 3 we obtain f = 3x2 y 3 + h(y),
h (y) = 4y 3 , and h(y) = y 4 . A solution of the differential equation is 3x2 y 3 + y 4 = c.
´
34. We note that (My − Nx )/N = − cot x, so an integrating factor is e− cot x dx = csc x. Let
M = cos x csc x = cot x and N = (1 + 2/y) sin x csc x = 1 + 2/y, so that My = 0 = Nx . From
fx = cot x we obtain f = ln (sin x) + h(y), h (y) = 1 + 2/y, and h(y) = y + ln y 2 . A solution
of the differential equation is ln (sin x) + y + ln y 2 = c.
´
35. We note that (My − Nx )/N = 3, so an integrating factor is e 3 dx = e3x . Let
M = (10 − 6y + e−3x )e3x = 10e3x − 6ye3x + 1 and N = −2e3x , so that My = −6e3x = Nx .
3x
3x
From fx = 10e3x − 6ye3x + 1 we obtain f = 10
3 e − 2ye + x + h(y), h (y) = 0, and h(y) = 0.
3x
3x
A solution of the differential equation is 10
3 e − 2ye + x = c.
´
36. We note that (Nx − My )/M = −3/y, so an integrating factor is e−3 dy/y = 1/y 3 . Let
M = (y 2 + xy 3 )/y 3 = 1/y + x and N = (5y 2 − xy + y 3 sin y)/y 3 = 5/y − x/y 2 + sin y, so that
My = −1/y 2 = Nx . From fx = 1/y+x we obtain f = x/y+ 12 x2 +h(y), h (y) = 5/y+sin y, and
h(y) = 5 ln |y| − cos y. A solution of the differential equation is x/y + 12 x2 + 5 ln |y| − cos y = c.
37. We note that (My − Nx )/N = 2x/(4 + x2 ), so an integrating factor is
´
2
e−2 x dx/(4+x ) = 1/(4 + x2 ). Let M = x/(4 + x2 ) and N = (x2 y + 4y)/(4 + x2 ) = y, so
that My = 0 = Nx . From fx = x(4 + x2 ) we obtain f = 12 ln (4 + x2 ) + h(y), h (y) = y, and
h(y) = 12 y 2 . A solution of the differential equation is 12 ln (4 + x2 )+ 12 y 2 = c. Multiplying both
79
80
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
2
sides by 2 the last equation can be written as ey x2 + 4 = c1 . Using the initial condition
2
y(4) = 0 we see that c1 = 20. A solution of the initial-value problem is ey x2 + 4 = 20.
´
38. We note that (My −Nx )/N = −3/(1+x), so an integrating factor is e−3 dx/(1+x) = 1/(1+x)3 .
Let M = (x2 + y 2 − 5)/(1 + x)3 and N = −(y + xy)/(1 + x)3 = −y/(1 + x)2 , so that
My = 2y/(1 + x)3 = Nx . From fy = −y/(1 + x)2 we obtain f = − 12 y 2 /(1 + x)2 + h(x),
h (x) = (x2 − 5)/(1 + x)3 , and h(x) = 2/(1 + x)2 + 2/(1 + x) + ln |1 + x|. A solution of the
differential equation is
−
2
2
y2
+
+
+ ln |1 + x| = c.
2
2
2(1 + x)
(1 + x)
(1 + x)
Using the initial condition y(0) = 1 we see that c = 7/2. A solution of the initial-value
problem is
2
2
7
y2
−
2 +
2 + 1 + x + ln |1 + x| = 2
2 (1 + x)
(1 + x)
39. (a) Implicitly differentiating x3 + 2x2 y + y 2 = c and solving for dy/dx we obtain
3x2 + 2x2
dy
dy
+ 4xy + 2y
= 0 and
dx
dx
dy
3x2 + 4xy
=− 2
.
dx
2x + 2y
By writing the last equation in differential form we get (4xy + 3x2 )dx + (2y + 2x2 )dy = 0.
(b) Setting x = 0 and y = −2 in x3 + 2x2 y + y 2 = c we find c = 4, and setting x = y = 1 we
also find c = 4. Thus, both initial conditions determine the same implicit solution.
(c) Solving x3 + 2x2 y + y 2 = 4 for y we get
y1 (x) = −x2 − 4 − x3 + x4
and
y2 (x) = −x2 +
4 − x3 + x4 .
Observe in the figure that y1 (0) = −2 and y2 (1) = 1.
4
2
–4
y
y2
2
–2
4
x
–2
y1
–4
–6
40. To see that the equations are not equivalent consider dx = −(x/y) dy. An integrating factor
is μ(x, y) = y resulting in y dx + x dy = 0. A solution of the latter equation is y = 0, but this
is not a solution of the original equation.
41. The explicit solution is y = (3 + cos2 x)/(1 − x2 ) . Since 3 + cos2 x > 0 for all x we must
have 1 − x2 > 0 or −1 < x < 1. Thus, the interval of definition is (−1, 1).
y
42. (a) Since fy = N (x, y) = xexy + 2xy + 1/x we obtain f = exy + xy 2 + + h(x) so that
x
y
y
fx = yexy + y 2 − 2 + h (x). Let M (x, y) = yexy + y 2 − 2 .
x
x
2.4 Exact Equations
−1
(b) Since fx = M (x, y) = y 1/2 x−1/2 + x x2 + y
we obtain
1
1 2
1/2 1/2
2
x +y
f = 2y x + ln x + y + g(y) so that fy = y −1/2 x1/2 +
2
2
1 2
−1
x +y
.
N (x, y) = y −1/2 x1/2 +
2
43. First note that
d
−1
+ g (y). Let
x
y
x2 + y 2 = dx + dy.
2
2
2
x +y
x + y2
x2 + y 2 dx becomes
x
y
dx + dy = d
x2 + y 2 = dx.
x2 + y 2
x2 + y 2
The left side is the total differential of x2 + y 2 and the right side is the total differential of
x + c. Thus x2 + y 2 = x + c is a solution of the differential equation.
Then x dx + y dy =
44. To see that the statement is true, write the separable equation as −g(x) dx + dy/h(y) = 0.
Identifying M = −g(x) and N = 1/h(y), we see that My = 0 = Nx , so the differential
equation is exact.
45. (a) In differential form
v 2 − gx dx + xv dv = 0
This is not an exact equation, but μ(x) = x is an integrating factor. The new equation
xv 2 − gx2 dx + x2 v dv = 0 is exact and solving yields 12 x2 v 2 − g3 x3 = c. When x = 1,
v = 0 and so c = − g3 . Solving 12 x2 v 2 − g3 x3 = − g3 for v yields the explicit solution
1
2g
x− 2 .
v(x) =
3
x
(b) The chain leaves the platform when x = 2.5, and so
2g
1
v(2.5) =
≈ 3.91 m/s
2.5 −
3
2.52
46. (a) Letting
M (x, y) =
(x2
2xy
+ y 2 )2
and
N (x, y) = 1 +
y 2 − x2
(x2 + y 2 )2
we compute
2x3 − 8xy 2
= Nx ,
(x2 + y 2 )3
so the differential equation is exact. Then we have
2xy
∂f
= M (x, y) = 2
= 2xy(x2 + y 2 )−2
∂x
(x + y 2 )2
My =
f (x, y) = −y(x2 + y 2 )−1 + g(y) = −
x2
y
+ g(y)
+ y2
∂f
y 2 − x2
y 2 − x2
= 2
+
g
(y)
=
N
(x,
y)
=
1
+
.
∂y
(x + y 2 )2
(x2 + y 2 )2
81
82
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
Thus, g (y) = 1 and g(y) = y. The solution is y −
x2
y
= c. When c = 0 the solution
+ y2
is x2 + y 2 = 1.
(b) The first graph below is obtained in Mathematica using f (x, y) = y − y/(x2 + y 2 ) and
ContourPlot[f[x, y], {x, -3, 3}, {y, -3, 3},
Axes−>True, AxesOrigin−>{0, 0}, AxesLabel−>{x, y},
Frame−>False, PlotPoints−>100, ContourShading−>False,
Contours−>{0, -0.2, 0.2, -0.4, 0.4, -0.6, 0.6, -0.8, 0.8}]
The second graph uses
x=−
y 3 − cy 2 − y
c−y
and
x=
y 3 − cy 2 − y
.
c−y
In this case the x-axis is vertical and the y-axis is horizontal. To obtain the third graph,
we solve y−y/(x2 +y 2 ) = c for y in a CAS. This appears to give one real and two complex
solutions. When graphed in Mathematica however, all three solutions contribute to the
graph. This is because the solutions involve the square root of expressions containing c.
For some values of c the expression is negative, causing an apparent complex solution to
actually be real.
y
3
2
2.5
–2
–1
3
2
2
1
1
–3
x
3
1
2
3
x
–1.5 –1 –0.5
y
1
0.5 1
y
1.5
–3
–2
–1
1
–1
–1
–1
–2
–2
–2
–3
–3
–3
Solutions by Substitutions
1. Letting y = ux we have
(x − ux) dx + x(u dx + x du) = 0
dx + x du = 0
dx
+ du = 0
x
ln |x| + u = c
x ln |x| + y = cx.
2
3
2.5
Solutions by Substitutions
2. Letting y = ux we have
(x + ux) dx + x(u dx + x du) = 0
(1 + 2u) dx + x du = 0
du
dx
+
=0
x
1 + 2u
ln |x| +
1
ln |1 + 2u| = c
2
y
x2 1 + 2
= c1
x
x2 + 2xy = c1 .
3. Letting x = vy we have
vy(v dy + y dv) + (y − 2vy) dy = 0
vy 2 dv + y v 2 − 2v + 1 dy = 0
dy
v dv
=0
+
2
(v − 1)
y
ln |v − 1| −
ln
1
+ ln |y| = c
v−1
x
1
−1 −
+ ln |y| = c
y
x/y − 1
(x − y) ln |x − y| − y = c(x − y).
4. Letting x = vy we have
y(v dy + y dv) − 2(vy + y) dy = 0
y dv − (v + 2) dy = 0
dy
dv
−
=0
v+2
y
ln |v + 2| − ln |y| = c
ln
x
+ 2 − ln |y| = c
y
x + 2y = c1 y 2 .
83
84
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
5. Letting y = ux we have
u2 x2 + ux2 dx − x2 (u dx + x du) = 0
u2 dx − x du = 0
dx du
− 2 =0
x
u
1
=c
u
x
ln |x| + = c
y
ln |x| +
y ln |x| + x = cy.
6. Letting y = ux and using partial fractions, we have
u2 x2 + ux2 dx + x2 (u dx + x du) = 0
x2 u2 + 2u dx + x3 du = 0
du
dx
+
=0
x
u(u + 2)
ln |x| +
1
1
ln |u| − ln |u + 2| = c
2
2
x2 u
= c1
u+2
y
y
+2
x2 = c1
x
x
x2 y = c1 (y + 2x).
7. Letting y = ux we have
(ux − x) dx − (ux + x)(u dx + x du) = 0
u2 + 1 dx + x(u + 1) du = 0
u+1
dx
+ 2
du = 0
x
u +1
1
ln u2 + 1 + tan−1 u = c
2
2
y
2 y
ln x
+ 1 + 2 tan−1 = c1
2
x
x
ln |x| +
ln x2 + y 2 + 2 tan−1
y
= c1
x
2.5
Solutions by Substitutions
8. Letting y = ux we have
(x + 3ux) dx − (3x + ux)(u dx + x du) = 0
u2 − 1 dx + x(u + 3) du = 0
u+3
dx
+
du = 0
x
(u − 1)(u + 1)
ln |x| + 2 ln |u − 1| − ln |u + 1| = c
x(u − 1)2
= c1
u+1
2
y
y
− 1 = c1
+1
x
x
x
(y − x)2 = c1 (y + x).
9. Letting y = ux we have
−ux dx + (x +
2
√
u x)(u dx + x du) = 0
2√
(x + x u ) du + xu3/2 dx = 0
1
dx
−3/2
=0
+
u
du +
u
x
−2u−1/2 + ln |u| + ln |x| = c
ln |y/x| + ln |x| = 2 x/y + c
y(ln |y| − c)2 = 4x.
10. Letting y = ux we have
x2 − u2 x2 dx − x2 du = 0
x 1 − u2 dx − x2 du = 0,
(x > 0)
du
dx
−√
=0
x
1 − u2
ln x − sin−1 u = c
sin−1 u = ln x + c1
sin−1
y
= ln x + c2
x
y
= sin (ln x + c2 )
x
y = x sin (ln x + c2 ).
See Problem 33 in this section for an analysis of the solution.
85
86
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
11. Letting y = ux we have
x3 − u3 x3 dx + u2 x3 (u dx + x du) = 0
dx + u2 x du = 0
dx
+ u2 du = 0
x
1
ln |x| + u3 = c
3
3x3 ln |x| + y 3 = c1 x3 .
Using y(1) = 2 we find c1 = 8. The solution of the initial-value problem is 3x3 ln |x|+y 3 = 8x3 .
12. Letting y = ux we have
(x2 + 2u2 x2 )dx − ux2 (u dx + x du) = 0
x2 (1 + u2 )dx − ux3 du = 0
u du
dx
−
=0
x
1 + u2
ln |x| −
1
ln(1 + u2 ) = c
2
x2
= c1
1 + u2
x4 = c1 (x2 + y 2 ).
Using y(−1) = 1 we find c1 = 1/2. The solution of the initial-value problem is 2x4 = y 2 + x2 .
13. Letting y = ux we have
(x + uxeu ) dx − xeu (u dx + x du) = 0
dx − xeu du = 0
dx
− eu du = 0
x
ln |x| − eu = c
ln |x| − ey/x = c.
Using y(1) = 0 we find c = −1. The solution of the initial-value problem is ln |x| = ey/x − 1.
2.5
Solutions by Substitutions
14. Letting x = vy we have
y(v dy + y dv) + vy(ln vy − ln y − 1) dy = 0
y dv + v ln v dy = 0
dy
dv
+
=0
v ln v
y
ln |ln |v|| + ln |y| = c
y ln
x
= c1 .
y
Using y(1) = e we find c1 = −e. The solution of the initial-value problem is y ln
x
= −e.
y
1
3
3
1
dw
y = y −2 and w = y 3 we obtain
+ w = . An integrating factor is x3 so
x
x
dx
x
x
that x3 w = x3 + c or y 3 = 1 + cx−3 .
15. From y +
16. From y − y = ex y 2 and w = y −1 we obtain
that ex w = − 12 e2x + c or y −1 = − 12 ex + ce−x .
dw
+ w = −ex . An integrating factor is ex so
dx
dw
− 3w = −3x. An integrating factor is e−3x so
dx
that e−3x w = xe−3x + 13 e−3x + c or y −3 = x + 13 + ce3x .
1
1
dw
2
−1
18. From y − 1 +
y = y and w = y we obtain
+ 1+
w = −1. An integrating
x
dx
x
c
1
factor is xex so that xex w = −xex + ex + c or y −1 = −1 + + e−x .
x x
17. From y + y = xy 4 and w = y −3 we obtain
1
1
1
dw 1
+ w = 2 . An integrating factor is t so
19. From y − y = − 2 y 2 and w = y −1 we obtain
t
t
dt
t
t
c
t
1
−1
= ln t + c, we see that the
that tw = ln t + c or y = ln t + . Writing this in the form
t
t
y
solution can also be expressed in the form et/y = c1 t.
2t
2t
2
dw
−2t
y =
y 4 and w = y −3 we obtain
−
w =
. An
2
2
2
3 (1 + t )
3 (1 + t )
dt
1+t
1 + t2
w
1
1
so that
=
+ c or y −3 = 1 + c 1 + t2 .
integrating factor is
1 + t2
1 + t2
1 + t2
20. From y +
2
3
dw
6
9
y = 2 y 4 and w = y −3 we obtain
+ w = − 2 . An integrating factor
x
x
dx
x
x
is x6 so that x6 w = − 95 x5 + c or y −3 = − 95 x−1 + cx−6 . If y(1) = 12 then c = 49
5 and
9 −1
49 −6
−3
y = −5x + 5 x .
21. From y −
dw 3
3
+ w = . An integrating factor is e3x/2 so
dx 2
2
= 1 + ce−3x/2 . If y(0) = 4 then c = 7 and y 3/2 = 1 + 7e−3x/2 .
22. From y + y = y −1/2 and w = y 3/2 we obtain
that e3x/2 w = e3x/2 + c or y 3/2
87
88
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
du
1
− 1 = u2 or
du = dx. Thus
dx
1 + u2
−1
tan u = x + c or u = tan (x + c), and x + y + 1 = tan (x + c) or y = tan (x + c) − x − 1.
23. Let u = x + y + 1 so that du/dx = 1 + dy/dx. Then
24. Let u = x + y so that du/dx = 1 + dy/dx. Then
1 2
2u
= x + c or u2 = 2x + c1 , and (x + y)2 = 2x + c1 .
1−u
du
−1 =
or u du = dx. Thus
dx
u
du
− 1 = tan2 u or cos2 u du = dx. Thus
dx
1
1
2 u + 4 sin 2u = x + c or 2u + sin 2u = 4x + c1 , and 2(x + y) + sin 2(x + y) = 4x + c1 or
2y + sin 2(x + y) = 2x + c1 .
25. Let u = x + y so that du/dx = 1 + dy/dx. Then
26. Let u = x + y so that du/dx = 1 + dy/dx. Then
1
du
− 1 = sin u or
du = dx.
dx
1 + sin u
1 − sin u
Multiplying by (1−sin u)/(1−sin u) we have
du = dx or (sec2 u−sec u tan u)du = dx.
cos2 u
Thus tan u − sec u = x + c or tan (x + y) − sec (x + y) = x + c.
√
1
du
+ 2 = 2 + u or √ du = dx. Thus
27. Let u = y − 2x + 3 so that du/dx = dy/dx − 2. Then
dx
u
√
√
2 u = x + c and 2 y − 2x + 3 = x + c.
28. Let u = y − x + 5 so that du/dx = dy/dx − 1. Then
−e−u = x + c and −e−y+x−5 = x + c.
29. Let u = x + y so that du/dx = 1 + dy/dx. Then
du
+ 1 = 1 + eu or e−u du = dx. Thus
dx
1
du
− 1 = cos u and
du = dx. Now
dx
1 + cos u
1 − cos u
1 − cos u
1
=
=
= csc2 u − csc u cot u
2
1 + cos u
1 − cos2 u
sin u
´
´
so we have (csc2 u − csc u cot u) du = dx and − cot u + csc u = x + c. Thus − cot (x + y) +
√
csc (x + y) = x + c. Setting x = 0 and y = π/4 we obtain c = 2 − 1. The solution is
csc (x + y) − cot (x + y) = x +
30. Let u = 3x + 2y so that du/dx = 3 + 2 dy/dx. Then
u+2
du = dx. Now by long division
5u + 6
√
2 − 1.
2u
5u + 6
du
= 3+
=
and
dx
u+2
u+2
1
4
u+2
= +
5u + 6
5 25u + 30
so we have
and
1
5u
+
4
25
ˆ 1
4
+
5 25u + 30
du = dx
ln |25u + 30| = x + c. Thus
4
1
(3x + 2y) +
ln |75x + 50y + 30| = x + c.
5
25
2.5
Setting x = −1 and y = −1 we obtain c =
4
25
Solutions by Substitutions
89
ln 95. The solution is
4
4
1
(3x + 2y) +
ln |75x + 50y + 30| = x +
ln 95
5
25
25
or
5y − 5x + 2 ln |75x + 50y + 30| = 2 ln 95
31. We write the differential equation M (x, y)dx + N (x, y)dy = 0 as dy/dx = f (x, y) where
f (x, y) = −
M (x, y)
.
N (x, y)
The function f (x, y) must necessarily be homogeneous of degree 0 when M and N are homogeneous of degree α. Since M is homogeneous of degree α, M (tx, ty) = tα M (x, y), and
letting t = 1/x we have
M (1, y/x) =
Thus
1
M (x, y)
xα
or M (x, y) = xα M (1, y/x).
y
xα M (1, y/x)
M (1, y/x)
dy
= f (x, y) = − α
=−
=F
.
dx
x N (1, y/x)
N (1, y/x)
x
32. Rewrite (5x2 − 2y 2 )dx − xy dy = 0 as
dy
= 5x2 − 2y 2
dx
xy
and divide by xy, so that
x
y
dy
=5 −2 .
dx
y
x
We then identify
F
y
x
=5
y −1
x
−2
y
x
.
33. (a) By inspection y = x and y = −x are solutions of the differential equation and not
members of the family y = x sin (ln x + c2 ).
(b) Letting x = 5 and y = 0 in sin−1 (y/x) = ln x + c2 we get
sin−1 0 = ln 5 + c2 or c2 = − ln 5. Then
sin−1 (y/x) = ln x − ln 5 = ln (x/5). Because the range of
the arcsine function is [−π/2, π/2] we must have
x
π
π
− ≤ ln ≤
2
5
2
x
e−π/2 ≤ ≤ eπ/2
5
−π/2
5e
≤ x ≤ 5e
π/2
y
20
15
10
5
5
The interval of definition of the solution is approximately [1.04, 24.05].
10
15
20
x
90
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
34. As x → −∞, e6x → 0 and y → 2x+3. Now write (1+ce6x )/(1−ce6x ) as (e−6x +c)/(e−6x −c).
Then, as x → ∞, e−6x → 0 and y → 2x − 3.
35. (a) The substitutions y = y1 + u and
dy1 du
dy
=
+
dx
dx
dx
lead to
dy1 du
+
= P + Q(y1 + u) + R(y1 + u)2
dx
dx
= P + Qy1 + Ry12 + Qu + 2y1 Ru + Ru2
or
du
− (Q + 2y1 R)u = Ru2 .
dx
This is a Bernoulli equation with n = 2 which can be reduced to the linear equation
dw
+ (Q + 2y1 R)w = −R
dx
by the substitution w = u−1 .
1
4
dw
+ − +
w = −1.
Q(x) = −1/x, and R(x) = 1. Then
(b) Identify P (x) =
dx
x x
−1
An integrating factor is x3 so that x3 w = − 14 x4 + c or u = − 14 x + cx−3 . Thus,
−4/x2 ,
2
y = +u
x
−1
2
1
−3
or y = + − x + cx
x
4
36. Write the differential equation in the form x(y /y) = ln x + ln y and let u = ln y. Then
du/dx = y /y and the differential equation becomes x(du/dx) = ln x + u or du/dx − u/x =
´
(ln x)/x, which is first-order and linear. An integrating factor is e− dx/x = 1/x, so that
(using integration by parts)
ln x
d 1
u = 2
dx x
x
and
1 ln x
u
=− −
+ c.
x
x
x
The solution is
ln y = −1 − ln x + cx or y =
37. Write the differential equation as
dv
1
+ v = 32v −1 ,
dx x
and let u = v 2 or v = u1/2 . Then
1
du
dv
= u−1/2 ,
dx
2
dx
ecx−1
.
x
2.5
Solutions by Substitutions
and substituting into the differential equation, we have
1 −1/2 du 1 1/2
u
+ u = 32u−1/2
2
dx x
or
du 2
+ u = 64.
dx x
´
The latter differential equation is linear with integrating factor e
(2/x) dx
= x2 , so
d 2 x u = 64x2
dx
and
x2 u =
64 3
64
c
x + c or v 2 =
x+ 2 .
3
3
x
38. Write the differential equation as dP/dt − aP = −bP 2 and let u = P −1 or P = u−1 . Then
du
dp
= −u−2
,
dt
dt
and substituting into the differential equation, we have
−u−2
du
− au−1 = −bu−2
dt
or
du
+ au = b.
dt
´
The latter differential equation is linear with integrating factor e
d at
[e u] = beat
dt
and
eat u =
b at
e +c
a
eat P −1 =
b at
e +c
a
P −1 =
P =
b
+ ce−at
a
1
a
=
.
b/a + ce−at
b + c1 e−at
a dt
= eat , so
91
92
CHAPTER 2
2.6
FIRST-ORDER DIFFERENTIAL EQUATIONS
A Numerical Method
1. We identify f (x, y) = 2x − 3y + 1. Then, for h = 0.1,
yn+1 = yn + 0.1(2xn − 3yn + 1) = 0.2xn + 0.7yn + 0.1,
and
y(1.1) ≈ y1 = 0.2(1) + 0.7(5) + 0.1 = 3.8
y(1.2) ≈ y2 = 0.2(1.1) + 0.7(3.8) + 0.1 = 2.98
For h = 0.05,
yn+1 = yn + 0.05(2xn − 3yn + 1) = 0.1xn + 0.85yn + 0.1,
and
y(1.05) ≈ y1 = 0.1(1) + 0.85(5) + 0.1 = 4.4
y(1.1) ≈ y2 = 0.1(1.05) + 0.85(4.4) + 0.1 = 3.895
y(1.15) ≈ y3 = 0.1(1.1) + 0.85(3.895) + 0.1 = 3.47075
y(1.2) ≈ y4 = 0.1(1.15) + 0.85(3.47075) + 0.1 = 3.11514
2. We identify f (x, y) = x + y 2 . Then, for h = 0.1,
yn+1 = yn + 0.1(xn + yn2 ) = 0.1xn + yn + 0.1yn2 ,
and
y(0.1) ≈ y1 = 0.1(0) + 0 + 0.1(0)2 = 0
y(0.2) ≈ y2 = 0.1(0.1) + 0 + 0.1(0)2 = 0.01
For h = 0.05,
yn+1 = yn + 0.05(xn + yn2 ) = 0.05xn + yn + 0.05yn2 ,
and
y(0.05) ≈ y1 = 0.05(0) + 0 + 0.05(0)2 = 0
y(0.1) ≈ y2 = 0.05(0.05) + 0 + 0.05(0)2 = 0.0025
y(0.15) ≈ y3 = 0.05(0.1) + 0.0025 + 0.05(0.0025)2 = 0.0075
y(0.2) ≈ y4 = 0.05(0.15) + 0.0075 + 0.05(0.0075)2 = 0.0150
3. Separating variables and integrating, we have
dy
= dx
y
and
ln |y| = x + c.
Thus y = c1 ex and, using y(0) = 1, we find c = 1, so y = ex is the solution of the initial-value
problem.
2.6
h = 0.1
A Numerical Method
h = 0.05
xn
yn
Actual
Value
0.00
1.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
1.0000
1.1000
1.2100
1.3310
1.4641
1.6105
1.7716
1.9487
2.1436
2.3579
2.5937
1.0000
1.1052
1.2214
1.3499
1.4918
1.6487
1.8221
2.0138
2.2255
2.4596
2.7183
Abs.
Error
%Rel.
Error
xn
yn
Actual
Value
Abs.
Error
%Rel.
Error
0.0000
0.0052
0.0114
0.0189
0.0277
0.0382
0.0506
0.0650
0.0820
0.1017
0.1245
0.00
0.47
0.93
1.40
1.86
2.32
2.77
3.23
3.68
4.13
4.58
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
0.55
0.60
0.65
0.70
0.75
0.80
0.85
0.90
0.95
1.00
1.0000
1.0500
1.1025
1.1576
1.2155
1.2763
1.3401
1.4071
1.4775
1.5513
1.6289
1.7103
1.7959
1.8856
1.9799
2.0789
2.1829
2.2920
2.4066
2.5270
2.6533
1.0000
1.0513
1.1052
1.1618
1.2214
1.2840
1.3499
1.4191
1.4918
1.5683
1.6487
1.7333
1.8221
1.9155
2.0138
2.1170
2.2255
2.3396
2.4596
2.5857
2.7183
0.0000
0.0013
0.0027
0.0042
0.0059
0.0077
0.0098
0.0120
0.0144
0.0170
0.0198
0.0229
0.0263
0.0299
0.0338
0.0381
0.0427
0.0476
0.0530
0.0588
0.0650
0.00
0.12
0.24
0.36
0.48
0.60
0.72
0.84
0.96
1.08
1.20
1.32
1.44
1.56
1.68
1.80
1.92
2.04
2.15
2.27
2.39
4. Separating variables and integrating, we have
dy
= 2x dx and
y
ln |y| = x2 + c.
Thus y = c1 ex and, using y(1) = 1, we find c = e−1 , so y = ex
initial-value problem.
2
h = 0.1
2 −1
is the solution of the
h = 0.05
xn
yn
Actual
Value
1.00
1.10
1.20
1.30
1.40
1.50
1.0000
1.2000
1.4640
1.8154
2.2874
2.9278
1.0000
1.2337
1.5527
1.9937
2.6117
3.4903
Abs.
Error
%Rel.
Error
xn
yn
Actual
Value
Abs.
Error
%Rel.
Error
0.0000
0.0337
0.0887
0.1784
0.3243
0.5625
0.00
2.73
5.71
8.95
12.42
16.12
1.00
1.05
1.10
1.15
1.20
1.25
1.30
1.35
1.40
1.45
1.50
1.0000
1.1000
1.2155
1.3492
1.5044
1.6849
1.8955
2.1419
2.4311
2.7714
3.1733
1.0000
1.1079
1.2337
1.3806
1.5527
1.7551
1.9937
2.2762
2.6117
3.0117
3.4903
0.0000
0.0079
0.0182
0.0314
0.0483
0.0702
0.0982
0.1343
0.1806
0.2403
0.3171
0.00
0.72
1.47
2.27
3.11
4.00
4.93
5.90
6.92
7.98
9.08
93
94
CHAPTER 2
5. x h = 0.1
FIRST-ORDER DIFFERENTIAL EQUATIONS
xn
yn
0.00
0.10
0.20
0.30
0.40
0.50
0.0000
0.1000
0.1905
0.2731
0.3492
0.4198
7. x h = 0.1
xn
yn
0.00
0.10
0.20
0.30
0.40
0.50
0.5000
0.5250
0.5431
0.5548
0.5613
0.5639
9. x h = 0.1
xn
yn
1.00
1.10
1.20
1.30
1.40
1.50
1.0000
1.0000
1.0191
1.0588
1.1231
1.2194
h = 0.05
xn
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
h = 0.05
xn
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
h = 0.05
xn
1.00
1.05
1.10
1.15
1.20
1.25
1.30
1.35
1.40
1.45
1.50
yn
0.0000
0.0500
0.0976
0.1429
0.1863
0.2278
0.2676
0.3058
0.3427
0.3782
0.4124
yn
0.5000
0.5125
0.5232
0.5322
0.5395
0.5452
0.5496
0.5527
0.5547
0.5559
0.5565
yn
1.0000
1.0000
1.0049
1.0147
1.0298
1.0506
1.0775
1.1115
1.1538
1.2057
1.2696
6. x h = 0.1
xn
yn
0.00
0.10
0.20
0.30
0.40
0.50
1.0000
1.1000
1.2220
1.3753
1.5735
1.8371
8. x h = 0.1
xn
yn
0.00
0.10
0.20
0.30
0.40
0.50
1.0000
1.1000
1.2159
1.3505
1.5072
1.6902
10. x h = 0.1
xn
yn
0.00
0.10
0.20
0.30
0.40
0.50
0.5000
0.5250
0.5499
0.5747
0.5991
0.6231
h = 0.05
xn
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
h = 0.05
xn
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
h = 0.05
xn
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
yn
1.0000
1.0500
1.1053
1.1668
1.2360
1.3144
1.4039
1.5070
1.6267
1.7670
1.9332
yn
1.0000
1.0500
1.1039
1.1619
1.2245
1.2921
1.3651
1.4440
1.5293
1.6217
1.7219
yn
0.5000
0.5125
0.5250
0.5375
0.5499
0.5623
0.5746
0.5868
0.5989
0.6109
0.6228
2.6
A Numerical Method
95
11. Tables of values were computed using the Euler and RK4 methods. The resulting points were
plotted and joined using ListPlot in Mathematica.
h = 0.25
h = 0.1
y
h = 0.05
y
7
6
5
4
3
2
1
7
6
5
4
3
2
1
RK4
Euler
2
4
6
y
8
10
7
6
5
4
3
2
1
RK4
Euler
x
2
4
6
8
10
RK4
Euler
x
2
4
6
8
10
x
12. See the comments in Problem 11 above.
h = 0.25
y
6
5
4
3
2
1
RK4
Euler
1
2
3
h = 0.1
y
6
5
4
3
2
1
4
5
RK4
Euler
x
1
2
3
4
h = 0.05
y
6
5
4
3
2
1
5
RK4
Euler
x
1
2
3
4
5
x
13. Using separation of variables we find that the solution of the differential equation is y =
1/(1 − x2 ), which is undefined at x = 1, where the graph has a vertical asymptote. Because
the actual solution of the differential equation becomes unbounded at x approaches 1, very
small changes in the inputs x will result in large changes in the corresponding outputs y. This
can be expected to have a serious effect on numerical procedures. The graphs below were
obtained as described in Problem 11.
10
h = 0.1
y
10
RK4
8
6
4
4
0.4
0.6
0.8
Euler
2
Euler
0.2
RK4
8
6
2
h = 0.05
y
1
x
14. (a) The graph to the right was obtained using RK4
and ListPlot in Mathematica\with h = 0.1.
0.2
0.4
0.6
0.8
1
x
96
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
(b) Writing the
differential equation in the form y + 2xy = 1 we see that an integrating
´
2
2xdx
= ex , so
factor is e
d x2
2
[e y] = ex
dx
and
ˆ x
2
2
−x2
et dt + ce−x .
y=e
0
This solution can also be expressed in terms of the inverse error function as
√
π −x2
2
y=
e erfi(x) + ce−x .
2
Letting x = 0 and y(0) = 0 we find c = 0, so the solution of the initial-value problem is
√
ˆ x
π −x2
−x2
t2
e erfi(x).
e dt =
y=e
2
0
(c) Using FindRoot in Mathematica we see that y (x) = 0 when x = 0.924139. Since
y(0.924139) = 0.541044, we see from the graph in part (a) that (0.924139, 0.541044) is
a relative maximum. Now, using the substitution u = −t in the integral below, we have
ˆ −x
ˆ x
ˆ x
2
−(−x)2
t2
−x2
(−u)2
−x2
e dt = e
e
(−du) = −e
eu du = −y(x).
y(−x) = e
0
0
0
Thus, y(x) is an odd function and (−0.924139, −0.541044) is a relative minimum.
Chapter 2 in Review
1. Writing the differential equation in the form y = k(y + A/k) we see that the critical point
−A/k is a repeller for k > 0 and an attractor for k < 0.
2. Separating variables and integrating we have
4
dy
= dx
y
x
ln y = 4 ln x + c = ln x4 + c
y = c1 x4 .
We see that when x = 0, y = 0, so the initial-value problem has an infinite number of solutions
for k = 0 and no solutions for k = 0.
3. True; y = k2 /k1 is always a solution for k1 = 0.
4. True; writing the differential equation as a1 (x) dy + a2 (x)y dx = 0 and separating variables
yields
a2 (x)
dy
=−
dx.
y
a1 (x)
Chapter 2 in Review
5.
d3 y
= x sin y
dx3
6. False:
(There are many answers.)
dr
= rθ + r + θ + 1 = (r + 1) (θ + 1).
dθ
7. True
8. Since the differential equation in the form y = 2 − |y| is seen to be autonomous, 2 − |y| = 0
has critical points 2 and −2 so y1 = 2 and y2 = −2 are constant (equilibrium) solutions.
9.
dy
= ex dx
y
ln y = ex + c
x +c
y = ee
10.
x
= ec ee
x
or y = c1 ee
y = |x| , y(−1) = 2
−x, x < 0
dy
=
dx
x,
x≥0
⎧
1 2
⎪
⎪
⎨− 2 x + c1 , x < 0
y=
⎪
⎪
⎩ 1 x2 + c2 ,
x≥0
2
5
1
The initial condition y(−1) = 2 implies 2 = − + c1 and thus c1 = . Now y(x) is supposed
2
2
to be differentiable and so continuous. At x = 0 the two parts of the functions must
5
agree and so c2 = c1 = . So,
2
y
10
⎧
1
2
⎪
⎪
⎨2 5 − x ,
y=
⎪
⎪
⎩ 1 x2 + 5 ,
2
5
x<0
4 2
x≥0
2
5
10
ˆ
11.
y = ecos x
x
te− cos t dt
0
dy
= ecos x xe− cos x + (− sin x) ecos x
dx
dy
= x − (sin x) y
dx
12.
dy
= y + 3,
dx
or
ˆ
x
te− cos t dt
0
dy
+ (sin x) y = x.
dx
dy
= (y + 3)2
dx
4
x
97
98
CHAPTER 2
13.
FIRST-ORDER DIFFERENTIAL EQUATIONS
dy
= (y − 1)2 (y − 3)2
dx
dy
= y(y − 2)2 (y − 4)
dx
15. When n is odd, xn < 0 for x < 0 and xn > 0 for x > 0. In this case 0 is unstable. When n is
even, xn > 0 for x < 0 and for x > 0. In this case 0 is semi-stable.
14.
When n is odd, −xn > 0 for x < 0 and −xn < 0 for x > 0. In this case 0 is asymptotically
stable. When n is even, −xn < 0 for x < 0 and for x > 0. In this case 0 is semi-stable.
16. Using a CAS we find that the zero of f occurs at approximately P = 1.3214. From the graph
we observe that dP/dt > 0 for P < 1.3214 and dP/dt < 0 for P > 1.3214, so P = 1.3214 is
an asymptotically stable critical point. Thus, lim P (t) = 1.3214.
t→∞
17. x
y
x
18.
(a) linear in y, homogeneous, exact
(b) linear in x
(c) separable, exact, linear in x and y
(d) Bernoulli in x
(e) separable
(f ) separable, linear in x, Bernoulli
(g) linear in x
(h) homogeneous
(i) Bernoulli
(j) homogeneous, exact, Bernoulli
(k) linear in x and y, exact, separable,
homogeneous
(l) exact, linear in y
(m) homogeneous
(n) separable
19. Separating variables and using the identity cos2 x = 12 (1 + cos 2x), we have
y
dy,
cos2 x dx = 2
y +1
1
1
1
x + sin 2x = ln y 2 + 1 + c,
2
4
2
Chapter 2 in Review
and
2x + sin 2x = 2 ln y 2 + 1 + c.
20. Write the differential equation in the form
x
y ln dx =
y
x
x ln − y dy.
y
This is a homogeneous equation, so let x = uy. Then dx = u dy + y du and the differential
equation becomes
y ln u(u dy + y du) = (uy ln u − y) dy
or y ln u du = −dy.
Separating variables, we obtain
ln u du = −
dy
y
u ln |u| − u = − ln |y| + c
x
x
x
ln
− = − ln |y| + c
y
y
y
x(ln x − ln y) − x = −y ln |y| + cy.
21. The differential equation
dy
2
3x2 −2
+
y=−
y
dx 6x + 1
6x + 1
is Bernoulli. Using w = y 3 , we obtain the linear equation
6
9x2
dw
+
w=−
.
dx
6x + 1
6x + 1
An integrating factor is 6x + 1, so
d
[(6x + 1)w] = −9x2 ,
dx
w=−
3x3
c
+
,
6x + 1 6x + 1
and
(6x + 1)y 3 = −3x3 + c.
(Note: The differential equation is also exact.)
22. Write the differential equation in the form (3y 2 + 2x)dx + (4y 2 + 6xy)dy = 0. Letting
M = 3y 2 + 2x and N = 4y 2 + 6xy we see that My = 6y = Nx , so the differential equation is
exact. From fx = 3y 2 +2x we obtain f = 3xy 2 +x2 +h(y). Then fy = 6xy +h (y) = 4y 2 +6xy
and h (y) = 4y 2 so h(y) = 43 y 3 . A one-parameter family of solutions is
4
3xy 2 + x2 + y 3 = c.
3
99
100
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
23. Write the equation in the form
dQ 1
+ Q = t3 ln t.
dt
t
An integrating factor is eln t = t, so
d
[tQ] = t4 ln t
dt
tQ = −
1 5 1 5
t + t ln t + c
25
5
and
Q=−
c
1 4 1 4
t + t ln t + .
25
5
t
24. Letting u = 2x + y + 1 we have
dy
du
=2+
,
dx
dx
and so the given differential equation is transformed into
du
du
2u + 1
u
− 2 = 1 or
=
.
dx
dx
u
Separating variables and integrating we get
u
du = dx
2u + 1
1 1
1
−
du = dx
2 2 2u + 1
1
1
u − ln |2u + 1| = x + c
2
4
2u − ln |2u + 1| = 4x + c1 .
Resubstituting for u gives the solution
4x + 2y + 2 − ln |4x + 2y + 3| = 4x + c1
or
2y + 2 − ln |4x + 2y + 3| = c1 .
25. Write the equation in the form
dy
8x
2x
+ 2
y= 2
.
dx x + 4
x +4
4
An integrating factor is x2 + 4 , so
d 2
x +4
dx
4
x2 + 4
y = 2x x2 + 4
4
y=
1 2
x +4
4
3
4
+c
Chapter 2 in Review
and
y=
1
+ c x2 + 4
4
−4
.
26. Letting M = 2r2 cos θ sin θ + r cos θ and N = 4r + sin θ − 2r cos2 θ we see that
Mr = 4r cos θ sin θ + cos θ = Nθ , so the differential equation is exact. From
fθ = 2r2 cos θ sin θ + r cos θ we obtain f = −r2 cos2 θ + r sin θ + h(r). Then
fr = −2r cos2 θ + sin θ + h (r) = 4r + sin θ − 2r cos2 θ and h (r) = 4r so h(r) = 2r2 . The
solution is
−r2 cos2 θ + r sin θ + 2r2 = c.
dy
1
dy
+ 4 (cos x) y = x in the standard form
+ 2 (cos x) y = x then
dx ´
dx
2
the integrating factor is e 2 cos x dx = e2 sin x . Therefore
27. We put the equation
ˆ
x
0
d 2 sin x e
y =
dx
d 2 sin t
y(t) dt =
e
dt
1 2 sin x
xe
2
ˆ
1 x 2 sin t
te
dt
2 0
1 ˆ x
y(x) − e y(0) =
te2 sin t dt
2 0
ˆ
1 x 2 sin t
2 sin x
y(x) − 1 =
te
dy
e
2 0
1
2 sin x
e
0
1
y(x) = e−2 sin x + e−2 sin x
2
ˆ
x
te2 sin t dt
0
dy
− 4xy = sin x2 is already in standard form so the integrating factor is
dx
´
d −2x2 2
2
y = e−2x sin x2 . Because of the initial condition
e− 4x dx = e−2x . Therefore
e
dx
y(0) = 7 we write
28. The equation
ˆ
x
0
ˆ x
d −2t2
2
e
y(t) dt =
e−2t sin t2 dt
dt
0
ˆ
y(x) − e y(0) =
7
−2x2
e
x
0
e−2t sin t2 dt
2
0
2x2
y(x) = 7e
2x2
ˆ
+e
0
x
e−2t sin t2 dt
2
101
102
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
2
dy
29. We put the equation x dx
+2y = xex into standard form
´
factor is e
2
x
dx
2
dy 2
2
+ y = ex . Then the integrating
dx x
= eln x = x2 . Therefore
dy
2
+ 2xy = x2 ex
dx
d 2 2
x y = x2 ex
dx
ˆ x
d 2
2
t2 et dt
t y(t) dt =
dt
1
x2
ˆ
x
1
ˆ
x y(x) − y(1) =
3
x
2
2
t2 et dt
1
y(x) =
1
3
+ 2
2
x
x
ˆ
x
2
t2 et dt
1
30.
x
dy
+ (sin x) y = 0
dx
sin x
dy
+
y=0
dx
x
´x
The integrating factor is e
sin t
t
0
ˆ
x
0
. Therefore,
d ´ x sin t dt e0 t y =0
dx
ˆ x
d ´ t sin u du
e0 u
y(t) dt =
0 dt = 0
dt
0
´x
e
dy
10
sin t
0
t
dt
y(x) − e y(0) = 0
0
y(x) = 10e−
´x
0
sin t
t
dt
31.
dy
+ y = f (x),
dx
y(0) = 5,
where
f (x) =
For 0 ≤ x < 1,
d x
[e y] = 1
dx
ex y = x + c1
y = xe−x + c1 e−x
−x
e , 0≤x<1
0,
x≥1
Chapter 2 in Review
Using y(0) = 5, we have c1 = 5. Therefore y = xe−x + 5e−x . Then for x ≥ 1,
d x
[e y] = 0
dx
ex y = c2
y = c2 e−x
Requiring that y(x) be continuous at x = 1 yields
c2 e−1 = e−1 + 5e−1
c2 = 6
Therefore
y(x) =
−x
xe + 5e−x , 0 ≤ x < 1
6e−x ,
x≥1
32.
dy
+ P (x)y = ex ,
dx
y(0) = −1,
where
P (x) =
1,
−1,
0≤x<1
x≥1
For 0 ≤ x < 1,
d x
[e y] = e2x
dx
1
ex y = e2x + c1
2
1
y = ex + c1 e−x
2
Using y(0) = −1, we have c1 = − 32 . Therefore y = 12 ex − 32 e−x . Then for x ≥ 1,
d −x e y =1
dx
e−x y = x + c2
y = xex + c2 ex
Requiring that y(x) be continuous at x = 1 yields
3
1
e + c2 e = e − e−1
2
2
1 3 −2
c2 = − − e
2 2
Therefore
⎧
3
1
⎪
⎨ ex − e−x ,
0≤x<1
2
2
y(x) =
⎪
⎩xex − 1 ex − 3 ex−2 , x ≥ 1
2
2
103
104
CHAPTER 2
FIRST-ORDER DIFFERENTIAL EQUATIONS
33. The differential equation has the form (d/dx) [(sin x)y] = 0. Integrating, we have (sin x)y = c
or y = c/ sin x. The initial condition implies c = −2 sin (7π/6) = 1. Thus, y = 1/ sin x, where
the interval (π, 2π) is chosen to include x = 7π/6.
34. Separating variables and integrating we have
dy
= −2(t + 1) dt
y2
−
1
= −(t + 1)2 + c
y
y=
1
,
(t + 1)2 + c1
where − c = c1
The initial condition y(0) = − 18 implies c1 = −9, so a solution of the initial-value problem is
y=
1
(t + 1)2 − 9
or
y=
t2
1
,
+ 2t − 8
where −4 < t < 2.
35. (a) For y < 0,
√
y is not a real number.
(b) Separating variables and integrating we have
dy
√ = dx
y
and
√
2 y = x + c.
√
Letting y(x0 ) = y0 we get c = 2 y0 − x0 , so that
1
√
and y = (x + 2 y0 − x0 )2 .
4
√
√
Since y > 0 for y = 0, we see that dy/dx = 12 (x + 2 y0 − x0 ) must be positive. Thus,
√
the interval on which the solution is defined is (x0 − 2 y0 , ∞).
√
√
2 y = x + 2 y0 − x0
36. (a) The differential equation is homogeneous and we let y = ux. Then
(x2 − y 2 ) dx + xy dy = 0
(x2 − u2 x2 ) dx + ux2 (u dx + x du) = 0
dx + ux du = 0
u du = −
dx
x
1 2
u = − ln |x| + c
2
y2
= −2 ln |x| + c1 .
x2
The initial condition gives c1 = 2, so an implicit solution is y 2 = x2 (2 − 2 ln |x|).
Chapter 2 in Review
(b) Solving for y in part (a) and being sure that the initial con√
dition is still satisfied, we have y = − 2 |x|(1 − ln |x|)1/2 ,
where −e ≤ x ≤ e so that 1 − ln |x| ≥ 0. The graph of
this function indicates that the derivative is not defined at
x = 0 and x = e. Thus, the solution of the initial-value
√
problem is y = − 2 x(1 − ln x)1/2 , for 0 < x < e.
105
y
2
1
–2
–1
1
2
x
–1
–2
37. The graph of y1 (x) is the portion of the closed blue curve lying in the fourth quadrant. Its
interval of definition is approximately (0.7, 4.3). The graph of y2 (x) is the portion of the
left-hand blue curve lying in the third quadrant. Its interval of definition is (−∞, 0).
38. The first step of Euler’s method gives y(1.1) ≈ 9 + 0.1(1 + 3) = 9.4. Applying Euler’s method
√
one more time gives y(1.2) ≈ 9.4 + 0.1(1 + 1.1 9.4 ) ≈ 9.8373.
39. Since the differential equation is autonomous, all lineal
elements on a given horizontal line have the same slope.
The direction field is then as shown in the figure at the
right. It appears from the figure that the differential
equation has critical points at −2 (an attractor) and
at 2 (a repeller). Thus, −2 is an aymptotically stable
critical point and 2 is an unstable critical point.
40. Since the differential equation is autonomous, all lineal
elements on a given horizontal line have the same slope.
The direction field is then as shown in the figure at the
right. It appears from the figure that the differential
equation has no critical points.
Chapter 3
Modeling with First-Order Differential Equations
3.1
Linear Models
1. Let P = P (t) be the population at time t, and P0 the initial population. From dP/dt = kP
we obtain P = P0 ekt . Using P (5) = 2P0 we find k = 15 ln 2 and P = P0 e(ln 2)t/5 . Setting
P (t) = 3P0 we have 3 = e(ln 2)t/5 , so
ln 3 =
(ln 2)t
5
and
t=
5 ln 3
≈ 7.9 years.
ln 2
Setting P (t) = 4P0 we have 4 = e(ln 2)t/5 , so
ln 4 =
(ln 2)t
5
and
t = 10 years.
2. From Problem 1 the growth constant is k = 15 ln 2. Then P = P0 e(1/5)(ln 2)t and
10,000 = P0 e(3/5) ln 2 . Solving for P0 we get P0 = 10,000e−(3/5) ln 2 = 6,597.5. Now
P (10) = P0 e(1/5)(ln 2)(10) = 6,597.5e2 ln 2 = 4P0 = 26,390.
The rate at which the population is growing is
1
P (10) = kP (10) = (ln 2)26,390 = 3658 persons/year.
5
3. Let P = P (t) be the population at time t. Then dP/dt = kP and P = cekt . From
P (0) = c = 500 we see that P = 500ekt . Since 15% of 500 is 75, we have
1
1
ln 575
P (10) = 500e10k = 575. Solving for k, we get k = 10
500 = 10 ln 1.15. When t = 30,
P (30) = 500e(1/10)(ln 1.15)30 = 500e3 ln 1.15 ≈ 760 years
and
P (30) = kP (30) ≈
1
(ln 1.15)760 ≈ 10.62 persons/year.
10
106
3.1
Linear Models
4. Let P = P (t) be bacteria population at time t and P0 the initial number. From dP/dt = kP
we obtain P = P0 ekt . Using P (3) = 400 and P (10) = 2000 we find 400 = P0 e3k or
ek = (400/P0 )1/3 . From P (10) = 2000 we then have 2000 = P0 e10k = P0 (400/P0 )10/3 , so
2000
−7/3
= P0
40010/3
and
P0 =
2000
40010/3
−3/7
≈ 201.
5. Let A = A(t) be the amount of lead present at time t. From dA/dt = kA and A(0) = 1
1
ln (1/2). When 90% of the lead has
we obtain A = ekt . Using A(3.3) = 1/2 we find k = 3.3
decayed, 0.1 grams will remain. Setting A(t) = 0.1 we have et(1/3.3) ln (1/2) = 0.1, so
1
t
ln = ln 0.1
3.3 2
and
t=
3.3 ln 0.1
≈ 10.96 hours.
ln (1/2)
6. Let A = A(t) be the amount present at time t. From dA/dt = kA and A(0) = 100 we obtain
A = 100ekt . Using A(6) = 97 we find k = 16 ln 0.97. Then A(24) = 100e(1/6)(ln 0.97)24 =
100(0.97)4 ≈ 88.5 mg.
7. Setting A(t) = 50 in Problem 6 we obtain 50 = 100ekt , so
kt = ln
1
2
and
t=
ln (1/2)
≈ 136.5 hours.
(1/6) ln 0.97
8. (a) The solution of dA/dt = kA is A(t) = A0 ekt . Letting A =
obtain the half-life T = −(ln 2)/k.
1
2 A0
and solving for t we
(b) Since k = −(ln 2)/T we have
A(t) = A0 e−(ln 2)t/T = A0 2−t/T .
(c) Writing 18 A0 = A0 2−t/T as 2−3 = 2−t/T and solving for t we get t = 3T . Thus, an initial
amount A0 will decay to 18 A0 in three half-lives.
9. Let I = I(t) be the intensity, t the thickness, and I(0) = I0 . If dI/dt = kI and I(1) = 0.25I0 ,
then I = I0 ekt , k = ln 0.25, and I(5) = 0.000977I0 .
10. From dS/dt = rS we obtain S = S0 ert where S(0) = S0 .
(a) If S0 = $5000 and r = 5.75% then S(5) = $6665.45.
(b) If S(t) =$10,000 then t = 12 years.
(c) S ≈ $6651.82
107
108
CHAPTER 3
MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS
11. Using 5730 years as the half-life of C-14 we have from Example 3 in the text A(t) =
A0 e−0.00012097t . Since 85.5% of the C-14 has decayed, 1 − 0.855 = 0.145 times the original amount is now present, so
0.145A0 = A0 e−0.00012097t ,
e−0.00012097t = 0.145,
and t = −
ln 0.145
≈ 15, 968 years
0.00012097
is the approximate age.
12. From Example 3 in the text, the amount of carbon present at time t is A(t) = A0 e−0.00012097t .
Letting t = 660 and solving for A0 we have A(660) = A0 e−0.00012097(660) = 0.923264A0 . Thus,
approximately 92% of the original amount of C-14 remained in the cloth as of 1988.
13. Assume that dT /dt = k(T − (−12)) so that T = −12 + cekt . If T (0) = 21◦ and T (1/2) = 10◦
then c = 33 and k = 2 ln (2/3) so that T (1) = 2.67◦ . If T (t) = −9◦ then t = 2.96 minutes.
14. Assume that dT /dt = k(T − (−15)) so that T = −15 + cekt . If T (1) = 13◦ and T (5) = −1◦
then k = − 14 ln 2 and c = 33.298 so that T (0) = 18◦ .
15. Assume that dT /dt = k(T − 100) so that T = 100 + cekt . If T (0) = 20◦ and T (1) = 22◦ , then
c = −80 and k = ln (39/40) so that T (t) = 90◦ , which implies t = 82.1 seconds. If T (t) = 98◦
then t = 145.7 seconds.
16. The differential equation for the first container is dT1 /dt = k1 (T1 − 0) = k1 T1 , whose solution
is T1 (t) = c1 ek1 t . Since T1 (0) = 100 (the initial temperature of the metal bar), we have
100 = c1 and T1 (t) = 100ek1 t . After 1 minute, T1 (1) = 100ek1 = 90◦ C, so k1 = ln 0.9 and
T1 (t) = 100et ln 0.9 . After 2 minutes, T1 (2) = 100e2 ln 0.9 = 100(0.9)2 = 81◦ C.
The differential equation for the second container is dT2 /dt = k2 (T2 − 100), whose solution
is T2 (t) = 100 + c2 ek2 t . When the metal bar is immersed in the second container, its initial
temperature is T2 (0) = 81, so
T2 (0) = 100 + c2 ek2 (0) = 100 + c2 = 81
and c2 = −19. Thus, T2 (t) = 100−19ek2 t . After 1 minute in the second tank, the temperature
of the metal bar is 91◦ C, so
T2 (1) = 100 − 19ek2 = 91
ek2 =
9
19
k2 = ln
9
19
3.1
Linear Models
109
and T2 (t) = 100 − 19et ln (9/19) . Setting T2 (t) = 99.9 we have
100 − 19et ln (9/19) = 99.9
et ln (9/19) =
t=
0.1
19
ln (0.1/19)
≈ 7.02.
ln (9/19)
Thus, from the start of the “double dipping” process, the total time until the bar reaches
99.9◦ C in the second container is approximately 9.02 minutes.
17. Using separation of variables to solve dT /dt = k(T − Tm ) we get T (t) = Tm + cekt . Using
T (0) = 21 we find c = 21 − Tm , so T (t) = Tm + (21 − Tm )ekt . Using the given observations,
we obtain
1
= Tm + (21 − Tm )ek/2 = 43
T
2
T (1) = Tm + (21 − Tm )ek = 63.
Then, from the first equation, ek/2 = (43 − Tm )/(21 − Tm ) and
ek = (ek/2 )2 =
43 − Tm
21 − Tm
2
63 − Tm
21 − Tm
=
2
2
1849 − 86Tm + Tm
= 1323 − 84Tm + Tm
Tm = 263.
The temperature in the oven is 263◦ .
18. (a) The initial temperature of the bath is Tm (0) = 16◦ , so in the short term the temperature
of the chemical, which starts at 27◦ , should decrease or cool. Over time, the temperature
of the bath will increase toward 38◦ since e−0.1t decreases from 1 toward 0 as t increases
from 0. Thus, in the long term, the temperature of the chemical should increase or warm
toward 38◦ .
(b) Adapting the model for Newton’s law of cooling,
we have
dT
= −0.1(T − 38 + 22e−0.1t ),
dt
40
T
35
T (0) = 27.
30
25
Writing the differential equation in the form
20
dT
+ 0.1T = 3.8 − 2.2e−0.1t
dt
15
´
we see that it is linear with integrating factor e
0.1 dt
= e0.1t .
10
20
30
40
50
t
110
CHAPTER 3
MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS
Thus
d 0.1t
[e T ] = 3.8e0.1t − 2.2
dt
e0.1t T = 38e0.1t − 2.2t + c
and
T (t) = 38 − 2.2te−0.1t + ce−0.1t .
Now T (0) = 27 so 38 + c = 27, c = −11 and
T (t) = 38 − 2.2te−0.1t − 11e−0.1t = 38 − (2.2t + 11)e−0.1t .
The thinner curve verifies the prediction of cooling followed by warming toward 38◦ .
The wider curve shows the temperature Tm of the liquid bath.
19. According to Newton’s Law of Cooling
dT
= k(T − Tm ).
dt
Separating variables we have
dT
= k dt
T − Tm
so
ln |T − Tm | = kt + c
and
T = Tm + c1 ekt .
Setting T (0) = T0 we find c1 = T0 − Tm . Thus
T (t) = Tm + (T0 − Tm )ekt .
In this problem we use T0 = 37 and Tm = 21. Now, let n denote the number of hours elapsed
before the body was found. Then T (n) = 29 and T (n + 1) = 27. Using this information, we
have
21 + (37 − 21)ekn = 29
and
21 + (37 − 21)ek(n+1) = 27
or
15ekn = 8
and
15ekn+k = 15ekn ek = 6.
The last equation is the same as 8ek = 6. Solving for k, we have k = ln 34 ≈ −0.288.
Finally, solving e−0.4055n = 15/28.6 for n, we have
8
−0.288n = ln
15
1
8
n=
ln
≈ 2.18.
−0.288
15
Thus, about 2.18 hours elapsed before the body was found.
3.1
Linear Models
20. Solving the differential equation dT /dt = kS(T − Tm ) subject to T (0) = T0 gives
T (t) = Tm + (T0 − Tm )ekSt .
The temperatures of the coffee in cups A and B are, respectively,
TA (t) = 21 + 45ekSt
and TB (t) = 21 + 45e2kSt .
Then TA (30) = 21 + 45e30kS = 38, which implies e30kS = 0.378. Hence
2
TB (30) = 21 + 45e60kS = 21 + 45 e30kS
= 21 + 45 (0.378)2 = 21 + 45 (0.143) = 27.43◦ C.
21. From dA/dt = 4 − A/50 we obtain A = 200 + ce−t/50 . If A(0) = 30 then c = −170 and
A = 200 − 170e−t/50 .
22. From dA/dt = 0 − A/50 we obtain A = ce−t/50 . If A(0) = 30 then c = 30 and A = 30e−t/50 .
23. From dA/dt = 5 − A/100 we obtain A = 500 + ce−t/100 . If A(0) = 0 then c = −500 and
A(t) = 500 − 500e−t/100 .
24. From Problem 23 the number of kilograms of salt in the tank at time t is A(t) = 500 −
500e−t/100 . The concentration at time t is c(t) = A(t)/2000 = 0.25 − 0.25e−t/100 . Therefore
c(5) = 0.25 − 0.25e−1/20 = 0.0122 kg/L and lim c(t) = 0.25. Solving c(t) = 0.125 = 0.25 −
0.25e−t/100 for t we obtain t ≈ 69.3 min.
25. From
t→∞
40A
2A
dA
=5−
=5−
dt
2000 − (40 − 20)t
100 − t
we obtain A = 500 − 5t + c(100 − t)2 . If A(0) = 0 then c = −0.05 . The tank is empty in 100
minutes.
26. With cin (t) = 2 + sin (t/4) kg/L, the initial-value problem is
1
t
dA
+
A = 6 + 3 sin ,
dt
100
4
A(0) = 25.
´
The differential equation is linear with integrating factor e
t
d t/100
[e
et/100
A(t)] = 6 + 3 sin
dt
4
et/100 A(t) = 600et/100 +
dt/100
= et/100 , so
3750 t/100
150 t/100
t
t
e
e
sin −
cos + c,
313
4
313
4
and
A(t) = 600 +
150
t
3750
t
sin −
cos + ce−t/100 .
313
4
313
4
111
112
CHAPTER 3
MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS
Letting t = 0 and A = 25 we have 600 − 3750/313 + c = 25 and c = −8/313. Then
A(t) = 600 +
t
3750
t
84200 −t/100
150
sin −
cos −
e
.
313
4
313
4
313
The graphs on [0, 300] and [0, 600] below show the effect of the sine function in the input
when compared with the graph in Figure 2.7.4(a) in the text.
A(t)
600
A(t)
600
500
500
400
400
300
300
200
200
100
100
50
27. From
100
150
200
250
t
300
100
200
300
400
500
t
600
15A
3A
dA
= 10 −
= 10 −
dt
400 + (20 − 15)t
80 + t
we obtain A = 80 + 2.5t + c(80 + t)−2 . If A(0) = 5 then c = −480,000 and A(30) = 64.38 kg.
28. (a) Initially the tank contains 300 gallons of solution. Since brine is pumped in at a rate of
3 gal/min and the mixture is pumped out at a rate of 2 gal/min, the net change is an
increase of 1 gal/min. Thus, in 100 minutes the tank will contain its capacity of 400
gallons.
(b) The differential equation describing the amount of salt in the tank is
A (t) = 6 − 2A/(300 + t) with solution
A(t) = 600 + 2t − (4.95 × 107 )(300 + t)−2 ,
0 ≤ t ≤ 100,
as noted in the discussion following Example 5 in the text. Thus, the amount of salt in
the tank when it overflows is
A(100) = 800 − (4.95 × 107 )(400)−2 = 490.625 lbs.
(c) When the tank is overflowing the amount of salt in the tank is governed by the differential
equation
A
dA
= (3 gal/min)(2 lb/gal) −
lb/gal (3 gal/min)
dt
400
=6−
3A
,
400
A(100) = 490.625.
3.1
Linear Models
Solving the equation, we obtain A(t) = 800 + ce−3t/400 . The initial condition yields
c = −654.947, so that
A(t) = 800 − 654.947e−3t/400 .
When t = 150, A(150) = 587.37 lbs.
(d) As t → ∞, the amount of salt is 800 lbs, which is to be expected since (400 gal)(2
lb/gal)= 800 lbs.
(e) x
A
800
600
400
200
200
400
600
t
29. Assume L di/dt + Ri = E(t), L = 0.1, R = 50, and E(t) = 50 so that i =
i(0) = 0 then c = −3/5 and lim i(t) = 3/5.
3
5
+ ce−500t . If
t→∞
30. Assume L di/dt + Ri = E(t), E(t) = E0 sin ωt, and i(0) = i0 so that
i=
E0 R
2
L ω 2 + R2
Since i(0) = i0 we obtain c = i0 +
sin ωt −
E0 Lω
2
L ω 2 + R2
E0 Lω
2
L ω 2 + R2
cos ωt + ce−Rt/L .
.
31. Assume R dq/dt + (1/C)q = E(t), R = 200, C = 10−4 , and E(t) = 100 so that
q = 1/100 + ce−50t . If q(0) = 0 then c = −1/100 and i = 12 e−50t .
32. Assume R dq/dt + (1/C)q = E(t), R = 1000, C = 5 × 10−6 , and E(t) = 200. Then
1
1
+ ce−200t and i = −200ce−200t . If i(0) = 0.4 then c = − 500
, q(0.005) = 0.0003
q = 1000
1
coulombs, and i(0.005) = 0.1472 amps. We have q → 1000 as t → ∞.
33. For 0 ≤ t ≤ 20 the differential equation is 20 di/dt+2i = 120. An integrating factor is et/10 , so
(d/dt)[et/10 i] = 6et/10 and i = 60 + c1 e−t/10 . If i(0) = 0 then c1 = −60 and i = 60 − 60e−t/10 .
For t > 20 the differential equation is 20 di/dt + 2i = 0 and i = c2 e−t/10 . At t = 20 we want
c2 e−2 = 60 − 60e−2 so that c2 = 60 e2 − 1 . Thus
0 ≤ t ≤ 20
60 − 60e−t/10 ,
i(t) =
−t/10
2
, t > 20
60 e − 1 e
113
114
CHAPTER 3
MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS
34. We first solve (1 − t/10)di/dt + 0.2i = 4. Separating variables we obtain
di/(40 − 2i) = dt/(10 − t). Then
√
1
40 − 2i = c1 (10 − t).
− ln |40 − 2i| = − ln |10 − t| + c or
2
√
20
10
10
20
Since i(0) = 0 we must have c1 = 2/ 10 . Solving for i we get i(t) = 4t − 15 t2 , 0 ≤ t < 10.
For t ≥ 10 the equation for the current becomes 0.2i = 4 or i = 20. Thus
⎧
1
⎪
⎨4t − t2 , 0 ≤ t < 10
5
i(t) =
⎪
⎩
20,
t ≥ 10.
The graph of i(t) is given in the figure.
35. (a) From m dv/dt = mg − kv we obtain v = mg/k + ce−kt/m . If v(0) = v0 then c = v0 − mg/k
and the solution of the initial-value problem is
mg −kt/m
mg + v0 −
e
.
v(t) =
k
k
(b) As t → ∞ the limiting velocity is mg/k.
(c) From ds/dt = v and s(0) = 0 we obtain
m
mg −kt/m m mg mg
t−
v0 −
e
v0 −
.
+
s(t) =
k
k
k
k
k
36. (a) Integrating d2 s/dt2 = −g we get v(t) = ds/dt = −gt + c. From v(0) = 90 we find c = 90,
and we are given g = 9.8, so the velocity is v(t) = −9t + 90.
(b) Integrating again and using s(0) = 0 we get s(t) = −4.9t2 + 90t. The maximum
height is attained when v = 0, that is, at ta = 9.184. The maximum height will be
s(9.184) = 413.27 m.
37. When air resistance is proportional to velocity, the model for the velocity is m dv/dt =
−mg − kv (using the fact that the positive direction is upward.) Solving the differential
equation using separation of variables we obtain v(t) = −mg/k + ce−kt/m . From v(0) = 90
we get
mg −kt/m
mg + 9+
e
.
v(t) = −
k
k
Integrating and using s(0) = 0 we find
s(t) = −
m
mg
mg t+
90 +
(1 − e−kt/m ).
k
k
k
Setting k = 0.0025, m = 75/9.8 = 7.653, and g = 9.8 we have
s(t) = 9.2117 − 30,000t − 9.2117 e−0.000327t
3.1
Linear Models
and
v(t) = −30,000 + 30,090e−0.000327t .
The maximum height is attained when v = 0, that is, at ta = 9.161. The maximum height
will be s(9.161) = 412.44 m, which is less than the maximum height in Problem 36.
38. Assuming that the air resistance is proportional to velocity and the positive direction is
downward with s(0) = 0, the model for the velocity is m dv/dt = mg − kv. Using separation
of variables to solve this differential equation, we obtain v(t) = mg/k + ce−kt/m . Then, using
v(0) = 0, we get v(t) = (mg/k)(1−e−kt/m ). Letting k = 7.25, m = (550+160)/9.8 = 72.5, and
g = 9.8, we have v(t) = 98(1 − e−0.1t ). Integrating, we find s(t) = 98t + 980e−0.1t + c1 . Solving
s(0) = 0 for c1 we find c1 = −980, therefore s(t) = 98t + 980e−0.1t − 980. At t = 15, when the
parachute opens, v(15) = 76.13 and s(15) = 708.67. At this time the value of k changes to k =
145 and the new initial velocity is v0 = 76.13. With the parachute open, the skydiver’s velocity
is vp (t) = mg/k +c2 e−kt/m , where t is reset to 0 when the parachute opens. Letting m = 72.5,
g = 9.8, and k = 145, this gives vp (t) = 4.9 + c2 e−2t . From v(0) = 76.13 we find c2 = 71.23, so
vp (t) = 4.9 + 71.23e−2t . Integrating, we get sp (t) = 4.9t − 35.615e−2t + c3 . Solving sp (0) = 0
for c3 , we find c3 = 35.615, so sp (t) = 4.9t−35.615e−2t +35.615. Twenty seconds after leaving
the plane is five seconds after the parachute opens. The skydiver’s velocity at this time is
vp (5) = 4.903 m/s and she has fallen a total of s(15) + sp (5) = 708.67 + 60.113 = 768.78 m.
Her terminal velocity is lim vp (t) = 4.9, so she has very nearly reached her terminal velocity
t→∞
five seconds after the parachute opens. When the parachute opens, the distance to the
ground is 4500 − s(15) = 4500 − 708.67 = 3791.33 m. Solving sp (t) = 3791.33 we get
t = 766.47 s = 12.77 min. Thus, it will take her approximately 12.77 minutes to reach the
ground after her parachute has opened and a total of (766 + 15)/60 = 13.02 minutes after she
exits the plane.
39. (a) With the values given in the text the initial-value problem becomes
2
2000
dv
+
v = −9.8 +
,
dt
200 − t
200 − t
v(0) = 0.
This is a linear differential equationwith integrating factor
´
e
[2/(200−t)] dt
= e−2 ln |200−t| = (200 − t)−2 .
Then
d
(200 − t)−2 v = −9.8(200 − t)−2 + 2000(200 − t)−3
dt
←− integrate
(200 − t)−2 v = −9.8(200 − t)−1 + 1000(200 − t)−2 + c
v = −9.8(200 − t) + 1000 + c(200 − t)2
= −960 + 9.8t + c(200 − t)2 .
←− multiply by(200 − t)2
115
116
CHAPTER 3
MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS
Using the initial condition we have
0 = v(0) = −960 + 40,000c
so
c=
960
= 0.024.
40,000
Thus
v(t) = −960 + 9.8t + 0.024(200 − t)2
= −960 + 9.8t + 0.024(40, 000 − 400t + t2 )
= −960 + 9.8t + 960 − 9.6t + 0.024t2
= 0.024t2 + 0.2t.
(b) Integrating both sides of
ds
= v(t) = 0.024t2 + 0.2t
dt
we find
s(t) = 0.008t3 + 0.1t2 + c1 .
We assume that the height of the rocket is measured from s = 0, so that s(0) = 0 and
c1 = 0. Then the height of the rocket at time t is s(t) = 0.008t3 + 0.1t2 .
40. (a) From Problem 22 in Exercises 1.3, tb = mf (0)/λ. In this case mf (0) = 50 and λ = 1, so
the time of burnout is 50 s.
(b) The velocity at burnout time is
v(50) = 0.024(50)2 + 0.2(50) = 70 m/s.
(c) The height of the rocket at burnout is
s(50) = 0.008(50)3 + 0.1(50)2 = 1250 m.
(d) At burnout the rocket will have upward momentum which will carry it higher.
(e) After burnout the total mass of the rocket is a constant 200 − 50 = 150 kg. By Problem
22 in Exercises 1.3 the velocity for a rocket with variable mass due to fuel consumption
is
dm
dv
+ kv = −mg + R.
m +v
dt
dt
Here m is the total mass of the rocket, and in this case m is constant after burnout, so
dm/dt = 0 and the velocity of the rocket satisfies
m
dv
+ kv = −mg + R.
dt
We identify m = 150, k = 3, g = 9.8 and R = 0, since the thrust is 0 after burnout. Then
150
dv
+ 3v = −150(9.8) = −1470 or
dt
dv
1
+ v = −9.8.
dt
50
3.1
Linear Models
117
41. (a) The differential equation is first-order and linear. Letting b = k/ρ, the integrating factor
´
is e 3b dt/(bt+r0 ) = (r0 + bt)3 . Then
d
[(r0 + bt)3 v] = g(r0 + bt)3
dt
and
(r0 + bt)3 v =
g
(r0 + bt)4 + c.
4b
The solution of the differential equation is v(t) = (g/4b)(r0 + bt) + c(r0 + bt)−3 . Using
v(0) = 0 we find c = −gr04 /4b, so that
gr04
gρ
g
=
v(t) = (r0 + bt) −
4b
4b(r0 + bt)3
4k
gρr04
k
r0 + t −
.
ρ
4k(r0 + kt/ρ)3
(b) Integrating dr/dt = k/ρ we get r = kt/ρ + c. Using r(0) = r0 we have c = r0 , so
r(t) = kt/ρ + r0 .
(c) If r = 2 mm when t = 10 s, then solving r(10) = 2 for k/ρ, we obtain k/ρ = −0.1 and
r(t) = 3 − 0.1t. Solving r(t) = 0 we get t = 30, so the raindrop will have evaporated
completely at 30 seconds.
42. Separating variables, we obtain dP/P = k cos t dt, so
P
ln |P | = k sin t + c
and
k sin t
P = c1 e
.
P0
If P (0) = P0 , then c1 = P0 and P = P0 ek sin t .
5
10
t
43. (a) From dP/dt = (k1 − k2 )P we obtain P = P0 e(k1 −k2 )t where P0 = P (0).
(b) If k1 > k2 then P → ∞ as t → ∞. If k1 = k2 then P = P0 for every t. If k1 < k2 then
P → 0 as t → ∞.
44. (a) The solution of the differential equation is P (t) = c1 ekt + h/k. If we let the initial
population of fish be P0 then P (0) = P0 which implies that
h
c1 = P0 −
k
h kt h
and P (t) = P0 −
e + .
k
k
(b) For P0 > h/k all terms in the solution are positive. In this case P (t) increases as time t
increases. That is, P (t) → ∞ as t → ∞. For P0 = h/k the population remains constant
for all time t:
P (t) =
h h
−
k k
ekt +
h
h
= .
k
k
For 0 < P0 < h/k the coefficient of the exponential function is negative and so the
function decreases as time t increases.
118
CHAPTER 3
MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS
(c) For 0 < P0 < h/k the function decreases and is concave down, therefore the graph of
P (t) crosses the t-axis. That is, there exists a time T > 0 such that P (T ) = 0. Solving
h kT h
e + =0
P0 −
k
k
for T shows that the time of extinction is
1
T = ln
k
h
h − kP0
.
45. (a) Solving r − kx = 0 for x we find the equilibrium solution x = r/k. When
x < r/k, dx/dt > 0 and when x > r/k, dx/dt < 0. From the phase
portrait we see that lim x(t) = r/k.
x
t→∞
r
k
(b) From dx/dt = r − kx and x(0) = 0 we obtain
x = r/k − (r/k)e−kt so that x → r/k as t → ∞. If
x(T ) = r/2k then T = (ln 2)/k.
x
r/k
t
46. (a) Solving k1 (M − A) − k2 A = 0 for A we find the equilibrium solution
A = k1 M/(k1 + k2 ). From the phase portrait we see that
lim A(t) = k1 M/(k1 + k2 ). Since k2 > 0, the material will never
t→∞
be completely memorized and the larger k2 is, the less the amount
of material will be memorized over time.
A
Mk1
k1 + k2
3.1
(b) Write the differential equation in the form
dA/dt + (k1 + k2 )A = k1 M . Then an integrating
factor is e(k1 +k2 )t , and
k1M
k1 + k2
Linear Models
119
A
d (k1 +k2 )t A = k1 M e(k1 +k2 )t
e
dt
e(k1 +k2 )t A =
A=
k1 M
+ ce−(k1 +k2 )t .
k1 + k2
Using A(0) = 0 we find c = −
A→
t
k1 M (k1 +k2 )t
e
+c
k1 + k2
k1 M
.
k1 + k2
k1 M
k1 M and A =
1 − e−(k1 +k2 )t . As t → ∞,
k1 + k2
k1 + k2
47. (a) For 0 ≤ t < 4, 6 ≤ t < 10 and 12 ≤ t < 16, no voltage is applied to the heart and
E(t) = 0. At the other times, the differential equation is dE/dt = −E/RC. Separating
variables, integrating, and solving for e, we get E = ke−t/RC , subject to E(4) = E(10) =
E(16) = 12. These intitial conditions yield, respectively, k = 12e4/RC , k = 12e10/RC ,
k = 12e16/RC , and k = 12e22/RC . Thus
⎧
0,
⎪
⎪
⎪
⎪
⎪
⎪
12e(4−t)/RC ,
⎪
⎨
E(t) = 12e(10−t)/RC ,
⎪
⎪
⎪
⎪12e(16−t)/RC ,
⎪
⎪
⎪
⎩ (22−t)/RC
,
12e
(b) x
0 ≤ t < 4, 6 ≤ t < 10, 12 ≤ t < 16
4≤t<6
10 ≤ t < 12
16 ≤ t < 18
22 ≤ t < 24
E
10
5
4
6
10 12
16 18
22 24
t
48. (a) (i) Using Newton’s second law of motion, F = ma = m dv/dt, the differential equation
for the velocity v is
m
dv
= mg sin θ
dt
or
dv
= g sin θ,
dt
where mg sin θ, 0 < θ < π/2, is the component of the weight along the plane in the
direction of motion.
120
CHAPTER 3
MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS
(ii) The model now becomes
dv
= mg sin θ − μmg cos θ,
dt
where μmg cos θ is the component of the force of sliding friction (which acts perpendicular
to the plane) along the plane. The negative sign indicates that this component of force
is a retarding force which acts in the direction opposite to that of motion.
m
(iii) If air resistance is taken to be proportional to the instantaneous velocity of the
body, the model becomes
dv
= mg sin θ − μmg cos θ − kv,
dt
where k is a constant of proportionality.
m
(b) (i) The differential equation is
dv
1
= (g) ·
dt
2
or
dv
= g/2.
dt
Integrating the last equation gives v(t) = gt/2 + c1 . Since v(0) = 0, we have c1 = 0 and
so v(t) = gt/2.
(ii) The differential equation is
√
√
1
3
3
dv
= (g) · −
· (g) ·
dt
2
4
2
or
dv
= g/8.
dt
In this case v(t) = gt/8.
(iii) When the retarding force due to air resistance is taken into account, the differential
equation for velocity v becomes
√
√
1
dv
1
3
3 1
dv
= (g) · −
· (g) ·
− v
or
= g/8 − v.
dt
2
4
2
4
dt
4
The last differential equation is linear and has solution v(t) = g/2 + c1 e−t/4 . Since
v(0) = 0, we find c1 = −g/2, so v(t) = g/2 − (g/2)e−t/4 .
49. (a) (i) If s(t) is distance measured down the plane from the highest point, then ds/dt = v.
Integrating ds/dt = (g/2)t gives s(t) = (g/4)t2 + c2 . Using s(0) = 0 then gives c2 = 0.
Now the length L of the plane is L = 15/ sin 30◦ = 30 ft. The time it takes the box to
slide completely down the plane is the solution of s(t) = 30 or t2 = 12.245, so t ≈ 3.5 s.
(ii) Integrating ds/dt = gt/8 gives s(t) = (g/16)t2 + c2 . Using s(0) = 0 gives c2 = 0, so
s(t) = (g/16)t2 and the solution of s(t) = 30 is now t ≈ 7 s.
(iii) Integrating ds/dt = g/2 − (g/2)e−t/4 and using s(0) = 0 to determine the constant
of integration, we obtain s(t) = (g/2)t + 2ge−t/4 − 2g. With the aid of a CAS we find
that the solution of s(t) = 30, or
30 = (g/2)t + 2ge−t/4 − 2g
is now t ≈ 9.78 s.92.1
3.2 Nonlinear Models
121
(b) The differential equation m dv/dt = mg sin θ − μmg cos θ can be written
m
dv
= mg cos θ(tan θ − μ).
dt
If tan θ = μ, dv/dt = 0 and v(0) = 0 implies that v(t) = 0. If tan θ < μ and v(0) = 0,
then integration implies v(t) = g cos θ(tan θ − μ)t < 0 for all time t.
√
(c) Since tan 23◦ = 0.4245 and μ = 3/4 = 0.4330, we see that tan 23◦ < 0.4330. The
√
differential equation is dv/dt = 9.8 cos 23◦ (tan 23◦ − 3/4) = −0.077. Integration and
the use of the initial condition gives v(t) = −0.077t+0.3. When the box stops, v(t) = 0 or
0 = −0.077t+0.3 or t = 3.896 s. From s(t) = −0.0385t2 +0.3t we find s(3.869) = 0.584 m.
(d) With v0 > 0, v(t) = −0.077t + v0 and s(t) = −0.0385t2 + v0 t. Because two real positive
solutions of the equation s(t) = 30, or 0 = −0.0385t2 + v0 t − 30, would be physically
meaningless, we use the quadratic formula and require that b2 − 4ac = 0 or v02 − 4.62 = 0.
From this last equality we find v0 ≈ 2.15 m/s. For the time it takes the box to traverse
the entire inclined plane, we must have 0 = −0.0385t2 + 2.15t − 30.
The roots are t = 27.2727 s and t = 28.5714 s. So if v0 > 2.15, we are guaranteed that
the box will slide completely down the plane.
50. (a) We saw in part (a) of Problem 36 that the ascent time is ta = 9.375. To find when the
cannonball hits the ground we solve s(t) = −16t2 + 300t = 0, getting a total time in
flight of t = 18.75 s. Thus, the time of descent is td = 18.75 − 9.375 = 9.375. The impact
velocity is vi = v(18.75) = −300, which has the same magnitude as the initial velocity.
(b) We saw in Problem 37 that the ascent time in the case of air resistance is ta = 9.162.
Solving s(t) = 1,340,000 − 6,400t − 1,340,000e−0.005t = 0 we see that the total time of
flight is 18.466 s. Thus, the descent time is td = 18.466 − 9.162 = 9.304. The impact
velocity is vi = v(18.466) = −290.91, compared to an initial velocity of v0 = 300.
3.2
Nonlinear Models
1. (a) Solving N (1 − 0.0005N ) = 0 for N we find the equilibrium solutions N = 0
and N = 2000. When 0 < N < 2000, dN/dt > 0. From the phase portrait we
see that lim N (t) = 2000. A graph of the solution is shown in part (b).
2000
t→∞
0
N
122
CHAPTER 3
MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS
(b) Separating variables and integrating we have
dN
1
1
=
−
dN = dt
N (1 − 0.0005N )
N
N − 2000
2000
N
1500
1000
500
and
5
ln N − ln |N − 2000| = t + c.
10
15
20
t
Solving for N we get N (t) = 2000ec+t /(1 + ec+t ) = 2000ec et /(1 + ec et ). Using N (0) = 1
and solving for ec we find ec = 1/1999 and so N (t) = 2000et /(1999 + et ). Then N (10) =
1833.59, so 1834 companies are expected to adopt the new technology when t = 10.
2. From dN/dt = N (a − bN ) and N (0) = 500 we obtain
N=
500a
.
500b + (a − 500b)e−at
Since lim N = a/b = 50, 000 and N (1) = 1000 we have a = 0.7033, b = 0.00014, and N =
t→∞
50,000/(1 + 99e−0.7033t ) .
3. From dP/dt = P 10−1 − 10−7 P and P (0) = 5000 we obtain P = 500/(0.0005+0.0995e−0.1t )
so that P → 1,000,000 as t → ∞. If P (t) = 500,000 then t = 52.9 months.
4. (a) We have dP/dt = P (a − bP ) with P (0) = 3.929 million. Using separation of variables we
obtain
P (t) =
=
a/b
3.929a
=
−at
3.929b + (a − 3.929b)e
1 + (a/3.929b − 1)e−at
c
,
1 + (c/3.929 − 1)e−at
where c = a/b. At t = 60(1850) the population is 23.192 million, so
23.192 =
c
1 + (c/3.929 − 1)e−60a
or c = 23.192 + 23.192(c/3.929 − 1)e−60a . At t = 120(1910),
91.972 =
c
1 + (c/3.929 − 1)e−120a
or c = 91.972 + 91.972(c/3.929 − 1)(e−60a )2 . Combining the two equations for c we get
(c − 23.192)/23.192
c/3.929 − 1
2 c − 91.972
c
−1 =
3.929
91.972
or
91.972(3.929)(c − 23.192)2 = (23.192)2 (c − 91.972)(c − 3.929).
The solution of this quadratic equation is c = 197.274. This in turn gives a = 0.0313.
Therefore,
197.274
.
P (t) =
1 + 49.21e−0.0313t
3.2 Nonlinear Models
(b) x
Year
Census
Population
Predicted
Population
Error
1790
1800
1810
1820
1830
1840
1850
1860
1870
1880
1890
1900
1910
1920
1930
1940
1950
3.929
5.308
7.240
9.638
12.866
17.069
23.192
31.433
38.558
50.156
62.948
75.996
91.972
105.711
122.775
131.669
150.697
3.929
5.334
7.222
9.746
13.090
17.475
23.143
30.341
39.272
50.044
62.600
76.666
91.739
107.143
122.140
136.068
148.445
0.000
–0.026
0.018
–0.108
–0.224
–0.406
0.049
1.092
–0.714
0.112
0.348
–0.670
0.233
–1.432
0.635
–4.399
2.252
123
%
Error
0.00
–0.49
0.24
–1.12
–1.74
–2.38
0.21
3.47
–1.85
0.22
0.55
–0.88
0.25
–1.35
0.52
–3.34
1.49
P
5. (a) The differential equation is dP/dt = P (5−P )−4. Solving P (5−P )−4 = 0 for P
we obtain equilibrium solutions P = 1 and P = 4. The phase portrait is shown
on the right and solution curves are shown in part (b). We see that for P0 > 4
and 1 < P0 < 4 the population approaches 4 as t increases. For 0 < P < 1 the
population decreases to 0 in finite time.
4
1
P
(b) The differential equation is
dP
= P (5−P )−4 = −(P 2 −5P +4) = −(P −4)(P −1).
dt
4
1
Separating variables and integrating, we obtain
dP
= −dt
(P − 4)(P − 1)
1/3
1/3
−
dP = −dt
P −4 P −1
1 P − 4 ln = −t + c
3
P − 1
P −4
= c1 e−3t .
P −1
Setting t = 0 and P = P0 we find c1 = (P0 − 4)/(P0 − 1). Solving for P we obtain
P (t) =
4(P0 − 1) − (P0 − 4)e−3t
.
(P0 − 1) − (P0 − 4)e−3t
3
t
124
CHAPTER 3
MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS
(c) To find when the population becomes extinct in the case 0 < P0 < 1 we set P = 0 in
P −4
P0 − 4 −3t
=
e
P −1
P0 − 1
from part (a) and solve for t. This gives the time of extinction
1 4(P0 − 1)
.
t = − ln
3
P0 − 4
5
5
6. Solving P (5 − P ) − 25
4 = 0 for P we obtain the equilibrium solution P = 2 . For P = 2 ,
dP/dt < 0. Thus, if P0 < 52 , the population becomes extinct (otherwise there would be
another equilibrium solution.) Using separation of variables to solve the initial-value problem,
we get
P (t) = [4P0 + (10P0 − 25)t]/[4 + (4P0 − 10)t].
To find when the population becomes extinct for P0 <
the time of extinction is t = 4P0 /5(5 − 2P0 ).
5
2
we solve P (t) = 0 for t. We see that
7. Solving P (5−P )−7 = 0 for P we obtain complex roots, so there are no equilibrium solutions.
Since dP/dt < 0 for all values of P , the population becomes extinct for any initial condition.
Using separation of variables to solve the initial-value problem, we get
√
√ 2P
3
−
5
3
5
0
√
−
tan tan−1
t .
P (t) = +
2
2
2
3
Solving P (t) = 0 for t we see that the time of extinction is
t=
√
√
√ 2 √
3 tan−1 (5/ 3 ) + 3 tan−1 (2P0 − 5)/ 3 .
3
8. (a) The differential equation is dP/dt = P (1 − ln P ), which has the
equilibrium solution P = e. When P0 > e, dP/dt < 0, and
when P0 < e, dP/dt > 0.
P
e
t
P
(b) The differential equation is dP/dt = P (1 + ln P ), which has the
equilibrium solution P = 1/e. When P0 > 1/e, dP/dt > 0, and
when P0 < 1/e, dP/dt < 0.
1/e
t
3.2 Nonlinear Models
125
9. Let X = X(t) be the amount of C at time t and dX/dt = k(120 − 2X)(150 − X). If X(0) = 0
and X(5) = 10, then
150 − 150e180kt
,
X(t) =
1 − 2.5e180kt
where k = .0001259 and X(20) = 29.3 g. Now by L’Hôpital’s rule, X → 60 as t → ∞, so that
the amount of A → 0 and the amount of B → 30 as t → ∞.
10. From dX/dt = k(150 − X)2 , X(0) = 0, and X(5) = 10 we obtain X = 150 − 150/(150kt + 1)
where k = .000095238. Then X(20) = 33.3 g and X → 150 as t → ∞ so that the amount of
A → 0 and the amount of B → 0 as t → ∞. If X(t) = 75 then t = 70 minutes.
11. (a) The initial-value problem is dh/dt = −4.427
√
Ah h /Aw , h(0) = H. Separating variables and
integrating we have
4.427Ah
dh
√ =−
dt
Aw
h
and
√
4.427Ah
2 h=−
t+c.
Aw
10
h
8
6
4
2
500
1000
1500
t
√
Using h(0) = H we find c = 2 H , so the solution of the initial-value problem is
√
√
h(t) = (Aw H − 2.214Ah t)/Aw , where Aw H − 2.214Ah t ≥ 0. Thus,
√
√
h(t) = (Aw H − 2.214Ah t)2 /A2w for 0 ≤ t ≤ Aw H /2.214Ah .
(b) Identifying H = 3, Aw = 0.36π, and Ah = π/6944 we have
h(t) = t2 /1.2756 − 0.00307t + 3. Solving h(t) = 0 we see that the tank empties in 1956
seconds or 32.6 minutes.
12. To obtain the solution of this differential equation we use h(t) from Problem 13 in Exercises
√
1.3. Then h(t) = (Aw H − 2.214cAh t)2 /A2w . Solving h(t) = 0 with c = 0.6 and the values
from Problem 11 we see that the tank empties in 3259.45 seconds or 54.3 minutes.
13. (a) Separating variables and integrating gives
h3/2 dh = −0.0266t
and
0.4h5/2 = −0.0266t + c.
Using h(0) = 20 we find c = 35.27, so the solution of the initial-value problem is
h(t) = (88.175 − 0.166t)2/5 . Solving h(t) = 0 we see that the tank empties in 531.17
seconds or 8.85 minutes.
(b) When the height of the water is h, the radius of the top of the water is
√
r = h tan 30◦ = h/ 3 and Aw = πh2 /3. The differential equation is
dh
0.02
Ah π(5/100)2 √
19.6h = − 3/2 .
= −c
2gh = −0.6
2
dt
Aw
πh /3
h
Separating variables and integrating gives
h3/2 dh = −0.02 dt and
0.4h5/2 = −0.02t + c.
126
CHAPTER 3
MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS
Using h(0) = 3 we find c = 6.235, so the solution of the initial-value problem is h(t) =
(15.588 − 0.05t)2/5 . Solving h(t) = 0 we see that the tank empties in 311.76 seconds or
5.196 minutes.
14. When the height of the water is h, the radius of the top of the water is 12 (6 − h) and Aw =
0.25π(6 − h)2 . The differential equation is
√
Ah π(5/100)2 √
h
dh
= −c
2gh = −0.6
19.6h = −0.0266
.
dt
Aw
0.25π(6 − h)2
(6 − h)2
Separating variables and integrating we have
(6 − h)2
√
dh = −0.0266 dt
h
and
√
2
72 h − 8h3/2 + h5/2 = −0.0266t + c.
5
Using h(0) = 6 we find c = 94.06, so an implicit solution of the initial-value problem is
√
2
72 h − 8h3/2 + h5/2 = −0.0266t + 94.06 .
5
To find the time it takes the tank to empty we set h = 0 and solve for t. The tank empties
in 3536 seconds or 58.93 minutes. Thus, the tank empties more slowly when the base of the
cone is on the bottom.
15. (a) After separating variables we obtain
m dv
= dt
mg − kv 2
dv
1
√
g 1 − ( k v/√mg )2
√
mg
k/mg dv
√
√
√
k g 1 − ( k v/ mg )2
√
kv
m
−1
tanh √
kg
mg
√
kv
tanh−1 √
mg
= dt
= dt
=t+c
=
kg
t + c1 .
m
Thus the velocity at time t is
v(t) =
mg
tanh
k
kg
t + c1 .
m
√
√
Setting t = 0 and v = v0 we find c1 = tanh−1 ( k v0 / mg ).
(b) Since tanh t → 1 as t → ∞, we have v →
mg/k as t → ∞.
3.2 Nonlinear Models
´
(c) Integrating the expression for v(t) in part (a) we obtain an integral of the form du/u:
ˆ
m
mg
kg
kg
+ c2 .
t + c1 dt =
ln cosh
t + c1
s(t) =
tanh
k
m
k
m
Setting t = 0 and s = 0 we find c2 = −(m/k) ln (cosh c1 ), where c1 is given in part (a).
16. The differential equation is m dv/dt = −mg − kv 2 . Separating variables and integrating, we
have
dv
dt
=−
mg + kv 2
m
√ 1
kv
1
√
=− t+c
tan−1 √
mg
m
mgk
√ k
v
gk
=−
t + c1
tan−1 √
mg
m
v(t) =
mg
gk
tan c1 −
t
k
m
75
= 7.653 , g = 9.8, and k = 0.0003, we find v(t) = 500 tan(c1 −
Setting v(0) = 90, m = 9.8
0.0196t) and c1 = 3.318. Integrating
v(t) = 500 tan (3.318 − 0.0196t)
we get
s(t) = 25510 ln | cos (3.318 − 0.0196t)| + c2 .
Using s(0) = 0 we find c2 = 399. Solving v(t) = 0 we see that the maximum height is attained
when t = 9. The maximum height is s(9) = 360.63 m.
17. (a) Let ρ be the weight density of the water and V the volume of the object. Archimedes’
principle states that the upward buoyant force has magnitude equal to the weight of the
water displaced. Taking the positive direction to be down, the differential equation is
m
dv
= mg − kv 2 − ρV.
dt
(b) Using separation of variables we have
m dv
= dt
(mg − ρV ) − kv 2
√
m
k dv
√ √
√
= dt
k ( mg − ρV )2 − ( k v)2
√
kv
m
1
−1
√ √
= t + c.
tanh √
mg − ρV
k mg − ρV
127
128
CHAPTER 3
MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS
Thus
√
kmg − kρV
t + c1 .
m
(c) Since tanh t → 1 as t → ∞, the terminal velocity is (mg − ρV )/k .
v(t) =
mg − ρV
tanh
k
18. (a) Writing the equation in the form (x − x2 + y 2 )dx + y dy = 0 we identify
M = x − x2 + y 2 and N = y. Since M and N are both homogeneous functions of
degree 1 we use the substitution y = ux. It follows that
x − x2 + u2 x2 dx + ux(u dx + x du) = 0
x 1 − 1 + u2 + u2 dx + x2 u du = 0
−
1+
u2
dx
u du
√
=
2
x
− 1+u
dx
u du
√
=
.
2
x
1+
− 1+u )
√
√
Letting w = 1 − 1 + u2 we have dw = −u du/ 1 + u2 so that
− ln 1 − 1 + u2 = ln |x| + c
√
u2 (1
1
√
= c1 x
1 − 1 + u2
c2
1 − 1 + u2 = −
x
y2
c2
= 1+ 2
1+
x
x
1+
(−c2 = 1/c1 )
2c2
c2
y2
+ 22 = 1 + 2 .
x
x
x
Solving for y 2 we have
y 2 = 2c2 x + c22 = 4
c 2
2
x+
c2 2
which is a family of parabolas symmetric with respect to the x-axis with vertex at
(−c2 /2, 0) and focus at the origin.
(b) Let u = x2 + y 2 so that
dy
du
= 2x + 2y
.
dx
dx
Then
1 du
dy
=
−x
dx
2 dx
and the differential equation can be written in the form
y
√
1 du
− x = −x + u
2 dx
or
1 du √
= u.
2 dx
3.2 Nonlinear Models
129
Separating variables and integrating gives
du
√ = dx
2 u
√
u=x+c
u = x2 + 2cx + c2
x2 + y 2 = x2 + 2cx + c2
y 2 = 2cx + c2 .
19. (a) From 2W 2 − W 3 = W 2 (2 − W ) = 0 we see that W = 0 and W = 2 are constant solutions.
(b) Separating variables and using a CAS to integrate we get
dW
√
= dx
W 4 − 2W
and
− tanh−1
1√
4 − 2W
2
= x + c.
Using the facts that the hyperbolic tangent is an odd function and 1 − tanh2 x = sech2 x
we have
1√
4 − 2W = tanh (−x − c) = − tanh (x + c)
2
1
(4 − 2W ) = tanh2 (x + c)
4
1
1 − W = tanh2 (x + c)
2
1
W = 1 − tanh2 (x + c) = sech2 (x + c)
2
Thus, W (x) = 2 sech2 (x + c).
(c) Letting x = 0 and W = 2 we find that sech2 (c) = 1
and c = 0.
W
2
–3
3
x
20. (a) Solving r2 + (10 − h)2 = 102 for r2 we see that r2 = 20h − h2 . Combining the rate of
input of water, π, with the rate of output due to evaporation,
kπr2 = kπ(20h − h2 ), we have dV /dt = π − kπ(20h − h2 ). Using V = 10πh2 − 13 πh3 , we
see also that dV /dt = (20πh − πh2 )dh/dt. Thus,
(20πh − πh2 )
dh
= π − kπ(20h − h2 )
dt
and
1 − 20kh + kh2
dh
=
.
dt
20h − h2
130
CHAPTER 3
MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS
(b) Letting k = 1/100, separating variables and integrating (with the help of a CAS), we get
and
10
100h(h − 20)
dh = dt
(h − 10)2
8
100(h2 − 10h + 100)
= t + c.
10 − h
4
h
6
2
Using h(0) = 0 we find c = 1000, and solving
for
√
2
t + 4000t − t , where
h we get h(t) = 0.005
the positive square root is chosen because h ≥ 0.
t
2000 4000 6000 8000 10000
(c) The volume of the tank is V = 23 π(10)3 m3 , so at a rate of π cubic meter per minute, the
tank will fill in 23 (10)3 ≈ 666.67 minutes ≈ 11.11 hours.
(d) At 666.67 minutes, the depth of the water is h(666.67) = 5.486 m. From the graph in
(b) we suspect that limt→∞ h(t) = 10, in which case the tank will never completely fill.
To prove this we compute the limit of h(t):
t2 + 4000t − t2
t2 + 4000t − t = 0.005 lim √
t→∞
t→∞
t2 + 4000t + t
lim h(t) = 0.005 lim
t→∞
4000
4000t
= 0.005(2000) = 10.
= 0.005
= 0.005 lim t→∞ t 1 + 4000/t + t
1+1
21. (a) With c = 0.01 the differential equation is dP/dt = kP 1.01 . Separating variables and
integrating we obtain
P −1.01 dP = k dt
P −0.01
= kt + c1
−0.01
P −0.01 = −0.01kt + c2
P (t) = (−0.01kt + c2 )−100
P (0) = c−100
= 10
2
c2 = 10−0.01 .
Then
P (t) =
1
(−0.01kt + 10−0.01 )100
and, since P doubles in 5 months from 10 to 20,
P (5) =
1
(−0.01k(5) + 10−0.01 )100
= 20
3.2 Nonlinear Models
131
so
−0.01k(5) + 10−0.01
100
=
−0.01k =
1
20
1 1/100
20
−
1 1/100 10
5
= −0.001350 .
100
.
Thus P (t) = 1/ −0.001350t + 10−0.01
(b) Define T =
P → ∞.
(c)
1 1/100
10
/0.001350 ≈ 724 months = 60 years. As t → 724 (from the left),
P (50) = 1/ −0.001350 (50) + 10−0.01
100
P (100) = 1/ −0.001350 (100) + 10−0.01
≈ 12, 839
100
and
≈ 28, 630, 966
P
22. (a) From the phase portrait we see that P = 0 is an attractor for 0 < P0 < K = a/b
and P = K is a repeller for P0 > K.
K
0
(b) Letting a = 0.1, b = 0.0005 and using separation of variables gives
b
1
dP = a dt
− +
P
bP − a
Integrating we have
− ln P + ln (bP − a) = at + c1
bP − a
ln
= at + c1
P
bP − a
= c2 eat
P
a
P =
.
b − c2 eat
Since P (0) = 300,
c2 =
300b − a
300
and
P (t)
=
300a
.
300b − (300b − a) eat
132
CHAPTER 3
MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS
Then, with b = 0.0005 and a = 0.1,
P (t) =
30
600
300(0.1)
=
=
0.1t
0.1t
00(0.0005) − [300(0.0005) − 0.1] e
0.15 − 0.05e
3 − e0.1t
and
3 − e0.1 = 0
implies
0.1t ln 3
so
t = 10 ln 3.
This is doomsday in finite time, since P (t) → ∞ as t → 10 ln 3 (from the left) ≈ 10.99.
(c) For P0 = 100
P (t) =
=
100a
100(0.1)
=
at
100b − (100b − a) e
100(0.0005) − [100(0.0005) − 0.1] e0.1t
10
200
=
,
0.05 + 0.05e0.1t
1 + e0.1t
and P (t) → 0 as t → ∞.
23. (a) x
t
P(t)
Q(t)
0
10
20
30
40
50
60
70
80
90
100
110
120
130
140
150
160
170
3.929
5.308
7.240
9.638
12.866
17.069
23.192
31.433
38.558
50.156
62.948
75.996
91.972
105.711
122.775
131.669
150.697
179.300
0.035
0.036
0.033
0.033
0.033
0.036
0.036
0.023
0.030
0.026
0.021
0.021
0.015
0.016
0.007
0.014
0.019
3.2 Nonlinear Models
(b) The regression line is Q = 0.0348391 − 0.000168222P .
Q
0.035
0.03
0.025
0.02
0.015
0.01
0.005
20
40
60
80
100
120
140
P
(c) The solution of the logistic equation is given in Equation (5) in the text. Identifying
a = 0.0348391 and b = 0.000168222 we have
P (t) =
aP0
.
bP0 + (a − bP0 )e−at
(d) With P0 = 3.929 the solution becomes
P (t) =
(e) x
175
0.136883
.
0.000660944 + 0.0341781e−0.0348391t
P
150
125
100
75
50
25
25
50
75
100
125
150
t
(f ) We identify t = 180 with 1970, t = 190 with 1980, and t = 200 with 1990. The model
predicts P (180) = 188.661, P (190) = 193.735, and P (200) = 197.485. The actual
population figures for these years are 203.303, 226.542, and 248.765 millions. As t → ∞,
P (t) → a/b = 207.102.
24. (a) Using a CAS to solve P (1 − P ) + 0.3e−P = 0 for P we see that P = 1.09216 is an
equilibrium solution.
133
134
CHAPTER 3
MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS
(b) Since f (P ) > 0 for 0 < P < 1.09216, the solution
P (t) of
dP/dt = P (1 − P ) + 0.3e−P ,
P (0) = P0 ,
f
2
1
is increasing for P0 < 1.09216. Since f (P ) < 0
for P > 1.09216, the solution P (t) is decreasing for
P0 > 1.09216. Thus P = 1.09216 is an attractor.
(c) The curves for the second initial-value problem are
thicker. The equilibrium solution for the logic model
is P = 1. Comparing 1.09216 and 1, we see that the
percentage increase is 9.216
0.5
1
1.5
2
p
2.5
3
8
t
10
–1
–2
2
p
1.5
1
0.5
2
4
6
25. To find td we solve
dv
= mg − kv 2 ,
dt
using separation of variables. This gives
m
v(t) =
mg
tanh
k
v(0) = 0
kg
t.
m
Integrating and using s(0) = 0 gives
m
kg
s(t) =
ln cosh
t .
k
m
To find the time of descent we solve s(t) = 360.63 and find td = 8.599. The impact velocity
is v(td ) = 83.481, which is positive because the positive direction is downward.
26. (a) Solving vt = mg/k for k we obtain k = mg/vt2 . The differential equation then becomes
mg 2
1 2
dv
dv
= mg − 2 v or
=g 1− 2 v .
m
dt
dt
vt
vt
Separating variables and integrating gives
vt tanh−1
v
= gt + c1 .
vt
The initial condition v(0) = 0 implies c1 = 0, so
v(t) = vt tanh
gt
.
vt
3.2 Nonlinear Models
We find the distance by integrating:
ˆ
gt
gt
vt2
+ c2 .
ln cosh
s(t) = vt tanh dt =
vt
g
vt
The initial condition s(0) = 0 implies c2 = 0, so
gt
vt2
ln cosh
.
s(t) =
g
vt
In 25 seconds she has fallen 6000 − 4500 = 1500 m. Using a CAS to solve
9.8(25)
2
1500 = (vt /9.8) ln cosh
vt
for vt gives vt ≈ 76.53 m/s. Then
gt
vt2
ln cosh
s(t) =
g
vt
(b) At t = 15, s(15) = 746.39 m and v(15) = s (15) = 142.9 m/sec.
27. While the object is in the air its velocity is modeled by the linear differential equation
m dv/dt = mg − kv. Using m = 71.43, k = 2.5, and g = 9.8, the differential equation
´
becomes dv/dt + (1/28.572)v = 9.8. The integrating factor is e dt/28.572 = et/28.572 and the
´
solution of the differential equation is et/28.572 v = 9.8et/28.572 dt = 280et/28.572 + c. Using
v(0) = 0 we see that c = −280 and v(t) = 280 − 280e−t/28.572 . Integrating we get s(t) =
280t + 8000e−t/28.572 + c. Since s(0) = 0, c = −8000 and s(t) = −8000 + 280t + 8000e−t/28.572 .
To find when the object hits the liquid we solve s(t) = 150 − 25 = 125, obtaining ta = 5.204.
The velocity at the time of impact with the liquid is va = v(ta ) = 46.623. When the object is
in the liquid its velocity is modeled by the nonlinear differential equation m dv/dt = mg −kv 2 .
Using m = 71.43, g = 9.8, and k = 0.15 this becomes dv/dt = (4666.76 − v 2 )/476.2. Separating variables and integrating we have
v − 68.3137 dt
dv
= 1 t + c.
and 0.007319 ln =
2
4666.76 − v
476.2
v + 68.3137 476.2
Solving v(0) = va = 46.623 we obtain c = −0.0122. Then, for v < 68.3137,
√
√
v − 68.3137 = e 2t/5−1.8443 or − v − 68.3137 = e 2t/5−1.8443 .
v + 67.3137 v + 68.3137
Solving for v we get
√
v(t) =
Integrating we find
1081 − 170.9e
15.82 +
2t/5
√
2.502e 2t/5
.
√
s(t) = 68.33t − 483.08 ln (6.323 + e
2t/5
) + c.
135
136
CHAPTER 3
MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS
Solving s(0) = 0 we see that c = 961.83, so
√
s(t) = 691.83 + 68.33t − 483.08 ln (6.323 + e
2t/5
).
To find when the object hits the bottom of the tank we solve s(t) = 25, obtaining tb =
0.5158. The time from when the object is dropped from the helicopter to when it hits the
bottom of the tank is ta + tb = 5.72 seconds.
28. The velocity vector of the swimmer is
v = vs + vr = (−vs cos θ, −vs sin θ) + (0, vr ) = (−vs cos θ, −vs sin θ + vr ) =
dx dy
,
dt dt
.
Equating components gives
dx
= −vs cos θ
dt
so
x
dx
= −vs 2
dt
x + y2
Thus,
dy
= vs sin θ + vr
dt
and
and
dy
y
+ vr .
= −vs 2
dt
x + y2
dy/dt
−vs y + vr x2 + y 2
vs y − vr x2 + y 2
dy
=
=
=
.
dx
dx/dt
−vs x
vs x
29. (a) With k = vr /vs ,
y − k x2 + y 2
dy
=
dx
x
is a first-order homogeneous differential equation (see Section 2.5). Substituting y = ux
into the differential equation gives
u+x
du
= u − k 1 + u2
dx
or
du
= −k 1 + u2 .
dx
Separating variables and integrating we obtain
ˆ
ˆ
du
√
= − k dx or ln u + 1 + u2 = −k ln x + ln c.
1 + u2
This implies
ln xk
u + 1 + u2 = ln c or xk
The condition y(1) = 0 gives c = 1 and so y +
y(x) =
y
+
x
x2 + y 2
= c.
x
x2 + y 2 = x1−k . Solving for y gives
1 1−k
− x1+k .
x
2
3.2 Nonlinear Models
(b) If k = 1, then vs = vr and y = 12 (1 − x2 ). Since y(0) = 12 , the swimmer lands on the
west beach at (0, 12 ). That is, 12 km north of (0, 0). If k > 1, then vr > vs and 1 − k < 0.
This means limx→0+ y(x) becomes infinite, since limx→0+ x1−k becomes infinite. The
swimmer never makes it to the west beach and is swept northward with the current. If
0 < k < 1, then vs > vr and 1 − k > 0. The value of y(x) at x = 0 is y(0) = 0. The
swimmer has made it to the point (0, 0).
30. The velocity vector of the swimmer is
v = vs + vr = (−vs , 0) + (0, vr ) =
dx dy
,
dt dt
.
Equating components gives
dx
= −vs
dt
so
and
dy
= vr
dt
dy/dt
vr
vr
dy
=
=
=− .
dx
dx/dt
−vs
vs
31. The differential equation
30x(1 − x)
dy
=−
dx
2
2
3
separates into dy = 15(−x + x2 )dx. Integration gives y(x) = − 15
2 x + 5x + c. The condition
5
1
2
3
y(1) = 0 gives c = 2 and so y(x) = 2 (−15x + 10x + 5). Since y(0) = 52 , the swimmer has
to walk 2.5 km back down the west beach to reach (0, 0).
32. This problem has a great many components, so we will consider the case in which air resistance
is assumed to be proportional to the velocity. By Problem 35 in Section 3.1 the differential
equation is
dv
= mg − kv,
m
dt
and the solution is
mg −kt/m
mg + v0 −
e
.
v(t) =
k
k
If we take the initial velocity to be 0, then the velocity at time t is
mg mg −kt/m
−
e
.
v(t) =
k
k
The mass of the raindrop is about m = 1000 × 0.0000000042 ≈ 4.2−6 and g = 9.8, so the
volocity at time t is
v(t) =
4.116 × 10−5 4.116 × 10−5 −238095kt
−
e
k
k
If we let k = 0.000009, then v(100) ≈ 4.57 m/s. In this case 100 is the time in seconds. Since
11 km/h ≈ 3 m/s, the assertion that the average velocity is 11 km/h is not unreasonable.
Of course, this assumes that the air resistance is proportional to the velocity, and, more
importantly, that the constant of proportionality is 0.000009. The assumption about the
constant is particularly suspect.
137
138
CHAPTER 3
MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS
33. (a) Letting c = 0.6, Ah = π(8−4 )2 , Aw = π · 0.32 = 0.09π, and g = 9.8, the differential
√
equation in Problem 12 becomes dh/dt = −0.00001889 h . Separating variables and
√
integrating, we get 2 h = −0.00001889t + c, so h = (c1 − 0.000009445t)2 . Setting
h(0) = 0.6, we find c = 0.7746, so h(t) = (0.7746 − 0.000009445t)2 , where h is measured
in m and t in seconds.
(b) One hour is 3,600 seconds, so the hour mark should be placed
at
h(3600) = [0.7746 − 0.000009445(3600)]2 ≈ 0.548 m ≈ 54.8 cm.
up from the bottom of the tank. The remaining marks corresponding to the passage of 2, 3, 4, . . . , 12 hours are placed at
the values shown in the table. The marks are not evenly spaced
because the water is not draining out at a uniform rate; that
is, h(t) is not a linear function of time.
Time
(seconds)
Height
(cm)
0
1
2
3
4
5
6
7
8
9
10
11
12
60
54.8
49.93
45.24
40.78
36.55
32.57
28.79
25.26
21.96
18.89
16.05
13.44
34. (a) In this case Aw = πh2 /4 and the differential equation is
1
dh
=−
h−3/2 .
dt
147, 055
Separating variables and integrating, we have
h3/2 dh = −
1
dt
147, 055
1
2 5/2
h =−
t + c1 .
5
147, 055
Setting h(0) = 0.6 we find c1 = 0.1115, so that
1
2 5/2
h =−
t + 0.1115,
5
147055
h5/2 = 0.2788 −
and
h = 0.2788 −
1
t,
58822
1
t
58822
2/5
.
(b) In this case h(4 hr) = h(14,400 s) = 25.86 cm and h(5 hr) = h(18,000 s) is not a real
number. Using a CAS to solve h(t) = 0, we see that the tank runs dry at t ≈ 16,400 s ≈
4.55 hr. Thus, this particular conical water clock can only measure time intervals of less
than 4.55 hours.
3.3
Modeling with Systems of First-Order DEs
35. If we let rh denote the radius of the hole and Aw = π[f (h)]2 , then
√
√
the differential equation dh/dt = −k h, where k = cAh 2g/Aw ,
becomes
√
√
cπrh2 2g √
8crh2 h
dh
=−
h=−
.
dt
π[f (h)]2
[f (h)]2
For the time marks to be equally spaced, the rate of change of the
height must be a constant; that is, dh/dt = −a. (The constant is
negative because the height is decreasing.) Thus
√
8crh2 h
−a = −
,
[f (h)]2
√
8crh2 h
[f (h)] =
,
a
2
and
r = f (h) = 2rh
2c 1/4
h .
a
Solving for h, we have
h=
a2
r4 .
64c2 rh4
The shape of the tank with c = 0.6, a = 2 ft/12 hr = 1 ft/21,600 s,
and rh = 1/32(12) = 1/384 is shown in the above figure.
3.3
Modeling with Systems of First-Order DEs
1. The linear equation dx/dt = −λ1 x can be solved by either separation of variables or by an
integrating factor. Integrating both sides of dx/x = −λ1 dt we obtain ln |x| = −λ1 t + c from
which we get x = c1 e−λ1 t . Using x(0) = x0 we find c1 = x0 so that x = x0 e−λ1 t . Substituting
this result into the second differential equation we have
dy
+ λ2 y = λ1 x0 e−λ1 t
dt
which is linear. An integrating factor is eλ2 t so that
d −λ2 t y + λ2 y = λ1 x0 e(λ2 −λ1 )t
e
dt
y=
λ1 x0 (λ2 −λ1 )t −λ2 t
λ1 x0 −λ1 t
e
e
+ c2 e−λ2 t =
e
+ c2 e−λ2 t .
λ2 − λ1
λ2 − λ1
Using y(0) = 0 we find c2 = −λ1 x0 /(λ2 − λ1 ). Thus
λ1 x0 −λ1 t
e
− e−λ2 t .
y=
λ2 − λ1
Substituting this result into the third differential equation we have
dz
λ1 λ2 x0 −λ1 t
e
− e−λ2 t .
=
dt
λ2 − λ1
Integrating we find
z=−
λ2 x0 −λ1 t
λ1 x0 −λ2 t
e
+
e
+ c3 .
λ2 − λ1
λ2 − λ1
139
140
CHAPTER 3
MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS
Using z(0) = 0 we find c3 = x0 . Thus
λ2
λ1
−λ1 t
−λ2 t
.
z = x0 1 −
e
+
e
λ2 − λ1
λ2 − λ1
2. We see from the graph that the half-life of A is
approximately 4.7 days. To determine the halflife of B we use t = 50 as a base, since at this
time the amount of substance A is so small that it
contributes very little to substance B. Now we see
from the graph that y(50) ≈ 16.2 and y(191) ≈ 8.1.
Thus, the half-life of B is approximately 141 days.
x, y, z
20
y(t)
15
10
5
x(t)
25
z(t)
50
75
100
125
150
t
3. The amounts x and y are the same at about t = 5 days. The amounts x and z are the same
at about t = 20 days. The amounts y and z are the same at about t = 147 days. The time
when y and z are the same makes sense because most of A and half of B are gone, so half of
C should have been formed.
4. Suppose that the series is described schematically by W =⇒ −λ1 X =⇒ −λ2 Y =⇒ −λ3 Z
where −λ1 , −λ2 , and −λ3 are the decay constants for W , X and Y , respectively, and Z is
a stable element. Let w(t), x(t), y(t), and z(t) denote the amounts of substances W , X, Y ,
and Z, respectively. A model for the radioactive series is
dw
= −λ1 w
dt
dx
= λ1 w − λ2 x
dt
dy
= λ2 x − λ 3 y
dt
dz
= λ3 y.
dt
5. (a) Since the third equation in the system is linear and containts only the variable K(t) we
have
dK
= − (λ1 + λ2 ) P
dt
K(t) = c1 e−(λ1 +λ2 )t
Using K(0) = K0 yields K(t) = K0 e−(λ1 +λ2 )t .
We can now solve for A(t) and C(t):
dA
= λ2 P (t) = λ2 K0 e−(λ1 +λ2 )t
dt
A(t) = −
λ2
K0 e−(λ1 +λ2 )t + c2
λ1 + λ2
3.3
Modeling with Systems of First-Order DEs
λ2
K0 . Therefore,
λ1 + λ2
Using A(0) = 0 implies c2 =
A(t) =
λ2
K0 1 − e−(λ1 +λ2 )t .
λ1 + λ2
We use the same approach to solve for C(t):
dC
= λ1 K(t) = λ1 K0 e−(λ1 +λ2 )t
dt
C(t) = −
Using C(0) = 0 implies c3 =
λ1
K0 e−(λ1 +λ2 )t + c3
λ1 + λ2
λ1
K0 . Therefore,
λ1 + λ2
C(t) =
λ1
K0 1 − e−(λ1 +λ2 )t .
λ1 + λ2
(b) It is known that λ1 = 4.7526 × 10−10 and λ2 = 0.5874 × 10−10 so
λ1 + λ2 = 5.34 × 10−10
K(t) = K0 e−0.000000000534t
1
K(t) = K0
2
t=
ln 12
≈ 1.3 × 109 years
−0.000000000534
or the half-life of K-40 is about 1.3 billion years.
(c) Using the solutions A(t), C(t), and the values of λ1 and λ2 from part (b) we see that
lim A(t) =
t→∞
λ2
lim K0 1 − e−(λ1 +λ2 )t
λ1 + λ2 t→∞
0.5874 × 10−10
λ2
K0 =
K0 = 0.11K0 or 11% of K0
λ1 + λ2
5.34 × 10−10
λ1
lim C(t) =
lim K0 1 − e−(λ1 +λ2 )t
t→∞
λ1 + λ2 t→∞
=
=
λ1
4.7526 × 10−10
K0 =
K0 = 0.89K0 or 89% of K0
λ1 + λ2
5.34 × 10−10
6. (a) From part (a) of Problem 5:
A(t)
=
K(t)
λ2
λ1 +λ2 K0
1 − e−(λ1 +λ2 )t
K0 e−(λ1 +λ2 )t
=
λ2 (λ1 +λ2 )t
−1
e
λ1 + λ2
141
142
CHAPTER 3
MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS
(b) Solving for t in part (a) we get
λ1 + λ2 A(t)
= e(λ1 +λ2 )t − 1
λ2
K(t)
λ1 + λ2 A(t)
(λ1 +λ2 )t
=1+
e
λ2
K(t)
1
λ1 + λ2 A(t)
t=
ln 1 +
λ1 + λ2
λ2
K(t)
(c) From part (b)
8.5 × 10−7
5.34 × 10−10
1
≈ 1.66 billion years
ln 1 +
t=
5.34 × 10−10
0.5874 × 10−10 5.4 × 10−6
7. The system is
x1 = 2 · 3 +
x2 =
1
1
2
1
x2 − x1 · 4 = − x1 + x2 + 6
50
50
25
50
1
1
1
2
2
x1 · 4 − x2 − x2 · 3 = x1 − x2 .
50
50
50
25
25
8. Let x1 , x2 , and x3 be the amounts of salt in tanks A, B, and C, respectively, so that
x1 =
1
1
1
3
x2 · 2 −
x1 · 6 = x2 − x1
100
100
50
50
x2 =
1
1
1
1
3
7
1
x1 · 6 +
x3 −
x2 · 2 −
x2 · 5 = x1 −
x2 +
x3
100
100
100
100
50
100
100
x3 =
1
1
1
1
1
x2 · 5 −
x3 −
x3 · 4 = x2 − x3 .
100
100
100
20
20
9. (a) A model is
x2
x1
dx1
=3·
−2·
,
dt
100 − t
100 + t
x1 (0) = 100
x1
x2
dx2
=2·
−3·
,
dt
100 + t
100 − t
x2 (0) = 50.
(b) Since the system is closed, no salt enters or leaves the system and x1 (t) + x2 (t) =
100 + 50 = 150 for all time. Thus x1 = 150 − x2 and the second equation in part (a)
becomes
2(150 − x2 )
3x2
300
2x2
3x2
dx2
=
−
=
−
−
dt
100 + t
100 − t
100 + t 100 + t 100 − t
or
2
3
300
dx2
+
+
x2 =
,
dt
100 + t 100 − t
100 + t
which is linear in x2 . An integrating factor is
e2 ln (100+t)−3 ln (100−t) = (100 + t)2 (100 − t)−3
3.3
Modeling with Systems of First-Order DEs
143
so
d
[(100 + t)2 (100 − t)−3 x2 ] = 300(100 + t)(100 − t)−3 .
dt
Using integration by parts, we obtain
−3
(100 + t) (100 − t)
2
1
1
−2
−1
x2 = 300 (100 + t)(100 − t) − (100 − t) + c .
2
2
Thus
300
1
1
3
2
c(100 − t) − (100 − t) + (100 + t)(100 − t)
x2 =
(100 + t)2
2
2
=
300
[c(100 − t)3 + t(100 − t)].
(100 + t)2
Using x2 (0) = 50 we find c = 5/3000. At t = 30, x2 = (300/1302 )(703 c + 30 · 70) ≈
47.4 kg.
10. A model is
1
dx1
= (4 L/min)(0 kg/gal) − (4 gal/min)
x1 kg/L
dt
200
1
1
dx2
= (4 L/min)
x1 kg/L − (4 gal/min)
x2 kg/L
dt
200
150
1
1
dx3
= (4 L/min)
x2 kg/L − (4 gal/min)
x3 kg/L
dt
150
100
or
1
dx1
= − x1
dt
50
1
2
dx2
= x1 − x2
dt
50
75
2
1
dx3
= x2 − x3 .
dt
75
25
Over a long period of time we would expect x1 , x2 , and x3 to approach 0 because the entering
pure water should flush the salt out of all three tanks.
11. Zooming in on the graph it can be seen that the
populations are first equal at about t = 5.6. The
approximate periods of x and y are both 45.
x, y
x
10
y
5
50
100
t
144
CHAPTER 3
MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS
12. (a) The population y(t) approaches 10,000, while the
population x(t) approaches extinction.
x, y
10
y
5
x
(b) The population x(t) approaches 5,000, while the population y(t) approaches extinction.
10
20
t
10
20
t
10
20
t
x, y
10
x
5
y
(c) The population y(t) approaches 10,000, while the
population x(t) approaches extinction.
x, y
10
y
5
x
(d) The population x(t) approaches 5,000, while the population y(t) approaches extinction.
x, y
10
x
5
y
10
13. (a) x
x, y
10
y
(b) x
x, y
10
x
5
(c) x
x, y
10
5
40 t
20
(d) x
y
x
5
20
x, y
10
t
y
x
40 t
20
y
x
5
40 t
20
20
40 t
In each case the population x(t) approaches 6,000, while the population y(t) approaches 8,000.
14. By Kirchhoff’s first law we have i1 = i2 + i3 . By Kirchhoff’s second law, on each loop we
have E(t) = Li1 + R1 i2 and E(t) = Li1 + R2 i3 + q/C so that q = CR1 i2 − CR2 i3 . Then
3.3
Modeling with Systems of First-Order DEs
i3 = q = CR1 i2 − CR2 i3 so that the system is
Li2 + Li3 + R1 i2 = E(t)
−R1 i2 + R2 i3 +
1
i3 = 0.
C
15. By Kirchhoff’s first law we have i1 = i2 + i3 . Applying Kirchhoff’s second law to each loop
we obtain
di2
+ i2 R2
E(t) = i1 R1 + L1
dt
and
di3
+ i3 R3 .
E(t) = i1 R1 + L2
dt
Combining the three equations, we obtain the system
L1
di2
+ (R1 + R2 )i2 + R1 i3 = E
dt
L2
di3
+ R1 i2 + (R1 + R3 )i3 = E.
dt
16. By Kirchhoff’s first law we have i1 = i2 + i3 . By Kirchhoff’s second law, on each loop we
have E(t) = Li1 + Ri2 and E(t) = Li1 + q/C so that q = CRi2 . Then i3 = q = CRi2 so that
system is
Li + Ri2 = E(t)
CRi2 + i2 − i1 = 0.
17. We first note that s(t) + i(t) + r(t) = n. Now the rate of change of the number of susceptible
persons, s(t), is proportional to the number of contacts between the number of people infected
and the number who are susceptible; that is, ds/dt = −k1 si. We use −k1 < 0 because s(t)
is decreasing. Next, the rate of change of the number of persons who have recovered is
proportional to the number infected; that is, dr/dt = k2 i where k2 > 0 since r is increasing.
Finally, to obtain di/dt we use
d
d
(s + i + r) =
n = 0.
dt
dt
This gives
dr ds
di
=− −
= −k2 i + k1 si.
dt
dt
dt
The system of differential equations is then
ds
= −k1 si
dt
di
= −k2 i + k1 si
dt
dr
= k2 i.
dt
145
146
CHAPTER 3
MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS
A reasonable set of initial conditions is i(0) = i0 , the number of infected people at time 0,
s(0) = n − i0 , and r(0) = 0.
18. (a) If we know s(t) and i(t) then we can determine r(t) from s + i + r = n.
(b) In this case the system is
ds
= −0.2si
dt
di
= −0.7i + 0.2si.
dt
We also note that when i(0) = i0 , s(0) = 10 − i0 since r(0) = 0 and i(t) + s(t) + r(t) = 0
for all values of t. Now k2 /k1 = 0.7/0.2 = 3.5, so we cconsider initial conditions s(0) = 2,
i(0) = 8; s(0) = 3.4, i(0) = 6.6; s(0) = 7, i(0) = 3; and s(0) = 9, i(0) = 1.
s, i
10
s, i
10
s, i
10
5
5
5
i
s, i
10
5
i
s
i
i
s
s
s
5
10 t
5
10 t
5
10 t
5
10 t
We see that an initial susceptible population greater than k2 /k1 results in an epidemic in
the sense that the number of infected persons increases to a maximum before decreasing
to 0. On the other hand, when s(0) < k2 /k1 , the number of infected persons decreases
from the start and there is no epidemic.
19. Since x0 > y0 > 0 we have x(t) > y(t) and y − x < 0. Thus
dx/dt < 0 and dy/dt > 0. We conclude that x(t) is decreasing
and y(t) is increasing. As t → ∞ we expect that x(t) → C
and y(t) → C, where C is a constant common equilibrium
concentration.
20. We write the system in the form
dx
= k1 (y − x)
dt
dy
= k2 (x − y),
dt
3.3
Modeling with Systems of First-Order DEs
where k1 = κ/VA and k2 = κ/VB . Letting z(t) = x(t) − y(t) we have
dx dy
−
= k1 (y − x) − k2 (x − y)
dt
dt
dz
= k1 (−z) − k2 z
dt
dz
+ (k1 + k2 )z = 0.
dt
This is a linear first-order differential equation with solution z(t) = c1 e−(k1 +k2 )t . Now
dx
= −k1 (y − x) = −k1 z = −k1 c1 e−(k1 +k2 )t
dt
and
x(t) = c1
k1
e−(k1 +k2 )t + c2 .
k1 + k2
Since y(t) = x(t) − z(t) we have
y(t) = −c1
k2
e−(k1 +k2 )t + c2 .
k1 + k2
The initial conditions x(0) = x0 and y(0) = y0 imply
c1 = x0 − y0
and
c2 =
x0 k2 + y0 k1
.
k1 + k2
The solution of the system is
x(t) =
(x0 − y0 )k1 −(k1 +k2 )t x0 k2 + y0 k1
e
+
k1 + k2
k1 + k2
y(t) =
(y0 − x0 )k2 −(k1 +k2 )t x0 k2 + y0 k1
e
+
.
k1 + k2
k1 + k2
As t → ∞, x(t) and y(t) approach the common limit
x0 k2 + y0 k1
x0 κ/VB + y0 κ/VA
x0 VA + y0 VB
=
=
k1 + k2
κ/VA + κ/VB
VA + VB
= x0
VA
VB
+ y0
.
VA + VB
VA + VB
This makes intuitive sense because the limiting concentration is seen to be a weighted average
of the two initial concentrations.
21. Since there are initially 25 pounds of salt in tank
A and none in tank B, and since furthermore
only pure water is being pumped into tank A, we
would expect that x1 (t) would steadily decrease
over time. On the other hand, since salt is being
added to tank B from tank A, we would expect
x2 (t) to increase over time. However, since pure
water is being added to the system at a constant
147
148
CHAPTER 3
MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS
rate and a mixed solution is being pumped out of the system, it makes sense that the amount
of salt in both tanks would approach 0 over time.
22. We assume here that the temperature, T (t), of the metal bar does not affect the temperature,
TA (t), of the medium in container A. By Newton’s law of cooling, then, the differential
equations for TA (t) and T (t) are
dTA
= kA (TA − TB ),
dt
dT
= k(T − TA ),
dt
kA < 0
k < 0,
subject to the initial conditions T (0) = T0 and TA (0) = T1 . Separating variables in the first
equation, we find TA (t) = TB + c1 ekA t . Using TA (0) = T1 we find c1 = T1 − TB , so
TA (t) = TB + (T1 − TB )ekA t .
Substituting into the second differential equation, we have
dT
= k(T − TA ) = kT − kTA = kT − k[TB + (T1 − TB )ekA t ]
dt
dT
− kT = −kTB − k(T1 − TB )ekA t .
dt
´
This is a linear differential equation with integrating factor e
−k dt
= e−kt . Then
d −kt
[e T ] = −kTB e−kt − k(T1 − TB )e(kA −k)t
dt
e−kt T = TB e−kt −
T = TB −
k
(T1 − TB )e(kA −k)t + c2
kA − k
k
(T1 − TB )ekA t + c2 ekt .
kA − k
k
(T1 − TB ), so
kA − k
k
k
kA t
(T1 − TB )e
(T1 − TB ) ekt .
+ T0 − TB +
T (t) = TB −
kA − k
kA − k
Using T (0) = T0 we find c2 = T0 − TB +
Chapter 3 in Review
1. The differential equation is dP/dt = 0.15P .
2. True. From dA/dt = kA, A(0) = A0 , we have A(t) = A0 ekt and A (t) = kA0 ekt , so A (0) =
kA0 . At T = −(ln 2)k,
1
A (−(ln 2)/k) = kA(−(ln 2)/k) = kA0 ek[−(ln 2)/k] = kA0 e− ln 2 = kA0 .
2
Chapter 3 in Review
3. From
dP
= 0.018P and P (0) = 4 billion we obtain P = 4e0.018t so that P (45) = 8.99 billion.
dt
4. Let A = A(t) be the volume of CO2 at time t. From dA/dt = 0.03 − A/4 and A(0) = 0.04 m3
we obtain A = 0.12 + 0.08e−t/4 . Since A(10) = 0.127 m3 , the concentration is 0.0254%. As
t → ∞ we have A → 0.12 m3 or 0.06%.
5. The starting point is A(t) = A0 e−0.00012097t . With A(t) = 0.53A0 we have
−0.00012097t = ln 0.53 or t =
ln 0.53
≈ 5248 years.
−0.00012097
This represents the iceman’s age in 1991, so the approximate date of his death would be
1991 − 5248 = −3257
or
3257 BC.
6. (a) We assume that the rate of disintegration of Iodine-131 is proportional to the amount
remaining. If A(t) is the amount of Iodine-131 remaining at time t then
dA
= kA, A(0) = A0 ,
dt
where k is the constant of proportionality and A0 is the initial amount. The solution of
the initial-value problem is A(t) = A0 ekt . After one day the amount of Iodine-131 is
(1 − 0.083)A0 = 0.917A0
so
A(1) = A0 ek = 0.917A0
and ek = 0.917. After eight days
A(8) = A0 e8k = A0 (ek )8 = A0 (0.917)8 ≈ 0.49998A0 .
(b) Since A(8) ≈ 12 A0 , the half-life of Iodine-131 is approximately 8 days.
7. Separating variables, we have
a2 − y 2
dy = −dx.
y
Substituting y = a sin θ, this becomes
a2 − a2 sin2 θ
(a cos θ) dθ
a sin θ
ˆ
cos2 θ
a
dθ
sin θ
ˆ
1 − sin2 θ
a
dθ
sin θ
ˆ
a (csc θ − sin θ) dθ
a
a ln y
= −dx
ˆ
=−
dx
= −x + c
= −x + c
a ln | csc θ − cot θ| + a cos θ = −x + c
a2 − y 2 a2 − y 2
−
= −x + c.
+a
y
a
149
150
CHAPTER 3
MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS
Letting a = 10, this is
10 100 − y 2 10 ln −
+ 100 − y 2 = −x + c.
y
y
Letting x = 0 and y = 10 we determine that c = 0, so the solution is
10 100 − y 2 10 ln −
+ 100 − y 2 = −x.
y
y
8. From V dC/dt = kA(Cs − C) and C(0) = C0 we obtain C = Cs + (C0 − Cs )e−kAt/V .
9. (a) The differential equation
dT
= k(T − Tm ) = k[T − T2 − B(T1 − T )]
dt
BT1 + T2
= k[(1 + B)T − (BT1 + T2 )] = k(1 + B) T −
1+B
is autonomous and has the single critical point (BT1 + T2 )/(1 + B). Since k < 0 and
B > 0, by phase-line analysis it is found that the critical point is an attractor and
lim T (t) =
t→∞
BT1 + T2
.
1+B
Moreover,
BT1 + T2
lim Tm (t) = lim [T2 + B(T1 − T )] = T2 + B T1 −
t→∞
t→∞
1+B
=
BT1 + T2
.
1+B
(b) The differential equation is
dT
= k(T − Tm ) = k(T − T2 − BT1 + BT )
dt
or
dT
− k(1 + B)T = −k(BT1 + T2 ).
dt
This is linear and has integrating factor e−
´
k(1+B)dt
= e−k(1+B)t . Thus,
d −k(1+B)t
T ] = −k(BT1 + T2 )e−k(1+B)t
[e
dt
e−k(1+B)t T =
T (t) =
Since T (0) = T1 we find, T (t) =
BT1 + T2 −k(1+B)t
e
+c
1+B
BT1 + T2
+ cek(1+B)t .
1+B
BT1 + T2 T1 − T2 k(1+B)t
+
e
.
1+B
1+B
Chapter 3 in Review
(c) The temperature T (t) decreases to (BT1 +T2 )/(1+B), whereas Tm (t) increases to (BT1 +
T2 )/(1 + B) as t → ∞. Thus, the temperature (BT1 + T2 )/(1 + B), (which is a weighted
average,
1
B
T1 +
T2 ,
1+B
1+B
of the two initial temperatures), can be interpreted as an equilibrium temperature. The
body cannot get cooler than this value whereas the medium cannot get hotter than this
value.
10. (a) By separation of variables and partial fractions,
T − Tm 3
− 2 tan−1 T
= 4Tm
kt + c.
ln T + Tm Tm
Then rewrite the right-hand side of the differential equation as
dT
4
4
= k(T 4 − Tm
) = [(Tm + (T − Tm ))4 − Tm
]
dt
T − Tm 4
4
1+
−1
= kTm
Tm
4
= kTm
T − Tm
1+4
+6
Tm
T − Tm
Tm
2
+ ···
− 1 ← binomial expansion
(b) When T − Tm is small compared to Tm , every term in the expansion after the first two
can be ignored, giving
dT
≈ k1 (T − Tm ),
dt
where
3
k1 = 4kTm
.
11. Separating variables, we obtain
dt
dq
=
E0 − q/C
k1 + k2 t
q 1
ln |k1 + k2 t| + c1
−C ln E0 − =
C
k2
(E0 − q/C)−C
= c2 .
(k1 + k2 t)1/k2
151
152
CHAPTER 3
MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS
Setting q(0) = q0 we find c2 = (E0 − q0 /C)−C /k1
1/k2
, so
(E0 − q/C)−C
(E0 − q0 /C)−C
=
1/k
(k1 + k2 t)1/k2
k 2
1
−1/k2
k1
q −C q0 −C
E0 −
= E0 −
C
C
k1 + k2 t
1/Ck2
q
q0 k1
= E0 −
E0 −
C
C
k1 + k2 t
1/Ck2
k1
q = E0 C + (q0 − E0 C)
.
k1 + k2 t
√ √
12. From y 1 + (y )2 = k we obtain dx = ( y/ k − y )dy. If y = k sin2 θ then
k
1 1
− cos 2θ dθ, and x = kθ − sin 2θ + c.
dy = 2k sin θ cos θ dθ, dx = 2k
2 2
2
If x = 0 when θ = 0 then c = 0.
13. From dx/dt = k1 x(α − x) we obtain
1/α
1/α
+
dx = k1 dt
x
α−x
so that x = αc1 eαk1 t /(1 + c1 eαk1 t ). From dy/dt = k2 xy we obtain
k2 /k1
k2 ln 1 + c1 eαk1 t + c or y = c2 1 + c1 eαk1 t
.
ln |y| =
k1
14. In tank A the salt input is
kg
L
x2 kg
1
kg
L
2
+ 1
= 14 +
x2
.
7
min
L
min
100 L
100
min
The salt output is
L
3
min
In tank B the salt input is
x1 kg
100 L
5
The salt output is
L
1
min
L
min
x2 kg
100 L
L
+ 5
min
x1 kg
100 L
L
+ 4
min
=
x1 kg
100 L
=
2
kg
x1
.
25
min
1
kg
x1
.
20
min
x2 kg
100 L
The system of differential equations is then
1
dx1
2
= 14 +
x2 − x1
dt
100
25
1
1
dx2
= x1 − x2 .
dt
20
20
=
1
kg
x2
.
20
min
Chapter 3 in Review
153
15.
4
y = c1 x
dy
= c1
dx
y
dy
=
dx
x
2
4
2
2
4
2
4
Therefore the differential equation of the orthogonal family is
x
dy
=−
dx
y
y dy + x dx = 0
x2 + y 2 = c2
which is a family of circles (c2 > 0) centered at the origin.
16.
4
x2 − 2y 2 = c1
2
dy
=0
2x − 4y
dx
x
dy
=
dx
2y
6
4
2
2
4
6
2
Therefore the differential equation of the orthogonal family is
4
2y
dy
=−
dx
x
2
1
dy = − dx
y
x
y=
c2
x2
17. From y = c1 ex we obtain y = y so that the differential equation of the orthogonal family is
4
1
dy
=− .
dx
y
Separating variables and integrating we get
y dy = − dx
1 2
y = −x + c
2
y 2 + 2x = c2
2
4
2
2
2
4
4
154
CHAPTER 3
MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS
18. Differentiating the family of curves, we have
y = −
1
= −y 2 .
(x + c1 )2
The differential equation for the family of orthogonal trajecto1
ries is then y = 2 . Separating variables and integrating we
y
get
y 2 dy = dx
1 3
y = x + c2
3
y 3 = 3x + c3 .
19. Critical points of the equation
dP
= rP
dt
P
P
1−
−1
K
A
K
r > 0,
A
are 0, A, and K. Here A is called the Allee threshold and satisfies
0 < A < K. From the accompanying phase portrait we see that K and 0 are
attractors, or asymptotically stable, but A is a repeller, or unstable. Thus, for an
initial value P0 < A the population decreases over time, that is, P → 0 as t → ∞.
20. (a) From the cross section on the right we see in this case that
√
w(x) = 2 4 − x2 and the initial-value problem is then
dx
2 4 − x2
= 1,
dt
x(0) = −2.
Solving the differential equation by separation of variables gives
√
x 4 − x2 + 4 sin−1 12 x = t + c. Using x(0) = −2 we have
π
c = 4 sin−1 (−1) = 4 −
= −2π .
2
√
Therefore an implicit solution is x 4 − x2 + 4 sin−1 12 x = t − 2π.
0
y
w(x)
x2+y2=4
x
Chapter 3 in Review
√
(b) The graph of t(x) = 2π + x 4 − x2 + 4 sin−1 12 x on the
x-interval [−2, 2] is given on the right. From the graph we
see that the time corresponding to x = 2 is approximately
t = 12.5. The exact time to cut throug the piece of wood
is
π t(2) = 2π + 2 4 − 22 + 4 sin−1 (1) = 2π + 4
2
or t = 4π. In other words, the solution x(t) defined
implicitely by the equation in part (a) is defined on the
t-interval [0, 4π].
21. The piecewise-defined function w(x) is now
⎧
√
⎪
2
⎪
⎪
0≤x≤
⎨x,
2
w(x) =
√
⎪
√
√
⎪
⎪
⎩ 2 − x,
<x≤ 2
2
First, we solve
dx
= 1,
x(0) = 0
dt
√
2t. The time interval corresponding to 0 ≤
by separation
of
variables.
This
yield
x(t)
=
√
2
1
x ≤ 2 is defined by 0 ≤ t ≤ 4 . Second, we solve
√
dx
√
1
2
= 1,
x
=
.
2−x
dt
4
2
√
√ √
the
quadratic
formula,
we
have
x(t)
=
2− 1 − 2t.
This gives x2 −2 2 x+2t+1 = 0. Using
√
√
2
1
1
The time interval corresponding to 2 < x ≤ 2 is defined by 4 ≤ t ≤ 2 . Thus,
⎧√
⎨ 2t,
0 ≤ t ≤ 14
x(t) = √
⎩ 2 − √1 − 2t, 1 < t ≤ 1 .
4
2
x
The time that it takes the saw to cut through the piece of wood is then t = 12 .
22. (a) By separation of variables
dA
= −kA2
dt
A−2 dA = −k dt
−A−1 = −kt + c1
A−1 = c2 + kt
A(t) =
1
c2 + kt
155
156
CHAPTER 3
MODELING WITH FIRST-ORDER DIFFERENTIAL EQUATIONS
Then A(0) =
1
1
A0
.
and thus c2 =
. Therefore A(t) =
c2
A0
1 + A0 kt
(b) A(t) + B(t) = A0
B(t) = A0 − A(t) = A0 −
B(t) =
(c) x
kA20 t
.
1 + A0 kt
A0
1 + A0 kt
Chapter 4
Higher-Order Differential Equations
4.1
Preliminary Theory - Linear Equations
1. From y = c1 ex + c2 e−x we find y = c1 ex − c2 e−x . Then y(0) = c1 + c2 = 0, y (0) = c1 − c2 = 1
so that c1 = 12 and c2 = − 12 . The solution is y = 12 ex − 12 e−x .
2. From y = c1 e4x + c2 e−x we find y = 4c1 e4x − c2 e−x . Then y(0) = c1 + c2 = 1, y (0) =
4c1 − c2 = 2 so that c1 = 35 and c2 = 25 . The solution is y = 35 e4x + 25 e−x .
3. From y = c1 x+c2 x ln x we find y = c1 +c2 (1+ln x). Then y(1) = c1 = 3, y (1) = c1 +c2 = −1
so that c1 = 3 and c2 = −4. The solution is y = 3x − 4x ln x.
4. From y = c1 + c2 cos x + c3 sin x we find y = −c2 sin x + c3 cos x and y = −c2 cos x − c3 sin x.
Then y(π) = c1 − c2 = 0, y (π) = −c3 = 2, y (π) = c2 = −1 so that c1 = −1, c2 = −1, and
c3 = −2. The solution is y = −1 − cos x − 2 sin x.
5. From y = c1 + c2 x2 we find y = 2c2 x. Then y(0) = c1 = 0, y (0) = 2c2 · 0 = 0 and hence
y (0) = 1 is not possible. Since a2 (x) = x is 0 at x = 0, Theorem 4.1.1 is not violated.
6. In this case we have y(0) = c1 = 0, y (0) = 2c2 · 0 = 0 so c1 = 0 and c2 is arbitrary. Two
solutions are y = x2 and y = 2x2 .
7. From x(0) = x0 = c1 we see that x(t) = x0 cos ωt+c2 sin ωt and x (t) = −x0 sin ωt+c2 ω cos ωt.
Then x (0) = x1 = c2 ω implies c2 = x1 /ω. Thus
x1
sin ωt.
x(t) = x0 cos ωt +
ω
8. Solving the system
x(t0 ) = c1 cos ωt0 + c2 sin ωt0 = x0
x (t0 ) = −c1 ω sin ωt0 + c2 ω cos ωt0 = x1
for c1 and c2 gives
c1 =
ωx0 cos ωt0 − x1 sin ωt0
ω
and c2 =
157
x1 cos ωt0 + ωx0 sin ωt0
.
ω
158
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
Thus
x1 cos ωt0 + ωx0 sin ωt0
ωx0 cos ωt0 − x1 sin ωt0
cos ωt +
sin ωt
ω
ω
x1
= x0 (cos ωt cos ωt0 + sin ωt sin ωt0 ) + (sin ωt cos ωt0 − cos ωt sin ωt0 )
ω
x1
sin ω(t − t0 ).
= x0 cos ω(t − t0 ) +
ω
x(t) =
9. Since a2 (x) = x − 2 and x0 = 0 the problem has a unique solution for −∞ < x < 2.
10. Since a0 (x) = tan x and x0 = 0 the problem has a unique solution for −π/2 < x < π/2.
11. (a) We have y(0) = c1 + c2 = 0, y (1) = c1 e + c2 e−1 = 1 so that c1 = e/ e2 − 1 and
c2 = −e/ e2 − 1 . The solution is y = e (ex − e−x ) / e2 − 1 .
(b) We have y(0) = c3 cosh 0 + c4 sinh 0 = c3 = 0 and y(1) = c3 cosh 1 + c4 sinh 1 = c4 sinh 1 =
1, so c3 = 0 and c4 = 1/ sinh 1. The solution is y = (sinh x)/(sinh 1).
(c) Starting with the solution in part (b) we have
y=
2
ex − e−x
e
ex − e−x
1
sinh x = 1
=
= 2
(ex − e−x ).
sinh 1
e − e−1
2
e − 1/e
e −1
12. In this case we have y(0) = c1 = 1, y (1) = 2c2 = 6 so that c1 = 1 and c2 = 3. The solution
is y = 1 + 3x2 .
13. From y = c1 ex cos x + c2 ex sin x we find y = c1 ex (− sin x + cos x) + c2 ex (cos x + sin x).
(a) We have y(0) = c1 = 1, y (0) = c1 + c2 = 0 so that c1 = 1 and c2 = −1. The solution is
y = ex cos x − ex sin x.
(b) We have y(0) = c1 = 1, y(π) = −eπ = −1, which is not possible.
(c) We have y(0) = c1 = 1, y(π/2) = c2 eπ/2 = 1 so that c1 = 1 and c2 = e−π/2 . The solution
is y = ex cos x + e−π/2 ex sin x.
(d) We have y(0) = c1 = 0, y(π) = c2 eπ sin π = 0 so that c1 = 0 and c2 is arbitrary. Solutions
are y = c2 ex sin x, for any real numbers c2 .
14. (a) We have y(−1) = c1 + c2 + 3 = 0, y(1) = c1 + c2 + 3 = 4, which is not possible.
(b) We have y(0) = c1 · 0 + c2 · 0 + 3 = 1, which is not possible.
(c) We have y(0) = c1 · 0 + c2 · 0 + 3 = 3, y(1) = c1 + c2 + 3 = 0 so that c1 is arbitrary and
c2 = −3 − c1 . Solutions are y = c1 x2 − (c1 + 3)x4 + 3.
4.1 Preliminary Theory - Linear Equations
159
(d) We have y(1) = c1 + c2 + 3 = 3, y(2) = 4c1 + 16c2 + 3 = 15 so that c1 = −1 and c2 = 1.
The solution is y = −x2 + x4 + 3.
15. Since (−4)x + (3)x2 + (1)(4x − 3x2 ) = 0 the set of functions is linearly dependent.
16. Since (1)0 + (0)x + (0)ex = 0 the set of functions is linearly dependent. A similar argument
shows that any set of functions containing f (x) = 0 will be linearly dependent.
17. Since (−1/5)5 + (1) cos2 x + (1) sin2 x = 0 the set of functions is linearly dependent.
18. Since (1) cos 2x + (1)1 + (−2) cos2 x = 0 the set of functions is linearly dependent.
19. Since (−4)x + (3)(x − 1) + (1)(x + 3) = 0 the set of functions is linearly dependent.
20. From the graphs of f1 (x) = 2 + x and
f2 (x) = 2 + |x| we see that the set of
functions is linearly independent since they
cannot be multiples of each other.
y
y
3
3
f1 = 2 + x
f2 = 2 + | x |
3 x
3 x
21. Suppose c1 (1 + x) + c2 x + c3 x2 = 0. Then c1 + (c1 + c2 )x + c3 x2 = 0 and so c1 = 0,
c1 + c2 = 0, and c3 = 0. Since c1 = 0 we also have c2 = 0. Thus, the set of functions is
linearly independent.
22. Since (−1/2)ex + (1/2)e−x + (1) sinh x = 0 the set of functions is linearly dependent.
23. The functions satisfy the differential equation and are linearly independent since
W e−3x , e4x = 7ex = 0
for −∞ < x < ∞. The general solution is
y = c1 e−3x + c2 e4x .
24. The functions satisfy the differential equation and are linearly independent since
W (cosh 2x, sinh 2x) = 2
for −∞ < x < ∞. The general solution is
y = c1 cosh 2x + c2 sinh 2x.
25. The functions satisfy the differential equation and are linearly independent since
W (ex cos 2x, ex sin 2x) = 2e2x = 0
for −∞ < x < ∞. The general solution is y = c1 ex cos 2x + c2 ex sin 2x.
160
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
26. The functions satisfy the differential equation and are linearly independent since
W ex/2 , xex/2 = ex = 0
for −∞ < x < ∞. The general solution is
y = c1 ex/2 + c2 xex/2 .
27. The functions satisfy the differential equation and are linearly independent since
W x3 , x4 = x6 = 0
for 0 < x < ∞. The general solution is
y = c1 x3 + c2 x4 .
28. The functions satisfy the differential equation and are linearly independent since
W (cos (ln x), sin (ln x)) = 1/x = 0
for 0 < x < ∞. The general solution is
y = c1 cos (ln x) + c2 sin (ln x).
29. The functions satisfy the differential equation and are linearly independent since
W x, x−2 , x−2 ln x = 9x−6 = 0
for 0 < x < ∞. The general solution is
y = c1 x + c2 x−2 + c3 x−2 ln x.
30. The functions satisfy the differential equation and are linearly independent since
W (1, x, cos x, sin x) = 1
for −∞ < x < ∞. The general solution is
y = c1 + c2 x + c3 cos x + c4 sin x.
31. The functions y1 = e2x and y2 = e5x form a fundamental set of solutions of the associated
homogeneous equation, and yp = 6ex is a particular solution of the nonhomogeneous equation.
32. The functions y1 = cos x and y2 = sin x form a fundamental set of solutions of the associated
homogeneous equation, and yp = x sin x + (cos x) ln (cos x) is a particular solution of the
nonhomogeneous equation.
4.1 Preliminary Theory - Linear Equations
33. The functions y1 = e2x and y2 = xe2x form a fundamental set of solutions of the associated
homogeneous equation, and yp = x2 e2x + x − 2 is a particular solution of the nonhomogeneous
equation.
34. The functions y1 = x−1/2 and y2 = x−1 form a fundamental set of solutions of the associated
1 2
x − 16 x is a particular solution of the nonhomogeneous
homogeneous equation, and yp = 15
equation.
35. (a) We have yp 1 = 6e2x and yp1 = 12e2x , so
yp1 − 6yp 1 + 5yp1 = 12e2x − 36e2x + 15e2x = −9e2x .
Also, yp 2 = 2x + 3 and yp2 = 2, so
yp2 − 6yp 2 + 5yp2 = 2 − 6(2x + 3) + 5(x2 + 3x) = 5x2 + 3x − 16.
(b) By the superposition principle for nonhomogeneous equations a particular solution of
y − 6y + 5y = 5x2 + 3x − 16 − 9e2x is yp = x2 + 3x + 3e2x . A particular solution of the
second equation is
1
1
yp = −2yp2 − yp1 = −2x2 − 6x − e2x .
9
3
36. (a) yp1 = 5
(b) yp2 = −2x
(c) yp = yp1 + yp2 = 5 − 2x
(d) yp = 12 yp1 − 2yp2 =
5
2
+ 4x
37. (a) Since D2 x = 0, x and 1 are solutions of y = 0. Since they are linearly independent, the
general solution is y = c1 x + c2 .
(b) Since D3 x2 = 0, x2 , x, and 1 are solutions of y = 0. Since they are linearly independent,
the general solution is y = c1 x2 + c2 x + c3 .
(c) Since D4 x3 = 0, x3 , x2 , x, and 1 are solutions of y (4) = 0. Since they are linearly
independent, the general solution is y = c1 x3 + c2 x2 + c3 x + c4 .
(d) By part (a), the general solution of y = 0 is yc = c1 x + c2 . Since D2 x2 = 2! = 2, yp = x2
is a particular solution of y = 2. Thus, the general solution is y = c1 x + c2 + x2 .
(e) By part (b), the general solution of y = 0 is yc = c1 x2 + c2 x + c3 . Since D3 x3 =
3! = 6, yp = x3 is a particular solution of y = 6. Thus, the general solution is
y = c1 x2 + c2 x + c3 + x3 .
161
162
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
(f ) By part (c), the general solution of y (4) = 0 is yc = c1 x3 + c2 x2 + c3 x + c4 . Since
D4 x4 = 4! = 24, yp = x4 is a particular solution of y (4) = 24. Thus, the general solution
is y = c1 x3 + c2 x2 + c3 x + c4 + x4 .
38. By the superposition principle, if y1 = ex and y2 = e−x are both solutions of a homogeneous
linear differential equation, then so are
ex + e−x
1
(y1 + y2 ) =
= cosh x and
2
2
39. (a) From the graphs of y1 = x3 and y2 = |x|3
we see that the functions are linearly independent since they cannot be multiples of
each other. It is easily shown that y1 = x3
is a solution of x2 y − 4xy + 6y = 0. To
show that y2 = |x|3 is a solution let y2 = x3
for x ≥ 0 and let y2 = −x3 for x < 0.
(b) If x ≥ 0 then y2 = x3 and
If x < 0 then y2 = −x3 and
1
ex − e−x
(y1 − y2 ) =
= sinh x.
2
2
y
y
3
3
y = x3
3 x
y = | x |3
3 x
3
x
x3 W (y1 , y2 ) = 2
=0
3x 3x2 3
3 x
−x
=0
W (y1 , y2 ) = 2
3x −3x2 This does not violate Theorem 4.1.3 since a2 (x) = x2 is zero at x = 0.
(c) The functions Y1 = x3 and Y2 = x2 are solutions of x2 y − 4xy + 6y = 0 on the interval
(−∞, ∞) because we have, in turn,
x2 Y1 − 4xY1 + 6Y1 = x2 (6x) − 4x 3x2 + 6x3 = 0
x2 Y2 − 4xY2 + 6Y2 = x2 (2) − 4x (2x) + 6x2 = 0.
The solutions Y1 = x3 and Y2 = x2 are also linearly independent on the interval (−∞, ∞).
In order for c1 x3 + c2 x2 = 0 for every real number x it is necessary that c1 = c2 = 0. To
see this, observe that if either c1 or c2 were not 0, then the equation c1 c3 + c2 x2 = 0 or
x2 (c1 x + c2 ) = 0 would hold for at most two real numbers.
(d) Since the linear differential equation is homogeneous, the superposition principle indicates that y = x3 + x2 is a solution of the equation. It is also clear that y = x3 + x2
satisfies the initial conditions y(0) = 0, y (0) = 0.
(e) Neither is the general solution on (−∞, ∞) since we form a general solution on an interval
for which a2 (x) = 0 for every x in the interval.
4.2
Reduction of Order
40. Since ex−3 = e−3 ex = (e−5 e2 )ex = e−5 ex+2 , we see that ex−3 is a constant multiple of ex+2
and the set of functions is linearly dependent.
41. Since 0y1 + 0y2 + · · · + 0yk + 1yk+1 = 0, the set of solutions is linearly dependent.
42. The set of solutions is linearly dependent. Suppose n of the solutions are linearly independent
(if not, then the set of n + 1 solutions is linearly dependent). Without loss of generality, let
this set be y1 , y2 , . . . , yn . Then y = c1 y1 + c2 y2 + · · · + cn yn is the general solution of
the nth-order differential equation and for some choice, c∗1 , c∗2 , . . . , c∗n , of the coefficients
yn+1 = c∗1 y1 + c∗2 y2 + · · · + c∗n yn . But then the set y1 , y2 , . . . , yn , yn+1 is linearly dependent.
4.2
Reduction of Order
In Problems 1–8 we use reduction of order to find a second solution. In Problems 9–16 we use
formula (5) from the text.
1. Define y = u(x)e2x so
y = 2ue2x + u e2x ,
y = e2x u + 4e2x u + 4e2x u,
and y − 4y + 4y = e2x u = 0.
Therefore u = 0 and u = c1 x + c2 . Taking c1 = 1 and c2 = 0 we see that a second solution
is y2 = xe2x .
2. Define y = u(x)xe−x so
y = (1 − x)e−x u + xe−x u ,
y = xe−x u + 2(1 − x)e−x u − (2 − x)e−x u,
and
y + 2y + y = e−x (xu + 2u ) = 0 or u +
If w = u we obtain the linear first-order equation w +
factor e2
´
dx/x
= x2 . Now
d 2
[x w] = 0 gives
dx
2 u = 0.
x
2
w = 0 which has the integrating
x
x2 w = c.
Therefore w = u = c/x2 and u = c1 /x. A second solution is y2 =
1 −x
xe = e−x .
x
3. Define y = u(x) cos 4x so
y = −4u sin 4x + u cos 4x,
y = u cos 4x − 8u sin 4x − 16u cos 4x
and
y + 16y = (cos 4x)u − 8(sin 4x)u = 0 or u − 8(tan 4x)u = 0.
163
164
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
If w = u we obtain the linear first-order equation w − 8(tan 4x)w = 0 which has the inte´
grating factor e−8 tan 4x dx = cos2 4x. Now
d
[(cos2 4x)w] = 0 gives (cos2 4x)w = c.
dx
Therefore w = u = c sec2 4x and u = c1 tan 4x. A second solution is
y2 = tan 4x cos 4x = sin 4x.
4. Define y = u(x) sin 3x so
y = 3u cos 3x + u sin 3x,
y = u sin 3x + 6u cos 3x − 9u sin 3x,
and
y + 9y = (sin 3x)u + 6(cos 3x)u = 0 or u + 6(cot 3x)u = 0.
If w = u we obtain the linear first-order equation w + 6(cot 3x)w = 0 which has the
´
integrating factor e6 cot 3x dx = sin2 3x. Now
d
[(sin2 3x)w] = 0 gives (sin2 3x)w = c.
dx
Therefore w = u = c csc2 3x and u = c1 cot 3x. A second solution is
y2 = cot 3x sin 3x = cos 3x.
5. Define y = u(x) cosh x so
y = u sinh x + u cosh x,
y = u cosh x + 2u sinh x + u cosh x
and
y − y = (cosh x)u + 2(sinh x)u = 0 or u + 2(tanh x)u = 0.
If w = u we obtain the linear first-order equation w + 2(tanh x)w = 0 which has the
´
integrating factor e2 tanh x dx = cosh2 x. Now
d
[(cosh2 x)w] = 0 gives (cosh2 x)w = c.
dx
Therefore w = u = c sech2 x and u = c tanh x. A second solution is
y2 = tanh x cosh x = sinh x.
6. Define y = u(x)e5x so
y = 5e5x u + e5x u ,
y = e5x u + 10e5x u + 25e5x u
and
y − 25y = e5x (u + 10u ) = 0 or u + 10u = 0.
4.2
Reduction of Order
If w = u we obtain the linear first-order equation w + 10w = 0 which has the integrating
´
factor e10 dx = e10x . Now
d 10x
[e w] = 0 gives
dx
e10x w = c.
Therefore w = u = ce−10x and u = c1 e−10x . A second solution is y2 = e−10x e5x = e−5x .
7. Define y = u(x)e2x/3 so
2
y = e2x/3 u + e2x/3 u ,
3
4
4
y = e2x/3 u + e2x/3 u + e2x/3 u
3
9
and
9y − 12y + 4y = 9e2x/3 u = 0.
Therefore u = 0 and u = c1 x + c2 . Taking c1 = 1 and c2 = 0 we see that a second solution
is y2 = xe2x/3 .
8. Define y = u(x)ex/3 so
1
y = ex/3 u + ex/3 u ,
3
and
2
1
y = ex/3 u + ex/3 u + ex/3 u
3
9
5
6y + y − y = ex/3 (6u + 5u ) = 0 or u + u = 0.
6
If w = u we obtain the linear first-order equation w + 56 w = 0 which has the integrating
´
factor e(5/6) dx = e5x/6 . Now
d 5x/6
[e
w] = 0 gives
dx
e5x/6 w = c.
Therefore w = u = ce−5x/6 and u = c1 e−5x/6 . A second solution is y2 = e−5x/6 ex/3 = e−x/2 .
9. Identifying P (x) = −7/x we have
ˆ − ´ (−7/x) dx
ˆ
e
1
4
4
dx = x4 ln |x|.
dx = x
y2 = x
8
x
x
A second solution is y2 = x4 ln |x|.
10. Identifying P (x) = 2/x we have
ˆ − ´ (2/x) dx
ˆ
e
1
2
2
x−6 dx = − x−3 .
dx = x
y2 = x
x4
5
A second solution is y2 = x−3 .
11. Identifying P (x) = 1/x we have
ˆ − ´ dx/x
ˆ
e
dx
1
= −1.
dx = ln x
= ln x −
y2 = ln x
(ln x)2
x(ln x)2
ln x
A second solution is y2 = 1.
165
166
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
12. Identifying P (x) = 0 we have
ˆ
1/2
y2 = x
ln x
´
e− 0 dx
1
1/2
= −x1/2 .
dx = x ln x −
x(ln x)2
ln x
A second solution is y2 = x1/2 .
13. Identifying P (x) = −1/x we have
´
ˆ
e− −dx/x
dx = x sin (ln x)
x2 sin2 (ln x)
y2 = x sin (ln x)
ˆ
ˆ
x
x2 sin2 (ln x)
dx
csc2 (ln x)
dx = [x sin (ln x)] [− cot (ln x)] = −x cos (ln x).
x
= x sin (ln x)
A second solution is y2 = x cos(ln x).
14. Identifying P (x) = −3/x we have
ˆ
2
y2 = x cos (ln x)
ˆ
2
= x cos (ln x)
´
e− −3 dx/x
dx = x2 cos(ln x)
x4 cos2 (ln x)
ˆ
x3
dx
x4 cos2 (ln x)
sec2 (ln x)
dx = x2 cos (ln x) tan (ln x) = x2 sin (ln x).
x
A second solution is y2 = x2 sin (ln x).
15. Identifying P (x) = 2(1 + x)/ 1 − 2x − x2 we have
ˆ
y2 = (x + 1)
e−
´
(x + 1)2
ˆ
= (x + 1)
ˆ
2(1+x) dx/(1−2x−x2 )
dx = (x + 1)
1 − 2x − x2
dx = (x + 1)
(x + 1)2
= (x + 1) −
ˆ
2
eln (1−2x−x )
dx
(x + 1)2
2
− 1 dx
(x + 1)2
2
− x = −2 − x2 − x.
x+1
A second solution is y2 = x2 + x + 2.
16. Identifying P (x) = 2x/ 1 − x2 we have
ˆ
y2 =
e−
´
(2x) dx/(1−x2 )
A second solution is y2 =
ˆ
dx =
1 3
x − x.
3
2
eln (1−x ) dx =
ˆ
1
1 − x2 dx = x − x3 .
3
4.2
Reduction of Order
17. Define y = u(x)e−2x so
y = −2ue−2x + u e−2x ,
and
y = u e−2x − 4u e−2x + 4ue−2x
y − 4y = e−2x u − 4e−2x u = u − 4u e−2x = 2.
If w = u we obtain the linear first-order equation w − 4w = 2e2x which has the integrating
´
factor e−4 dx = e−4x . Now
d −4x
[e w] = 2e−2x
dx
gives
e−4x w = −e−2x + c1 .
Therefore w = u = −e2x + c1 e4x and
1
1
u = − e2x + c1 e4x + c2
2
4
1 1
y = − + c1 e2x + c2 e−2x .
2 4
From the last equation we see that a second solution is y2 = e2x and yp = − 12 .
18. Define y = u(x) · 1 = u(x) so
y = u ,
y = u
and y + y = u + u = 1.
If w = u we obtain the linear first-order equation w + w = 1 which has the integrating factor
´
e dx = ex . Now
d x
[e w] = ex gives ex w = ex + c1 .
dx
Therefore w = u = 1 + c1 e−x and
y = u = x − c1 e−x + c2
From the last equation we see that a second solution is y2 = e−x and yp = x.
19. Define y = u(x)ex so
y = uex + u ex ,
y = u ex + 2u ex + uex
and
y − 3y + 2y = ex u − ex u = 5e3x .
If w = u we obtain the linear first-order equation w − w = 5e2x which has the integrating
´
factor e− dx = e−x . Now
d −x
[e w] = 5ex
dx
gives
e−x w = 5ex + c1 .
167
168
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
Therefore w = u = 5e2x + c1 ex and
5
u = e2x + c1 ex + c2
2
5
y = e3x + c1 e2x + c2 ex
2
From the last equation we see that a second solution is y2 = e2x and yp = 52 e3x .
20. Define y = u(x)ex so
y = uex + u ex ,
y = u ex + 2u ex + uex
and
y − 4y + 3y = ex u − ex u = x.
If w = u we obtain the linear first-order equation w − 2w = xe−x which has the integrating
´
factor e− 2dx = e−2x . Now
d −2x
[e w] = xe−3x
dx
1
1
gives e−2x w = − xe−3x − e−3x + c1 .
3
9
Therefore w = u = − 13 xe−x − 19 e−x + c1 e2x and
4
1
1
u = xe−x + e−x + c1 e2x + c2
3
9
2
1
4 1
y = x + + c1 e3x + c2 ex
3
9 2
From the last equation we see that a second solution is y2 = e3x and yp = 13 x + 49 .
21. Dividing by x2 we have
x2 y + x2 − x y + (1 − x) y = 0
1
1
1
y + 1−
−
y +
y=0
x
x2 x
Using P (x) = 1 −
ˆ
1
and formula (5) in the text we have
x
x
y2 (x) = x
x0
e−(1−1/t) dt
dt = x
t2
ˆ
x
Therefore y2 (x) = x
x0
ˆ
x
x0
e−t
dt, x + 0 > 0.
t
e−t+ln t
dt = x
t2
ˆ
x
x0
e−t eln t
dt = x
t2
ˆ
x
x0
e−t t
dt
t2
4.2
Reduction of Order
22. Dividing by 2x we have
2xy − (2x + 1) y + y = 0
1
1
y +
=0
y + −1 −
2x
2x
1
and formula (5) in the text we have
2x
Using P (x) = −1 −
ˆ
x
x
y2 (x) = e
x0
e−(1+1/2t) dt
dt = ex
e2t
ˆ
x
Therefore y2 (x) = e
x√
ˆ
x
x0
1
et+ 2 ln t
dt = ex
e2t
ˆ
x
x0
1/2
ˆ x t√
et eln t
e t
x
dt = e
dt
2t
2t
e
x0 e
t e−t dt, x0 ≥ 0.
x0
23. (a) For m1 constant, let y1 = em1 x . Then y1 = m1 em1 x and y1 = m21 em1 x . Substituting into
the differential equation we obtain
ay1 + by1 + cy1 = am21 em1 x + bm1 em1 x + cem1 x
= em1 x (am21 + bm1 + c) = 0.
Thus, y1 = em1 x will be a solution of the differential equation whenever am21 +bm1 +c = 0.
Since a quadratic equation always has at least one real or complex root, the differential
equation must have a solution of the form y1 = em1 x .
(b) Write the differential equation in the form
b
c
y + y + y = 0,
a
a
and let y1 = em1 x be a solution. Then a second solution is given by
ˆ
y2 = em1 x
ˆ
e−bx/a
dx
e2m1 x
=e
e−(b/a+2m1 )x dx
=−
1
em1 x e−(b/a+2m1 )x
b/a + 2m1
=−
1
e−(b/a+m1 )x .
b/a + 2m1
m1 x
(m1 = −b/2a)
Thus, when m1 = −b/2a, a second solution is given by y2 = em2 x where m2 = −b/a−m1 .
When m1 = −b/2a a second solution is given by
ˆ
y2 = em1 x
dx = xem1 x .
169
170
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
(c) The functions
sin x =
1 ix
(e − e−ix )
2i
1
cos x = (eix + e−ix )
2
1
1
cosh x = (ex + e−x )
sinh x = (ex − e−x )
2
2
are all expressible in terms of exponential functions.
24. We have y1 = 1 and y1 = 0, so xy1 − xy1 + y1 = 0 − x + x = 0 and y1 (x) = x is a solution of
the differential equation. Letting y = u(x)y1 (x) = xu(x) we get
y = xu (x) + u(x)
and y = xu (x) + 2u (x).
Then xy − xy + y = x2 u + 2xu − x2 u − xu + xu = x2 u − (x2 − 2x)u = 0. If we make the
substitution w = u , the linear first-order differential equation becomes x2 w − (x2 − x)w = 0,
which is separable:
1
dw
= 1−
w
dx
x
dw
1
= 1−
dx
w
x
ln w = x − ln x + c
w = c1
ex
.
x
´
Then u = c1 ex /x and u = c1 ex dx/x. To integrate ex /x we use the series representation
for ex . Thus, a second solution is
ˆ x
e
dx
y2 = xu(x) = x
x
ˆ
1 2
1
1 3
1 + x + x + x + · · · dx
=x
x
2!
3!
ˆ 1
1
1 2
+ 1 + x + x + · · · dx
=x
x
2!
3!
1 3
1 2
x +
x + ···
= x ln x + x +
2(2!)
3(3!)
= x ln x + x2 +
1 3
1 4
x +
x + ··· .
2(2!)
3(3!)
An interval of definition is probably (0, ∞) because of the ln x term.
25. (a) We have y = y = ex , so
xy − (x + 10)y + 10y = xex − (x + 10)ex + 10ex = 0,
and y = ex is a solution of the differential equation.
4.3
Homogeneous Linear Equations with Constant Coefficients
171
(b) By (5) in the text a second solution is
ˆ
y 2 = y1
=e
´
ˆ
P (x) dx
x
dx = e
y12
ˆ
x
e−
ex+ln x
e2x
ˆ
10
x
dx = e
´
e
x+10
x
ˆ
dx
x
dx = e
e2x
´
e
(1+10/x) dx
dx
e2x
x10 e−x dx
= ex (−3,628,800 − 3,628,800x − 1,814,400x2 − 604,800x3 − 151,200x4
− 30,240x5 − 5,040x6 − 720x7 − 90x8 − 10x9 − x10 )e−x
= −3,628,800 − 3,628,800x − 1,814,400x2 − 604,800x3 − 151,200x4
− 30,240x5 − 5,040x6 − 720x7 − 90x8 − 10x9 − x10 .
1
y2 =
(c) By Corollary (a) of Theorem 3.1.2, −
10!
4.3
10
n=0
1 n
x is a solution.
n!
Homogeneous Linear Equations with Constant Coefficients
1. From 4m2 + m = 0 we obtain m1 = 0 and m2 = −1/4 so that y = c1 + c2 e−x/4 .
2. From m2 − 36 = 0 we obtain m1 = 6 and m1 = −6 so that y = c1 e6x + c2 e−6x .
3. From m2 − m − 6 = 0 we obtain m1 = 3 and m2 = −2 so that y = c1 e3x + c2 e−2x .
4. From m2 − 3m + 2 = 0 we obtain m1 = 1 and m2 = 2 so that y = c1 ex + c2 e2x .
5. From m2 + 8m + 16 = 0 we obtain m1 = −4 and m2 = −4 so that y = c1 e−4x + c2 xe−4x .
6. From m2 − 10m + 25 = 0 we obtain m1 = 5 and m2 = 5 so that y = c1 e5x + c2 xe5x .
7. From 12m2 − 5m − 2 = 0 we obtain m1 = −1/4 and m2 = 2/3 so that y = c1 e−x/4 + c2 e2x/3 .
8. From m2 + 4m − 1 = 0 we obtain m = −2 ±
√
5 so that y = c1 e(−2+
√
5 )x
+ c2 e(−2−
√
9. From m2 + 9 = 0 we obtain m1 = 3i and m2 = −3i so that y = c1 cos 3x + c2 sin 3x.
√
√
10. From 3m2 + 1 = 0 we obtain m1 = i/ 3 and m2 = −i/ 3 so that
√
√
y = c1 cos (x/ 3 ) + c2 sin (x/ 3).
11. From m2 − 4m + 5 = 0 we obtain m = 2 ± i so that y = e2x (c1 cos x + c2 sin x).
12. From 2m2 + 2m + 1 = 0 we obtain m = −1/2 ± i/2 so that
y = e−x/2 [c1 cos (x/2) + c2 sin (x/2)].
5 )x
.
172
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
√
13. From 3m2 + 2m + 1 = 0 we obtain m = −1/3 ± 2 i/3 so that
√
√
y = e−x/3 [c1 cos ( 2 x/3) + c2 sin ( 2 x/3)].
√
14. From 2m2 − 3m + 4 = 0 we obtain m = 3/4 ± 23 i/4 so that
√
√
y = e3x/4 [c1 cos ( 23 x/4) + c2 sin ( 23 x/4)].
15. From m3 − 4m2 − 5m = 0 we obtain m1 = 0, m2 = 5, and m3 = −1 so that
y = c1 + c2 e5x + c3 e−x .
√
16. From m3 − 1 = 0 we obtain m1 = 1 and m2 = −1/2 ± 3 i/2 so that
√
√
y = c1 ex + e−x/2 [c2 cos ( 3 x/2) + c3 sin ( 3 x/2)].
17. From m3 − 5m2 + 3m + 9 = 0 we obtain m1 = −1, m2 = 3, and m3 = 3 so that
y = c1 e−x + c2 e3x + c3 xe3x .
18. From m3 + 3m2 − 4m − 12 = 0 we obtain m1 = −2, m2 = 2, and m3 = −3 so that
y = c1 e−2x + c2 e2x + c3 e−3x .
19. From m3 + m2 − 2 = 0 we obtain m1 = 1 and m2 = −1 ± i so that
u = c1 et + e−t (c2 cos t + c3 sin t).
√
20. From m3 − m2 − 4 = 0 we obtain m1 = 2 and m2 = −1/2 ± 7 i/2 so that
√
√
x = c1 e2t + e−t/2 [c2 cos ( 7 t/2) + c3 sin ( 7 t/2)].
21. From m3 + 3m2 + 3m + 1 = 0 we obtain m1 = −1, m2 = −1, and m3 = −1 so that
y = c1 e−x + c2 xe−x + c3 x2 e−x .
22. From m3 − 6m2 + 12m − 8 = 0 we obtain m1 = 2, m2 = 2, and m3 = 2 so that
y = c1 e2x + c2 xe2x + c3 x2 e2x .
√
23. From m4 + m3 + m2 = 0 we obtain m1 = 0, m2 = 0, and m3 = −1/2 ± 3 i/2 so that
√
√
y = c1 + c2 x + e−x/2 [c3 cos ( 3 x/2) + c4 sin ( 3 x/2)].
24. From m4 − 2m2 + 1 = 0 we obtain m1 = 1, m2 = 1, m3 = −1, and m4 = −1 so that
y = c1 ex + c2 xex + c3 e−x + c4 xe−x .
4.3
Homogeneous Linear Equations with Constant Coefficients
√
√
25. From 16m4 + 24m2 + 9 = 0 we obtain m1 = ± 3 i/2 and m2 = ± 3 i/2 so that
√
√
√
√
y = c1 cos ( 3 x/2) + c2 sin ( 3 x/2) + c3 x cos ( 3 x/2) + c4 x sin ( 3 x/2).
√
26. From m4 − 7m2 − 18 = 0 we obtain m1 = 3, m2 = −3, and m3 = ± 2 i so that
y = c1 e3x + c2 e−3x + c3 cos
√
√
2 x + c4 sin 2 x.
27. From m5 + 5m4 − 2m3 − 10m2 + m + 5 = 0 we obtain m1 = −1, m2 = −1, m3 = 1, and
m4 = 1, and m5 = −5 so that
u = c1 e−r + c2 re−r + c3 er + c4 rer + c5 e−5r .
28. From 2m5 − 7m4 + 12m3 + 8m2 = 0 we obtain m1 = 0, m2 = 0, m3 = −1/2, and m4 = 2 ± 2i
so that
x = c1 + c2 s + c3 e−s/2 + e2s (c4 cos 2s + c5 sin 2s).
29. From m2 + 16 = 0 we obtain m = ±4i so that y = c1 cos 4x + c2 sin 4x. If y(0) = 2 and
y (0) = −2 then c1 = 2, c2 = −1/2, and y = 2 cos 4x − 12 sin 4x.
30. From m2 + 1 = 0 we obtain m = ±i so that y = c1 cos θ + c2 sin θ. If y(π/3) = 0 and
y (π/3) = 2 then
√
1
3
c1 +
c2 = 0
2
2
√
1
3
c1 + c2 = 2,
−
2
2
√
√
so c1 = − 3, c2 = 1, and y = − 3 cos θ + sin θ.
31. From m2 − 4m − 5 = 0 we obtain m1 = −1 and m1 = 5, so that y = c1 e−t + c2 e5t . If y(1) = 0
and y (1) = 2, then c1 e−1 + c2 e5 = 0, −c1 e−1 + 5c2 e5 = 2, so c1 = −e/3, c2 = e−5 /3, and
y = − 13 e1−t + 13 e5t−5 .
32. From 4m2 − 4m − 3 = 0 we obtain m1 = −1/2 and m2 = 3/2 so that y = c1 e−x/2 + c2 e3x/2 .
If y(0) = 1 and y (0) = 5 then c1 + c2 = 1, − 12 c1 + 12 32c2 = 5, so c1 = −7/4, c2 = 11/4, and
3x/2 .
y = − 74 e−x/2 + 11
4 e
√
33. From m2 + m + 2 = 0 we obtain m = −1/2 ± 7 i/2 so that
√
√
y = e−x/2 [c1 cos ( 7 x/2) + c2 sin ( 7 x/2)]. If y(0) = 0 and y (0) = 0 then c1 = 0 and c2 = 0
so that y = 0.
34. From m2 − 2m + 1 = 0 we obtain m1 = 1 and m2 = 1 so that y = c1 ex + c2 xex . If y(0) = 5
and y (0) = 10 then c1 = 5, c1 + c2 = 10 so c1 = 5, c2 = 5, and y = 5ex + 5xex .
173
174
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
35. From m3 + 12m2 + 36m = 0 we obtain m1 = 0, m2 = −6, and m3 = −6 so that
y = c1 + c2 e−6x + c3 xe−6x . If y(0) = 0, y (0) = 1, and y (0) = −7 then
c1 + c2 = 0,
−6c2 + c3 = 1,
so c1 = 5/36, c2 = −5/36, c3 = 1/6, and y =
5
36
−
36c2 − 12c3 = −7,
5 −6x
36 e
+ 16 xe−6x .
36. From m3 + 2m2 − 5m − 6 = 0 we obtain m1 = −1, m2 = 2, and m3 = −3 so that
y = c1 e−x + c2 e2x + c3 e−3x .
If y(0) = 0, y (0) = 0, and y (0) = 1 then
c1 + c2 + c3 = 0,
−c1 + 2c2 − 3c3 = 0,
c1 + 4c2 + 9c3 = 1,
so c1 = −1/6, c2 = 1/15, c3 = 1/10, and
1
1
1
y = − e−x + e2x + e−3x .
6
15
10
37. From m2 − 10m + 25 = 0 we obtain m1 = 5 and m2 = 5 so that y = c1 e5x + c2 xe5x . If y(0) = 1
and y(1) = 0 then c1 = 1, c1 e5 + c2 e5 = 0, so c1 = 1, c2 = −1, and y = e5x − xe5x .
38. From m2 +4 = 0 we obtain m = ±2i so that y = c1 cos 2x+c2 sin 2x. If y(0) = 0 and y(π) = 0
then c1 = 0 and y = c2 sin 2x.
39. From m2 + 1 = 0 we obtain m = ±i so that y = c1 cos x + c2 sin x and y = −c1 sin x + c2 cos x.
From y (0) = c1 (0) + c2 (1) = c2 = 0 and y (π/2) = −c1 (1) = 0 we find c1 = c2 = 0. A solution
of the boundary-value problem is y = 0.
40. From m2 − 2m + 2 = 0 we obtain m = 1 ± i so that y = ex (c1 cos x + c2 sin x). If y(0) = 1
and y(π) = 1 then c1 = 1 and y(π) = eπ cos π = −eπ . Since −eπ = 1, the boundary-value
problem has no solution.
√
√
41. The auxiliary equation
is √
m2 − 3 = 0 which has roots − 3 and 3 . By (10) the general solu√
√
√
tion is y = c√1 e 3 x + c2√e− 3 x . By (11) the general solution is y = c1 cosh 3 x + c2 sinh 3 x.
√
√
For y = c1 e 3 x + c2 e− 3 x the initial conditions imply c1 + c2 = 1, 3 c1 − 3 c2 = 5. Solving
√
√
c1 = 12 (1 + 5 3) and
c2 = 12 (1 − 5 3) so
for c1 and c2 we find
√
√
√
√
√
√
y = 12 (1 + 5 3)e 3 x + 12 (1 − 5 3 )e− 3 x . For y = c1 cosh 3 x + c2 sinh 3 x the initial
√
√
conditions imply c1 = 1, 3 c2 = 5. Solving for c1 and c2 we find c1 = 1 and c2 = 53 3 so
√
√
√
y = cosh 3 x + 53 3 sinh 3 x.
42. The auxiliary equation is m2 −1 = 0 which has roots −1 and 1. By (10) the general solution is
y = c1 ex +c2 e−x . By (11) the general solution is y = c1 cosh x+c2 sinh x. For y = c1 ex +c2 e−x
the boundary conditions imply c1 + c2 = 1, c1 e − c2 e−1 = 0. Solving for c1 and c2 we find
c1 = 1/(1 + e2 ) and c2 = e2 /(1 + e2 ) so y = ex /(1 + e2 ) + e2 e−x /(1 + e2 ). For
y = c1 cosh x + c2 sinh x the boundary conditions imply c1 = 1, c2 = − tanh 1, so
y = cosh x − (tanh 1) sinh x.
4.3
Homogeneous Linear Equations with Constant Coefficients
43. The auxiliary equation should have two positive roots, so that the solution has the form
y = c1 ek1 x + c2 ek2 x . Thus, the differential equation is (f).
44. The auxiliary equation should have one positive and one negative root, so that the solution
has the form y = c1 ek1 x + c2 e−k2 x . Thus, the differential equation is (a).
45. The auxiliary equation should have a pair of complex roots α ± βi where α < 0, so that the
solution has the form eαx (c1 cos βx + c2 sin βx). Thus, the differential equation is (e).
46. The auxiliary equation should have a repeated negative root, so that the solution has the
form y = c1 e−x + c2 xe−x . Thus, the differential equation is (c).
47. The differential equation should have the form y + k 2 y = 0 where k = 1 so that the period
of the solution is 2π. Thus, the differential equation is (d).
48. The differential equation should have the form y + k 2 y = 0 where k = 2 so that the period
of the solution is π. Thus, the differential equation is (b).
49. We have (m − 1)(m − 5) = m2 − 6m + 5, so the differential equation is y − 6y + 5y = 0.
50. We have (m + 4)(m + 3) = m2 + 7m + 12, so the differential equation is y + 7y + 12y = 0.
51. We have m(m − 2) = m2 − 2m, so the differential equation is y − 2y = 0.
52. We have (m − 10)2 = m2 − 20m + 100, so the differential equation is y − 20y + 100y = 0.
53. We have (m − 3i)(m + 3i) = m2 + 9, so the differential equation is y + 9y = 0.
54. We have (m − 7)(m + 7) = m2 − 49, so the differential equation is y − 49y = 0.
55. We have [m − (−1 + i)] [m − (−1 − i)] = m2 + 2m + 2, so the differential equation is
y + 2y + 2y = 0.
56. We have m [m − (2 + 5i)] [m − (2 − 5i)] = m3 − 4m2 + 29m, so the differential equation is
y − 4y + 29y = 0.
57. We have m2 (m − 8) = m3 − 8m2 , so the differential equation is y − 8y = 0.
58. We have (m2 + 1)(m2 + 4) = m4 + 5m2 + 29, so the differential equation is y (4) + 5y + 4y = 0.
59. A third root must be m3 = 3 − i and the auxiliary equation is
1
1
11
m+
[m − (3 + i)][m − (3 − i)] = m +
(m2 − 6m + 10) = m3 − m2 + 7m + 5.
2
2
2
The differential equation is
y −
11 y + 7y + 5y = 0.
2
175
176
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
60. The auxiliary equation of 2y + 7y + 4y = 0 is 2m3 + 7m2 + 4m − 4 = 0. Because m1 = 12
is a root of the equation it follows from the Factor Theorem of algebra that m − 12 is a factor
of 2m3 + 7m2 + 4m − 4. By synthetic division we find that
1 2
3
2
2m + 7m + 4m − 4 = m −
2m + 8m + 8
2
or
2m3 + 7m2 + 4m − 4 = (2m − 1) m2 + 4m + 4 = (2m − 1) (m + 2)2
Thus the roots of the auxiliary equation are m1 = 12 , m2 = m3 = −2, and the general solution
of the differential equation is
y = c1 ex/2 + c2 e−2x + c3 xe−2x .
61. From the solution y1 = e−4x cos x we conclude that m1 = −4 + i and m2 = −4 − i are roots
of the auxiliary equation. Hence another solution must be y2 = e−4x sin x. Now dividing the
polynomial m3 + 6m2 + m − 34 by [ m − (−4 + i)] [m − (−4 − i)] = m2 + 8m + 17 gives m − 2.
Therefore m3 = 2 is the third root of the auxiliary equation, and the general solution of the
differential equation is
y = c1 e−4x cos x + c2 e−4x sin x + c3 e2x .
62. Factoring the difference of two squares we obtain
m4 + 1 = (m2 + 1)2 − 2m2 = (m2 + 1 −
√
2 m)(m2 + 1 +
√
2 m) = 0.
√
√
Using the quadratic formula on each factor we get m = ± 2/2 ± 2 i/2. The solution of the
differential equation is
√
y(x) = e
√
2 x/2
√
2
2
c1 cos
x + c2 sin
x
2
2
√
− 2 x/2
+e
√
√
2
2
c3 cos
x + c4 sin
x .
2
2
63. Using the definition of sinh x and the formula for the cosine of the sum of two angles, we have
y = sinh x − 2 cos (x + π/6)
1
π
π 1
− (sin x) sin
= ex − e−x − 2 (cos x) cos
2
2
6
6
√
1
1
3
1
cos x − sin x
= ex − e−x − 2
2
2
2
2
√
1
1
= ex − e−x − 3 cos x + sin x.
2
2
This form of the solution can be obtained from the general solution
√
y = c1 ex + c2 e−x + c3 cos x + c4 sin x by choosing c1 = 12 , c2 = − 12 , c3 = − 3 , and c4 = 1.
4.3
Homogeneous Linear Equations with Constant Coefficients
64. The auxiliary equation is m2 + λ = 0 and we consider three cases where λ = 0, λ = α2 > 0,
and λ = −α2 < 0:
Case I
When α = 0 the general solution of the differential equation is y = c1 + c2 x. The boundary
conditions imply 0 = y(0) = c1 and 0 = y(π/2) = c2 π/2, so that c1 = c2 = 0 and the problem
possesses only the trivial solution.
Case II
When λ = −α2 < 0 the general solution of the differential equation is y = c1 eαx +c2 e−αx , or alternatively, y = c1 cosh (αx) + c2 sinh (αx). Again, y(0) = 0 implies c1 = 0 so
y = c2 sinh (αx). The second boundary condition implies 0 = y(π/2) = c2 sinh (απ/2) or
c2 = 0. In this case also, the problem possesses only the trivial solution.
Case III
When λ = α2 > 0 the general solution of the differential equation is
y = c1 cos (αx) + c2 sin (αx). In this case also, y(0) = 0 yields c1 = 0, so that y = c2 sin (αx).
The second boundary condition implies 0 = c2 sin (απ/2). When α π/2 is an integer multiple
of π, that is, when α = 2k for k a nonzero integer, the problem will have nontrivial solutions. Thus, for λ = α2 = 4k 2 the boundary-value problem will have nontrivial solutions
y = c2 sin (2kx), where k is a nonzero integer. On the other hand, when α is not an even
integer, the boundary-value problem will have only the trivial solution.
65. Using a CAS to solve the auxiliary equation m3 − 6m2 + 2m + 1 we find m1 = −0.270534,
m2 = 0.658675, and m3 = 5.61186. The general solution is
y = c1 e−0.270534x + c2 e0.658675x + c3 e5.61186x .
66. Using a CAS to solve the auxiliary equation 6.11m3 + 8.59m2 + 7.93m + 0.778 = 0 we find
m1 = −0.110241, m2 = −0.647826+0.857532i, and m3 = −0.647826−0.857532i. The general
solution is
y = c1 e−0.110241x + e−0.647826x (c2 cos 0.857532x + c3 sin 0.857532x).
67. Using a CAS to solve the auxiliary equation 3.15m4 − 5.34m2 + 6.33m − 2.03 = 0 we find
m1 = −1.74806, m2 = 0.501219, m3 = 0.62342 + 0.588965i, and m4 = 0.62342 − 0.588965i.
The general solution is
y = c1 e−1.74806x + c2 e0.501219x + e0.62342x (c3 cos 0.588965x + c4 sin 0.588965x).
√
68. Using a CAS to solve the auxiliary equation m4 + 2m2 − m + 2 = 0 we find m1 = 1/2 + 3 i/2,
√
√
√
m2 = 1/2 − 3 i/2, m3 = −1/2 + 7 i/2, and m4 = −1/2 − 7 i/2. The general solution is
√
√
√
√
3
3
7
7
x/2
−x/2
c1 cos
c3 cos
x + c2 sin
x +e
x + c4 sin
x .
y=e
2
2
2
2
177
178
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
69. From 2m4 + 3m3 − 16m2 + 15m − 4 = 0 we obtain m1 = −4, m2 = 12 , m3 = 1, and m4 = 1, so
that y = c1 e−4x + c2 ex/2 + c3 ex + c4 xex . If y(0) = −2, y (0) = 6, y (0) = 3, and y (0) = 12 ,
then
c1 + c2 + c3 = −2
1
−4c1 + c2 + c3 + c4 = 6
2
1
16c1 + c2 + c3 + 2c4 = 3
4
1
1
−64c1 + c2 + c3 + 3c4 = ,
8
2
4
, c2 = − 116
so c1 = − 75
3 , c3 =
918
25
y=−
, c4 = − 58
5 , and
4 −4x 116 x/2 918 x 58 x
e
e +
e − xe .
−
75
3
25
5
70. From m4 − 3m3 + 3m2 − m = 0 we obtain m1 = 0, m2 = 1, m3 = 1, and m4 = 1 so that
y = c1 + c2 ex + c3 xex + c4 x2 ex . If y(0) = 0, y (0) = 0, y (0) = 1, and y (0) = 1 then
c1 + c2 = 0,
c2 + c3 = 0,
c2 + 2c3 + 2c4 = 1,
c2 + 3c3 + 6c4 = 1,
so c1 = 2, c2 = −2, c3 = 2, c4 = −1/2, and
1
y = 2 − 2ex + 2xex − x2 ex .
2
4.4
Undetermined Coefficients - Superposition Approach
1. From m2 + 3m + 2 = 0 we find m1 = −1 and m2 = −2. Then yc = c1 e−x + c2 e−2x and we
assume yp = A. Substituting into the differential equation we obtain 2A = 6. Then A = 3,
yp = 3 and
y = c1 e−x + c2 e−2x + 3.
2. From 4m2 + 9 = 0 we find m1 = − 32 i and m2 = 32 i. Then yc = c1 cos 32 x + c2 sin 32 x and we
assume yp = A. Substituting into the differential equation we obtain 9A = 15. Then A = 53 ,
yp = 53 and
3
3
5
y = c1 cos x + c2 sin x + .
2
2
3
3. From m2 − 10m + 25 = 0 we find m1 = m2 = 5. Then yc = c1 e5x + c2 xe5x and we
assume yp = Ax + B. Substituting into the differential equation we obtain 25A = 30 and
−10A + 25B = 3. Then A = 65 , B = 35 , yp = 65 x + 35 , and
3
6
y = c1 e5x + c2 xe5x + x + .
5
5
4.4
Undetermined Coefficients - Superposition Approach
4. From m2 + m − 6 = 0 we find m1 = −3 and m2 = 2. Then yc = c1 e−3x + c2 e2x and we assume
yp = Ax + B. Substituting into the differential equation we obtain −6A = 2 and A − 6B = 0.
1
1
, yp = − 13 x − 18
, and
Then A = − 13 , B = − 18
1
1
.
y = c1 e−3x + c2 e2x − x −
3
18
5. From 14 m2 + m + 1 = 0 we find m1 = m2 = −2. Then yc = c1 e−2x + c2 xe−2x and we assume
yp = Ax2 +Bx+C. Substituting into the differential equation we obtain A = 1, 2A+B = −2,
and 12 A + B + C = 0. Then A = 1, B = −4, C = 72 , yp = x2 − 4x + 72 , and
y = c1 e−2x + c2 xe−2x + x2 − 4x +
7
.
2
6. From m2 −8m+20 = 0 we find m1 = 4+2i and m2 = 4−2i. Then yc = e4x (c1 cos 2x+c2 sin 2x)
and we assume yp = Ax2 + Bx + C + (Dx + E)ex . Substituting into the differential equation
we obtain
2A − 8B + 20C = 0
−6D + 13E = 0
−16A + 20B = 0
13D = −26
20A = 100.
11
2
, D = −2, E = − 12
13 , yp = 5x + 4x + 10 + −2x −
12 x
11
4x
2
y = e (c1 cos 2x + c2 sin 2x) + 5x + 4x +
+ −2x −
e .
10
13
Then A = 5, B = 4, C =
11
10
12
13
ex and
√
√
√
√
7. From m2 + 3 = 0 we find m1 = 3 i and m2 = − 3 i. Then yc = c1 cos 3 x + c2 sin 3 x
and we assume yp = (Ax2 + Bx + C)e3x . Substituting into the differential equation we obtain
2A + 6B + 12C = 0, 12A + 12B = 0, and 12A = −48. Then A = −4, B = 4, C = − 43 ,
yp = −4x2 + 4x − 43 e3x and
√
√
4 3x
2
y = c1 cos 3 x + c2 sin 3 x + −4x + 4x −
e .
3
8. From 4m2 − 4m − 3 = 0 we find m1 = 32 and m2 = − 12 . Then yc = c1 e3x/2 + c2 e−x/2 and
we assume yp = A cos 2x + B sin 2x. Substituting into the differential equation we obtain
19
8
19
8
, B = − 425
, yp = − 425
cos 2x − 425
sin 2x,
−19 − 8B = 1 and 8A − 19B = 0. Then A = − 425
and
19
8
cos 2x −
sin 2x.
y = c1 e3x/2 + c2 e−x/2 −
425
425
9. From m2 − m = 0 we find m1 = 1 and m2 = 0. Then yc = c1 ex + c2 and we assume yp = Ax.
Substituting into the differential equation we obtain −A = −3. Then A = 3, yp = 3x and
y = c1 ex + c2 + 3x.
179
180
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
10. From m2 + 2m = 0 we find m1 = −2 and m2 = 0. Then yc = c1 e−2x + c2 and we assume
yp = Ax2 + Bx + Cxe−2x . Substituting into the differential equation we obtain 2A + 2B = 5,
4A = 2, and −2C = −1. Then A = 12 , B = 2, C = 12 , yp = 12 x2 + 2x + 12 xe−2x , and
1
1
y = c1 e−2x + c2 + x2 + 2x + xe−2x .
2
2
11. From m2 − m + 14 = 0 we find m1 = m2 = 12 . Then yc = c1 ex/2 + c2 xex/2 and we assume
yp = A + Bx2 ex/2 . Substituting into the differential equation we obtain 14 A = 3 and 2B = 1.
Then A = 12, B = 12 , yp = 12 + 12 x2 ex/2 , and
1
y = c1 ex/2 + c2 xex/2 + 12 + x2 ex/2 .
2
12. From m2 − 16 = 0 we find m1 = 4 and m2 = −4. Then yc = c1 e4x + c2 e−4x and we assume
yp = Axe4x . Substituting into the differential equation we obtain 8A = 2. Then A = 14 ,
yp = 14 xe4x and
1
y = c1 e4x + c2 e−4x + xe4x .
4
13. From m2 + 4 = 0 we find m1 = 2i and m2 = −2i. Then yc = c1 cos 2x + c2 sin 2x and
we assume yp = Ax cos 2x + Bx sin 2x. Substituting into the differential equation we obtain
4B = 0 and −4A = 3. Then A = − 34 , B = 0, yp = − 34 x cos 2x, and
3
y = c1 cos 2x + c2 sin 2x − x cos 2x.
4
14. From m2 − 4 = 0 we find m1 = 2 and m2 = −2. Then yc = c1 e2x + c2 e−2x and we assume
that yp = (Ax2 + Bx + C) cos 2x + (Dx2 + Ex + F ) sin 2x. Substituting into the differential
equation we obtain
−8A = 0
−8B + 8D = 0
2A − 8C + 4E = 0
−8D = 1
−8A − 8E = 0
−4B + 2D − 8F = −3
Then A = 0, B = − 18 , C = 0, D = − 18 , E = 0, F =
yp = − 18 x cos 2x + − 18 x2 + 13
32 sin 2x, and
2x
y = c1 e
−2x
+ c2 e
13
32
, so
1 2 13
1
sin 2x.
− x cos 2x + − x +
8
8
32
4.4
Undetermined Coefficients - Superposition Approach
15. From m2 + 1 = 0 we find m1 = i and m2 = −i. Then yc = c1 cos x + c2 sin x and we assume
yp = (Ax2 +Bx) cos x+(Cx2 +Dx) sin x. Substituting into the differential equation we obtain
4C = 0, 2A + 2D = 0, −4A = 2, and −2B + 2C = 0. Then A = − 12 , B = 0, C = 0, D = 12 ,
yp = − 12 x2 cos x + 12 x sin x, and
1
1
y = c1 cos x + c2 sin x − x2 cos x + x sin x.
2
2
16. From m2 − 5m = 0 we find m1 = 5 and m2 = 0. Then yc = c1 e5x + c2 and we assume
yp = Ax4 + Bx3 + Cx2 + Dx. Substituting into the differential equation we obtain −20A = 2,
1
53
, B = 14
12A − 15B = −4, 6B − 10C = −1, and 2C − 5D = 6. Then A = − 10
75 , C = 250 ,
1 4
14 3
53 2
697
D = − 697
625 , yp = − 10 x + 75 x + 250 x − 625 x, and
y = c1 e5x + c2 −
1 4 14 3
53 2 697
x + x +
x −
x.
10
75
250
625
17. From m2 −2m+5 = 0 we find m1 = 1+2i and m2 = 1−2i. Then yc = ex (c1 cos 2x+c2 sin 2x)
and we assume yp = Axex cos 2x + Bxex sin 2x. Substituting into the differential equation we
obtain 4B = 1 and −4A = 0. Then A = 0, B = 14 , yp = 14 xex sin 2x, and
1
y = ex (c1 cos 2x + c2 sin 2x) + xex sin 2x.
4
18. From m2 − 2m + 2 = 0 we find m1 = 1 + i and m2 = 1 − i. Then yc = ex (c1 cos x + c2 sin x)
and we assume yp = Ae2x cos x + Be2x sin x. Substituting into the differential equation we
obtain A + 2B = 1 and −2A + B = −3. Then A = 75 , B = − 15 , yp = 75 e2x cos x − 15 e2x sin x
and
7
1
y = ex (c1 cos x + c2 sin x) + e2x cos x − e2x sin x.
5
5
19. From m2 + 2m + 1 = 0 we find m1 = m2 = −1. Then yc = c1 e−x + c2 xe−x and we assume
yp = A cos x + B sin x + C cos 2x + D sin 2x. Substituting into the differential equation we
obtain 2B = 0, −2A = 1, −3C + 4D = 3, and −4C − 3D = 0. Then A = − 12 , B = 0,
9
1
9
12
, D = 12
C = − 25
25 , yp = − 2 cos x − 25 cos 2x + 25 sin 2x, and
y = c1 e−x + c2 xe−x −
9
12
1
cos x −
cos 2x +
sin 2x.
2
25
25
20. From m2 + 2m − 24 = 0 we find m1 = −6 and m2 = 4. Then yc = c1 e−6x + c2 e4x and
we assume yp = A + (Bx2 + Cx)e4x . Substituting into the differential equation we obtain
1
19
, C = − 100
,
−24A = 16, 2B + 10C = −2, and 20B = −1. Then A = − 23 , B = − 20
2
1 2
19
4x
yp = − 3 − 20 x + 100 x e , and
−6x
y = c1 e
4x
+ c2 e
2
− −
3
1 2
19
x +
x e4x .
20
100
181
182
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
21. From m3 − 6m2 = 0 we find m1 = m2 = 0 and m3 = 6. Then yc = c1 + c2 x + c3 e6x
and we assume yp = Ax2 + B cos x + C sin x. Substituting into the differential equation we
6
1
, C = 37
,
obtain −12A = 3, 6B − C = −1, and B + 6C = 0. Then A = − 14 , B = − 37
1 2
6
1
yp = − 4 x − 37 cos x + 37 sin x, and
1
6
1
y = c1 + c2 x + c3 e6x − x2 −
cos x +
sin x.
4
37
37
22. From m3 − 2m2 − 4m + 8 = 0 we find m1 = m2 = 2 and m3 = −2. Then
yc = c1 e2x + c2 xe2x + c3 e−2x and we assume yp = (Ax3 + Bx2 )e2x . Substituting into the
3
,
differential equation we obtain 24A = 6 and 6A + 8B = 0. Then A = 14 , B = − 16
1 3
3 2
2x
yp = 4 x − 16 x e , and
2x
y = c1 e
2x
+ c2 xe
−2x
+ c3 e
+
1 3
3 2 2x
x − x e .
4
16
23. From m3 − 3m2 + 3m − 1 = 0 we find m1 = m2 = m3 = 1. Then yc = c1 ex + c2 xex + c3 x2 ex
and we assume yp = Ax + B + Cx3 ex . Substituting into the differential equation we obtain
−A = 1, 3A − B = 0, and 6C = −4. Then A = −1, B = −3, C = − 23 , yp = −x − 3 − 23 x3 ex ,
and
2
y = c1 ex + c2 xex + c3 x2 ex − x − 3 − x3 ex .
3
24. From m3 − m2 − 4m + 4 = 0 we find m1 = 1, m2 = 2, and m3 = −2. Then
yc = c1 ex + c2 e2x + c3 e−2x and we assume yp = A + Bxex + Cxe2x . Substituting into
the differential equation we obtain 4A = 5, −3B = −1, and 4C = 1. Then A = 54 , B = 13 ,
C = 14 , yp = 54 + 13 xex + 14 xe2x , and
y = c1 ex + c2 e2x + c3 e−2x +
5 1 x 1 2x
+ xe + xe .
4 3
4
25. From m4 + 2m2 + 1 = 0 we find m1 = m3 = i and m2 = m4 = −i. Then
yc = c1 cos x + c2 sin x + c3 x cos x + c4 x sin x and we assume yp = Ax2 + Bx + C. Substituting into the differential equation we obtain A = 1, B = −2, and 4A + C = 1. Then
A = 1, B = −2, C = −3, yp = x2 − 2x − 3, and
y = c1 cos x + c2 sin x + c3 x cos x + c4 x sin x + x2 − 2x − 3.
26. From m4 − m2 = 0 we find m1 = m2 = 0, m3 = 1, and m4 = −1. Then
yc = c1 + c2 x + c3 ex + c4 e−x and we assume yp = Ax3 + Bx2 + (Cx2 + Dx)e−x . Substituting into the differential equation we obtain −6A = 4, −2B = 0, 10C − 2D = 0, and
−4C = 2. Then A = − 23 , B = 0, C = − 12 , D = − 52 , yp = − 23 x3 − 12 x2 + 52 x e−x , and
x
−x
y = c 1 + c2 x + c 3 e + c4 e
2
− x3 −
3
1 2 5
x + x e−x .
2
2
4.4
Undetermined Coefficients - Superposition Approach
27. We have yc = c1 cos 2x + c2 sin 2x and we assume yp = A. Substituting into the differential
equation we find A = − 12 . Thus y = c1 cos 2x + c2 sin 2x − 12 . From the initial conditions we
√
obtain c1 = 0 and c2 = 2 , so
√
1
y = 2 sin 2x − .
2
28. We have yc = c1 e−2x + c2 ex/2 and we assume yp = Ax2 + Bx + C. Substituting into the
differential equation we find A = −7, B = −19, and C = −37. Thus y = c1 e−2x + c2 ex/2 −
7x2 − 19x − 37. From the initial conditions we obtain c1 = − 15 and c2 = 186
5 , so
1
186 x/2
y = − e−2x +
e − 7x2 − 19x − 37.
5
5
29. We have yc = c1 e−x/5 + c2 and we assume yp = Ax2 + Bx. Substituting into the differential
equation we find A = −3 and B = 30. Thus y = c1 e−x/5 + c2 − 3x2 + 30x. From the initial
conditions we obtain c1 = 200 and c2 = −200, so
y = 200e−x/5 − 200 − 3x2 + 30x.
30. We have yc = c1 e−2x + c2 xe−2x and we assume yp = (Ax3 + Bx2 )e−2x . Substituting into the
differential equation we find A = 16 and B = 32 . Thus y = c1 e−2x +c2 xe−2x + 16 x3 + 32 x2 e−2x .
From the initial conditions we obtain c1 = 2 and c2 = 9, so
1 3 3 2 −2x
−2x
−2x
x + x e .
+ 9xe
+
y = 2e
6
2
31. We have yc = e−2x (c1 cos x + c2 sin x) and we assume yp = Ae−4x . Substituting into the
differential equation we find A = 7. Thus y = e−2x (c1 cos x + c2 sin x) + 7e−4x . From the
initial conditions we obtain c1 = −10 and c2 = 9, so
y = e−2x (−10 cos x + 9 sin x) + 7e−4x .
32. We have yc = c1 cosh x + c2 sinh x and we assume yp = Ax cosh x + Bx sinh x. Substituting
into the differential equation we find A = 0 and B = 12 . Thus
1
y = c1 cosh x + c2 sinh x + x sinh x.
2
From the initial conditions we obtain c1 = 2 and c2 = 12, so
1
y = 2 cosh x + 12 sinh x + x sinh x.
2
33. We have xc = c1 cos ωt + c2 sin ωt and we assume xp = At cos ωt + Bt sin ωt. Substituting into
the differential equation we find A = −F0 /2ω and B = 0. Thus x = c1 cos ωt + c2 sin ωt −
(F0 /2ω)t cos ωt. From the initial conditions we obtain c1 = 0 and c2 = F0 /2ω 2 , so
x = (F0 /2ω 2 ) sin ωt − (F0 /2ω)t cos ωt.
183
184
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
34. We have xc = c1 cos ωt + c2 sin ωt and we assume xp = A cos γt + B sin γt, where γ = ω.
Substituting into the differential equation we find A = F0 /(ω 2 − γ 2 ) and B = 0. Thus
x = c1 cos ωt + c2 sin ωt +
ω2
F0
cos γt.
− γ2
From the initial conditions we obtain c1 = −F0 /(ω 2 − γ 2 ) and c2 = 0, so
x=−
ω2
F0
F0
cos ωt + 2
cos γt.
2
−γ
ω − γ2
35. We have yc = c1 + c2 ex + c3 xex and we assume yp = Ax + Bx2 ex + Ce5x . Substituting into
the differential equation we find A = 2, B = −12, and C = 12 . Thus
1
y = c1 + c2 ex + c3 xex + 2x − 12x2 ex + e5x .
2
From the initial conditions we obtain c1 = 11, c2 = −11, and c3 = 9, so
1
y = 11 − 11ex + 9xex + 2x − 12x2 ex + e5x .
2
√
√
36. We have yc = c1 e−2x + ex (c2 cos 3 x + c3 sin 3 x) and we assume yp = Ax + B + Cxe−2x .
Substituting into the differential equation we find A = 14 , B = − 58 , and C = 23 . Thus
√
√
5 2
1
3 x) + x − + xe−2x .
4
8 3
√
59
17
From the initial conditions we obtain c1 = − 23
12 , c2 = − 24 , and c3 = 72 3 , so
√
√
5 2
59
17 √
1
23
3 sin 3 x + x − + xe−2x .
y = − e−2x + ex − cos 3 x +
12
24
72
4
8 3
y = c1 e−2x + ex (c2 cos
3 x + c3 sin
37. We have yc = c1 cos x + c2 sin x and we assume yp = Ax2 + Bx + C. Substituting into the
differential equation we find A = 1, B = 0, and C = −1. Thus y = c1 cos x + c2 sin x + x2 − 1.
From y(0) = 5 and y(1) = 0 we obtain
c1 − 1 = 5
(cos 1)c1 + (sin 1)c2 = 0.
Solving this system we find c1 = 6 and c2 = −6 cot 1. The solution of the boundary-value
problem is
y = 6 cos x − 6(cot 1) sin x + x2 − 1.
38. We have yc = ex (c1 cos x + c2 sin x) and we assume yp = Ax + B. Substituting into the
differential equation we find A = 1 and B = 0. Thus y = ex (c1 cos x + c2 sin x) + x. From
y(0) = 0 and y(π) = π we obtain
c1 = 0
π − eπ c1 = π.
4.4
Undetermined Coefficients - Superposition Approach
Solving this system we find c1 = 0 and c2 is any real number. The solution of the boundaryvalue problem is
y = c2 ex sin x + x.
√
√
39. The general solution of the differential equation y +3y = 6x is y = c1 cos 3x+c2 sin 3x+2x.
√
The condition y(0) = 0 implies c1 = 0 and so y = c2 sin 3x + 2x. The condition
√
√
√
√
√
√
y(1) + y (1) = 0 implies c2 sin 3 + 2 + c2 3 cos 3 + 2 = 0 so c2 = −4/(sin 3 + 3 cos 3 ).
The solution is
√
−4 sin 3x
√
√
√ + 2x.
y=
sin 3 + 3 cos 3
√
√
40. Using the general solution y = c1 cos 3x + c2 sin 3x + 2x, the boundary conditions y(0) +
y (0) = 0, y(1) = 0 yield the system
√
c 1 + 3 c2 + 2 = 0
√
√
c1 cos 3 + c2 sin 3 + 2 = 0.
Solving gives
Thus,
√ √
2 − 3 + sin 3
√
√
c1 = √
3 cos 3 − sin 3
and
√ 2 1 − cos 3
√
√ .
c2 = √
3 cos 3 − sin 3
√ √ √
√
√
2 − 3 + sin 3 cos 3 x 2 1 − cos 3 sin 3 x
√
√
√
√
√
+ √
+ 2x.
y=
3 cos 3 − sin 3
3 cos 3 − sin 3
41. We have yc = c1 cos 2x+c2 sin 2x and we assume yp = A cos x+B sin x on [0, π/2]. Substituting
into the differential equation we find A = 0 and B = 13 . Thus y = c1 cos 2x+c2 sin 2x+ 13 sin x
on [0, π/2]. On (π/2, ∞) we have y = c3 cos 2x + c4 sin 2x. From y(0) = 1 and y (0) = 2 we
obtain
c1 = 1
1
+ 2c2 = 2.
3
Solving this system we find c1 = 1 and c2 =
Now continuity of y at x = π/2 implies
cos π +
or −1 +
1
3
= −c3 . Hence c3 =
2
3
−2 sin π +
or − 53 = −2c4 . Then c4 =
5
6
5
6
. Thus y = cos 2x + 56 sin 2x + 13 sin x on [0, π/2].
1
π
5
sin π + sin = c3 cos π + c4 sin π
6
3
2
. Continuity of y at x = π/2 implies
1
π
5
cos π + cos = −2c3 sin π + 2c4 cos π
3
3
2
and so
⎧
5
1
⎪
⎪
⎨cos 2x + 6 sin 2x + 3 sin x, 0 ≤ x ≤ π/2
y(x) =
⎪2
⎪
⎩ cos 2x + 5 sin 2x,
x > π/2
3
6
185
186
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
42. We have yc = ex (c1 cos 3x + c2 sin 3x) and we assume yp = A on [0, π]. Substituting into the
differential equation we find A = 2. Thus, y = ex (c1 cos 3x + c2 sin 3x) + 2 on [0, π]. On (π, ∞)
we have y = ex (c3 cos 3x + c4 sin 3x). From y(0) = 0 and y (0) = 0 we obtain
c1 = −2,
c1 + 3c2 = 0.
Solving this system, we find c1 = −2 and c2 =
[0, π]. Now, continuity of y at x = π implies
eπ (−2 cos 3π +
2
3
. Thus y = ex (−2 cos 3x +
2
3
sin 3x) + 2 on
2
sin 3π) + 2 = eπ (c3 cos 3π + c4 sin 3π)
3
or 2 + 2eπ = −c3 eπ or c3 = −2e−π (1 + eπ ). Continuity of y at π implies
20 π
e sin 3π = eπ [(c3 + 3c4 ) cos 3π + (−3c3 + c4 ) sin 3π]
3
or −c3 eπ − 3c4 eπ = 0. Since c3 = −2e−π (1 + eπ ) we have c4 = 23 e−π (1 + eπ ). Therefore
⎧
2
⎪
⎨ex (−2 cos 3x + sin 3x) + 2,
0≤x≤π
3
y(x) =
⎪
⎩(1 + eπ )ex−π (−2 cos 3x + 2 sin 3x), x > π.
3
43. (a) From yp = Aekx we find yp = Akekx and yp = Ak 2 ekx . Substituting into the differential
equation we get
aAk 2 ekx + bAkekx + cAekx = (ak 2 + bk + c)Aekx = ekx ,
so (ak 2 + bk + c)A = 1. Since k is not a root of am2 + bm + c = 0, A = 1/(ak 2 + bk + c).
(b) From yp = Axekx we find yp = Akxekx + Aekx and yp = Ak 2 xekx + 2Akekx . Substituting
into the differential equation we get
aAk 2 xekx + 2aAkekx + bAkxekx + bAekx + cAxekx
= (ak 2 + bk + c)Axekx + (2ak + b)Aekx
= (0)Axekx + (2ak + b)Aekx = (2ak + b)Aekx = ekx
where ak 2 + bk + c = 0 because k is a root of the auxiliary equation. Now, the roots of
√
the auxiliary equation are −b/2a ± b2 − 4ac /2a, and since k is a root of multiplicity
one, k = −b/2a and 2ak + b = 0. Thus (2ak + b)A = 1 and A = 1/(2ak + b).
(c) If k is a root of multiplicity two, then, as we saw in part (b), k = −b/2a and 2ak + b = 0.
From yp = Ax2 ekx we find yp = Akx2 ekx + 2Axekx and yp = Ak 2 x2 ekx + 4Akxekx =
2Aekx . Substituting into the differential equation, we get
aAk 2 x2 ekx + 4aAkxekx + 2aAekx + bAkx2 ekx + 2bAxekx + cAx2 ekx
= (ak 2 + bk + c)Ax2 ekx + 2(2ak + b)Axekx + 2aAekx
= (0)Ax2 ekx + 2(0)Axekx + 2aAekx = 2aAekx = ekx .
Since the differential equation is second order, a = 0 and A = 1/(2a).
4.4
Undetermined Coefficients - Superposition Approach
44. Using the double-angle formula for the cosine, we have
sin x cos 2x = sin x(cos2 x − sin2 x) = sin x(1 − 2 sin2 x) = sin x − 2 sin3 x.
Since sin x is a solution of the related homogeneous differential equation we look for a particular solution of the form yp = Ax sin x + Bx cos x + C sin3 x.
Substituting into the differential equation we obtain
2A cos x + (6C − 2B) sin x − 8C sin3 x = sin x − 2 sin3 x.
Equating coefficients we find A = 0, C =
1
4
, and B =
1
4
. Thus, a particular solution is
1
1
yp = x cos x + sin3 x.
4
4
45. f (x) = ex sin x. We see that yp → ∞ as x → ∞ and yp → 0 as x → −∞.
46. f (x) = e−x . We see that yp → 0 as x → ∞ and yp → ∞ as x → −∞.
47. f (x) = sin 2x. We see that yp is sinusoidal.
48. f (x) = 1. We see that yp is constant and simply translates yc vertically.
49. The complementary function is yc = e2x (c1 cos 2x+c2 sin 2x). We assume a particular solution
of the form yp = (Ax3 + Bx2 + Cx)e2x cos 2x + (Dx3 + Ex2 + F )e2x sin 2x. Substituting into
the differential equation and using a CAS to simplify yields
[12Dx2 + (6A + 8E)x + (2B + 4F )]e2x cos 2x
+ [−12Ax2 + (−8B + 6D)x + (−4C + 2E)]e2x sin 2x
= (2x2 − 3x)e2x cos 2x + (10x2 − x − 1)e2x sin 2x
This gives the system of equations
12D = 2,
−12A = 10,
6A + 8E = −3
2B + 4F = 0,
−8B + 6D = −1
−4C + 2E = −1,
from which we find A = − 56 , B = 14 , C = 38 , D = 16 , E = 14 , and F = − 18 . Thus, a
particular solution of the differential equation is
5 3 1 2 3
1 3 1 2 1
2x
x + x − x e2x sin 2x.
yp = − x + x + x e cos 2x +
6
4
8
6
4
8
50. The complementary function is yc = c1 cos x + c2 sin x + c3 x cos x + c4 x sin x. We assume a
particular solution of the form yp = Ax2 cos x + Bx3 sin x. Substituting into the differential
equation and using a CAS to simplify yields
(−8A + 24B) cos x + 3Bx sin x = 2 cos x − 3x sin x.
This implies −8A + 24B = 2 and −24B = −3. Thus B =
yp = 18 x2 cos x + 18 x3 sin x.
1
8
,A=
1
8
, and
187
188
CHAPTER 4
4.5
HIGHER-ORDER DIFFERENTIAL EQUATIONS
Undetermined Coefficients - Superposition Approach
1. (9D2 − 4)y = (3D − 2)(3D + 2)y = sin x
√
√
2. (D2 − 5)y = (D − 5 )(D + 5 )y = x2 − 2x
3. (D2 − 4D − 12)y = (D − 6)(D + 2)y = x − 6
4. (2D2 − 3D − 2)y = (2D + 1)(D − 2)y = 1
5. (D3 + 10D2 + 25D)y = D(D + 5)2 y = ex
6. (D3 + 4D)y = D(D2 + 4)y = ex cos 2x
7. (D3 + 2D2 − 13D + 10)y = (D − 1)(D − 2)(D + 5)y = xe−x
8. (D3 + 4D2 + 3D)y = D(D + 1)(D + 3)y = x2 cos x − 3x
9. (D4 + 8D)y = D(D + 2)(D2 − 2D + 4)y = 4
10. (D4 − 8D2 + 16)y = (D − 2)2 (D + 2)2 y = (x3 − 2x)e4x
11. D4 y = D4 (10x3 − 2x) = D3 (30x2 − 2) = D2 (60x) = D(60) = 0
12. (2D − 1)y = (2D − 1)4ex/2 = 8Dex/2 − 4ex/2 = 4ex/2 − 4ex/2 = 0
13. (D − 2)(D + 5)(e2x + 3e−5x ) = (D − 2)(2e2x − 15e−5x + 5e2x + 15e−5x ) = (D − 2)7e2x =
14e2x − 14e2x = 0
14. (D2 + 64)(2 cos 8x − 5 sin 8x) = D(−16 sin 8x − 40 cos 8x) + 64(2 cos 8x − 5 sin 8x)
= −128 cos 8x + 320 sin 8x + 128 cos 8x − 320 sin 8x = 0
15. D4 because of x3
16. D5 because of x4
17. D(D − 2) because of 1 and e2x
18. D2 (D − 6)2 because of x and xe6x
19. D2 + 4 because of cos 2x
20. D(D2 + 1) because of 1 and sin x
21. D3 (D2 + 16) because of x2 and sin 4x
22. D2 (D2 + 1)(D2 + 25) because of x, sin x, and cos 5x
23. (D + 1)(D − 1)3 because of e−x and x2 ex
24. D(D − 1)(D − 2) because of 1, ex , and e2x
25. D(D2 − 2D + 5) because of 1 and ex cos 2x
26. (D2 + 2D + 2)(D2 − 4D + 5) because of e−x sin x and e2x cos x
4.5
Undetermined Coefficients - Superposition Approach
27. 1, x, x2 , x3 , x4
28. D2 + 4D = D(D + 4);
1, e−4x
29. e6x , e−3x/2
30. D2 − 9D − 36 = (D − 12)(D + 3);
√
√
31. cos 5 x, sin 5 x
e12x , e−3x
32. D2 − 6D + 10 = D2 − 2(3)D + (32 + 12 );
33. D3 − 10D2 + 25D = D(D − 5)2 ;
e3x cos x, e3x sin x
1, e5x , xe5x
34. 1, x, e5x , e7x
35. Applying D to the differential equation we obtain
D(D2 − 9)y = 0.
Then
y = c1 e3x + c2 e−3x + c3
yc
and yp = A. Substituting yp into the differential equation yields −9A = 54 or A = −6. The
general solution is
y = c1 e3x + c2 e−3x − 6.
36. Applying D to the differential equation we obtain
D(2D2 − 7D + 5)y = 0.
Then
y = c1 e5x/2 + c2 ex + c3
yc
and yp = A. Substituting yp into the differential equation yields 5A = −29 or A = −29/5.
The general solution is
29
.
y = c1 e5x/2 + c2 ex −
5
37. Applying D to the differential equation we obtain
D(D2 + D)y = D2 (D + 1)y = 0.
Then
y = c1 + c2 e−x + c3 x
yc
and yp = Ax. Substituting yp into the differential equation yields A = 3. The general solution
is
y = c1 + c2 e−3x + 3x.
189
190
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
38. Applying D to the differential equation we obtain
D(D3 + 2D2 + D)y = D2 (D + 1)2 y = 0.
Then
y = c1 + c2 e−x + c3 xe−x + c4 x
yc
and yp = Ax. Substituting yp into the differential equation yields A = 10. The general
solution is
y = c1 + c2 e−x + c3 xe−x + 10x.
39. Applying D2 to the differential equation we obtain
D2 (D2 + 4D + 4)y = D2 (D + 2)2 y = 0.
Then
y = c1 e−2x + c2 xe−2x + c3 + c4 x
yc
and yp = Ax+B. Substituting yp into the differential equation yields 4Ax+(4A+4B) = 2x+6.
Equating coefficients gives
4A = 2
4A + 4B = 6.
Then A = 1/2, B = 1, and the general solution is
1
y = c1 e−2x + c2 xe−2x + x + 1.
2
40. Applying D2 to the differential equation we obtain
D2 (D2 + 3D)y = D3 (D + 3)y = 0.
Then
y = c1 + c2 e−3x + c3 x2 + c4 x
yc
Ax2
+ Bx. Substituting yp into the differential equation yields 6Ax + (2A + 3B) =
and yp =
4x − 5. Equating coefficients gives
6A = 4
2A + 3B = −5.
Then A = 2/3, B = −19/9, and the general solution is
2
19
y = c1 + c2 e−3x + x2 − x.
3
9
4.5
Undetermined Coefficients - Superposition Approach
41. Applying D3 to the differential equation we obtain
D3 (D3 + D2 )y = D5 (D + 1)y = 0.
Then
y = c1 + c2 x + c3 e−x + c4 x4 + c5 x3 + c6 x2
yc
and yp = Ax4 + Bx3 + Cx2 . Substituting yp into the differential equation yields
12Ax2 + (24A + 6B)x + (6B + 2C) = 8x2 .
Equating coefficients gives
12A = 8
24A + 6B = 0
6B + 2C = 0.
Then A = 2/3, B = −8/3, C = 8, and the general solution is
2
8
y = c1 + c2 x + c3 e−x + x4 − x3 + 8x2 .
3
3
42. Applying D4 to the differential equation we obtain
D4 (D2 − 2D + 1)y = D4 (D − 1)2 y = 0.
Then
y = c1 ex + c2 xex + c3 x3 + c4 x2 + c5 x + c6
yc
and yp = Ax3 + Bx2 + Cx + E. Substituting yp into the differential equation yields
Ax3 + (B − 6A)x2 + (6A − 4B + C)x + (2B − 2C + E) = x3 + 4x.
Equating coefficients gives
A=1
B − 6A = 0
6A − 4B + C = 4
2B − 2C + E = 0.
Then A = 1, B = 6, C = 22, E = 32 , and the general solution is
y = c1 ex + c2 xex + x3 + 6x2 + 22x + 32.
191
192
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
43. Applying D − 4 to the differential equation we obtain
(D − 4)(D2 − D − 12)y = (D − 4)2 (D + 3)y = 0.
Then
y = c1 e4x + c2 e−3x + c3 xe4x
yc
and yp = Axe4x . Substituting yp into the differential equation yields 7Ae4x = e4x . Equating
coefficients gives A = 1/7. The general solution is
1
y = c1 e4x + c2 e−3x + xe4x .
7
44. Applying D − 6 to the differential equation we obtain
(D − 6)(D2 + 2D + 2)y = 0.
Then
y = e−x (c1 cos x + c2 sin x) + c3 e6x
yc
and yp = Ae6x . Substituting yp into the differential equation yields 50Ae6x = 5e6x . Equating
coefficients gives A = 1/10. The general solution is
y = e−x (c1 cos x + c2 sin x) +
1 6x
e .
10
45. Applying D(D − 1) to the differential equation we obtain
D(D − 1)(D2 − 2D − 3)y = D(D − 1)(D + 1)(D − 3)y = 0.
Then
y = c1 e3x + c2 e−x + c3 ex + c4
yc
and yp = Aex + B. Substituting yp into the differential equation yields −4Aex − 3B = 4ex − 9.
Equating coefficients gives A = −1 and B = 3. The general solution is
y = c1 e3x + c2 e−x − ex + 3.
46. Applying D2 (D + 2) to the differential equation we obtain
D2 (D + 2)(D2 + 6D + 8)y = D2 (D + 2)2 (D + 4)y = 0.
Then
y = c1 e−2x + c2 e−4x + c3 xe−2x + c4 x + c5
yc
4.5
Undetermined Coefficients - Superposition Approach
and yp = Axe−2x + Bx + C. Substituting yp into the differential equation yields
2Ae−2x + 8Bx + (6B + 8C) = 3e−2x + 2x.
Equating coefficients gives
2A = 3
8B = 2
6B + 8C = 0.
Then A = 3/2, B = 1/4, C = −3/16 , and the general solution is
3
1
3
y = c1 e−2x + c2 e−4x + xe−2x + x − .
2
4
16
47. Applying D2 + 1 to the differential equation we obtain
(D2 + 1)(D2 + 25)y = 0.
Then
y = c1 cos 5x + c2 sin 5x + c3 cos x + c4 sin x
yc
and yp = A cos x + B sin x. Substituting yp into the differential equation yields
24A cos x + 24B sin x = 6 sin x.
Equating coefficients gives A = 0 and B = 1/4. The general solution is
y = c1 cos 5x + c2 sin 5x +
1
sin x.
4
48. Applying D(D2 + 1) to the differential equation we obtain
D(D2 + 1)(D2 + 4)y = 0.
Then
y = c1 cos 2x + c2 sin 2x + c3 cos x + c4 sin x + c5
yc
and yp = A cos x + B sin x + C. Substituting yp into the differential equation yields
3A cos x + 3B sin x + 4C = 4 cos x + 3 sin x − 8.
Equating coefficients gives A = 4/3, B = 1, and C = −2. The general solution is
y = c1 cos 2x + c2 sin 2x +
4
cos x + sin x − 2.
3
193
194
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
49. Applying (D − 4)2 to the differential equation we obtain
(D − 4)2 (D2 + 6D + 9)y = (D − 4)2 (D + 3)2 y = 0.
Then
y = c1 e−3x + c2 xe−3x + c3 xe4x + c4 e4x
yc
and yp = Axe4x + Be4x . Substituting yp into the differential equation yields
49Axe4x + (14A + 49B)e4x = −xe4x .
Equating coefficients gives
49A = −1
14A + 49B = 0.
Then A = −1/49, B = 2/343, and the general solution is
y = c1 e−3x + c2 xe−3x −
1 4x
2 4x
xe +
e .
49
343
50. Applying D2 (D − 1)2 to the differential equation we obtain
D2 (D − 1)2 (D2 + 3D − 10)y = D2 (D − 1)2 (D − 2)(D + 5)y = 0.
Then
y = c1 e2x + c2 e−5x + c3 xex + c4 ex + c5 x + c6
yc
and yp =
Axex
+
Bex
+ Cx + E. Substituting yp into the differential equation yields
−6Axex + (5A − 6B)ex − 10Cx + (3C − 10E) = xex + x.
Equating coefficients gives
−6A = 1
5A − 6B = 0
−10C = 1
3C − 10E = 0.
Then A = −1/6, B = −5/36, C = −1/10, E = −3/100, and the general solution is
3
1
5
1
.
y = c1 e2x + c2 e−5x − xex − ex − x −
6
36
10
100
4.5
Undetermined Coefficients - Superposition Approach
51. Applying D(D − 1)3 to the differential equation we obtain
D(D − 1)3 (D2 − 1)y = D(D − 1)4 (D + 1)y = 0.
Then
y = c1 ex + c2 e−x + c3 x3 ex + c4 x2 ex + c5 xex + c6
yc
and yp = Ax3 ex + Bx2 ex + Cxex + E. Substituting yp into the differential equation yields
6Ax2 ex + (6A + 4B)xex + (2B + 2C)ex − E = x2 ex + 5.
Equating coefficients gives
6A = 1
6A + 4B = 0
2B + 2C = 0
−E = 5.
Then A = 1/6, B = −1/4, C = 1/4, E = −5, and the general solution is
1
1
1
y = c1 ex + c2 e−x + x3 ex − x2 ex + xex − 5.
6
4
4
52. Applying (D + 1)3 to the differential equation we obtain
(D + 1)3 (D2 + 2D + 1)y = (D + 1)5 y = 0.
Then
y = c1 e−x + c2 xe−x + c3 x4 e−x + c4 x3 e−x + c5 x2 e−x
yc
and yp = Ax4 e−x + Bx3 e−x + Cx2 e−x . Substituting yp into the differential equation yields
12Ax2 e−x + 6Bxe−x + 2Ce−x = x2 e−x .
Equating coefficients gives A =
1
12 ,
B = 0, and C = 0. The general solution is
y = c1 e−x + c2 xe−x +
1 4 −x
x e .
12
53. Applying D2 − 2D + 2 to the differential equation we obtain
(D2 − 2D + 2)(D2 − 2D + 5)y = 0.
Then
y = ex (c1 cos 2x + c2 sin 2x) + ex (c3 cos x + c4 sin x)
yc
195
196
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
and yp = Aex cos x + Bex sin x. Substituting yp into the differential equation yields
3Aex cos x + 3Bex sin x = ex sin x.
Equating coefficients gives A = 0 and B = 1/3. The general solution is
1
y = ex (c1 cos 2x + c2 sin 2x) + ex sin x.
3
54. Applying D2 − 2D + 10 to the differential equation we obtain
1
1 2
2
2
2
y = (D − 2D + 10) D +
y = 0.
(D − 2D + 10) D + D +
4
2
Then
y = c1 e−x/2 + c2 xe−x/2 + c3 ex cos 3x + c4 ex sin 3x
yc
and yp =
Aex cos 3x
+
Bex sin 3x.
Substituting yp into the differential equation yields
(9B − 27A/4)ex cos 3x − (9A + 27B/4)ex sin 3x = −ex cos 3x + ex sin 3x.
Equating coefficients gives
−
27
A + 9B = −1
4
−9A −
27
B = 1.
4
Then A = −4/225, B = −28/225, and the general solution is
y = c1 e−x/2 + c2 xe−x/2 −
4 x
28 x
e cos 3x −
e sin 3x.
225
225
55. Applying D2 + 25 to the differential equation we obtain
(D2 + 25)(D2 + 25) = (D2 + 25)2 = 0.
Then
y = c1 cos 5x + c2 sin 5x + c3 x cos 5x + c4 x sin 5x
yc
and yp = Ax cos 5x + Bx sin 5x. Substituting yp into the differential equation yields
10B cos 5x − 10A sin 5x = 20 sin 5x.
Equating coefficients gives A = −2 and B = 0. The general solution is
y = c1 cos 5x + c2 sin 5x − 2x cos 5x.
4.5
Undetermined Coefficients - Superposition Approach
56. Applying D2 + 1 to the differential equation we obtain
(D2 + 1)(D2 + 1) = (D2 + 1)2 = 0.
Then
y = c1 cos x + c2 sin x + c3 x cos x + c4 x sin x
yc
and yp = Ax cos x + Bx sin x. Substituting yp into the differential equation yields
2B cos x − 2A sin x = 4 cos x − sin x.
Equating coefficients gives A = 1/2 and B = 2. The general solution is
1
y = c1 cos x + c2 sin x + x cos x + 2x sin x.
2
57. Applying (D2 + 1)2 to the differential equation we obtain
(D2 + 1)2 (D2 + D + 1) = 0.
Then
−x/2
y=e
√ 3
3
x + c2 sin
x + c3 cos x + c4 sin x + c5 x cos x + c6 x sin x
c1 cos
2
2
√
yc
and yp = A cos x + B sin x + Cx cos x + Ex sin x. Substituting yp into the differential equation
yields
(B + C + 2E) cos x + Ex cos x + (−A − 2C + E) sin x − Cx sin x = x sin x.
Equating coefficients gives
B + C + 2E = 0
E=0
−A − 2C + E = 0
−C = 1.
Then A = 2, B = 1, C = −1, and E = 0, and the general solution is
√
√ 3
3
−x/2
c1 cos
x + c2 sin
x + 2 cos x + sin x − x cos x.
y=e
2
2
197
198
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
58. Writing cos2 x = 12 (1 + cos 2x) and applying D(D2 + 4) to the differential equation we obtain
D(D2 + 4)(D2 + 4) = D(D2 + 4)2 = 0.
Then
y = c1 cos 2x + c2 sin 2x + c3 x cos 2x + c4 x sin 2x + c5
yc
and yp = Ax cos 2x + Bx sin 2x + C. Substituting yp into the differential equation yields
−4A sin 2x + 4B cos 2x + 4C =
1 1
+ cos 2x.
2 2
Equating coefficients gives A = 0, B = 1/8, and C = 1/8. The general solution is
1
1
y = c1 cos 2x + c2 sin 2x + x sin 2x + .
8
8
59. Applying D3 to the differential equation we obtain
D3 (D3 + 8D2 ) = D5 (D + 8) = 0.
Then
y = c1 + c2 x + c3 e−8x + c4 x2 + c5 x3 + c6 x4
yc
and yp = Ax2 + Bx3 + Cx4 . Substituting yp into the differential equation yields
16A + 6B + (48B + 24C)x + 96Cx2 = 2 + 9x − 6x2 .
Equating coefficients gives
16A + 6B = 2
48B + 24C = 9
96C = −6.
Then A = 11/256, B = 7/32, and C = −1/16, and the general solution is
y = c1 + c2 x + c3 e−8x +
11 2
7
1
x + x3 − x4 .
256
32
16
60. Applying D(D − 1)2 (D + 1) to the differential equation we obtain
D(D − 1)2 (D + 1)(D3 − D2 + D − 1) = D(D − 1)3 (D + 1)(D2 + 1) = 0.
Then
y = c1 ex + c2 cos x + c3 sin x + c4 + c5 e−x + c6 xex + c7 x2 ex
yc
4.5
Undetermined Coefficients - Superposition Approach
and yp = A + Be−x + Cxex + Ex2 ex . Substituting yp into the differential equation yields
4Exex + (2C + 4E)ex − 4Be−x − A = xex − e−x + 7.
Equating coefficients gives
4E = 1
2C + 4E = 0
−4B = −1
−A = 7.
Then A = −7, B = 1/4, C = −1/2, and E = 1/4, and the general solution is
1
1
1
y = c1 ex + c2 cos x + c3 sin x − 7 + e−x − xex + x2 ex .
4
2
4
61. Applying D2 (D − 1) to the differential equation we obtain
D2 (D − 1)(D3 − 3D2 + 3D − 1) = D2 (D − 1)4 = 0.
Then
y = c1 ex + c2 xex + c3 x2 ex + c4 + c5 x + c6 x3 ex
yc
and yp = A + Bx +
Cx3 ex .
Substituting yp into the differential equation yields
(−A + 3B) − Bx + 6Cex = 16 − x + ex .
Equating coefficients gives
−A + 3B = 16
−B = −1
6C = 1.
Then A = −13, B = 1, and C = 1/6, and the general solution is
1
y = c1 ex + c2 xex + c3 x2 ex − 13 + x + x3 ex .
6
62. Writing (ex +e−x )2 = 2+e2x +e−2x and applying D(D −2)(D + 2) to the differential equation
we obtain
D(D − 2)(D + 2)(2D3 − 3D2 − 3D + 2) = D(D − 2)2 (D + 2)(D + 1)(2D − 1) = 0.
Then
y = c1 e−x + c2 e2x + c3 ex/2 + c4 + c5 xe2x + c6 e−2x
yc
199
200
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
and yp = A + Bxe2x + Ce−2x . Substituting yp into the differential equation yields
2A + 9Be2x − 20Ce−2x = 2 + e2x + e−2x .
Equating coefficients gives A = 1, B = 1/9, and C = −1/20. The general solution is
1
1
y = c1 e−x + c2 e2x + c3 ex/2 + 1 + xe2x − e−2x .
9
20
63. Applying D(D − 1) to the differential equation we obtain
D(D − 1)(D4 − 2D3 + D2 ) = D3 (D − 1)3 = 0.
Then
y = c1 + c2 x + c3 ex + c4 xex + c5 x2 + c6 x2 ex
yc
and yp = Ax2 +Bx2 ex . Substituting yp into the differential equation yields 2A+2Bex = 1+ex .
Equating coefficients gives A = 1/2 and B = 1/2. The general solution is
1
1
y = c1 + c2 x + c3 ex + c4 xex + x2 + x2 ex .
2
2
64. Applying D3 (D − 2) to the differential equation we obtain
D3 (D − 2)(D4 − 4D2 ) = D5 (D − 2)2 (D + 2) = 0.
Then
y = c1 + c2 x + c3 e2x + c4 e−2x + c5 x2 + c6 x3 + c7 x4 + c8 xe2x
yc
and yp = Ax2 + Bx3 + Cx4 + Exe2x . Substituting yp into the differential equation yields
(−8A + 24C) − 24Bx − 48Cx2 + 16Ee2x = 5x2 − e2x .
Equating coefficients gives
−8A + 24C = 0
−24B = 0
−48C = 5
16E = −1.
Then A = −5/16, B = 0, C = −5/48, and E = −1/16, and the general solution is
y = c1 + c2 x + c3 e2x + c4 e−2x −
5 2
5
1
x − x4 − xe2x .
16
48
16
4.5
Undetermined Coefficients - Superposition Approach
65. The complementary function is yc = c1 e8x + c2 e−8x . Using D to annihilate 16 we find yp = A.
Substituting yp into the differential equation we obtain −64A = 16. Thus A = −1/4 and
y = c1 e8x + c2 e−8x −
1
4
y = 8c1 e8x − 8c2 e−8x .
The initial conditions imply
c1 + c2 =
5
4
8c1 − 8c2 = 0.
Thus c1 = c2 = 5/8 and
5
1
5
y = e8x + e−8x − .
8
8
4
66. The complementary function is yc = c1 + c2 e−x . Using D2 to annihilate x we find
yp = Ax + Bx2 . Substituting yp into the differential equation we obtain (A + 2B) + 2Bx = x.
Thus A = −1 and B = 1/2, and
1
y = c1 + c2 e−x − x + x2
2
y = −c2 e−x − 1 + x.
The initial conditions imply
c1 + c2 = 1
−c2 = 1.
Thus c1 = 2 and c2 = −1, and
1
y = 2 − e−x − x + x2 .
2
67. The complementary function is yc = c1 + c2 e5x . Using D2 to annihilate x − 2 we find
yp = Ax+Bx2 . Substituting yp into the differential equation we obtain (−5A+2B)−10Bx =
−2 + x. Thus A = 9/25 and B = −1/10, and
y = c1 + c2 e5x +
y = 5c2 e5x +
1
9
x − x2
25
10
1
9
− x.
25 5
The initial conditions imply
c1 + c 2 = 0
c2 =
41
.
125
201
202
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
Thus c1 = −41/125 and c2 = 41/125, and
y=−
41 5x
1
41
9
+
e + x − x2 .
125 125
25
10
68. The complementary function is yc = c1 ex + c2 e−6x . Using D − 2 to annihilate 10e2x we find
yp = Ae2x . Substituting yp into the differential equation we obtain 8Ae2x = 10e2x . Thus
A = 5/4 and
5
y = c1 ex + c2 e−6x + e2x
4
5
y = c1 ex − 6c2 e−6x + e2x .
2
The initial conditions imply
c 1 + c2 = −
1
4
3
c1 − 6c2 = − .
2
Thus c1 = −3/7 and c2 = 5/28, and
3
5
5
y = − ex + e−6x + e2x
7
28
4
69. The complementary function is yc = c1 cos x + c2 sin x. Using (D2 + 1)(D2 + 4) to annihilate
8 cos 2x−4 sin x we find yp = Ax cos x+Bx sin x+C cos 2x+E sin 2x. Substituting yp into the
differential equation we obtain 2B cos x − 3C cos 2x − 2A sin x − 3E sin 2x = 8 cos 2x − 4 sin x.
Thus A = 2, B = 0, C = −8/3, and E = 0, and
y = c1 cos x + c2 sin x + 2x cos x −
8
cos 2x
3
y = −c1 sin x + c2 cos x + 2 cos x − 2x sin x +
16
sin 2x.
3
The initial conditions imply
8
= −1
3
−c1 − π = 0.
c2 +
Thus c1 = −π and c2 = −11/3, and
y = −π cos x −
8
11
sin x + 2x cos x − cos 2x.
3
3
4.5
Undetermined Coefficients - Superposition Approach
70. The complementary function is yc = c1 + c2 ex + c3 xex . Using D(D − 1)2 to annihilate xex + 5
we find yp = Ax + Bx2 ex + Cx3 ex . Substituting yp into the differential equation we obtain
A + (2B + 6C)ex + 6Cxex = xex + 5. Thus A = 5, B = −1/2, and C = 1/6, and
1
1
y = c1 + c2 ex + c3 xex + 5x − x2 ex + x3 ex
2
6
1
y = c2 ex + c3 (xex + ex ) + 5 − xex + x3 ex
6
1
1
y = c2 ex + c3 (xex + 2ex ) − ex − xex + x2 ex + x3 ex .
2
6
The initial conditions imply
c1 + c2 = 2
c2 + c3 + 5 = 2
c2 + 2c3 − 1 = −1.
Thus c1 = 8, c2 = −6, and c3 = 3, and
1
1
y = 8 − 6ex + 3xex + 5x − x2 ex + x3 ex .
2
6
71. The complementary function is yc = e2x (c1 cos 2x + c2 sin 2x). Using D4 to annihilate x3 we
find yp = A + Bx + Cx2 + Ex3 . Substituting yp into the differential equation we obtain
(8A − 4B + 2C) + (8B − 8C + 6E)x + (8C − 12E)x2 + 8Ex3 = x3 . Thus A = 0, B = 3/32,
C = 3/16, and E = 1/8, and
y = e2x (c1 cos 2x + c2 sin 2x) +
3
3
1
x + x2 + x3
32
16
8
y = e2x [c1 (2 cos 2x − 2 sin 2x) + c2 (2 cos 2x + 2 sin 2x)] +
3
3
3
+ x + x2 .
32 8
8
The initial conditions imply
c1 = 2
3
= 4.
2c1 + 2c2 +
32
Thus c1 = 2, c2 = −3/64, and
y = e2x (2 cos 2x −
3
3
3
1
sin 2x) + x + x2 + x3 .
64
32
16
8
72. The complementary function is yc = c1 + c2 x + c3 x2 + c4 ex . Using D2 (D − 1) to annihilate
x + ex we find yp = Ax3 + Bx4 + Cxex . Substituting yp into the differential equation we
203
204
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
obtain (−6A + 24B) − 24Bx + Cex = x + ex . Thus A = −1/6, B = −1/24, and C = 1, and
1
1
y = c1 + c2 x + c3 x2 + c4 ex − x3 − x4 + xex
6
24
1
1
y = c2 + 2c3 x + c4 ex − x2 − x3 + ex + xex
2
6
1
y = 2c3 + c4 ex − x − x2 + 2ex + xex .
2
y = c4 ex − 1 − x + 3ex + xex
The initial conditions imply
c1 + c4 = 0
c 2 + c4 + 1 = 0
2c3 + c4 + 2 = 0
2 + c4 = 0.
Thus c1 = 2, c2 = 1, c3 = 0, and c4 = −2, and
1
1
y = 2 + x − 2ex − x3 − x4 + xex .
6
24
73. To see in this case that the factors of L do not commute consider the operators (xD−1)(D+4)
and (D + 4)(xD − 1). Applying the operators to the function x we find
(xD − 1)(D + 4)x = (xD2 + 4xD − D − 4)x
= xD2 x + 4xDx − Dx − 4x
= x(0) + 4x(1) − 1 − 4x = −1
and
(D + 4)(xD − 1)x = (D + 4)(xDx − x)
= (D + 4)(x · 1 − x) = 0.
Thus, the operators are not the same.
4.6 Variation of Parameters
4.6
Variation of Parameters
The particular solution, yp = u1 y1 + u2 y2 , in the following problems can take on a variety of
forms, especially where trigonometric functions are involved. The validity of a particular form
can best be checked by substituting it back into the differential equation.
1. The auxiliary equation is m2 + 1 = 0, so yc = c1 cos x + c2 sin x and
cos x sin x =1
W =
− sin x cos x
Identifying f (x) = sec x we obtain
sin x sec x
= − tan x
1
cos x sec x
= 1.
u2 =
1
u1 = −
Then u1 = ln | cos x|, u2 = x, and
y = c1 cos x + c2 sin x + cos x ln | cos x| + x sin x.
2. The auxiliary equation is m2 + 1 = 0, so yc = c1 cos x + c2 sin x and
cos x sin x =1
W =
− sin x cos x
Identifying f (x) = tan x we obtain
u1 = − sin x tan x =
cos2 x − 1
= cos x − sec x
cos x
u2 = sin x.
Then u1 = sin x − ln | sec x + tan x|, u2 = − cos x, and
y = c1 cos x + c2 sin x + cos x (sin x − ln | sec x + tan x|) − cos x sin x
= c1 cos x + c2 sin x − cos x ln | sec x + tan x|.
3. The auxiliary equation is m2 + 1 = 0, so yc = c1 cos x + c2 sin x and
cos x sin x =1
W = − sin x cos x
Identifying f (x) = sin x we obtain
u1 = − sin2 x
u2 = cos x sin x.
205
206
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
Then
u1 =
1
1
1
1
sin 2x − x = sin x cos x − x
4
2
2
2
1
u2 = − cos2 x.
2
and
y = c1 cos x + c2 sin x +
1
1
1
sin x cos2 x − x cos x − cos2 x sin x
2
2
2
1
= c1 cos x + c2 sin x − x cos x.
2
4. The auxiliary equation is m2 + 1 = 0, so yc = c1 cos x + c2 sin x and
cos x sin x =1
W = − sin x cos x
Identifying f (x) = sec x tan x we obtain
u1 = − sin x(sec x tan x) = − tan2 x = 1 − sec2 x
u2 = cos x(sec x tan x) = tan x.
Then u1 = x − tan x, u2 = − ln | cos x|, and
y = c1 cos x + c2 sin x + x cos x − sin x − sin x ln | cos x|
= c1 cos x + c3 sin x + x cos x − sin x ln | cos x|.
5. The auxiliary equation is m2 + 1 = 0, so yc = c1 cos x + c2 sin x and
cos x sin x =1
W =
− sin x cos x
Identifying f (x) = cos2 x we obtain
u1 = − sin x cos2 x
u2 = cos3 x = cos x 1 − sin2 x .
Then u1 =
1
3
cos3 x, u2 = sin x − 13 sin3 x, and
1
1
cos4 x + sin2 x − sin4 x
3
3
1 2
cos x + sin2 x cos2 x − sin2 x + sin2 x
= c1 cos x + c2 sin x +
3
y = c1 cos x + c2 sin x +
= c1 cos x + c2 sin x +
1
2
cos2 x + sin2 x
3
3
= c1 cos x + c2 sin x +
1 1
+ sin2 x.
3 3
4.6 Variation of Parameters
6. The auxiliary equation is m2 + 1 = 0, so yc = c1 cos x + c2 sin x and
cos x sin x =1
W =
− sin x cos x
Identifying f (x) = sec2 x we obtain
u1 = −
sin x
cos2 x
u2 = sec x.
Then
u1 = −
1
= − sec x
cos x
u2 = ln | sec x + tan x|
and
y = c1 cos x + c2 sin x − cos x sec x + sin x ln | sec x + tan x|
= c1 cos x + c2 sin x − 1 + sin x ln | sec x + tan x|.
7. The auxiliary equation is m2 − 1 = 0, so yc = c1 ex + c2 e−x and
x
e
e−x = −2
W = x
e −e−x Identifying f (x) = cosh x = 12 (e−x + ex ) we obtain
1
1
u1 = e−2x +
4
4
1 1
u2 = − − e2x .
4 4
Then
1
1
u1 = − e−2x + x
8
4
1
1
u2 = − e2x − x
8
4
and
1
1
1
1
y = c1 ex + c2 e−x − e−x + xex − ex − xe−x
8
4
8
4
1
= c3 ex + c4 e−x + x(ex − e−x )
4
1
= c3 ex + c4 e−x + x sinh x.
2
207
208
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
8. The auxiliary equation is m2 − 1 = 0, so yc = c1 ex + c2 e−x and
x
−x e
e
= −2
W = x
e −e−x Identifying f (x) = sinh 2x we obtain
1
1
u1 = − e−3x + ex u2
4
4
1
1
= e−x − e3x .
4
4
Then
u1 =
1 −3x 1 x
e
+ e
12
4
1
1
u2 = − e−x − e3x .
4
12
and
1 −2x 1 2x 1 −2x
1
e
+ e − e
− e2x
12
4
4
12
1 2x
= c1 ex + c2 e−x +
e − e−2x
6
y = c1 ex + c2 e−x +
= c1 ex + c2 e−x +
1
sinh 2x.
3
9. The auxiliary equation is m2 − 9 = 0, so yc = c1 e3x + c2 e−3x and
3x
−3x e
e
= −6
W = 3x
3e
−3e−3x Identifying f (x) = 9x/e3x we obtain u1 = 32 xe−6x and u2 = − 32 x. Then
u1 = −
1 −6x 1 −6x
e
− xe ,
24
4
3
u2 = − x2
4
and
y = c1 e3x + c2 e−3x −
1 −3x 1 −3x 3 2 −3x
e
− xe
− x e
24
4
4
1
= c1 e3x + c3 e−3x − xe−3x (1 + 3x).
4
4.6 Variation of Parameters
10. The auxiliary equation is 4m2 − 1 = 0, so yc = c1 e−x/2 + c2 ex/2 and
−x/2
e
ex/2 W = 1
1 x/2 = 1
− e−x/2
e 2
2
3
3
1
1
Dividing by 4 we now identify f (x) = ex/2 + and obtain u1 = − ex/2 − ex and u2 =
4
4
4
4
3 −x/2 1
e
+ . Then
4
4
3
1
u1 = − ex/2 − ex ,
2
4
3
1
u2 = − e−x/2 + x
2
4
and
1
3 1
3
y = c1 e−x/2 + c2 ex/2 − ex/2 − + xex/2 −
4
2 4
2
1
= c1 e−x/2 + c3 ex/2 + xex/2 − 3
4
11. The auxiliary equation is m2 + 3m + 2 = (m + 1)(m + 2) = 0, so yc = c1 e−x + c2 e−2x and
−x
e
e−2x W = −x
= −e−3x
−e
−2e−2x Identifying f (x) = 1/(1 + ex ) we obtain
u1 =
ex
1 + ex
u2 = −
e2x
ex
=
− ex .
1 + ex
1 + ex
Then u1 = ln (1 + ex ), u2 = ln (1 + ex ) − ex , and
y = c1 e−x + c2 e−2x + e−x ln (1 + ex ) + e−2x ln (1 + ex ) − e−x
= c3 e−x + c2 e−2x + (1 + e−x )e−x ln (1 + ex ).
12. The auxiliary equation is m2 − 2m + 1 = (m − 1)2 = 0, so yc = c1 ex + c2 xex and
x
e
xex W = x
= e2x
e xex + ex Identifying f (x) = ex / 1 + x2 we obtain
u1 = −
u2 =
xex ex
x
=−
e2x (1 + x2 )
1 + x2
1
ex ex
=
.
2x
2
e (1 + x )
1 + x2
209
210
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
Then u1 = − 12 ln 1 + x2 , u2 = tan−1 x, and
1
y = c1 ex + c2 xex − ex ln 1 + x2 + xex tan−1 x.
2
13. The auxiliary equation is m2 + 3m + 2 = (m + 1)(m + 2) = 0, so yc = c1 e−x + c2 e−2x and
−x
e
e−2x W =
= −e−3x
−e−x = 2e−2 Identifying f (x) = sin ex we obtain
u1 =
e−2x sin ex
= ex sin ex
e−3x
u2 =
e−x sin ex
= −e2x sin ex .
−e−3x
Then u1 = − cos ex , u2 = ex cos x − sin ex , and
y = c1 e−x + c2 e−2x − e−x cos ex + e−x cos ex − e−2x sin ex
= c1 e−x + c2 e−2x − e−2x sin ex .
14. The auxiliary equation is m2 − 2m + 1 = (m − 1)2 = 0, so yc = c1 et + c2 tet and
t
e
tet W =
= e2t
et tet + et Identifying f (t) = et tan−1 t we obtain
u1 = −
u2 =
tet et tan−1 t
= −t tan−1 t
e2t
et et tan−1 t
= tan−1 t.
e2t
Then
1 + t2
t
tan−1 t +
2
2
1 u2 = t tan−1 t − ln 1 + t2
2
u1 = −
and
1 + t2
t
tan−1 t +
y = c1 e + c2 te + −
2
2
t
t
1 −1
2
tet
e + t tan t − ln 1 + t
2
t
1 = c1 et + c3 tet + et t2 − 1 tan−1 t − ln 1 + t2 .
2
4.6 Variation of Parameters
15. The auxiliary equation is m2 + 2m + 1 = (m + 1)2 = 0, so yc = c1 e−t + c2 te−t and
−t
e
te−t
W =
= e−2t
−e−t −te−t + e−t Identifying f (t) = e−t ln t we obtain
u1 = −
u2 =
te−t e−t ln t
= −t ln t
e−2t
e−t e−t ln t
= ln t.
e−2t
Then
1
1
u1 = − t2 ln t + t2
2
4
u2 = t ln t − t
and
1
1
y = c1 e−t + c2 te−t − t2 e−t ln t + t2 e−t + t2 e−t ln t − t2 e−t
2
4
1
3
= c1 e−t + c2 te−t + t2 e−t ln t − t2 e−t .
2
4
16. The auxiliary equation is 2m2 + m = m(2m + 1) = 0, so yc = c1 + c2 e−x/2 and
−x/2 1
e
1
W =
= − e−x/2
1
0 − e−x/2 2
2
Dividing by 2 we identify f (t) = 3x and obtain
u1 = 6x
u2 = −6xex/2 .
Then
u1 = 3x2
u2 = −12xex/2 + 24ex/2
and
y = c1 + c2 e−x/2 + 3x2 − 12x + 24
= c3 + c2 e−x/2 + 3x2 − 12x
211
212
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
17. The auxiliary equation is 3m2 − 6m + 6 = 0, so yc = ex (c1 cos x + c2 sin x) and
x cos x
x sin x
e
e
W =
= e2x
ex cos x − ex sin x ex cos x + ex sin x
Identifying f (x) = 13 ex sec x we obtain
u1 = −
u2 =
Then u1 =
1
3
(ex sin x)(ex sec x)/3
1
= − tan x
e2x
3
(ex cos x)(ex sec x)/3
1
= .
2x
e
3
ln |cos x|, u2 = 13 x, and
y = c1 ex cos x + c2 ex sin x +
1
1
ln |cos x|ex cos x + xex sin x.
3
3
18. The auxiliary equation is 4m2 − 4m + 1 = (2m − 1)2 = 0, so yc = c1 ex/2 + c2 xex/2 and
x/2
e
W =
1 ex/2
2
= ex
1
x/2
x/2
xe + e xex/2
2
√
Identifying f (x) = 14 ex/2 1 − x2 we obtain
√
xex/2 ex/2 1 − x2
1 =−
= − x 1 − x2
x
4e
4
√
ex/2 ex/2 1 − x2
1
u2 =
=
1 − x2 .
4ex
4
u1
To find u1 and u2 we use the substitution v = 1 − x2 and the trig substitution x = sin θ,
respectively:
3/2
1 1 − x2
12
x
1
u2 =
1 − x2 + sin−1 x.
8
8
u1 =
Thus
y = c1 ex/2 + c2 xex/2 +
3/2 1 2 x/2 1 x/2 1
1 − x2
e
+ x e
1 − x2 + xex/2 sin−1 x.
12
8
8
4.6 Variation of Parameters
19. The auxiliary equation is 4m2 − 1 = (2m − 1)(2m + 1) = 0, so yc = c1 ex/2 + c2 e−x/2 and
x/2
e
e−x/2 W =
= −1
1 ex/2 − 1 e−x/2 2
2
Identifying f (x) = xex/2 /4 we obtain u1 = x/4 and u2 = −xex /4. Then u1 = x2 /8 and
u2 = −xex /4 + ex /4. Thus
1
1
1
y = c1 ex/2 + c2 e−x/2 + x2 ex/2 − xex/2 + ex/2
8
4
4
1
1
= c3 ex/2 + c2 e−x/2 + x2 ex/2 − xex/2
8
4
and
1
1
1
1
1
y = c3 ex/2 − c2 e−x/2 + x2 ex/2 + xex/2 − ex/2 .
2
2
16
8
4
The initial conditions imply
c 3 + c2
=0
1
1
1
c3 − c2 − = 0
2
2
4
Thus c3 = 3/4 and c2 = 1/4, and
3
1
1
1
y = ex/2 + e−x/2 + x2 ex/2 − xex/2 .
4
4
8
4
20. The auxiliary equation is 2m2 + m − 1 = (2m − 1)(m + 1) = 0, so yc = c1 ex/2 + c2 e−x and
x/2
−x e
e
3
W =
= − e−x/2
1
ex/2 −e−x 2
2
Identifying f (x) = (x + 1)/2 we obtain
1
u1 = e−x/2 (x + 1)
3
1
u2 = − ex (x + 1).
3
Then
−x/2
u1 = −e
2
x−2
3
1
u2 = − xex .
3
Thus
y = c1 ex/2 + c2 e−x − x − 2
213
214
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
and
1
y = c1 ex/2 − c2 e−x − 1.
2
The initial conditions imply
c 1 − c2 − 2 = 1
1
c1 − c2 − 1 = 0.
2
Thus c1 = 8/3 and c2 = 1/3, and
8
1
y = ex/2 + e−x − x − 2.
3
3
21. The auxiliary equation is m2 + 2m − 8 = (m − 2)(m + 4) = 0, so yc = c1 e2x + c2 e−4x and
2x
e
e−4x W =
= −6e−2x
2e2x −4e−4x Identifying f (x) = 2e−2x − e−x we obtain
1
1
u1 = e−4x − e−3x
3
6
1
1
u2 = e3x − e2x .
6
3
Then
u1 = −
u2 =
1 −4x
1
+ e−3x
e
12
18
1 3x 1 2x
e − e .
18
6
Thus
y = c1 e2x + c2 e−4x −
1 −2x
1
1
1
e
+ e−x + e−x − e−2x
12
18
18
6
1
1
= c1 e2x + c2 e−4x − e−2x + e−x
4
9
and
1
1
y = 2c1 e2x − 4c2 e−4x + e−2x − e−x .
2
9
The initial conditions imply
c1 + c 2 −
5
=1
36
2c1 − 4c2 +
7
=0
18
Thus c1 = 25/36 and c2 = 4/9, and
y=
25 2x 4 −4x 1 −2x 1 −x
e + e
− e
+ e .
36
9
4
9
4.6 Variation of Parameters
22. The auxiliary equation is m2 − 4m + 4 = (m − 2)2 = 0, so yc = c1 e2x + c2 xe2x and
2x
e
xe2x
W =
= e4x
2e2x 2xe2x + e2x Identifying f (x) = 12x2 − 6x e2x we obtain
u1 = 6x2 − 12x3
u2 = 12x2 − 6x.
Then
u1 = 2x3 − 3x4
u2 = 4x3 − 3x2 .
Thus
y = c1 e2x + c2 xe2x + 2x3 − 3x4 e2x + 4x3 − 3x2 xe2x
= c1 e2x + c2 xe2x + e2x x4 − x3
and
y = 2c1 e2x + c2 2xe2x + e2x + e2x 4x3 − 3x2 + 2e2x x4 − x3 .
The initial conditions imply
c1
=0
2c1 + c2 = 0
Thus c1 = 1 and c2 = −2, and
y = e2x − 2xe2x + e2x x4 − x3 = e2x x4 − x3 − 2x + 1 .
23. The auxiliary equation is m2 + 1 = 0, so yc = c1 cos x + c2 sin x and
cos x sin x =1
W =
− sin x cos x
Identifying f (x) = ex we obtain u1 = −ex sin x and u2 = ex cos x. Then
ˆ x
2
et sin t dt
u1 = −
2
2
2
x0
ˆ
x
u2 =
2
et cos t dt
x0
ˆ
and
y = c1 cos x + c2 sin x − cos x
x
t2
ˆ
x
e sin t dt + sin x
x0
x0
2
et cos t dt.
215
216
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
24. The auxiliary equation is m2 − 4 = 0, so yc = c1 e2x + c2 e−2x and
2x
−2x e
e
W =
= −4
2e2x −2e−2x Identifying f (x) = e2x /x we obtain u1 = 1/4x and u2 = −e4x /4x. Then
1
ln |x|,
4
ˆ
1 x e4t
u2 = −
dt
4 x0 t
u1 =
and
y = c1 e2x + c2 e−2x +
1
4
ˆ
e2x ln |x| − e−2x
x
x0
e4t
dt ,
t
x0 > 0.
25. The auxiliary equation is m2 + m − 2 = 0, so yc = c1 e−2x + c2 ex and
−2x
e
ex W =
= 3e−x
−2e−2x ex 1
1
Identifying f (x) = ln x we obtain u1 = − e2x ln x and u2 = e−x ln x. Then
3
3
ˆ
1 x 2t
e ln t dt,
u1 = −
3 x0
ˆ
1 x −t
e ln t dt
u2 =
3 x0
and
−2x
y = c1 e
1
+ c2 e − e−2x
3
x
ˆ
x
x0
1
e ln t dt + ex
3
2t
ˆ
x
e−t ln t dt,
x0 > 0.
x0
26. The auxiliary equation is 2m2 + 2m + 1 = 0, so yc = e−x/2 [c1 cos (x/2) + c2 sin (x/2)] and
x
x
−x/2
−x/2
cos
sin
e
e
1
2
2
W = 1
= e−x
x
1
x
1
x
1
x
e−x/2 cos − e−x/2 sin
−x/2
x/2
e
cos − e sin 2
2
2 2
2 2
2 2
2
√
Identifying f (x) = 2 x we obtain
√
√
e−x/2 sin (x/2)2 x
x
=−
= −4ex/2 x sin
1 −x/2
2
2e
√
√
e−x/2 cos (x/2)2 x
x
= 4ex/2 x cos .
u2 = −
1 −x/2
2
2e
u1
4.6 Variation of Parameters
Then
ˆ
u1 = −4
ˆ
x0
x
u2 = 4
x0
and
−x/2
y=e
x
√
t
et/2 t sin dt
2
√
t
et/2 t cos dt
2
ˆ
x
x
x x t/2 √
t
−x/2
cos
e
t sin dt
c1 cos + c2 sin
− 4e
2
2
2 x0
2
ˆ
x x t/2 √
t
−x/2
+ 4e
sin
e
t cos dt.
2 x0
2
27. Write the equation in the form
1 1
y + y + 1 − 2 y = x−1/2
x
4x
and identify f (x) = x−1/2 . From y1 = x−1/2 cos x and y2 = x−1/2 sin x we compute
−1/2 cos x
−1/2 sin x
x
x
1
=
W =
1
1
−1/2
−3/2
−1/2
−3/2
−x
sin x − x
cos x x
cos x − x
sin x x
2
2
Now
u1 = − sin x so
u1 = cos x,
and
u2 = cos x so
u2 = sin x.
Thus a particular solution is
yp = x−1/2 cos2 x + x−1/2 sin2 x,
and the general solution is
y = c1 x−1/2 cos x + c2 x−1/2 sin x + x−1/2 cos2 x + x−1/2 sin2 x
= c1 x−1/2 cos x + c2 x−1/2 sin x + x−1/2 .
28. Write the equation in the form
y +
1 1
sec (ln x)
y + 2y =
x
x
x2
and identify f (x) = sec (ln x)/x2 . From y1 = cos (ln x) and y2 = sin (ln x) we compute
cos (ln x)
sin (ln x) 1
W = sin (ln x) cos (ln x) =
x
−
x
x
217
218
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
Now
u1 = −
tan (ln x)
x
so
u1 = ln | cos (ln x)|,
1
x
so
u2 = ln x.
and
u2 =
Thus, a particular solution is
yp = cos (ln x) ln | cos (ln x)| + (ln x) sin (ln x),
and the general solution is
y = c1 cos (ln x) + c2 sin (ln x) + cos (ln x) ln | cos (ln x)| + (ln x) sin (ln x).
29. The auxiliary equation is m3 + m = m(m2 + 1) = 0, so yc = c1 + c2 cos x + c3 sin x and
1 cos x
sin x W = 0 − sin x cos x = 1
0 − cos x − sin x
Identifying f (x) = tan x we obtain
0
cos x
sin x − sin x cos x = tan x
u1 = W1 = 0
tan x − cos x − sin x
1
0
sin
x
0
cos x = − sin x
u 2 = W 2 = 0
0 tan x − sin x
1 cos x
0 cos2 x − 1
0 = − sin x tan x =
= cos x − sec x
u3 = W3 = 0 − sin x
cos x
0 − cos x tan x
Then
u1 = − ln | cos x|
u2 = cos x
u3 = sin x − ln | sec x + tan x|
and
y = c1 + c2 cos x + c3 sin x − ln | cos x| + cos2 x
+ sin2 x − sin x ln | sec x + tan x|
= c4 + c2 cos x + c3 sin x − ln | cos x| − sin x ln | sec x + tan x|
for −π/2 < x < π/2.
4.6 Variation of Parameters
30. The auxiliary equation is m3 + 4m = m m2 + 4 = 0, so yc = c1 + c2 cos 2x + c3 sin 2x and
1
cos
2x
sin
2x
W = 0 −2 sin 2x 2 cos 2x = 8
0 −4 cos 2x −4 sin 2x
Identifying f (x) = sec 2x we obtain
0
cos 2x
sin 2x 1
1
1
−2 sin 2x 2 cos 2x = sec 2x
u1 = W1 = 0
8
8
4
sec 2x −4 cos 2x −4 sin 2x
1
0
sin
2x
1
1 1
0
2 cos 2x = −
u 2 = W 2 = 0
8
8
4
0 sec 2x −4 sin 2x
1
cos 2x
0 1
1
1
0 = − tan 2x
u3 = W3 = 0 −2 sin 2x
8
8
4
0 −4 cos 2x sec 2x
Then
u1 =
1
ln | sec 2x + tan 2x|
8
1
u2 = − x
4
u3 =
1
ln | cos 2x|
8
and
y = c1 + c2 cos 2x + c3 sin 2x +
1
1
1
ln | sec 2x + tan 2x| − x cos 2x + sin 2x ln | cos 2x|
8
4
8
for −π/4 < x < π/4.
31. The auxiliary equation is m3 − 2m2 − m + 2 = 0 so yc = c1 ex + c2 e−x + c3 e2x
1
e−x e2x ex
x
= −6e2x
W = e −e−x 2e2x
ex e−x 4e2x
Identifying f (x) = e4x we obtain
1
1
W1 =
u1 =
2x
−6e
−6e2x
u2 =
1
1
W2 =
2x
−6e
−6e2x
u3 =
1
1
W3 =
−6e2x
−6e2x
0
e2x e−x
0 −e−x 2e2x = 1 · 3e5x = − 1 e3x
2x
2
e4x e−x 4e2x −6e
x
e
e2x x 0
1
1
e
0 2e2x =
· −e7x = e5x
2x
6
ex e4x 4e2x −6e
x
−x
e
0 x e
e −e−x 0 = 1 · −2e4x = 1 e4x
−6e2x
3
ex e−x e4x 219
220
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
Then
1
u1 = − e3x
6
u2 =
1 5x
e
30
u3 =
1 2x
e
6
1
1 5x −x 1 2x 2x
1 4x
1 4x
e ·e + e ·e =
e and y = c1 ex + c2 e−x + c3 e2x +
e
Thus yp = − e3x · ex +
6
30
6
30
30
on (−∞, ∞).
32. The auxiliary equation is m3 − 3m2 + 2m = m (m − 1) (m − 2) = 0 so yc = c1 + c2 ex + c3 e2x
1 ex e2x e2x = 2e3x
W = 0 ex 2e2x
0 ex 4e2x
e2x
we obtain
1 + ex
0
ex e2x 1
1
e5x
1 e2x
1 x 1 ex
1
ex 2e2x = 3x ·
e −
=
=
u1 = 3x W1 = 3x 0
2e
2e e2x
2e
1 + ex
2 1 + ex
2
2 1 + ex
x 4e2x e
1+ex
2x 1
0
e
ex
1
1 2e4x
1
2x
0
2e = 3x · −
=
−
u2 = 3x W2 = 3x 0
2x
2e
2e 2e
1 + ex
1 + ex
0 e x 4e2x Identifying f (x) =
1+e
1 ex
1
1
u3 = 3x W3 = 3x 0 ex
2e
2e 0 ex
3x
1 1
1 e−x
= 1 · e
=
=
2e3x 1 + ex
2 1 + ex
2 1 + e−x
e2x x
0
0
1+e
Then
u1 =
1 x 1
e − ln (1 + ex )
2
2
u2 = − ln 1 + ex
u3 = −
Thus
yp = 1 ·
=
1
ln (1 + e−x )
2
1 x 1
1 x
x
x
2x
−x
e − ln (1 + e ) + e · (− ln (1 + e )) + e · − ln 1 + e
2
2
2
1 x 1
1
e − ln (1 + ex ) − ex ln (1 + ex ) − e2x ln 1 + e−x
2
2
2
4.6 Variation of Parameters
33. The auxiliary equation is 3m2 − 6m + 30 = 0, which has roots 1 ± 3i, so yc = ex (c1 cos 3x +
c2 sin 3x). We consider first the differential equation 3y − 6y + 30y = 15 sin x, which can be
solved using undetermined coefficients. Letting yp1 = A cos x + B sin x and substituting into
the differential equation we get
(27A − 6B) cos x + (6A + 27B) sin x = 15 sin x.
Then
27A − 6B = 0 and
6A + 27B = 15,
2
9
2
9
and B = 17
. Thus, yp1 = 17
cos x + 17
sin x. Next, we consider the differential
so A = 17
equation 3y − 6y + 30y, for which a particular solution yp2 can be found using variation of
parameters. The Wronskian is
ex sin 3x
ex cos 3x
= 3e2x
W = x
x
x
x
e cos 3x − 3e sin 3x 3e cos 3x + e sin 3x
Identifying f (x) = 13 ex tan x we obtain
u1
1
1
= − sin 3x tan 3x = −
9
9
so
u1 = −
Next
u2 =
sin2 3x
cos 3x
1
=−
9
1 − cos2 3x
cos 3x
1
= − (sec 3x − cos 3x)
9
1
1
ln | sec 3x + tan 3x| +
sin 3x.
27
27
1
sin 3x so
9
u2 = −
1
cos 3x.
27
Thus
yp 2 = −
=−
1 x
1 x
e cos 3x(ln | sec 3x + tan 3x| − sin 3x) −
e sin 3x cos 3x
27
27
1 x
e (cos 3x) ln | sec 3x + tan 3x|
27
and the general solution of the original differential equation is
y = ex (c1 cos 3x + c2 sin 3x) + yp1 (x) + yp2 (x).
34. The auxiliary equation is m2 − 2m + 1 = (m − 1)2 = 0, which has repeated root 1, so
yc = c1 ex + c2 xex . We consider first the differential equation y − 2y + y = 4x2 − 3, which
can be solved using undetermined coefficients. Letting yp1 = Ax2 + Bx + C and substituting
into the differential equation we get
Ax2 + (−4A + B)x + (2A − 2B + C) = 4x2 − 3.
Then
A = 4,
−4A + B = 0,
and
2A − 2B + C = −3,
221
222
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
so A = 4, B = 16, and C = 21. Thus, yp1 = 4x2 + 16x + 21. Next we consider the differential
equation y − 2y + y = x−1 ex , for which a particular solution yp2 can be found using variation
of parameters. The Wronskian is
x
e
xex W = x
= e2x
e xex + ex Identifying f (x) = ex /x we obtain u1 = −1 and u2 = 1/x. Then u1 = −x and u2 = ln x, so
that
yp2 = −xex + xex ln x,
and the general solution of the original differential equation is
y = yc + yp1 + yp2 = c1 ex + c2 xex + 4x2 + 16x + 21 − xex + xex ln x
= c1 ex + c3 xex + 4x2 + 16x + 21 + xex ln x
35. The interval of definition for Problem 1 is (−π/2, π/2), for Problem 7 is (−∞, ∞), for Problem
15 is (0, ∞), and for Problem 18 is (−1, 1). In Problem 28 the general solution is
y = c1 cos (ln x) + c2 sin (ln x) + cos (ln x) ln | cos (ln x)| + (ln x) sin (ln x)
for −π/2 < ln x < π/2 or e−π/2 < x < eπ/2 . The bounds on ln x are due to the presence of
sec (ln x) in the differential equation.
36. We are given that y1 = x2 is a solution of x4 y + x3 y − 4x2 y = 0. To find a second solution
we use reduction of order. Let y = x2 u(x). Then the product rule gives
y = x2 u + 2xu and y = x2 u + 4xu + 2u,
so
x4 y + x3 y − 4x2 y = x5 (xu + 5u ) = 0.
Letting w = u , this becomes xw + 5w = 0. Separating variables and integrating we have
5
dw
= − dx and ln |w| = −5 ln x + c.
w
x
Thus, w = x−5 and u = − 14 x−4 . A second solution is then y2 = x2 x−4 = 1/x2 , and the
general solution of the homogeneous differential equation is yc = c1 x2 + c2 /x2 . To find a
particular solution, yp , we use variation of parameters. The Wronskian is
2
1 x
4
2 x
W =
=−
2 x
2x − 3 x
Identifying f (x) = 1/x4 we obtain u1 =
u2 = − 14 ln x, so
yp = −
1 −5
4x
1 −4
and u2 = − 14 x−1 . Then u1 = − 16
x and
1 −4 2 1
1
1
x x − (ln x)x−2 = − x−2 − x−2 ln x.
16
4
16
4
The general solution is
y = c1 x2 +
c2
1
1
−
− 2 ln x.
2
2
x
16x
4x
4.7 Cauchy–Euler Equation
4.7
Cauchy–Euler Equation
1. The auxiliary equation is m2 − m − 2 = (m + 1)(m − 2) = 0 so that y = c1 x−1 + c2 x2 .
2. The auxiliary equation is 4m2 − 4m + 1 = (2m − 1)2 = 0 so that y = c1 x1/2 + c2 x1/2 ln x.
3. The auxiliary equation is m2 = 0 so that y = c1 + c2 ln x.
4. The auxiliary equation is m2 − 4m = m(m − 4) = 0 so that y = c1 + c2 x4 .
5. The auxiliary equation is m2 + 4 = 0 so that y = c1 cos (2 ln x) + c2 sin (2 ln x).
6. The auxiliary equation is m2 + 4m + 3 = (m + 1)(m + 3) = 0 so that y = c1 x−1 + c2 x−3 .
7. The auxiliary equation is m2 − 4m − 2 = 0 so that y = c1 x2−
√
8. The auxiliary equation is m2 + 2m − 4 = 0 so that y = c1 x−1+
9. The auxiliary equation is 25m2 + 1 = 0 so that y = c1 cos
1
5
6
√
+ c2 x2+
5
√
6.
+ c2 x−1−
√
5.
ln x + c2 sin 15 ln x .
10. The auxiliary equation is 4m2 − 1 = (2m − 1)(2m + 1) = 0 so that y = c1 x1/2 + c2 x−1/2 .
11. The auxiliary equation is m2 + 4m + 4 = (m + 2)2 = 0 so that y = c1 x−2 + c2 x−2 ln x.
12. The auxiliary equation is m2 + 7m + 6 = (m + 1)(m + 6) = 0 so that y = c1 x−1 + c2 x−6 .
13. The auxiliary equation is 3m2 + 3m + 1 = 0 so that
√
3
−1/2
c1 cos
ln x + c2 sin
y=x
6
√
3
ln x
6
.
14. The auxiliary equation is m2 − 8m + 41 = 0 so that y = x4 [c1 cos (5 ln x) + c2 sin (5 ln x)].
15. Assuming that y = xm and substituting into the differential equation we obtain
m(m − 1)(m − 2) − 6 = m3 − 3m2 + 2m − 6 = (m − 3)(m2 + 2) = 0.
Thus
y = c1 x3 + c2 cos
√
√
2 ln x + c3 sin
2 ln x .
16. Assuming that y = xm and substituting into the differential equation we obtain
m(m − 1)(m − 2) + m − 1 = m3 − 3m2 + 3m − 1 = (m − 1)3 = 0.
Thus
y = c1 x + c2 x ln x + c3 x(ln x)2 .
223
224
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
17. Assuming that y = xm and substituting into the differential equation we obtain
m(m−1)(m−2)(m−3)+6m(m−1)(m−2) = m4 −7m2 +6m = m(m−1)(m−2)(m+3) = 0.
Thus
y = c1 + c2 x + c3 x2 + c4 x−3 .
18. Assuming that y = xm and substituting into the differential equation we obtain
m(m−1)(m−2)(m−3)+6m(m−1)(m−2)+9m(m−1)+3m+1 = m4 +2m2 +1 = (m2 +1)2 = 0.
Thus
y = c1 cos (ln x) + c2 sin (ln x) + c3 (ln x) cos (ln x) + c4 (ln x) sin (ln x).
19. The auxiliary equation is m2 − 5m = m(m − 5) = 0 so that yc = c1 + c2 x5 and
1 x5 5
= 5x4 .
W 1, x = 0 5x4 1 5
x , u2 =
Identifying f (x) = x3 we obtain u1 = − 15 x4 and u2 = 1/5x. Then u1 = − 25
and
1
1
1
y = c1 + c2 x5 − x5 + x5 ln x = c1 + c3 x5 + x5 ln x.
25
5
5
1
5
ln x,
20. The auxiliary equation is 2m2 + 3m + 1 = (2m + 1)(m + 1) = 0 so that yc = c1 x−1 + c2 x−1/2
and
−1
x−1/2 x
= 1 x−5/2 .
W x−1 , x−1/2 = 1
−x−2 − x−3/2 2
2
1
we obtain u1 = x − x2 and u2 = x3/2 − x1/2 . Then u1 = 12 x2 − 13 x3 ,
Identifying f (x) = 12 − 2x
u2 = 25 x5/2 − 23 x3/2 , and
1
1
1
2
2
1
y = c1 x−1 + c2 x−1/2 + x − x2 + x2 − x = c1 x−1 + c2 x−1/2 − x + x2 .
2
3
5
3
6
15
21. The auxiliary equation is m2 − 2m + 1 = (m − 1)2 = 0 so that yc = c1 x + c2 x ln x and
x x ln x 2
=x
W x, x = 1 1 + ln x
Identifying f (x) = 2/x we obtain u1 = −2(ln x)/x and u2 = 2/x. Then u1 = −(ln x)2 ,
u2 = 2 ln x, and
y = c1 x + c2 x ln x − x(ln x)2 + 2x(ln x)2
= c1 x + c2 x ln x + x(ln x)2 ,
x > 0.
4.7 Cauchy–Euler Equation
22. The auxiliary equation is m2 − 3m + 2 = (m − 1)(m − 2) = 0 so that yc = c1 x + c2 x2 and
x x2 2
= x2
W x, x = 1 2x
Identifying f (x) = x2 ex we obtain u1 = −x2 ex and u2 = xex . Then u1 = −x2 ex + 2xex − 2ex ,
u2 = xex − ex , and
y = c1 x + c2 x2 − x3 ex + 2x2 ex − 2xex + x3 ex − x2 ex
= c1 x + c2 x2 + x2 ex − 2xex .
23. The auxiliary equation m(m − 1) + m − 1 = m2 − 1 = 0 has roots m1 = −1, m2 = 1, so
yc = c1 x−1 + c2 x. With y1 = x−1 , y2 = x, and the identification f (x) = ln x/x2 , we get
W = 2x−1 ,
W1 = − (ln x) /x,
and
W2 = (ln x) /x3 .
Then u1 = W1 /W = −(ln x)/2, u2 = W2 /W = (ln x)/2x2 , and integration by parts gives
1
1
u1 = x − x ln x
2
2
so
1
1
u2 = − x−1 ln x − x−1 ,
2
2
1
1
1 −1
1 −1
−1
x = − ln x
x − x ln x x + − x ln x − x
y p = u 1 y1 + u 2 y2 =
2
2
2
2
and
y = yc + yp = c1 x−1 + c2 x − ln x,
x > 0.
24. The auxiliary equation m(m − 1) + m − 1 = m2 − 1 = 0 has roots m1 = −1, m2 = 1, so
yc = c1 x−1 + c2 x. With y1 = x−1 , y2 = x, and the identification f (x) = 1/x2 (x + 1), we get
W = 2x−1 ,
W1 = −1/x(x + 1),
Then u1 = W1 /W = −1/2(x + 1),
fractions for u2 ) gives
and
W2 = 1/x3 (x + 1).
u2 = W2 /W = 1/2x2 (x + 1), and integration (by partial
1
u1 = − ln (x + 1)
2
1
1
1
u2 = − x−1 − ln x + ln (x + 1),
2
2
2
so
1
1
1
1
yp = u1 y1 + u2 y2 = − ln (x + 1) x−1 + − x−1 − ln x + ln (x + 1) x
2
2
2
2
1
ln (x + 1)
1 1
1
1 1
ln (x + 1)
= − + x ln 1 +
= − − x ln x + x ln (x + 1) −
−
2 2
2
2x
2 2
x
2x
and
−1
y = y c + yp = c 1 x
1
ln (x + 1)
1 1
−
,
+ c2 x − + x ln 1 +
2 2
x
2x
x > 0.
225
226
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
25. The auxiliary equation is m2 + 2m = m(m + 2) = 0, so that
y = c1 + c2 x−2 and y = −2c2 x−3 . The initial conditions imply
y
5 x
c1 + c2 = 0
−2c2 = 4.
Thus, c1 = 2, c2 = −2, and y = 2 − 2x−2 . The graph is given to the
right.
–10
–20
26. The auxiliary equation is m2 − 6m + 8 = (m − 2)(m − 4) = 0, so that
2
4
y = c1 x + c2 x
y
40
3
and y = 2c1 x + 4c2 x .
30
20
The initial conditions imply
10
4 2
10
4c1 + 16c2 = 32
and
1
1
y = −c1 sin (ln x) + c2 cos (ln x).
x
x
x
30
40
Thus, c1 = 16, c2 = −2, and y = 16x2 − 2x4 . The graph is given to
the right.
y = c1 cos (ln x) + c2 sin (ln x)
4
20
4c1 + 32c2 v = 0.
27. The auxiliary equation is m2 + 1 = 0, so that
2
y
3
50
100 x
–3
The initial conditions imply c1 = 1 and c2 = 2. Thus y = cos (ln x) + 2 sin (ln x). The graph
is given to the right.
4.7 Cauchy–Euler Equation
28. The auxiliary equation is m2 − 4m + 4 = (m − 2)2 = 0, so that
227
y
y = c1 x2 + c2 x2 ln x and y = 2c1 x + c2 (x + 2x ln x).
5
x
The initial conditions imply c1 = 5 and c2 + 10 = 3.
y = 5x2 − 7x2 ln x. The graph is given to the right.
Thus
–10
–20
–30
29. The auxiliary equation is m2 = 0 so that yc = c1 + c2 ln x and
1 ln x 1
W (1, ln x) = 1 =
x
0
x
y
15
10
Identifying f (x) = 1 we obtain u1 = −x ln x and u2 = x. Then
u1 = 14 x2 − 12 x2 ln x, u2 = 12 x2 , and
5
1
1
1
1
y = c1 + c2 ln x + x2 − x2 ln x + x2 ln x = c1 + c2 ln x + x2 .
4
2
2
4
The initial conditions imply c1 + 14 = 1 and c2 + 12 = − 12 . Thus,
c1 = 34 , c2 = −1, and y = 34 − ln x + 14 x2 . The graph is given to the
right.
30. The auxiliary equation is m2 −6m+8 = (m−2)(m−4) = 0,
so that yc = c1 x2 + c2 x4 and
2 4 x2 x4 = 2x5
W x ,x = 2x 4x3 5
y
0.05
–1
1
x
Identifying f (x) = 8x4 we obtain u1 = −4x3 and u2 = 4x. Then u1 = −x4 , u2 = 2x2 , and
y = c1 x2 + c2 x4 + x6 . The initial conditions imply
1
1
1
c1 + c 2 = −
4
16
64
1
3
c 1 + c2 = − .
2
16
Thus c1 =
1
16
, c2 = − 12 , and y =
1 2
16 x
− 12 x4 + x6 . The graph is given to the right.
31. Substituting x = et into the differential equation we obtain
dy
d2 y
− 20y = 0.
+8
2
dt
dt
228
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
The auxiliary equation is m2 + 8m − 20 = (m + 10)(m − 2) = 0 so that
y = c1 e−10t + c2 e2t
or y = c1 x−10 + c2 x2 .
32. Substituting x = et into the differential equation we obtain
dy
d2 y
+ 25y = 0.
− 10
2
dt
dt
The auxiliary equation is m2 − 10m + 25 = (m − 5)2 = 0 so that
y = c1 e5t + c2 te5t
or y = c1 x5 + c2 x5 ln x.
33. Substituting x = et into the differential equation we obtain
dy
d2 y
+ 8y = e2t .
+9
dt2
dt
The auxiliary equation is m2 +9m+8 = (m+1)(m+8) = 0 so that yc = c1 e−t +c2 e−8t . Using
undetermined coefficients we try yp = Ae2t . This leads to 30Ae2t = e2t , so that A = 1/30 and
y = c1 e−t + c2 e−8t +
1 2t
e
30
or y = c1 x−1 + c2 x−8 +
1 2
x .
30
34. Substituting x = et into the differential equation we obtain
dy
d2 y
+ 6y = 2t.
−5
2
dt
dt
The auxiliary equation is m2 − 5m + 6 = (m − 2)(m − 3) = 0 so that yc = c1 e2t + c2 e3t . Using
undetermined coefficients we try yp = At + B. This leads to (−5A + 6B) + 6At = 2t, so that
A = 1/3, B = 5/18, and
1
5
y = c1 e2t + c2 e3t + t +
3
18
or y = c1 x2 + c2 x3 +
1
5
ln x + .
3
18
35. Substituting x = et into the differential equation we obtain
d2 y
dy
+ 13y = 4 + 3et .
−4
dt2
dt
The auxiliary equation is m2 − 4m + 13 = 0 so that yc = e2t (c1 cos 3t + c2 sin 3t). Using
undetermined coefficients we try yp = A + Bet . This leads to 13A + 10Bet = 4 + 3et , so that
A = 4/13, B = 3/10, and
y = e2t (c1 cos 3t + c2 sin 3t) +
4
3
+ et
13 10
or
y = x2 [c1 cos (3 ln x) + c2 sin (3 ln x)] +
3
4
+ x.
13 10
4.7 Cauchy–Euler Equation
36. From
d2 y
1
= 2
2
dx
x
it follows that
1
x2
=
1
x2
=
1
x3
d2 y dy
−
dt2
dt
d2 y dy
2 d2 y dy
− 3
−
−
dt2
dt
x
dt2
dt
2 1 d dy
2 d2 y
2 dy
d d y
−
−
+ 3
2
2
3
2
dx dt
x dx dt
x dt
x dt
1 d2 y 1
2 d2 y
2 dy
d3 y 1
−
−
+ 3
dt3 x
x2 dt2 x
x3 dt2
x dt
3
d y
d2 y
dy
.
−3 2 +2
dt3
dt
dt
1 d
d3 y
= 2
3
dx
x dx
=
229
Substituting into the differential equation we obtain
2
d3 y
d2 y
dy
d y dy
dy
−3
+6
− 6y = 3 + 3t
−3 2 +2
−
3
2
dt
dt
dt
dt
dt
dt
or
d3 y
d2 y
dy
− 6y = 3 + 3t.
−
6
+ 11
3
2
dt
dt
dt
The auxiliary equation is m3 − 6m2 + 11m − 6 = (m − 1)(m − 2)(m − 3) = 0 so that
yc = c1 et + c2 e2t + c3 e3t . Using undetermined coefficients we try yp = A + Bt. This leads to
(11B − 6A) − 6Bt = 3 + 3t, so that A = −17/12, B = −1/2, and
y = c1 et + c2 e2t + c3 e3t −
17 1
− t
12 2
or y = c1 x + c2 x2 + c3 x3 −
17 1
− ln x.
12 2
In the next two problems we use the substitution t = −x since the initial conditions are on the
interval (−∞, 0). In this case
dy dx
dy
dy
=
=−
dt
dx dt
dx
and
d
dy
d d2 y dx
d2 y
d2 y
d dy
dy dx
=
−
=
−
(y
=
−
=
=
)
=
−
.
dt2
dt dt
dt
dx
dt
dx dt
dx2 dt
dx2
37. The differential equation and initial conditions become
d2 y
= 2, y (t)
= −4.
4t2 2 + y = 0; y(t)
dt
t=1
t=1
The auxiliary equation is 4m2 − 4m + 1 = (2m − 1)2 = 0, so that
1 −1/2
1 −1/2
1/2
1/2
−1/2
y = c1 t + c2 t ln t and y = c1 t
+ c2 t
+ t
ln t .
2
2
The initial conditions imply c1 = 2 and 1 + c2 = −4. Thus
y = 2t1/2 − 5t1/2 ln t
or y = 2(−x)1/2 − 5(−x)1/2 ln (−x),
x < 0.
230
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
38. The differential equation and initial conditions become
dy
d2 y
+ 6y = 0; y(t)
= 8,
t2 2 − 4t
dt
dt
t=2
y (t)
t=2
= 0.
The auxiliary equation is m2 − 5m + 6 = (m − 2)(m − 3) = 0, so that
y = c 1 t 2 + c2 t 3
and y = 2c1 t + 3c2 t2 .
The initial conditions imply
4c1 + 8c2 = 8
4c1 + 12c2 = 0
from which we find c1 = 6 and c2 = −2. Thus
y = 6t2 − 2t3
or y = 6x2 + 2x3 ,
x < 0.
39. If we force the function y = (x − (−3))m = (x + 3)m into the equation we get
(x + 3)2 y − 8(x + 3)y + 14y = 0
(x + 3)2 m(m − 1)(x + 3)m−2 − 8(x + 3)m(x + 3)m−1 + 14(x + 3)m = 0
(x + 3)m [m2 − 9m + 14] = 0
(x + 3)m (m − 2)(m − 7) = 0
The solutions to the auxiliary equation are therefore m = 2 and m = 7 from which we get
the general solution y = c1 (x + 3)2 + c2 (x + 3)7 .
40. If we force the function y = (x − 1)m into the equation we get
(x − 1)2 y − (x − 1)y + 5y = 0
(x − 1)2 m(m − 1)(x − 1)m−2 − (x − 1)m(x − 1)m−1 + 5(x − 1)m = 0
(x − 1)m [m2 − 2m + 5] = 0
The solutions to the auxiliary equation are therefore the complex roots m = 1±2i from which
we get the general solution y = c1 (x − 1) cos (2 ln (x − 1)) + c2 (x − 1) sin (2 ln (x − 1)).
41. Letting t = x + 2 we obtain dy/dx = dy/dt and, using the Chain Rule,
d dy
d2 y dt
d2 y
d2 y
d2 y
=
=
=
(1)
=
.
dx2
dx dt
dt2 dx
dt2
dt2
Substituting into the differential equation we obtain
t2
dy
d2 y
+ y = 0.
+t
2
dt
dt
The auxiliary equation is m2 + 1 = 0 so that
y = c1 cos (ln 2t) + c2 sin (ln 2t) = c2 cos [ln (x + 2)] + c2 sin [ln (x + 2)].
4.7 Cauchy–Euler Equation
42. Letting t = x − 4 we obtain dy/dx = dy/dt and, using the Chain Rule,
d
d2 y
=
2
dx
dx
dy
dt
=
d2 y dt
d2 y
d2 y
=
(1)
=
.
dt2 dx
dt2
dt2
Substituting into the differential equation we obtain
t2
dy
d2 y
− 5t
+ 9y = 0.
dt2
dt
The auxiliary equation is m(m − 1) − 5m + 9 = m2 − 6m + 9 = (m − 3)2 = 0 so that
m1 = m2 = 3. Then
y = c1 t3 + c2 t3 ln t = c1 (x − 4)3 + c2 (x − 4)3 ln (x − 4).
43. Since the leading coefficient a2 (x) = (x − 4)2 = 0 for every value of x satisfying x > 4 we may
take the interval of definition of the general solution to be (4, ∞).
44. If 1 − i is a root of the auxiliary equation then so is 1 + i, and the auxiliary equation is
(m − 2)[m − (1 + i)][m − (1 − i)] = m3 − 4m2 + 6m − 4 = 0.
We need m3 − 4m2 + 6m − 4 to have the form m(m − 1)(m − 2) + bm(m − 1) + cm + d.
Expanding this last expression and equating coefficients we get b = −1, c = 3, and d = −4.
Thus, the differential equation is
x3 y − x2 y + 3xy − 4y = 0.
45. For x2 y = 0 the auxiliary equation is m(m − 1) = 0 and the general solution is y = c1 + c2 x.
The initial conditions imply c1 = y0 and c2 = y1 , so y = y0 + y1 x. The initial conditions are
satisfied for all real values of y0 and y1 .
For x2 y − 2xy + 2y = 0 the auxiliary equation is m2 − 3m + 2 = (m − 1)(m − 2) = 0 and
the general solution is y = c1 x + c2 x2 . The initial condition y(0) = y0 implies 0 = y0 and the
condition y (0) = y1 implies c1 = y1 . Thus, the initial conditions are satisfied for y0 = 0 and
for all real values of y1 .
For x2 y − 4xy + 6y = 0 the auxiliary equation is m2 − 5m + 6 = (m − 2)(m − 3) = 0 and the
general solution is y = c1 x2 + c2 x3 . The initial conditions imply y(0) = 0 = y0 and y (0) = 0.
Thus, the initial conditions are satisfied only for y0 = y1 = 0.
√
46. The function y(x) = − x cos (2 ln x) is defined for x > 0 and has x-intercepts where
2 ln x = π/2 + kπ for k an integer or where x = eπ/4+kπ/2 . Solving π/4 + kπ/2 = 0.5
we get k ≈ −0.18, so eπ/2+kπ < 0.5 for all negative integers and the graph has infinitely many
x-intercepts in the interval (0, 0.5).
231
232
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
47. (a) Multiplying by r3 we have
1 dw
q
d3 w 1 d2 w
=
r
+
− 2
3
2
dr
r dr
r dr
2D
r3
2
q 4
dw
d3 w
2 d w
=
r
+
r
−r
3
2
dr
dr
dr
2D
This is a nonhomogeneous third-order Cauchy-Euler differential equation. We first solve
the homogenous equation. The auxiliary equation is
m (m − 1) (m − 2) + m (m − 1) − m = 0
m3 − 2m2 = 0
m2 (m − 2) = 0
Therefore m1 = m2 = 0 and m3 = 2. Then the complementary function is
wc (r) = c1 + c2 ln r + c3 r2 . To use variation of parameters we must use the first form of
the differential equation and identify f (r) = qr/2D. By (10) of Section 3.5 we identify
y1 = 1, y2 = ln r, and y3 = r2 , and expand each of the four determinants by the first
column:
2
0
ln
r
r
ln r r2 1
q
= q 2r2 ln r − r2
2r =
r 1
W1 = 0
r
q
2D 2r 2D
1
r
r − 2 2
2D
r
1
0
r2 0
2r
0
2r = 1 · q
= − q r2
W 2 = 0
D
r 2
0 q r 2 2D
2D
1 ln r
0
1
0
1
q
r
0 =1·
W3 = 0
=
q 2D
r
1
r
− 2
1
q r
2D
r
0 − 2
r
2D
1 ln r
1
W = 0
r
1
0 − 2
r
r2 1
2r = 1 · r
1
− 2
r
2
2r 2 2 4
=
= − −
r
r
r
2
4.7 Cauchy–Euler Equation
Then
q
r (2r ln r − r)
q 3
W
q 3
1
=
= 2D
r ln r −
r
u1 =
4
W
4D
8D
r
q
− r2
q 3
W2
D
=−
=
r
u2 =
4
W
4D
r
q
q
W3
u3 =
= 2D =
r
4
W
8D
r
and
q 4
3q 4
r ln r −
r
16D
64D
q 4
r
u2 = −
16D
q 2
r
u3 =
16D
u1 =
(integration by parts)
Thus a particular solution is
q 4
4q 2 2
q 4
3q 4
q 4
r ln r −
r ·1+ −
r ln r +
r r =
r
wp (r) =
16D
64D
16D
64D
64D
Therefore the general solution is w(r) = wc (r) + wp (r) or
w(r) = c1 + c2 ln r + c3 r2 +
q 4
r
64D
(b) Taking the derivative of w we have
w (r) −
c2
q 3
+ 2c3 r +
r .
r
16D
The condition w (0) = 0 requires c2 = 0. Using the condition w (a) = 0 gives
0 = 2c3 a +
c3 = −
q
a3
16D
q
a2 .
32D
So,
w(r) = c1 −
q
q 4
a2 r2 +
r .
32D
64D
Using the condition w(a) = 0 gives
0 = c1 −
c1 =
q
q
a4 +
a4
32D
64D
q
a4 .
64D
233
234
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
Therefore
w(r) =
2
q
q 4
q 2
q
a4 −
a2 r2 +
r =
a − r2 .
64D
32D
64D
64D
48. (a) Using the product rule twice, the left-hand side becomes
2
dw
d 1
d w dw
d d2 w 1 dw
d 1 d
r
=
r 2 +
=
+
dr r dr
dr
dr r
dr
dr
dr dr2
r dr
=
1 dw
d3 w 1 d2 w
+
− 2
2
dr3
r dr
r dr
This is the left-hand side of equation (8).
(b) Integrating both sides with respect to r gives
d 1 d
dw
q
r
r
=
dr r dr
dr
2D
1 d
dw
q 2
r + C1
r
=
r dr
dr
4D
dw
1 3
d
r
=
r + C1 r
dr
dr
4D
r
q 4 C1 2
dw
=
r +
r + C2
dr
16D
2
dw
1 3 C1
C2
=
r +
r+
dr
16D
2
r
w(r) =
q 2 C1 2
r +
r + C2 ln r + C3 .
64D
4
Relabeling c1 = C3 , c2 = C2 , and c3 = C1 /4 shows this is the same as the general
solution obtained in part (a) of Problem 47.
49. The auxiliary equation is 2m(m − 1)(m − 2) − 10.98m(m − 1) + 8.5m + 1.3 = 0, so that
m1 = −0.053299, m2 = 1.81164, m3 = 6.73166, and
y = c1 x−0.053299 + c2 x1.81164 + c3 x6.73166 .
50. The auxiliary equation is m(m − 1)(m − 2) + 4m(m − 1) + 5m − 9 = 0, so that m1 = 1.40819
and the two complex roots are −1.20409 ± 2.22291i. The general solution of the differential
equation is
y = c1 x1.40819 + x−1.20409 [c2 cos (2.22291 ln x) + c3 sin (2.22291 ln x)] .
51. The auxiliary equation is
m(m − 1)(m − 2)(m − 3) + 6m(m − 1)(m − 2) + 3m(m − 1) − 3m + 4 = 0,
4.7 Cauchy–Euler Equation
so that m1 = m2 =
equation is
√
√
2 and m3 = m4 = − 2 . The general solution of the differential
√
y = c1 x
2
√
2
+ c2 x
ln x + c3 x−
√
2
+ c4 x−
√
2
ln x.
52. The auxiliary equation is m(m − 1)(m − 2)(m − 3) − 6m(m − 1)(m − 2) + 33m(m − 1) −
105m + 169 = 0, so that m1 = m2 = 3 + 2i and m3 = m4 = 3 − 2i. The general solution of
the differential equation is
y = x3 [c1 cos (2 ln x) + c2 sin (2 ln x)] + x3 ln x [c3 cos (2 ln x) + c4 sin (2 ln x)] .
53. First solve the associated homogeneous equation. From the auxiliary equation
(m − 3)(m − 2)(m + 1) = 0 we find yc = c1 x3 + c2 x2 + c3 x−1 . Before finding the particular
solution using variation of parameters, first put the differential equation in standard form by
dividing through by x3 . We therefore identify f (x) = x−1 . Now with y1 = x3 , y2 = x2 , and
y3 = x−1 we get
3
2
x
2 x
W = 3x 2x
6x 2
3
x
0
0
W2 = 3x2
6x x−1
x−1 −x−2 = −12x,
2x−3 x−1 −x−2 = 4
2x−3 0
x2 x−1 W1 = 0 2x −x−2 = −3x,
x−1 2 2x−3 3
2
x
x
0
2
W3 = 3x 2x 0 = −x3
6x 2 x−1 1
−1
x2
W1
W2
W3
= 2 , u2 =
=
, and u3 =
=
. The integrals of these three
W
4x
W
3x
W
12
functions are straight forward: u1 = −1/4x, u2 = − (ln x) /3, u3 = x3 /36. The particular
solution is
Therefore u1 =
y p = u 1 y1 + u 2 y2 + u 3 y3 = −
=−
x2 x2 ln x x2
−
+
4
3
36
2x2 x2 ln x
−
9
3
Finally, because −2x2 /9 can be absorbed in the c2 x2 term the general solutions is
y = yc + yp = c1 x3 + c2 x2 + c3 x−1 −
x2 ln x
,
3
x > 0.
235
236
CHAPTER 4
4.8
HIGHER-ORDER DIFFERENTIAL EQUATIONS
Green’s Functions
1.
y − 16y = f (x)
y − 16y = 0
y1 = e−4x ,
y2 = e4x
e4x −4x 4x e−4x
=
W e ,e
=8
−4e−4x 4e4x 1 4(x−t)
e−4t e4x − e−4x e4t
=
− e−4(x−t)
e
8
8
ˆ x
ˆ
1
1 x
yp (x) =
sinh 4(x − t) f (t) dt
e4(x−t) − e−4(x−t) f (t) dt =
8 x0
4 x0
G(x, t) =
2.
y + 3y − 10y = f (x)
y + 3y − 10y = 0
y2 = e−5x
−5x 2x −5x e2x
e
= −7e−3x
W e ,e
= 2x
2e
−5e−5x y1 = e2x ,
e2t e−5x − e2x e−5t
1 2(x−t)
−5(x−t)
=
−
e
e
−7e−3t
7
ˆ x
1
yp (x) =
e2(x−t) − e−5(x−t) f (t) dt
7 x0
G(x, t) =
3.
y + 2y + y = f (x)
y + 2y + y = 0
y1 = e−x ,
y2 = xe−x
−x
e−x
−x
xe
−x
= e−2x
= −x
W e , xe
−x
−x
−e
−xe + e e−t xe−x − e−x te−t
= (x − t)e−(x−t)
e−2t
ˆ x
yp (x) =
(x − t)e−(x−t) f (t) dt
G(x, t) =
x0
4.8
Green’s Functions
4.
4y − 4y + y = f (x)
4y − 4y + y = 0
y1 = ex/2 ,
y2 = xex/2
ex/2
x/2
x/2
W e , xe
= 1 x/2
2e
xex/2
= ex
1
x/2
x/2
xe
+
e
2
et/2 xex/2 − ex/2 tet/2
= (x − t)e(x−t)/2
et
ˆ x
1
yp (x) =
(x − t)e(x−t)/2 f (t) dt
4
x0
G(x, t) =
5.
y + 9y = f (x)
y + 9y = 0
y1 = cos 3x,
y2 = sin 3x
cos 3x
sin 3x W (cos 3x, sin 3x) = =3
−3 sin 3x 3 cos 3x
1
cos 3t sin 3x − cos 3x sin 3t
= sin 3(x − t)
3
3
ˆ x
1
sin 3(x − t)f (t) dt
yp (x) =
3 x0
G(x, t) =
6.
y − 2y + 2y = f (x)
y − 2y + 2y = 0
y1 = ex cos x,
y2 = ex sin x
x cos x
x sin x
e
e
x
x
= e2x
W (e cos x, e sin x) = x
x
x
x
−e sin x + e cos x e cos x + e sin x
et cos tex sin x − ex cos xet sin t
= ex−t sin (x − t)
e2t
ˆ x
ex−t sin (x − t)f (t) dt
yp (x) =
G(x, t) =
x0
237
238
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
7.
y − 16y = xe−2x
y = yc + yp
y = c1 e−4x + c2 e4x + yp
y = c1 e−4x + c2 e4x +
1
4
ˆ
x
sinh 4(x − t)te−2t dt
x0
8.
y + 3y − 10y = x2
y = yc + yp
y = c1 e2x + c2 e−5x + yp
2x
y = c1 e
−5x
+ c2 e
1
+
7
ˆ
x
e2(x−t) − e−5(x−t) t2 dt
x0
9.
y + 2y + y = e−x
y = y c + yp
y = c1 e−x + c2 xe−x + yp
ˆ x
−x
−x
y = c1 e + c2 xe +
(x − t)e−(x−t) e−t dt
x0
10.
4y − 4y + y = arctan x
y = y c + yp
y = c1 ex/2 + c2 xex/2 + yp
ˆ x
1
x/2
x/2
y = c1 e + c2 xe +
(x − t)e(x−t)/2 arctan t dt
4
x0
11.
y + 9y = x + sin x
y = yc + yp
y = c1 cos 3x + c2 sin 3x + yp
1
y = c1 cos 3x + c2 sin 3x +
3
ˆ
x
x0
sin 3(x − t)(t + sin t) dt
4.8
Green’s Functions
12.
y − 2y + 2y = cos2 x
y = y c + yp
y = c1 ex cos x + c2 ex sin x + yp
ˆ x
y = c1 ex cos x + c2 ex sin x =
e(x−t) sin (x − t) cos2 t dt
x0
13. The initial-value problem is y − 4y = e2x ,
y(0) = 0, y (0) = 0. Then we find that
y1 = e−2x ,
y2 = e2x
2x −2x 2x e−2x
e
= 4.
= W e ,e
−2e−2x 2e2x Then
1 2(x−t)
e−2t e2x − e−2x e2t
=
e
− e−2(x−t)
4
4
and the solution of the initial-value problem is
ˆ
1 x 2(x−t)
e
− e−2(x−t) e2t dt
yp (x) =
4 0
ˆ
ˆ
1 −2x x 4t
1 2x x
dt − e
e dt
= e
4
4
0
0
G(x, t) =
1
1
1
= xe2x − e2x + e−2x .
4
16
16
14. The initial-value problem is y − y = 1,
y(0) = 0, y (0) = 0. Then we find that
y2 = ex
1 ex x
= ex .
W (1, e ) = 0 ex y1 = 1,
Then
ex − et
= ex−t − 1
et
and the solution of the initial-value problem is
ˆ x
(ex−t − 1) dt
yp (x) =
G(x, t) =
0
ˆ
x
= ex
e−t dt −
0
= ex − x − 1.
ˆ
x
dt
0
239
240
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
15. The initial-value problem is y − 10y + 25y = e5x ,
y(0) = 0, y (0) = 0. Then we find that
y1 = e5x ,
y2 = xe5x
5x
e5x
xe5x
5x
= e10x .
W e , xe
= 5x
5x
5x
5e
5xe + e Then
e5t xe5x − e5x te5t
= (x − t)e5(x−t)
e10t
and the solution of the initial-value problem is
ˆ x
(x − t)e5(x−t) e5t dt
yp (x) =
G(x, t) =
0
ˆ
5x
ˆ
x
dt − e
5x
= xe
0
x
t dt
0
1
= x2 e5x − x2 e5x
2
1 2 5x
= x e .
2
16. The initial-value problem is y + 6y + 9y = x,
y(0) = 0, y (0) = 0. Then we find that
y1 = e−3x ,
y2 = xe−3x
−3x
−3x
e−3x
xe
−3x
= e−6x .
W e , xe
= −3x
−3x
−3x
−3e
−3xe
+e Then
e−3t xe−3x − e−3x te−3t
= (x − t)e−3(x−t)
e−6t
and the solution of the initial-value problem is
ˆ x
(x − t)e−3(x−t) t dt
yp (x) =
G(x, t) =
0
−3x
ˆ
x
= xe
−3x
ˆ
te dt − e
3t
0
x
t2 e3t dt
0
1 −3x 1 2
2 −3x 2
1 2
2
1
− e
+ x −
− x+ x
= − x + xe
9
9
3
27 27
9
3
=
1
2 −3x 1 −3x
2
e
+ x.
+ xe
−
27
9
27 9
17. The initial-value problem is y + y = csc x cot x,
that
y1 = cos x,
y(π/2) = 0, y (π/2) = 0. Then we find
y2 = sin x
cos x sin x = 1.
W (cos x, sin x) = − sin x cos x
4.8
Green’s Functions
Then
G(x, t) = cos t sin x − cos x sin t = sin (x − t)
and the solution of the initial-value problem is
ˆ
x
(cos t sin x − cos x sin t) csc t cot t dt
yp (x) =
π/2
ˆ
ˆ
x
= sin x
cot t dt
π/2
ˆ
x
cot t dt − cos x
2
π/2
ˆ
x
(csc t − 1) dt − cos x
2
= sin x
π/2
x
cot t dt
π/2
π
= sin x − cot x − x +
− cos x ln | sin x|
2
π
= − cos x − x sin x + sin x − cos x ln | sin x|
2
π
= − cos x + sin x − x sin x − cos x ln | sin x|.
2
18. The initial-value problem is y + y = sec2 x,
y(π) = 0, y (π) = 0. Then we find that
y1 = cos x,
y2 = sin x
cos x sin x = 1.
W (cos x, sin x) = − sin x cos x
Then
G(x, t) = cos t sin x − cos x sin t = sin (x − t)
and the solution of the initial-value problem is
ˆ
x
yp (x) =
(cos t sin x − cos x sin t) sec2 t dt
π
ˆ
x
= sin x
ˆ
sec t dt − cos x
π
x
sec t tan t dt
π
= − cos x − 1 + sin x ln | sec x + tan x|.
19. The initial-value problem is y − 4y = e2x ,
The initial conditions give
y(0) = 1, y (0) = −4, so y(x) = c1 e−2x + c2 e2x .
c1 + c2 =
4
−2c1 + 2c2 = −4,
241
242
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
so c1 = 32 and c2 = − 12 , which implies that yh = 32 e−2x − 12 e2x . Now, yp found in the solution
of Problem 13 in this section gives
1 2x
3
1
1
1
xe − e2x + e−2x
y = yh + yp = e−2x − e2x +
2
2
4
16
16
=
25 −2x
9
1
e
− e2x + xe2x .
16
16
4
20. The initial-value problem is y − y = 1,
initial conditions give
y(0) = 10, y (0) = 1, so y(x) = c1 + c2 ex . The
c1 + c2 = 10
c2 = 1,
so c1 = 9 and c2 = 1, which implies that yh = 9 + ex . Now, yp found in the solution of
Problem 14 in this section gives
y = yh + yp = 9 + ex + (ex − x − 1) = 8 + 2ex − x.
21. The initial-value problem is y − 10y + 25y = e5x ,
y(x) = c1 e5x + c2 xe5x . The initial conditions give
y(0) = −1, y (0) = 1, so
c1 = −1
5c1 + c2 =
1,
so c1 = −1 and c2 = 6, which implies that yh = −e5x + 6xe5x . Now, yp found in the solution
of Problem 15 in this section gives
1
y = yh + yp = −e5x + 6xe5x + x2 e5x .
2
22. The initial-value problem is y + 6y + 9y = x, y(0) = 1, y (0) = −3, so
y(x) = c1 e−3x + c2 xe−3x . The initial conditions give
c1 =
1
−3c1 + c2 = −3,
so c1 = 1 and c2 = 0, which implies that yh = e−3x . Now, yp found in the solution of Problem
16 in this section gives
2 −3x 1 −3x
2
1
29
1
2
1
−3x
e
+
= e−3x + xe−3x −
+ x.
+
+ xe
−
y = y h + yp = e
27
9
27 9
27
9
27 9
23. The initial-value problem is y + y = csc x cot x,
y(x) = c1 cos x + c2 sin x. The initial conditions give
π
2
−c1 = −1,
c2 = −
y(π/2) = − π2 , y (π/2) = −1, so
4.8
Green’s Functions
so c1 = 1 and c2 = − π2 , which implies that yh = cos x − π2 sin x. Now, yp found in the solution
of Problem 17 in this section gives
π
π
y = yh + yp = cos x − sin x + − cos x + sin x − x sin x − cos x ln | sin x|
2
2
= −x sin x − cos x ln | sin x|.
24. The initial-value problem is y + y = sec2 x, y(π) = 12 , y (π) = −1, so
y(x) = c1 cos x + c2 sin x. The initial conditions give
−c1 =
1
2
−c2 = −1,
so c1 = − 12 and c2 = 1, which implies that yh = − 12 cos x + sin x. Now, yp found in the
solution of Problem 18 in this section gives
1
y = yh + yp = − cos x + sin x + (− cos x − 1 + sin x ln | sec x + tan x|)
2
3
= − cos x + sin x − 1 + sin x ln | sec x + tan x|.
2
25. The initial-value problem is y + 3y + 2y = sin ex ,
y(x) = c1 e−x + c2 e−2x . The initial conditions give
y(0) = −1, y (0) = 0, so
c1 + c2 = −1
−c1 − 2c2 =
0,
so c1 = −2 and c2 = 1, which implies that yh = −2e−x + e−2x . The Wronskian is
−x −2x e−x
e−2x = −x
W e ,e
= −e−3x .
−e
−2e−2x e−t e−2x − e−x e−2t
= e−x et − e2t e−2x so
−e−3t
ˆ x
(e−x et − e2t e−2x ) sin et dt
yp (x) =
Then G(x, t) =
0
−x
ˆ
x
=e
0
−2x
ˆ
x
e sin e dt − e
t
t
e2t sin et dt
0
= e−x (− cos ex + cos 1) − e−2x (−ex cos ex + sin ex + cos 1 − sin 1)
= e−x cos 1 + (sin 1 − cos 1)e−2x − e−2x sin ex .
The solution of the initial-value problem is
y = yh + yp = −2e−x + e−2x + e−x cos 1 + e−2x (sin 1 − cos 1) − e−2x sin ex
= (cos 1 − 2)e−x + (1 + sin 1 − cos 1)e−2x − e−2x sin ex .
243
244
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
26. The initial-value problem is y + 3y + 2y =
c1 e−x + c2 e−2x . The initial conditions give
1
,
1 + ex
y(0) = 0, y (0) = −2, so y(x) =
c1 + c 2 = 0
−c1 − 2c2 = 1,
so c1 = 1 and c2 = −1, which implies that yh = e−x − e−2x . The Wronskian is
−x −2x e−x
e−2x W e ,e
= −x
= −e−3x .
−e
−2e−2x e−t e−2x − e−x e−2t
= e−x et − e2t e−2x so
−e−3t
ˆ x
ˆ x 2t
ˆ x
1
et
e
−x t
2t −2x
−x
−2x
(e e − e e )
dt = e
dt − e
dt
yp (x) =
t
t
t
1+e
0
0 1+e
0 1+e
Then G(x, t) =
= e−x [ln (1 + ex ) − ln 2] − e−2x [ex − ln (1 + ex ) − 1 + ln 2]
= e−x (−1 − ln 2) + (1 − ln 2)e−2x + e−x ln (1 + ex ) + e−2x ln (1 + ex ).
The solution of the initial-value problem is
y = yh + yp = e−x − e−2x + e−x (−1 − ln 2) + (1 − ln 2)e−2x + e−x ln (1 + ex ) + e−2x ln (1 + ex )
= −(ln 2)e−x − (ln 2)e−2x + e−x ln (1 + ex ) + e−2x ln (1 + ex ).
27. The Cauchy-Euler initial-value problem x2 y − 2xy + 2y = x, y(1) = 2, y (1) = −1, has
auxiliary equation m(m − 1) − 2m + 2 = m2 − 3m + 2 = (m − 1)(m − 2) = 0 so m1 = 1,
m2 = 2, y(x) = c1 x + c2 x2 , and y = c1 + 2c2 x. The initial conditions give
c1 + c2 =
2
c1 + 2c2 = −1,
so c1 = 5 and c2 = −3, which implies that yh = 5x − 3x2 . The Wronskian is
x x2 2
= x2 .
W x, x = 1 2x x(x − t)
tx2 − xt2
. From the standard form of the differential equation
=
t2
t
1
we identify the forcing function f (t) = . Then, for x > 1,
t
ˆ x
x(x − t) 1
dt
yp (x) =
t
t
1
ˆ x
ˆ x
1
1
dt
−
x
dt
= x2
2
t
1
1 t
1
2
= x − + 1 − x(ln x − ln 1)
x
Then G(x, t) =
= x2 − x − x ln x.
4.8
Green’s Functions
The solution of the initial-value problem is
y = yh + yp = (5x − 3x2 ) + (x2 − x − x ln x) = 4x − 2x2 − x ln x.
28. The Cauchy-Euler initial-value problem x2 y − 2xy + 2y = x ln x, y(1) = 1, y (1) = 0, has
auxiliary equation m(m − 1) − 2m + 2 = m2 − 3m + 2 = (m − 1)(m − 2) = 0 so m1 = 1,
m2 = 2, y(x) = c1 x + c2 x2 , and y = c1 + 2c2 x. The initial conditions give
c1 + c2 = 1
c1 + 2c2 = 0,
so c1 = 2 and c2 = −1, which implies that yh = 2x − x2 . The Wronskian is
2
W x, x
x x2 = x2 .
= 1 2x x(x − t)
tx2 − xt2
. From the standard form of the differential equation
=
2
t
t
ln t
. Then, for x > 1,
we identify the forcing function f (t) =
t
Then G(x, t) =
ˆ
ˆ x
ˆ x
x(x − t) ln t
ln t
dt = x2
dt
t−2 ln t dt − x
t
t
t
1
1
1
ln x 1
1
1
2
2
− +1 −x
(ln x) = x2 − x − x ln x − x(ln x)2 .
=x −
x
x
2
2
x
yp (x) =
The solution of the initial-value problem is
1
1
2
2
y = yh + yp = (2x − x ) + x − x − x ln x − x(ln x) = x − x ln x − x(ln x)2 .
2
2
2
29. The Cauchy-Euler initial-value problem x2 y − 6y = ln x, y(1) = 1, y (1) = 3, has auxiliary
equation m(m − 1) − 6 = m2 − m − 6 = (m − 3)(m + 2) = 0 so m1 = 3, m2 = −2,
y(x) = c1 x3 + c2 x−2 , and y = 3c1 x2 − 2c2 x−3 . The initial conditions give
c1 + c2 = 1
3c1 − 2c2 = 3,
so c1 = 1 and c2 = 0, which implies that yh = x3 . The Wronskian is
3 −2 x3
x−2 = 2
= −5.
W x ,x
3x −2x−3 245
246
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
t3
x3
− 2 . From the standard form of the differential
x2
t
ln t
equation we identify the forcing function f (t) = 2 . Then, for x > 1,
t
ˆ x 3
x3 ln t
1 t
− 2
dt
yp (x) =
x2
t
t2
1 5
ˆ
ˆ
1 3 x −4
1 −2 x
t ln t dt + x
t ln t dt
=− x
5
5
1
1
2
x
1
1 2
1 3
1 −2
1
ln x 1
− + x ln x +
+ x − 3− 3 +
=− x
5
4
2
4
5
9x
3x
9
1
t3 x−2 − x3 t−2
=−
Then G(x, t) =
−5
5
=
1
1
1
1
1
1
−
ln x − x−2 −
−
ln x + x3
20 10
20
45 15
45
=
1
1
1
1
− ln x + x3 − x−2 .
36 6
45
20
The solution of the initial-value problem is
1
1 3
1
46
1
1 −2
1
1
3
− ln x + x − x
− ln x.
= x3 − x−2 +
y = y h + yp = x +
36 6
45
20
45
20
36 6
30. The Cauchy-Euler initial-value problem x2 y − xy + y = x2 , y(1) = 4, y (1) = 3, has
auxiliary equation m(m − 1) − m + 1 = m2 − 2m + 1 = (m − 1)2 = 0 so m1 = m2 = 1,
y(x) = c1 x + c2 x ln x, and y = c1 + c2 (1 + ln x). The initial conditions give
c1 = 4
c1 + c2 = 3,
so c1 = 4 and c2 = −1, which implies that yh = 4x − x ln x. The Wronskian is
x x ln x = x.
W (x, x ln x) = 1 1 + ln x
tx ln x − xt ln t
= x ln x − x ln t. From the standard form of the differential
Then G(x, t) =
t
equation we identify the forcing function f (t) = 1. Then, for x > 1,
ˆ x
ˆ x
ˆ x
(x ln x − x ln t) dt = x ln x
dt − x
ln t dt
yp (x) =
1
1
1
= x ln x(x − 1) − x(x ln x − x + 1) = x − x − x ln x.
2
The solution of the initial-value problem is
y = yh + yp = 4x − x ln x + x2 − x − x ln x = x2 + 3x − 2x ln x.
4.8
31. The initial-value problem is y − y = f (x),
f (x) =
y(0) = 8, y (0) = 2, where
−1,
x<0
1,
x ≥ 0.
We first find
−x
x
yh (x) = 5e + 3e
Then for x < 0,
1
yp (x) = −
2
ˆ
x
x−t
[e
1
yp (x) =
2
and
−(x−t)
−e
0
Green’s Functions
ˆ
x
ex−t − e−(x−t) f (t) dt.
0
1
] dt = − ex
2
ˆ
x
−t
e
0
1
dt + e−x
2
ˆ
x
et dt
0
1
1
1 1 1
1
= − ex (1 − e−x ) + e−x (ex − 1) = − ex + + − e−x
2
2
2
2 2 2
1
1
= 1 − ex − e−x ,
2
2
and for x ≥ 0
1
yp (x) =
2
ˆ
x
x−t
[e
−(x−t)
−e
0
ˆ
1
] dt = ex
2
x
−t
e
0
1
dt − e−x
2
ˆ
x
et dt
0
1
1
1 1 1
1
= ex (1 − e−x ) − e−x (ex − 1) = ex − − + e−x
2
2
2
2 2 2
1
1
= −1 + ex + e−x .
2
2
The solution is
y(x) = yh (x) + yp (x) = 5ex + 3e−x + yp (x),
where
⎧
1 x 1 −x
⎪
⎪
⎨1 − 2 e − 2 e ,
yp (x) =
⎪
⎪
⎩−1 + 1 ex + 1 e−x ,
2
2
32. The initial-value problem is y − y = f (x),
f (x) =
x<0
=
x≥0
and
Then for x < 0,
1
yp (x) =
2
ˆ
0
x
x<0
⎪
⎩
−1 + cosh x,
x≥0
y(0) = 3, y (0) = 2, where
0, x < 0
x,
x ≥ 0.
We first find
5
1
yh (x) = ex + e−x
2
2
⎧
⎪
⎨1 − cosh x,
yp (x) =
1
2
ˆ
x
[ex−t − e−(x−t) f (t) dt.
0
[ex−t − e−(x−t) ]0 dt = 0,
247
248
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
and for x ≥ 0
1
yp (x) =
2
ˆ
x
0
1
= ex
2
[ex−t − e−(x−t) ]t dt
ˆ
x
−t
te
0
1
dt −
2
ˆ
x
tet dt
0
1
1
= ex (1 − e−x − xe−x ) − e−x (1 − ex + xex )
2
2
1
1
1
1 1
1 1
1
= ex − − x − e−x + − x = ex − e−x − x.
2
2 2
2
2 2
2
2
The solution is
where
5
1
y(x) = yh (x) + yp (x) = ex + e−x + yp (x),
2
2
⎧
x<0
⎨0,
0,
x<0
=
yp (x) = 1 x 1 −x
⎩ e − e − x, x ≥ 0
sinh x − x, x ≥ 0
2
2
33. The initial-value problem is y + y = f (x), y(0) = 1, y (0) = −1, where
⎧
⎪
0, x < 0
⎪
⎨
f (x) = 10, 0 ≤ x ≤ 3π
⎪
⎪
⎩ 0, x > 3π.
We first find
ˆ
yh (x) = cos x − sin x
and
x
yp (x) =
sin (x − t)f (t) dt.
0
ˆ
Then for x < 0
x
yp (x) =
sin (x − t)0 dt = 0,
0
for 0 ≤ x ≤ 3π
ˆ
x
yp (x) = 10
sin (x − t) dt = 10 − 10 cos x,
0
and for x > 3π
ˆ
yp (x) = 10
3π
ˆ
sin (x − t) dt +
0
x
sin (x − t)0 dt = −20 cos x.
3π
The solution is
y(x) = yh (x) + yp (x) = cos x − sin x + yp (x),
where
⎧
⎪
⎪
⎨
yp (x) =
0,
x<0
10 − 10 cos x, 0 ≤ x ≤ 3π
⎪
⎪
⎩−20 cos x,
x > 3π.
4.8
Green’s Functions
34. The initial-value problem is y + y = f (x), y(0) = 0, y (0) = 1, where
⎧
⎪
0,
x<0
⎪
⎨
f (x) = cos x, 0 ≤ x ≤ 4π
⎪
⎪
⎩0,
x > 4π
ˆ
We first find
yh (x) = sin x
and
yp (x) =
x
sin (x − t)f (t) dt.
0
ˆ
Then for x < 0
yp (x) =
x
sin (x − t)0 dt = 0,
0
for 0 ≤ x ≤ 4π
ˆ
yp (x) =
0
and for x > 4π
ˆ
4π
yp (x) =
x
1
sin (x − t) cos t dt = x sin x,
2
ˆ
sin (x − t) cos t dt +
0
x
sin (x − t)0 dt = 2π sin x.
4π
The solution is
y(x) = yh (x) + yp (x) = sin x + yp (x),
where
⎧
0,
x<0
⎪
⎪
⎪
⎪
⎨
1
yp (x) =
x sin x, 0 ≤ x ≤ 4π
⎪
2
⎪
⎪
⎪
⎩
2π sin x, x > 4π
To evaluate the integral of
´
sin (x − t) cos t dt we use the identities
sin (A − B) = sin A cos B − cos A sin B
and
cos (A − B) = cos A cos B + sin A sin B.
35. The boundary-value problem is y = f (x), y(0) = 0, y(1) = 0. The solution of the
associated homogeneous equation is y = c1 + c2 x.
(a) To satisfy y(0) = 0 we take y1 (x) = x and to satisfy y(1) = 0 we take y2 (x) = x − 1. The
Wronskian of y1 and y2 is
x x − 1
= 1,
W (y1 , y2 ) = 1
1 so
t(x − 1), 0 ≤ t ≤ x
G(x, t) =
x(t − 1), x ≤ t ≤ 1.
Therefore
ˆ
1
yp (x) =
0
ˆ
ˆ
x
G(x, t)f (t) dt = (x − 1)
tf (t) dt + x
0
x
1
(t − 1)f (t) dt.
249
250
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
(b) By the Product Rule and the Fundamental Theorem of Calculus, the first two derivative
of yp (x) are
yp (x)
ˆ
ˆ
x
= (x − 1)xf (x) +
1
tf (t) dt + x[−(x − 1)f (x)] +
0
(t − 1)f (t) dt
x
yp (x) = (x − 1)[xf (x) + f (x)] + xf (x) + xf (x)
− [(x2 − x)f (x) + (2x − 1)f (x)] − (x − 1)f (x)
= x2 f (x) + xf (x) − xf (x) − f (x) + xf (x) + xf (x) − x2 f (x)
+ xf (x) − 2xf (x) + f (x) − xf (x) + f (x) = f (x)
Thus, yp (x) satisfies the differential equation. To see that the boundary conditions are
satisfied we compute
ˆ
0
yp (0) = (0 − 1)
ˆ
0
and
ˆ
yp (1) = (1 − 1)
1
1
tf (t) dt + 0 ·
(t − 1)f (t) dt = 0
0
ˆ
1
tf (t) dt + 1 ·
0
(t − 1)f (t) dt = 0.
1
36. The boundary-value problem is y = f (x), y(0) = 0, y(1) + y (1) = 0. The solution of the
associated homogeneous equation is y = c1 + c2 x.
(a) To satisfy y(0) = 0 we take y1 (x) = x and to satisfy y(1) + y (1) = 0 we note that
y(1) + y (1) = (c1 + c2 ) + c2 = 0 which implies that c1 = −2c2 . Taking c2 = 1, we find
that c1 = −2, so we have y2 (x) = −2 + x. The Wronskian of y1 and y2 is
x −2 + x
= 2,
W (y1 , y2 ) = 1
1 so
G(x, t) =
⎧
1
⎪
⎪
⎨ 2 t(x − 2),
0≤t≤x
⎪
⎪
⎩ 1 x(t − 2),
2
x≤t≤1
Therefore
ˆ
yp (x) =
0
1
1
G(x, t)f (t) dt = (x − 2)
2
ˆ
x
0
1
tf (t) dt + x
2
ˆ
1
(t − 2)f (t) dt.
x
(b) By the Product Rule and the Fundamental Theorem of Calculus, the first two derivative
4.8
Green’s Functions
of yp (x) are
yp (x)
1
1
= (x − 2)xf (x) +
2
2
ˆ
x
0
1
1
tf (t) dt + x[−(x − 2)f (x)] +
2
2
ˆ
1
(t − 2)f (t) dt
x
1
1
1
yp (x) = (x − 2)[xf (x) + f (x)] + xf (x) + xf (x)
2
2
2
1
1
1
− (x2 − 2x)f (x) − (2x − 2)f (x) − (x − 2)f (x)
2
2
2
1 2 =
x f (x) + xf (x) − 2xf (x) − 2f (x) + xf (x) + xf (x) − x2 f (x)
2
+ 2xf (x) − 2xf (x) + 2f (x) − xf (x) + 2f (x) = f (x)
Thus, yp (x) satisfies the differential equation. To see that the boundary conditions are
satisfied we first compute
ˆ 0
ˆ 1
1
1
tf (t) dt + (0)
(t − 2)f (t) dt = 0.
yp (0) = (0 − 2)
2
2
0
0
Next we use yp (x) found at the beginning of this part of the solution to compute
ˆ 1
ˆ 1
1
1
1
yp (1) + yp (1) = (1 − 2)
tf (t) dt + (1)
(t − 2)f (t) dt + (1 − 2)f (x)
2
2
2
0
1
ˆ 1
ˆ 1
1
1
1
tf (t) dt + (1)[−(1 − 2)f (x)] + (1)
(1 − 2)f (t) dt
+
2 0
2
2
1
ˆ 1
ˆ
1
1
1 1
1
tf (t) dt + (−1)f (1) +
tf (t) dt + f (1) = 0.
= (−1)
2
2
2 0
2
0
37. If f (x) = 1 in Problem 35, then
ˆ
ˆ
x
yp (x) = (x − 1)
1
x
1
1
(t − 1) dt = x2 − x.
2
2
ˆ
1
t dt + x
0
38. If f (x) = x in Problem 36, then
1
yp (x) = (x − 2)
2
ˆ
x
0
1
t dt + x
2
2
x
1
1
(t − 2)t dt = x3 − x.
6
3
39. The boundary-value problem is y + y = 1, y(0) = 0, y(1) = 0. The solution of the
associated homogeneous equation is y = c1 cos x + c2 sin x. Since y(0) = c1 cos 0 + c2 sin 0 =
c1 = 0, we take y1 (x) = sin x. To satisfy y(1) = 0 we note that y(1) = c1 cos 1 + c2 sin 1 = 0
which implies that c1 = −c2 sin 1/ cos 1 so
y(x) = −c2
c2
sin 1
cos x + c2 sin x = −
(sin x cos 1 − cos x sin 1).
cos 1
cos 1
Taking c2 = − cos 1, we have
y2 (x) = sin x cos 1 − cos x sin 1 = sin (x − 1).
251
252
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
The Wronskian of y1 and y2 is
sin x sin (x − 1) = sin x cos (x − 1) − cos x sin (x − 1) = sin [x − (x − 1)] = sin 1,
W (y1 , y2 ) = cos x cos (x − 1)
so
G(x, t) =
⎧
sin t sin (x − 1)
⎪
⎪
, 0≤t≤x
⎨
sin 1
⎪
⎪
⎩ sin x sin (t − 1) ,
sin 1
x ≤ t ≤ 1.
Therefore, taking f (t) = 1,
ˆ
ˆ
ˆ 1
sin (x − 1) x
sin x 1
yp (x) =
G(x, t) dt =
sin t dt +
sin (t − 1) dt
sin 1
sin 1 x
0
0
=
sin x
sin (x − 1) sin x
sin (x − 1)
(− cos x + 1) +
[−1 + cos (x − 1)] =
−
+ 1.
sin 1
sin 1
sin 1
sin 1
40. The boundary-value problem is y + 9y = 1, y(0) = 0, y (π) = 0. The solution of the
associated homogeneous equation is y = c1 cos 3x + c2 sin 3x. Since
y(0) = c1 cos 0+c2 sin 0 = c1 = 0, we take y1 (x) = sin 3x. Since y (x) = −3c1 sin 3x+3c2 cos 3x
we see that y (π) = −3c2 . Then y (π) = 0, implies that −3c2 = 0 or c2 = 0. Taking c1 = 1
we have y2 (x) = cos 3x. The Wronskian of y1 and y2 is
sin 3x
cos 3x W (y1 , y2 ) = = −3,
3 cos 3x −3 sin 3x
so
⎧
1
⎪
⎪
⎨− 3 sin 3t cos 3x, 0 ≤ t ≤ x
G(x, t) =
⎪
⎪
⎩− 1 sin 3x cos 3t, x ≤ t ≤ π
3
Therefore, taking f (t) = 1,
ˆ x
ˆ π
ˆ π
1
1
G(x, t) dt = − cos 3x
sin 3t dt − sin 3x
cos 3t dt
yp (x) =
3
3
0
0
x
1 1
1
1
1 1
1
− cos 3x − sin 3x − sin 3x = − cos 3x.
= − cos 3x
3
3 3
3
3
9 9
41. The boundary-value problem is y − 2y + 2y = ex , y(0) = 0, y(π/2) = 0. The auxiliary
equation is
√
2± 4−8
2
= 1 ± i.
m − 2m + 2 = 0 so m =
2
The solution of the associated homogeneous equation is then y = ex (c1 cos x + c2 sin x). It is
easily seen that
and
y2 (x) = ex cos x
y1 (x) = ex sin x
4.8
Green’s Functions
satisfy the boundary conditions. The Wronskian of y1 and y2 is
ex cos x
ex sin x
W (y1 , y2 ) = x
x
x
x
e cos x + e sin x −e sin x + e cos x
= e2x − sin2 x + sin x cos x − (cos2 x + sin x cos x) = −e2x ,
so
⎧ t
e sin tex cos x
⎪
⎪
, 0≤t≤x
⎪
⎨
−e2t
G(x, t) =
⎪ ex sin xet cos t
⎪
⎪
⎩
, x ≤ t ≤ π/2
−e2t
Therefore, taking f (t) = et ,
ˆ
yp (x) =
π/2
ˆ
G(x, t)et dt = −ex cos x
0
x
ˆ
sin t dt − ex sin x
0
π/2
cos t dt
x
= −ex cos x(1 − cos x) − ex sin x(1 − sin x) = −ex cos x − ex sin x + ex .
42. The boundary-value problem is y − y = e2x , y(0) = 0, y(1) = 0. The solution of the
associated homogeneous equation is y = c1 + c2 ex . Since y(0) = c1 + c2 = 0, we see that
c1 = −c2 and we take y1 (x) = 1 − ex . To satisfy y(1) = 0 we note that y(1) = c1 + c2 e = 0
which implies that c1 = −c2 e so
y(x) = −c2 e + c2 ex = −c2 e(1 − ex−1 ).
Taking c2 = −1/e, we have y2 (x) = 1 − ex−1 . The Wronskian of y1 and y2 is
1 − ex 1 − ex−1 = −ex−1 + e2x−1 + ex − e2x−1 = ex − ex−1 = ex (1 − e−1 ),
W (y1 , y2 ) = −ex
−ex−1 so
G(x, t) =
⎧
(1 − et )(1 − ex−1 )
⎪
⎪
, 0≤t≤x
⎪
⎨
et (1 − e−1 )
⎪
⎪
(1 − ex )(1 − et−1 )
⎪
⎩
,
et (1 − e−1 )
x≤t≤1
Therefore, taking f (t) = e2t ,
ˆ 1
ˆ
ˆ 1
1 − ex−1 x t
1 − ex
2t
2t
G(x, t)e dt =
(e − e ) dt +
(et − e2t−1 ) dt
yp (x) =
−1
−1
1
−
e
1
−
e
0
0
x
x−1
x
1
1 2x 1
1−e
1 2x−1
1−e
x
x
e − e −
+
e−e + e
=
1 − e−1
2
2
1 − e−1 2
2
1
1
1
1
= e2x − ex − ex+1 + e.
2
2
2
2
253
254
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
43. The Cauchy-Euler boundary-value problem x2 y + xy = 1, y(e−1 ) = 0, y(1) = 0 has
auxiliary equation m(m−1)+m = m2 = 0 so y(x) = c1 +c2 ln x. Since y(e−1 ) = c1 +c2 ln e−1 =
c1 −c2 = 0, c1 = c2 and y(x) = c2 +c2 ln x = c2 (1+ln x). Taking c2 = 1 we have y1 (x) = 1+ln x.
To satisfy y(1) = 0 we note that y(1) = c1 + c2 ln 1 = c1 = 0 which implies that y(x) = c2 ln x.
Taking c2 = 1 we find y2 (x) = ln x. The Wronskian of y1 and y2 is
1 + ln x ln x 1
= ,
W (y1 , y2 ) = 1/x
1/x x
so
⎧
(1 + ln t)(ln x)
⎪
⎪
, 0≤t≤x
⎪
⎨
1/t
G(x, t) =
⎪
⎪
(1 + ln x)(ln t)
⎪
⎩
, x ≤ t ≤ 1.
1/t
1
From the standard form of the differential equation we identify the forcing function f (t) = 2 .
t
Then,
ˆ 1
ˆ 1
ˆ x 1 ln t
ln t
1
+
dt + (1 + ln x)
dt
G(x, t) 2 dt = ln x
yp (x) =
t
t
t
t
e−1
e−1
x
1
1
1
1
1
2
2
= (ln x) ln x + (ln x) +
+ (1 + ln x) − (ln x) = (ln x)2 + ln x.
2
2
2
2
2
44. The Cauchy-Euler boundary-value problem x2 y −4xy +6y = x4 , y(1)−y (1) = 0, y(3) = 0
has auxiliary equation m(m − 1) − 4m + 6 = m2 − 5m + 6 = (m − 2)(m − 3) = 0 so
y(x) = c1 x2 + c2 x3 . Since y(1) − y (1) = (c1 + c2 ) − (2c1 + 3c2 ) = −c1 − 2c2 = 0 we have
c1 = −2c2 and y(x) = −2c2 x2 + c2 x3 . Taking c2 = −1 we have y1 (x) = 2x2 − x3 . From
y(3) = 9c1 + 27c2 = 0 we have c1 = −3c2 , so y(x) = −3c2 x2 + c2 x3 = −c2 (3x2 − x3 ). Again,
letting c2 = −1 we have y2 (x) = 3x2 − x3 . The Wronskian of y1 and y2 is
2
2x − x3 3x2 − x3 = x4 ,
W (y1 , y2 ) = 4x − 3x2 6x − 3x2 so
⎧
(2t2 − t3 )(3x2 − x3 )
⎪
⎪
⎪
, 0≤t≤x
⎨
t4
G(x, t) =
⎪
⎪
(2x2 − x3 )(3t2 − t3 )
⎪
⎩
, x ≤ t ≤ 3.
t4
From the standard form of the differential equation we identify the forcing function f (t) = t2 .
Then,
ˆ 3
ˆ x
ˆ 3
G(x, t)t2 dt = (3x2 − x3 )
(2 − t) dt + (2x2 − x3 )
(3 − t) dt
yp (x) =
1
1
x
1 2 3
9
1 2
2
3
2
3
+ (2x − x )
− 3x + x
= (3x − x ) 2x − x −
2
2
2
2
9
1
= x2 − 3x3 + x4 .
2
2
4.9
4.9
Solving Systems of Linear Equations
Solving Systems of Linear Equations
1. From Dx = 2x − y and Dy = x we obtain y = 2x − Dx, Dy = 2Dx − D2 x, and
(D2 − 2D + 1)x = 0. The solution is
x = c1 et + c2 tet
y = (c1 − c2 )et + c2 tet .
2. From Dx = 4x + 7y and Dy = x − 2y we obtain y = 17 Dx − 47 x, Dy = 17 D2 x − 47 Dx, and
(D2 − 2D − 15)x = 0. The solution is
x = c1 e5t + c2 e−3t
1
y = c1 e5t − c2 e−3t .
7
3. From Dx = −y +t and Dy = x−t we obtain y = t−Dx, Dy = 1−D2 x, and (D2 +1)x = 1+t.
The solution is
x = c1 cos t + c2 sin t + 1 + t
y = c1 sin t − c2 cos t + t − 1.
4. From Dx − 4y = 1 and x + Dy = 2 we obtain y = 14 Dx − 14 , Dy = 14 D2 x, and (D2 + 1)x = 2.
The solution is
x = c1 cos 2t + c2 sin 2t + 2
1
1
1
y = c2 cos 2t − c1 sin 2t − .
4
4
4
5. From (D2 + 5)x − 2y = 0 and −2x + (D2 + 2)y = 0 we obtain y =
1
4
2
2
2
2 (D + 5D )x, and (D + 1)(D + 6)x = 0. The solution is
1
2
2 (D
+ 5)x, D2 y =
√
√
x = c1 cos t + c2 sin t + c3 cos 6 t + c4 sin 6 t
√
√
1
1
y = 2c1 cos t + 2c2 sin t − c3 cos 6 t − c4 sin 6 t.
2
2
6. From (D + 1)x + (D − 1)y = 2 and 3x + (D + 2)y = −1 we obtain x = − 13 − 13 (D + 2)y,
Dx = − 13 (D2 + 2D)y, and (D2 + 5)y = −7. The solution is
√
7
5 t + c2 sin 5 t −
5
√
√
5
2
c2 cos 5 t +
− c1 −
3
3
y = c1 cos
x=
√
√
2
5
c1 − c2
3
3
√
3
sin 5 t + .
5
255
256
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
7. From D2 x = 4y + et and D2 y = 4x − et we obtain y = 14 D2 x − 14 et , D2 y = 14 D4 x − 14 et , and
(D2 + 4)(D − 2)(D + 2)x = −3et . The solution is
1
x = c1 cos 2t + c2 sin 2t + c3 e2t + c4 e−2t + et
5
1
2t
−2t
y = −c1 cos 2t − c2 sin 2t + c3 e + c4 e
− et .
5
8. From (D2 + 5)x + Dy = 0 and (D + 1)x + (D − 4)y = 0 we obtain (D − 5)(D2 + 4)x = 0 and
(D − 5)(D2 + 4)y = 0. The solution is
x = c1 e5t + c2 cos 2t + c3 sin 2t
y = c4 e5t + c5 cos 2t + c6 sin 2t.
Substituting into (D + 1)x + (D − 4)y = 0 gives
(6c1 + c4 )e5t + (c2 + 2c3 − 4c5 + 2c6 ) cos 2t + (−2c2 + c3 − 2c5 − 4c6 ) sin 2t = 0
so that c4 = −6c1 , c5 = 12 c3 , c6 = − 12 c2 , and
1
1
y = −6c1 e5t + c3 cos 2t − c2 sin 2t.
2
2
9. From Dx + D2 y = e3t and (D + 1)x + (D − 1)y = 4e3t we obtain D(D2 + 1)x = 34e3t and
D(D2 + 1)y = −8e3t . The solution is
y = c1 + c2 sin t + c3 cos t −
4 3t
e
15
x = c4 + c5 sin t + c6 cos t +
17 3t
e .
15
Substituting into (D + 1)x + (D − 1)y = 4e3t gives
(c4 − c1 ) + (c5 − c6 − c3 − c2 ) sin t + (c6 + c5 + c2 − c3 ) cos t = 0
so that c4 = c1 , c5 = c3 , c6 = −c2 , and
x = c1 − c2 cos t + c3 sin t +
17 3t
e .
15
10. From D2 x − Dy = t and (D + 3)x + (D + 3)y = 2 we obtain D(D + 1)(D + 3)x = 1 + 3t and
D(D + 1)(D + 3)y = −1 − 3t. The solution is
1
x = c1 + c2 e−t + c3 e−3t − t + t2
2
1
y = c4 + c5 e−t + c6 e−3t + t − t2 .
2
4.9
Solving Systems of Linear Equations
Substituting into (D + 3)x + (D + 3)y = 2 and D2 x − Dy = t gives
3(c1 + c4 ) + 2(c2 + c5 )e−t = 2
and
(c2 + c5 )e−t + 3(3c3 + c6 )e−3t = 0
so that c4 = 2/3 − c1 , c5 = −c2 , c6 = −3c3 , and
y=
2
1
− c1 − c2 e−t − 3c3 e−3t + t − t2 .
3
2
11. From (D2 − 1)x − y = 0 and (D − 1)x + Dy = 0 we obtain y = (D2 − 1)x, Dy = (D3 − D)x,
and (D − 1)(D2 + D + 1)x = 0. The solution is
√
√ 3
3
t + c3 sin
t
x = c1 et + e−t/2 c2 cos
2
2
y=
√
3
3
c3
− c2 −
2
2
√
−t/2
e
3
t+
cos
2
√
3
3
c2 − c3
2
2
√
−t/2
e
sin
3
t.
2
12. From (2D2 − D − 1)x − (2D + 1)y = 1 and (D − 1)x + Dy = −1 we obtain
(2D + 1)(D − 1)(D + 1)x = −1 and (2D + 1)(D + 1)y = −2. The solution is
x = c1 e−t/2 + c2 e−t + c3 et + 1
y = c4 e−t/2 + c5 e−t − 2.
Substituting into (D − 1)x + Dy = −1 gives
3
1
− c1 − c4 e−t/2 + (−2c2 − c5 )e−t = 0
2
2
so that c4 = −3c1 , c5 = −2c2 , and
y = −3c1 e−t/2 − 2c2 e−t − 2.
13. From (2D − 5)x + Dy = et and (D − 1)x + Dy = 5et we obtain Dy = (5 − 2D)x + et and
(4 − D)x = 4et . Then
4
x = c1 e4t + et
3
4t
t
and Dy = −3c1 e + 5e so that
3
y = − c1 e4t + c2 + 5et .
4
14. From Dx + Dy = et and (−D2 + D + 1)x + y = 0 we obtain y = (D2 − D − 1)x,
Dy = (D3 − D2 − D)x, and D2 (D − 1)x = et . The solution is
x = c1 + c2 t + c3 et + tet
y = −c1 − c2 − c2 t − c3 et − tet + et .
257
258
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
15. Multiplying the first equation by D + 1 and the second equation by D2 + 1 and subtracting
we obtain (D4 − D2 )x = 1. Then
1
x = c1 + c2 t + c3 et + c4 e−t − t2 .
2
Multiplying the first equation by D + 1 and subtracting we obtain D2 (D + 1)y = 1. Then
1
y = c5 + c6 t + c7 e−t − t2 .
2
Substituting into (D − 1)x + (D2 + 1)y = 1 gives
(−c1 + c2 + c5 − 1) + (−2c4 + 2c7 )e−t + (−1 − c2 + c6 )t = 1
so that c5 = c1 − c2 + 2, c6 = c2 + 1, and c7 = c4 . The solution of the system is
1
x = c1 + c2 t + c3 et + c4 e−t − t2
2
1
y = (c1 − c2 + 2) + (c2 + 1)t + c4 e−t − t2 .
2
16. From D2 x − 2(D2 + D)y = sin t and x + Dy = 0 we obtain x = −Dy, D2 x = −D3 y, and
D(D2 + 2D + 2)y = − sin t. The solution is
y = c1 + c2 e−t cos t + c3 e−t sin t +
2
1
cos t + sin t
5
5
x = (c2 + c3 )e−t sin t + (c2 − c3 )e−t cos t +
2
1
sin t − cos t.
5
5
17. From Dx = y, Dy = z. and Dz = x we obtain x = D2 y = D3 x so that
(D − 1)(D2 + D + 1)x = 0,
t
−t/2
x = c1 e + e
t
y = c1 e +
√ 3
3
t + c3 cos
t ,
c2 sin
2
2
√
√
3
1
c3
− c2 −
2
2
√
−t/2
e
3
t+
sin
2
√
3
1
c2 − c3
2
2
√
−t/2
e
cos
3
t,
2
and
t
z = c1 e +
√
1
3
c3
− c2 +
2
2
√
−t/2
e
3
t+
sin
2
√
1
3
−
c2 − c3
2
2
√
−t/2
e
cos
3
t.
2
4.9
Solving Systems of Linear Equations
18. From Dx + z = et , (D − 1)x + Dy + Dz = 0, and x + 2y + Dz = et we obtain z = −Dx + et ,
Dz = −D2 x + et , and the system (−D2 + D − 1)x + Dy = −et and (−D2 + 1)x + 2y = 0.
Then y = 12 (D2 − 1)x, Dy = 12 D(D2 − 1)x, and (D − 2)(D2 + 1)x = −2et so that the solution
is
x = c1 e2t + c2 cos t + c3 sin t + et
3
y = c1 e2t − c2 cos t − c3 sin t
2
z = −2c1 e2t − c3 cos t + c2 sin t.
19. Write the system in the form
Dx − 6y = 0
x − Dy + z = 0
x + y − Dz = 0.
Multiplying the second equation by D and adding to the third equation we obtain (D + 1)x −
(D2 − 1)y = 0. Eliminating y between this equation and Dx − 6y = 0 we find
(D3 − D − 6D − 6)x = (D + 1)(D + 2)(D − 3)x = 0.
Thus
x = c1 e−t + c2 e−2t + c3 e3t ,
and, successively substituting into the first and second equations, we get
1
1
1
y = − c1 e−t − c2 e−2t + c3 e3t
6
3
2
5
1
1
z = − c1 e−t − c2 e−2t + c3 e3t .
6
3
2
20. Write the system in the form
(D + 1)x − z = 0
(D + 1)y − z = 0
x − y + Dz = 0.
Multiplying the third equation by D + 1 and adding to the second equation we obtain (D +
1)x + (D2 + D − 1)z = 0. Eliminating z between this equation and (D + 1)x − z = 0 we find
D(D + 1)2 x = 0. Thus
x = c1 + c2 e−t + c3 te−t ,
and, successively substituting into the first and third equations, we get
y = c1 + (c2 − c3 )e−t + c3 te−t
z = c1 + c3 e−t .
259
260
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
21. From (D + 5)x + y = 0 and 4x − (D + 1)y = 0 we obtain y = −(D + 5)x so that Dy =
−(D2 + 5D)x. Then 4x + (D2 + 5D)x + (D + 5)x = 0 and (D + 3)2 x = 0. Thus
x = c1 e−3t + c2 te−3t
y = −(2c1 + c2 )e−3t − 2c2 te−3t .
Using x(1) = 0 and y(1) = 1 we obtain
c1 e−3 + c2 e−3 = 0
−(2c1 + c2 )e−3 − 2c2 e−3 = 1
or
c1 + c2 = 0
2c1 + 3c2 = −e3 .
Thus c1 = e3 and c2 = −e3 . The solution of the initial-value problem is
x = e−3t+3 − te−3t+3
y = −e−3t+3 + 2te−3t+3 .
22. From Dx − y = −1 and 3x + (D − 2)y = 0 we obtain x = − 13 (D − 2)y so that
Dx = − 13 (D2 − 2D)y. Then − 13 (D2 − 2D)y = y − 1 and (D2 − 2D + 3)y = 3. Thus
√
√ t
y = e c1 cos 2 t + c2 sin 2 t + 1
and
√
√
√ 2
√
1 2 c1 + c2 sin 2 t + .
x = et c1 − 2 c2 cos 2 t +
3
3
Using x(0) = y(0) = 0 we obtain
c1 + 1 = 0
2
√
1
c1 − 2 c2 + = 0.
3
3
Thus c1 = −1 and c2 =
√
2/2. The solution of the initial-value problem is
√
√
√
2
2
2
t
sin 2 t +
x = e − cos 2 t −
3
6
3
t
y=e
− cos
√
√
2t +
√
2
sin 2 t
2
+ 1.
23. Equating Newton’s law with the net forces in the x- and y-directions gives m d2 x/dt2 = 0
and m d2 y/dt2 = −mg, respectively. From mD2 x = 0 we obtain x(t) = c1 t + c2 , and from
mD2 y = −mg or D2 y = −g we obtain y(t) = − 12 gt2 + c3 t + c4 .
4.9
Solving Systems of Linear Equations
24. From Newton’s second law in the x-direction we have
m
dx
d2 x
1 dx
= −|c| .
= −k cos θ = −k
dt2
v dt
dt
In the y-direction we have
m
1 dy
dy
d2 y
= −mg − |c| .
= −mg − k sin θ = −mg − k
2
dt
v dt
dt
From mD2 x + |c|Dx = 0 we have D(mD + |c|)x = 0 so that (mD + |c|)x = c1 or
(D + |c|/m)x = c2 . This is a linear first-order differential equation. An integrating factor is
´
e |c|dt/m = e|c|t/m so that
d |c|t/m x = c2 e|c|t/m
e
dt
and e|c|t/m x = (c2 m/|c|)e|c|t/m + c3 . The general solution of this equation is
x(t) = c4 + c3 e−|c|t/m . From (mD2 + |c|D)y = −mg we have D(mD + |c|)y = −mg so that
(mD + |c|)y = −mgt + c1 or (D + |c|/m)y = −gt + c2 . This is a linear first-order differential
´
equation with integrating factor e |c| dt/m = e|c|t/m . Thus
d |c|t/m e
y = (−gt + c2 )e|c|t/m
dt
e|c|t/m y = −
mg |c|t/m m2 g |c|t/m
te
+ 2 e
+ c3 e|c|t/m + c4
|c|
c
and
y(t) = −
mg
m2 g
t + 2 + c3 + c4 e−|c|t/m .
|c|
c
25. Multiplying the first equation by D + 1 and the second equation by D we obtain
D (D + 1) x − 2D (D + 1) y = 2t + t2
D (D + 1) x − 2D (D + 1) y = 0.
This leads to 2t + t2 = 0, so the system has no solutions.
26. The FindRoot application of Mathematica gives a solution of x1 (t) = x2 (t) as approximately
t = 13.73 minutes. So tank B contains more salt than tank A for t > 13.73 minutes.
27. (a) Separating variables in the first equation, we have dx1 /x1 = −dt/50, so x1 = c1 e−t/50 .
From x1 (0) = 15 we get c1 = 15. The second differential equation then becomes
15
2
dx2
= e−t/50 − x2
dt
50
75
or
dx2
2
3
+ x2 = e−t/50 .
dt
75
10
´
This differential equation is linear and has the integrating factor e
3
3
d 2t/75 e
x2 = e−t/50+2t/75 = et/150
dt
10
10
2 dt/75
= e2t/75 . Then
261
262
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
so
e2t/75 x2 = 45et/150 + c2
and
x2 = 45e−t/50 + c2 e−2t/75 .
From x2 (0) = 10 we get c2 = −35. The third differential equation then becomes
90
70
1
dx3
= e−t/50 − e−2t/75 − x3
dt
75
75
25
or
dx3
1
6
14
+ x3 = e−t/50 − e−2t/75 .
dt
25
5
15
´
This differential equation is linear and has the integrating factor e
dt/25
= et/25 . Then
d t/25 6 −t/50+t/25 14 −2t/75+t/25 6 t/50 14 t/75
e x3 = e
− e
= e
− e ,
dt
5
15
5
15
so
et/25 x3 = 60et/50 − 70et/75 + c3
and
x3 = 60e−t/50 − 70e−2t/75 + c3 e−t/25 .
From x3 (0) = 5 we get c3 = 15. The solution of the initial-value problem is
x1 (t) = 15e−t/50
x2 (t) = 45e−t/50 − 35e−2t/75
x3 (t) = 60e−t/50 − 70e−2t/75 + 15e−t/25 .
(b) x kg salt
14
12
10
8
6
4
2
x1
x2
x3
50
100
150
200
time
(c) Solving x1 (t) = 12 , x2 (t) = 12 , and x3 (t) = 12 , FindRoot gives, respectively,
t1 = 170.06 min, t2 = 214.7 min, and t3 = 224.4 min. Thus, all three tanks will contain less than or equal to 0.5 kg of salt after 224.4 minutes.
4.10
4.10
Nonlinear Differential Equations
Nonlinear Differential Equations
1. We have y1 = y1 = ex , so
(y1 )2 = (ex )2 = e2x = y12 .
Also, y2 = − sin x and y2 = − cos x, so
(y2 )2 = (− cos x)2 = cos2 x = y22 .
However, if y = c1 y1 + c2 y2 , we have (y )2 = (c1 ex − c2 cos x)2 and y 2 = (c1 ex + c2 cos x)2 .
Thus (y )2 = y 2 .
2. We have y1 = y1 = 0, so
1
1
y1 y1 = 1 · 0 = 0 = (0)2 = (y1 )2 .
2
2
Also, y2 = 2x and y2 = 2, so
1
1
y2 y2 = x2 (2) = 2x2 = (2x)2 = (y2 )2 .
2
2
However, if y = c1 y1 + c2 y2 , we have yy = (c1 · 1 + c2 x2 )(c1 · 0 + 2c2 ) = 2c2 (c1 + c2 x2 ) and
1 2
1
1 2
2
2 2
2 (y ) = 2 [c1 · 0 + c2 (2x)] = 2c2 x . Thus yy = 2 (y ) .
3. Let u = y so that u = y . The equation becomes u = −u − 1 which is separable. Thus
du
= −dx
+1
u2
tan−1 u = −x + c1
y = tan (c1 − x)
y = ln | cos (c1 − x)| + c2 .
4. Let u = y so that u = y . The equation becomes u = 1 + u2 . Separating variables we obtain
du
= dx
1 + u2
tan−1 u = x + c1
u = tan (x + c1 )
y = − ln | cos (x + c1 )| + c2 .
263
264
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
5. Let u = y so that u = y . The equation becomes x2 u + u2 = 0. Separating variables we
obtain
dx
du
=− 2
2
u
x
−
1
c1 x + 1
1
= + c1 =
u
x
x
1
1
x
=
−1
u=−
c1 x + 1
c1 c1 x + 1
y=
1
1
ln |c1 x + 1| − x + c2 .
2
c1
c1
6. Let u = y so that u = y . The equation becomes e−x u = u2 . Separating variables we obtain
du
= ex dx
u2
−
1
= ex + c1
u
u=−
y=
1
e−x
1 −c1 e−x
=
−
=
ex + c1
1 + c1 e−x
c1 1 + c1 e−x
1 ln 1 + c1 e−x + c2
c1
7. Let u = y so that y = u du/dy. The equation becomes yu du/dy + u2 + 1 = 0. Separating
variables we obtain
dy
u du
+
=0
2
u +1
y
1 2
ln u + 1 + ln |y| = ln c1
2
2
u + 1 y 2 = c21
c21 − y 2
y2
c21 − y 2
dy
=±
dx
y
u2 =
y
± 2
dy = dx
c1 − y 2
∓ c21 − y 2 = x + c2
(x + c2 )2 + y 2 = c21
4.10
Nonlinear Differential Equations
8. Let u = y so that y = u du/dy. The equation becomes (y + 1)u du/dy = u2 . Separating
variables we obtain
dy
du
=
u
y+1
ln |u| = ln |y + 1| + ln c1
u = c1 (y + 1)
dy
= c1 dx
y+1
ln |y + 1| = c1 x + c2
y + 1 = c3 ec1 x .
9. Let u = y so that y = u du/dy. The equation becomes u du/dy + 2yu3 = 0. Separating
variables we obtain
du
+ 2y dy = 0
u2
−
1
+ y2 = c
u
u=
y2
2
y + c1 dy = dx
1
+ c1
1 3
y + c1 y = x + c2 .
3
10. Let u = y so that y = u du/dy. The equation becomes y 2 u du/dy = u. Separating variables
we obtain
du =
dy
y2
1
u = − + c1
y
y =
c1 y − 1
y
y
dy = dx
c1 y − 1
1
1
1+
dy = dx
c1
c1 y − 1
(for c1 = 0)
1
1
y + 2 ln |y − 1| = x + c2 .
c1
c1
If c1 = 0, then y dy = −dx and another solution is 12 y 2 = −x + c2 .
265
266
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
11. Letting u = y we have
y =
du dy
du
du
=
=u
dx
dy du
dy
2y y = 1
so
becomes
2u2
du
= 1.
dy
Then separating variables, integrating, and simplifying, we have
2u2 du = dy
2 3
u = y + c1
3
1/3
3
dy
u=
=
y + c2
2
dx
−1/3
3
y + c2
dy = dx
2
2/3
3
y + c2
= x + c3
2
3
y + c2 = (x + c3 )3/2 .
2
Now
3/2
y(0) = 2 implies 3 + c2 = c3
Thus c2 = −2 and
and y (0) = 1 implies c3 = 1.
3
y − 2 = (x + 1)3/2 . The solution of the initial-value problem is
2
y=
4
2
(x + 1)3/2 + .
3
3
12. Letting u = y the differential equation becomes u + xu2 = 0. Separating variables, integrating, and simplifying we have
du
= −xu2
dx
u−2 du = −x dx
−
1
1
= − x2 + c2
u
2
y =
x2
2
.
+ c2
Using the initial conditions y (1) = 2 we have 2 = 2/(1 + c2 ), so c2 = 0 and y = 2x−2 .
INtegrating we find y = −2x−1 + c3 . Using the other initial condition, y(1) = 4, we have
4 = −2 + c3 so c3 = 6 and the solution of the intial-value problem is
y =6−
2
.
x
4.10
13. (a) x
Nonlinear Differential Equations
y
10
–π /2
3π /2 x
–10
(b) Let u = y so that y = u du/dy. The equation becomes u du/dy + yu = 0. Separating
variables we obtain
du = −y dy
1
u = − y 2 + c1
2
1
y = − y 2 + c1 .
2
When x = 0, y = 1 and y = −1 so −1 = −1/2 + c1 and c1 = −1/2. Then
1
1
dy
= − y2 −
dx
2
2
1
dy
= − dx
y2 + 1
2
1
tan−1 y = − x + c2
2
1
y = tan − x + c2 .
2
When x = 0, y = 1 so 1 = tan c2 and c2 = π/4. The solution of the initial-value problem
is
π 1
− x .
y = tan
4 2
The graph is shown in part (a).
(c) The interval of definition is −π/2 < π/4 − x/2 < π/2 or −π/2 < x < 3π/2.
267
268
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
14. Let u = y so that u = y . The equation becomes
√
(u )2 + u2 = 1 which results in u = ± 1 − u2 . To solve
√
u = 1 − u2 we separate variables:
√
du
= dx
1 − u2
y
2
2π x
–2π
sin−1 u = x + c1
u = sin (x + c1 )
y = sin (x + c1 ).
When x = π/2, y =
√
3/2, so
√
3/2 = sin (π/2 + c1 ) and c1 = −π/6. Thus
π
y = sin x −
6
π
y = − cos x −
+ c2 .
6
When x = π/2, y = 1/2, so 1/2 = − cos (π/2 − π/6) + c2 = −1/2 + c2 and c2 = 1. The
solution of the initial-value problem is y = 1 − cos (x − π/6).
√
To solve u = − 1 − u2 we separate variables:
√
y
1
du
= −dx
1 − u2
–2π
cos−1 u = x + c1
u = cos (x + c1 )
2π
x
–1
y = cos (x + c1 ).
When x = π/2, y =
√
3/2, so
√
3/2 = cos (π/2 + c1 ) and c1 = −π/3. Thus
π
y = cos x −
3
π
+ c2 .
y = sin x −
3
When x = π/2, y = 1/2, so 1/2 = sin (π/2 − π/3) + c2 = 1/2 + c2 and c2 = 0. The solution
of the initial-value problem is y = sin (x − π/3).
4.10
Nonlinear Differential Equations
15. Let u = y so that u = y . The equation becomes u − (1/x)u = (1/x)u3 , which is Bernoulli.
Using w = u−2 we obtain dw/dx + (2/x)w = −2/x. An integrating factor is x2 , so
d 2
[x w] = −2x
dx
x2 w = −x2 + c1
c1
x2
c1
u−2 = −1 + 2
x
x
u= √
c1 − x2
w = −1 +
x
dy
=√
dx
c1 − x2
y = − c1 − x2 + c2
c1 − x2 = (c2 − y)2
x2 + (c2 − y)2 = c1 .
16. Let u = y so that u = y . The equation becomes u − (1/x)u = u2 , which is a Bernoulli
differential equation. Using the substitution w = u−1 we obtain dw/dx + (1/x)w = −1. An
integrating factor is x, so
d
[xw] = −x
dx
1
1
w =− x+ c
2
x
1
c1 − x2
=
u
2x
2x
c1 − x2
y = − ln c1 − x2 + c2 .
u=
269
270
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
In Problems 17–20 the thinner curve is obtained using a numerical solver, while the thicker curve
is the graph of the Taylor polynomial.
17. We look for a solution of the form
y(x) = y(0) + y (0)x +
1 1
1
1
y (0)x2 + y (0)x3 + y (4) (0)x4 + y (5) (0)x5 .
2!
3!
4!
5!
From y (x) = x + y 2 we compute
y
y (x) = 1 + 2yy
40
y (4) (x) = 2yy + 2(y )2
y (5) (x) = 2yy + 6y y .
30
Using y(0) = 1 and y (0) = 1 we find
y (0) = 1,
y (0) = 3,
y (4) (0) = 4,
y (5) (0) = 12.
An approximate solution is
20
10
1
1
1
1
y(x) = 1 + x + x2 + x3 + x4 + x5 .
2
2
6
10
0.5 1 1.5 2 2.5 3
x
18. We look for a solution of the form
y(x) = y(0) + y (0)x +
1 1
1
1
y (0)x2 + y (0)x3 + y (4) (0)x4 + y (5) (0)x5 .
2!
3!
4!
5!
From y (x) = 1 − y 2 we compute
y
10
y (x) = −2yy y (4) (x) = −2yy − 2(y )2
y (5) (x) = −2yy − 6y y .
5
Using y(0) = 2 and y (0) = 3 we find
y (0) = −3,
y (0) = −12,
y
(4)
(0) = −6,
y
(5)
0.5 1 1.5 2 2.5 3
(0) = 102.
An approximate solution is
3
1
17
y(x) = 2 + 3x − x2 − 2x3 − x4 + x5 .
2
4
20
–5
–10
x
4.10
Nonlinear Differential Equations
271
19. We look for a solution of the form
y(x) = y(0) + y (0)x +
1 1
1
1
y (0)x2 + y (0)x3 + y (4) (0)x4 + y (5) (0)x5 .
2!
3!
4!
5!
From y (x) = x2 + y 2 − 2y we compute
40
y (x) = 2x + 2yy − 2y y
y (4) (x) = 2 + 2(y )2 + 2yy − 2y y (5) (x) = 6y y + 2yy − 2y (4) .
30
Using y(0) = 1 and y (0) = 1 we find
y (0) = −1,
y (0) = 4,
20
y (4) (0) = −6,
y (5) (0) = 14.
An approximate solution is
10
1
2
1
7
y(x) = 1 + x − x2 + x3 − x4 + x5 .
2
3
4
60
x
0.5 1 1.5 2 2.5 3 3.5
20. We look for a solution of the form
y(x) = y(0) + y (0)x +
1 1
1
1
1
y (0)x2 + y (0)x3 + y (4) (0)x4 + y (5) (0)x5 + y (6) (0)x6 .
2!
3!
4!
5!
6!
From y (x) = ey we compute
y
y (x) = ey y 10
y (4) (x) = ey (y )2 + ey y y
3
y 8
y y
(5)
(x) = e (y ) + 3e y y + e y
y
(6)
(x) = ey (y )4 + 6ey (y )2 y + 3ey (y )2 + 4ey y y + ey y (4) .
Using y(0) = 0 and y (0) = −1 we find
y (0) = 1, y (0) = −1, y (4) (0) = 2, y (5) (0) = −5, y (6) (0) = 16.
6
4
2
An approximate solution is
1
1
1
1
1
y(x) = −x + x2 − x3 + x4 + x5 + x6 .
2
6
12
24
45
1
–2
2
3
4
5
x
272
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
21. We need to solve [1 + (y )2 ]3/2 = y . Let u = y so that u = y . The equation becomes
(1 + u2 )3/2 = u or (1 + u2 )3/2 = du/dx. Separating variables and using the substitution
u = tan θ we have
du
(1 + u2 )3/2
ˆ
= dx
sec2 θ
dθ = x
3/2
(1 + tan2 θ)
ˆ
sec2 θ
dθ = x
sec3 θ
ˆ
cos θ dθ = x
sin θ = x
√
u
=x
1 + u2
y
=x
1 + (y )2
(y )2 = x2 1 + (y )2 = x2 + x2 (y )2
x2
1 − x2
x
y = √
( for x > 0)
1 − x2
y = − 1 − x2 .
(y )2 =
22. When y = sin x, y = cos x, y = − sin x, and
(y )2 − y 2 = sin2 x − sin2 x = 0.
When y = e−x , y = −e−x , y = e−x , and
(y )2 − y 2 = e−2x − e−2x = 0.
From (y )2 − y 2 = 0 we have y = ±y, which can be treated as two linear equations. Since
linear combinations of solutions of linear homogeneous differential equations are also solutions,
we see that y = c1 ex + c2 e−x and y = c3 cos x + c4 sin x must satisfy the differential equation.
However, linear combinations that involve both exponential and trigonometric functions will
not be solutions since the differential equation is not linear and each type of function satisfies
a different linear differential equation that is part of the original differential equation.
4.10
Nonlinear Differential Equations
23. Letting u = y , separating variables, and integrating we have
du = 1 + u2 ,
dx
√
du
= dx,
1 + u2
and
sinh−1 u = x + c1 .
Then
u = y = sinh (x + c1 ),
y = cosh (x + c1 ) + c2 ,
and y = sinh (x + c1 ) + c2 x + c3 .
24. If the constant −c21 is used instead of c21 , then, using partial fractions,
ˆ
ˆ x + c1 dx
1
1
1
1
+ c2 .
dx =
=−
−
ln
y=−
2c1
x − c1 x + c1
2c1 x − c1 x2 − c21
Alternatively, the inverse hyperbolic tangent can be used.
25. Let u = dx/dt so that d2 x/dt2 = u du/dx. The equation becomes u du/dx = −k 2 /x2 .
Separating variables we obtain
u du = −
k2
dx
x2
1 2 k2
u =
+c
2
x
1 2 k2
v =
+ c.
2
x
When t = 0, x = x0 and v = 0 so 0 = (k 2 /x0 ) + c and c = −k 2 /x0 . Then
√
1
dx
1 2
x0 − x
2 1
v =k
−
= −k 2
and
.
2
x x0
dt
xx0
Separating variables we have
√
xx0
dx = k 2 dt
−
x0 − x
ˆ x
1 x0
dx.
t=−
k
2
x0 − x
Using Mathematica to integrate we obtain
1 x0
x0
x
−1 (x0 − 2x)
− x(x0 − x) −
tan
t=−
k
2
2
2x
x0 − x
x0
x0 − 2x
1 x0 −1
tan
x(x0 − x) +
.
=
k
2
2
2 x(x0 − x)
273
274
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
26. x
x
x
x
2
2
2
10
–2
20 t
10
–2
x1 = 0
20 t
10
–2
x1 = 1
20 t
x1 = –1.5
For d2 x/dt2 + sin x = 0 the motion appears to be periodic with amplitude 1 when x1 = 0.
The amplitude and period are larger for larger magnitudes of x1 .
x
x
x
1
1
1
–1
x1 = 0
10 t
–1
x1 = 1
10 t
–1
x1 = –2.5
10 t
For d2 x/dt2 + dx/dt + sin x = 0 the motion appears to be periodic with decreasing amplitude.
The dx/dt term could be said to have a damping effect.
Chapter 4 in Review
1. Since, simply by substitution, y = 0 is seen to be a solution of the given initial-value problem
by Theorem 4.1.1 in the text, it is the only solution
2. Since yc = c1 ex + c2 e−x , a particular solution for y − y = 1 + ex is yp = A + Bxex .
3. False; it is not true unless the differential equation is homogeneous. For example, y1 = x is a
solution of y + y = x, but y2 = 5x is not.
4. False; Theorem 4.1.3 in the text requires that f1 and f2 be solutions of a homogeneous
linear differential equation. For example, f1 (x) = x and f2 (x) = |x| are defined and linearly
independent on [−1, 1], but the Wronskian does not exist at x = 0, since f2 (x) is not defined
there.
5. The auxiliary equation is second-order and has 5i as a root. Thus, the other root of the
auxiliary equation must be −5i, since complex roots of polynomials with real coefficients
must occur in conjugate pairs. Thus, a second solution of the differential equation must be
cos 5x and the general solution is y = c1 cos 5x + c2 sin 5x.
6. The roots are m1 = m2 = m3 = 0 and m4 = 1. To see this, note that the general solution of a
homogeneous linear fourth-order differential equation with auxiliary equation m3 (m − 1) = 0
is y = c1 + c2 x + c3 x2 + c4 ex .
Chapter 4 in Review
7. If y = c1 x2 + c2 x2 ln x, x > 0, is the general solution of a Cauchy-Euler differential equation,
then the roots of its auxiliary equation are m1 = m2 = 2 and the auxiliary equation is
m2 − 4m + 4 = m(m − 1) − 3m + 4 = 0.
Thus, the Cauchy-Euler differential equation is x2 y − 3xy + 4y = 0.
8. For yp = Ax2 we have yp = 2Ax, yp = 2A, and yp = 0. Thus, substituting yp into the
differential equation, we have 0 + 2A = 1 and A = 12 .
9. By the superposition principle for nonhomogeneous equations a particular solution is
yp = yp1 + yp2 = x + x2 − 2 = x2 + x − 2.
10. True, by the superposition for homogeneous equations.
11. The set is linearly independent over (−∞, 0) and linearly dependent over (0, ∞).
12. (a) Since f2 (x) = 2 ln x = 2f1 (x), the set of functions is linearly dependent.
(b) Since xn+1 is not a constant multiple of xn , the set of functions is linearly independent.
(c) Since x + 1 is not a constant multiple of x, the set of functions is linearly independent.
(d) Since f1 (x) = cos x cos (π/2) − sin x sin (π/2) = − sin x = −f2 (x), the set of functions is
linearly dependent.
(e) Since f1 (x) = 0 · f2 (x), the set of functions is linearly dependent.
(f ) Since 2x is not a constant multiple of 2, the set of functions is linearly independent.
(g) Since 3(x2 ) + 2(1 − x2 ) − (2 + x2 ) = 0, the set of functions is linearly dependent.
(h) Since xex+1 + 0(4x − 5)ex − exex = 0, the set of functions is linearly dependent.
13. (a) The auxiliary equation is (m − 3)(m + 5)(m − 1) = m3 + m2 − 17m + 15 = 0, so the
differential equation is y + y − 17y + 15y = 0.
(b) The form of the auxiliary equation is
m(m − 1)(m − 2) + bm(m − 1) + cm + d = m3 + (b − 3)m2 + (c − b + 2)m + d = 0.
Since (m−3)(m+5)(m−1) = m3 +m2 −17m+15 = 0, we have b−3 = 1, c−b+2 = −17,
and d = 15. Thus, b = 4 and c = −15, so the differential equation is
y + 4y − 15y + 15y = 0.
275
276
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
14. Variation of parameters will work for all choices of g(x), although the integral involved may not
always be able to be expressed in terms of elementary functions. The method of undetermined
coefficients will work for the functions in (b), (c), and (e).
15. The auxiliary equation is am(m − 1) + bm + c = am2 + (b − a)m + c = 0. If the roots are
3 and −1, then we want (m − 3)(m + 1) = m2 − 2m − 3 = 0. Thus, let a = 1, b = −1, and
c = −3, so that the differential equation is x2 y − xy − 3y = 0.
16. In this case we want the auxiliary equation to be m2 + 1 = 0, so let a = 1, b = 1, and c = 1.
Then the differential equation is x2 y + xy + y = 0.
√
17. From m2 − 2m − 2 = 0 we obtain m = 1 ± 3 so that
y = c1 e(1+
√
3 )x
+ c2 e(1−
√
3 )x
.
√
18. From 2m2 + 2m + 3 = 0 we obtain m = −1/2 ± ( 5/2)i so that
√
√
5
5
x + c2 sin
x .
c1 cos
2
2
−x/2
y=e
19. From m3 + 10m2 + 25m = 0 we obtain m = 0, m = −5, and m = −5 so that
y = c1 + c2 e−5x + c3 xe−5x .
20. From 2m3 + 9m2 + 12m + 5 = 0 we obtain m = −1, m = −1, and m = −5/2 so that
y = c1 e−5x/2 + c2 e−x + c3 xe−x .
√
21. From 3m3 + 10m2 + 15m + 4 = 0 we obtain m = −1/3 and m = −3/2 ± ( 7/2)i so that
√
−x/3
y = c1 e
−3x/2
+e
√
7
7
x + c3 sin
x .
c2 cos
2
2
√
22. From 2m4 + 3m3 + 2m2 + 6m − 4 = 0 we obtain m = 1/2, m = −2, and m = ± 2 i so that
y = c1 ex/2 + c2 e−2x + c3 cos
√
√
2 x + c4 sin 2 x.
23. Applying D4 to the differential equation we obtain D4 (D2 − 3D + 5) = 0. Then
√
3x/2
y=e
√
11
11
x + c2 sin
x +c3 + c4 x + c5 x2 + c6 x3
c1 cos
2
2
yc
and yp = A + Bx + Cx2 + Dx3 . Substituting yp into the differential equation yields
(5A − 3B + 2C) + (5B − 6C + 6D)x + (5C − 9D)x2 + 5Dx3 = −2x + 4x3 .
Chapter 4 in Review
Equating coefficients gives A = −222/625, B = 46/125, C = 36/25, and D = 4/5. The
general solution is
√
3x/2
y=e
√
11
11
x + c2 sin
x
c1 cos
2
2
−
222
46
36
4
+
x + x2 + x3 .
625 125
25
5
24. Applying (D − 1)3 to the differential equation we obtain (D − 1)3 (D − 2D + 1) = (D − 1)5 = 0.
Then
y = c1 ex + c2 xex +c3 x2 ex + c4 x3 ex + c5 x4 ex
yc
and yp = Ax2 ex + Bx3 ex + Cx4 ex . Substituting yp into the differential equation yields
12Cx2 ex + 6Bxex + 2Aex = x2 ex .
Equating coefficients gives A = 0, B = 0, and C = 1/12. The general solution is
y = c1 ex + c2 xex +
1 4 x
x e .
12
25. Applying D(D2 + 1) to the differential equation we obtain
D(D2 + 1)(D3 − 5D2 + 6D) = D2 (D2 + 1)(D − 2)(D − 3) = 0.
Then
y = c1 + c2 e2x + c3 e3x +c4 x + c5 cos x + c6 sin x
yc
and yp = Ax + B cos x + C sin x. Substituting yp into the differential equation yields
6A + (5B + 5C) cos x + (−5B + 5C) sin x = 8 + 2 sin x.
Equating coefficients gives A = 4/3, B = −1/5, and C = 1/5. The general solution is
1
1
4
y = c1 + c2 e2x + c3 e3x + x − cos x + sin x.
3
5
5
26. Applying D to the differential equation we obtain D(D3 − D2 ) = D3 (D − 1) = 0. Then
y = c1 + c2 x + c3 ex +c4 x2
yc
and yp = Ax2 . Substituting yp into the differential equation yields −2A = 6. Equating
coefficients gives A = −3. The general solution is
y = c1 + c2 x + c3 ex − 3x2 .
277
278
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
27. The auxiliary equation is m2 − 2m + 2 = [m − (1 + i)][m − (1 − i)] = 0, so
yc = c1 ex sin x + c2 ex cos x and
x sin x
x cos x
e
e
= −e2x
W = x
x
x
x
e cos x + e sin x −e sin x + e cos x
Identifying f (x) = ex tan x we obtain
u1 = −
u2 =
(ex cos x)(ex tan x)
= sin x
−e2x
(ex sin x)(ex tan x)
sin2 x
= cos x − sec x.
=
−
−e2x
cos x
Then u1 = − cos x, u2 = sin x − ln | sec x + tan x|, and
y = c1 ex sin x + c2 ex cos x − ex sin x cos x + ex sin x cos x − ex cos x ln | sec x + tan x|
= c1 ex sin x + c2 ex cos x − ex cos x ln | sec x + tan x|.
28. The auxiliary equation is m2 − 1 = 0, so yc = c1 ex + c2 e−x and
x
e
e−x W = x
= −2
e −e−x Identifying f (x) = 2ex /(ex + e−x ) we obtain
u1 =
1
ex
=
ex + e−x
1 + e2x
u2 = −
ex
e2x
e3x
ex
=−
= −ex +
.
−x
2x
+e
1+e
1 + e2x
Then u1 = tan−1 ex , u2 = −ex + tan−1 ex , and
y = c1 ex + c2 e−x + ex tan−1 ex − 1 + e−x tan−1 ex .
29. The auxiliary equation is 6m2 − m − 1 = 0 so that
y = c1 x1/2 + c2 x−1/3 .
30. The auxiliary equation is 2m3 + 13m2 + 24m + 9 = (m + 3)2 (m + 1/2) = 0 so that
y = c1 x−3 + c2 x−3 ln x + c3 x−1/2 .
31. The auxiliary equation is m2 − 5m + 6 = (m − 2)(m − 3) = 0 and a particular solution is
yp = x4 − x2 ln x so that
y = c1 x2 + c2 x3 + x4 − x2 ln x.
Chapter 4 in Review
32. The auxiliary equation is m2 − 2m + 1 = (m − 1)2 = 0 and a particular solution is yp = 14 x3
so that
1
y = c1 x + c2 x ln x + x3 .
4
33. The auxiliary equation is m2 + ω 2 = 0, so yc = c1 cos ωt + c2 sin ωt. When ω = α, yp =
A cos αt + B sin αt and
y = c1 cos ωt + c2 sin ωt + A cos αt + B sin αt.
When ω = α, yp = At cos ωt + Bt sin ωt and
y = c1 cos ωt + c2 sin ωt + At cos ωt + Bt sin ωt.
34. The auxiliary equation is m2 − ω 2 = 0, so yc = c1 eωt + c2 e−ωt . When ω = α, yp = Aeαt and
y = c1 eωt + c2 e−ωt + Aeαt .
When ω = α, yp = Ateωt and
y = c1 eωt + c2 e−ωt + Ateωt .
35. If y = sin x is a solution then so is y = cos x and m2 + 1 is a factor of the auxiliary equation
m4 + 2m3 + 11m2 + 2m + 10 = 0. Dividing by m2 + 1 we get m2 + 2m + 10, which has roots
−1 ± 3i. The general solution of the differential equation is
y = c1 cos x + c2 sin x + e−x (c3 cos 3x + c4 sin 3x).
36. The auxiliary equation is m(m + 1) = m2 + m = 0, so the associated homogeneous differential
equation is y + y = 0. Letting y = c1 + c2 e−x + 12 x2 − x and computing y + y we get x.
Thus, the differential equation is y + y = x.
37. (a) The auxiliary equation is m4 − 2m2 + 1 = (m2 − 1)2 = 0, so the general solution of the
differential equation is
y = c1 sinh x + c2 cosh x + c3 x sinh x + c4 x cosh x.
(b) Since both sinh x and x sinh x are solutions of the associated homogeneous differential
equation , a particular solution of y (4) − 2y + y = sinh x has the form
yp = Ax2 sinh x + Bx2 cosh x.
38. Since y1 = 1 and y1 = 0, x2 y1 − (x2 + 2x)y1 + (x + 2)y1 = −x2 − 2x + x2 + 2x = 0, and
y1 = x is a solution of the associated homogeneous equation. Using the method of reduction
of order, we let y = ux. Then y = xu + u and y = xu + 2u , so
x2 y − (x2 + 2x)y + (x + 2)y = x3 u + 2x2 u − x3 u − 2x2 u − x2 u − 2xu + x2 u + 2xu
= x3 u − x3 u = x3 (u − u ).
279
280
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
To find a second solution of the homogeneous equation we note that u = ex is a solution of
u − u = 0. Thus, yc = c1 x + c2 xex . To find a particular solution we set x3 (u − u ) = x3
so that u − u = 1. This differential equation has a particular solution of the form Ax.
Substituting, we find A = −1, so a particular solution of the original differential equation is
yp = −x2 and the general solution is y = c1 x + c2 xex − x2 .
39. The auxiliary equation is m2 − 2m + 2 = 0 so that m = 1 ± i and y = ex (c1 cos x + c2 sin x).
Setting y(π/2) = 0 and y(π) = −1 we obtain c1 = e−π and c2 = 0. Thus, y = ex−π cos x.
40. The auxiliary equation is m2 + 2m + 1 = (m + 1)2 = 0, so that y = c1 e−x + c2 xe−x . Setting
y(−1) = 0 and y (0) = 0 we get c1 e − c2 e = 0 and −c1 + c2 = 0. Thus c1 = c2 and
y = c1 (e−x + xe−x ) is a solution of the boundary-value problem for any real number c1 .
41. The auxiliary equation is m2 − 1 = (m − 1)(m + 1) = 0 so that m = ±1 and y = c1 ex + c2 e−x .
Assuming yp = Ax+B+C sin x and substituting into the differential equation we find A = −1,
B = 0, and C = − 12 . Thus yp = −x − 12 sin x and
y = c1 ex + c2 e−x − x −
1
sin x.
2
Setting y(0) = 2 and y (0) = 3 we obtain
c1 + c 2 = 2
3
c1 − c2 − = 3.
2
5
Solving this system we find c1 = 13
4 and c2 = − 4 . The solution of the initial-value problem
is
5
1
13
y = ex − e−x − x − sin x.
4
4
2
42. The auxiliary equation is m2 + 1 = 0, so yc = c1 cos x + c2 sin x and
cos x sin x = 1.
W = − sin x cos x
Identifying f (x) = sec3 x we obtain
u1 = − sin x sec3 x = −
sin x
cos3 x
u2 = cos x sec3 x = sec2 x.
Then
u1 = −
1
1 1
= − sec2 x
2
2 cos x
2
u2 = tan x.
Chapter 4 in Review
Thus
y = c1 cos x + c2 sin x −
1
cos x sec2 x + sin x tan x
2
= c1 cos x + c2 sin x −
1 − cos2 x
1
sec x +
2
cos x
= c3 cos x + c2 sin x +
1
sec x.
2
and
y = −c3 sin x + c2 cos x +
1
sec x tan x.
2
The initial conditions imply
c3 +
1
=1
2
1
c2 = .
2
Thus c3 = c2 = 1/2 and
y=
1
1
1
cos x + sin x + sec x.
2
2
2
43. Let u = y so that u = y . The equation becomes u du/dx = 4x. Separating variables we
obtain
u du = 4x dx
1 2
u = 2x2 + c1
2
u2 = 4x2 + c2 .
When x = 1, y = u = 2, so 4 = 4 + c2 and c2 = 0. Then
u2 = 4x2
dy
= 2x or
dx
y = x2 + c3
dy
= −2x
dx
or y = −x2 + c4 .
When x = 1, y = 5, so 5 = 1+c3 and 5 = −1+c4 . Thus c3 = 4 and c4 = 6. We have y = x2 +4
and y = −x2 + 6. Note however that when y = −x2 + 6, y = −2x and y (1) = −2 = 2. Thus,
the solution of the initial-value problem is y = x2 + 4.
44. Let u = y so that y = u du/dy. The equation becomes 2u du/dy = 3y 2 . Separating variables
we obtain
2u du = 3y 2 dy
u 2 = y 3 + c1 .
281
282
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
When x = 0, y = 1 and y = u = 1 so 1 = 1 + c1 and c1 = 0. Then
u2 = y 3
dy
dx
2
= y3
dy
= y 3/2
dx
y −3/2 dy = dx
−2y −1/2 = x + c2
y=
4
.
(x + c2 )2
When x = 0, y = 1, so 1 = 4/c22 and c2 = ±2. Thus, y = 4/(x + 2)2 and y = 4/(x − 2)2 .
Note, however, that when y = 4/(x + 2)2 , y = −8/(x + 2)3 and y (0) = −1 = 1. Thus, the
solution of the initial-value problem is y = 4/(x − 2)2 .
45. (a) The auxiliary equation is 12m4 + 64m3 + 59m2 − 23m − 12 = 0 and has roots −4, − 32 ,
− 13 , and 12 . The general solution is
y = c1 e−4x + c2 e−3x/2 + c3 e−x/3 + c4 ex/2 .
(b) The system of equations is
c1 + c2 + c3 + c4 = −1
3
1
1
−4c1 − c2 − c3 + c4 = 2
2
3
2
9
1
1
16c1 + c2 + c3 + c4 = 5
4
9
4
−64c1 −
27
1
1
c2 − c3 + c4 = 0.
8
27
8
73
3726
257
, c2 = 109
Using a CAS we find c1 = − 495
35 , c3 = − 385 , and c4 = 45 . The solution of
the initial-value problem is
73 −4x 109 −3x/2 3726 −x/3 257 x/2
e
e
e
e .
+
−
+
y=−
495
35
385
45
46. Consider xy + y = 0 and look for a solution of the form
y
1
2
3
4
5
y = xm . Substituting into the differential equation we have
x
xy + y = m(m − 1)xm−1 + mxm−1 = m2 xm−1 .
Thus, the general solution of xy + y = 0 is yc = c1 + c2 ln x.
√
To find a particular solution of xy + y = − x we use
variation of parameters.
–1
–2
–3
–4
–5
Chapter 4 in Review
The Wronskian is
1 ln x 1
=
W =
0 1/x x
Identifying f (x) = −x−1/2 we obtain
u1 =
√
x−1/2 ln x √
−x−1/2
= x ln x and u2 =
= − x,
1/x
1/x
so that
3/2
u1 = x
4
2
ln x −
3
9
2
and u2 = − x3/2 .
3
4
2
2
4
ln x −
− x3/2 ln x = − x3/2
yp = x
3
9
3
9
and the general solution of the differential equation is
Then
3/2
4
y = c1 + c2 ln x − x3/2 .
9
The initial conditions are y(1) = 0 and y (1) = 0. These imply that c1 =
solution of the initial-value problem is
y=
4
9
and c2 =
2
3
. The
4
4 2
+ ln x − x3/2 .
9 3
9
The graph is shown above.
47. From (D − 2)x + (D − 2)y = 1 and Dx + (2D − 1)y = 3 we obtain (D − 1)(D − 2)y = −6 and
Dx = 3 − (2D − 1)y. Then
3
y = c1 e2t + c2 et − 3 and x = −c2 et − c1 e2t + c3 .
2
Substituting into (D − 2)x + (D − 2)y = 1 gives c3 =
5
2
so that
3
5
x = −c2 et − c1 e2t + .
2
2
48. From (D − 2)x − y = t − 2 and −3x + (D − 4)y = −4t we obtain (D − 1)(D − 5)x = 9 − 8t.
Then
3
8
x = c1 et + c2 e5t − t −
5
25
and
16 11
+ t.
y = (D − 2)x − t + 2 = −c1 et + 3c2 e5t +
25 25
49. From (D − 2)x − y = −et and −3x + (D − 4)y = −7et we obtain (D − 1)(D − 5)x = −4et so
that
x = c1 et + c2 e5t + tet .
Then
y = (D − 2)x + et = −c1 et + 3c2 e5t − tet + 2et .
283
284
CHAPTER 4
HIGHER-ORDER DIFFERENTIAL EQUATIONS
50. From (D + 2)x + (D + 1)y = sin 2t and 5x + (D + 3)y = cos 2t we obtain
(D2 + 5)y = 2 cos 2t − 7 sin 2t. Then
y = c1 cos t + c2 sin t −
7
2
cos 2t + sin 2t
3
3
and
1
1
x = − (D + 3)y + cos 2t
5
5
1
5
3
3
1
1
c1 − c2 sin t + − c2 − c1 cos t − sin 2t − cos 2t.
=
5
5
5
5
3
3
Chapter 5
Modeling with Higher-Order Differential Equations
5.1
Linear Models: Initial-Value Problems
1. From 0.408x + 16x = 0 we obtain
x = c1 cos 6.261 t + c2 sin 6.261 t
so that the period of motion is 2π/6.261 = 1.004 seconds.
2. From 20x + kx = 0 we obtain
1 k
1 k
t + c2 sin
t
x = c1 cos
2 5
2 5
so that the frequency 2/π =
k/20 /2π and k = 320 N/m. If 80x + 320x = 0 then
x = c1 cos 2t + c2 sin 2t so that the frequency is 2/2π = 1/π cycles/s.
3. From 2.449x + 6x = 0, x(0) = −3, and x (0) = 0 we obtain x = −3 cos 1.565 t.
4. From 2.449x + 6x = 0, x(0) = 0, and x (0) = 2 we obtain x = 3.13 sin 1.565 t.
5. From 20x + 160x = 0, x(0) = 1/2, and x (0) = 0 we obtain x =
1
2
cos 8t.
(a) x(π/12) = −1/4, x(π/8) = −1/2, x(π/6) = −1/4, x(π/4) = 1/2, x(9π/32) =
√
2/4.
(b) x = −4 sin 8t so that x (3π/16) = 4 m/s directed downward.
(c) If x =
1
2
cos 8t = 0 then t = (2n + 1)π/16 for n = 0, 1, 2, . . . .
6. From 50x +200x = 0, x(0) = 0, and x (0) = −10 we obtain x = −5 sin 2t and x = −10 cos 2t.
7. From 20x + 20x = 0, x(0) = 0, and x (0) = −10 we obtain x = −10 sin t and x = −10 cos t.
(a) The 20 kg mass has the larger amplitude.
√
(b) 20 kg: x (π/4) = −5 2 m/s, x (π/2) = 0 m/s; 50 kg: x (π/4) = 0 m/s, x (π/2) = 10
m/s
285
286
CHAPTER 5
MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS
(c) If −5 sin 2t = −10 sin t then 2 sin t(cos t − 1) = 0 so that t = nπ for n = 0, 1, 2, . . .,
placing both masses at the equilibrium position. The 50 kg mass is moving upward; the
20 kg mass is moving upward when n is even and downward when n is odd.
8. From x + 16x = 0, x(0) = −1 , and x (0) = −2, we get
√
5
1
cos (4t − 3.605)
x(t) = − cos 4t − sin 4t =
2
2
The period is T = 2π/ω = π/2 seconds and the amplitude is
√
5/2 feet.
In 4π seconds it will make 8 complete cycles. To see this, note that the frequency is
f=
ω
4
1
=
=
T
2π
2π
This means the mass will make 4 complete cycles every 2π seconds.
9. (a) From x + 16x = 0, x(0) = 1/2, and x (0) = 3/2, we get
3
1
cos 2t + sin 2t
2
4
√
(b) We use x(t) = A sin (ωt + φ) where A = c21 + c22 = (1/2)2 + (3/4)2 = 13 /4. From
this we get
x(t) =
2
c1
=
c2
3
2
= 0.588
φ = tan−1
3
tan φ =
√
Therefore x(t) =
13
4
sin (2t + 0.588).
(c) We use x(t) = A cos (ωt − φ) where A =
time we get
tan φ =
c21 + c22 =
3
c2
=
c1
2
−1
φ = tan
√
Therefore x(t) =
13
4
√
(1/2)2 + (3/4)2 = 13 /4. This
3
= 0.983
2
cos(2t − 0.983).
10. (a) From 1.6x + 40x = 0, x(0) = −1/3, and x (0) = 5/4, we get
1
1
x(t) = − cos 5t + sin 5t
3
4
5.1
We use x(t) = A sin (ωt + φ) where A =
this we get
Linear Models: Initial-Value Problems
c21 + c22 =
(−1/3)2 + (1/4)2 = 5/12. From
4
c1
=−
c2
3
4
= −0.927
φ = tan−1 −
3
tan φ =
Therefore x(t) =
5
12
sin (5t − 0.927).
(b) We use x(t) = A cos (ωt − φ) where A =
cos φ < 0 so
c21 + c22 = 5/12. Note that sin φ > 0 and
4
c2
=−
c1
3
4
= −0.927
φ = tan−1 −
3
tan φ =
Therefore x(t) =
5
12
cos (5t − 2.489).
11. From x + 100x = 0, x(0) = −2/3, and x (0) = 5 we obtain
(a) x = − 23 cos 10t + 12 sin 10t =
5
6
sin (10t − 0.927).
(b) The amplitude is 5/6 ft and the period is 2π/10 = π/5
(c) 3π = πk/5 and k = 15 cycles.
(d) If x = 0 and the weight is moving downward for the second time, then 10t − 0.927 = 2π
or t = 0.721 s.
(e) If x = 25
3 cos (10t − 0.927) = 0 then 10t − 0.927 = π/2 + nπ or t = (2n + 1)π/20 + 0.0927
for n = 0, 1, 2, . . . .
(f ) x(3) = −0.597 ft
(g) x (3) = −5.814 ft/s
(h) x (3) = 59.702 ft/s2
(i) If x = 0 then t =
x = ±8.33 ft/s.
1
10 (0.927
(j) If x = 5/12 then t =
2, . . . .
+ nπ) for n = 0, 1, 2, . . .. The velocity at these times is
1
10 (π/6 + 0.927 + 2nπ)
and t =
1
10 (5π/6 + 0.927 + 2nπ)
for n = 0, 1,
287
288
CHAPTER 5
MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS
(k) If x = 5/12 and x < 0 then t =
1
10 (5π/6
+ 0.927 + 2nπ) for n = 0, 1, 2, . . ..
√
12. From x + 9x = 0, x(0) = −1, and x (0) = − 3 we obtain
√
2
3
4π
sin 3t = √ sin 3t +
x = − cos 3t −
3
3
3
√
and x = 2 3 cos (3t + 4π/3). If x = 3 then t = −7π/18 + 2nπ/3 and t = −π/2 + 2nπ/3 for
n = 1, 2, 3, . . . .
13. The springs are parallel as in Figure 3.8.5 so the effective spring constant is keff = k1 + k2 .
From Hooke’s Law we find that a mass weighing 20 N determines the spring constants k1 = 40
and k2 = 120. Then we have
keff = k1 + k2 = 40 + 120 = 160 N/ft
20 x + 160x = 0
9.8
x + 78.4x = 0
x(t) = c1 cos 8.854t + c2 sin 8.854t
Using the initial condition x(0) = 0 we have c1 = 0 and therefore x(t) = c2 sin 8.854t. Then
x (t) = 8.854c2 cos 8.854t. Using the condition x (0) = 2 we have c2 = 1/8. Thus
x(t) = 0.226 sin 8.854t
14. Hooke’s Law applied to the same force W (weight) for the two springs gives
W = k1
1
1
= k2
3
2
2k1 = 3k2
3
k1 = k2
2
When the mass weighing 8 N slug is put on the parallel-spring system its period is given by
2π/ω = π/15 or ω = 30. Now
ω2 =
keff
m
keff = mω 2 =
8
· 302 = 734.7
9.8
k1 + k2 = 734.7
5
k2 = 734.7
2
2
· 734.7 = 293.9
5
3
k1 = · 293.9 = 440.8
2
k2 =
5.1
Linear Models: Initial-Value Problems
Therefore W = mg = 12 k2 = 146.9 N. Alternatively, W = mg = 13 k1 = 146.9 N.
15. Using k1 = 40 and k2 = 120 we have
keff =
40 · 120
40 · 120
=
= 30 N/m
40 + 120
160
20 x + 30x = 0
9.8
x + 14.7x = 0
x(t) = c1 cos 3.834 t + c2 sin 3.834 t
Using the initial condition x(0) = 0 we have c1 = 0 and therefore x(t) = c2 sin 3.834 t. Then
x (t) = 3.834 c2 cos 3.834 t. Using the condition x (0) = 2 we have c2 = 1.917. Thus
x(t) = 1.917 sin 3.834 t
16. As is Problem 14,
W = k1
1
1
= k2
3
2
2k1 = 3k2
3
k1 = k2
2
When the mass weighing 8 N is put on the series-spring system its period is given as 2π/ω =
π/15 or ω = 30. Now
ω 2 = keff /m
keff = mω 2 =
8
· 302 = 734.7
9.8
k1 k2
= 734.7
k1 + k2
3 2
2 k2
3
2 k2
+ k2
=
3/2
k2 = 734.7
5/2
k2 =
5
· 734.7 = 1224.5
3
k1 =
3
· 1224.5 = 1
2
Therefore W = mg = 12 k2 = 612.2 N. Alternatively, W = mg = 13 k1 = 612.2 N.
17. For parallel springs the effective spring constant is keff = k1 + k2 . When k = k1 = k2 , the
effective spring constant is keff = 2k. Compared to a single spring with constant k, the
parallel-spring system is more stiff.
289
290
CHAPTER 5
MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS
18. For springs attached in series the effective spring constant is
keff =
k1 k2
.
k1 + k2
When k = k1 = k2 , the effective spring constant is
keff =
1
k2
= k
k+k
2
Compared to a spring with spring constant k, the series-spring system is less stiff.
19. For large values of t the differential equation is approximated by x = 0. The solution of this
equation is the linear function x = c1 t + c2 . Thus, for large time, the restoring force will have
decayed to the point where the spring is incapable of returning the mass, and the spring will
simply keep on stretching.
20. As t becomes larger the spring constant increases; that is, the spring is stiffening. It would
seem that the oscillations would become periodic and the spring would oscillate more rapidly.
It is likely that the amplitudes of the oscillations would decrease as t increases.
21. (a) above
(b) heading upward
22. (a) below
(b) from rest
23. (a) below
(b) heading upward
24. (a) above
(b) heading downward
25. From 18 x + x + 2x = 0, x(0) = −1, and x (0) = 8 we obtain x = 4te−4t − e−4t and
x = 8e−4t − 16te−4t . If x = 0 then t = 1/4 second. If x = 0 then t = 1/2 second and the
extreme displacement is x = e−2 m.
√
√
26. From 14 x√+ 2 x + 2x = 0, x(0) = 0, and x (0) = 5 we obtain x = 5te−2 2 t and
√
√ x = 5e−2 2 t 1 − 2 2 t . If x = 0 then t = 2/4 second and the extreme displacement
√
is x = 5 2 e−1 /4 m.
27. (a) From x + 10x + 16x = 0, x(0) = 1, and x (0) = 0 we obtain x = 43 e−2t − 13 e−8t .
(b) From x + 10x + 16x = 0, x(0) = 1, and x (0) = −12 then x = − 23 e−2t + 53 e−8t .
28. (a) x = 13 e−8t 4e6t − 1 is not zero for t ≥ 0; the extreme displacement is x(0) = 1 meter.
(b) x = 13 e−8t 5 − 2e6t = 0 when t = 16 ln 52 ≈ 0.153 second; if x = 43 e−8t e6t − 10 = 0
then t = 16 ln 10 ≈ 0.384 second and the extreme displacement is x = −0.232 meter.
5.1
Linear Models: Initial-Value Problems
29. (a) From 0.1x + 0.4x + 2x = 0, x(0) = −1, and x (0) = 0 we obtain
x = e−2t − cos 4t − 12 sin 4t .
√
5 −2t
e sin (4t + 4.25)
2
(b) x =
(c) If x = 0 (and t > 0) then 4t + 4.25 = 2π, 3π, 4π, . . . so that the first time heading upward
is t = 1.294 seconds.
30. (a) From 14 x +x +5x = 0, x(0) = 1/2, and x (0) = 1 we obtain x = e−2t
1
2
cos 4t + 12 sin 4t .
π
1
(b) x = √ e−2t sin 4t +
.
4
2
(c) If x = 0 then 4t + π/4 = π, 2π, 3π, . . . so that the times heading downward are t =
(7 + 8n)π/16 for n = 0, 1, 2, . . . .
(d) x
1
x
0.5
0.5
2
t
–0.5
–1
5 31. From 16
x + βx + 5x = 0 we find that the roots of the auxiliary equation are
m = − 85 β ± 45 4β 2 − 25 .
(a) If 4β 2 − 25 > 0 then β > 5/2.
(b) If 4β 2 − 25 = 0 then β = 5/2.
(c) If 4β 2 − 25 < 0 then 0 < β < 5/2.
√
32. From 0.75x + βx + 6x = 0 and β > 3 2 we find that the roots of the auxiliary equation are
m = − 23 β ± 23 β 2 − 18 and
−2βt/3
x=e
2
2
2
2
c1 cosh
β − 18 t + c2 sinh
β − 18 t
.
3
3
If x(0) = 0 and x (0) = −2 then c1 = 0 and c2 = −3/ β 2 − 18.
291
292
CHAPTER 5
MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS
33. If 12 x + 12 x + 6x = 10 cos 3t, x(0) = −2, and x (0) = 0 then
√
−t/2
c1 cos
xc = e
10
3 (cos 3t
and xp =
47
t
2
√
+ c2 sin
47
t
2
+ sin 3t) so that the equation of motion is
−t/2
x=e
4
− cos
3
√
64
− √ sin
3 47
47
t
2
√
47
t
2
+
10
(cos 3t + sin 3t).
3
34. (a) If x + 2x + 5x = 12 cos 2t + 3 sin 2t, x(0) = 1, and x (0) = 5 then xc = e−t (c1 cos 2t +
c2 sin 2t) and xp = 3 sin 2t so that the equation of motion is
x = e−t cos 2t + 3 sin 2t.
(b) x
x
3
2
–3
(c) x
steady-state
x
3
4
6
t
x = xc + xp
2
4
6
t
–3
transient
35. From x + 8x + 16x = 8 sin 4t, x(0) = 0, and x (0) = 0 we obtain xc = c1 e−4t + c2 te−4t and
xp = − 14 cos 4t so that the equation of motion is
1
1
x = e−4t + te−4t − cos 4t.
4
4
36. From x + 8x + 16x = e−t sin 4t, x(0) = 0, and x (0) = 0 we obtain xc = c1 e−4t + c2 te−4t and
24 −t
7 −t
e cos 4t − 625
e sin 4t so that
xp = − 625
x=
1 −t
1 −4t
e (24 + 100t) −
e (24 cos 4t + 7 sin 4t).
625
625
As t → ∞ the displacement x → 0.
37. From 2x + 32x = 68e−2t cos 4t, x(0) = 0, and x (0) = 0 we obtain xc = c1 cos 4t + c2 sin 4t
and xp = 12 e−2t cos 4t − 2e−2t sin 4t so that
9
1
1
x = − cos 4t + sin 4t + e−2t cos 4t − 2e−2t sin 4t.
2
4
2
√
38. Since x =
t → ∞.
85
4
sin (4t − 0.219) −
√
17 −2t
sin (4t
2 e
+ 2.897), the amplitude approaches
√
85/4 as
5.1
Linear Models: Initial-Value Problems
39. (a) By Hooke’s law the external force is F (t) = kh(t) so that mx + βx + kx = kh(t).
(b) From 12 x + 2x + 4x = 20 cos t, x(0) = 0, and x (0) = 0 we obtain xc = e−2t (c1 cos 2t +
32
c2 sin 2t) and xp = 56
13 cos t + 13 sin t so that
72
56
32
56
sin 2t +
cos t +
sin t.
x = e−2t − cos 2t −
13
13
13
13
40. (a) From 100x + 1600x = 1600 sin 8t, x(0) = 0, and x (0) = 0 we obtain
xc = c1 cos 4t + c2 sin 4t and xp = − 13 sin 8t so that by a trigonometric identity
x=
(b) If x =
1
3
1
2
2
2
sin 4t − sin 8t = sin 4t − sin 4t cos 4t.
3
3
3
3
sin 4t(2 − 2 cos 4t) = 0 then t = nπ/4 for n = 0, 1, 2, . . . .
(c) If x = 83 cos 4t − 83 cos 8t = 83 (1 − cos 4t)(1 + 2 cos 4t) = 0 then t = π/3 + nπ/2 and
t = π/6 + nπ/2 for n = 0, 1, 2, . . . at the extreme values. Note: There are other
values of t for which x = 0. However these other values are multiples of π/2 and the
displacement is zero from part (b).
(d) x(π/6 + nπ/2) =
(e) x
√
√
3/2 cm and x(π/3 + nπ/2) = − 3/2 cm.
x
1
1
2
3
t
–1
41. From x +4x = −5 sin 2t+3 cos 2t, x(0) = −1, and x (0) = 1 we obtain xc = c1 cos 2t+c2 sin 2t,
xp = 34 t sin 2t + 54 t cos 2t, and
x = − cos 2t −
3
5
1
sin 2t + t sin 2t + t cos 2t.
8
4
4
42. From x + 9x = 5 sin 3t, x(0) = 2, and x (0) = 0 we obtain xc = c1 cos 3t + c2 sin 3t,
xp = − 56 t cos 3t, and
5
5
x = 2 cos 3t +
sin 3t − t cos 3t.
18
6
43. (a) From x + ω 2 x = F0 cos γt, x(0) = 0, and x (0) = 0 we obtain xc = c1 cos ωt + c2 sin ωt
and xp = (F0 cos γt)/ ω 2 − γ 2 so that
x=−
F0
F0
cos ωt + 2
cos γt.
ω2 − γ 2
ω − γ2
293
294
CHAPTER 5
(b) lim
γ→ω
MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS
F0
F0
−F0 t sin γt
=
t sin ωt.
(cos γt − cos ωt) = lim
γ→ω
ω2 − γ 2
−2γ
2ω
44. From x + ω 2 x = F0 cos ωt, x(0) = 0, and x (0) = 0 we obtain xc = c1 cos ωt + c2 sin ωt and
xp = (F0 t/2ω) sin ωt so that x = (F0 t/2ω) sin ωt.
45. (a) From cos (u − v) = cos u cos v + sin u sin v and cos (u + v) = cos u cos v − sin u sin v we
obtain sin u sin v = 12 [cos (u − v) − cos (u + v)]. Letting u = 12 (γ − ω)t and v = 12 (γ + ω)t,
the result follows.
(b) If = 12 (γ − ω) then γ ≈ ω so that x = (F0 /2γ) sin t sin γt.
46. See the article “Distinguished Oscillations of a Forced Harmonic Oscillator” by T.G. Procter
in The College Mathematics Journal, March, 1995. In this article the author illustrates that
for F0 = 1, λ = 0.01, γ = 22/9, and ω = 2 the system exhibits beats oscillations on the
interval [0, 9π], but that this phenomenon is transient as t → ∞.
x
1
t
–1
π
9π
47. (a) The general solution of the homogeneous equation is
xc (t) = c1 e−λt cos ( ω 2 − λ2 t) + c2 e−λt sin ( ω 2 − λ2 t)
−λt
2
2
= Ae
sin
ω −λ t+φ ,
where A =
c21 + c22 , sin φ = c1 /A, and cos φ = c2 /A. Now
xp (t) =
where
F0 (ω 2 − γ 2 )
F0 (−2λγ)
sin γt + 2
cos γt = A sin (γt + θ),
2
2
2
2
2
(ω − γ ) + 4λ γ
(ω − γ 2 )2 + 4λ2 γ 2
F0 (−2λγ)
−2λγ
− γ 2 )2 + 4λ2 γ 2
=
sin θ =
2
F0
(ω − γ 2 )2 + 4λ2 γ 2
ω 2 − γ 2 + 4λ2 γ 2
(ω 2
and
F0 (ω 2 − γ 2 )
ω2 − γ 2
(ω 2 − γ 2 )2 + 4λ2 γ 2
=
.
cos θ =
2 − γ 2 )2 + 4λ2 γ 2
F0
(ω
(ω 2 − γ 2 )2 + 4λ2 γ 2
5.1
Linear Models: Initial-Value Problems
295
√
(b) If g (γ) = 0 then γ γ 2 + 2λ2 − ω 2 = 0 so that γ = 0 or γ = ω 2 − 2λ2 . The first
√
derivative test shows that g has a maximum value at γ = ω 2 − 2λ2 . The maximum
value of g is
ω 2 − 2λ2 = F0 /2λ ω 2 − λ2 .
g
√
(c) We identify ω 2 = k/m = 4, λ = β/2, and γ1 = ω 2 − 2λ2 = 4 − β 2 /2 . As β → 0,
γ1 → 2 and the resonance curve grows without bound at γ1 = 2. That is, the system
approaches pure resonance.
β
γ1
2.00
1.00
0.75
0.50
0.25
4
g
1.41
1.87
1.93
1.97
1.99
g
β = 1/4
3
0.58
1.03
1.36
2.02
4.01
β
β
β
β
2
1
1
2
= 1/2
= 3/4
=1
=2
4
3
γ
48. (a) For n = 2, sin2 γt = 12 (1−cos 2γt). The system is in pure resonance when 2γ1 /2π = ω/2π,
or when γ1 = ω/2.
(b) Note that
sin3 γt = sin γt sin2 γt =
1
[sin γt − sin γt cos 2γt] .
2
Now
sin (A + B) + sin (A − B) = 2 sin A cos B
so
sin γt cos 2γt =
and
sin3 γt =
1
[sin 3γt − sin γt]
2
3
1
sin γt − sin 3γt.
4
4
Thus
x + ω 2 x =
1
3
sin γt − sin 3γt.
4
4
The frequency of free vibration is ω/2π. Thus, when γ1 /2π = ω/2π or γ1 = ω, and when
3γ2 /2π = ω/2π or 3γ2 = ω or γ2 = ω/3, the system will be in pure resonance.
(c) x
γ 1 = 1/2
x
γ1 = 1
x
n=2
10
γ 2 = 1/3
x
n=3
10
5
5
5
10
20
30
t
n=3
10
10
20
30
t
20
–5
–5
–5
–10
–10
–10
40
t
296
CHAPTER 5
MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS
1 49. Solving 20
q + 2q + 100q = 0 we obtain q(t) = e−20t (c1 cos 40t + c2 sin 40t). The initial
conditions q(0) = 5 and q (0) = 0 imply c1 = 5 and c2 = 5/2. Thus
−20t
q(t) = e
√
5
5 5 −20t
5 cos 40t + sin 40t =
e
sin (40t + 1.1071)
2
2
and q(0.01) ≈ 4.5676 coulombs. The charge is zero for the first time when 40t + 1.1071 = π
or t ≈ 0.0509 second.
50. Solving 14 q + 20q + 300q = 0 we obtain q(t) = c1 e−20t + c2 e−60t . The initial conditions
q(0) = 4 and q (0) = 0 imply c1 = 6 and c2 = −2. Thus
q(t) = 6e−20t − 2e−60t .
Setting q = 0 we find e40t = 1/3 which implies t < 0. Therefore the charge is not 0 for t ≥ 0.
51. Solving 53 q + 10q + 30q = 300 we obtain q(t) = e−3t (c1 cos 3t + c2 sin 3t) + 10. The initial
conditions q(0) = q (0) = 0 imply c1 = c2 = −10. Thus
q(t) = 10 − 10e−3t (cos 3t + sin 3t) and i(t) = 60e−3t sin 3t.
Solving i(t) = 0 we see that the maximum charge occurs when t = π/3 and q(π/3) ≈ 10.432.
52. Solving q + 100q + 2500q = 30 we obtain q(t) = c1 e−50t + c2 te−50t + 0.012. The initial
conditions q(0) = 0 and q (0) = 2 imply c1 = −0.012 and c2 = 1.4. Thus, using i(t) = q (t)
we get
q(t) = −0.012e−50t + 1.4te−50t + 0.012 and i(t) = 2e−50t − 70te−50t .
Solving i(t) = 0 we see that the maximum charge occurs when t = 1/35 second and
q(1/35) ≈ 0.01871 coulomb.
√ √
53. Solving q + 2q + 4q = 0 we obtain qc = e−t cos 3 t + sin 3 t . The steady-state charge
has the form qp = A cos t + B sin t. Substituting into the differential equation we find
(3A + 2B) cos t + (3B − 2A) sin t = 50 cos t.
Thus, A = 150/13 and B = 100/13. The steady-state charge is
qp (t) =
100
150
cos t +
sin t
13
13
and the steady-state current is
ip (t) = −
100
150
sin t +
cos t.
13
13
5.1
54. From
E0
ip (t) =
Z
and Z =
√
Linear Models: Initial-Value Problems
R
X
sin γt −
cos γt
Z
Z
X 2 + R2 we see that the amplitude of ip (t) is
A=
E0 E0
E02 R2 E02 X 2
+
= 2 R2 + X 2 =
.
4
4
Z
Z
Z
Z
55. The differential equation is 12 q + 20q + 1000q = 100 sin 60t. To use Example 10 in the text
we identify E0 = 100 and γ = 60. Then
1
1
1
= (60) −
≈ 13.3333,
cγ
2
0.001(60)
Z = X 2 + R2 = X 2 + 400 ≈ 24.0370,
X = Lγ −
and
100
E0
=
≈ 4.1603.
Z
Z
From Problem 54, then
ip (t) ≈ 4.1603 sin (60t + φ)
where sin φ = −X/Z and cos φ = R/Z. Thus tan φ = −X/R ≈ −0.6667 and φ is a fourth
quadrant angle. Now φ ≈ −0.5880 and
ip (t) = 4.1603 sin (60t − 0.5880).
56. Solving 12 q +20q +1000q = 0 we obtain qc (t) = e−20t (c1 cos 40t+c2 sin 40t). The steady-state
charge has the form qp (t) = A sin 60t + B cos 60t + C sin 40t + D cos 40t. Substituting into the
differential equation we find
(−1600A − 2400B) sin 60t + (2400A − 1600B) cos 60t
+ (400C − 1600D) sin 40t + (1600C + 400D) cos 40t
= 200 sin 60t + 400 cos 40t.
Equating coefficients we obtain A = −1/26, B = −3/52, C = 4/17, and D = 1/17. The
steady-state charge is
qp (t) = −
3
4
1
1
sin 60t −
cos 60t +
sin 40t +
cos 40t
26
52
17
17
and the steady-state current is
ip (t) = −
45
160
40
30
cos 60t +
sin 60t +
cos 40t −
sin 40t.
13
13
17
17
297
298
CHAPTER 5
MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS
57. Solving 12 q + 10q + 100q = 150 we obtain q(t) = e−10t (c1 cos 10t + c2 sin 10t) + 3/2. The
initial conditions q(0) = 1 and q (0) = 0 imply c1 = c2 = −1/2. Thus
1
3
q(t) = − e−10t (cos 10t + sin 10t) + .
2
2
As t → ∞, q(t) → 3/2.
58. In Problem 54 it is shown that the amplitude of the steady-state current is E0 /Z, where
√
Z = X 2 + R2 and X = Lγ − 1/Cγ. Since E0 is constant the amplitude will be a maximum
when Z is a minimum. Since R is constant, Z will be a minimum when X = 0. Solving
√
Lγ − 1/Cγ = 0 for γ we obtain γ = 1/ LC . The maximum amplitude will be E0 /R.
√
59. By Problem 54 the amplitude of the steady-state current is E0 /Z, where Z = X 2 + R2
and X = Lγ − 1/Cγ. Since E0 is constant the amplitude will be a maximum when Z is a
minimum. Since R is constant, Z will be a minimum when X = 0. Solving Lγ − 1/Cγ = 0
for C we obtain C = 1/Lγ 2 .
60. Solving 0.1q + 10q = 100 sin γt we obtain
q(t) = c1 cos 10t + c2 sin 10t + qp (t)
where qp (t) = A sin γt + B cos γt. Substituting qp (t) into the differential equation we find
(100 − γ 2 )A sin γt + (100 − γ 2 )B cos γt = 100 sin γt.
100
sin γt.
100 − γ 2
The initial conditions q(0) = q (0) = 0 imply c1 = 0 and c2 = −10γ/(100 − γ 2 ). The charge
is
10
(10 sin γt − γ sin 10t)
q(t) =
100 − γ 2
and the current is
100γ
i(t) =
(cos γt − cos 10t).
100 − γ 2
Equating coefficients we obtain A = 100/(100−γ 2 ) and B = 0. Thus, qp (t) =
61. In an LC-series circuit there is no resistor, so the differential equation is
L
√
Then q(t) = c1 cos t/ LC
1
d2 q
+ q = E(t).
2
dt
C
√
+ c2 sin t/ LC
+ qp (t) where qp (t) = A sin γt + B cos γt.
Substituting qp (t) into the differential equation we find
1
1
− Lγ 2 A sin γt +
− Lγ 2 B cos γt = E0 cos γt.
C
C
Equating coefficients we obtain A = 0 and B = E0 C/(1 − LCγ 2 ). Thus, the charge is
q(t) = c1 cos √
E0 C
1
1
t+
cos γt.
t + c2 sin √
1 − LCγ 2
LC
LC
5.2
Linear Models: Boundary-Value Problems
The initial conditions q(0) = q0 and q (0) = i0 imply c1 = q0 − E0 C/(1 − LCγ 2 ) and
√
c2 = i0 LC . The current is i(t) = q (t) or
c1
1
c2
1
E0 Cγ
sin γt
i(t) = − √
sin √
t+ √
cos √
t−
1 − LCγ 2
LC
LC
LC
LC
1
E0 Cγ
1
1
E0 C
t− √
q0 −
t−
sin √
sin γt.
= i0 cos √
2
1 − LCγ
1 − LCγ 2
LC
LC
LC
62. When the circuit is in resonance the form of qp (t) is qp (t) = At cos kt + Bt sin kt where
√
k = 1/ LC . Substituting qp (t) into the differential equation we find
qp + k 2 qp = −2kA sin kt + 2kB cos kt =
E0
cos kt.
L
Equating coefficients we obtain A = 0 and B = E0 /2kL. The charge is
q(t) = c1 cos kt + c2 sin kt +
E0
t sin kt.
2kL
The initial conditions q(0) = q0 and q (0) = i0 imply c1 = q0 and c2 = i0 /k. The current is
i(t) = −c1 k sin kt + c2 k cos kt +
=
5.2
E0
(kt cos kt + sin kt)
2kL
E0
E0
− q0 k sin kt + i0 cos kt +
t cos kt.
2kL
2L
Linear Models: Boundary-Value Problems
1. (a) The general solution is
y(x) = c1 + c2 x + c3 x2 + c4 x3 +
w0 4
x .
24EI
The boundary conditions are y(0) = 0, y (0) = 0, y (L) = 0, y (L) = 0. The first two
conditions give c1 = 0 and c2 = 0. The conditions at x = L give the system
w0 2
w0
L = 06c4 +
L
= 0.
2c3 + 6c4 L +
2EI
EI
Solving, we obtain c3 = w0 L2 /4EI and c4 = −w0 L/6EI. The deflection is
w0
(6L2 x2 − 4Lx3 + x4 ).
y(x) =
24EI
(b) x
0.2
1
2
3 y
0.4
0.6
0.8
1
x
299
300
CHAPTER 5
MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS
2. (a) The general solution is
y(x) = c1 + c2 x + c3 x2 + c4 x3 +
w0 4
x .
24EI
The boundary conditions are y(0) = 0, y (0) = 0, y(L) = 0, y (L) = 0. The first two
conditions give c1 = 0 and c3 = 0. The conditions at x = L give the system
w0 4
L =0
c2 L + c4 L3 +
24EI
w0 2
L = 0.
6c4 L +
2EI
Solving, we obtain c2 = w0 L3 /24EI and c4 = −w0 L/12EI. The deflection is
w0
(L3 x − 2Lx3 + x4 ).
y(x) =
24EI
(b) x
0.2
0.4
0.6
0.8
1
x
1 y
3. (a) The general solution is
y(x) = c1 + c2 x + c3 x2 + c4 x3 +
w0 4
x .
24EI
The boundary conditions are y(0) = 0, y (0) = 0, y(L) = 0, y (L) = 0. The first two
conditions give c1 = 0 and c2 = 0. The conditions at x = L give the system
w0 4
L =0
c3 L2 + c4 L3 +
24EI
w0 2
L = 0.
2c3 + 6c4 L +
2EI
Solving, we obtain c3 = w0 L2 /16EI and c4 = −5w0 L/48EI. The deflection is
w0
(3L2 x2 − 5Lx3 + 2x4 ).
y(x) =
48EI
(b) x
0.2
1
y
0.4
0.6
0.8
1
x
5.2
Linear Models: Boundary-Value Problems
4. (a) The general solution is
w0 L4
π
sin x.
4
EIπ
L
The boundary conditions are y(0) = 0, y (0) = 0, y(L) = 0, y (L) = 0. The first two
conditions give c1 = 0 and c2 = −w0 L3 /EIπ 3 . The conditions at x = L give the system
w0 4
L =0
c3 L2 + c4 L3 +
EIπ 3
y(x) = c1 + c2 x + c3 x2 + c4 x3 +
2c3 + 6c4 L = 0.
Solving, we obtain c3 = 3w0 L2 /2EIπ 3 and c4 = −w0 L/2EIπ 3 . The deflection is
π
2L3
w0 L
2
2
3
sin x .
−2L x + 3Lx − x +
y(x) =
2EIπ 3
π
L
(b) x
0.2
1
0.4
0.6
1
0.8
x
y
(c) Using a CAS we find the maximum deflection to be 0.270806 when x = 0.572536.
5. (a) The general solution is
w0
x5 .
120EI
The boundary conditions are y(0) = 0, y (0) = 0, y(L) = 0, y (L) = 0. The first two
conditions give c1 = 0 and c3 = 0. The conditions at x = L give the system
w0
L5 = 0
c2 L + c4 L3 +
120EI
w0 3
L = 0.
6c4 L +
6EI
y(x) = c1 + c2 x + c3 x2 + c4 x3 +
Solving, we obtain c2 = 7w0 L4 /360EI and c4 = −w0 L2 /36EI. The deflection is
w0
(7L4 x − 10L2 x3 + 3x5 ).
y(x) =
360EI
(b) x
0.2
1 y
0.4
0.6
0.8
1
x
301
302
CHAPTER 5
MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS
(c) Using a CAS we find the maximum deflection to be 0.234799 when x = 0.51933.
6. (a) ymax = y(L) = w0 L4 /8EI
(b) Replacing both L and x by L/2 in y(x) we obtain w0 L4 /128EI, which is 1/16 of the
maximum deflection when the length of the beam is L.
(c) ymax = y(L/2) = 5w0 L4 /384EI
(d) The maximum deflection in Example 1 is y(L/2) = (w0 /24EI)L4 /16 = w0 L4 /384EI,
which is 1/5 of the maximum displacement of the beam in part (c).
7. The general solution of the differential equation is
w0 2 w0 EI
P
P
x + c2 sinh
x+
x +
.
y = c1 cosh
EI
EI
2P
P2
Setting y(0) = 0 we obtain c1 = −w0 EI/P 2 , so that
w0 2 w0 EI
P
P
w0 EI
x + c2 sinh
x+
x +
cosh
.
y=−
2
P
EI
EI
2P
P2
Setting y (L) = 0 we find
c2 =
P w0 EI
sinh
EI P 2
w0 L
P
L−
EI
P
P
cosh
EI
P
L.
EI
8. The general solution of the differential equation is
P
P
w0 2 w0 EI
y = c1 cos
x + c2 sin
x−
x +
.
EI
EI
2P
P2
Setting y(0) = 0 we obtain c1 = −w0 EI/P 2 , so that
w0 2 w0 EI
P
P
w0 EI
x + c2 sin
x−
x +
cos
.
y=−
2
P
EI
EI
2P
P2
Setting y (L) = 0 we find
c2 =
−
P w0 EI
sin
EI P 2
P
w0 L
L+
EI
P
P
cos
EI
P
L.
EI
9. This is Example 2 in the text with L = π. The eigenvalues are λn = n2 π 2 /π 2 = n2 , n =
1, 2, 3, . . . and the corresponding eigenfunctions are yn = sin(nπx/π) = sin nx, n = 1, 2, 3, . . ..
10. This is Example 2 in the text with L = π/4. The eigenvalues are λn = n2 π 2 /(π/4)2 = 16n2 ,
n = 1, 2, 3, . . . and the eigenfunctions are yn = sin (nπx/(π/4)) = sin 4nx, n = 1, 2, 3, . . . .
5.2
Linear Models: Boundary-Value Problems
11. For λ ≤ 0 the only solution of the boundary-value problem is y = 0. For λ = α2 > 0 we have
y = c1 cos αx + c2 sin αx.
Now
y (x) = −c1 α sin αx + c2 α cos αx
and y (0) = 0 implies c2 = 0, so
y(L) = c1 cos αL = 0
gives
αL =
(2n − 1)π
2
or λ = α2 =
(2n − 1)2 π 2
, n = 1, 2, 3, . . . .
4L2
The eigenvalues (2n − 1)2 π 2 /4L2 correspond to the eigenfunctions cos
1, 2, 3, . . . .
(2n − 1)π
x for n =
2L
12. For λ ≤ 0 the only solution of the boundary-value problem is y = 0. For λ = α2 > 0 we have
y = c1 cos αx + c2 sin αx.
Since y(0) = 0 implies c1 = 0, y = c2 sin x dx. Now
y
gives
α
(2n − 1)π
π
=
2
2
π
2
= c2 α cos α
π
=0
2
or λ = α2 = (2n − 1)2 , n = 1, 2, 3, . . . .
The eigenvalues λn = (2n − 1)2 correspond to the eigenfunctions yn = sin (2n − 1)x.
13. For λ = −α2 < 0 the only solution of the boundary-value problem is y = 0. For λ = 0 we
have y = c1 x + c2 . Now y = c1 and y (0) = 0 implies c1 = 0. Then y = c2 and y (π) = 0.
Thus, λ = 0 is an eigenvalue with corresponding eigenfunction y = 1. For λ = α2 > 0 we
have
y = c1 cos αx + c2 sin αx.
Now
y (x) = −c1 α sin αx + c2 α cos αx
and y (0) = 0 implies c2 = 0, so
y (π) = −c1 α sin απ = 0
gives
απ = nπ
or λ = α2 = n2 , n = 1, 2, 3, . . . .
The eigenvalues n2 correspond to the eigenfunctions cos nx for n = 0, 1, 2, . . ..
303
304
CHAPTER 5
MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS
14. For λ ≤ 0 the only solution of the boundary-value problem is y = 0. For λ = α2 > 0 we have
y = c1 cos αx + c2 sin αx.
Now y(−π) = y(π) = 0 implies
c1 cos απ − c2 sin απ = 0
c1 cos απ + c2 sin απ = 0.
(1)
This homogeneous system will have a nontrivial solution when
cos απ − sin απ = 2 sin απ cos απ = sin 2απ = 0.
cos απ
sin απ Then
2απ = nπ
or
λ = α2 =
n2
;
4
n = 1, 2, 3, . . . .
When n = 2k − 1 is odd, the eigenvalues are (2k − 1)2 /4. Since cos (2k − 1)π/2 = 0 and
sin (2k − 1)π/2 = 0, we see from either equation in ((1)) that c2 = 0. Thus, the eigenfunctions
corresponding to the eigenvalues (2k − 1)2 /4 are y = cos (2k − 1)x/2 for k = 1, 2, 3, . . ..
Similarly, when n = 2k is even, the eigenvalues are k 2 with corresponding eigenfunctions
y = sin kx for k = 1, 2, 3, . . ..
15. The auxiliary equation has solutions
m=
1
−2 ± 4 − 4(λ + 1) = −1 ± α.
2
For λ = −α2 < 0 we have
y = e−x (c1 cosh αx + c2 sinh αx) .
The boundary conditions imply
y(0) = c1 = 0y(5)
= c2 e−5 sinh 5α = 0
so c1 = c2 = 0 and the only solution of the boundary-value problem is y = 0.
For λ = 0 we have
y = c1 e−x + c2 xe−x
and the only solution of the boundary-value problem is y = 0.
For λ = α2 > 0 we have
y = e−x (c1 cos αx + c2 sin αx) .
Now y(0) = 0 implies c1 = 0, so
y(5) = c2 e−5 sin 5α = 0
5.2
Linear Models: Boundary-Value Problems
gives
or λ = α2 =
5α = nπ
The eigenvalues λn =
n = 1, 2, 3, . . . .
n2 π 2
, n = 1, 2, 3, . . . .
25
n2 π 2
nπ
correspond to the eigenfunctions yn = e−x sin
x for
25
5
16. For λ < −1 the only solution of the boundary-value problem is y = 0. For λ = −1 we have
y = c1 x + c2 . Now y = c1 and y (0) = 0 implies c1 = 0. Then y = c2 and y (1) = 0.
Thus, λ = −1 is an eigenvalue with corresponding eigenfunction y = 1. For λ > −1 or
λ + 1 = α2 > 0 we have
y = c1 cos αx + c2 sin αx.
Now
y = −c1 α sin αx + c2 α cos αx
and y (0) = 0 implies c2 = 0, so
y (1) = −c1 α sin α = 0
gives
α = nπ,
λ + 1 = α 2 = n2 π 2 ,
or λn = n2 π 2 − 1, n = 1, 2, 3, . . . .
The eigenvalues n2 π 2 − 1 correspond to the eigenfunctions yn = cos nπx for n = 0, 1, 2, . . . .
17. For λ = α2 > 0 a general solution of the given differential equation is
y = c1 cos (α ln x) + c2 sin (α ln x).
Since ln 1 = 0, the boundary condition y(1) = 0 implies c1 = 0. Therefore
y = c2 sin (α ln x).
Using ln eπ = π we find that y (eπ ) = 0 implies
c2 sin απ = 0
or απ = nπ, n = 1, 2, 3, . . . . The eigenvalues and eigenfunctions are, in turn,
λn = α 2 = n 2 ,
n = 1, 2, 3, . . .
and yn = sin (n ln x).
For λ ≤ 0 the only solution of the boundary-value problem is y = 0.
18. For λ = 0 the general solution is y = c1 + c2 ln x. Now y = c2 /x, so y (e−1 ) = c2 e = 0 implies
c2 = 0. Then y = c1 and y(1) = 0 gives c1 = 0. Thus y(x) = 0.
For λ = −α2 < 0, y = c1 x−α + c2 xα . The boundary conditions give c2 = c1 e2α and c1 = 0, so
that c2 = 0 and y(x) = 0.
305
306
CHAPTER 5
MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS
For λ = α2 > 0, y = c1 cos (α ln x) + c2 sin (α ln x). From y(1) = 0 we obtain c1 = 0 and
y = c2 sin (α ln x). Now y = c2 (α/x) cos (α ln x), so y (e−1 ) = c2 eα cos α = 0 implies cos α = 0
or α = (2n − 1)π/2 and λn = α2 = (2n − 1)2 π 2 /4 for n = 1, 2, 3, . . .. The corresponding
eigenfunctions are
2n − 1
π ln x .
yn = sin
2
19. For λ = 0 the general solution is y = c1 + c2 ln x. Now y = c2 /x, so y (1) = c2 = 0 implies
c2 = 0. Then y = c1 and y (e2 ) = 0 is satisfied for any c1 .
For λ = −α2 < 0, y = c1 x−α + c2 xα . The boundary conditions give c1 = c2 e4α and c1 = 0, so
that c2 = 0 and y(x) = 0.
For λ = α2 > 0, y = c1 cos (α ln x) + c2 sin (α ln x). From y (1) = 0 we obtain c2 = 0 and
y = c1 cos (α ln x). Now y = −c1 (α/x) sin (α ln x), so y (e2 ) = −c1 (α/e2 ) sin 2α = 0 implies
sin 2α = 0 or α = nπ/2 and λn = α2 = n2 π 2 /4 for n = 0, 1, 2, . . .. The corresponding
eigenfunctions are
nπ
ln x .
yn = cos
2
Note n = 0 in yn gives y = 1.
20. For λ = 0 the general solution is y = c1 + c2 ln x. Now y = c2 /x, so y (e) = c2 /e = 0 implies
c2 = 0. Then y = c1 and y(1) = 0 gives c1 = 0. Thus y(x) = 0.
For λ = −α2 < 0, y = c1 x−α + c2 xα . The boundary conditions give c1 = c2 e2α and c2 = 0, so
that c1 = 0 and y(x) = 0.
For λ = α2 > 0, y = c1 cos (α ln x) + c2 sin (α ln x). From y(1) = 0 we obtain c1 = 0
and y = c2 sin (α ln x). Now y = c2 (α/x) cos (α ln x), so y (e) = c2 (α/e) cos α = 0 implies
cos α = 0 or α = (2n − 1)π/2 and λn = α2 = (2n − 1)2 π 2 /4 for n = 1, 2, 3, . . .. The
corresponding eigenfunctions are
2n − 1
π ln x .
yn = sin
2
21. For λ = α4 , α > 0, the general solution of the boundary-value problem
y (4) − λy = 0,
y(0) = 0, y (0) = 0, y(1) = 0, y (1) = 0
is
y = c1 cos αx + c2 sin αx + c3 cosh αx + c4 sinh αx.
The boundary conditions y(0) = 0, y (0) = 0 give c1 + c3 = 0 and −c1 α2 + c3 α2 = 0, from
which we conclude c1 = c3 = 0. Thus, y = c2 sin αx + c4 sinh αx. The boundary conditions
y(1) = 0, y (1) = 0 then give
c2 sin α + c4 sinh α = 0
−c2 α2 sin α + c4 α2 sinh α = 0.
5.2
Linear Models: Boundary-Value Problems
307
In order to have nonzero solutions of this system, we must have the determinant of the
coefficients equal zero, that is,
sin α
sinh α 2
−α2 sin α α2 sinh α = 0 or 2α sinh α sin α = 0
But since α > 0, the only way that this is satisfied is to have sin α = 0 or α = nπ. The system
is then satisfied by choosing c2 = 0, c4 = 0, and α = nπ. The eigenvalues and corresponding
eigenfunctions are then
λn = α4 = (nπ)4 , n = 1, 2, 3, . . .
and
yn = sin nπx.
22. For λ = α4 , α > 0, the general solution of the differential equation is
y = c1 cos αx + c2 sin αx + c3 cosh αx + c4 sinh αx.
The boundary conditions y (0) = 0, y (0) = 0 give c2 α + c4 α = 0 and −c2 α3 + c4 α3 = 0 from
which we conclude c2 = c4 = 0. Thus, y = c1 cos αx + c3 cosh αx. The boundary conditions
y(π) = 0, y (π) = 0 then give
c1 cos απ + c3 cosh απ = 0
−c1 α2 cos απ + c3 α2 cosh απ = 0.
The determinant of the coefficients is 2α2 cosh απ cos απ = 0. But since α > 0, the only way
that this is satisfied is to have c1 = 0, c3 = 0, and cos απ = 0 or α = (2n−1)/2, n = 1, 2, 3, . . ..
The eigenvalues and corresponding eigenfunctions are
2n − 1 4
2n − 1
4
x.
, n = 1, 2, 3, . . .
and
yn = cos
λn = α =
2
2
23. If restraints are put on the column at x = L/4, x = L/2, and x = 3L/4, then the
critical load will be P4 .
y
L
x
24. (a) The general solution of the differential equation is
P
P
x + c2 sin
x + δ.
y = c1 cos
EI
EI
Since the column is embedded at x = 0, the boundary conditions are y(0) = y (0) = 0.
If δ = 0 this implies that c1 = c2 = 0 and y(x) = 0. That is, there is no deflection.
308
CHAPTER 5
MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS
(b) If δ = 0, the boundary conditions give, in turn, c1 = −δ and c2 = 0. Then
P
x .
y = δ 1 − cos
EI
In order to satisfy the boundary condition y(L) = δ we must have
P
P
L
or cos
L = 0.
δ = δ 1 − cos
EI
EI
This gives P/EI L = (2n − 1)π/2 for n = 1, 2, 3, . . .. The smallest value of Pn , the
Euler load, is then
π
1 π 2 EI
P1
L=
or P1 =
.
EI
2
4
L2
25. If λ = α2 = P/EI, then the solution of the differential equation is
y = c1 cos αx + c2 sin αx + c3 x + c4 .
The conditions y(0) = 0, y (0) = 0 yield, in turn, c1 + c4 = 0 and c1 = 0. With c1 = 0 and
c4 = 0 the solution is y = c2 sin αx + c3 x. The conditions y(L) = 0, y (L) = 0, then yield
c2 sin αL + c3 L = 0
and
− c2 α2 sin αL = 0.
Hence, nontrivial solutions of the problem exist only if sin αL = 0. From this point on, the
analysis is the same as in Example 3 in the text.
26. (a) The boundary-value problem is
d4 y
d2 y
+
λ
= 0,
dx4
dx2
y(0) = 0, y (0) = 0, y(L) = 0, y (L) = 0,
where λ = α2 = P/EI. The solution of the differential equation is y = c1 cos αx +
c2 sin αx + c3 x + c4 and the conditions y(0) = 0, y (0) = 0 yield c1 = 0 and c4 = 0. Next,
by applying y(L) = 0, y (L) = 0 to y = c2 sin αx + c3 x we get the system of equations
c2 sin αL + c3 L = 0
αc2 cos αL + c3
= 0.
To obtain nontrivial solutions c2 , c3 , we must have the determinant of the coefficients
equal to zero:
sin αL L
α cos αL 1 = 0 or tan β = β
where β = αL. If βn denotes the positive roots of the last equation, then the eigenvalues
√
are found from βn = αn L = λn L or λn = (βn /L)2 . From λ = P/EI we see that
5.2
Linear Models: Boundary-Value Problems
the critical loads are Pn = βn2 EI/L2 . With the aid of a CAS we find that the first
positive root of tan β = β is (approximately) β1 = 4.4934, and so the Euler load is
(approximately) P1 = 20.1907EI/L2 . Finally, if we use c3 = −c2 α cos αL, then the
deflection curves are
βn
βn
x −
cos βn x .
yn (x) = c2 sin αn x + c3 x = c2 sin
L
L
(b) With L = 1 and c2 appropriately chosen, the general shape of the first buckling mode,
4.4934
4.4934
x −
cos (4.4934) x ,
y1 (x) = c2 sin
L
L
is shown below.
y1
0.2
0.4
27. The general solution is
0.6
1
x
ρ
ωx.
T
From y(0) = 0 we obtain c1 = 0. Setting y(L) = 0 we find ρ/T ωL = nπ, n = 1, 2, 3, . . . .
√
√
Thus, critical speeds are ωn = nπ T /L ρ , n = 1, 2, 3, . . . . The corresponding deflection
curves are
nπ
x, n = 1, 2, 3, . . . ,
y(x) = c2 sin
L
where c2 = 0.
y = c1 cos
ρ
ωx + c2 sin
T
0.8
28. (a) When T (x) = x2 the given differential equation is the Cauchy-Euler equation
x2 y + 2xy + ρω 2 y = 0.
The solutions of the auxiliary equation
m(m − 1) + 2m + ρω 2 = m2 + m + ρω 2 = 0
are
1 1
4ρω 2 − 1 i,
m1 = − −
2 2
when ρω 2 > 0.25. Thus
1 1
m2 = − +
4ρω 2 − 1 i
2 2
y = c1 x−1/2 cos (λ ln x) + c2 x−1/2 sin (λ ln x)
where λ =
1
2
4ρω 2 − 1. Applying y(1) = 0 gives c1 = 0 and consequently
y = c2 x−1/2 sin (λ ln x).
309
310
CHAPTER 5
MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS
The condition y(e) = 0 requires c2 e−1/2 sin λ = 0. We obtain a nontrivial solution when
λn = nπ, n = 1, 2, 3, . . .. But
λn =
1
4ρωn2 − 1 = nπ.
2
ωn =
1 2 2
(4n π + 1)/ρ .
2
Solving for ωn gives
The corresponding solutions are
yn (x) = c2 x−1/2 sin (nπ ln x).
(b) x
y
y
y
1
1
1
n=2
n=1
e
1
x
n=3
e
1
x
–1
–1
e
1
x
–1
29. The auxiliary equation is m2 + m = m(m + 1) = 0 so that u(r) = c1 r−1 + c2 . The boundary
conditions u(a) = u0 and u(b) = u1 yield the system c1 a−1 + c2 = u0 , c1 b−1 + c2 = u1 . Solving
gives
u0 − u1
u1 b − u0 a
ab and c2 =
.
c1 =
b−a
b−a
Thus
u(r) =
u0 − u1
b−a
ab u1 b − u0 a
+
.
r
b−a
30. The auxiliary equation is m2 = 0 so that u(r) = c1 + c2 ln r. The boundary conditions
u(a) = u0 and u(b) = u1 yield the system c1 + c2 ln a = u0 , c1 + c2 ln b = u1 . Solving gives
c1 =
Thus
u(r) =
u1 ln a − u0 ln b
ln (a/b)
and c2 =
u0 − u1
.
ln (a/b)
u0 ln (r/b) − u1 ln (r/a)
u1 ln a − u0 ln b u0 − u1
+
ln r =
.
ln (a/b)
ln (a/b)
ln(a/b)
31. (a) This is similar to Problem 21 with 1 replaced by L. Thus the eigenvalues and eigenfunctions are, in turn,
λn = αn4 =
n4 π 4
,
L4
n = 1, 2, 3, . . .
and
(b)
λn = αn4 = ρωn2 /EI = n4 π 4 /L4
ωn2 =
n4 π 4 EI
.
L4 ρ
yn (x) = sin
nπx
L
5.2
Linear Models: Boundary-Value Problems
311
Therefore the critical speeds are
n2 π 2
ωn =
L2
EI
,
ρ
n = 1, 2, 3, . . . .
Then the fundamental critical speed is
π2
ω1 = 2
L
EI
.
ρ
Noe that ωn = n2 ω1 .
32. (a) For λ = α4 , α > 0, the solution of the differential
equation is
y
1
y = c1 cos αx + c2 sin αx + c3 cosh αx + c4 sinh αx.
The boundary conditions y(0) = 0,
y (1) = 0 give, in turn,
y (0)
2
4
6
8
10
12
x
= 0, y(1) = 0,
c1 + c3 = 0
αc2 + αc4 = 0,
c1 cos α + c2 sin α + c3 cosh α + c4 sinh α = 0
−c1 α sin α + c2 α cos α + c3 α sinh α + c4 α cosh α = 0.
The first two equations enable us to write
c1 (cos α − cosh α) + c2 (sin α − sinh α) = 0
c1 (− sin α − sinh α) + c2 (cos α − cosh α) = 0.
The determinant
cos α − cosh α sin α − sinh α − sin α − sinh α cos α − cosh α = 0
simplifies to cos α cosh α = 1. From the figure showing the graphs of 1/ cosh x and cos x,
we see that this equation has an infinite number of positive roots.
(b) Using the third equation in the system to eliminate c2 , we find the eigenfunctions are
yn = (− sin αn + sinh αn )(cos αn x − cosh αn x) + (cos αn − cosh αn )(sin αn x − sinh αn x).
33. The solution of the initial-value problem
x + ω 2 x = 0,
x(0) = 0, x (0) = v0 , ω 2 = 10/m
312
CHAPTER 5
MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS
is x(t) = (v0 /ω) sin ωt. To satisfy the additional boundary condition x(1) = 0 we require that
ω = nπ, n = 1, 2, 3, . . .. The eigenvalues λ = ω 2 = n2 π 2 and eigenfunctions of the problem
are then x(t) = (v0 /nπ) sin nπt. Using ω 2 = 10/m we find that the only masses that can pass
through the equilibrium position at t = 1 are mn = 10/n2 π 2 . Note for n = 1, the heaviest
mass m1 = 10/π 2 will not pass through the equilibrium position on the interval 0 < t < 1
(the period of x(t) = (v0 /π) sin πt is T = 2, so on 0 ≤ t ≤ 1 its graph passes through x = 0
only at t = 0 and t = 1). Whereas for n > 1, masses of lighter weight will pass through the
equilibrium position n − 1 times prior to passing through at t = 1. For example, if n = 2,
the period of x(t) = (v0 /2π) sin 2πt is 2π/2π = 1, the mass will pass through x = 0 only once
(t = 12 ) prior to t = 1; if n = 3, the period of x(t) = (v0 /3π) sin 3πt is 23 , the mass will pass
through x = 0 twice (t = 13 and t = 23 ) prior to t = 1; and so on.
34. The initial-value problem is
2 k
x + x = 0,
m
m
x +
x(0) = 0, x (0) = v0 .
√
With k = 10, the auxiliary equation has roots γ = −1/m ± 1 − 10m/m. Consider the three
cases:
(i) m =
1
10 .
The roots are γ1 = γ2 = −10 and the solution of the differential equation is x(t) = c1 e−10t +
c2 te−10t . The initial conditions imply c1 = 0 and c2 = v0 and so x(t) = v0 te−10t . The
condition x(1) = 0 implies v0 e−10 = 0 which is impossible because v0 = 0.
(ii) 1 − 10m > 0 or 0 < m <
1
10
.
The roots are
γ1 = −
1√
1
−
1 − 10m
m m
and
γ2 = −
1√
1
+
1 − 10m
m m
and the solution of the differential equation is x(t) = c1 eγ1 t + c2 eγ2 t . The initial conditions
imply
c1 + c2 = 0
γ 1 c 1 + γ 2 c 2 = v0
so c1 = v0 /(γ1 − γ2 ), c2 = −v0 /(γ1 − γ2 ), and
x(t) =
v0
(eγ1 t − eγ2 t ).
γ1 − γ2
Again, x(1) = 0 is impossible because v0 = 0.
(iii) 1 − 10m < 0 or m >
1
10
.
The roots of the auxiliary equation are
γ1 = −
1√
1
−
10m − 1 i
m m
and
γ2 = −
1√
1
+
10m − 1 i
m m
5.2
Linear Models: Boundary-Value Problems
and the solution of the differential equation is
x(t) = c1 e−t/m cos
1√
1√
10m − 1 t + c2 e−t/m sin
10m − 1 t.
m
m
√
The initial conditions imply c1 = 0 and c2 = mv0 / 10m − 1, so that
x(t) = √
mv0
e−t/m sin
10m − 1
1 √
10m − 1 t ,
m
The condition x(1) = 0 implies
√
1√
mv0
e−1/m sin
10m − 1 = 0
m
10m − 1
sin
1√
10m − 1 = 0
m
1√
10m − 1 = nπ
m
10m − 1
= n2 π 2 , n = 1, 2, 3, . . .
m2
(n2 π 2 )m2 − 10m + 1 = 0
√
√
10 100 − 4n2 π 2
5 ± 25 − n2 π 2
m=
=
.
2n2 π 2
n2 π 2
Since m is real, 25 − n2 π 2 ≥ 0. If 25 − n2 π 2 = 0, then n2 = 25/π 2 , and n is not an integer.
Thus, 25 − n2 π 2 = (5 − nπ)(5 + nπ) > 0 and since n > 0, 5 + nπ > 0, so 5 − nπ > 0 also. Then
n < 5/π, and so n = 1. Therefore, the mass m will pass through the equilibrium position
when t = 1 for
√
√
5 + 25 − π 2
5 − 25 − π 2
and
m2 =
.
m1 =
π2
π2
35. (a) The general solution of the differential equation is y = c1 cos 4x + c2 sin 4x. From y0 =
y(0) = c1 we see that y = y0 cos 4x + c2 sin 4x. From y1 = y(π/2) = y0 we see that any
solution must satisfy y0 = y1 . We also see that when y0 = y1 , y = y0 cos 4x + c2 sin 4x is
a solution of the boundary-value problem for any choice of c2 . Thus, the boundary-value
problem does not have a unique solution for any choice of y0 and y1 .
(b) Whenever y0 = y1 there are infinitely many solutions.
(c) When y0 = y1 there are no solutions.
(d) The boundary-value problem has the trivial solution when y0 = y1 = 0. The solution is
not unique.
313
314
CHAPTER 5
MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS
36. (a) The general solution of the differential equation is y = c1 cos 4x + c2 sin 4x. From 1 =
y(0) = c1 we see that y = cos 4x + c2 sin 4x. From 1 = y(L) = cos 4L + c2 sin 4L we see
that c2 = (1 − cos 4L)/ sin 4L. Thus,
1 − cos 4L
y = cos 4x +
sin 4x
sin 4L
will be a unique solution when sin 4L = 0; that is, when L = kπ/4 where k = 1, 2, 3, . . ..
(b) There will be infinitely many solutions when sin 4L = 0 and 1 − cos 4L = 0; that is, when
L = kπ/2 where k = 1, 2, 3, . . ..
(c) There will be no solution when sin 4L = 0 and 1 − cos 4L = 0; that is, when L = kπ/4
where k = 1, 3, 5, . . ..
(d) There can be no trivial solution since it would fail to satisfy the boundary conditions.
37. (a) A solution curve has the same y-coordinate at both ends of the interval [−π, π] and the
tangent lines at the endpoints of the interval are parallel.
(b) For λ = 0 the solution of y = 0 is y = c1 x + c2 . From the first boundary condition we
have
y(−π) = −c1 π + c2 = y(π) = c1 π + c2
or 2c1 π = 0. Thus, c1 = 0 and y = c2 . This constant solution is seen to satisfy the
boundary-value problem.
For λ = −α2 < 0 we have y = c1 cosh αx + c2 sinh αx. In this case the first boundary
condition gives
y(−π) = c1 cosh (−απ) + c2 sinh (−απ)
= c1 cosh απ − c2 sinh απ
= y(π) = c1 cosh απ + c2 sinh απ
or 2c2 sinh απ = 0. Thus c2 = 0 and y = c1 cosh αx. The second boundary condition
implies in a similar fashion that c1 = 0. Thus, for λ < 0, the only solution of the
boundary-value problem is y = 0.
For λ = α2 > 0 we have y = c1 cos αx + c2 sin αx. The first boundary condition implies
y(−π) = c1 cos (−απ) + c2 sin (−απ)
= c1 cos απ − c2 sin απ
= y(π) = c1 cos απ + c2 sin απ
or 2c2 sin απ = 0. Similarly, the second boundary condition implies 2c1 α sin απ = 0.
If c1 = c2 = 0 the solution is y = 0. However, if c1 = 0 or c2 = 0, then sin απ = 0,
5.2
Linear Models: Boundary-Value Problems
315
which implies that α must be an integer, n. Therefore, for c1 and c2 not both 0,
y = c1 cos nx + c2 sin nx is a nontrivial solution of the boundary-value problem. Since
cos (−nx) = cos nx and sin (−nx) = − sin nx, we may assume without loss of generality
that the eigenvalues are λn = α2 = n2 , for n a positive integer. The corresponding
eigenfunctions are yn = cos nx and yn = sin nx.
(c) x
y
y
3
3
–π
π
–3
y = 2 sin 3x
x
–π
π
x
–3
y = sin 4x − 2 cos 3x
√
√
38. For λ = α2 > 0 the general solution is y = c1 cos α x + c2 sin α x. Setting y(0) = 0 we find
√
c1 = 0, so that y = c2 sin α x. The boundary condition y(1) + y (1) = 0 implies
√
√
√
c2 sin α + c2 α cos α = 0.
√
√
Taking c2 = 0, this equation is equivalent to tan α = − α . Thus, the eigenvalues are λn =
√
√
αn2 = x2n , n = 1, 2, 3, . . . , where the xn are the consecutive positive roots of tan α = − α .
39. We see from the graph that tan x = −x has infinitely
roots. Since λn = αn2 , there are no new eigenvalues
αn < 0. For λ = 0, the differential equation y =
general solution y = c1 x + c2 . The boundary conditions
c1 = c2 = 0, so y = 0.
tan x
5
many
when
0 has
imply
2.5
2
4
6
8
10
12
x
–2.5
–5
–7.5
–10
40. Using a CAS we find that the first four nonnegative roots of tan x = −x are approximately
2.02876, 4.91318, 7.97867, and 11.0855. The corresponding eigenvalues are 4.11586, 24.1393,
63.6591, and 122.889, with eigenfunctions sin (2.02876x), sin (4.91318x), sin(7.97867x), and
sin(11.0855x).
41. In the case when λ = −α2 < 0, the solution of the
differential equation is y = c1 cosh αx + c2 sinh αx. The
condition y(0) = 0 gives c1 = 0. The condition
y(1) − 12 y (1) = 0 applied to y = c2 sinh αx gives
c2 (sinh α − 12 α cosh α) = 0 or tanh α = 12 α. As can be
seen from the figure, the graphs of y = tanh x and y = 12 x
intersect at a single point with approximate x-coordinate
y
1
1
2
x
316
CHAPTER 5
MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS
α1 = 1.915. Thus, there is a single negative eigenvalue λ1 = −α12 ≈ −3.667 and the corresponding eigenfuntion is y1 = sinh 1.915x. For λ = 0 the only solution of the boundaryvalue problem is y = 0. For λ = α2 > 0 the solution of the differential equation is
y = c1 cos αx + c2 sin αx. The condition y(0) = 0 gives c1 = 0, so y = c2 sin αx. The
condition y(1) − 12 y (1) = 0 gives c2 (sin α − 12 αcos α) = 0, so the eigenvalues are λn = αn2
when αn , n = 2, 3, 4, . . . , are the positive roots of tan α = 12 α. Using a CAS we find that the
first three values of α are α2 = 4.27487, α3 = 7.59655, and α4 = 10.8127. The first three
eigenvalues are then λ2 = α22 = 18.2738, λ3 = α32 = 57.7075, and λ4 = α42 = 116.9139 with
corresponding eigenfunctions y2 = sin 4.27487x, y3 = sin 7.59655x, and y4 = sin 10.8127x.
42. From the figure in Problem 32 showing the graphs of 1/ cosh x and cos x, we see that this
equation has an infinite number of positive roots. With the aid of a CAS the first four
roots are found to be α1 = 4.73004, α2 = 7.8532, α3 = 10.9956, and α4 = 14.1372, and
the corresponding eigenvalues are λ1 = 500.5636, λ2 = 3803.5281, λ3 = 14,617.5885, and
λ4 = 39,944.1890.
5.3
Nonlinear Models
1. The period corresponding to x(0) = 1, x (0) = 1
is approximately 5.6. The period corresponding to
x(0) = 1/2, x (0) = −1 is approximately 6.2.
2
x
1
4
2
6
8
–1
–2
2. The solutions are not periodic.
10
x
8
6
4
2
t
–2
3. The period corresponding to x(0) = 1, x (0) = 1 is
approximately 5.8. The second initial-value problem
does not have a periodic solution.
10
x
8
6
4
2
2
–2
4
6
8
10
t
5.3 Nonlinear Models
317
x
4. Both solutions have periods of approximately 6.3.
3
2
1
2
4
6
8
10
t
–1
–2
–3
5. From the graph we see that |x1 | ≈ 1.2.
x
4
3
x1 = 1.2
2
1
–1
x1 = 1.1
5
10
t
5
10
t
x
6. From the graphs we see that the interval is
approximately (−0.8, 1.1).
3
2
1
–1
7. Since
xe0.01x = x[1 + 0.01x +
for small values of x, a linearization is
1
(0.01x)2 + · · · ] ≈ x
2!
d2 x
+ x = 0.
dt2
318
CHAPTER 5
MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS
8. x
x
3
5
10
15
t
–3
For x(0) = 1 and x (0) = 1 the oscillations are symmetric about the line x = 0 with amplitude
slightly greater than 1.
For x(0) = −2 and x (0) = 0.5 the oscillations are symmetric about the line x = −2 with
small amplitude.
√
For x(0) = 2 and x (0) = 1 the oscillations are symmetric about the line x = 0 with
amplitude a little greater than 2.
For x(0) = 2 and x (0) = 0.5 the oscillations are symmetric about the line x = 2 with small
amplitude.
For x(0) = −2 and x (0) = 0 there is no oscillation; the solution is constant.
√
For x(0) = − 2 and x (0) = −1 the oscillations are symmetric about the line x = 0 with
amplitude a little greater than 2.
x
9. This is a damped hard spring, so x will approach 0 as
t approaches ∞.
2
2
–2
4
6
8
t
5.3 Nonlinear Models
319
x
10. This is a damped soft spring, so we might expect no oscillatory
solutions. However, if the initial conditions are sufficiently small
the spring can oscillate.
5
4
3
2
1
t
4
2
–1
–2
11. x
15
k1 = 0.01
x
k1 = 1
x
3
10
2
5
1
10
20
t
30
10
–5
–1
–10
–2
20
t
–3
–15
k1 = 20
x
k1 = 100
x
3
3
2
2
1
1
5
10
t
1
–1
–1
–2
–2
–3
–3
2
t
3
When k1 is very small the effect of the nonlinearity is greatly diminished, and the system is
close to pure resonance.
12. (a) x
40
x
40
20
20
20
40
60
80
–20
–40
x
100
t
20
40
60
80
100
–20
k = –0.000465
–40
k = –0.000466
t
320
CHAPTER 5
MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS
The system appears to be oscillatory for −0.000465 ≤ k1 < 0 and nonoscillatory for
k1 ≤ −0.000466.
(b) x
x
x
3
3
2
2
1
1
20
40
60
80
100
120
140
t
20
40
60
80
100
120
140
t
–1
–1
k = –0.3493
–2
k = –0.3494
–2
–3
–3
The system appears to be oscillatory for −0.3493 ≤ k1 < 0 and nonoscillatory for
k1 ≤ −0.3494.
θ
13. For λ2 − ω 2 > 0 we choose λ = 2 and
ω = 1 with θ(0) = 1 and θ (0) = 2.
For λ2 − ω 2 < 0 we choose λ = 1/3 and
ω = 1 with θ(0) = −2 and θ (0) = 4.
In both cases the motion corresponds to
the overdamped and underdamped cases
for spring/mass systems.
3
λ = 1/3, ω = 1
λ = 2, ω = 1
5
10
15
t
–3
14. (a) Setting dy/dt = v, the differential equation in (13) becomes dv/dt = −gR2 /y 2 . But, by
the chain rule, dv/dt = (dv/dy)(dy/dt) = v dv/dy, so v dv/dy = −gR2 /y 2 . Separating
variables and integrating we obtain
v dv = −gR2
dy
y2
and
1 2 gR2
v =
+ c.
2
y
Setting v = v0 and y = R we find c = −gR + 12 v02 and
v 2 = 2g
R2
− 2gR + v02 .
y
(b) As y → ∞ we assume that v → 0+ . Then v02 = 2gR and v0 =
√
2gR .
(c) Using g = 9.8 ft/s and R = 6500 km we find
v0 =
(d) v0 =
2(9.8)(6500)(1000) ≈ 11287 m/s ≈ 40633 km/hr.
2(0.165)(9.8)(1738 × 103 ) ≈ 2371 m/s ≈ 8536 km/hr
5.3 Nonlinear Models
15. (a) Intuitively, one might expect that only half of a 3-N chain could be lifted by a 1.5-newton
vertical force.
(b) Since x = 0 when t = 0, and v = dx/dt =
1 − 19.6x/3 , we have v(0) =
√
196 ≈ 14 m/s.
(c) Since x should always be positive, we solve x(t) = 0, getting t = 0 and t ≈ 8.572. Since
the graph of x(t) is a parabola, the maximum value occurs at tm = 4.27. (This can also
be obtained by solving x (t) = 0.) At this time the height of the chain is x(tm ) ≈ 30 m.
This is higher than predicted because of the momentum generated by the force. When
the chain is 25 m high it still has a positive velocity of about 5.72 m/s, which keeps it
going higher for a while.
(d) As discussed in the solution to part (c) of this problem, the chain has momentum generated by the force applied to it that will cause it to go higher than expected. It will
then fall back to below the expected maximum height, again due to momentum. This,
in turn, will cause it to next go higher than expected, and so on.
16. (a) Setting dx/dt = v, the differential equation becomes (L − x)dv/dt − v 2 = Lg. But,
by the Chain Rule, dv/dt = (dv/dx)(dx/dt) = v dv/dx, so (L − x)v dv/dx − v 2 = Lg.
Separating variables and integrating we obtain
v2
v
1
dv =
dx
+ Lg
L−x
and
1
ln(v 2 + Lg) = − ln(L − x) + ln c,
2
√
so v 2 + Lg = c/(L − x). When x = 0, v = 0, and c = L Lg . Solving for v and
simplifying we get
Lg(2Lx − x2 )
dx
= v(x) =
.
dt
L−x
Again, separating variables and integrating we obtain
√
L−x
dx = dt
Lg(2Lx − x2 )
Since x(0) = 0, we have c1 = 0 and
x(t) = L −
√
and
2Lx − x2
√
= t + c1 .
Lg
√
2Lx − x2 / Lg = t. Solving for x we get
L2 − Lgt2
√
dx
Lgt
=
and v(t) =
.
dt
L − gt2
(b) The chain will be completely on the ground when x(t) = L or t =
L/g .
(c) The predicted velocity of the upper end of the chain when it hits the ground is infinity.
321
322
CHAPTER 5
MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS
17. (a) Let (x, y) be the coordinates of S2 on the curve C. The slope at (x, y) is then
v1 t − y
y − v1 t
dy
=
=
dx
0−x
x
or xy − y = −v1 t.
(b) Differentiating with respect to x and using r = v1 /v2 gives
xy + y − y = −v1
dt
dx
xy = −v1
dt ds
ds dx
xy = −v1
1
− 1 + (y )2
v2
xy = r
1 + (y )2 .
Letting u = y and separating variables, we obtain
du
= r 1 + u2
x
dx
√
r
du
= dx
2
x
1+u
sinh−1 u = r ln x + ln c = ln (cxr )
u = sinh (ln cxr )
1
1
dy
r
=
cx − r .
dx
2
cx
At t = 0, dy/dx = 0 and x = a, so 0 = car − 1/car . Thus c = 1/ar and
1 x r
a r 1
x r
x −r
dy
=
.
=
−
−
dx
2 a
x
2
a
a
If r > 1 or r < 1, integrating gives
1
a
y=
2 1+r
x
a
1+r
1
−
1−r
x
a
1−r
+ c1 .
When t = 0, y = 0 and x = a, so 0 = (a/2)[1/(1+r)−1/(1−r)]+c1 . Thus c1 = ar/(1−r2 )
and
ar
1
x 1+r
1
x 1−r
a
+
−
.
y=
2 1+r a
1−r a
1 − r2
If r = 1, then integration gives
y=
1 x2
1
− ln x + c2 .
2 2a a
When t = 0, y = 0 and x = a, so 0 = (1/2)[a/2 − (1/a) ln a] + c2 . Thus
c2 = −(1/2)[a/2 − (1/a) ln a] and
1
1 a 1
1 1 2
1 x2
1 a
2
− ln x −
− ln a =
(x − a ) + ln
.
y=
2 2a a
2 2 a
2 2a
a x
5.3 Nonlinear Models
(c) To see if the paths ever intersect we first note that if r > 1, then v1 > v2 and y → ∞ as
x → 0+ . In other words, S2 always lags behind S1 . Next, if r < 1, then v1 < v2 and
y = ar/(1 − r2 ) when x = 0. In other words, when the submarine’s speed is greater than
the ship’s, their paths will intersect at the point (0, ar/(1 − r2 )). Finally, if r = 1, then
y → ∞ as x → 0+ , meaning S2 will never catch up with S1 .
18. (a) Let (r, θ) denote the polar coordinates of the destroyer S1 . When S1 travels the 6 miles
from (9, 0) to (3, 0) it stands to reason, since S2 travels half as fast as S1 , that the polar
coordinates of S2 are (3, θ2 ), where θ2 is unknown. In other words, the distances of the
ships from (0, 0) are the same and r(t) = 15t then gives the radial distance of both ships.
This is necessary if S1 is to intercept S2 .
(b) The differential of arc length in polar coordinates is (ds)2 = (r dθ)2 + (dr)2 , so that
ds
dt
2
=r
2
dθ
dt
2
+
dr
dt
2
.
Using ds/dt = 30 and dr/dt = 15 then gives
900 = 225t
2
675 = 225t
2
dθ
dt
dθ
dt
2
+ 225
2
√
dθ
3
=
dt
t
√
√
r
θ(t) = 3 ln t + c = 3 ln
+ c.
15
√
When r = 3, θ = 0, so c = − 3 ln 15 and
√
√
r
1
r
− ln
= 3 ln .
θ(t) = 3 ln
15
5
3
√
Thus r = 3eθ/
3,
whose graph is a logarithmic spiral.
(c) The time for S1 to go from (9, 0) to (3, 0) = 15 hour. Now S1 must intercept the path
angle β, where
0 < β < 2π. At the time of interception t2 we have
of S2 for some
√
√
1 β/ 3
3
β/
or t = 5 e
. The total time is then
15t2 = 3e
t=
√
1 1 β/√3 1
+ e
< (1 + e2π/ 3 ).
5 5
5
19. (a) The auxiliary equation is m2 +g/l = 0, so the general solution of the differential equation
is
g
g
t + c2 sin
t.
θ(t) = c1 cos
l
l
323
324
CHAPTER 5
MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS
The initial condtion θ(0) = 0 implies c1 = 0 and θ (0) = ω0 implies c2 = ω0
l
g
sin
t.
θ(t) = ω0
g
l
(b) At θmax , sin
g/l t = 1, so
l
vb
mb
=
g
mw + mb l
θmax = ω0
and
vb =
l
v
mb
√b
=
g
mw + mb lg
mw + mb lg θmax .
mb
(c) We have cos θmax = (l − h)/l = 1 − h/l. Then
cos θmax ≈ 1 −
1 2
h
θmax = 1 −
2
l
and
2
θmax
Thus
2h
=
l
or
mw + mb lg
vb =
mb
θmax =
2h
.
l
mw + mb 2h
=
2gh .
l
mb
(d) When mb = 5 g, mw = 1 kg, and h = 6 cm, we have
vb =
1005 2 (980) (6) ≈ 21, 797 cm/s.
5
20. (a) Substituting u = dx/dt into
d2 x
m 2 = −k
dt
dx
dt
2
,
x(0) = 0, x (0) = v0
gives
k
k
du
= − u2 or u−1 =
t + c1 .
dt
m
m
The initial conditions imply c1 = 1/v0 , so
u=
dx
1
v0
=
.
=
dt
kt/m + 1/v0
kv0 t/m + 1
Integrating and applying the intial conditions gives
m kv0
m kv0
ln t + 1 + c2 =
ln t + 1.
x(t) =
k
m
k
m
Similarly, substituting u = dy/dt into
2
dy
d2 y
,
m 2 = mg − k
dt
dt
y(0) = 0, y (0) = 0
l/g . Thus,
5.3 Nonlinear Models
325
gives
k
du
= g − u2
dt
m
or
m
du
= dt.
k mg/k − u2
Integrating, we obtain
1
u
m
tanh−1 = t + c3 .
k
mg/k
mg/k
The intial conditions imply c3 = 0 so
dy
=u=
dt
mg
tanh
k
kg
t.
m
Integrating and applying the initial conditions gives
m
mg
m
kg
kg
ln cosh
t + c4 =
ln cosh
t .
y(t) =
k
kg
m
k
m
(b) Using the fact that 1000 N is equivalent to 102 kg of mass, we solve (using numerical
procedure)
0.0053(9.8)
102
ln cosh
t ,
300 =
0.0053
102
which gives t = 7.84 seconds. Then x(7.84) = 1089 m is the horizontal distance travelled
by the supply pack.
21. Since (dx/dt)2 is always positive, it is necessary to use |dx/dt| (dx/dt) in order to account
for the fact that the motion is oscillatory and the velocity (or its square) should be negative
when the spring is contracting.
y
22. (a) From the graph we see that the approximations
appears to be quite good for 0 ≤ x ≤ 0.4.
Using an equation solver to solve sin x − x =
0.05 and sin x − x = 0.005, we find that the
approximation is accurate to one decimal place
for θ1 = 0.67 and to two decimal places for
θ1 = 0.31.
(b) x
1.5
1.25
1
0.75
0.5
0.25
0.25
θ
0.5
0.75
1
1.25
1.5
5
6
θ
1
1
θ1 = 1,3,5
0.5
1
2
3
θ1 = 7,9,11
0.5
4
5
6
t
1
–0.5
–0.5
–1
–1
2
3
4
x
326
CHAPTER 5
MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS
23. (a) Write the differential equation as
d2 θ
+ ω 2 sin θ = 0,
dt2
θ
2
moon
earth
where ω 2 = g/l. To determine the differences between the Earth and the Moon
–2
we take l = 1, θ(0) = 1, and θ (0) = 2.
Using g = 9.8 on the Earth and g =
9.8 × 0.165
on the Moon we obtain the graphs shown in the figure.
5
t
(b) Comparing the apparent periods of the graphs, we see that the pendulum oscillates faster
on the Earth than on the Moon and the amplitude is greater on the Moon than on the
Earth.
θ
24. The linear model is
2
d2 θ
+ ω 2 θ = 0,
dt2
moon
earth
where ω 2 = g/l. When g = 9.8, l = 1,
θ(0) = 1, and θ (0) = 2, the solution is
5
t
–2
θ(t) = cos 3.13t + 0.639 sin 3.13t.
When g = 9.8 × 0.1652 the solution is
θ(t) = cos 1.272t + 1.572 sin 1.272t.
As in the nonlinear case, the pendulum oscillates faster on the Earth than on the Moon and
still has greater amplitude on the Moon.
25. (a) The general solution of
d2 θ
+θ =0
dt2
is θ(t) = c1 cos t + c2 sin t. From θ(0) = π/12 and θ (0) = −1/3 we find
π
1
θ(t) =
cos t −
sin t.
12
3
Setting θ(t) = 0 we have tan t = π/4 which implies t1 = tan−1 (π/4) ≈ 0.66577.
(b) We set θ(t) = θ(0) + θ (0)t + 12 θ (0)t2 + 16 θ (0)t3 + · · · and use θ (t) = − sin θ(t) together
with θ(0) = π/12 and θ (0) = −1/3. Then
√
√ ( 3 − 1)
π
=− 2
θ (0) = − sin
12
4
5.3 Nonlinear Models
and
π
θ (0) = − cos θ(0) · θ (0) = − cos
12
Thus
π
1
θ(t) =
− t−
12 3
√
327
√
√
1
3+1
−
= 2
.
3
12
√ √
√
2 ( 3 − 1) 2
2 ( 3 + 1) 3
t +
t + ··· .
8
72
(c) Setting π/12 − t/3 = 0 we obtain t1 = π/4 ≈ 0.785398.
(d) Setting
1
π
− t−
12 3
√
√
2 ( 3 − 1) 2
t =0
8
and using the positive root we obtain t1 ≈ 0.63088.
(e) Setting
1
π
− t−
12 3
√
√ √
√
2 ( 3 − 1) 2
2 ( 3 + 1) 3
t +
t =0
8
72
we find with the help of a CAS that t1 ≈ 0.661973 is the first positive root.
(f ) From the output we see that y(t) is an interpolating
function on the interval 0 ≤ t ≤ 5, whose graph is shown.
The positive root of y(t) = 0 near t = 1 is t1 = 0.666404.
0.4
0.2
1
2
3
4
5
2
4
6
8
10
–0.2
–0.4
(f ) To find the next two positive roots we change the interval
used in NDSolve and Plot from {t, 0, 5} to {t, 0, 10}.
We see from the graph that the second and third positive
roots are near 4 and 7, respectively. Replacing {t, 1} in
FindRoot with {t, 4} and then {t, 7} we obtain
t2 = 3.84411 and t3 = 7.0218.
0.4
0.2
–0.2
–0.4
26. From the table below we see that the pendulum first passes the vertical position between
1.6 and 1.7 seconds. To refine our estimate of t1 we estimate the solution of the differential
equation on [1.6, 1.7] using a step size of h = 0.01. From the resulting table we see that
t1 is between 1.63 and 1.64 seconds. Repeating the process with h = 0.001 we conclude
that t1 ≈ 1.634. Then the period of the pendulum is approximately 4t1 = 6.536. The
error when using t1 = 2π is 6.536 − 6.283 = 0.253 and the percentage relative error is
(0.253/6.536)100 = 3.87.
328
CHAPTER 5
MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS
h = 0.1
tn
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
1.10
1.20
1.30
1.40
1.50
1.60
1.70
1.80
1.90
2.00
θn
0.78540
0.78187
0.77129
0.75375
0.72937
0.69833
0.66086
0.61726
0.56788
0.51313
0.45347
0.38943
0.32161
0.25062
0.17716
0.10194
0.02570
−0.05080
−0.12678
−0.20151
−0.27423
h = 0.01
tn
1.60
1.61
1.62
1.63
1.64
1.65
1.66
1.67
1.68
1.69
1.70
θn
0.02570
0.01805
0.01040
0.00274
−0.00491
−0.01256
−0.02021
−0.02786
−0.03551
−0.04315
−0.05080
h = 0.001
1.630
1.631
1.632
1.633
1.634
1.635
1.636
1.637
0.00274
0.00198
0.00121
0.00045
−0.00032
−0.00108
−0.00185
−0.00261
Chapter 5 in Review
1. 8 m, since k = 4
2. 0.441, since 0.816x + 6.25x = 0
3. 5/4 m, since x = − cos 4t + 34 sin 4t
4. True
5. False; since an external force may exist
6. False; since the equation of motion in this case is x(t) = e−λt (c1 + c2 t) and x(t) = 0 can have
at most one real solution
7. overdamped
√
√
8. From x(0) = ( 2/2) sin φ = −1/2 we see that sin φ = −1/ 2 , so φ is an angle in the third
√
√
or fourth quadrant. Since x (t) = 2 cos(2t + φ), x (0) = 2 cos φ = 1 and cos φ > 0. Thus
φ is in the fourth quadrant and φ = −π/4.
9. y = 0 because λ = 8 is not an eigenvalue
10. y = cos 6x because λ = (6)2 = 36 is an eigenvalue
11. The period of a spring/mass system is given by T = 2π/ω where ω 2 = k/m = kg/W ,
where k is the spring constant, W is the weight of the mass attached to the spring, and g
Chapter 5 in Review
√
√
is the acceleration due to gravity. Thus, the period of oscillation is T = (2π/ kg ) W . If
√
√
√
√
the weight of the original mass is W , then (2π/ kg ) W = 3 and (2π/ kg ) W − 8 = 2.
√
√
Dividing, we get W / W − 8 = 3/2 or W = 94 (W − 8). Solving for W we find that the
weight of the original mass was 14.4 N.
12. (a) Solving 38 x + 6x = 0 subject to x(0) = 1 and x (0) = −4 we obtain
x = cos 4t − sin 4t =
(b) The amplitude is
√
√
2 sin (4t + 3π/4).
2, period is π/2, and frequency is 2/π.
(c) If x = 1 then t = nπ/2 and t = −π/8 + nπ/2 for n = 1, 2, 3, . . ..
(d) If x = 0 then t = π/16 + nπ/4 for n = 0, 1, 2, . . .. The motion is upward for n even and
downward for n odd.
(e) x (3π/16) = 0
(f ) If x = 0 then 4t + 3π/4 = π/2 + nπ or t = 3π/16 + nπ.
13. We assume that the spring is initially compressed by 10 cm and that the positive direction
on the x-axis is in the direction of elongation of the spring. Then, from 14 x + 32 x + 2x = 0,
x(0) = −0.1, and x (0) = 0 we obtain x = −0.2e−2t + 0.1e−4t .
14. From x + βx + 100x = 0 we see that oscillatory motion results if β 2 − 400 < 0 or 0 ≤ β < 10.
15. From mx +4x +2x = 0 we see that nonoscillatory motion results if 16−8m ≥ 0 or 0 < m ≤ 2.
16. From 14 x + x + x = 0, x(0) = 4, and x (0) = 2 we obtain x = 4e−2t + 10te−2t . If x (t) = 0,
then t = 1/10, so that the maximum displacement is x = 5e−0.2 ≈ 4.094.
γt + 8 sin γt we identify
17. Writing 18 x + 83 x = cos γt + sin γt in the form x + 64
3 x = 8 cos
√
64
2
ω = 3 . The system is in a state of pure resonance when γ = ω = 64/3 = 8/ 3 .
18. Clearly xp = A/ω 2 suffices.
19. From 18 x + x + 3x = e−t , x(0) = 2, and x (0) = 0 we obtain
√
√
xc = e−4t c1 cos 2 2 t + c2 sin 2 2 t ,
−4t
x=e
xp =
√
√
√
26
28 2
cos 2 2 t +
sin 2 2 t
17
17
8 −t
e ,
17
+
8 −t
e .
17
and
329
330
CHAPTER 5
MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS
20. (a) Let k be the effective spring constant and x1 and x2 the elongation of springs k1 and k2 .
The restoring forces satisfy k1 x1 = k2 x2 so x2 = (k1 /k2 )x1 . From k(x1 + x2 ) = k1 x1 we
have
k1
x2 = k1 x1
k x1 +
k2
k2 + k1
= k1
k
k2
k=
k1 k2
k1 + k2
1
1
1
=
+
.
k
k1 k2
From k1 = 2W and k2 = 4W we find 1/k = 1/2W + 1/4W = 3/4W . Then
k = 4W/3 = 4mg/3. The differential equation mx + kx = 0 then becomes
x + (4g/3)x = 0. The solution is
g
g
t + c2 sin 2
t.
x(t) = c1 cos 2
3
3
The initial conditions x(0) = 0.3 and x (0) = 0.2 imply c1 = 0.3 and c2 = 0.1/
3
g
.
(b) To compute the maximum velocity of the mass we compute
3
g
g
g
3
g
sin 2
t + 0.1
cos 2
t and |x (t)| = 4 + 0.01 .
x (t) = 2
3
3
g
3
3
g
21. From q + 104 q = 100 sin 50t, q(0) = 0, and q (0) = 0 we obtain qc = c1 cos 100t + c2 sin 100t,
1
sin 50t, and
qp = 75
1
sin 100t +
(a) q(t) = − 150
1
75
sin 50t,
(b) i(t) = − 23 cos 100t + 23 cos 50t, and
(c) q(t) = 0 when sin 50t(1 − cos 50t) = 0 or t = nπ/50 for n = 0, 1, 2, . . ..
22. By Kirchhoff’s second law,
L
dq
1
d2 q
+ q = E(t).
+R
2
dt
dt
C
Using q (t) = i(t) we can write the differential equation in the form
L
1
di
+ Ri + q = E(t).
dt
C
Then differentiating we obtain
L
d2 i
1
di
+ R + i = E (t).
2
dt
dt C
Chapter 5 in Review
23. For λ = α2 > 0 the general solution is y = c1 cos αx + c2 sin αx. Now
y(0) = c1
and
y(2π) = c1 cos 2πα + c2 sin 2πα,
so the condition y(0) = y(2π) implies
c1 = c1 cos 2πα + c2 sin 2πα
which is true when α =
√
λ = n or λ = n2 for n = 1, 2, 3, . . .. Since
y = −αc1 sin αx + αc2 cos αx = −nc1 sin nx + nc2 cos nx,
we see that y (0) = nc2 = y (2π) for n = 1, 2, 3, . . . Thus, the eigenvalues are n2 for
n = 1, 2, 3, . . ., with corresponding eigenfunctions cos nx and sin nx. When λ = 0, the general
solution is y = c1 x + c2 and the corresponding eigenfunction is y = 1.
For λ = −α2 < 0 the general solution is y = c1 cosh αx + c2 sinh αx. In this case y(0) = c1
and y(2π) = c1 cosh 2πα + c2 sinh 2πα, so y(0) = y(2π) can only be valid for α = 0. Thus,
there are no eigenvalues corresponding to λ < 0.
24. (a) The differential equation is d2 r/dt2 − ω 2 r = −g sin ωt. The auxiliary equation is
m2 − ω 2 = 0, so rc = c1 eωt + c2 e−ωt . A particular solution has the form
rp = A sin ωt + B cos ωt. Substituting into the differential equation we find
−2Aω 2 sin ωt − 2Bω 2 cos ωt = −g sin ωt. Thus, B = 0, A = g/2ω 2 , and
rp = (g/2ω 2 ) sin ωt. The general solution of the differential equation is r(t) = c1 eωt +
c2 e−ωt + (g/2ω 2 ) sin ωt. The initial conditions imply c1 + c2 = r0 and g/2ω − ωc1 + ωc2 =
v0 . Solving for c1 and c2 we get
c1 = (2ω 2 r0 + 2ωv0 − g)/4ω 2
and c2 = (2ω 2 r0 − 2ωv0 + g)/4ω 2 ,
so that
r(t) =
g
2ω 2 r0 + 2ωv0 − g ωt 2ω 2 r0 − 2ωv0 + g −ωt
e +
e
+ 2 sin ωt.
2
2
4ω
4ω
2ω
(b) The bead will exhibit simple harmonic motion when the exponential terms are missing.
Solving c1 = 0, c2 = 0 for r0 and v0 we find r0 = 0 and v0 = g/2ω.
To find the minimum length of rod that will accommodate simple harmonic motion we
determine the amplitude of r(t) and double it. Thus L = g/ω 2 .
(c) As t increases, eωt approaches infinity and e−ωt approaches 0. Since sin ωt is bounded,
the distance, r(t), of the bead from the pivot point increases without bound and the
distance of the bead from P will eventually exceed L/2.
331
332
CHAPTER 5
(d) x
MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS
r
17
20
16.1
16
10
4
2
8
6
10
12
14
t
–10
–20
0
10
15
(e) For each v0 we want to find the smallest value of t for which r(t) = ±20. Whether we
look for r(t) = −20 or r(t) = 20 is determined by looking at the graphs in part (d). The
total times that the bead stays on the rod is shown in the table below.
v0
0
10
15
16.1
17
r
–20
–20
–20
20
20
t
1.55007
2.35494
3.43088
6.11627
4.22339
When v0 = 16 the bead never leaves the rod.
25. Unlike the derivation given in Section 3.8 in the text, the weight mg of the mass m does not
appear in the net force since the spring is not stretched by the weight of the mass when it is
in the equilibrium position (i.e., there is no mg − ks term in the net force). The only force
acting on the mass when it is in motion is the restoring force of the spring. By Newton’s
second law,
d2 x
k
d2 x
or
+ x = 0.
m 2 = −kx
dt
dt2
m
26. By Newton’s second law of motion:
m
d2 x
d2 x
=
−k
x
−
k
x
or
m
+ (k1 + k2 ) x = 0.
1
2
dt2
dt2
Here −k1 x is the force to the left due to the elongation x of the spring with constant k1 and
−k2 x is the force to the left due to the compression x of the spring with constant k2 .
27. The force of kinetic friction opposing the motion of the mass in μN , where μ is the coefficient
of sliding friction and N is the normal component of the weight. Since friction is a force
opposite to the direction of motion and since N is pointed directly downward (it is simply
the weight of the mass), Newton’s second law gives, for motion to the right (x > 0),
m
d2 x
= −kx − μmg,
dt2
m
d2 x
= −kx + μmg.
dt2
and for motion to the left (x < 0),
Chapter 5 in Review
Traditionally, these two equations are written as one expression
m
d2 x
+ fk sgn(x ) + kx = 0,
dt2
where fk = μmg and
1,
sgn(x ) =
−1,
x > 0
x < 0
28. (a) The differential equation is x + sgn(x ) + x = 0 or
x + x =
1,
motion to the left
−1, motion to the right.
Correspondingly
c3 cos t + c4 sin t + 1, motion to the left
x(t) =
c1 cos t + c2 sin t − 1, motion to the right
For motion to the left x + x = 1, x(0) = 5.5, x (0) = 0 gives x(t) = 4.5 cos t + 1. From
x (t) = −4.5 sin t we see that the mass is at rest (x (t) = 0) at t = π so the interval of
definition is [0, π]. Note that x (t) < 0 for 0 < t < π. The mass is now on the left at
x(π) = −3.5.
(b) For motion to the right we solve the initial value problem x + x = −1, x(π) = −3.5,
x (π) = 0 gives x(t) = 2.5 cos t − 1. From x (t) = −2.5 sin t we see that the velocity is 0
next at t = 2π so the interval of definition is [π, 2π]. Note that x (t) > 0 for π < t < 2π.
The mass is once again on the right at x(2π) = 1.5.
(c) For motion to the left we solve the initial-value problem x + x = 1, x(2π) = 1.5,
x (2π) = 0 gives x(t) = 0.5 cos t + 1. From x (t) = −0.5 sin t we see that the velocity
is 0 next at t = 3π so the interval of definition is [2π, 3π]. Note that x (t) < 0 for
2π < t < 3π. The mass is still on the right at x(3π) = 0.5.
(d) Now if there is further motion to the right we solve the initial-value problem x +x = −1,
x(3π) = 0.5, x (3π) = 0, which gives x(t) = −1.5 cos t − 1. From x (t) = 1.5 sin t we
see that the velocity is 0 next at t = 4π so the interval of definition is [3π, 4π]. But for
3π < t < 4π, x (t) < 0 which contradicts the assumption of motion to the right. This
indicates that the solution x(t) = c1 cos t + c2 sin t − 1 is no longer applicable. The other
solution for motion to the left is also not applicable since it implies x (t) > 0. The mass
has undoubtedly stopped at x(3π) = 0.5.
333
334
CHAPTER 5
MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS
x
Motion on the interval [0, 3π]:
(e)
x(t) =
6
⎧
4.5 cos t + 1, 0 ≤ t < π
⎪
⎪
⎪
⎪
⎨2.5 cos t − 1, π ≤ t < 2π
4
⎪
0.5 cos t + 1, 2π ≤ t < 3π
⎪
⎪
⎪
⎩
0.5,
t ≥ 3π
2
π
2π
2
3π
t
29. The given initial-value problem is
d2 θ g
+ sin θ = 0,
dt2
l
θ(0) = π/6,
θ (0) = 0
Differentiating with respect to time t the differential equation yields
g
d2 θ
= − sin θ,
2
dt
l
d3 θ
dθ
g
= − cos θ ,
3
dt
l
dt
d4 θ
d2 θ g
g
cos
θ
=
−
+ sin θ
dt4
l
dt2
l
and
dθ
dt
2
, ...
From the differential equation, the initial conditions, and the foregoing derivatives we see at
√
t = 0 that sin (θ(0)) = sin π/6 = 1/2 and cos (θ(0)) = cos π/6 = 3 /2 and so:
π
θ(0) = ,
6
θ (0) = 0,
g
θ (0) = − ,
2l
√
θ (0) = 0,
θ
(4)
(0) =
3 g2
, ...
4l2
The Maclaurin expansion
θ(t) = θ(0) + θ (0)t +
is the
θ (0) 2 θ (0) 3 θ(4) (0) 4
t +
t +
t + ···
2!
3!
4!
√ 2
g 2
3g 4
π
t + ··· .
θ(t) = − t +
6 4l
96l2
30. (a) If m and l are constant, then (1) in the text is simply
d
ml
dt
2
ml2
dθ
dt
= −mgl sin θ
d2 θ
= −mgl sin θ
dt2
d2 θ g
+ sin θ = 0
dt2
l
Chapter 5 in Review
(b) If m is constant and l is a function of time t then by the product rule of differentiation,
dθ
d
ml2
= −mgl sin θ
dt
dt
2
dl dθ
2 d θ
ml
+ mgl sin θ = 0
+ m 2l
dt2
dt dl
ml2
dl dθ
d2 θ
+ mgl sin θ = 0
+ 2ml
2
dt
dt dt
l
d2 θ
dl dθ
+ g sin θ = 0
+2
2
dt
dt dt
When l(t) = l0 + x(t) the differential equation becomes
(l0 + x)
d2 θ
dx dθ
+ g sin θ = 0
+2
2
dt
dt dt
31. (a) The linear initial-value problem is
d2 θ g
+ θ1 = 0,
dt2
l
θ1 (0) = θ0 , θ1 (0) = 0.
Applying the initial conditions to θ1 (t) = c1 cos
c2 = 0. Therefore the solution is
θ1 (t) = θ0 cos
The string hits the nail when θ1 (t1 ) = 0 or cos
g/l t + c2 sin g/l t gives c1 = θ0 and
g/l t.
g/l t1 = 0. Thus
π
g
t1 =
l
2
π
t1 =
2
l
seconds
g
The interval over which the solution θ1 (t) is defined is 0, π2
l
g .
(b) Now θ2 (t1 ) = θ1 (t1 ) = 0. Also, the initial angular speed when the string hits the nail
satisfies θ2 (t1 ) = θ1 (t1 ). Using t1 found in part (a)
θ1 (t1 )
= −θ0
g
sin
l
g π
l 2
l
g
= −θ0
g
π
sin = −θ0
l
2
Since the length of the string is l/4 the initial-value problem for θ2 (t) is
g
g
d2
θ2 = 0, θ2 (T ) = 0, θ2 (T ) = −θ0
+
2
dt
l/4
l
g
l
335
336
CHAPTER 5
MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS
Applying the initial condition θ2 (t1 ) = 0 to θ2 (t) = c3 cos 2
c3 = 0. Therefore
g/l t + c2 sin 2 g/l t gives
g
t
θ2 (t) = c4 sin 2
l
g
g
θ2 (t) = c4 2
cos 2
t
l
l
and so
π
g
g π
l
g
g
cos 2
= −2c4
= −θ0
θ2 (t1 ) = c4 2
l
l 2 g
l
l
So c4 = θ0 /2 and
1
g
t.
θ2 (t) = θ0 sin 2
2
l
At the time t2 we have θ2 (t) = 0 for the second time:
1
g
t2 = 0
θ(t2 ) = θ0 sin 2
2
l
g
t2 = 2π
2
l
t2 = π
l
g
π l
l
,π
.
The interval over which θ2 (t) is defined is
2 g
g
32. (a) The solution of the differential equation
d2 x
+ 4x = sin 4t
dt2
1
sin 4t. The intial conditions x(0) = 0, x (0) = v0 give
is x(t) = c1 cos 2t + c2 sin 2t − 12
1
sin 4t.
c1 = 0 and c2 = 16 + 12 v0 . Therefore x(t) = 16 + 12 v0 sin 2t − 12
Using sin 2θ = 2 sin theta cos θ we can write sin 4t = 2 sin 2t cos 2t and so
x(t) =
1 1
1
1
1 1
+ v0 sin 2t − sin 2t cos 2t = sin 2t
+ v0 − cos 2t
6 2
6
6 2
6
The first positive solution t1 of x(t) = 0 is then obtained from sin 2t = 0 or t1 = π/2.
Thes the solution of the initial-value problemis defined on [0, π/2].
Chapter 5 in Review
(b) The solution of the differential equation
d2 x
+ x = sin 4t
dt2
1
sin 4t. The initial conditions x(π/2) = 0, x (π/2) = − 23 −v0
is x(t) = c3 cos t+c4 sin t− 15
give c4 = 0 and c3 = 25 + v0 . So
1
2
+ v0 cos t −
sin 4t.
x(t) =
5
15
By rewriting:
2
+ v0 cos t −
x(t) =
5
2
=
+ v0 cos t −
5
2
2
+ v0 cos t −
sin 2t cos 2t
5
15
2
4
4
sin t cos t cos 2t = cos t
+ v0 −
sin t cos 2t
15
5
15
1
sin 4t =
15
we see x(t) = 0 for t = 3π/2. Thus the solution is defined on [π/2, 3π/2].
(c) Because the mass is again below the equilibrium position x = 0 the differential equation
in the next initial-value problem is again
d2 x
+ 4x = sin 4t
dt2
1
sin 4t. The intial condition are x(3π/2) =
and its solution is x(t) = c5 cos 2t+c6 sin 2t− 12
2
7
− 12 v0 . So the solution is
0, x (3π/2) = 15 + v0 and give c5 = 0 and c6 = − 30
1
1
7
+ v0 sin 2t −
sin 4t.
x(t) = −
30 2
12
From the alternative form
1
7
1
1
1
7
sin 4t = − − v0 sin 2t − sin 2t cos 2t
x(t) = − − v0 sin 2t −
30 2
12
30 2
6
1
7
1
= sin 2t − − v0 − cos 2t
30 2
6
we see that x(t) = 0 for t3 = 2π. The the solution is defined on [3π/2, 2π].
1
sin 4t The initial conditions
(d) Proceeding as in part (b) we find x(t) = c7 cos t + c8 sin t − 15
4
8
x(2π) = 0, x (2π) = − 5 − v0 give c7 = 0 and c8 = − 15 − v0 . So
8
1
+ v0 sin t −
sin 4t.
x(t) = −
15
15
8
− v0 −
From the form x(t) = sin − 15
solution is defined on [2π, 3π].
4
15
sin t cos 2t we see x(t) = 0 when t4 = 3π. The
337
338
CHAPTER 5
MODELING WITH HIGHER-ORDER DIFFERENTIAL EQUATIONS
(e) The velocity of the mass at the beginning of the first cycle is x (0) = v0 , the velocity at
2
+ v0 , the velocity at the beginning of
the beginning of the second cycle is x (3π/2) = 15
4
the third cycle is x (3π) = 15 + v0 . and so on. It stands to reason that the amplitudes
2
at the
of oscillation will increase because the velocity of the mass is increasing by 15
beginning of each cycle.
(f ) For v0 = 0.01 the graph of the piecewise-defined function
⎧
1
1
⎪
⎪
+ 0.005 sin 2t −
sin 4t,
⎪
⎪
6
12
⎪
⎪
⎪
⎪
⎪
⎪
⎪
1
2
⎪
⎪
+ 0.01 cos t −
sin 4t,
⎪
⎨ 5
15
x(t) =
⎪
⎪
1
7
⎪
⎪
+
0.005
sin
2t
−
sin 4t,
−
⎪
⎪
30
12
⎪
⎪
⎪
⎪
⎪
⎪
⎪
1
8
⎪
⎩−
+ 0.01 sin t −
sin 4t,
15
15
0 ≤ t ≤ π/2
π/2 < t ≤ 3π/2
3π/2 < t ≤ 2π
2π < t ≤ 3π
is shown below.
x
0.5
Π
2
0.5
Π
3Π
2
2Π
5Π
2
3Π
t
Chapter 6
Series Solutions of Linear Equations
6.1
Review of Power Series
n+1
x
an+1
/(n + 1) n
|x| = |x|
1. lim
= lim lim
= n→∞
n→∞ an
n→∞
xn /n
n+1
The series is absolutely convergent on (−1, 1). At x = −1, the series
series which diverges. At x = 1, the series
∞
(−1)n
∞
1
is the harmonic
n
n=1
converges by the alternating series test.
n
Thus, the given series converges on (−1, 1], and the radius of converegence is 1.
n=1
n+1
x
an+1
/(n + 1)2 n
2. lim
|x| = |x|
= lim = lim
n
2
n→∞ an
n→∞
n→∞
x /n
n+1
The series is absolutely convergent on (−1, 1). At x = −1, the series
∞
(−1)n
n=1
n2
converges by
∞
1
is a convergent p-series. Thus, the
n2
n=1
given series converges on [−1, 1], and the radius of converegence is 1.
the alternating series test. At x = 1, the series
n+1 n+1
2
an+1
x
/(n + 1) 2n
|x| = 2|x|
= lim 3. lim
= lim
n
n
n→∞ an
n→∞
n→∞ n + 1
2 x /n
The series is absolutely convergent for 2|x| < 1 or |x| < 12 . The radius of convergence is
∞
(−1)n
1
1
converges by the alternating series test. At x = 12 ,
R = 2 . At x = − 2 , the series
n
n=1
∞
1
is the harmonic series which diverges. Thus, the given series converges on
the series
n
n=1
[− 12 , 12 ), and the radius of convergence is 12 .
n+1 n+1
5
an+1
x
/(n
+
1)!
5
= lim
= lim 4. lim
n→∞ n + 1 |x| = 0
n
n
n→∞ an
n→∞
5 x /n!
The series is absolutely convergent on (−∞, ∞), and the radius of convergence is R = ∞.
339
340
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
(x − 5)k+1 /10k+1 ak+1
= lim 1 |x − 5| = 1 |x − 5|
5. lim
= lim k→∞ 10
k
k
k→∞ ak
k→∞
(x − 5) /10
10
1
|x − 5| < 1, |x − 5| < 10, or on (−5, 15). At
The series is absolutely convergent for 10
∞
∞
n
(−10)
(−1)k
=
1 diverges by the nth term test. At x = 15, the
x = −5, the series
10n
k=1
k=1
∞
∞
10k
series
(−1)k k =
(−1)k diverges by the nth term test. Thus, the series converges on
10
k=1
k=1
(−5, 15), and the radius of convergence is 10.
∞,
(k + 1)!(x − 1)k+1 ak+1
= lim (k + 1)|x − 1| =
6. lim
= lim k→∞ ak
k→∞
k→∞
k!(x − 1)k
0,
x = 1
x=1
The radius of convergence is R = 0 and the series converges only for x = 1.
(3x − 1)n+1 / (n + 1)2 + (n + 1) ak+1
n2 + n
|3x − 1| = |3x − 1|
= lim =
lim
7. lim
k→∞ n2 + 3n + 2
k→∞ ak
k→∞ (3x − 1)n /(n2 + n)
The series is absolutely convergent for |3x − 1| < 1 or on (0, 2/3). At x = 0, the series
∞
∞
(−1)k
1
converges
by
the
alternating
series
test.
At
x
=
2/3,
the
series
converges
2
2
k +k
k +k
k=1
k=1
∞
1
by comparison with the p-series
. Thus, the series converges on [0, 2/3], and the radius
k2
k=1
of convergence is 1/3.
(4x − 5)n+1 /3n+1 ak+1
= lim 1 |4x − 5| = 1 |4x − 5|
= lim 8. lim
k→∞ 3
n
n
k→∞ ak
k→∞
(4x − 5) /3
3
The series is absolutely convergent for 13 |4x − 5| < 1, |4x − 5| < 3 or on (1/2, 2). At x = 12 ,
∞
∞
(−3)k
=
(−1)k diverges by the nth term test. At x = 2, the series
the series
3k
k=0
k=0
∞
3k
=
1 diverges by nth term test. Thus, the series converges on (1/2, 2), and the
3k
k=0
k=0
radius of convergence is 5/4.
9. Write the series as
∞ 32 k
k=1
ak+1
k→∞ ak
lim
75
xk .Then
(32/75)k+1 xk+1 = lim 32 |x| = 32 |x|
= lim k→∞ 75
k
k
k→∞
(32/75) x
75
The series is absolutely convergent for 32
75 |x| < 1, or on (−75/32, 75/32). At x = −75/32, the
∞
(−1)k diverges by the nth term test. At x = 75/32, the series
1 diverges by nth
series
k=1
k=1
term test. Thus, the series converges on (−75/32, 75/32), and the radius of convergence is
75/32.
6.1
Review of Power Series
x2(k+1)+1 /9k+1 1
ak+1
1
10. lim
= lim = lim x2 = x2
k→∞ ak
k→∞ x2n+1 /9k k→∞ 9
9
The series is absolutely convergent for 19 x2 < 1, or on (−3, 3). At x = −3, the series
∞
(−1)k (−3) diverges by the nth term test. At x = 3, the series
(−1)k 3 diverges by nth
k=1
k=1
term test. Thus, the series converges on (−3, 3), and the radius of convergence is 3.
11. We replace x by −x/2 in the Maclaurin series of ex .
∞
∞
1 −x n (−1)n n
−x/2
=
=
x
e
n!
2
n!2n
n=0
n=0
12. We replace x by 3x in the Maclaurin series of ex and multiply the result by x.
xe3x = x ·
∞
∞
1
3n n+1
(3x)n =
x
n!
n!
n=0
13. We factor out a
n=0
1
x
1
and replace x by − in the Maclaurin series of
.
2
2
1−x
1
1
1
x
x
1
= ·
=
1+ −
+ −
2+x
2 1 − (−x/2)
2
2
2
2
+ ··· =
1
x x2
1 − + 2 − ···
2
2
2
∞
=
(−1)n
x
x2
1
− 2 + 3 − ··· =
xn
2 2
2
2n+1
n=0
14. We replace x by −x2 in the Maclaurin series of
1
1−x
and multiply the result by x.
2 2 2 2 3
1
x
=
x
1
+
−x
=
x
·
+
−x
+
·
·
·
+
−x
1 + x2
1 − (−x2 )
= x − x3 + x5 − x7 + · · · =
∞
(−1)n x2n+1
n=0
15. We replace x by −x in the Maclaurin series of ln (1 + x).
∞
ln (1 − x) = −x −
−1
x2 x3
(−x)2 (−x)3
+
+ · · · = −x −
−
− ··· =
xn
2
3
2
3
n
n=1
16. We replace x by x2 in the Maclaurin series of sin x.
∞
∞
(−1)n 2 2n+1 (−1)n
x
x4n+2
=
sin x =
(2n + 1)!
(2n + 1)!
2
n=0
n=0
17. By periodicity sin x = sin [(x − 2π) + 2π] = sin (x − 2π), so
sin x = sin (x − 2π) =
∞
(−1)n
(x − 2π)2n+1
(2n + 1)!
n=0
341
342
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
18. We first note that
x−2
ln x = ln 2 1 +
2
x−2
= ln 2 + ln 1 +
.
2
x−2
.
2
x−2 1 x−2 2 1 x−2 3 1 x−2 4
ln x = ln 2 +
−
+
−
+ ···
2
2
2
3
2
4
2
Next use the Maclaurin series of ln (1 + x) with x replaces by
= ln 2 +
∞
(−1)n+1
n2n
n=1
19. sin x cos x =
x3
x5
x7
x2 x4
x6
+
−
+ ···
1−
+
−
+ ···
x−
6
120 5040
2
24 720
=x−
20. e−x cos x =
21. sec x =
(x − 2)n
2x3 2x5 4x7
+
−
+ ···
3
15
315
1−x+
1
=
cos x
x2 x3 x4
−
+
− ···
2
6
24
1
1−
=1+
x2 x4
+
− ···
2
24
=1−x+
x3 x4
−
+ ···
3
6
x2 5x4 61x6
+
+
+ ···
2!
4!
6!
x2 x4 x6
+
−
+ ···
2
4!
6!
Since cos (π/2) = cos (−π/2) = 0, the series converges on (−π/2, π/2).
1−
x5
x7
x3
+
−
+ ···
2 5
17 7
1
sin x
6
120 5040
=
x +
x + ···
= x + x3 +
22. tan x =
2
4
6
cos x
3
15
315
x
x
x
+
−
+ ···
1−
2
24 720
Since cos (π/2) = cos (−π/2) = 0, the series converges on (−π/2, π/2).
x−
23. Let k = n + 2 so that n = k − 2 and
∞
ncn xn+2 =
n=1
∞
(k − 2)ck−2 xk .
k=3
24. Let k = n − 3 so that n = k + 3 and
∞
(2n − 1)cn xn−3 =
n=3
∞
(2k + 5)ck+3 xk .
k=0
25. In the first summation let k = n − 1 so that n = k + 1 and
∞
n=1
ncn xn−1 −
∞
n=0
cn xn =
∞
k=0
(k + 1) ck+1 xk −
∞
k=0
ck xk =
∞
n=0
[(k + 1)ck+1 − ck ] xk .
6.1
Review of Power Series
26. In the first summation let k = n − 1 and in the second summation let k = n + 2. Then
n = k + 1 in the first summation, n = k − 2 in the second summation, and
∞
ncn xn−1 + 3
n=1
∞
cn xn+2 =
n=0
∞
(k + 1)ck+1 xk + 3
n=0
∞
ck−2 xk
k=2
= c2 + 2c2 x +
∞
k
(k + 1)ck+1 x + 3
k=2
= c1 + 2c2 x +
∞
∞
ck−2 xk
k=2
[(k + 1)ck+1 + 3ck−2 ] xk .
k=2
27. In the first summation let k = n − 1 and in the second summation let k = n + 1. Then
n = k + 1 in the first summation, n = k − 1 in the second summation, and
∞
2ncn xn−1 +
n=1
∞
6cn xn+1 = 2 · 1 · c1 x0 +
n=0
∞
2ncn xn−1 +
n=2
= 2c1 +
∞
= 2c1 +
6cn xn+1
n=0
2(k + 1)ck+1 xk +
k=1
∞
∞
∞
6ck−1 xk
k=1
[2(k + 1)ck+1 + 6ck−1 ]xk
k=1
28. In the first summation let k = n − 2 and in the second summation let k = n + 2. Then
n = k + 2 in the first summation, n = k − 2 in the second summation, and
∞
n=2
n(n − 1)cn x
n−2
+
∞
n=0
n+2
cn x
=
∞
k
(k + 2)(k + 1)ck+2 x +
k=0
= 2c2 + 6c3 x +
ck−2 xk
k=2
∞
k=2
= 2c2 + 6c3 x +
∞
∞
k=2
(k + 2)(k + 1)ck+2 xk +
∞
ck−2 xk
k=2
[(k + 2)(k + 1)ck+2 + ck−2 ]xk
343
344
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
29. In the first summation let k = n − 2 and in the second and third summations let k = n. Then
n = k + 2 in the first summation, n = k in the second and third summations, and
∞
n(n − 1)cn xn−2 − 2
n=2
∞
ncn xn +
n=1
∞
cn xn
n=0
=
∞
(k + 2)(k + 1)ck+2 x −
k
k=0
∞
k
2kck x +
k=1
= 2c2 + 6c3 x +
∞
∞
ck xk
k=0
(k + 2)(k + 1)ck+2 xk +
k=2
= c0 + 2c2 +
∞
∞
ck−2 xk
k=2
[(k + 2)(k + 1)ck+2 − (2k − 1)ck ]xk = 0
k=1
30. In the first and third summations let k = n and in the second summation let k = n − 2. Then
n = k in the first and third summations, n = k + 2 in the second summation, and
∞
n(n − 1)cn x + 2
n
n=2
∞
n(n − 1)cn x
n−2
+3
n=2
∞
ncn xn
n=1
= 2 · 2 · 1c2 x0 + 2 · 3 · 2c3 x1 + 3 · 1 · c1 x1
+
∞
n(n − 1)cn xn + 2
n=2
∞
n(n − 1)cn xn−2 + 3
n=4
= 4c2 + (3c1 + 12c3 )x +
∞
∞
ncn xn
n=2
k(k − 1)ck x + 2
k
k=2
= 4c2 + (3c1 + 12c3 )x +
∞
∞
k
(k + 2)(k + 1)ck+2 x + 3
k=2
∞
kck xk
k=2
[(k(k − 1) + 3k) ck + 2(k + 2)(k + 1)ck+2 ] xk
k=2
= 4c2 + (3c1 + 12c3 )x +
∞
[k(k + 2)ck + 2(k + 1)(k + 2)ck+2 ] xk
k=2
31. Since y =
∞
(−1)n 2n
n!
n=1
y + 2xy =
∞
(−1)n 2n
n=1
n!
x2n−1 , we have
x2n−1 + 2x
∞
(−1)n
n=0
n!
∞
∞
(−1)n 2n−1 (−1)n 2n+1
=2
+
x
x
(n − 1)!
n!
n=1
n=0
x2n
k=n
k=n+1
∞
∞
∞
(−1)k−1 2k−1
(−1)k
(−1)k 2k−1 (−1)k−1 2k−1
=2
x
x
+
x
+
=2
(k − 1)!
(k − 1)!
(k − 1)!
(k − 1)!
k=1
=2
∞
k=1
k=1
(−1)k
(−1)k
−
x2k−1 = 0
(k − 1)! (k − 1)!
k=1
6.1
32. Since y =
∞
Review of Power Series
(−1)n 2nx2n−1 , we have
n=1
1 + x2 y +2xy
∞
∞
(−1)n 2nx2n−1 + 2x
(−1)n x2n
= 1 + x2
n=1
=
∞
n=0
n
2n−1
(−1) 2nx
∞
+
n=1
=
∞
n 2n−1
(−1) x
+
∞
(−1) 2nx
(−1) x
+
∞
(−1)k x2k−1 +
∞
n=1
= −2x +
k
k=n+1
(−1)k−1 (2k − 2)x2k−1 +
∞
2(−1)k−1 x2k−1
k=1
2k−1
(−1) 2kx
2(−1) x
k=2
∞
(−1)n x2n
n 2n+1
k=n+1
k=1
∞
+
k=2
=
+ 2x
∞
n=0
n 2n+1
n=1
k=n
∞
2n+1
n=1
n=1
=
n
k−1
(−1)
(2k − 2)x
2k−1
+ 2x +
k=2
∞
2(−1)k−1 x2k−1
k=2
∞
∞
(−1)k 2k + (−1)k−1 2k x2k−1 =
(−1)k 2k − (−1)k 2k x2k−1 = 0
k=2
k=2
33. In this problem we must take special care with starting values for the indices of summation.
Normally when a power series is given in summation notation, successive derivatives of the
power series of the unknown function start with an index that is one higher than the preceding
one. In this case, the power series starts with n = 1 to avoid division by zero. From the first
derivative on this is no longer necessary and the index of summation starts again with n = 1
for y . To justify this to yourself you could simply write out the first few term of the power
series for y.
Since y =
∞
(−1)n+1 xn−1
and
∞
y =
n=1
(−1)n+1 (n − 1)xn−2
n=2
(x + 1)y + y = (x + 1)
∞
(−1)n+1 (n − 1)xn−2 +
n=2
=
∞
n+1
(−1)
(n − 1)x
n−1
+
n=2
= −x0 + x0 +
∞
(−1)n+1 xn−1
n=1
∞
n+1
(−1)
(n − 1)x
n−2
+
n=2
∞
n=2
(−1)n+1 xn−1
n=1
(−1)n+1 (n − 1)xn−1 +
k=n−1
∞
∞
n=3
(−1)n+1 (n − 1)xn−2 +
k=n−2
∞
n=2
(−1)n+1 xn−1
k=n−1
345
346
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
=
∞
(−1)k+2 kxk +
k=1
=
∞
∞
∞
(−1)k+3 (k + 1)xk +
k=1
(−1)k+2 xk
k=1
(−1)k+2 k − (−1)k+2 k − (−1)k+2 + (−1)k+2 xk = 0
k=1
34. SInce y =
∞
(−1)n 2n
n=1
22n (n!)2
xy + y + xy =
x2n−1
and
y =
∞
(−1)n 2n(2n − 1)
22n (n!)2
n=1
∞
(−1)n 2n(2n − 1)
22n (n!)2
n=1
2n−1
x
+
∞
(−1)n 2n
n=1
22n (n!)2
k=n
=
x2n−2
∞
(−1)n 2n+1
+
x
22n (n!)2
n=0
2n−1
x
k=n
k=n+1
∞
(−1)k 2k(2k − 1) (−1)k 2k
(−1)k−1
x2k−1
+
+
22k (k!)2
22k (k!)2
22k−2 [(k − 1)!]2
k=1
∞
(−1)k (2k)2
(−1)k
x2k−1
=
−
22k (k!)2
22k−2 [(k − 1)!]2
k=1
=
∞
(−1)k
k=1
(2k)2 − 22 k 2 2k−1
x
=0
22k (k!)2
In Problems 3538 we start with the assumption that y =
∞
cn xn , substitute into the differential
n=0
equation, and finally find some values of cn . The solution is then written in terms of elementary
functions. (One of the points of power series solutions of differential equations however is that it
wont always be possible to express the power series in terms of elementary functions.)
35. Substituting into the differential equation we have
y − 5y =
∞
n=1
=
∞
k=0
n−1
ncn x
−
∞
n=0
n
5cn x =
∞
k=0
[(k + 1)ck+1 − 5ck ] xk = 0.
(k + 1)ck+1 x −
k
∞
k=0
5ck xk
6.1
Thus ck+1 =
Review of Power Series
5
ck , for k = 01, 1, 2, . . ., and
k+1
c1 =
5
c0 = 5c0
1
c2 =
5
52
c1 =
c0
2
2
c3 =
5
53
c2 =
c0
3
3·2
c4 =
5
54
c3 =
c0
4
4·3·2
..
.
Hence,
y = c0 + 5c0 x +
52
53
54
c0 x2 +
c0 x3 +
c0 x4 + · · ·
2
3·2
4·3·2
and
y = c0
∞
1
(5x)k = c0 e5x .
k!
k=0
36. Substituting into the differential equation we have
4y + y = 4
∞
ncn xn−1 +
n=1
=
∞
∞
cn xn = 4
n=0
∞
(k + 1)ck+1 xk +
k=0
∞
ck xk
k=0
[4(k + 1)ck+1 + ck ] xk = 0.
k=0
Thus ck+1 = −
1
ck , for k = 0, 1, 2, . . ., and
4(k + 1)
c1 = −
1
c0
4·1
c2 = −
1
1
c1 = 2
c0
4·2
4 ·1·2
c3 = −
1
1
c2 = − 3
c0
4·3
4 ·1·2·3
c4 = −
1
1
c3 = 4
c0
4·4
4 ·1·2·3·4
..
.
Hence,
y = c0 −
1
1
1
1
c0 x + 2
c0 x2 − 3
c0 x3 + 4
c0 x4 − · · ·
4·1
4 ·1·2
4 ·1·2·3
4 ·1·2·3·4
347
348
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
and
y = c0
∞
(−1)k
4k k!
k=0
xk = c0
∞
(−1)k
x
4
k!
k=0
k
= c0 e−x/4 .
37. Substituting into the differential equation we have
y − xy =
∞
n−1
ncn x
n=1
= c1 +
−
∞
n+1
cn x
n=0
∞
∞
(k + 1)ck+1 x −
k
k=0
(k + 1)ck+1 xk −
k=1
= c1 +
=
∞
∞
∞
ck−1 xk
k=1
ck+1 xk
k=1
[(k + 1)ck+1 − ck−1 ] xk = 0.
k=1
Thus c1 = 0 and ck+1 = −
1
ck , for k = 0, 1, 2, . . ., and
4(k + 1)
c2 =
1
c0
2
c3 =
1
c1 = 0
3
1
1
c4 = c2 =
4
4
c5 =
1
c3 = 0
5
1
1
c6 = c4 =
6
6
c7 =
1
c5 = 0
7
1
1
c8 = c6 =
8
8
..
.
1
c0
2
=
1
c0
2
2 2!
1
c0
3
2 3!
1
c0
22 2!
1
=
c0
23 3!
=
c0
24 4!
1
Hence,
1
1
1
1
c0 x2 + 2 c0 x4 + 3 c0 x6 + 4 c0 x8 + · · ·
2
2 2!
2 3!
2 4!
2
2 2
2 3
4
x
1 x
1 x
1 x2
+
+
+
+ ··· ,
= c0 1 +
2
2! 2
3! 2
4! 2
y = c0 +
and
k
∞
1 x2
2
= c0 ex /2 .
y = c0
k! 2
k=0
6.1
Review of Power Series
38. Substituting into the differential equatoin we have
(1 + x)y + y =
=
∞
ncn xn−1 +
∞
n=1
n=1
∞
k
cn nxn +
(k + 1)ck+1 x +
k=0
∞
n=0
∞
k
ck kx +
k=1
∞
= c1 + c0 +
= c1 + c0 +
∞
ck xk
k=0
(k + 1)ck+1 xk +
k=1
∞
cn xn
∞
ck kxk +
k=1
∞
ck xk
k=1
[(k + 1)ck+1 + (k + 1)ck ] xk = 0
k=1
Thus c1 + c0 = 0 and ck+1 = −ck , for k = 1, 2, 3 . . ., so
c1 = −c0
c2 = −c1 = c0
c3 = −c2 = −c0
c4 = −c3 = c0
..
.
Hence,
y = c0 − c0 x + c0 x2 − c0 x3 + c0 x4 − · · · = c0 1 − x + x2 − x3 + x4 − · · · ,
and
y = c0
∞
(−1)l xl =
k=0
c0
.
1+x
39. From the double-angle formula
sin 2x = 2 sin x cos x
and
sin x cos x =
1
sin 2x.
2
Therefore we replace x by 2x in the Maclaurin series for sin x. This gives
∞
∞
n=0
n=0
(−4)n
1
1
(−1)n
x2n+1
sin x cos x = sin 2x =
=
2
2
(2n + 1)!(2x)2n+1
(2n + 1)!
=x−
2 3
2 5
x +
x − ··· .
3
15
40. Even though the interval of convergence of cos x is (−∞, ∞), the power series for sec x cannot
converge on that interval because sec x = 1/ cos x is discontinuous at odd integer multiples of
π/2. Since the series is centered at 0 it makes sense, and can be proved, that the interval of
convergence is the open interval (−π/2, π/2).
349
350
CHAPTER 6
6.2
SERIES SOLUTIONS OF LINEAR EQUATIONS
Solutions About Ordinary Points
1. The singular points of (x2 − 25)y + 2xy + y = 0 are −5 and 5. The distance from 0 to either
of these points is 5. The distance from 1 to the closest of these points is 4.
2. The singular points of (x2 − 2x + 10)y + xy − 4y = 0 are 1 + 3i and 1 − 3i. The distance
√
from 0 to either of these points is 10 . The distance from 1 to either of these points is 3.
In Problems 3-6 we use
y=
∞
cn xn ,
y=
n=0
∞
ncn xn−1 ,
and
y =
n=1
∞
n(n − 1)cn xn−2 .
n=2
3. We have
y + y =
∞
n=2
n(n − 1)cn xn−2 +
∞
n=0
k=n−2
=
∞
cn xn =
∞
(k + 2)(k + 1)ck+2 xk +
k=0
∞
k=0
k=n
[(k + 2)(k + 1)ck+2 + ck ] xk = 0.
k=0
Thus ck+2 = −
ck
, for k = 0, 1, 2, . . ., and for k = 0, 2, 4, 6, . . ., we get
(k + 2)(k + 1)
c2 = −
c0
2!
c0
4!
c0
c6 = −
6!
..
.
c4 =
For k = 1, 3, 5, 7, . . . we get
c3 = −
c1
3!
c1
5!
c1
c7 = −
7!
..
.
c5 =
Hence,
y1 (x) = c0 1 −
1 2
1
1
x + x4 − x6 + · · ·
2!
4!
6!
ck xk
6.2
Solutions About Ordinary Points
and
y2 (x) = c1 x −
1
1
1 3
x + x5 − x7 + · · · .
3!
5!
7!
The solution y1 (x) is recognized as y1 (x) = c0 cos x, and the solution y2 (x) is recognized as
y2 (x) = c1 sin x.
4. We have
y −y =
∞
n=2
n(n − 1)cn x
n−2
−
∞
∞
cn x =
n=0
k=n−2
=
n
∞
(k + 2)(k + 1)ck+2 x −
k=0
k
∞
ck xk
k=0
k=n
[(k + 2)(k + 1)ck+2 − ck ] xk = 0.
k=0
Thus ck+2 =
ck
, for k = 0, 1, 2, . . ., and for k = 0, 2, 4, 6, . . ., we get
(k + 2)(k + 1)
c0
2!
c0
c4 =
4!
c0
c6 =
6!
..
.
c2 =
For k = 1, 3, 5, 7, . . . we get
c1
3!
c1
c5 =
5!
c1
c7 =
7!
..
.
c3 =
Hence,
y1 (x) = c0 1 +
1
1
1 2
x + x4 + x6 + · · ·
2!
4!
6!
y2 (x) = c1 x +
1 3
1
1
x + x5 + x7 + · · · .
3!
5!
7!
and
The solution y1 (x) is recognized as y1 (x) = c0 cosh x, and the solution y2 (x) is recognized as
y2 (x) = c1 sinh x.
351
352
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
5. We have
y − y =
∞
n(n − 1)cn xn−2 −
n=2
∞
n=0
k=n−2
=
∞
ncn xn−1 =
∞
(k + 2)(k + 1)ck+2 xk −
k=0
∞
(k + 1)ck+1 xk
k=0
k=n−1
[(k + 2)(k + 1)ck+2 − (k + 1)ck+1 ] xk = 0.
k=0
Thus ck+2 =
ck+1
(k + 1)ck+1
=
, for k = 0, 1, 2, . . ., so
(k + 2)(k + 1)
k+2
c1
2!
c1
c2
c3 =
=
3
3!
c3
c1
c7 =
=
4
4!
..
.
c2 =
Hence, the solution of the differential equation is
y(x) = y1 (x) + y2 (x) = c0 + c1 x +
= c0 + c1 −1 + 1 + x +
= c 0 − c1 + c1
1 2
1
1
x + x3 + x4 + · · ·
2!
3!
4!
1 2
1
1
x + x3 + x4 + · · ·
2!
3!
4!
∞
1 k
x .
k!
k=0
The solutions y1 (x) and y2 (x) are recognized as
y1 (x) = c0
y2 (x) = −c1 + c1 ex
and
6. We have
y + 2y =
∞
n=2
n(n − 1)cn x
n−2
k=n−2
=
∞
k=0
+2
∞
n=0
n−1
ncn x
=
∞
k
(k + 2)(k + 1)ck+2 x +
k=0
k=n−1
[(k + 2)(k + 1)ck+2 + 2(k + 1)ck+1 ] xk = 0.
∞
k=0
2(k + 1)ck+1 xk
6.2
Thus ck+2 = −
Solutions About Ordinary Points
2ck+1
2(k + 1)ck+1
=−
, for k = 0, 1, 2, . . ., so
(k + 2)(k + 1)
k+2
c2 = −
2c1
2!
22 c2
22 c1
=
3
3!
c3 =
c7 = −
23 c1
23 c3
=−
4
4!
..
.
Hence, the solution of the differential equation is
y(x) = y1 (x) + y2 (x) = c0 + c1 x −
2 2 22 3 23 4
x +
x −
x + ···
2!
3!
4!
= c0 +
22 2 23 3 24 4
1
c1 2x −
x +
x −
x + ···
2
2!
3!
4!
= c0 +
22 2 23 3 24 4
1
c1 1 − 1 + 2x −
x +
x −
x + ···
2
2!
3!
4!
∞
1
1 1
(2x)k .
= c0 + c1 − c1
2
2
k!
k=0
The solutions y1 (x) and y2 (x) are recognized as
y1 (x) = c0
7. Substituting y =
∞
and
y2 (x) =
1
1
c1 − c1 e−2x
2
2
cn xn into the differential equation we have
n=0
y + xy =
∞
n(n − 1)cn x
n=2
n−2
k=n−2
= 2c2 +
∞
+
∞
n=0
n+1
cn x
=
∞
(k + 2)(k + 1)ck+2 x +
k=0
k=n+1
[(k + 2)(k + 1)ck+2 + ck−1 ]xk = 0.
k=1
Thus
c2 = 0
(k + 2)(k + 1)ck+2 + ck−1 = 0
and
ck+2 = −
ck−1
,
(k + 2)(k + 1)
k
k = 1, 2, 3, . . . .
∞
k=1
ck−1 xk
353
354
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
Choosing c0 = 1 and c1 = 0 we find
c3 = −
1
6
c4 = c5 = 0
c6 =
1
180
and so on. For c0 = 0 and c1 = 1 we obtain
c4 = −
c3 = 0
1
12
c5 = c6 = 0
c7 =
1
504
and so on. Thus, two solutions are
1
1 6
x − ···
y1 = 1 − x3 +
6
180
8. Substituting y =
∞
y2 = x −
and
1 4
1 7
x +
x − ··· .
12
504
cn xn into the differential equation we have
n=0
2
y +x y =
∞
n(n − 1)cn x
n−2
n=2
+
∞
n=0
k=n−2
= 2c2 + 6c3 x +
n+2
cn x
=
∞
∞
k
(k + 2)(k + 1)ck+2 x +
k=0
ck−2 xk
k=2
k=n+2
∞
[(k + 2)(k + 1)ck+2 + ck−2 ]xk = 0.
k=2
Thus
c 2 = c3 = 0
(k + 2)(k + 1)ck+2 + ck−2 = 0
and
ck+2 = −
1
ck−2 ,
(k + 2)(k + 1)
k = 2, 3, 4, . . . .
Choosing c0 = 1 and c1 = 0 we find
c4 = −
1
12
c 5 = c6 = c7 = 0
c8 =
1
672
and so on. For c0 = 0 and c1 = 1 we obtain
c4 = 0
c5 = −
1
20
c6 = c 7 = c 8 = 0
c9 =
1
1440
and so on. Thus, two solutions are
y1 = 1 −
1 4
1 8
x +
x − ···
12
672
and
y2 = x −
1 5
1 9
x +
x − ··· .
20
1440
6.2
9. Substituting y =
∞
Solutions About Ordinary Points
cn xn into the differential equation we have
n=0
y − 2xy + y =
∞
n(n − 1)cn xn−2 −2
n=2
∞
n=1
k=n−2
=
∞
ncn xn +
∞
n=0
k=n
(k + 2)(k + 1)ck+2 x − 2
k
k=0
cn xn
k=n
∞
k
kck x +
k=1
= 2c2 + c0 +
∞
∞
ck xk
k=0
[(k + 2)(k + 1)ck+2 − (2k − 1)ck ]xk = 0.
k=1
Thus
2c2 + c0 = 0
(k + 2)(k + 1)ck+2 − (2k − 1)ck = 0
and
1
c 2 = − c0
2
ck+2 =
2k − 1
ck ,
(k + 2)(k + 1)
k = 1, 2, 3, . . . .
Choosing c0 = 1 and c1 = 0 we find
1
c3 = c 5 = c 7 = · · · = 0
2
and so on. For c0 = 0 and c1 = 1 we obtain
1
c3 =
c 2 = c4 = c6 = · · · = 0
6
and so on. Thus, two solutions are
c2 = −
1
1
7 6
x − ···
y1 = 1 − x2 − x4 −
2
8
240
10. Substituting y =
∞
c4 = −
c5 =
1
8
c6 = −
1
24
7
240
c7 = 1112
1
1
1 7
x + ··· .
y2 = x + x3 + x5 +
6
24
112
and
cn xn into the differential equation we have
n=0
y − xy + 2y =
∞
n=2
n(n − 1)cn xn−2 −
∞
n=1
k=n−2
=
∞
ncn xn +2
(k + 2)(k + 1)ck+2 x −
= 2c2 + 2c0 +
∞
k=1
∞
k=1
cn xn
n=0
k=n
k
k=0
∞
k=n
k
kck x + 2
∞
ck xk
k=0
[(k + 2)(k + 1)ck+2 − (k − 2)ck ]xk = 0.
355
356
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
Thus
2c2 + 2c0 = 0
(k + 2)(k + 1)ck+2 − (k − 2)ck = 0
and
c2 = −c0
ck+2 =
k−2
ck ,
(k + 2)(k + 1)
k = 1, 2, 3, . . . .
Choosing c0 = 1 and c1 = 0 we find
c2 = −1
c3 = c 5 = c 7 = · · · = 0
c6 = c8 = c10 = · · · = 0.
c4 = 0
For c0 = 0 and c1 = 1 we obtain
c 2 = c4 = c6 = · · · = 0
c3 = −
1
6
c5 = −
1
120
and so on. Thus, two solutions are
y1 = 1 − x2
11. Substituting y =
∞
1
1 5
x − ··· .
y2 = x − x3 −
6
120
and
cn xn into the differential equation we have
n=0
2 y + x y + xy =
∞
n(n − 1)cn x
n−2
n=2
+
∞
=
ncn x
n=1
k=n−2
n+1
+
k=n+1
∞
n=0
cn xn+1
k=n+1
∞
∞
∞
(k + 2)(k + 1)ck+2 xk +
(k − 1)ck−1 xk +
ck−1 xk
k=0
k=2
= 2c2 + (6c3 + c0 )x +
∞
[(k + 2)(k + 1)ck+2 + kck−1 ]xk = 0.
k=2
Thus
c2 = 0
6c3 + c0 = 0
(k + 2)(k + 1)ck+2 + kck−1 = 0
and
c2 = 0
1
c 3 = − c0
6
ck+2 = −
k=1
k
ck−1 ,
(k + 2)(k + 1)
k = 2, 3, 4, . . . .
6.2
Solutions About Ordinary Points
Choosing c0 = 1 and c1 = 0 we find
c3 = −
1
6
c4 = c5 = 0
c6 =
1
45
and so on. For c0 = 0 and c1 = 1 we obtain
c4 = −
c3 = 0
1
6
c5 = c6 = 0
c7 =
5
252
and so on. Thus, two solutions are
1
1
y1 = 1 − x3 + x6 − · · ·
6
45
12. Substituting y =
∞
1
5 7
x − ··· .
y2 = x − x4 +
6
252
and
cn xn into the differential equation we have
n=0
y + 2xy + 2y =
∞
n(n − 1)cn xn−2 +2
n=2
n=1
k=n−2
=
∞
∞
ncn xn +2
∞
n=0
k=n
(k + 2)(k + 1)ck+2 xk + 2
k=0
k=n
∞
kck xk + 2
k=1
= 2c2 + 2c0 +
∞
cn xn
∞
ck xk
k=0
[(k + 2)(k + 1)ck+2 + 2(k + 1)ck ]xk = 0.
k=1
Thus
2c2 + 2c0 = 0
(k + 2)(k + 1)ck+2 + 2(k + 1)ck = 0
and
c2 = −c0
ck+2 = −
2
ck ,
k+2
k = 1, 2, 3, . . . .
Choosing c0 = 1 and c1 = 0 we find
c2 = −1
c3 = c 5 = c 7 = · · · = 0
c4 =
1
2
c6 = −
1
6
and so on. For c0 = 0 and c1 = 1 we obtain
c2 = c4 = c6 = · · · = 0
c3 = −
2
3
c5 = 415
c7 = −8105
and so on. Thus, two solutions are
1
1
y1 = 1 − x2 + x4 − x6 + · · ·
2
6
2
4
8 7
x + ··· .
and y2 = x − x3 + x5 −
3
15
105
357
358
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
13. Substituting y =
∞
cn xn into the differential equation we have
n=0
(x − 1)y + y =
∞
n(n − 1)cn xn−1 −
n=2
∞
n(n − 1)cn xn−2 +
n=2
k=n−1
=
∞
∞
n=1
k=n−2
(k + 1)kck+1 xk −
k=1
∞
ncn xn−1
k=n−1
(k + 2)(k + 1)ck+2 xk +
k=0
∞
(k + 1)ck+1 xk
k=0
∞
= −2c2 + c1 +
[(k + 1)kck+1 − (k + 2)(k + 1)ck+2 + (k + 1)ck+1 ]xk = 0.
k=1
Thus
−2c2 + c1 = 0
(k + 1) ck+1 − (k + 2)(k + 1)ck+2 = 0
2
and
1
c2 = c1
2
ck+2 =
k+1
ck+1 ,
k+2
k = 1, 2, 3, . . . .
Choosing c0 = 1 and c1 = 0 we find c2 = c3 = c4 = · · · = 0. For c0 = 0 and c1 = 1 we obtain
1
c2 = ,
2
1
c3 = ,
3
1
c4 = ,
4
and so on. Thus, two solutions are
y1 = 1
14. Substituting y =
∞
and
1
1
1
y2 = x + x2 + x3 + x4 + · · · .
2
3
4
cn xn into the differential equation we have
n=0
(x + 2)y + xy − y =
∞
n=2
n(n − 1)cn xn−1 +
∞
n=2
k=n−1
=
∞
(k + 1)kck+1 x +
= 4c2 − c0 +
∞
k=0
∞
k=1
∞
n=1
k=n−2
k
k=1
2n(n − 1)cn xn−2 +
ncn xn −
∞
n=0
k=n
k
2(k + 2)(k + 1)ck+2 x +
∞
k=1
cn xn
k=n
kck x −
k
∞
ck xk
k=0
[(k + 1)kck+1 + 2(k + 2)(k + 1)ck+2 + (k − 1)ck ] xk = 0.
6.2
Solutions About Ordinary Points
Thus
4c2 − c0 = 0
(k + 1)kck+1 + 2(k + 2)(k + 1)ck+2 + (k − 1)ck = 0,
k = 1, 2, 3, . . .
and
1
c 2 = c0
4
(k + 1)kck+1 + (k − 1)ck
ck+2 = −
,
2(k + 2)(k + 1)
k = 1, 2, 3, . . . .
Choosing c0 = 1 and c1 = 0 we find
1
c2 = ,
4
c1 = 0,
c3 = −
1
,
24
c4 = 0,
c5 =
1
480
and so on. For c0 = 0 and c1 = 1 we obtain
c2 = 0
c4 = c5 = c6 = · · · = 0.
c3 = 0
Thus, two solutions are
1
1
1 5
x + ···
y1 = c0 1 + x2 − x3 +
4
24
480
15. Substituting y =
∞
and
y2 = c1 x.
cn xn into the differential equation we have
n=0
y − (x + 1)y − y =
∞
n=2
n(n − 1)cn x
n−2
k=n−2
=
∞
−
∞
n=1
ncn x −
n
k=n
(k + 2)(k + 1)ck+2 xk −
k=0
∞
∞
n−1
ncn x
n=1
−
k=n−1
kck xk −
k=1
∞
∞
cn xn
n=0
k=n
(k + 1)ck+1 xk −
k=0
∞
ck xk
k=0
∞
= 2c2 − c1 − c0 +
[(k + 2)(k + 1)ck+2 − (k + 1)ck+1 − (k + 1)ck ]xk = 0.
k=1
Thus
2c2 − c1 − c0 = 0
(k + 2)(k + 1)ck+2 − (k + 1)(ck+1 + ck ) = 0
and
c1 + c0
2
ck+1 + ck
,
=
k+2
c2 =
ck+2
k = 1, 2, 3, . . . .
359
360
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
Choosing c0 = 1 and c1 = 0 we find
1
c2 = ,
2
1
c3 = ,
6
1
c4 = ,
6
and so on. For c0 = 0 and c1 = 1 we obtain
1
c2 = ,
2
1
c3 = ,
2
1
c4 = ,
4
and so on. Thus, two solutions are
1
1
1
y1 = 1 + x2 + x3 + x4 + · · ·
2
6
6
16. Substituting y =
∞
and
1
1
1
y2 = x + x2 + x3 + x4 + · · · .
2
2
4
cn xn into the differential equation we have
n=0
∞
∞
∞
x2 + 1 y − 6y =
n(n − 1)cn xn +
n(n − 1)cn xn−2 −6
cn xn
n=2
n=2
k=n
=
∞
k=n−2
n=0
k=n
∞
∞
k
k(k − 1)ck x +
(k + 2)(k + 1)ck+2 x − 6
ck xk
k
k=2
k=0
k=0
= 2c2 − 6c0 + (6c3 − 6c1 )x +
∞
2
k − k − 6 ck + (k + 2)(k + 1)ck+2 xk = 0.
k=2
Thus
2c2 − 6c0 = 0
6c3 − 6c1 = 0
(k − 3)(k + 2)ck + (k + 2)(k + 1)ck+2 = 0
and
c2 = 3c0
ck+2 = −
c3 = c1
k−3
ck ,
k+1
k = 2, 3, 4, . . . .
Choosing c0 = 1 and c1 = 0 we find
c3 = c5 = c 7 = · · · = 0
c2 = 3
c4 = 1
c6 = −
1
5
and so on. For c0 = 0 and c1 = 1 we obtain
c 2 = c4 = c6 = · · · = 0
c5 = c7 = c9 = · · · = 0.
c3 = 1
Thus, two solutions are
1
y1 = 1 + 3x2 + x4 − x6 + · · ·
5
and
y2 = x + x3 .
6.2
17. Substituting y =
∞
Solutions About Ordinary Points
cn xn into the differential equation we have
n=0
∞
∞
∞
∞
n
n−2
n
n(n − 1)cn x +2
n(n − 1)cn x
+3
ncn x −
cn xn
x + 2 y + 3xy − y =
2
n=2
n=2
k=n
=
∞
k(k − 1)ck xk + 2
k=2
n=1
k=n−2
n=0
k=n
k=n
∞
∞
∞
(k + 2)(k + 1)ck+2 xk + 3
kck xk −
ck xk
k=0
k=1
k=0
∞
= (4c2 − c0 ) + (12c3 + 2c1 )x +
2(k + 2)(k + 1)ck+2 + k 2 + 2k − 1 ck xk = 0.
k=2
Thus
4c2 − c0 = 0
12c3 + 2c1 = 0
2(k + 2)(k + 1)ck+2 + k + 2k − 1 ck = 0
2
and
1
c 2 = c0
4
1
c3 = − c1
6
ck+2 = −
k 2 + 2k − 1
ck ,
2(k + 2)(k + 1)
k = 2, 3, 4, . . . .
Choosing c0 = 1 and c1 = 0 we find
1
c3 = c5 = c7 = · · · = 0
4
and so on. For c0 = 0 and c1 = 1 we obtain
c4 = −
c2 =
c 2 = c4 = c6 = · · · = 0
c3 = −
1
6
c5 =
7
96
7
120
and so on. Thus, two solutions are
1
7
y1 = 1 + x2 − x4 + · · ·
4
96
18. Substituting y =
∞
and
1
7 5
x − ··· .
y2 = x − x3 +
6
120
cn xn into the differential equation we have
n=0
∞
∞
∞
∞
n(n − 1)cn xn −
n(n − 1)cn xn−2 +
ncn xn −
cn xn
x2 − 1 y + xy − y =
n=2
n=2
k=n
=
∞
k=2
n=1
k=n−2
k(k − 1)ck x −
k
∞
n=0
k=n
k
(k + 2)(k + 1)ck+2 x +
k=0
= (−2c2 − c0 ) − 6c3 x +
∞
k=1
k=n
kck x −
k
∞
ck xk
k=0
∞
−(k + 2)(k + 1)ck+2 + k 2 − 1 ck xk = 0.
k=2
361
362
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
Thus
−2c2 − c0 = 0
−6c3 = 0
−(k + 2)(k + 1)ck+2 + (k − 1)(k + 1)ck = 0
and
1
c 2 = − c0
2
c3 = 0
ck+2 =
k−1
ck ,
k+2
k = 2, 3, 4, . . . .
Choosing c0 = 1 and c1 = 0 we find
c2 = −
1
2
c3 = c 5 = c7 = · · · = 0
c4 = −
1
8
and so on. For c0 = 0 and c1 = 1 we obtain
c 2 = c4 = c6 = · · · = 0
c3 = c5 = c7 = · · · = 0.
Thus, two solutions are
1
1
y1 = 1 − x2 − x4 − · · ·
2
8
19. Substituting y =
∞
and
y2 = x.
cn xn into the differential equation we have
n=0
(x − 1)y − xy + y =
∞
n(n − 1)cn xn−1 −
n=2
∞
n=2
k=n−1
=
∞
n(n − 1)cn xn−2 −
n=1
k=n−2
(k + 1)kck+1 x −
k
k=1
∞
∞
ncn xn +
k
∞
k=1
cn xn
n=0
k=n
k
kck x +
∞
ck xk
k=0
[−(k + 2)(k + 1)ck+2 + (k + 1)kck+1 − (k − 1)ck ]xk = 0.
k=1
Thus
−2c2 + c0 = 0
−(k + 2)(k + 1)ck+2 + (k − 1)kck+1 − (k − 1)ck = 0
and
1
c 2 = c0
2
ck+2 =
∞
k=n
(k + 2)(k + 1)ck+2 x −
k=0
= −2c2 + c0 +
∞
(k − 1)ck
kck+1
−
,
k+2
(k + 2)(k + 1)
k = 1, 2, 3, . . . .
6.2
Solutions About Ordinary Points
363
Choosing c0 = 1 and c1 = 0 we find
1
c2 = ,
2
1
c3 = ,
6
c4 = 0,
and so on. For c0 = 0 and c1 = 1 we obtain c2 = c3 = c4 = · · · = 0. Thus,
1 2 1 3
y = C1 1 + x + x + · · · + C2 x
2
6
1 2
y = C1 x + x + · · · + C2 .
2
and
The initial conditions imply C1 = −2 and C2 = 6, so
1 2 1 3
y = −2 1 + x + x + · · · + 6x = 8x − 2ex .
2
6
20. Substituting y =
∞
cn xn into the differential equation we have
n=0
(x + 1)y − (2 − x)y + y
=
∞
n=2
n(n − 1)cn xn−1 +
∞
n=2
k=n−1
=
∞
n(n − 1)cn xn−2 −2
∞
n=1
k=n−2
k
(k + 1)kck+1 x +
k=1
∞
∞
k=n−1
(k + 2)(k + 1)ck+2 x − 2
k
k=0
= 2c2 − 2c1 + c0 +
ncn xn−1 +
∞
∞
n=1
ncn xn +
∞
n=0
k=n
k
(k + 1)ck+1 x +
k=0
cn xn
k=n
∞
kck x +
k=1
[(k + 2)(k + 1)ck+2 + (k − 2)(k + 1)ck+1 + (k + 1)ck ]xk = 0.
k=1
Thus
2c2 − 2c1 + c0 = 0
(k + 2)(k + 1)ck+2 − (k − 2)(k + 1)ck+1 + (k + 1)ck = 0
and
1
c 2 = c1 − c0
2
ck+2 =
k−2
1
ck+1 −
ck ,
k+2
k+2
k = 1, 2, 3, . . . .
Choosing c0 = 1 and c1 = 0 we find
1
c2 = − ,
2
1
c3 = ,
6
1
c4 = ,
8
and so on. For c0 = 0 and c1 = 1 we obtain
c2 = 1,
k
2
c3 = − ,
3
1
c4 = − ,
4
∞
k=0
ck xk
364
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
and so on. Thus,
1 2 1 3 1 4
2 3 1 4
2
y = C1 1 − x + x + x + · · · + C2 x + x − x − x + · · ·
2
6
8
3
4
and
1 2 1 3
y = C1 −x + x + x + · · · + C2 1 + 2x − 2x2 − x3 + · · · .
2
2
The initial conditions imply C1 = 2 and C2 = −1, so
1
1
1
2
1
y = 2 1 − x2 + x3 + x4 + · · · − x + x2 − x3 − x4 + · · ·
2
6
8
3
4
= 2 − x − 2x2 + x3 +
21. Substituting y =
∞
1 4
x + ··· .
2
cn xn into the differential equation we have
n=0
y − 2xy + 8y =
∞
n(n − 1)cn xn−2 −2
n=2
∞
n=1
k=n−2
=
∞
ncn xn + 8
(k + 2)(k + 1)ck+2 x − 2
k=0
∞
k=n
k
kck x + 8
k=1
∞
cn xn
n=0
k=n
k
= 2c2 + 8c0 +
∞
∞
ck xk
k=0
[(k + 2)(k + 1)ck+2 + (8 − 2k)ck ]xk = 0.
k=1
Thus
2c2 + 8c0 = 0
(k + 2)(k + 1)ck+2 + (8 − 2k)ck = 0
and
c2 = −4c0
ck+2 =
2(k − 4)
ck ,
(k + 2)(k + 1)
k = 1, 2, 3, . . . .
Choosing c0 = 1 and c1 = 0 we find
c2 = −4
c3 = c5 = c7 = · · · = 0
c4 =
4
3
c6 = c8 = c10 = · · · = 0.
For c0 = 0 and c1 = 1 we obtain
c2 = c4 = c6 = · · · = 0
c3 = −1
c5 =
1
10
6.2
and so on. Thus,
Solutions About Ordinary Points
4 4
1 5
2
3
y = C1 1 − 4x + x + C2 x − x + x + · · ·
3
10
16 3
1 4
2
y = C1 −8x + x + C2 1 − 3x + x + · · · .
3
2
and
The initial conditions imply C1 = 3 and C2 = 0, so
4 4
2
y = 3 1 − 4x + x = 3 − 12x2 + 4x4 .
3
22. Substituting y =
∞
cn xn into the differential equation we have
n=0
2
(x + 1)y + 2xy =
∞
n=2
n(n − 1)cn x +
n
∞
n(n − 1)cn x
n−2
n=2
k=n
=
∞
k(k − 1)ck xk +
k=2
+
∞
n=1
k=n−2
∞
(k + 2)(k + 1)ck+2 xk +
∞
k=n
k=0
= 2c2 + (6c3 + 2c1 )x +
2ncn xn
∞
2kck xk
k=1
[k(k + 1)ck + (k + 2)(k + 1)ck+2 ] xk = 0.
k=2
Thus
2c2 = 0
6c3 + 2c1 = 0
k(k + 1)ck + (k + 2)(k + 1)ck+2 = 0
and
1
c3 = − c1
3
c2 = 0
ck+2 = −
k
ck ,
k+2
k = 2, 3, 4, . . . .
Choosing c0 = 1 and c1 = 0 we find c3 = c4 = c5 = · · · = 0. For c0 = 0 and c1 = 1 we obtain
c3 = −
and so on. Thus
and
1
3
c4 = c 6 = c 8 = · · · = 0
c5 = −
1
5
1 3 1 5 1 7
y = C0 + C1 x − x + x − x + · · ·
3
5
7
y = c1 1 − x2 + x4 − x6 + · · · .
The initial conditions imply c0 = 0 and c1 = 1, so
1
1
1
y = x − x3 + x5 − x7 + · · · .
3
5
7
c7 =
1
7
365
366
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
23. Substituting y =
∞
cn xn into the differential equation we have
n=0
y + (sin x)y =
∞
n=2
1
1 5
x − ···
n(n − 1)cn xn−2 + x − x3 +
c0 + c1 x + c2 x2 + · · ·
6
120
1
2
3
2
= 2c2 + 6c3 x + 12c4 x + 20c5 x + · · · + c0 x + c1 x + c2 − c0 x3 + · · ·
6
1
= 2c2 + (6c3 + c0 )x + (12c4 + c1 )x2 + 20c5 + c2 − c0 x3 + · · · = 0.
6
Thus
2c2 = 0
6c3 + c0 = 0
c2 = 0
1
c 3 = − c0
6
1
20c5 + c2 − c0 = 0
6
12c4 + c1 = 0
and
c4 = −
1
c1
12
c5 = −
1
1
c2 +
c0 .
20
120
Choosing c0 = 1 and c1 = 0 we find
1
c3 = − ,
6
c2 = 0,
c4 = 0,
c5 =
1
120
and so on. For c0 = 0 and c1 = 1 we obtain
c2 = 0,
c3 = 0,
c4 = −
1
,
12
c5 = 0
and so on. Thus, two solutions are
1
1 5
x + ···
y1 = 1 − x3 +
6
120
24. Substituting y =
∞
and
y2 = x −
1 4
x + ··· .
12
cn xn into the differential equation we have
n=0
y + ex y − y =
∞
n(n − 1)cn xn−2
n=2
∞
1 2 1 3
2
3
+ 1 + x + x + x + ···
cn xn
c1 + 2c2 x + 3c3 x + 4c4 x + · · · −
2
6
n=0
= 2c2 + 6c3 x + 12c4 x2 + 20c5 x3 + · · ·
1
+ c1 + (2c2 + c1 )x + 3c3 + 2c2 + c1 x2 + · · · − [c0 + c1 x + c2 x2 + · · · ]
2
1
= (2c2 + c1 − c0 ) + (6c3 + 2c2 )x + 12c4 + 3c3 + c2 + c1 x2 + · · · = 0.
2
6.2
Solutions About Ordinary Points
Thus
2c2 + c1 − c0 = 0
6c3 + 2c2 = 0
1
12c4 + 3c3 + c2 + c1 = 0
2
and
1
1
c2 = c0 − c1
2
2
1
c3 = − c2
3
1
1
1
c4 = − c3 + c 2 − c 1 .
4
12
24
Choosing c0 = 1 and c1 = 0 we find
1
c2 = ,
2
1
c3 = − ,
6
c4 = 0
and so on. For c0 = 0 and c1 = 1 we obtain
1
c2 = − ,
2
1
c3 = ,
6
c4 = −
1
24
and so on. Thus, two solutions are
1
1
y1 = 1 + x2 − x3 + · · ·
2
6
1
1
1
y2 = x − x2 + x3 − x4 + · · · .
2
6
24
and
25. The singular points of (cos x)y + y + 5y = 0 are odd integer multiples of π/2. The distance
from 0 to either ±π/2 is π/2. The singular point closest to 1 is π/2. The distance from 1 to
the closest singular point is then π/2 − 1.
26. Substituting y =
∞
cn xn into the first differential equation leads to
n=0
y − xy =
∞
n(n − 1)cn xn−2 −
n=2
k=n−2
= 2c2 +
∞
∞
n=0
cn xn+1 =
∞
(k + 2)(k + 1)ck+2 xk −
k=0
k=1
k=n+1
[(k + 2)(k + 1)ck+2 − ck−1 ]xk = 1.
k=1
Thus
2c2 = 1
(k + 2)(k + 1)ck+2 − ck−1 = 0
and
c2 =
1
2
ck+2 =
ck−1
,
(k + 2)(k + 1)
∞
k = 1, 2, 3, . . . .
ck−1 xk
367
368
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
Let c0 and c1 be arbitrary and iterate to find
c2 =
1
2
1
c3 = c 0
6
c4 =
1
c1
12
c5 =
1
1
c2 =
20
40
and so on. The solution is
1
1
1
1
y = c0 + c1 x + x2 + c0 x3 + c1 x4 + c5 + · · ·
2
6
12
40
1 3
1 4
1
1
= c0 1 + x + · · · + c1 x + x + · · · + x2 + x5 + · · · .
6
12
2
40
Substituting y =
∞
cn xn into the second differential equation leads to
n=0
y − 4xy − 4y =
∞
n=2
n(n − 1)cn xn−2 −
∞
n=1
k=n−2
=
∞
4ncn xn −
n=0
k=n
(k + 2)(k + 1)ck+2 x −
k
k=0
∞
∞
4cn xn
4kck x −
k
∞
= ex = 1 +
k=1
[(k + 2)(k + 1)ck+2 − 4(k + 1)ck ] xk
1 k
x .
k!
Thus
2c2 − 4c0 = 1
(k + 2)(k + 1)ck+2 − 4(k + 1)ck =
1
k!
and
1
+ 2c0
2
4
1
+
ck ,
=
(k + 2)! k + 2
c2 =
ck+2
4ck xk
k=0
k=1
∞
k=n
k=1
= 2c2 − 4c0 +
∞
k = 1, 2, 3, . . . .
6.2
Solutions About Ordinary Points
Let c0 and c1 be arbitrary and iterate to find
c2 =
1
+ 2c0
2
c3 =
4
4
1
1
+ c1 = + c1
3! 3
3! 3
c4 =
4
1
1
1
13
+ c2 = + + 2c0 =
+ 2c0
4! 4
4! 2
4!
c5 =
4
4
16
1
1
17 16
+ c3 = +
+ c1 =
+ c1
5! 5
5! 5 · 3! 15
5!
15
c6 =
4
4 · 13 8
1
1
261 4
+ c4 = +
+ c0 =
+ c0
6! 6
6!
6 · 4!
6
6!
3
c7 =
1
4
1
4 · 17
64
409
64
+ c5 = +
+
c1 =
+
c1
7! 7
7!
7 · 5!
105
7!
105
and so on. The solution is
4
1
1
13
17 16
2
3
4
+ 2c0 x +
+ c1 x +
+ 2c0 x +
+ c1 x5
y = c0 + c1 x +
2
3! 3
4!
5!
15
64
261 4
409
+ c0 x6 +
+
c1 x7 + · · ·
+
6!
3
7!
105
4
4
16
64 7
x + ···
= c0 1 + 2x2 + 2x4 + x6 + · · · + c1 x + x3 + x5 +
3
3
15
105
1
13
17
261 6 409 7
1
x +
x + ··· .
+ x2 + x3 + x4 + x5 +
2
3!
4!
5!
6!
7!
27. We identify P (x) = 0 and Q(x) = sin x/x. The Taylor series representation for sin x/x is
1 − x2 /3! + x4 /5! − · · · , for |x| < ∞. Thus, Q(x) is analytic at x = 0 and x = 0 is an ordinary
point of the differential equation.
√
28. Since x is continuous at x = 0, but derivatives of all orders are discontinuous at this point,
x = 0 is a singular point of the differential equation; not an ordinary point.
29. (a) Substituting y =
∞
cn xn into the differential equation we have
n=0
y + xy + y =
∞
n=2
n(n − 1)cn xn−2 +
∞
n=1
k=n−2
=
∞
ncn xn +
k
k=0
∞
k=1
∞
k=1
cn xn
n=0
k=n
(k + 2)(k + 1)ck+2 x +
= (2c2 + c0 ) +
∞
k=n
k
kck x +
∞
ck xk
k=0
[(k + 2)(k + 1)ck+2 + (k + 1)ck ] xk = 0.
369
370
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
Thus
2c2 + c0 = 0
(k + 2)(k + 1)ck+2 + (k + 1)ck = 0
and
1
c 2 = − c0
2
ck+2 = −
1
ck ,
k+2
k = 1, 2, 3, . . . .
Choosing c0 = 1 and c1 = 0 we find
1
2
c 3 = c5 = c7 = · · · = 0
1
1
1
−
= 2
c4 = −
4
2
2 ·2
1
1
1
=− 3
c6 = −
6 22 · 2
2 · 3!
c2 = −
and so on. For c0 = 0 and c1 = 1 we obtain
c 2 = c4 = c6 = · · · = 0
2
1
c3 = − = −
3 3!
4·2
1
1
1
=
c5 = −
−
=
5
3
5·3
5!
6·4·2
1 4·2
=−
c7 = −
7
5!
7!
and so on. Thus, two solutions are
y1 =
∞
(−1)k
k=0
x2k
2k · k!
and
y2 =
∞
(−1)k 2k k!
k=0
x2k+1 .
(2k + 1)!
(b) For y1 , S3 = S2 and S5 = S4 , so we plot S2 , S4 , S6 , S8 , and S10 .
y
y
4
2
–4 –2
N=2
2 4
x
4
2
2
–4 –2
2
–2
–2
–4
–4
4
x
N=6
–4 –2
N=4
y
y
y
4
2 4
x
4
4
2
2
–4 –2
2
–2
–2
–4
–4
4
x
N=8
–4
N = 10
–2
–2
2
4
2
4
x
–4
For y2 , S3 = S4 and S5 = S6 , so we plot S2 , S4 , S6 , S8 , and S10 .
y
y
4
4
2
2
–4 –2
2
–2
–4
4
N=2
x
–4 –2
y
N=4
2 4
x
y
4
4
2
2
–4 –2
2
–2
–2
–4
–4
4
N=6
x
–4 –2
y
4
N=8
2
4
2
x
–4 –2
–2
–2
–4
–4
N = 10
x
6.2
(c)
–4
y1
y2
4
4
2
2
–2
2
x
4
–2
–4
Solutions About Ordinary Points
2
–2
–2
–4
–4
4
371
x
The graphs of y1 and y2 obtained from a numerical solver are shown. We see that the
partial sum representations indicate the even and odd natures of the solution, but don’t
really give a very accurate representation of the true solution. Increasing N to about 20
gives a much more accurate representation on [−4, 4].
x
(d) From e =
∞
−x2 /2
k
x /k! we see that e
k=0
=
∞
2
k
(−x /2) /k! =
k=0
∞
(−1)k x2k /2k k! . From
k=0
(5) of Section 3.2 we have
ˆ
e−
y 2 = y1
=
´
x dx
−x2 /2
dx = e
y12
ˆ
e−x /2
−x2 /2
2 /2 2 dx = e
−x
(e
)
2
ˆ
e−x /2
−x2 /2
2 dx = e
−x
e
2
ˆ
ex
2 /2
dx
∞ ˆ
∞
ˆ ∞
(−1)k
1
1
x2k
x2k dx =
x2k
x2k dx
2k k!
2k k!
2k k!
2k k!
∞
(−1)k
k=0
=
k=0
∞
(−1)k
k=0
2k k!
2k
x
∞
k=0
k=0
1
x2k+1
(2k + 1)2k k!
k=0
1
1
1 3
1
1
1 2
4
6
5
7
x + ···
x +
x +
x + ···
x+
= 1− x + 2 x − 3
2
2 ·2
2 · 3!
3·2
5 · 22 · 2
7 · 23 · 3!
∞
=x−
(−1)k 2k k!
2 3 4·2 5 6·4·2 7
x +
x −
x + ··· =
x2k+1 .
3!
5!
7!
(2k + 1)!
k=0
30. (a) We have
y + (cos x)y = 2c2 + 6c3 x + 12c4 x2 + 20c5 x3 + 30c6 x4 + 42c7 x5 + · · ·
x2 x4 x6
+ 1−
+
−
+ · · · (c0 + c1 x + c2 x2 + c3 x3 + c4 x4 + c5 x5 + · · · )
2!
4!
6!
1
1
2
= (2c2 + c0 ) + (6c3 + c1 )x + 12c4 + c2 − c0 x + 20c5 + c3 − c1 x3
2
2
1
1
1
1
+ 30c6 + c4 + c0 − c2 x4 + 42c7 + c5 + c1 − c3 x5 + · · · .
24
2
24
2
372
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
Then
1
1
1
1
c 0 − c2 = 0
and
42c7 + c5 + c1 − c3 = 0,
24
2
24
2
which gives c6 = −c0 /80 and c7 = −19c1 /5040. Thus
1
1
1
y1 (x) = 1 − x2 + x4 − x6 + · · ·
2
12
80
and
1
1
19 7
x + ··· .
y2 (x) = x − x3 + x5 −
6
30
5040
30c6 + c4 +
(b) From part (a) the general solution of the differential equation is y = c1 y1 + c2 y2 . Then
y(0) = c1 + c2 · 0 = c1 and y (0) = c1 · 0 + c2 = c2 , so the solution of the initial-value
problem is
1
1
1
1
1
19 7
x + ··· .
y = y1 + y2 = 1 + x − x2 − x3 + x4 + x5 − x6 −
2
6
12
30
80
5040
(c)
y
y
4
4
4
2
2
2
–6 –4 –2
2
4
6
x
–6 –4
–2
4
6
x
–6 –4 –2
–2
–2
–4
–4
–4
y
4
4
2
2
2
2
4
6
x
–6 –4 –2
2
4
6
x
–6 –4
–2
–2
–2
–2
–4
–4
–4
(d)
y
6
4
2
–2
2
–2
–4
–6
4
6
x
2
4
6
2
4
6
x
y
4
–6 –4 –2
–4
2
–2
y
–6
y
x
6.3
6.3
Solutions About Singular Points
Solutions About Singular Points
1. Irregular singular point: x = 0
2. Regular singular points: x = 0, −3
3. Irregular singular point: x = 3; regular singular point: x = −3
4. Irregular singular point: x = 1; regular singular point: x = 0
5. Regular singular points: x = 0, ±2i
6. Irregular singular point: x = 5; regular singular point: x = 0
7. Regular singular points: x = −3, 2
8. Regular singular points: x = 0, ±i
9. Irregular singular point: x = 0; regular singular points: x = 2, ±5
10. Irregular singular point: x = −1; regular singular points: x = 0, 3
11. Writing the differential equation in the form
y +
5
x
y +
y=0
x−1
x+1
we see that x0 = 1 and x0 = −1 are regular singular points. For x0 = 1 the differential
equation can be put in the form
(x − 1)2 y + 5(x − 1)y +
x(x − 1)2
y = 0.
x+1
In this case p(x) = 5 and q(x) = x(x − 1)2 /(x + 1). For x0 = −1 the differential equation can
be put in the form
(x + 1)2 y + 5(x + 1)
x+1 y + x(x + 1)y = 0.
x−1
In this case p(x) = (x + 1)/(x − 1) and q(x) = x(x + 1).
12. Writing the differential equation in the form
y +
x+3 y + 7xy = 0
x
we see that x0 = 0 is a regular singular point. Multiplying by x2 , the differential equation
can be put in the form
x2 y + x(x + 3)y + 7x3 y = 0.
We identify p(x) = x + 3 and q(x) = 7x3 .
373
374
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
13. We identify P (x) = 5/3x + 1 and Q(x) = −1/3x2 , so that p(x) = xP (x) =
q(x) = x2 Q(x) = − 13 . Then a0 = 53 , b0 = − 13 , and the indicial equation is
5
3
+ x and
1
1
1
5
2
1
r(r − 1) + r − = r2 + r − = (3r2 + 2r − 1) = (3r − 1)(r + 1) = 0.
3
3
3
3
3
3
The indicial roots are 13 and −1. Since these do not differ by an integer we expect to find two
series solutions using the method of Frobenius.
14. We identify P (x) = 1/x and Q(x) = 10/x, so that p(x) = xP (x) = 1 and q(x) = x2 Q(x) =
10x. Then a0 = 1, b0 = 0, and the indicial equation is
r(r − 1) + r = r2 = 0.
The indicial roots are 0 and 0. Since these are equal, we expect the method of Frobenius to
yield a single series solution.
15. Substituting y =
∞
cn xn+r into the differential equation and collecting terms, we obtain
n=0
∞
[2(k + r − 1)(k + r)ck − (k + r)ck + 2ck−1 ] xk+r−1
2xy − y + 2y = 2r2 − 3r c0 xr−1 +
k=1
= 0,
which implies
2r2 − 3r = r(2r − 3) = 0
and
(k + r)(2k + 2r − 3)ck + 2ck−1 = 0.
The indicial roots are r = 0 and r = 3/2. For r = 0 the recurrence relation is
ck = −
2ck−1
,
k(2k − 3)
k = 1, 2, 3, . . . ,
and
c2 = −2c0 ,
c1 = 2c0 ,
4
c3 = c0 ,
9
and so on. For r = 3/2 the recurrence relation is
ck = −
2ck−1
,
(2k + 3)k
k = 1, 2, 3, . . . ,
and
2
c 1 = − c0 ,
5
c2 =
2
c0 ,
35
c3 = −
4
c0 ,
945
and so on. The general solution on (0, ∞) is
2 2
2
4 3
4 3
2
3/2
x + ··· .
1− x+ x −
y = C1 1 + 2x − 2x + x + · · · + C2 x
9
5
35
945
6.3
∞
16. Substituting y =
Solutions About Singular Points
cn xn+r into the differential equation and collecting terms, we obtain
n=0
2xy + 5y + xy = 2r2 + 3r c0 xr−1 + 2r2 + 7r + 5 c1 xr
∞
+
[2(k + r)(k + r − 1)ck + 5(k + r)ck + ck−2 ]xk+r−1
k=2
= 0,
which implies
2r2 + 3r = r(2r + 3) = 0,
2
2r + 7r + 5 c1 = 0,
and
(k + r)(2k + 2r + 3)ck + ck−2 = 0.
The indicial roots are r = −3/2 and r = 0, so c1 = 0 . For r = −3/2 the recurrence relation
is
ck−2
, k = 2, 3, 4, . . . ,
ck = −
(2k − 3)k
and
1
c 2 = − c0 ,
2
c3 = 0,
c4 =
1
c0 ,
40
c4 =
1
c0 ,
616
and so on. For r = 0 the recurrence relation is
ck = −
ck−2
,
k(2k + 3)
k = 2, 3, 4, . . . ,
and
c2 = −
1
c0 ,
14
c3 = 0,
and so on. The general solution on (0, ∞) is
1 2
1 2
1 4
1 4
−3/2
1 − x + x + · · · + C2 1 − x +
x + ··· .
y = C1 x
2
40
14
616
∞
17. Substituting y =
cn xn+r into the differential equation and collecting terms, we obtain
n=0
1
4xy + y + y =
2
∞
7
1
2
4(k + r)(k + r − 1)ck + (k + r)ck + ck−1 xk+r−1
4r − r c0 xr−1 +
2
2
k=1
= 0,
which implies
7
7
=0
4r − r = r 4r −
2
2
2
375
376
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
and
1
(k + r)(8k + 8r − 7)ck + ck−1 = 0.
2
The indicial roots are r = 0 and r = 7/8. For r = 0 the recurrence relation is
ck = −
2ck−1
,
k(8k − 7)
k = 1, 2, 3, . . . ,
and
2
c2 = c0 ,
9
c1 = −2c0 ,
c3 = −
4
c0 ,
459
and so on. For r = 7/8 the recurrence relation is
ck = −
2ck−1
,
(8k + 7)k
k = 1, 2, 3, . . . ,
and
c1 = −
2
c0 ,
15
c2 = 2345c0 ,
c3 = −
4
c0 ,
32, 085
and so on. The general solution on (0, ∞) is
2 2
2
4 3
2 2
4
7/8
3
1− x+
y = C1 1 − 2x + x −
x + · · · + C2 x
x −
x + ··· .
9
459
15
345
32,085
18. Substituting y =
∞
cn xn+r into the differential equation and collecting terms, we obtain
n=0
2x2 y − xy + x2 + 1 y = 2r2 − 3r + 1 c0 xr + 2r2 + r c1 xr+1
∞
+
[2(k + r)(k + r − 1)ck − (k + r)ck + ck + ck−2 ]xk+r
k=2
= 0,
which implies
2r2 − 3r + 1 = (2r − 1)(r − 1) = 0,
2
2r + r c1 = 0,
and
[(k + r)(2k + 2r − 3) + 1]ck + ck−2 = 0.
The indicial roots are r = 1/2 and r = 1, so c1 = 0. For r = 1/2 the recurrence relation is
ck = −
ck−2
,
k(2k − 1)
k = 2, 3, 4, . . . ,
and
1
c 2 = − c0 ,
6
c3 = 0,
c4 =
1
c0 ,
168
6.3
Solutions About Singular Points
and so on. For r = 1 the recurrence relation is
ck = −
ck−2
,
k(2k + 1)
k = 2, 3, 4, . . . ,
and
c2 = −
1
c0 ,
10
c3 = 0,
c4 =
1
c0 ,
360
and so on. The general solution on (0, ∞) is
1 2
1 4
1 2
1 4
1/2
1− x +
x + · · · + C2 x 1 − x +
x + ··· .
y = C1 x
6
168
10
360
19. Substituting y =
∞
cn xn+r into the differential equation and collecting terms, we obtain
n=0
3xy + (2 − x)y − y = 3r2 − r c0 xr−1
∞
+
[3(k + r − 1)(k + r)ck + 2(k + r)ck − (k + r)ck−1 ]xk+r−1
k=1
= 0,
which implies
3r2 − r = r(3r − 1) = 0
and
(k + r)(3k + 3r − 1)ck − (k + r)ck−1 = 0.
The indicial roots are r = 0 and r = 1/3. For r = 0 the recurrence relation is
ck =
ck−1
,
3k − 1
k = 1, 2, 3, . . . ,
and
1
c 1 = c0 ,
2
c2 =
1
c0 ,
10
c3 =
1
c0 ,
80
and so on. For r = 1/3 the recurrence relation is
ck =
ck−1
,
3k
k = 1, 2, 3, . . . ,
and
1
c1 = c 0 ,
3
c2 =
1
c0 ,
18
c3 =
1
c0 ,
162
and so on. The general solution on (0, ∞) is
1 2
1 2
1
1
1 3
1 3
1/3
x + ··· .
1+ x+ x +
y = C1 1 + x + x + x + · · · + C2 x
2
10
80
3
18
162
377
378
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
20. Substituting y =
∞
cn xn+r into the differential equation and collecting terms, we obtain
n=0
∞
2
2
2
2
r
y = r −r+
c0 x +
(k + r)(k + r − 1)ck + ck − ck−1 xk+r
x y − x−
9
9
9
2 k=1
= 0,
which implies
2
r −r+ =
9
2
2
1
r−
r−
=0
3
3
and
(k + r)(k + r − 1) +
2
ck − ck−1 = 0.
9
The indicial roots are r = 2/3 and r = 1/3. For r = 2/3 the recurrence relation is
ck =
3ck−1
,
3k 2 + k
k = 1, 2, 3, . . . ,
and
3
c1 = c 0 ,
4
c2 =
9
c0 ,
56
c3 =
9
c0 ,
560
c3 =
9
c0 ,
160
and so on. For r = 1/3 the recurrence relation is
ck =
3ck−1
,
3k 2 − k
k = 1, 2, 3, . . . ,
and
3
c1 = c 0 ,
2
c2 =
9
c0 ,
20
and so on. The general solution on (0, ∞) is
3
3
9 3
9 3
9 2
9 2
2/3
1/3
1+ x+ x +
1+ x+ x +
x + · · · + C2 x
x + ··· .
y = C1 x
4
56
560
2
20
160
21. Substituting y =
∞
cn xn+r into the differential equation and collecting terms, we obtain
n=0
∞
2xy − (3 + 2x)y + y = 2r2 − 5r c0 xr−1 +
[2(k + r)(k + r − 1)ck
k=1
− 3(k + r)ck − 2(k + r − 1)ck−1 + ck−1 ]xk+r−1
= 0,
which implies
2r2 − 5r = r(2r − 5) = 0
6.3
Solutions About Singular Points
and
(k + r)(2k + 2r − 5)ck − (2k + 2r − 3)ck−1 = 0.
The indicial roots are r = 0 and r = 5/2. For r = 0 the recurrence relation is
ck =
(2k − 3)ck−1
,
k(2k − 5)
k = 1, 2, 3, . . . ,
and
1
c 1 = c0 ,
3
1
c2 = − c0 ,
6
1
c3 = − c0 ,
6
and so on. For r = 5/2 the recurrence relation is
ck =
2(k + 1)ck−1
,
k(2k + 5)
k = 1, 2, 3, . . . ,
and
4
c1 = c0 ,
7
c2 =
4
c0 ,
21
c3 =
32
c0 ,
693
and so on. The general solution on (0, ∞) is
1
4
1 2 1 3
4 2
32 3
5/2
1+ x+ x +
x + ··· .
y = C1 1 + x − x − x + · · · + C2 x
3
6
6
7
21
693
22. Substituting y =
∞
cn xn+r into the differential equation and collecting terms, we obtain
n=0
4
x y + xy + x −
9
2 2
y=
4
5
r
2
r −
c0 x + r + 2r +
c1 xr+1
9
9
∞
4
(k + r)(k + r − 1)ck + (k + r)ck − ck + ck−2 xk+r
+
9
2
k=2
= 0,
which implies
4
2
2
r−
= 0,
r − = r+
9
3
3
5
2
c1 = 0,
r + 2r +
9
2
and
(k + r)2 −
4
ck + ck−2 = 0.
9
The indicial roots are r = −2/3 and r = 2/3, so c1 = 0. For r = −2/3 the recurrence relation
is
9ck−2
, k = 2, 3, 4, . . . ,
ck = −
3k(3k − 4)
379
380
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
and
3
c2 = − c0 ,
c3 = 0,
4
and so on. For r = 2/3 the recurrence relation is
ck = −
9ck−2
,
3k(3k + 4)
c4 =
9
c0 ,
128
k = 2, 3, 4, . . . ,
and
c2 = −
3
c0 ,
20
c3 = 0,
c4 =
9
c0 ,
1,280
and so on. The general solution on (0, ∞) is
3 2
3 2
9 4
9
−2/3
2/3
4
1− x +
1− x +
y = C1 x
x + · · · + C2 x
x + ··· .
4
128
20
1,280
23. Substituting y =
∞
cn xn+r into the differential equation and collecting terms, we obtain
n=0
9x y + 9x y + 2y = 9r2 − 9r + 2 c0 xr
∞
+
[9(k + r)(k + r − 1)ck + 2ck + 9(k + r − 1)ck−1 ]xk+r
2 2 k=1
= 0,
which implies
9r2 − 9r + 2 = (3r − 1)(3r − 2) = 0
and
[9(k + r)(k + r − 1) + 2]ck + 9(k + r − 1)ck−1 = 0.
The indicial roots are r = 1/3 and r = 2/3. For r = 1/3 the recurrence relation is
ck = −
(3k − 2)ck−1
,
k(3k − 1)
k = 1, 2, 3, . . . ,
and
1
1
c 1 = − c0 ,
c2 = c0 ,
2
5
and so on. For r = 2/3 the recurrence relation is
ck = −
(3k − 1)ck−1
,
k(3k + 1)
c3 = −
7
c0 ,
120
k = 1, 2, 3, . . . ,
and
1
5
1
c 1 = − c0 ,
c2 = c0 ,
c3 = − c0 ,
2
28
21
and so on. The general solution on (0, ∞) is
1 2
5 2
1
1
7 3
1 3
1/3
2/3
x + · · · + C2 x
1− x+ x −
1 − x + x − x + ··· .
y = C1 x
2
5
120
2
28
21
6.3
24. Substituting y =
∞
Solutions About Singular Points
cn xn+r into the differential equation and collecting terms, we obtain
n=0
2x2 y + 3xy + (2x − 1)y = 2r2 + r − 1 c0 xr
∞
+
[2(k + r)(k + r − 1)ck + 3(k + r)ck − ck + 2ck−1 ]xk+r
k=1
= 0,
which implies
2r2 + r − 1 = (2r − 1)(r + 1) = 0
and
[(k + r)(2k + 2r + 1) − 1]ck + 2ck−1 = 0.
The indicial roots are r = −1 and r = 1/2. For r = −1 the recurrence relation is
ck = −
2ck−1
,
k(2k − 3)
k = 1, 2, 3, . . . ,
and
c2 = −2c0 ,
c1 = 2c0 ,
4
c3 = c0 ,
9
and so on. For r = 1/2 the recurrence relation is
ck = −
2ck−1
,
k(2k + 3)
k = 1, 2, 3, . . . ,
and
2
c 1 = − c0 ,
5
c2 =
2
c0 ,
35
c3 = −
4
c0 ,
945
and so on. The general solution on (0, ∞) is
2
4 3
2 2
4 3
−1
2
1/2
1 + 2x − 2x + x + · · · + C2 x
1− x+ x −
y = C1 x
x + ··· .
9
5
35
945
25. Substituting y =
∞
cn xn+r into the differential equation and collecting terms, we obtain
n=0
xy + 2y − xy = r2 + r c0 xr−1 + r2 + 3r + 2 c1 xr
∞
+
[(k + r)(k + r − 1)ck + 2(k + r)ck − ck−2 ]xk+r−1
k=2
= 0,
which implies
r2 + r = r(r + 1) = 0,
2
r + 3r + 2 c1 = 0,
381
382
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
and
(k + r)(k + r + 1)ck − ck−2 = 0.
The indicial roots are r1 = 0 and r2 = −1, so c1 = 0. For r1 = 0 the recurrence relation is
ck =
ck−2
,
k(k + 1)
k = 2, 3, 4, . . . ,
and
1
c0
3!
c2 =
c3 = c5 = c7 = · · · = 0
c4 =
1
c0
5!
c2n =
1
c0 .
(2n + 1)!
For r2 = −1 the recurrence relation is
ck =
ck−2
,
k(k − 1)
k = 2, 3, 4, . . . ,
and
c2 =
1
c0
2!
c3 = c5 = c7 = · · · = 0
c4 =
1
c0
4!
c2n =
1
c0 .
(2n)!
The general solution on (0, ∞) is
y = C1
∞
n=0
∞
1
1
x2n + C2 x−1
x2n
(2n + 1)!
(2n)!
n=0
∞
∞
n=0
n=0
1
1
1
C1
x2n+1 + C2
x2n
=
x
(2n + 1)!
(2n)!
=
26. Substituting y =
∞
1
[C1 sinh x + C2 cosh x] .
x
cn xn+r into the differential equation and collecting terms, we obtain
n=0
1
x y + xy + x −
4
2 2
y=
1
r −
4
2
+
∞
k=2
3
c0 x + r + 2r +
4
r
2
c1 xr+1
1
(k + r)(k + r − 1)ck + (k + r)ck − ck + ck−2 xk+r
4
= 0,
which implies
r2 −
1
=
4
1
1
3
r−
r+
= 0, r2 + 2r +
c1 = 0,
2
2
4
and
(k + r)2 −
1
ck + ck−2 = 0.
4
6.3
Solutions About Singular Points
The indicial roots are r1 = 1/2 and r2 = −1/2, so c1 = 0. For r1 = 1/2 the recurrence
relation is
ck−2
, k = 2, 3, 4, . . . ,
ck = −
k(k + 1)
and
1
1
(−1)n
c 2 = − c0
c0 .
c3 = c5 = c7 = · · · = 0
c4 = c 0
c2n =
3!
5!
(2n + 1)!
For r2 = −1/2 the recurrence relation is
ck−2
,
ck = −
k(k − 1)
k = 2, 3, 4, . . . ,
and
1
c 2 = − c0
2!
c3 = c5 = c7 = · · · = 0
c4 =
1
c0
4!
c2n =
(−1)n
c0 .
(2n)!
The general solution on (0, ∞) is
y = C1 x1/2
∞
∞
(−1)n 2n
(−1)n 2n
x + C2 x−1/2
x
(2n + 1)!
(2n)!
n=0
−1/2
= C1 x
∞
n=0
n=0
∞
(−1)n
(−1)n 2n+1
x
x2n
+ C2 x−1/2
(2n + 1)!
(2n)!
n=0
= x−1/2 [C1 sin x + C2 cos x].
27. Substituting y =
∞
cn xn+r into the differential equation and collecting terms, we obtain
n=0
∞
[(k + r + 1)(k + r)ck+1 − (k + r)ck + ck ]xk+r = 0
xy − xy + y = r2 − r c0 xr−1 +
k=0
which implies
r2 − r = r(r − 1) = 0
and
(k + r + 1)(k + r)ck+1 − (k + r − 1)ck = 0.
The indicial roots are r1 = 1 and r2 = 0. For r1 = 1 the recurrence relation is
kck
ck+1 =
, k = 0, 1, 2, . . . ,
(k + 2)(k + 1)
and one solution is y1 = c0 x. A second solution is
ˆ − ´ −1 dx
ˆ x
ˆ
1 2
e
e
1
1 3
1 + x + x + x + · · · dx
dx = x
dx = x
y2 = x
x2
x2
x2
2
3!
ˆ
1
1 1
1
1
1
1
1
1
=x
+ + + x + x2 + · · · dx = x − + ln x + x + x2 + x3 + · · ·
2
x
x 2 3!
4!
x
2
12
72
1
1
1
= x ln x − 1 + x2 + x3 + x4 + · · · .
2
12
72
383
384
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
The general solution on (0, ∞) is
y = C1 x + C2 y2 (x).
28. Substituting y =
∞
cn xn+r into the differential equation and collecting terms, we obtain
n=0
y +
3 y − 2y = r2 + 2r c0 xr−2 + r2 + 4r + 3 c1 xr−1
x
+
∞
[(k + r)(k + r − 1)ck + 3(k + r)ck − 2ck−2 ] xk+r−2
k=2
= 0,
which implies
r2 + 2r = r(r + 2) = 0
r2 + 4r + 3 c1 = 0
(k + r)(k + r + 2)ck − 2ck−2 = 0.
The indicial roots are r1 = 0 and r2 = −2, so c1 = 0. For r1 = 0 the recurrence relation is
ck =
2ck−2
,
k(k + 2)
k = 2, 3, 4, . . . ,
and
1
c 2 = c0
4
The result is
c3 = c5 = c7 = · · · = 0
c4 =
1
c0
48
c6 =
1
c0 .
1,152
1 2
1 4
1
x6 + · · · .
y1 = c 0 1 + x + x +
4
48
1,152
A second solution is
ˆ − ´ (3/x)dx
ˆ
e
dx
dx = y1
y 2 = y1
2
1 2
1 4
3
y12
x 1 + 4 x + 48
x + ···
ˆ
ˆ
dx
1
1 2
7 4
19 6
= y1
1− x + x +
x + · · · dx
= y1
5 4
7 6
x3
2
48
576
x3 1 + 12 x2 + 48
x + 576
x + ···
ˆ 1
7
19 3
7
1
1
1
19 4
+ x−
x + · · · dx = y1 − 2 − ln x + x2 −
x + ···
−
= y1
3
x
2x 48
576
2x
2
96
2,304
1
1
7
19 4
x + ··· .
= − y1 ln x + y − 2 + x2 −
2
2x
96
2,304
The general solution on (0, ∞) is
y = C1 y1 (x) + C2 y2 (x).
6.3
29. Substituting y =
∞
n=0
Solutions About Singular Points
cn xn+r into the differential equation and collecting terms, we obtain
xy + (1 − x)y − y = r c0 x
2
r−1
+
∞
[(k + r)(k + r − 1)ck + (k + r)ck − (k + r)ck−1 ]xk+r−1 = 0,
k=1
which implies r2 = 0 and
(k + r)2 ck − (k + r)ck−1 = 0.
The indicial roots are r1 = r2 = 0 and the recurrence relation is
ck =
ck−1
,
k
k = 1, 2, 3, . . . .
1 2
1 3
y1 = c0 1 + x + x + x + · · · = c0 ex .
2
3!
One solution is
A second solution is
ˆ − ´ (1/x−1) dx
ˆ x
ˆ
e
e /x
1 −x
x
x
e dx
dx = e
dx = e
y 2 = y1
e2x
e2x
x
ˆ
ˆ 1
1
1
1
1
1
1 − x + x2 − x3 + · · · dx = ex
− 1 + x − x2 + · · · dx
= ex
x
2
3!
x
2
3!
∞
= ex ln x − x +
(−1)n+1
1 2
1 3
x −
x + · · · = ex ln x − ex
xn .
2·2
3 · 3!
n · n!
n=1
The general solution on (0, ∞) is
x
x
y = C1 e + C2 e
ln x −
∞
(−1)n+1
n=1
30. Substituting y =
∞
n · n!
n
x
.
cn xn+r into the differential equation and collecting terms, we obtain
n=0
2
r−1
xy + y + y = r c0 x
∞
+
[(k + r)(k + r − 1)ck + (k + r)ck + ck−1 ]xk+r−1 = 0
k=1
which implies r2 = 0 and
(k + r)2 ck + ck−1 = 0.
The indicial roots are r1 = r2 = 0 and the recurrence relation is
ck = −
ck−1
,
k2
k = 1, 2, 3, . . . .
One solution is
∞
1 2
1 3
1 4
(−1)n n
x
+
x
−
·
·
·
=
c
x .
y 1 = c0 1 − x + 2 x −
0
2
(3!)2
(4!)2
(n!)2
n=0
385
386
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
A second solution is
ˆ
y 2 = y1
e−
´
ˆ
(1/x) dx
y12
dx = y1
ˆ
dx
1 2
x 1 − x + 4x −
1 3
36 x
+ ···
2
dx
− 59 x3 +
35 4
x 1 − 2x +
288 x − · · ·
ˆ
1
5 2 23 3 677 4
1 + 2x + x + x +
x + · · · dx
= y1
x
2
9
288
ˆ 1
5
23 2 677 3
+2+ x+ x +
x + · · · dx
= y1
x
2
9
288
= y1
3 2
2x
5
23
677 4
x + ···
= y1 ln x + 2x + x2 + x3 +
4
27
1,152
The general solution on (0, ∞) is
y = C1 y1 (x) + C2 y2 (x).
31. Substituting y =
∞
n+r
n=0 cn x
into the differential equation and collecting terms, we obtain
xy + (x − 6)y − 3y = (r − 7r)c0 x
2
r−1
+
∞
[(k + r)(k + r − 1)ck + (k + r − 1)ck−1
k=1
−6(k + r)ck − 3ck−1 ] xk+r−1 = 0,
which implies
r2 − 7r = r(r − 7) = 0
and
(k + r)(k + r − 7)ck + (k + r − 4)ck−1 = 0.
The indicial roots are r1 = 7 and r2 = 0. For r1 = 7 the recurrence relation is
(k + 7)kck + (k + 3)ck−1 = 0,
or
ck = −
k+3
ck−1 ,
k(k + 7)
k = 1, 2, 3, . . . ,
k = 1, 2, 3, . . . .
Taking c0 = 0 we obtain
1
c1 = − c0
2
c2 =
5
c0
18
1
c3 = − c0 ,
6
and so on. Thus, the indicial root r1 = 7 yields a single solution. Now, for r2 = 0 the
recurrence relation is
k(k − 7)ck + (k − 4)ck−1 = 0,
k = 1, 2, 3, . . . .
6.3
Solutions About Singular Points
Then
−6c1 − 3c0 = 0
−10c2 − 2c1 = 0
−12c3 − c2 = 0
and
−12c4 + 0c3 = 0
−10c5 + c4 = 0
−c6 + 2c5 = 0
c4 = 0
c5 = 0
c6 = 0
and
ck = −
k−4
ck−1 ,
k(k − 7)
0c7 + 3c6 = 0
c7 is arbitrary
k = 8, 9, 10, . . . .
Taking c0 = 0 and c7 = 0 we obtain
1
c 1 = − c0
2
c2 =
1
c0
10
c3 = −
1
c0
120
c4 = c5 = c6 = · · · = 0.
Taking c0 = 0 and c7 = 0 we obtain
c 1 = c2 = c3 = c4 = c5 = c6 = 0
1
c8 = − c7
2
c9 =
5
c7
36
c10 = −
1
c7 ,
36
and so on. In this case we obtain the two solutions
1
1
1 3
x
y1 = 1 − x + x2 −
2
10
120
32. Substituting y =
∞
n+r
n=0 cn x
and
1
5
1
y2 = x7 − x8 + x9 − x10 + · · · .
2
36
36
into the differential equation and collecting terms, we obtain
x(x − 1)y + 3y − 2y = 4r − r2 c0 xr−1
∞
+
[(k + r − 1)(k + r − 12)ck−1 − (k + r)(k + r − 1)ck
k=1
+3(k + r)ck − 2ck−1 ] xk+r−1
= 0,
which implies
4r − r2 = r(4 − r) = 0
and
−(k + r)(k + r − 4)ck + [(k + r − 1)(k + r − 2) − 2]ck−1 = 0.
The indicial roots are r1 = 4 and r2 = 0. For r1 = 4 the recurrence relation is
−(k + 4)kck + [(k + 3)(k + 2) − 2]ck−1 = 0
or
ck =
k+1
ck−1 ,
k
k = 1, 2, 3, . . . .
387
388
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
Taking c0 = 0 we obtain
c1 = 2c0
c2 = 3c0
c3 = 4c0 ,
and so on. Thus, the indicial root r1 = 4 yields a single solution. For r2 = 0 the recurrence
relation is
−k(k − 4)ck + k(k − 3)ck−1 = 0,
k = 1, 2, 3, . . . ,
or
−(k − 4)ck + (k − 3)ck−1 = 0,
k = 1, 2, 3, . . . .
Then
3c1 − 2c0 = 0
2c2 − c1 = 0
c3 + 0c2 = 0
c3 = 0
and
ck =
(k − 3)ck−1
,
k−4
0c4 + c3 = 0
c4 arbitrary
k = 5, 6, 7, . . . .
Taking c0 = 0 and c4 = 0 we obtain
2
c 1 = c0
3
1
c2 = c0
3
c3 = c4 = c5 = · · · = 0.
Taking c0 = 0 and c4 = 0 we obtain
c1 = c2 = c3 = 0
c5 = 2c4
c6 = 3c4
c7 = 4c4 ,
and so on. In this case we obtain the two solutions
1
2
y1 = 1 + x + x2
3
3
and
y2 = x4 + 2x5 + 3x6 + 4x7 + · · · .
33. (a) From t = 1/x we have dt/dx = −1/x2 = −t2 . Then
dy dt
dy
dy
=
= −t2
dx
dt dx
dt
and
d
d2 y
=
dx2
dx
Now
dy
dx
=
d
dx
dy
d2 y dt
d2 y
dy
dy
dt
−t2
= −t2 2
−
2t
= t4 2 + 2t3
.
dt
dt dx
dt
dx
dt
dt
1
d2 y
x
+ λy = 4
2
dx
t
4
2
d2 y 2 dy
4 d y
3 dy
+
λy
=
+ λy = 0
t
+
2t
+
dt2
dt
dt2
t dt
becomes
t
d2 y
dy
+ λty = 0.
+2
2
dt
dt
6.3
(b) Substituting y =
∞
Solutions About Singular Points
cn tn+r into the differential equation and collecting terms, we obtain
n=0
t
d2 y
dt2
+2
dy
+ λty = (r2 + r)c0 tr−1 + (r2 + 3r + 2)c1 tr
dt
+
∞
[(k + r)(k + r − 1)ck + 2(k + r)ck + λck−2 ]tk+r−1
k=2
= 0,
which implies
r2 + r = r(r + 1) = 0,
2
r + 3r + 2 c1 = 0,
and
(k + r)(k + r + 1)ck + λck−2 = 0.
The indicial roots are r1 = 0 and r2 = −1, so c1 = 0. For r1 = 0 the recurrence relation
is
λck−2
, k = 2, 3, 4, . . . ,
ck = −
k(k + 1)
and
λ
c 2 = − c0
3!
c3 = c5 = c7 = · · · = 0
c4 =
λ2
c0
5!
c2n = (−1)n
λn
c0 .
(2n + 1)!
For r2 = −1 the recurrence relation is
ck = −
λck−2
,
k(k − 1)
k = 2, 3, 4, . . . ,
and
λ
c 2 = − c0
2!
c3 = c5 = c7 = · · · = 0
c4 =
λ2
c0
4!
c2n = (−1)n
λn
c0 .
(2n)!
The general solution on (0, ∞) is
y(t) = c1
∞
∞
(−1)n √
(−1)n √
( λ t)2n + c2 t−1
( λ t)2n
(2n + 1)!
(2n)!
n=0
n=0
∞
∞
1
(−1)n √
(−1)n √
C1
( λ t)2n+1 + C2
( λ t)2n
=
t
(2n + 1)!
(2n)!
n=0
=
n=0
√
√ 1
C1 sin λ t + C2 cos λ t .
t
(c) Using t = 1/x, the solution of the original equation is
√
√
λ
λ
+ C2 x cos
.
y(x) = C1 x sin
x
x
389
390
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
34. (a) From the boundary conditions y(a) = 0, y(b) = 0 we find
√
√
λ
λ
+ C2 cos
=0
C1 sin
a
a
√
√
λ
λ
+ C2 cos
= 0.
C1 sin
b
b
Since this is a homogeneous system of linear equations, it will have nontrivial solutions
for C1 and C2 if
√ √
sin λ cos λ √
√
√
√
λ
λ
λ
λ
a
a cos
− cos
sin
= sin
√
√
a
b
a
b
sin λ cos λ b
b
√
√ √ b−a
λ
λ
−
= sin
= 0.
λ
= sin
a
b
ab
This will be the case if
√
b−a
= nπ
λ
ab
√
or
λ=
nπab
nπab
=
, n = 1, 2, . . . ,
b−a
L
or, if
λn =
n2 π 2 a2 b2
Pn b4
.
=
L2
EI
The critical loads are then Pn = n2 π 2 (a/b)2 EI0 /L2 . Using
√
√
C2 = −C1 sin ( λ/a)/ cos ( λ/a) we have
√
√ √
λ sin ( λ/a)
λ
√
−
cos
y = C1 x sin
x
x
cos ( λ/a)
√
√
√
√ λ
λ
λ
λ
cos
− cos
sin
= C3 x sin
x
a
x
a
= C3 x sin
and
nπab
yn (x) = C3 x sin
L
1 1
−
x a
√
λ
1 1
−
,
x a
= C3 x sin
(b) When n = 1, b = 11, and a = 1, we have,
for C4 = 1,
1
y1 (x) = x sin 1.1π 1 −
.
x
a
nπab a
nπb
− 1 = C4 x sin
1−
.
La x
L
x
y
2
1
1
3
5
7
9
11
x
6.3
Solutions About Singular Points
35. Express the differential equation in standard form:
y + P (x)y + Q(x)y + R(x)y = 0.
Suppose x0 is a singular point of the differential equation. Then we say that x0 is a regular
singular point if (x − x0 )P (x), (x − x0 )2 Q(x), and (x − x0 )3 R(x) are analytic at x = x0 .
36. Substituting y =
∞
cn xn+r into the first differential equation and collecting terms, we obtain
n=0
3 r
x y + y = c0 x +
∞
[ck + (k + r − 1)(k + r − 2)ck−1 ] xk+r = 0.
k=1
It follows that c0 = 0 and
ck = −(k + r − 1)(k + r − 2)ck−1 .
The only solution we obtain is y(x) = 0.
Substituting y =
∞
cn xn+r into the second differential equation and collecting terms, we
n=0
obtain
∞
x y + (3x − 1)y + y = −rc0 +
(k + r + 1)2 ck − (k + r + 1)ck+1 xk+r = 0,
2 k=0
which implies
−rc0 = 0
(k + r + 1)2 ck − (k + r + 1)ck+1 = 0.
If c0 = 0, then the solution of the differential equation is y = 0. Thus, we take r = 0, from
which we obtain
ck+1 = (k + 1)ck , k = 0, 1, 2, . . . .
Letting c0 = 1 we get c1 = 2, c2 = 3!, c3 = 4!, and so on. The solution of the differential
∞
(n + 1)!xn , which converges only at x = 0.
equation is then y =
n=0
37. We write the differential equation in the form x2 y + (b/a)xy + (c/a)y = 0 and identify
a0 = b/a and b0 = c/a as in (14) in the text. Then the indicial equation is
r(r − 1) +
c
b
r+ =0
a
a
or
ar2 + (b − a)r + c = 0,
which is also the auxiliary equation of ax2 y + bxy + cy = 0.
391
392
CHAPTER 6
6.4
SERIES SOLUTIONS OF LINEAR EQUATIONS
Special Functions
1. Since ν 2 = 1/9 the general solution is y = c1 J1/3 (x) + c2 J−1/3 (x).
2. Since ν 2 = 1 the general solution is y = c1 J1 (x) + c2 Y1 (x).
3. Since ν 2 = 25/4 the general solution is y = c1 J5/2 (x) + c2 J−5/2 (x).
4. Since ν 2 = 1/16 the general solution is y = c1 J1/4 (x) + c2 J−1/4 (x).
5. Since ν 2 = 0 the general solution is y = c1 J0 (x) + c2 Y0 (x).
6. Since ν 2 = 4 the general solution is y = c1 J2 (x) + c2 Y2 (x).
7. We identify α = 3 and ν = 2. Then the general solution is y = c1 J2 (3x) + c2 Y2 (3x).
8. We identify α = 6 and ν =
1
2
. Then the general solution is y = c1 J1/2 (6x) + c2 J−1/2 (6x).
9. We identify α = 4 and ν = 2/3 . Then the general solution is y = c1 I2/3 (4x) + c2 K2/3 (4x)
10. We identify α =
√
√
√
2 and ν = 8. Then the general solution is y = c1 I8 ( 2 x) + c2 K8 ( 2 x).
11. If y = x−1/2 v(x) then
1
y = x−1/2 v (x) − x−3/2 v(x),
2
3
y = x−1/2 v (x) − x−3/2 v (x) + x−5/2 v(x),
4
and
2 2 2
3/2 x y + 2xy + α x y = x
1 −1/2
2 3/2
v(x) = 0.
v (x) + α x − x
4
1/2 v (x) + x
Multiplying by x1/2 we obtain
1
2 2
v(x) = 0,
x v (x) + xv (x) + α x −
4
2 whose solution is v = c1 J1/2 (αx) + c2 J−1/2 (αx).
Then y = c1 x−1/2 J1/2 (αx) + c2 x−1/2 J−1/2 (αx).
6.4 Special Functions
12. If y =
√
x v(x) then
1
y = x1/2 v (x) + x−1/2 v(x)
2
1
y = x1/2 v (x) + x−1/2 v (x) − x−3/2 v(x)
4
and
1
1 1/2
1
2 2
2
5/2 3/2 2 2
2
x y + α x −ν +
y = x v (x) + x v (x) − x v(x) + α x − ν +
x1/2 v(x)
4
4
4
2 = x5/2 v (x) + x3/2 v (x) + (α2 x5/2 − ν 2 x1/2 )v(x) = 0.
Multiplying by x−1/2 we obtain
x2 v (x) + xv (x) + (α2 x2 − ν 2 )v(x) = 0,
√
√
whose solution is v(x) = c1 Jν (αx) + c2 Yν (αx). Then y = c1 x Jν (αx) + c2 x Yν (αx).
13. Write the differential equation in the form y + (2/x)y + (4/x)y = 0. This is the form of (18)
in the text with a = − 12 , c = 12 , b = 4, and p = 1, so, by (19) in the text, the general solution
is
y = x−1/2 [c1 J1 (4x1/2 ) + c2 Y1 (4x1/2 )].
14. Write the differential equation in the form y + (3/x)y + y = 0. This is the form of (18) in
the text with a = −1, c = 1, b = 1, and p = 1, so, by (19) in the text, the general solution is
y = x−1 [c1 J1 (x) + c2 Y1 (x)].
15. Write the differential equation in the form y − (1/x)y + y = 0. This is the form of (18) in
the text with a = 1, c = 1, b = 1, and p = 1, so, by (19) in the text, the general solution is
y = x[c1 J1 (x) + c2 Y1 (x)].
16. Write the differential equation in the form y − (5/x)y + y = 0. This is the form of (18) in
the text with a = 3, c = 1, b = 1, and p = 2, so, by (19) in the text, the general solution is
y = x3 [c1 J3 (x) + c2 Y3 (x)].
17. Write the differential equation in the form y + (1 − 2/x2 )y = 0. This is the form of (18) in
the text with a = 12 , c = 1, b = 1, and p = 32 , so, by (19) in the text, the general solution is
y = x1/2 [c1 J3/2 (x) + c2 Y3/2 (x)] = x1/2 [C1 J3/2 (x) + C2 J−3/2 (x)].
18. Write the differential equation in the form y + (4 + 1/4x2 )y = 0. This is the form of (18) in
the text with a = 12 , c = 1, b = 2, and p = 0, so, by (19) in the text, the general solution is
y = x1/2 [c1 J0 (2x) + c2 Y0 (2x)].
393
394
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
19. Write the differential equation in the form y + (3/x)y + x2 y = 0. This is the form of (18) in
the text with a = −1, c = 2, b = 12 , and p = 12 , so, by (19) in the text, the general solution is
1 2
1 2
x + c2 Y1/2
x
y = x−1 c1 J1/2
2
2
or
1 2
1 2
−1
x + C2 J−1/2
x
C1 J1/2
.
y=x
2
2
20. Write the differential equation in the form y + (1/x)y + ( 19 x4 − 4/x2 )y = 0. This is the form
of (18) in the text with a = 0, c = 3, b = 19 , and p = 23 , so, by (19) in the text, the general
solution is
1 3
1 3
x + c2 Y2/3
x
y = c1 J2/3
9
9
or
1 3
1 3
x + C2 J−2/3
x .
y = C1 J2/3
9
9
21. Using the fact that i2 = −1, along with the definition of Jν (x) in (7) in the text, we have
2n+ν
∞
ix
(−1)n
−ν
−ν
Iν (x) = i Jν (ix) = i
n!Γ(1 + ν + n) 2
n=0
=
∞
n=0
=
∞
n=0
=
∞
n=0
=
∞
n=0
x
(−1)n
i2n+ν−ν
n!Γ(1 + ν + n)
2
x
(−1)n
(i2 )n
n!Γ(1 + ν + n)
2
x
(−1)2n
n!Γ(1 + ν + n) 2
2n+ν
1
x
n!Γ(1 + ν + n) 2
2n+ν
2n+ν
2n+ν
,
which is a real function.
22. (a) The differential equation has the form of (20) in the text with
1 − 2a = 0
2c − 2 = 2
1
2
c=2
a=
Then, by (21) in the text,
b2 c2 = −β 2 c2 = −1
1
β= ,
2
1/2
y=x
c1 J1/4
1 2
ix
2
a2 − p2 c2 = 0
1
b= i
2
+ c2 J−1/4
1 2
ix
2
p=
.
In terms of real functions the general solution can be written
1 2
1 2
1/2
y=x
x + C2 K1/4
x
C1 I1/4
.
2
2
1
4
6.4 Special Functions
(b) Write the differential equation in the form y + (1/x)y − 7x2 y = 0. This is the form of
(18) in the text with
1 − 2a = 1
2c − 2 = 2
a=0
c=2
b2 c2 = −β 2 c2 = −7
β=
1√
7,
2
b=
a2 − p2 c2 = 0
1√
7i
2
p = 0.
Then, by (19) in the text,
y = c1 J0
1√
7 ix2
2
+ c2 Y0
1√
2
7 ix .
2
In terms of real functions the general solution can be written
1√ 2
1√ 2
y = C1 I0
7 x + C2 K0
7x .
2
2
23. The differential equation has the form of (20) in the text with
1 − 2a = 0
2c − 2 = 0
b2 c 2 = 1
1
2
c=1
b=1
a=
a2 − p2 c2 = 0
p=
1
.
2
Then, by (21) in the text,
y = x1/2 [c1 J1/2 (x) + c2 J−1/2 (x)] = x1/2 c1
2
sin x + c2
πx
2
cos x = C1 sin x + C2 cos x.
πx
24. Write the differential equation in the form y + (4/x)y + (1 + 2/x2 )y = 0. This is the form
of (20) in the text with
1 − 2a = 4
a=−
3
2
2c − 2 = 0
b2 c2 = 1
c=1
b=1
a2 − p2 c2 = 2
p=
1
.
2
Then, by (21), (26), and (27) in the text,
−3/2
y=x
= C1
−3/2
[c1 J1/2 (x) + c2 J−1/2 (x)] = x
c1
2
sin x + c2
πx
2
cos x
πx
1
1
sin x + C2 2 cos x.
2
x
x
1 2
x − 3/4x2 )y = 0. This is the
25. Write the differential equation in the form y + (2/x)y + ( 16
form of (20) in the text with
1 − 2a = 2
a=−
1
2
2c − 2 = 2
b 2 c2 =
c=2
b=
1
16
1
8
a2 − p2 c2 = −
p=
3
4
1
.
2
395
396
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
Then, by (21) in the text,
1 2
c1 J1/2
+ c2 J−1/2
x
y=x
8
1
1
16
16
= x−1/2 c1
x2 + c2
x2
sin
cos
πx2
8
πx2
8
−1/2
−3/2
= C1 x
sin
1 2
x
8
1 2
x
8
−3/2
+ C2 x
cos
1 2
x .
8
26. Write the differential equation in the form y − (1/x)y + (4 + 3/4x2 )y = 0. This is the form
of (20) in the text with
1 − 2a = −1
2c − 2 = 0
b2 c2 = 4
a2 − p2 c2 =
c=1
b=2
p=
a=1
3
4
1
.
2
Then, by (21) in the text,
2
2
y = x c1 J1/2 (2x) + c2 J−1/2 (2x) = x c1
sin 2x + c2
cos 2x
π2x
π2x
= C1 x1/2 sin 2x + C2 x1/2 cos 2x.
27. (a) The recurrence relation follows from
−νJν (x) + xJν−1 (x) = −
∞
n=0
=−
∞
n=0
=
(−1)n ν
x
n!Γ(1 + ν + n) 2
2n+ν
(−1)n ν
x
n!Γ(1 + ν + n) 2
2n+ν
∞
(−1)n (2n + ν) x
n!Γ(1 + ν + n) 2
+x
∞
(−1)n
x
n!Γ(ν + n) 2
n=0
2n+ν−1
∞
x
(−1)n (ν + n)
·2
+
n!Γ(1 + ν + n)
2
n=0
2n+ν
x
2
= xJν (x).
n=0
(b) The formula in part (a) is a linear first-order differential equation in Jν (x). An integrating
factor for this equation is xν , so
d ν
[x Jν (x)] = xν Jν−1 (x).
dx
28. Subtracting the formula in part (a) of Problem 27 from the formula in Example 5 we obtain
0 = 2νJν (x) − xJν+1 (x) − xJν−1 (x)
or
29. Letting ν = 1 in (21) in the text we have
d
[xJ1 (x)]
xJ0 (x) =
dx
ˆ
so
0
x
2νJν (x) = xJν+1 (x) + xJν−1 (x).
r=x
rJ0 (r) dr = rJ1 (r)
= xJ1 (x).
r=0
2n+ν−1
6.4 Special Functions
30. From (20) we obtain J0 (x) = −J1 (x), and from (23) we obtain J0 (x) = J−1 (x). Thus
J0 (x) = J−1 (x) = −J1 (x).
(2n + 1)! √
1
π we get the following
31. Using Γ 1 + + n = 2n+1
2
2
n!
1
Γ 1+ +n =
2
1
1
+n Γ
+n =
2
2
1
1
1− +n Γ 1− +n =
2
2
1
Γ 1− +n =
2
1
Γ 1− +n =
2
1
Γ 1− +n =
2
(2n + 1)! √
π
22n+1 n!
(2n + 1)! √
π
22n+1 n!
(2n + 1)! √
π
22n+1 n!
√
(2n + 1)!
π
1 − + n 22n+1 n!
1
2
1
2
√
(2n + 1)!
π
2n+1
(1 + 2n) 2
n!
(2n)! √
π
22n n!
From the last result we obtain
∞
x
(−1)n
J−1/2 (x) =
1
n!Γ(1 − 2 + n) 2
n=0
=
2n−1/2
=
∞
n=0
(−1)n
(2n)! √
n! 22n n! π
x
2
2n
x
2
∞
2 (−1)n 2n
·
x
πx
(2n)!
n=0
The last series is the Maclaurin series for the cosine therefore
∞
2 (−1)n 2n
2
·
x =
cos x
J−1/2 (x) =
πx
(2n)!
πx
n=0
32. From Problem 28, 2νJν (x) = xJν+1 (x) + xJν−1 (x) and so with ν = 1/2 we get
J1/2 (x) = xJ3/2 (x) + xJ−1/2 (x)
1
J (x) − J−1/2 (x)
x 1/2
2
2
1
sin x −
cos x
=
x πx
πx
2 sin x
− cos x
=
πx
x
J3/2 (x) =
−1/2
397
398
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
From the last result and using ν = 3/2 we obtain
3J3/2 (x) = xJ5/2 (x) + xJ1/2 (x)
3
J (x) − J1/2 (x)
x 3/2
2 sin x
2
3
− cos x −
sin x
=
x πx
x
πx
3
3 cos x
2
=
− 1 sin x −
πx
x2
x
J5/2 (x) =
From the last result and using ν = 5/2 we obtain
5J5/2 (x) = xJ7/2 (x) + xJ3/2 (x)
5
J (x) − J3/2 (x)
x 5/2
2 3 sin x 3 cos x
2 sin x
5
− sin x −
− cos x
−
=
x πx
x2
x
πx
x
15 6
15
2
sin x −
−
− 1 cos x
=
πx
x3 x
x2
J7/2 (x) =
33. (a) To find the spherical Bessel functions j1 (x) and j2 (x) we use the first formula in (30),
π
J
jn (x) =
2x n+1/2
with n = 1 and n = 2,
j1 (x) =
π
J (x)
2x 3/2
and
Then from Problem 32 we have
√
sin x
− cos x
J3/2 (x) = 2πx
x
and
J5/2 (x) =
√
2πx
3 sin x 3 cos x
− sin x
−
x2
x
(b) Using a graphing utility to plot the graphs of
j1 (x) and j2 (x), we get the red and blue graphcs
in the figure to the right.
j2 (x) =
so
π
J (x) .
2x 5/2
j1 (x) =
sin x cos x
−
x2
x
so
j2 (x) =
3
1
−
3
x
x
sin x −
3 cos x
x2
6.4 Special Functions
399
34. (a) To find the spherical Bessel functions y1 (x) and y2 (x), we use the second formula in (30),
π
Y
(x)
yn (x) =
2x n+1/2
with n = 1 and n = 2,
y1 (x) =
π
Y (x)
2x 3/2
and
y2 (x) =
π
Y (X) .
2x 5/2
From the formula Yn+1/2 (x) = (−1)n+1 J−(n+1/2) (x) we get
Y3/2 (x) = J−3/2 (x),
and
Then from Problem 32 we have
2 cos x
+ sin x
J−3/2 (x) = −
πx
x
and
J−5/2 (x) = −
2
πx
3 cos x 3 sin x
− cos x
+
x2
x
Y5/2 (x) = −J−5/2 (x).
y1 (x) = −
so
so
cos x sin x
−
x2
x
3
1
3 sin x
y2 (x) = − 3 +
cos x −
x
x
x2
(b) Using a graphing utility to plot the graphs
of y1 (x) and y2 (x), we get the red and blue
graphcs in the figure to the right.
35. Letting
we have
k −αt/2
,
e
m
dx
dx ds
dx 2 k
k −αt/2
dx
α −αt/2
=
=
=
−
e
−
e
dt
ds dt
dt α m
2
ds
m
2
s=
α
and
d2 x
d
=
2
dt
dt
dx
dt
dx
=
ds
α k −αt/2
d dx
k −αt/2
e
−
e
+
2 m
dt ds
m
dx
=
ds
α k −αt/2
d2 x ds
k −αt/2
e
−
e
+ 2
2 m
ds dt
m
dx
=
ds
d2 x k −αt
α k −αt/2
+ 2
.
e
e
2 m
ds
m
400
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
Then
d2 x
d2 x mα
m 2 + ke−αt x = ke−αt 2 +
dt
ds
2
k −αt/2 dx
e
+ ke−αt x = 0.
m
ds
Multiplying by 22 /α2 m we have
2
22 k −αt d2 x
e
+
2
2
α m
ds
α
or, since s = (2/α) k/m e−αt/2 ,
s2
36. (a) Differentiating y = x1/2 w
y =x
and
y = αxw
3/2
3 αx
w
22 k −αt
k −αt/2 dx
e
+ 2
e x=0
m
ds α m
dx
d2 x
+ s2 x = 0.
+s
2
ds
ds
2
1/2
with respect to 23 αx3/2 we obtain
2 3/2
1 −1/2
2 3/2
1/2
αx
αx
αx + x
w
3
2
3
2 3/2
1/2
2
3/2
αx
αx
αx + αw
3
3
1
+ αw
2
2 3/2
αx
3
1
− x−3/2 w
4
2 3/2
αx
.
3
Then, after combining terms and simplifying, we have
3 1
2
3/2 3/2
w = 0.
y + α xy = α αx w + w + αx −
2
4αx3/2
Letting t = 23 αx3/2 or αx3/2 = 32 t this differential equation becomes
1
3 α 2 2
t w (t) + tw (t) + t −
w(t) = 0,
2 t
9
t > 0.
(b) Using the same substitution y = x1/2 w 23 αx3/2 the differential equation is now
y − α2 xy = 0:
1 −1/2
2 3/2
1/2 2
3/2
1/2
αx
αx
αx + x
y =x w
w
3
2
3
1 −1/2
2 3/2
2
3/2
αx
αx
+ x
= αxw
w
3
2
3
1 2 3/2
1 −3/2
2 3/2
2
3/2
1/2
2
3/2
αx
αx
αx
αx
αx + αw
+ αw
− x
y = αxw
w
3
3
2
3
4
3
3 2 3/2
1 −3/2
2 3/2
2 3/2 2
3/2
αx
αx
αx
+ αw
− x
=α x w
w
3
2
3
4
3
6.4 Special Functions
Then
y − α xy = α x
2
2 3/2
w
2 3/2
αx
3
3
+ αw
2
2 3/2
αx
3
3 1 −3/2
2 3/2
w
= α x w + αw − α x + x
2
4
3 1
2 3/2 2 3/2
= α x w + αw − α x + 3/2 w
2
4x
3 1
3/2 3/2
w
= α αx w + w = αx +
2
4αx3/2
3
3
3
1
←−
= α tw + w −
t + t−1 w
2
2
2
6
3α 2 1
t w + tw − t2 +
w =0
=
2t
9
when w is a solution of t2 w + tw − t2 + 19 w = 0.
2 3/2
2 3/2
αx
3
2 3/2
2 3/2
αx
−α x w
3
1
− x−3/2 w
4
2
t = αx3/2
3
37. (a) By part (a) of Problem 34, a solution of Airy’s equation is y = x1/2 w( 23 αx3/2 ), where
w(t) = c1 J1/3 (t) + c2 J−1/3 (t)
is a solution of Bessel’s equation of order 13 . Thus, the general solution of Airy’s equation
for x > 0 is
2 3/2
2 3/2
2 3/2
1/2
1/2
1/2
= c1 x J1/3
+ c2 x J−1/3
.
αx
αx
αx
y=x w
3
3
3
(b) By part (b) of Problem 34, a solution of Airy’s equation is y = x1/2 w( 23 αx3/2 ), where
w(t) = c1 I1/3 (t) + c2 I−1/3 (t)
is a solution of modified Bessel’s equation of order 13 . Thus, the general solution of
Airy’s equation for x > 0 is
2 3/2
2 3/2
2 3/2
1/2
1/2
1/2
αx
αx
αx
y=x w
= c1 x I1/3
+ c2 x I−1/3
.
3
3
3
38. The general solution of the differential equation is
y(x) = c1 J0 (αx) + c2 Y0 (αx).
In order to satisfy the conditions that lim y(x) and lim y (x) are finite we are forced to
x→0+
x→0+
define c2 = 0. Thus, y(x) = c1 J0 (αx). The second boundary condition, y(2) = 0, implies
401
402
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
c1 = 0 or J0 (2α) = 0. In order to have a nontrivial solution we require that J0 (2α) = 0. From
Table 5.3.1, the first three positive zeros of J0 are found to be
2α1 = 2.4048, 2α2 = 5.5201, 2α3 = 8.6537
and so α1 = 1.2024, α2 = 2.7601, α3 = 4.3269. The eigenfunctions corresponding to the
eigenvalues λ1 = α12 , λ2 = α22 , λ3 = α32 are J0 (1.2024x), J0 (2.7601x), and J0 (4.3269x).
39. (a) The differential equation y + (λ/x)y = 0 has the form of (20) in the text with
1 − 2a = 0
2c − 2 = −1
1
2
a=
c=
1
2
b2 c 2 = λ
a2 − p 2 c2 = 0
√
b=2 λ
p = 1.
Then, by (21) in the text,
√
√
y = x1/2 c1 J1 (2 λx ) + c2 Y1 (2 λx ) .
(b) We first note that y = J1 (t) is a solution of Bessel’s equation, t2 y + ty + (t2 − 1)y = 0,
with ν = 1. That is,
t2 J1 (t) + tJ1 (t) + (t2 − 1)J1 (t) = 0,
√
or, letting t = 2 x ,
√ √ √ √
4xJ1 2 x + 2 xJ1 2 x + (4x − 1)J1 2 x = 0.
Now, if y =
y =
and
√
√
xJ1 (2 x ), we have
√
√ 1
√ √ 1
√ 1
x J1 2 x √ + √ J1 2 x = J1 2 x + x−1/2 J1 2 x
2
x 2 x
√ 1 √ 1 −3/2 √ J 2 x − x
y = x−1/2 J1 2 x +
J1 2 x .
2x 1
4
Then
√ √
√ √
1 √ 1
x J1 2 x + J1 2 x − x−1/2 J1 2 x + x J 2 x
2
4
√ √ √ √ √
1 = √ 4xJ1 2 x + 2 x J1 2 x − J1 2 x + 4xJ 2 x
4 x
xy + y =
√
= 0,
and y =
√
√
x J1 (2 x ) is a solution of Airy’s differential equation.
6.4 Special Functions
40.
41. (a) We identify m = 4, k = 1, and α = 0.1. Then
x(t) = c1 J0 (10e−0.05t ) + c2 Y0 (10e−0.05t )
and
x (t) = −0.5c1 J0 (10e−0.05t ) − 0.5c2 Y0 (10e−0.05t ).
Now x(0) = 1 and x (0) = −1/2 imply
c1 J0 (10) + c2 Y0 (10) = 1
c1 J0 (10) + c2 Y0 (10) = 1.
Using Cramer’s rule we obtain
c1 =
Y0 (10) − Y0 (10)
J0 (10)Y0 (10) − J0 (10)Y0 (10)
and
c2 =
J0 (10) − J0 (10)
.
J0 (10)Y0 (10) − J0 (10)Y0 (10)
Using Y0 = −Y1 and J0 = −J1 and Table 5.2 we find c1 = −4.7860 and c2 = −3.1803.
Thus
x(t) = −4.7860J0 (10e−0.05t ) − 3.1803Y0 (10e−0.05t ).
x
(b)
10
5
t
−5
42. (a) Identifying α =
50
1
2
100
150
200
, the general solution of x + 14 tx = 0 is
1 3/2
1 3/2
1/2
1/2
x
x
+ c2 x J−1/3
.
x(t) = c1 x J1/3
3
3
Solving the system x(0.1) = 1, x (0.1) = − 12 we find c1 = −0.809264 and c2 = 0.782397.
403
404
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
x
(b)
1
t
50
100
150
200
−1
43. (a) Letting t = L − x, the boundary-value problem becomes
d2 θ
+ α2 tθ = 0,
dt2
θ (0) = 0,
θ(L) = 0,
where α2 = δg/EI. This is Airy’s differential equation, so by Problem 35 its solution is
2 3/2
2 3/2
1/2
1/2
+ c2 t J−1/3
= c1 θ1 (t) + c2 θ2 (t).
y = c1 t J1/3
αt
αt
3
3
(b) Looking at the series forms of θ1 and θ2 we see that θ1 (0) = 0, while θ2 (0) = 0. Thus,
the boundary condition θ (0) = 0 implies c1 = 0, and so
√
2 3/2
.
αt
θ(t) = c2 t J−1/3
3
From θ(L) = 0 we have
√
c2 L J−1/3
2 3/2
αL
3
= 0,
so either c2 = 0, in which case θ(t) = 0, or J−1/3 ( 23 αL3/2 ) = 0. The column will just
start to bend when L is the length corresponding to the smallest positive zero of J−1/3 .
(c) Using Mathematica, the first positive root of J−1/3 (x) is x1 ≈ 1.86635.
2
3/2 = 1.86635 implies
3 αL
L=
3(1.86635)
2α
2/3
=
9EI
(1.86635)2
4δg
Thus
1/3
1/3
9(180 × 109 )π(0.0013)4 /4
=
(1.86635)2
4(7.8)π(0.0013)2
≈ 4.243 m
44. (a) Writing the differential equation in the form xy +(P L/M )y = 0, we identify λ = P L/M .
From Problem 37 the solution of this differential equation is
√
y = c1 x J1 2 P Lx/M
√
+ c2 x Y1 2 P Lx/M
Now J1 (0) = 0, so y(0) = 0 implies c2 = 0 and
√
y = c1 x J1 2 P Lx/M
.
.
6.4 Special Functions
405
√
(b) From y(L) = 0 we have y = J1 (2L P M ) = 0. The first positive zero of J1 is 3.8317 so,
solving 2L P1 /M = 3.8317, we find P1 = 3.6705M/L2 . Therefore,
√
√
3.8317 √
3.6705x
√
= c1 x J1
x .
y1 (x) = c1 x J1 2
L
L
(c) For c1 = 1 and L = 1 the graph of
√
√
y1 = x J1 (3.8317 x ) is shown.
y
0.3
0.2
0.1
0.2
0.4
0.6
0.8
1
x
45. (a) Since l = v, we integrate to obtain l(t) = vt + c. Now l(0) = l0 implies c = l0 , so
l(t) = vt + l0 . Using sin θ ≈ θ in l d2 θ/dt2 + 2l dθ/dt + g sin θ = 0 gives
(l0 + vt)
d2 θ
dθ
+ gθ = 0.
+ 2v
2
dt
dt
(b) Dividing by v, the differential equation in part (a) becomes
l0 + vt d2 θ
dθ g
+2
+ θ = 0.
2
v
dt
dt
v
Letting x = (l0 + vt)/v = t + l0 /v we have dx/dt = 1, so
dθ
dθ dx
dθ
=
=
dt
dx dt
dx
and
d2 θ
d(dθ/dt)
d(dθ/dx) dx
d2 θ
=
=
=
.
dt2
dt
dx
dt
dx2
Thus, the differential equation becomes
x
g
dθ
d2 θ
+ θ=0
+2
2
dx
dx v
or
d2 θ
g
2 dθ
+
θ = 0.
+
2
dx
x dx vx
(c) The differential equation in part (b) has the form of (20) in the text with
g
a2 − p2 c2 = 0
1 − 2a = 2
2c − 2 = −1
b 2 c2 =
v
1
g
1
c=
b=2
p = 1.
a=−
2
2
v
Then, by (21) in the text,
−1/2
θ(x) = x
or
θ(t) =
g 1/2
g 1/2
x
x
c1 J1 2
+ c2 Y1 2
v
v
v
c1 J1
l0 + vt
2
g(l0 + vt)
v
+ c2 Y1
2
g(l0 + vt)
v
.
406
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
(d) To simplify calculations, let
2
g 1/2
u=
x ,
g(l0 + vt) = 2
v
v
√
and at t = 0 let u0 = 2 gl0 /v. The general solution for θ(t) can then be written
θ = C1 u−1 J1 (u) + C2 u−1 Y1 (u).
(1)
Before applying the initial conditions, note that
dθ
dθ du
=
dt
du dt
so when dθ/dt = 0 at t = 0 we have dθ/du = 0 at u = u0 . Also,
d −1
d −1
dθ
= C1
[u J1 (u)] + C2
[u Y1 (u)]
du
du
du
which, in view of (20) in the text, is the same as
dθ
= −C1 u−1 J2 (u) − C2 u−1 Y2 (u).
du
(2)
Now at t = 0, or u = u0 , (1) and (2) give the system
−1
C1 u−1
0 J1 (u0 ) + C2 u0 Y1 (u0 ) = θ0
−1
C1 u−1
0 J2 (u0 ) + C2 u0 Y2 (u0 ) = 0
whose solution is easily obtained using Cramer’s rule:
C1 =
u0 θ0 Y2 (u0 )
,
J1 (u0 )Y2 (u0 ) − J2 (u0 )Y1 (u0 )
C2 =
−u0 θ0 J2 (u0 )
.
J1 (u0 )Y2 (u0 ) − J2 (u0 )Y1 (u0 )
In view of the given identity these results simplify to
π
C1 = − u20 θ0 Y2 (u0 )
2
and
C2 =
π 2
u θ0 J2 (u0 ).
2 0
The solution is then
J1 (u)
Y1 (u)
π 2
u0 θ0 −Y2 (u0 )
+ J2 (u0 )
.
2
u
u
√
Returning to u = (2/v) g(l0 + vt) and u0 = (2/v) gl0 , we have
⎡
2
√
J1
+
vt)
g(l
0
π gl0 θ0 ⎢
v
⎢−Y2 2 gl0
√
θ(t) =
⎣
v
v
l0 + vt
θ=
+J2
2
gl0
v
Y1
2
g(l0 + vt)
v
√
l0 + vt
⎤
⎥
⎥.
⎦
6.4 Special Functions
(e) When l0 = 3.265 m, θ0 =
θ(t) = −1.6892
1
10
radian, and v =
1
60
407
m/s, the above function is
Y1 (375.66(3.265 + t/60))
J1 (375.66(3.265 + t/60))
− 2.7923
.
3.265 + t/60
3.265 + t/60
The plots of θ(t) on [0, 10], [0, 30], and [0, 60] are
θ (t)
0.1
θ (t)
0.1
0.05
θ (t)
0.1
0.05
2
4
6
8
10
t
0.05
5
10
15
20
25
30
t
10
–0.05
–0.05
–0.05
–0.1
–0.1
–0.1
(f ) The graphs indicate that θ(t) decreases as t
increases. The graph of θ(t) on [0, 300] is shown.
20
30
40
50
60
t
θ (t)
0.1
0.05
50
100
150
200
250
300
t
–0.05
–0.1
46. (a) From (32) in the text, we have
6 · 7 2 4 · 6 · 7 · 9 4 2 · 4 · 6 · 7 · 9 · 11 6
x +
x −
x ,
P6 (x) = c0 1 −
2!
4!
6!
where
c0 = (−1)3
5
1·3·5
=− .
2·4·6
16
Thus,
P6 (x) = −
5
16
1
231 6
1 − 21x2 + 63x4 −
x = (231x6 − 315x4 + 105x2 − 5).
5
16
Also, from (26) in the text we have
6 · 9 3 4 · 6 · 9 · 11 5 2 · 4 · 6 · 9 · 11 · 13 7
x +
x −
x
P7 (x) = c1 x −
3!
5!
7!
where
c1 = (−1)3
Thus
35
1·3·5·7
=− .
2·4·6
16
1
35
99 5 429 7
3
x − 9x + x −
x = (429x7 − 693x5 + 315x3 − 35x).
P7 (x) = −
16
5
35
16
(b) P6 (x) satisfies 1 − x2 y −2xy +42y = 0 and P7 (x) satisfies 1 − x2 y −2xy +56y = 0.
408
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
47. The recurrence relation can be written
2k + 1
k
xPk (x) −
Pk−1 (x),
k = 2, 3, 4, . . . .
Pk+1 (x) =
k+1
k+1
3
1
k = 1 : P2 (x) = x2 −
2
2
5
3
3 2 1
2
5
x −
− x = x3 − x
k = 2 : P3 (x) = x
3
2
2
3
2
2
7
30
3
5 3 3
3 3 2 1
35
x − x −
x −
= x4 − x2 +
k = 3 : P4 (x) = x
4
2
2
4 2
2
8
8
8
9
35 4 30 2 3
4 5 3 3
63
35
15
x − x +
−
x − x = x5 − x3 + x
k = 4 : P5 (x) = x
5
8
8
8
5 2
2
8
4
8
11
63 5 35 3 15
5 35 4 30 2 3
x − x + x −
x − x +
k = 5 : P6 (x) = x
6
8
4
8
6 8
8
8
5
231 6 315 4 105 2
x −
x +
x −
16
16
16
16
231 6 315 4 105 2
6 63 5 35 3 15
13
5
P7 (x) = x
x −
x +
x −
−
x − x + x
7
16
16
16
16
7 8
4
8
=
k=6:
=
429 7 693 5 315 3 35
x −
x +
x − x
16
16
16
16
48. If x = cos θ then
dy
dy
= − sin θ
,
dθ
dx
2
d2 y
dy
2 d y
,
=
sin
θ
− cos θ
2
2
dθ
dx
dx
and
sin θ
d2 y
dy
dy
d2 y
2
+
n(n
+
1)(sin
θ)y
=
sin
θ
+ n(n + 1)y = 0.
+
cos
θ
θ
− 2 cos θ
1
−
cos
2
2
dθ
dθ
dx
dx
That is,
d2 y
dy
+ n(n + 1)y = 0.
− 2x
2
dx
dx
49. The only solutions bounded on [−1, 1] are y = cPn (x), c a constant and n = 0, 1, 2, . . . . By
(iv) of the properties of the Legendre polynomials, y(0) = 0 or Pn (0) = 0 implies n must be
odd. Thus the first three positive eigenvalues correspond to n = 1, 3, and 5 or λ1 = 1 · 2,
λ2 = 3 · 4 = 12, and λ3 = 5 · 6 = 30. We can take the eigenfunctions to be y1 = P1 (x),
y2 = P3 (x), and y3 = P5 (x).
1 − x2
50. Using a CAS we find
P1 (x) =
1 d
(x2 − 1)1 = x
2 dx
P2 (x) =
1 d2
1
(x2 − 1)2 = (3x2 − 1)
22 2! dx2
2
P3 (x) =
1 d3
1
(x2 − 1)3 = (5x3 − 3x)
3
3
2 3! dx
2
P4 (x) =
1 d4
1
(x2 − 1)4 = (35x4 − 30x2 + 3)
4
4
2 4! dx
8
6.4 Special Functions
P5 (x) =
1
25 5!
d5
1
(x2 − 1)5 = (63x5 − 70x3 + 15x)
dx5
8
P6 (x) =
1 d6
1
(x2 − 1)6 = (231x6 − 315x4 + 105x2 − 5)
6
6
2 6! dx
16
P7 (x) =
1 d7
1
(x2 − 1)7 = (429x7 − 693x5 + 315x3 − 35x)
7
7
2 7! dx
16
P1
1
51.
P2
1
0.5
–1
–0.5
1
x
–1 –0.5
P4
1
P3
1
0.5
0.5
0.5
0.5
1
x
–1
–0.5
0.5
0.5
1
x
–1
–0.5
0.5
–0.5
–0.5
–0.5
–0.5
–1
–1
–1
–1
P5
1
P6
1
0.5
–1 –0.5
1
x
–1
–0.5
1
x
P7
1
0.5
0.5
409
0.5
0.5
1
x
–1
–0.5
–0.5
–0.5
–0.5
–1
–1
–1
0.5
1
x
52. Zeros of Legendre polynomials for n ≥ 1 are
P1 (x) : 0
P2 (x) : ± 0.57735
P3 (x) : 0, ±0.77460
P4 (x) : ± 0.33998, ±0.86115
P5 (x) : 0, ±0.53847, ±0.90618
P6 (x) : ± 0.23862, ±0.66121, ±0.93247
P7 (x) : 0, ±0.40585, ±0.74153 , ±0.94911
P10 (x) : ± 0.14887, ±0.43340, ±0.67941, ±0.86506, ±0.097391
The zeros of any Legendre polynomial are in the interval (−1, 1) and are symmetric with
respect to 0.
53. Letting y =
∞
ck xk we have
k=0
y =
∞
k=1
ncn xn−1
and
y =
∞
n=2
n(n + 1)cn xn−2
410
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
Then, with appropriate substitutions, we have
∞
[(k + 2)(k + 1)ck+2 − 2kck + 2αck ]xk = 0
k=0
This leads to the recurrence relation
(2α − 2k)
ck ,
ck+2 = −
(k + 2)(k + 1)
for k = 0, 1, 2, 3, . . .
Thus
c2 = −
(2α)
2α
c0 = − c 0
(2)(1)
2!
(2α − 2)
2(α − 1)
c1 = −
c1
(3)(2)
3!
2α
22 α(α − 2)
(2α − 4)
2(α − 2)
c2 = −
· − c0 =
c0
c4 = −
(4)(3)
4·3
2!
4!
2(α − 1)
22 (α − 1)(α − 3)
(2α − 6)
2(α − 3)
c3 = −
· −
c1 =
c1
c5 = −
(5)(4)
5·4
3!
5!
..
.
c3 = −
Then two solutions are
∞
y1 (x) = 1 −
(−1)k 2k α(α − 2)(α − 4) . . . (α − (2k − 2))
2α 2 22 α(α − 2) 4
x +
x − ··· = 1 +
x2k
2!
4!
(2k)!
k=1
and
y2 (x) = x −
=1+
2α − 1 3 22 (α − 1)(α − 3) 5
x +
x − ···
3!
5!
∞
(−1)k 2k (α − 1)(α − 3) . . . (α − (2k − 1))
k=1
(2k + 1)!
x2k+1
and the general solution is y(x) = c0 y1 (x) + c1 y2 (x).
54. (a) From Problem 53 we know that
y1 (x) = 1 −
2α 2 22 α(α − 2) 4 23 α(α − 2)(α − 4) 6
x +
x −
x + ...
2!
4!
6!
2(α − 1) 3 22 (α − 1)(α − 3) 5
x +
x + ...
3!
5!
Therefore we have the following
y2 (x) = x −
When α = n = 0 : y1 (x) = 1 − 0x2 + 0x4 − 0x6 + . . . = 1
When α = n = 2 : y1 (x) = 1 −
4 2
x + 0x4 − 0x6 + . . . = 1 − 2x2
2!
When α = n = 4 : y1 (x) = 1 −
8 2 32 4
4
x + x − 0x6 + . . . = 1 − 4x2 + x4
2!
4!
3
6.4 Special Functions
Also
When α = n = 1 : y2 (x) = x − 0x3 + 0x5 + . . . = x
When α = n = 3 : y2 (x) = x −
4 3
2
x + 0x5 + . . . = x − x3
3!
3
When α = n = 5 : y2 (x) = x −
8 3 32 5
4
4
x + x + 0x7 + . . . = x − x3 + x5
3!
5!
3
15
(b) Using the results of part (a) we get
H0 (x) = 20 · (1) = 1
H1 (x) = 21 · (x) = 2x
H2 (x) = −21 · (1 − 2x2 ) = 22 x2 − 2 = 4x2 − 2
2 3
2
H3 (x) = −2 · 3 x − x = 23 x3 − 22 · 3x = 8x3 − 12x
3
4 4
2
2
H4 (x) = 2 · 3 1 − 4x + x = 24 x4 − 4 · 22 · 3x2 + 22 · 3 = 16x4 − 48x2 + 12
3
4 3
4 5
3
H5 (x) = 2 · 15 x − x + x = 25 x5 − 23 · 5 · 4x3 + 23 · 15x = 32x5 − 160x3 + 120x
3
15
55. Substitute the assumed solution y =
∞
ck xk into the equation to get
k=0
(1 − x2 ) ·
∞
k(k − 1)ck xk−2 − x ·
k=2
∞
kck xk−1 + α2
k=1
∞
ck xk = 0
k=0
From this we get
(2c2 + α2 c0 ) + (6c3 − c1 + α2 c1 )x +
∞
[(k + 2)(k + 1)ck+2 − k(k − 1)ck − kck + α2 ck ]xk = 0
k=2
Therefore we have
c2 = −
α2
c0 ,
2!
c3 =
1 − α2
c1 ,
3!
and
ck+2 =
k 2 − α2
ck for k = 2, 3, 4, . . .
(k + 2)(k + 1)
411
412
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
Therefore we get
y=
∞
ck xk = c0 + c1 x + c2 x2 + c3 x3 + c4 x4 + c5 x5 + c6 x6 + · · ·
k=0
= c0 + c1 x −
α2
1 − α2
(4 − α2 )α2
(9 − α2 )(1 − α2 )
c0 x2 +
c1 x3 −
c0 x4 +
c1 x5
2!
3!
4!
5!
−
= c0 1 −
(16 − α2 )(4 − α2 )α2
c0 x6 + · · ·
6!
α2 2 α2 (4 − α2 ) 4 (16 − α2 )(4 − α2 )α2 6
x −
x −
x + ···
2!
4!
6!
+ c1 x +
1 − α2 3 (9 − α2 )(1 − α2 ) 5
x +
x + ···
3!
5!
= c0 y1 (x) + c1 y2 (x)
where we take
y1 (x) = 1 −
α2 2 (4 − α2 )α2 4 (16 − α2 )(4 − α2 )α2 6
x −
x −
x + ...
2!
4!
6!
y2 (x) = x +
1 − α2 3 (9 − α2 )(1 − α2 ) 5
x +
x + ...
3!
5!
When α = n is a nonnegative even integer y1 (x) terminates at xn , and when α = n is a
positive odd integer y2 (x) terminates at xn . With α = n = 5, y2 (x) yields the fifth degree
5
polynomial solution y = x − 4x3 + 16
5 x .
56. With R(x) = (αx)−1/2 Z(z) the product rule gives:
1 −3/2
−1/2
−1/2 x
R =α
Z − x
Z
2
and
1 −3/2 1 −3/2 3 −5/2
−1/2 x
Z − x
Z − x
Z + x
Z
R =α
2
2
4
3 −5/2
−1/2
−1/2 −3/2 x
=α
Z −x
Z + x
Z .
4
−1/2
The differential equation then becomes
3
1
α−1/2 x2 x−1/2 Z − x−3/2 Z + x−5/2 Z + α−1/2 2x x−1/2 Z − x−3/2 Z
4
2
+ α2 x2 − n(n + 1) α−1/2 x−1/2 Z(x) = 0
or
1
2
x Z + xZ + α x − n + n +
Z = 0.
4
2
2 2
Chapter 6 in Review
This is equivalent to
1
x2 Z + xZ + α2 x2 − n +
2
2 Z = 0.
which is the parametric Bessel equation, so
Z(x) = C1 Jn+1/2 (αx) + C2 Yn+1/2 (αx),
and
R(x) = α−1/2 x1/2 Z(x) = α−1/2 x−1/2 C1 Jn+1/2 (αx) + C2 Yn+1/2 (αx) .
Renaming C1 and C2 this becomes
π Jn+1/2 (αx)
π Yn+1/2 (αx)
√
√
R(x) = c1
+ c2
2
2
αx
αx
π
π
Jn+1/2 (αx) + c2
Y
(αx)
= c1
2αx
2αx n+1/2
= c1 jn (αx) + c2 yn (αx),
where jn (αx) and yn (αx) are the spherical Bessel functions of the first and second kind defined
in the text.
Chapter 6 in Review
1. False; J1 (x) and J−1 (x) are not linearly independent when ν is a positive integer. (In this
case ν = 1). The general solution of x2 y + xy + (x2 − 1)y = 0 is y = c1 J1 (x) + c2 Y1 (x).
2. False; y = x is a solution that is analytic at x = 0.
3. x = −1 is the nearest singular point to the ordinary point x = 0. Theorem 5.1.1 guarantees
∞
cn xn of the differential equation that
the existence of two power series solutions y =
n=1
converge at least for −1 < x < 1. Since − 12 ≤ x ≤ 12 is properly contained in −1 < x < 1,
both power series must converge for all points contained in − 12 ≤ x ≤ 12 .
4. The easiest way to solve the system
2c2 + 2c1 + c0 = 0
6c3 + 4c2 + c1 = 0
1
12c4 + 6c3 − c1 + c2 = 0
3
2
20c5 + 8c4 − c2 + c3 = 0
3
413
414
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
is to choose, in turn, c0 = 0, c1 = 0 and c0 = 0, c1 = 0. Assuming that c0 = 0, c1 = 0, we
have
1
c 2 = − c0
2
2
1
c 3 = − c2 = c 0
3
3
1
1
1
c 4 = − c3 − c 2 = − c 0
2
12
8
2
1
1
1
c 5 = − c4 + c 2 − c 3 = c 0 ;
5
30
20
60
whereas the assumption that c0 = 0, c1 = 0 implies
c2 = −c1
2
1
1
c 3 = − c 2 − c1 = c 1
3
6
2
1
1
1
5
c4 = − c3 + c 1 − c 2 = − c1
2
36
12
36
2
1
1
1
c1 .
c5 = − c4 + c 2 − c 3 = −
5
30
20
360
five terms of two power series solutions are then
1
1
1
1
y1 (x) = c0 1 − x2 + x3 − x4 + x5 + · · ·
2
3
8
60
and
1
5
1 5
x + ··· .
y2 (x) = c1 x − x2 + x3 − x4 −
2
36
360
5. The interval of convergence is centered at 4. Since the series converges at −2, it converges
at least on the interval [−2, 10). Since it diverges at 13, it converges at most on the interval
[−5, 13). Thus, at −7 it does not converge, at 0 and 7 it does converge, and at 10 and 11 it
might converge.
6. We have
x5
x3
+
− ···
x3 2x5
sin x
6
120
=
x
+
=
+
+ ··· .
f (x) =
cos x
3
15
x2 x4
+
− ···
1−
2
24
x−
7. The differential equation (x3 − x2 )y + y + y = 0 has a regular singular point at x = 1 and
an irregular singular point at x = 0.
8. The differential equation (x − 1)(x + 3)y + y = 0 has regular singular points at x = 1 and
x = −3.
Chapter 6 in Review
9. Substituting y =
∞
cn xn+r into the differential equation we obtain
n=0
∞
[2(k + r)(k + r − 1)ck + (k + r)ck + ck−1 ]xk+r−1 = 0
2xy + y + y = 2r2 − r c0 xr−1 +
k=1
which implies
2r2 − r = r(2r − 1) = 0
and
(k + r)(2k + 2r − 1)ck + ck−1 = 0.
The indicial roots are r = 0 and r = 1/2. For r = 0 the recurrence relation is
ck = −
ck−1
,
k(2k − 1)
k = 1, 2, 3, . . . ,
so
1
c2 = c0 ,
6
c1 = −c0 ,
c3 = −
1
c0 .
90
c3 = −
1
c0 .
630
For r = 1/2 the recurrence relation is
ck = −
ck−1
,
k(2k + 1)
k = 1, 2, 3, . . . ,
so
1
c 1 = − c0 ,
3
c2 =
1
c0 ,
30
Two linearly independent solutions are
1
1
y1 = 1 − x + x2 − x3 + · · ·
6
90
and
1/2
y2 = x
10. Substituting y =
∞
1
1 2
1 3
1− x+ x −
x + ··· .
3
30
630
cn xn into the differential equation we have
n=0
y − xy − y =
∞
n=2
n(n − 1)cn xn−2 −
∞
n=1
k=n−2
=
∞
ncn xn −
(k + 2)(k + 1)ck+2 x −
= 2c2 − c0 +
∞
k=1
∞
k=1
cn xn
n=0
k=n
k
k=0
∞
k=n
kck x −
k
∞
ck xk
k=0
[(k + 2)(k + 1)ck+2 − (k + 1)ck ]xk = 0.
415
416
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
Thus
2c2 − c0 = 0(k + 2)(k + 1)ck+2 − (k + 1)ck = 0
and
1
c2 = c0
2
ck+2 =
1
ck ,
k+2
k = 1, 2, 3, . . . .
Choosing c0 = 1 and c1 = 0 we find
c2 =
1
2
c3 = c 5 = c 7 = · · · = 0
c4 =
1
8
c6 =
1
48
and so on. For c0 = 0 and c1 = 1 we obtain
c 2 = c4 = c6 = · · · = 0
c3 =
1
3
c5 =
1
15
c7 =
1
105
and so on. Thus, two solutions are
1
1
1
y1 = 1 + x2 + x4 + x6 + · · ·
2
8
48
and
1
1
1 7
x + ··· .
y2 = x + x3 + x5 +
3
15
105
11. Substituting y =
∞
cn xn into the differential equation we obtain
n=0
(x − 1)y + 3y = (−2c2 + 3c0 ) +
∞
[(k + 1)kck+1 − (k + 2)(k + 1)ck+2 + 3ck ]xk = 0
k=1
which implies c2 = 3c0 /2 and
ck+2 =
(k + 1)kck+1 + 3ck
,
(k + 2)(k + 1)
k = 1, 2, 3, . . . .
Choosing c0 = 1 and c1 = 0 we find
3
c2 = ,
2
1
c3 = ,
2
c4 =
5
8
c4 =
1
4
and so on. For c0 = 0 and c1 = 1 we obtain
1
c3 = ,
2
c2 = 0,
and so on. Thus, two solutions are
3
1
5
y1 = 1 + x2 + x3 + x4 + · · ·
2
2
8
and
1
1
y2 = x + x3 + x4 + · · · .
2
4
Chapter 6 in Review
12. Substituting y =
∞
cn xn into the differential equation we obtain
n=0
y − x2 y + xy = 2c2 + (6c3 + c0 )x +
∞
[(k + 3)(k + 2)ck+3 − (k − 1)ck ]xk+1 = 0
k=1
which implies c2 = 0, c3 = −c0 /6, and
ck+3 =
k−1
ck ,
(k + 3)(k + 2)
k = 1, 2, 3, . . . .
Choosing c0 = 1 and c1 = 0 we find
c3 = −
1
6
c4 = c7 = c10 = · · · = 0
c5 = c8 = c11 = · · · = 0
c6 = −
1
90
and so on. For c0 = 0 and c1 = 1 we obtain
c 3 = c6 = c9 = · · · = 0
c4 = c7 = c10 = · · · = 0
c5 = c8 = c11 = · · · = 0
and so on. Thus, two solutions are
1
1
y1 = 1 − x3 − x6 − · · ·
6
90
13. Substituting y =
∞
and y2 = x.
cn xn+r into the differential equation, we obtain
n=0
xy −(x+2)y +2y = (r2 −3r)c0 xr−1 +
∞
[(k + r)(k + r − 3)ck − (k + r − 3)ck−1 ]xk+r−1 = 0,
k=1
which implies
r2 − 3r = r(r − 3) = 0
and
(k + r)(k + r − 3)ck − (k + r − 3)ck−1 = 0.
The indicial roots are r1 = 3 and r2 = 0. For r2 = 0 the recurrence relation is
k(k − 3)ck − (k − 3)ck−1 = 0,
k = 1, 2, 3, . . . .
Then
c 1 − c0 = 0
2c2 − c1 = 0
Therefore c3 is arbitrary and
ck =
1
ck−1 ,
k
k = 4, 5, 6, . . . .
0c3 − 0c2 = 0
417
418
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
Taking c0 = 0 and c3 = 0 we obtain
1
c2 = c0
2
c1 = c0
c3 = c4 = c5 = · · · = 0.
Taking c0 = 0 and c3 = 0 we obtain
1
6
1
6
1
6
c 3 = c3
c3 = c 3 ,
c4 = c3 = c3
c5 =
c6 =
c 0 = c1 = c2 = 0
4
4!
5·4
5!
6·5·4
6!
and so on. In this case we obtain the two solutions
1
y1 = 1 + x + x2
2
and
6 4
6 5
6 6
1 2
3
x
y2 = x + x + x + x + · · · = 6e − 6 1 + x + x .
4!
5!
6!
2
14. Substituting y =
∞
cn xn into the differential equation we have
n=0
1 2
1 4
1 6
x + · · · (2c2 + 6c3 x + 12c4 x2 + 20c5 x3
(cos x)y + y = 1 − x + x −
2
24
720
+ 30c6 x + · · · ) +
4
= 2c2 + 6c3 x + (12c4 − c2 )x2 + (20c5 − 3c3 )x3 + 30c6 − 6c4 +
∞
cn xn
n=0
1
c2 x4 + · · ·
12
+ c0 + c1 x + c2 x2 + c3 x3 + c4 x4 + · · ·
1
2
3
= (c0 + 2c2 ) + (c1 + 6c3 )x + 12c4 x + (20c5 − 2c3 )x + 30c6 − 5c4 + c2 x4 + · · ·
12
= 0.
Thus
c0 + 2c2 = 0
c1 + 6c3 = 0
12c4 = 0
20c5 − 2c3 = 0
30c6 − 5c4 +
1
c2 = 0
12
and
1
c 2 = − c0
2
1
c3 = − c1
6
c4 = 0
Choosing c0 = 1 and c1 = 0 we find
1
c3 = 0,
c4 = 0,
c2 = − ,
2
and so on. For c0 = 0 and c1 = 1 we find
1
c2 = 0,
c4 = 0,
c3 = − ,
6
and so on. Thus, two solutions are
1
1 6
x + ···
and
y1 = 1 − x2 +
2
720
c5 =
1
c3
10
1
1
c2 .
c6 = c4 −
6
360
c5 = 0,
c5 = −
1
,
60
c6 =
1
720
c6 = 0
1
1
y2 = x − x3 − x5 + · · · .
6
60
Chapter 6 in Review
15. Substituting y =
∞
cn xn into the differential equation we have
n=0
y + xy + 2y =
∞
n(n − 1)cn x
n−2
n=2
+
n=1
k=n−2
=
∞
∞
n
ncn x +2
∞
n=0
k=n
(k + 2)(k + 1)ck+2 xk +
k=0
∞
k=n
kck xk + 2
k=1
= 2c2 + 2c0 +
∞
cn xn
∞
ck xk
k=0
[(k + 2)(k + 1)ck+2 + (k + 2)ck ]xk = 0.
k=1
Thus
2c2 + 2c0 = 0
(k + 2)(k + 1)ck+2 + (k + 2)ck = 0
and
c2 = −c0
ck+2 = −
1
ck ,
k+1
k = 1, 2, 3, . . . .
Choosing c0 = 1 and c1 = 0 we find
c2 = −1
c3 = c5 = c 7 = · · · = 0
c4 =
1
3
c6 = −
1
15
c5 =
1
8
c7 = −
1
48
and so on. For c0 = 0 and c1 = 1 we obtain
c 2 = c4 = c6 = · · · = 0
c3 = −
1
2
and so on. Thus, the general solution is
1 3 1 5
1 4
1 6
1 7
2
y = C0 1 − x + x − x + · · · + C1 x − x + x − x + · · ·
3
15
2
8
48
and
4
3
2
5
7
y = C0 −2x + x3 − x5 + · · · + C1 1 − x2 + x4 − x6 + · · · .
3
5
2
8
48
Setting y(0) = 3 and y (0) = −2 we find c0 = 3 and c1 = −2. Therefore, the solution of the
initial-value problem is
1
1
1
y = 3 − 2x − 3x2 + x3 + x4 − x5 − x6 + x7 + · · · .
4
5
24
419
420
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
16. Substituting y =
∞
cn xn into the differential equation we have
n=0
(x + 2)y + 3y =
∞
n(n − 1)cn xn−1 +2
n=2
∞
n=2
k=n−1
=
∞
n(n − 1)cn xn−2 +3
∞
n=0
k=n−2
k
(k + 1)kck+1 x + 2
k=1
∞
k=n
k
(k + 2)(k + 1)ck+2 x + 3
k=0
= 4c2 + 3c0 +
∞
cn xn
∞
ck xk
k=0
[(k + 1)kck+1 + 2(k + 2)(k + 1)ck+2 + 3ck ]xk = 0.
k=1
Thus
4c2 + 3c0 = 0
(k + 1)kck+1 + 2(k + 2)(k + 1)ck+2 + 3ck = 0
and
3
c 2 = − c0
4
ck+2 = −
k
3
ck+1 −
ck ,
2(k + 2)
2(k + 2)(k + 1)
k = 1, 2, 3, . . . .
Choosing c0 = 1 and c1 = 0 we find
c2 = −
3
4
c3 =
1
8
c4 =
1
16
c5 = −
9
320
and so on. For c0 = 0 and c1 = 1 we obtain
c2 = 0
c3 = −
1
4
c4 =
1
16
c5 = 0
and so on. Thus, the general solution is
3 2 1 3
1 3
1 4
9 5
1 4
x + · · · + C1 x − x + x + · · ·
y = C0 1 − x + x + x −
4
8
16
320
4
16
and
3
3
3
1
9
1
y = C0 − x + x2 + x3 − x4 + · · · + C1 1 − x2 + x3 + · · · .
2
8
4
64
4
4
Setting y(0) = 0 and y (0) = 1 we find c0 = 0 and c1 = 1. Therefore, the solution of the
initial-value problem is
1
1
y = x − x3 + x4 + · · · .
4
16
17. The singular point of (1 − 2 sin x)y + xy = 0 closest to x = 0 is π/6. Hence a lower bound
is π/6.
Chapter 6 in Review
18. While we can find two solutions of the form
y1 = c0 [1 + · · · ]
and y2 = c1 [x + · · · ],
the initial conditions at x = 1 give solutions for c0 and c1 in terms of infinite series. Letting
t = x − 1 the initial-value problem becomes
d2 y
dy
+ y = 0,
+ (t + 1)
2
dt
dt
Substituting y =
∞
y(0) = −6, y (0) = 3.
cn tn into the differential equation, we have
n=0
∞
∞
∞
∞
d2 y
dy
n−2
n
n−1
+
y
=
+
(t
+
1)
n(n
−
1)c
t
+
nc
t
+
nc
t
+
cn t n
n
n
n
dt2
dt
n=2
n=1
n=1
n=0
k=n−2
=
∞
k=n
(k + 2)(k + 1)ck+2 tk +
k=0
∞
k=n−1
kck tk +
k=1
= 2c2 + c1 + c0 +
∞
∞
k=n
(k + 1)ck+1 tk +
k=0
[(k + 2)(k + 1)ck+2 + (k + 1)ck+1 + (k + 1)ck ] tk = 0.
Thus
2c2 + c1 + c0 = 0
(k + 2)(k + 1)ck+2 + (k + 1)ck+1 + (k + 1)ck = 0
and
c1 + c0
2
ck+1 + ck
,
=−
k+2
c2 = −
k = 1, 2, 3, . . . .
Choosing c0 = 1 and c1 = 0 we find
1
c2 = − ,
2
c3 =
1
,
6
c4 =
1
,
12
and so on. For c0 = 0 and c1 = 1 we find
1
c2 = − ,
2
ck t k
k=0
k=1
ck+2
∞
1
c3 = − ,
6
c4 =
1
,
6
and so on. Thus, the general solution is
1
1
1
1
1
1
y = c0 1 − t 2 + t 3 + t 4 + · · · + c 1 t − t 2 − t 3 + t 4 + · · · .
2
6
12
2
6
6
421
422
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
The initial conditions then imply c0 = −6 and c1 = 3. Thus the solution of the initial-value
problem is
1
1
1
y = −6 1 − (x − 1)2 + (x − 1)3 + (x − 1)4 + · · ·
2
6
12
1
1
1
+ 3 (x − 1) − (x − 1)2 − (x − 1)3 + (x − 1)4 + · · · .
2
6
6
19. Writing the differential equation in the form
1
−
cos
x
y + xy = 0,
y +
x
and noting that
x x3
x5
1 − cos x
= −
+
− ···
x
2 24 720
is analytic at x = 0, we conclude that x = 0 is an ordinary point of the differential equation.
20. Writing the differential equation in the form
x
y=0
y + x
e −1−x
and noting that
2 2
x
x2
x
=
−
+
+
− ···
ex − 1 − x
x 3 18 270
we see that x = 0 is a singular point of the differential equation. Since
x
2x2 x3
x4
2
=
2x
−
+
+
− ··· ,
x
ex − 1 − x
3
18 270
we conclude that x = 0 is a regular singular point.
21. Substituting y =
∞
cn xn into the differential equation we have
n=0
2 y + x y + 2xy =
∞
n=2
n(n − 1)cn x
n−2
+
∞
=
∞
ncn x
n=1
k=n−2
n+1
k=n+1
(k + 2)(k + 1)ck+2 xk +
k=0
∞
+2
∞
cn xn+1
n=0
(k − 1)ck−1 xk + 2
k=2
= 2c2 + (6c3 + 2c0 )x +
k=n+1
∞
ck−1 xk
k=1
∞
[(k + 2)(k + 1)ck+2 + (k + 1)ck−1 ]xk = 5 − 2x + 10x3 .
k=2
Thus, equating coefficients of like powers of x gives
2c2 = 5
6c3 + 2c0 = −2
12c4 + 3c1 = 0
20c5 + 4c2 = 10
Chapter 6 in Review
and
(k + 2)(k + 1)ck+2 + (k + 1)ck−1 = 0,
k = 4, 5, 6, . . . ,
Therefore
5
c2 =
2
1
1
c 3 = − c0 −
3
3
1
c 4 = − c1
4
and
ck+2 = −
1 1
1 1
c5 = − c2 = −
2 5
2 5
5
=0
2
1
ck−1 .
k+2
Using the recurrence relation, we find
1
1
1
1
(c0 + 1) = 2
c0 + 2
c 6 = − c3 =
6
3·6
3 · 2!
3 · 2!
1
1
c1
c 7 = − c4 =
7
4·7
c8 = c11 = c14 = · · · = 0
1
1
1
c0 − 3
c9 = − c6 = − 3
9
3 · 3!
3 · 3!
c10 = −
1
1
c7 = −
c1
10
4 · 7 · 10
c12 = −
1
1
1
c9 = 4
c0 + 4
12
3 · 4!
3 · 4!
c13 = −
1
1
c0 =
c1
13
4 · 7 · 10 · 13
and so on. Thus
1
1
1
1
x6 − 3
x9 + 4
x12 − · · ·
y = c0 1 − x3 + 2
3
3 · 2!
3 · 3!
3 · 4!
1 7
1
1
1
x −
x10 +
x13 − · · ·
+ c1 x − x4 +
4
4·7
4 · 7 · 10
4 · 7 · 10 · 13
+
22. (a) From y = −
1
1
1
5 2 1 3
x − x + 2
x6 − 3
x9 + 4
x12 − · · · .
2
3
3 · 2!
3 · 3!
3 · 4!
1 du
we obtain
u dx
dy
1 d2 u
1
=−
+ 2
2
dx
u dx
u
du
dx
2
.
Then dy/dx = x2 + y 2 becomes
1
1 d2 u
+ 2
−
2
u dx
u
so
d2 u
+ x2 u = 0.
dx2
du
dx
2
1
=x + 2
u
2
du
dx
2
,
423
424
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
(b) The differential equation u + x2 u = 0 has the form (20) in the text with
1 − 2a = 0
2c − 2 = 2
1
2
c=2
a=
a2 − p2 c2 = 0
b 2 c2 = 1
b=
1
2
p=
1
4
Then, by (19) in the text,
u = x1/2 c1 J1/4
(c) Letting t =
y=−
=−
1
2
1 2
x
2
+ c2 J−1/4
1 2
x
2
.
x2 and w(t) = c1 J1/4 (t) + c2 J−1/4 (t), we have
d 1/2
1
dw dt
1
1
1 du
x1/2
= − 1/2
x w(t) = − 1/2
+ x−1/2 w
u dx
dt dx 2
x w(t) dx
x w
1
x1/2 w
x3/2
1
1
1
dw
dw
dw
+ 1/2 w = −
+w =−
+w .
2x2
4t
dt
2xw
dt
2xw
dt
2x
Now
4t
dw
d
+ w = 4t [c1 J1/4 (t) + c2 J−1/4 (t)] + c1 J1/4 (t) + c2 J−1/4 (t)
dt
dt
1
1
= 4t c1 J−3/4 (t) − J1/4 (t) + c2 − J−1/4 (t) − J3/4 (t)
4t
4t
= 4c1 tJ−3/4 (t) − 4c2 tJ3/4 (t) = 2c1 x2 J−3/4
+ c1 J1/4 (t) + c2 J−1/4 (t)
1 2
1 2
2
x − 2c2 x J3/4
x ,
2
2
so
y=−
2c1 x2 J−3/4 ( 12 x2 ) − 2c2 x2 J3/4 ( 12 x2 )
2x[c1 J1/4 ( 12 x2 ) + c2 J−1/4 ( 12 x2 )]
=x
−c1 J−3/4 ( 12 x2 ) + c2 J3/4 ( 12 x2 )
c1 J1/4 ( 12 x2 ) + c2 J−1/4 ( 12 x2 )
.
Letting c = c1 /c2 we have
y=x
J3/4 ( 12 x2 ) − cJ−3/4 ( 12 x2 )
cJ1/4 ( 12 x2 ) + J−1/4 ( 12 x2 )
.
23. (a) From (10) of Section 6.4, with n = 32 , we have
Y3/2 (x) =
−J−3/2 (x)
= J−3/2 (x).
−1
Then from the solutions of Problems 28 and 32 in Section 6.4 we have
cos x
2 cos x
2
J−3/2 (x) = −
+ sin x
so
Y3/2 (x) = −
x
+ sin x
πx
x
π
x
Chapter 6 in Review
(b) From (15) of Section 6.4 in the text
I1/2 (x) = i−1/2 J1/2 (ix)
so
I1/2 (x) =
and
I−1/2 (x) = i1/2 J−1/2 (ix)
∞
2 1
x2n+1 =
πx
(2n + 1)!
n=0
and
I−1/2 (x) =
∞
2 1
x2n =
πx
(2n)!
n=0
2
sinh x
πx
2
cosh x.
πx
(c) Equation (16) of Section 6.4 in the text and part (b) imply
π I−1/2 (x) − I1/2 (x)
π
2
2
cosh x −
sinh x
=
K1/2 (x) =
2
sin π2
2
πx
πx
=
π ex + e−x ex − e−x
−
=
2x
2
2
π −x
e .
2x
24. (a) Using formula (5) of Section 4.2 in the text, we find that a second solution of
(1 − x2 )y − 2xy = 0 is
ˆ ´ 2x dx/(1−x2 )
ˆ
e
2
y2 (x) = 1 ·
dx
=
e− ln(1−x ) dx
12
ˆ
1
1+x
dx
,
= ln
=
1 − x2
2
1−x
where partial fractions was used to obtain the last integral.
(b) Using formula (5) of Section 4.2 in the text, we find that a second solution of
(1 − x2 )y − 2xy + 2y = 0 is
ˆ ´ 2x dx/(1−x2 )
ˆ − ln (1−x2 )
e
e
y2 (x) = x ·
dx
=
x
dx
x2
x2
ˆ
1
1+x
1
x
1+x
dx
=x
ln
−
= ln
− 1,
=x
x2 (1 − x2 )
2
1−x
x
2
1−x
where partial fractions was used to obtain the last integral.
(c) x
y2
2
y2
2
1
1
–1
1
x
–1
1
–1
–1
–2
–2
x
425
426
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
1
y2 (x) = ln
2
1+x
1−x
x
y2 = ln
2
1+x
1−x
−1
25. (a) By the binomial theorem we have
−1/2
1 + t2 − 2xt
=1−
(−1/2)(−3/2) 2
1 2
(−1/2)(−3/2)(−5/2) 2
(t − 2xt)2 +
(t − 2xt)3 + · · ·
t − 2xt +
2
2!
3!
3
5
1
= 1 − (t2 − 2xt) + (t2 − 2xt)2 − (t2 − 2xt)3 + · · ·
2
8
16
∞
1
1
Pn (x)tn .
= 1 + xt + (3x2 − 1)t2 + (5x3 − 3x)t3 + · · · =
2
2
n=0
−1/2
, we have
(b) Letting x = 1 in 1 − 2xt + t2
1 − 2t + t2
−1/2
= (1 − t)−1 =
∞
=
1
= 1 + t + t2 + t3 + · · ·
1−t
(|t| < 1)
tn .
n=0
From part (a) we have
∞
∞
2 −1/2
Pn (1)t = 1 − 2t + t
=
tn .
n
n=0
n=0
Equating the coefficients of corresponding terms in the two series, we see that Pn (1) = 1.
SImilarly, letting x = −1 we have
1 + 2t + t2
−1/2
= (1 + t)−1 =
=
∞
1
= 1 − t + t2 − 3t3 + · · ·
1+t
(−1)n tn =
n=0
∞
(|t| < 1)
Pn (−1)tn ,
n=0
so that Pn (−1) = (−1)n .
26. (a) Rewrite the equation as
25
2
y = 0.
x y + xy + 16x −
4
2 The new equation has the form of (12) from Section 5.3 with α = 4 and υ = 52 . Thus
the general solution is given by
y = c1 J5/2 (4x) + c2 Y5/2 (4x)
Chapter 6 in Review
(b) This equation has the parametric form of the modified Bessel equation of order υ = 3
√
with α = 3 2. Therefore the general solution is
√
√
y = c1 I3 (3 2 x) + c2 K3 (3 2 x)
27. Using the Product rule gives
d
dr
r
r2
dT
r
= a2 r (T − Tm )
dr
d2 T
dT
− a2 r (T − Tm ) = 0
+
dr2
dr
dT
d2 T
− a2 r2 (T − Tm ) = 0
+r
2
dr
dr
The given boundary-value problem is
dT
d2 T
− a2 r2 (T − 70) = 0,
r2 2 + r
dr
dr
T (1) = 160,
dT dr =0
r=3
If we change the dependent variable by means of the substitution w(r) = T (r) − 70 the
differential equation becomes
r2
dw
d2 w
− a2 r2 w = 0.
+r
dr2
dr
This differential equation is the parametric form of the modified Bessel’s equation of order
ν = 0 so:
w(r) = c1 I0 (ar) + c2 K0 (ar)
T (r) = 70 + w(r)
T (r) = 70 + c1 I0 (ar) + c2 K0 (ar)
The derivatives
d
d
I0 (x) = I1 (x) and
K0 (x) = −K1 (x) and so the Chain rule gives
dx
dx
u
d dI0 du
d
I0 (ar) =
= I0 (ar) ar = aI1 (ar) and
dr
du dr
dr
d
K0 (ar) = −aK1 (ar)
dr
Therefore
T (r) = c1
d
d
I0 (ar) + c2 K0 (ar) or T (r) = c1 aI1 (ar) − c2 aK1 (ar)
dr
dr
The boundary conditions T (1) = 160 and T (3) = 0 then give the linear system of equations
c1 I0 (a) + c2 K0 (a) = 90
c1 I1 (3a) − c2 K1 (3a) = 0.
427
428
CHAPTER 6
SERIES SOLUTIONS OF LINEAR EQUATIONS
By Cramer’s Rule, the solution of the system is
90
K0 (a) 0 −K1 (3a)
90K1 (3a)
=
,
c1 = I0 (a)K1 (3a) + I1 (3a)K0 (a)
K0 (a) I0 (a)
I1 (3a) −K1 (3a)
I0 (a) 90
I1 (3a) 0 90I1 (3a)
=
.
c2 = I0 (a)K1 (3a) + I1 (3a)K0 (a)
K0 (a) I0 (a)
I1 (3a) −K1 (3a)
Hence the solution of the boundary-value problem is
T (r) = 70 + c1 I0 (ar) + c2 K0 (ar)
= 70 +
90I1 (3a)K0 (ar)
90K1 (3a)I0 (ar)
+
I0 (a)K1 (3a) + I1 (3a)K0 (a) I0 (a)K1 (3a) + I1 (3a)K0 (a)
= 70 + 90
K1 (3a)I0 (ar) + I1 (3a)K0 (ar)
.
K1 (3a)I0 (a) + I1 (3a)K0 (a)
28. The differential equation is the same as in Problem 27 so the substitution w(r) = T (r) − 70
again yields
r2
dw
d2 w
− a2 r2 w = 0
+r
2
dr
dr
and so
w(r) = c1 I0 (ar) + c2 K0 (ar)
T (r) = 70 + w(r)
T (r) = 70 + c1 I0 (ar) + c2 K0 (ar)
The boundary conditions T (1) = 160 and T (3) = 90 then give the linear system of equations
c1 I0 (a) + c2 K0 (a) = 90
c1 I0 (3a) + c2 K0 (3a) = 20
Chapter 6 in Review
By Cramer’s Rule, the solution of the system is
90 K0 (a) 20 K0 (3a)
90K0 (3a) − 20K0 (a)
=
c1 = ,
I0 (a)K0 (3a) − I0 (3a)K0 (a)
I0 (a) K0 (a) I0 (3a) K0 (3a)
I0 (a) 90
I0 (3a) 20
20I0 (a) − 90I0 (3a)
=
.
c2 = I0 (a)K0 (3a) − I0 (3a)K0 (a)
I0 (a) K0 (a) I0 (3a) K0 (3a)
Hence the solution of the boundary-value problem is
T (r) = 70 +
90K0 (3a) − 20K0 (a)
20I0 (a) − 90I0 (3a)
I0 (ar) +
K0 (ar) .
I0 (a)K0 (3a) − I0 (3a)K0 (a)
I0 (a)K0 (3a) − I0 (3a)K0 (a)
429
Chapter 7
The Laplace Transform
7.1
Definition of the Laplace Transform
ˆ
1. L {f (t)} =
1
−e−st dt +
ˆ
0
∞
1
1
∞
1
1
e−st dt = e−st − e−st s
s
0
1
2
1 −s
1 −s 1
1
= e−s − , s > 0
= e − − 0− e
s
s
s
s
s
2
ˆ 2
4 −st 4
−st
4e dt = − e = − e−2s − 1 , s > 0
2. L {f (t)} =
s
s
0
0
ˆ
3. L {f (t)} =
1
te−st dt +
0
ˆ
∞
e−st dt =
1
1
1
− te−st − 2 e−st
s
s
∞
1
1 −st − e s
0
1
1
1
1 −s
1 −s
1
− 0 − 2 − (0 − e−s ) = 2 (1 − e−s ),
= − e − 2e
s
s
s
s
s
1
ˆ 1
2
2
1
(2t + 1)e−st dt = − te−st − 2 e−st − e−st 4. L {f (t)} =
s
s
s
0
s>0
0
2
2 −s
2 −s 1 −s
1
1
2
− 0− 2 −
= (1 − 3e−s ) + 2 (1 − e−s ), s > 0
= − e − 2e − e
s
s
s
s
s
s
s
π
ˆ π
s
1
−st
−st
−st
5. L {f (t)} =
e sin t − 2
e cos t (sin t)e dt = − 2
s
+
1
s
+
1
0
= 0+
ˆ
6. L {f (t)} =
∞
s2
1
e−πs
+1
− 0−
−st
(cos t)e dt = −
π/2
=0− 0+
1
e−πs/2
s2 + 1
s2
1
+1
0
=
s2
1
(e−πs + 1),
+1
s
1
e−st cos t + 2
e−st sin t
2
s +1
s +1
=−
1
e−πs/2 ,
s2 + 1
430
s>0
s>0
∞
π/2
7.1
7. f (t) =
0, 0 < t < 1
t,
t>1
ˆ
∞
L {f (t)} =
1
8. f (t) =
∞
1
1
1
1
te−st dt = − te−st − 2 e−st = e−s + 2 e−s ,
s
s
s
s
0,
0<t<1
2t − 2,
t>1
∞
L {f (t)} = 2
−st
(t − 1)e
1
1
1
dt = 2 − (t − 1)e−st − 2 e−st
s
s
∞
2
= 2 e−s ,
s
1
1 − t, 0 < t < 1
0,
t>1
ˆ
1
L {f (t)} =
−st
(1 − t)e
ˆ
∞
dt +
−st
0e
0
ˆ
1
dt =
1
(1 − t)e−st dt
0
1
1 −st 1
1
1
1
−st
+ 2e
= − (1 − t)e
= 2 e−s + − 2 ,
s
s
s
s s
s>0
0
10. f (t) =
0, 0 < t < a
c,
a<t<b
ˆ
b
L {f (t)} =
a
ˆ
b
c
c
ce−st dt = − e−st = (e−sa − e−sb ),
s
s
s>0
a
∞
11. L {f (t)} =
t+7 −st
e
e
ˆ
∞
7
dt = e
0
e(1−s)t dt
0
∞
e7
e7
e7 (1−s)t e
=
,
=
=0−
1−s
1−s
s−1
s>1
0
ˆ
∞
12. L {f (t)} =
e−2t−5 e−st dt = e−5
0
ˆ
∞
e−(s+2)t dt
0
∞
e−5
e−5 −(s+2)t e
,
=−
=
s+2
s+2
s > −2
0
ˆ
∞
13. L {f (t)} =
4t −st
te e
0
=
s>0
1
ˆ
9. f (t) =
Definition of the Laplace Transform
ˆ
dt =
∞
te(4−s)t dt
0
1
1
te(4−s)t −
e(4−s)t
4−s
(4 − s)2
∞
1
,
=
(4 − s)2
0
s>4
s>0
431
432
CHAPTER 7
THE LAPLACE TRANSFORM
ˆ
∞
14. L {f (t)} =
2 −2t −st
t e
e
ˆ
∞
dt =
0
t2 e−(s+2)t dt
0
∞
2
2
1 2 −(s+2)t
−(s+2)t
−(s+2)t −
te
−
e
t e
= −
s+2
(s + 2)2
(s + 2)3
0
2
, s > −2
(s + 2)3
ˆ
ˆ ∞
−t
−st
e (sin t)e dt =
15. L {f (t)} =
=
0
=
∞
(sin t)e−(s+1)t dt
0
1
−(s + 1) −(s+1)t
e
e−(s+1)t cos t
sin t −
2
(s + 1) + 1
(s + 1)2 + 1
∞
0
1
1
= 2
, s > −1
(s + 1)2 + 1
s + 2s + 2
ˆ ∞
ˆ ∞
t
−st
16. L {f (t)} =
e (cos t)e dt =
(cos t)e(1−s)t dt
=
0
=
=−
ˆ
0
1
1−s
e(1−s)t cos t +
e(1−s)t sin t
(1 − s)2 + 1
(1 − s)2 + 1
s−1
1−s
= 2
,
2
(1 − s) + 1
s − 2s + 2
∞
17. L {f (t)} =
=
0
s>1
t(cos t)e−st dt
0
∞
s2 − 1
st
−
− 2
s + 1 (s2 + 1)2
−st
(cos t)e
+
2s
t
+
2
s + 1 (s2 + 1)2
(sin t)e−st
∞
0
s2 − 1
=
, s>0
(s2 + 1)2
ˆ ∞
18. L {f (t)} =
t(sin t)e−st dt
0
=
=
19. L {2t4 } = 2
2s
t
−
− 2
2
s + 1 (s + 1)2
2s
(s2
+ 1)2
,
4
10
−
2
s
s
23. L {t2 + 6t − 3} =
−st
(cos t)e
−
st
s2 − 1
+
s2 + 1 (s2 + 1)2
−st
∞
(sin t)e
0
s>0
4!
48
= 5
5
s
s
21. L {4t − 10} =
2
6
3
+ 2−
3
s
s
s
20. L {t5 } =
5!
120
= 6
6
s
s
22. L {7t + 3} =
7
3
+
2
s
s
24. L {−4t2 + 16t + 9} = −
8
16 9
+ 2 +
3
s
s
s
7.1
Definition of the Laplace Transform
6
6
3
1
+
+
+
s4 s3 s2
s
48 24
6
1
26. L {8t3 − 12t2 + 6t − 1} = 4 − 3 + 2 −
s
s
s
s
25. L {t3 + 3t2 + 3t + 1} =
27. L {1 + e4t } =
1
1
+
s s−4
2
1
1
+
+
s s−2 s−4
29. L {1 + 2e2t + e4t } =
31. L {4t2 − 5 sin 3t} =
28. L {t2 − e−9t + 5} =
30. L {e2t − 2 + e−2t } =
8
15
− 2
3
s
s +9
32. L {cos 5t + sin 2t} =
1
1
1
k
1
kt
−kt
33. L {sinh kt} = L {e − e } =
−
= 2
2
2 s−k s+k
s − k2
2
1
5
+
−
3
s
s+9 s
2
1
1
− +
s−2 s s+2
s2
s
2
+ 2
+ 25 s + 4
1
s
34. L {cosh kt} = L {ekt + e−kt } = 2
2
s − k2
35. L {et sinh t} = L
et
36. L {e−t cosh t} = L
37. L {sin 2t cos 2t} = L
38. L {cos2 t} = L
et − e−t
2
e−t
=L
et + e−t
2
1
sin 4t
2
1 1
+ cos 2t
2 2
=L
=
=
1 2t 1
e −
2
2
=
1 1 −2t
+ e
2 2
1
1
−
2(s − 2) 2s
=
1
1
+
2s 2(s + 2)
2
s2 + 16
1
1 s
+
2s 2 s2 + 4
39. From the addition formula for the sine function, sin (4t + 5) = sin 4t cos 5 + cos 4t sin 5 so
L {sin (4t + 5)} = (cos 5) L {sin 4t} + (sin 5) L {cos 4t}
= (cos 5)
s
4 cos 5 + (sin 5)s
4
+ (sin 5) 2
=
.
s2 + 16
s + 16
s2 + 16
40. From the addition formula for the cosine function,
π
cos t −
6
so
√
π
1
3
π
cos t + sin t
= cos t cos + sin t sin =
6
6
2
2
√
π π 1
3
L 10 cos t −
= 10L cos t −
= 10 ·
L {cos t} + 10 · L {sin t}
6
6
2
2
√
√
s
3s + 1
1
+5 2
=5· 2
.
=5 3 2
s +1
s +1
s +1
41. Use integration by parts for α > 0 to get
∞
ˆ ∞
ˆ ∞
α −t
α −t Γ(α + 1) =
t e dt = −t e + α
tα−1 e−t dt = αΓ(α)
0
0
0
433
434
CHAPTER 7
THE LAPLACE TRANSFORM
42. Let u = st so that du = s dt and we get
ˆ ∞
ˆ ∞
ˆ ∞
u α −u 1
1
1 α −u
du =
tα e−st dt =
e
u e du = α+1 Γ(α + 1)
L {tα } =
α+1
s
s
s
s
0
0
0
Γ 12
1
π
Γ − + 1 = 1/2 =
−1/2+1
2
s
s
s
1 1
√
Γ 32
1
π
1
1/2
2 Γ 2
+ 1 = 3/2 =
44. L t
= 3/2
= 1/2+1 Γ
3/2
2
s
s
s
2s
3 3
√
3
1
1
3
3
1
3 π
3/2
2 Γ 2
+1 =
+ 1 = 5/2 Γ
45. L t
= 5/2 Γ
= 3/2+1 Γ
= 5/2
2
2
2
s
s5/2
2s
4s
4s
Γ 52 + 1
Γ 12 + 1
1/2
5/2
46. L {f (t)} = 2L t
+8·
+ 8L t
=2·
s3/2
s7/2
√
√
8
5
5
8
5 3
3
π
π
= 3/2 + 7/2 · · Γ
= 3/2 + 7/2 · Γ
2
2
2 2
2
s
s
s
s
√
√
1
8
5 3 1
8
5 3 1√
π
π
= 3/2 + 7/2 · · ·
π
= 3/2 + 7/2 · · · Γ
2 2 2
2
2 2 2
s
s
s
s
√
√
π
15 π
= 3/2 + 7/2
s
s
43. L t−1/2 =
1
47. The relation will be valid when s is greater than the maximum of c1 and c2 .
2
48. Since et is an increasing function and t2 > ln M + ct for M > 0 we have et > eln M +ct = M ect
2
for t sufficiently large and for any c. Thus, et is not of exponential order.
49. Assuming that (c) of Theorem 7.1.1 is applicable with a complex exponent, we have
L {e(a+ib)t } =
1
(s − a) + ib
s − a + ib
1
=
=
.
s − (a + ib)
(s − a) − ib (s − a) + ib
(s − a)2 + b2
By Euler’s formula, eiθ = cos θ + i sin θ, so
L {e(a+ib)t } = L {eat eibt } = L {eat (cos bt + i sin bt)}
= L {eat cos bt} + iL {eat sin bt}
=
b
s−a
+i
.
2
2
(s − a) + b
(s − a)2 + b2
Equating real and imaginary parts we get
L {eat cos bt} =
s−a
(s − a)2 + b2
and L {eat sin bt} =
b
.
(s − a)2 + b2
7.1
Definition of the Laplace Transform
50. We want f (αx + βy) = αf (x) + βf (y) or
m(αx + βy) + b = α(mx + b) + β(my + b) = m(αx + βy) + (α + β)b
for all real numbers α and β. Taking α = β = 1 we see that b = 2b, so b = 0. Thus,
f (x) = mx + b will be a linear transformation when b = 0.
51. As written, the function is not defined at x = 5. As should be written, f (x) = 1/(t − 5),
t ≥ 5, the function f is not bounded on, say, the interval [4, 6] because it has an infinite
discontiunity at x = 5.
52. L
1
t2
ˆ
∞
=
0
e−st
dt =
t2
ˆ
0
1
e−st
dt +
t2
ˆ
∞
1
e−st
dt = I1 + I2 .
t2
Then we use the comparison test in three cases: s > 0, s = 0, and s < 0. For s > 0,
ˆ
1
I1 =
0
e−st
dt > e−s
t2
ˆ
1
0
1
dt .
t2
´ 1 −st
dt diverges, the integral 0 e t2 dt diverges.
´1
If s = 0, then the integral I1 = 0 t12 dt diverges. If s < 0, then
Because
´1
1
0 t2
ˆ
1
I1 =
0
Again, because
´1
1
0 t2
e−st
dt >
t2
dt diverges, the integral
´1
0
ˆ
1
0
e−st
t2
1
dt .
t2
dt diverges.
53. Applying the definition of the Laplace transform gives us
ˆ ∞
2
2
t2
t2
=
e−st 2tet cos et dt
L 2te cos e
0
= e−st
∞
ˆ ∞
2
sin e + s
e−st sin et dt
0
t2
0
ˆ
= − sin 1 + s
∞
e−st sin et dt
2
0
The last integral exists for s > 0 since it’s piecewise continuous on (0, ∞) and of exponential
order. Alternatively, for s > 0
ˆ ∞
ˆ ∞
1
−st
t2 e−st dt =
e sin e dt ≤ 1 ·
s
0
0
The absolute convergence of the integral
integral.
´∞
0
e−st sin et dt implies the convergence of the
2
435
436
CHAPTER 7
THE LAPLACE TRANSFORM
54. Working backwards we get
1
F
a
s
1
= ·
a
a
ˆ
1
= ·a
a
∞
e−(s/a)t f (t) dt
0
ˆ
∞
e−su f (au) du
←−
t = at,
dt = a du
0
= L {f (at)}
1
1
1
=
55. L eat = ·
a (s/a − 1)
s−a
56. L {sin kt} =
1
1
k2
k
1
·
=
·
= 2
k (s/k)2 + 1
k s2 + k 2
s + k2
57. L {1 − cos kt} =
1
1
1
k3
k2
·
=
·
=
k (s/k) [(s/k)2 + 1]
k s (s2 + k 2 )
s (s2 + k 2 )
58. L {sin kt sinh kt} =
7.2
2(s/k)
1
2sk 3
2k 3 s
1
·
=
·
=
k (s/k)4 + 4
k s4 + 4k 4
s4 + 4k 4
The Inverse Transform and Transforms of Derivatives
1. L −1
1
s3
1
= L −1
2
2
s3
1
= t2
2
2. L −1
1
s4
1
= L −1
6
3!
s4
1
= t3
6
3. L −1
1
48
− 5
2
s
s
4. L −1
2
1
− 3
s s
1
48 4!
·
−
2
s
24 s5
= L −1
2 = L −1 4 ·
= t − 2t4
1
5!
4 3!
1
· 6
− · 4+
2
s
6 s
120 s
5. L −1
(s + 1)3
s4
= L −1
1
3 2
1 3!
1
+3· 2 + · 3 + · 4
s
s
2 s
6 s
6. L −1
(s + 2)2
s3
= L −1
1
1
2
+4· 2 +2· 3
s
s
s
7. L −1
1
1
1
− +
2
s
s s−2
= t − 1 + e2t
8. L −1
6
4
1
+ 5−
s s
s+8
= L −1 4 ·
9. L −1
1
4s + 1
1
= L −1
4
1
s + 1/4
1
3
= 1 + 3t + t2 + t3
2
6
= 1 + 4t + 2t2
1 1 4!
1
+ · 5−
s 4 s
s+8
1
= e−t/4
4
2
1 5
= 4t − t3 +
t
3
120
1
= 4 + t4 − e−8t
4
7.2
10. L −1
1
5s − 2
11. L −1
5
s2 + 49
12. L −1
13. L −1
14. L −1
= L −1
= L −1
1
1
·
5 s − 2/5
= 10 cos 4t
4s
+1
= L −1
1
4s2 + 1
= L −1
4s2
2s − 6
s2 + 9
1
= e2t/5
5
7
5
·
7 s2 + 49
10s
+ 16
s2
The Inverse Transform and Transforms of Derivatives
s2
s
+ 1/4
=
5
sin 7t
7
1
= cos t
2
1
1/2
·
2 s2 + 1/4
=
1
1
sin t
2
2
s
3
−2· 2
= 2 cos 3t − 2 sin 3t
+9
s +9
√
√ √
√
1
s
2
2
−1
2
−1
√
f racs + 1s + 2 = L
+
=
cos
sin
16. L
2
t
+
2t
s2 + 2
2
2 s2 + 2
15. L −1
17. L −1
18. L −1
s2
1
+ 3s
s+1
s2 − 4s
= L −1 2 ·
= L −1
s2
1 1 1
1
· − ·
3 s 3 s+3
1
1 1 5
= L −1 − · + ·
4 s 4 s−4
=
1 1 −3t
− e
3 3
1 5
= − + e4t
4 4
s
+ 2s − 3
= L −1
1
3
1
1
·
+ ·
4 s−1 4 s+3
1
3
= et + e−3t
4
4
20. L −1
1
s2 + s − 20
= L −1
1
1
1
1
·
− ·
9 s−4 9 s+5
1
1
= e4t − e−5t
9
9
21. L −1
0.9s
(s − 0.1)(s + 0.2)
22. L −1
s−3
√
√
(s − 3 )(s + 3 )
23. L −1
s
(s − 2)(s − 3)(s − 6)
24. L −1
s2 + 1
s(s − 1)(s + 1)(s − 2)
19. L −1
s2
1
1
+ (0.6) ·
= 0.3e0.1t + 0.6e−0.2t
s − 0.1
s + 0.2
√ √
√
√
√
s
3
−1
=L
−
=
cosh
3
·
3
t
−
3
sinh
3t
s2 − 3
s2 − 3
= L −1 (0.3) ·
= L −1
=
25. L −1
1
s3 + 5s
= L −1
1
1
1
1
1
·
−
+ ·
2 s−2 s−3 2 s−6
= L −1
=
1 2t
1
e − e3t + e6t
2
2
1 1
1
1
1
5
1
· −
− ·
+ ·
2 s s−1 3 s+1 6 s−2
1
5
1
− et − e−t + e2t
2
3
6
1
s(s2 + 5)
= L −1
1 1 1 s
· −
5 s 5 s2 + 5
=
√
1 1
− cos 5 t
5 5
437
438
CHAPTER 7
26. L −1
(s2
THE LAPLACE TRANSFORM
s
+ 4)(s + 2)
=
27. L −1
(s2
s
1
2
1
1
1
· 2
+ · 2
− ·
4 s +4 4 s +4 4 s+2
= L −1
2s − 4
+ s)(s2 + 1)
1
1
1
cos 2t + sin 2t − e−2t
4
4
4
2s − 4
s(s + 1)(s2 + 1)
= L −1
4
3
s
3
+ 2
+ 2
= L −1 − +
s s+1 s +1 s +1
= −4 + 3e−t + cos t + 3 sin t
28. L
−1
29. L −1
30. L −1
1
s4 − 9
(s2
=L
−1
1
+ 1)(s2 + 4)
6s + 3
(s2 + 1)(s2 + 4)
√
√ √
√
1
1
1
1
3
3
√ · 2
− √ · 2
= √ sinh 3 t − √ sin 3 t
6 3 s −3 6 3 s +3
6 3
6 3
= L −1
1
1
1
1
· 2
− · 2
3 s +1 3 s +4
= L −1
1
1
1
2
·
− ·
3 s2 + 1 6 s2 + 4
= L −1 2 ·
31. Since f (t) = eat sinh bt = eat 21 ebt − e−bt =
(c) of Theorem 7.1.1
1 (a+b)t 1 (a−b)t
e
− e
2
2
L −1
1
(s − a)2 − b2
1
1
sin t − sin 2t
3
6
1
s
1
2
s
+
−2· 2
− ·
s2 + 1 s2 + 1
s + 4 2 s2 + 4
= 2 cos t + sin t − 2 cos 2t −
L
=
1
2
1
sin 2t
2
e(1+b)t −
1
2
e(a−b)t , then by linearity and part
=
1
1
1
1
b
−
=
2 s − (a + b) 2 s − (a − b)
((s − a) − b) ((s − a) + b)
=
b
(s − a)2 − b2
=
1 at
e sinh bt
b
32. By linearity and part (d) of Theorem 7.1.1:
L {at − sin at} =
L −1
1
+ a2 )
s2 (s2
=
a
a3
a
−
=
s2 s2 + b2
s2 (s2 + a2 )
at − sin at
a3
7.2
The Inverse Transform and Transforms of Derivatives
33. By linearity and part (d) of Theorem 7.1.1:
a
a3 b − ab3
b
−b 2
= 2
2
2
+b
s +a
(s + a2 ) (s2 + b2 )
ab a2 − b2
= 2
(s + a2 ) (s2 + b2 )
L {a sin bt − b sin at} = a
L −1
1
(s2
+
a2 ) (s2
+
=
b2 )
s2
a sin bt − b sin at
ab (a2 − b2 )
34. By linearity and part (e) of Theorem 7.1.1:
s a2 − b2
s
s
−
= 2
L {cos bt − cos at} = 2
s + b2 s2 + a2
(s + a2 ) (s2 + b2 )
L −1
s
(s2 + a2 ) (s2 + b2 )
=
cos bt − cos at
a2 − b2
35. The Laplace transform of the differential equation is
sL {y} − y(0) − L {y} =
Solving for L {y} we obtain
1
.
s
1
1
.
L {y} = − +
s s−1
Thus
y = −1 + et .
36. The Laplace transform of the differential equation is
2sL {y} − 2y(0) + L {y} = 0.
Solving for L {y} we obtain
L {y} =
−3
−6
=
.
2s + 1
s + 1/2
Thus
y = −3e−t/2 .
37. The Laplace transform of the differential equation is
sL {y} − y(0) + 6L {y} =
1
.
s−4
Solving for L {y} we obtain
L {y} =
2
1
1
19
1
1
+
=
·
+
·
.
(s − 4)(s + 6) s + 6
10 s − 4 10 s + 6
Thus
y=
1 4t 19 −6t
e + e .
10
10
439
440
CHAPTER 7
THE LAPLACE TRANSFORM
38. The Laplace transform of the differential equation is
sL {y} − L {y} =
s2
2s
.
+ 25
Solving for L {y} we obtain
L {y} =
1
1
1
s
5
5
2s
=
·
−
+
· 2
.
2
2
(s − 1)(s + 25)
13 s − 1 13 s + 25 13 s + 25
Thus
y=
5
1
1 t
e −
cos 5t +
sin 5t.
13
13
13
39. The Laplace transform of the differential equation is
s2 L {y} − sy(0) − y (0) + 5 [sL {y} − y(0)] + 4L {y} = 0.
Solving for L {y} we obtain
L {y} =
s+5
4 1
1 1
=
−
.
s2 + 5s + 4
3 s+1 3 s+4
Thus
1
4
y = e−t − e−4t .
3
3
40. The Laplace transform of the differential equation is
s2 L {y} − sy(0) − y (0) − 4 [sL {y} − y(0)] =
3
6
−
.
s−3 s+1
Solving for L {y} we obtain
L {y} =
=
3
s−5
6
−
+
(s − 3)(s2 − 4s) (s + 1)(s2 − 4s) s2 − 4s
2
3
1
11
1
5 1
· −
− ·
+
·
.
2 s s − 3 5 s + 1 10 s − 4
Thus
y=
3
11
5
− 2e3t − e−t + e4t .
2
5
10
41. The Laplace transform of the differential equation is
s2 L {y} − sy(0) − y (0) + L {y} =
s2
2
.
+2
Solving for L {y} we obtain
L {y} =
(s2
10s
10s
2
2
2
+ 2
= 2
+ 2
− 2
.
2
+ 1)(s + 2) s + 1
s +1 s +1 s +2
Thus
y = 10 cos t + 2 sin t −
√
√
2 sin 2 t.
7.2
The Inverse Transform and Transforms of Derivatives
42. The Laplace transform of the differential equation is
s2 L {y} + 9L {y} =
1
.
s−1
Solving for L {y} we obtain
L {y} =
1
1
1
1
1
1
s
=
·
−
·
−
·
.
(s − 1)(s2 + 9)
10 s − 1 10 s2 + 9 10 s2 + 9
Thus
y=
1 t
1
1
e −
sin 3t −
cos 3t.
10
30
10
43. The Laplace transform of the differential equation is
2 s3 L {y} − s2 (0) − sy (0) − y (0) + 3 s2 L {y} − sy(0) − y (0) − 3[sL {y} − y(0)] − 2L {y}
=
1
.
s+1
Solving for L {y} we obtain
L {y} =
1 1
5 1
8
1
1 1
2s + 3
=
+
−
+
.
(s + 1)(s − 1)(2s + 1)(s + 2)
2 s + 1 18 s − 1 9 s + 1/2 9 s + 2
Thus
1
5
8
1
y = e−t + et − e−t/2 + e−2t .
2
18
9
9
44. The Laplace transform of the differential equation is
s3 L {y} − s2 (0) − sy (0) − y (0) + 2 s2 L {y} − sy(0) − y (0) − [sL {y} − y(0)] − 2L {y}
=
3
.
s2 + 9
Solving for L {y} we obtain
L {y} =
=
s2 + 12
(s − 1)(s + 1)(s + 2)(s2 + 9)
13 1
16 1
3
s
1
3
13 1
−
+
+
−
.
2
2
60 s − 1 20 s + 1 39 s + 2 130 s + 9 65 s + 9
Thus
y=
1
3
13 t 13 −t 16 −2t
e − e + e
cos 3t −
sin 3t.
+
60
20
39
130
65
441
442
CHAPTER 7
THE LAPLACE TRANSFORM
45. The Laplace transform of the differential equation is
sL {y} + L {y} =
s2
s+3
.
+ 6s + 13
Solving for L {y} we obtain
1
1
1
s+1
s+3
= ·
− · 2
2
(s + 1)(s + 6s + 13)
4 s + 1 4 s + 6s + 13
1
1
s+3
2
1
−
−
.
= ·
4 s + 1 4 (s + 3)2 + 4 (s + 3)2 + 4
L {y} =
Thus
1
1
1
y = e−t − e−3t cos 2t + e−3t sin 2t.
4
4
4
46. The Laplace transform of the differential equation is
s2 L {y} − s · 1 − 3 − 2[sL {y} − 1] + 5L {y} = (s2 − 2s + 5)L {y} − s − 1 = 0.
Solving for L {y} we obtain
L {y} =
s2
s−1+2
s−1
2
s+1
=
=
+
.
2
2
2
2
− 2s + 5
(s − 1) + 2
(s − 1) + 2
(s − 1)2 + 22
Thus
y = et cos 2t + et sin 2t.
47. The Laplace transform of the differential equation in the initial-value problem is
2
s + 4 Y (s) =
Y (s) =
10s
+ 25
s2
10s
(s2 + 4) (s2 + 25)
with the identifications a2 = 4 and b2 = 25 or a2 − b2 = −21 we have from Problem 34:
L −1
s
(s2 + a2 ) (s2 + b2 )
=
cos bt − cos at
a2 − b2
y(t) = 10L −1
y(t) =
s
(s2 + 4) (s2 + 25)
= 10 ·
cos 5t − cos 2t
−21
10
10
cos 2t −
cos 5t
21
21
48. The Laplace transform of the differential equation in the initial-value problem is
2
4
s + 2 Y (s) = 2
s
Y (s) =
1
s2 (s2
+ 2)
7.2
The Inverse Transform and Transforms of Derivatives
√
with the identification a2 = 2 and a3 = 2 2 we have from problem 44:
L −1
1
+ a2 )
s2 (s2
=
at − sin at
a3
y(t) = 4L
1
−1
s2 (s2 + 2)
√
√
y(t) = 2t − 2 sin 2 t.
√
=4·
2 t − sin
√
2 2
√
2t
49. (a) Differentiating f (t) = teat we get f (t) = ateat +eat so L {ateat +eat } = sL {teat }, where
we have used f (0) = 0. Writing the equation as
aL {teat } + L eat = sL teat
and solving for L {teat } we get
L teat =
1
1
L eat =
.
s−a
(s − a)2
(b) Starting with f (t) = t sin kt we have
f (t) = kt cos kt + sin kt
f (t) = −k 2 t sin kt + 2k cos kt.
Then
L −k 2 t sin t + 2k cos kt = s2 L {t sin kt}
where we have used f (0) = 0 and f (0) = 0. Writing the above equation as
−k 2 L {t sin kt} + 2kL {cos kt} = s2 L {t sin kt}
and solving for L {t sin kt} gives
L {t sin kt} =
s2
2k
2ks
s
2k
L {cos kt} = 2
= 2
.
2
2
2
2
+k
s +k s +k
(s + k 2 )2
50. Let f1 (t) = 1 and
f2 (t) =
1,
0,
t ≥ 0, t = 1
t=1
Then L {f1 (t)} = L {f2 (t)} = 1/s, but f1 (t) = f2 (t).
51. For y −4y = 6e3t −3e−t the transfer function is W (s) = 1/(s2 −4s). The zero-input response
is
1
5 1
s−5
5 1 1
= L −1
· − ·
= − e4t ,
y0 (t) = L −1
s2 − 4s
4 s 4 s−4
4 4
443
444
CHAPTER 7
THE LAPLACE TRANSFORM
and the zero-state response is
y1 (t) = L −1
= L −1
=
3
6
−
2
(s − 3)(s − 4s) (s + 1)(s2 − 4s)
1
2
5 1 3
1
27
·
−
+ · − ·
20 s − 4 s − 3 4 s 5 s + 1
27 4t
5 3
e − 2e3t + − e−t .
20
4 5
52. From Theorem 7.2.2, if f and f are continuous and of exponential order,
L {f (t)} = sF (s) − f (0). From Theorem 7.2.3, lim L {f (t)} = 0 so
s→∞
lim [sF (s) − f (0)] = 0
and
s→∞
For f (t) = cos kt,
lim sF (s) = lim s
s→∞
7.3
s→∞
s2
lim F (s) = f (0).
s→∞
s
= 1 = f (0).
+ k2
Operational Properties I
1. L te
10t
2. L te−6t =
1
=
(s − 10)2
1
(s + 6)2
3. L t3 e−2t =
3!
10!
4. L t10 e−7t =
4
(s + 2)
(s + 7)11
2 2
1
1
= L te2t + 2te3t + te4t =
5. L t et + e2t
+
+
2
2
(s − 2)
(s − 3)
(s − 4)2
6. L e2t (t − 1)2 = L t2 e2t − 2te2t + e2t =
2
1
2
−
+
3
2
(s − 2)
(s − 2)
s−2
3
s+2
8. L e−2t cos 4t =
2
(s − 1) + 9
(s + 2)2 + 16
9. L (1 − et + 3e−4t ) cos 5t = L cos 5t − et cos 5t + 3e−4t cos 5t
7. L et sin 3t =
=
10. L
t
9 − 4t + 10 sin
e
2
3t
s2
=L
=
11. L −1
1
(s + 2)3
= L −1
s−1
3(s + 4)
s
−
+
2
+ 25 (s − 1) + 25 (s + 4)2 + 25
9e3t − 4te3t + 10e3t sin
t
2
4
5
9
−
+
s − 3 (s − 3)2 (s − 3)2 + 1/4
2
1
2 (s + 2)3
1
= t2 e−2t
2
7.3
12. L −1
1
(s − 1)4
13. L −1
1
s2 − 6s + 10
14. L −1
15. L −1
16. L −1
1
= L −1
6
3!
(s − 1)4
= L −1
1
= t3 et
6
1
(s − 3)2 + 12
= e3t sin t
s2
1
+ 2s + 5
= L −1
2
1
2 (s + 1)2 + 22
s2
s
+ 4s + 5
= L −1
s+2
1
−2
2
2
(s + 2) + 1
(s + 2)2 + 12
2s + 5
s2 + 6s + 34
= L −1 2
Operational Properties I
1
= e−t sin 2t
2
= e−2t cos t − 2e−2t sin t
5
(s + 3)
1
−
(s + 3)2 + 52 5 (s + 3)2 + 52
1
= 2e−3t cos 5t − e−3t sin 5t
5
17. L −1
s
(s + 1)2
= L −1
s+1−1
(s + 1)2
18. L −1
5s
(s − 2)2
= L −1
5(s − 2) + 10
(s − 2)2
19. L −1
2s − 1
+ 1)3
s2 (s
= L −1
1
1
−
s + 1 (s + 1)2
= L −1
= e−t − te−t
5
10
+
s − 2 (s − 2)2
= 5e2t + 10te2t
5
1
4
2
5
3
− 2−
−
−
2
s s
s + 1 (s + 1)
2 (s + 1)3
= L −1
3
= 5 − t − 5e−t − 4te−t − t2 e−t
2
20. L −1
(s + 1)2
(s + 2)4
= L −1
3!
1
2
1
−
+
2
3
(s + 2)
(s + 2)
6 (s + 2)4
1
= te−2t − t2 e−2t + t3 e−2t
6
21. The Laplace transform of the differential equation is
sL {y} − y(0) + 4L {y} =
Solving for L {y} we obtain
L {y} =
1
.
s+4
2
1
.
+
2
(s + 4)
s+4
Thus
y = te−4t + 2e−4t .
22. The Laplace transform of the differential equation is
1
1
sL {y} − L {y} = +
.
s (s − 1)2
Solving for L {y} we obtain
L {y} =
Thus
1
1
1
1
1
+
+
=− +
.
3
s(s − 1) (s − 1)
s s − 1 (s − 1)3
1
y = −1 + et + t2 et .
2
445
446
CHAPTER 7
THE LAPLACE TRANSFORM
23. The Laplace transform of the differential equation is
s2 L {y} − sy(0) − y (0) + 2 [sL {y} − y(0)] + L {y} = 0.
Solving for L {y} we obtain
L {y} =
2
1
s+3
+
=
.
2
(s + 1)
s + 1 (s + 1)2
Thus
y = e−t + 2te−t .
24. The Laplace transform of the differential equation is
s2 L {y} − sy(0) − y (0) − 4 [sL {y} − y(0)] + 4L {y} =
6
.
(s − 2)4
Solving for L {y} we obtain
L {y} =
Thus, y =
5!
1
.
20 (s − 2)6
1 5 2t
20 t e .
25. The Laplace transform of the differential equation is
s2 L {y} − sy(0) − y (0) − 6 [sL {y} − y(0)] + 9L {y} =
1
.
s2
Solving for L {y} we obtain
L {y} =
10
1
2 1 1 1
2 1
1 + s2
+
+
=
−
.
2
2
2
s (s − 3)
27 s 9 s
27 s − 3
9 (s − 3)2
Thus
y=
1
2
10
2
+ t − e3t + te3t .
27 9
27
9
26. The Laplace transform of the differential equation is
s2 L {y} − sy(0) − y (0) − 4 [sL {y} − y(0)] + 4L {y} =
6
.
s4
Solving for L {y} we obtain
L {y} =
13
1
31 9 1
3 2
1 3!
1 1
s5 − 4s4 + 6
+
−
=
+
+
+
.
4
2
2
3
4
s (s − 2)
4 s 8s
4s
4s
4 s−2
8 (s − 2)2
Thus
y=
3
1
1
13
3 9
+ t + t2 + t3 + e2t − te2t .
4 8
4
4
4
8
7.3
Operational Properties I
27. The Laplace transform of the differential equation is
s2 L {y} − sy(0) − y (0) − 6 [sL {y} − y(0)] + 13L {y} = 0.
Solving for L {y} we obtain
L {y} = −
3
3
2
.
=−
s2 − 6s + 13
2 (s − 3)2 + 22
Thus
3
y = − e3t sin 2t.
2
28. The Laplace transform of the differential equation is
2 s2 L {y} − sy(0) + 20 [sL {y} − y(0)] + 51L {y} = 0.
Solving for L {y} we obtain
L {y} =
2s2
4s + 40
2s + 20
2(s + 5)
10
=
=
+
.
2
2
+ 20s + 51
(s + 5) + 1/2
(s + 5) + 1/2 (s + 5)2 + 1/2
Thus
√
√
√
y = 2e−5t cos (t/ 2 ) + 10 2 e−5t sin (t/ 2 ).
29. The Laplace transform of the differential equation is
s2 L {y} − sy(0) − y (0) − [sL {y} − y(0)] =
s−1
.
(s − 1)2 + 1
Solving for L {y} we obtain
L {y} =
11 1
s−1
1
1
1
=
−
+
.
s(s2 − 2s + 2)
2 s 2 (s − 1)2 + 1 2 (s − 1)2 + 1
Thus
y=
1
1 1 t
− e cos t + et sin t.
2 2
2
30. The Laplace transform of the differential equation is
s2 L {y} − sy(0) − y (0) − 2 [sL {y} − y(0)] + 5L {y} =
1
1
+ 2.
s s
Solving for L {y} we obtain
L {y} =
=
7 1 1 1
−7s/25 + 109/25
4s2 + s + 1
=
+
+
s2 (s2 − 2s + 5)
25 s 5 s2
s2 − 2s + 5
s−1
2
7
51
7 1 1 1
+
−
+
.
2
2
2
25 s 5 s
25 (s − 1) + 2
25 (s − 1)2 + 22
Thus
y=
1
7
51
7
+ t − et cos 2t + et sin 2t.
25 5
25
25
447
448
CHAPTER 7
THE LAPLACE TRANSFORM
31. Taking the Laplace transform of both sides of the differential equation and letting c = y(0)
we obtain
L {y } + L {2y } + L {y} = 0
s2 L {y} − sy(0) − y (0) + 2sL {y} − 2y(0) + L {y} = 0
s2 L {y} − cs − 2 + 2sL {y} − 2c + L {y} = 0
2
s + 2s + 1 L {y} = cs + 2c + 2
L {y} =
2c + 2
cs
+
(s + 1)2 (s + 1)2
=c
=
s+1−1
2c + 2
+
2
(s + 1)
(s + 1)2
c
c+2
+
.
s + 1 (s + 1)2
Therefore,
y(t) = cL −1
1
s+1
+ (c + 2)L −1
1
(s + 1)2
= ce−t + (c + 2)te−t .
To find c we let y(1) = 2. Then 2 = ce−1 + (c + 2)e−1 = 2(c + 1)e−1 and c = e − 1. Thus
y(t) = (e − 1)e−t + (e + 1)te−t .
32. Taking the Laplace transform of both sides of the differential equation and letting c = y (0)
we obtain
L {y } + L {8y } + L {20y} = 0
s2 L {y} − y (0) + 8sL {y} + 20L {y} = 0
s2 L {y} − c + 8sL {y} + 20L {y} = 0
(s2 + 8s + 20)L {y} = c
L {y} =
c
c
=
.
s2 + 8s + 20
(s + 4)2 + 4
Therefore,
y(t) = L −1
c
(s + 4)2 + 4
=
c −4t
e sin 2t = c1 e−4t sin 2t.
2
To find c1 we let y (π) = 0. Then 0 = y (π) = c1 e−4π and c1 = 0. Thus, y(t) = 0. (Since
the differential equation is homogeneous and both boundary conditions are 0, we can see
immediately that y(t) = 0 is a solution. We have shown that it is the only solution.)
33. Recall from Section 3.8 that mx = −kx−βx . Now β = 7/8, m = W/g = 1.225/9.8 = 0.125 kg,
and 1.225 = 0.6125k so that k = 2 N/m. Thus, the differential equation is x + 7x + 16x = 0.
7.3
Operational Properties I
The initial conditions are x(0) = −3/2 and x (0) = 0. The Laplace transform of the differential
equation is
3
21
+ 16L {x} = 0.
s2 L {x} + s + 7sL {x} +
2
2
Solving for L {x} we obtain
√
√
−3s/2 − 21/2
3
s + 7/2
7 15
15/2
√
√
L {x} = 2
=−
−
.
2
2
2
s + 7s + 16
2 (s + 7/2) + ( 15/2)
10 (s + 7/2) + ( 15/2)2
Thus
√
√
√
3 −7t/2
7 15 −7t/2
15
15
x=− e
t−
e
t.
cos
sin
2
2
10
2
34. The differential equation is
d2 q
dq
+ 20 + 200q = 150,
2
dt
dt
q(0) = q (0) = 0.
The Laplace transform of this equation is
s2 L {q} + 20sL {q} + 200L {q} =
150
.
s
Solving for L {q} we obtain
L {q} =
s(s2
31 3
s + 10
3
10
150
=
−
−
.
2
2
+ 20s + 200)
4 s 4 (s + 10) + 10
4 (s + 10)2 + 102
Thus
q(t) =
3
3 3 −10t
− e
cos 10t − e−10t sin 10t
4 4
4
and
i(t) = q (t) = 15e−10t sin 10t.
449
450
CHAPTER 7
THE LAPLACE TRANSFORM
35. The differential equation is
d2 q
E0
,
+ 2λdqdt + ω 2 q =
2
dt
L
q(0) = q (0) = 0.
The Laplace transform of this equation is
s2 L {q} + 2λsL {q} + ω 2 L {q} =
or
E0 1
L s
2
E0 1
.
s + 2λs + ω 2 L {q} =
L s
Solving for L {q} and using partial fractions we obtain
E0
L {q} =
L
1/ω 2 (1/ω 2 )s + 2λ/ω 2
− 2
s
s + 2λs + ω 2
E0
=
Lω 2
1
s + 2λ
− 2
s s + 2λs + ω 2
.
For λ > ω we write s2 + 2λs + ω 2 = (s + λ)2 − λ2 − ω 2 , so (recalling that ω 2 = 1/LC)
L {q} = E0 C
1
s+λ
λ
−
−
2
2
2
2
s (s + λ) − (λ − ω ) (s + λ) − (λ2 − ω 2 )
.
Thus for λ > ω,
λ
−λt
cosh λ2 − ω 2 t − √
q(t) = E0 C 1 − e
sinh λ2 − ω 2 t .
λ2 − ω 2
For λ < ω we write s2 + 2λs + ω 2 = (s + λ)2 + ω 2 − λ2 , so
L {q} = E0 C
s+λ
λ
1
−
−
s (s + λ)2 + (ω 2 − λ2 ) (s + λ)2 + (ω 2 − λ2 )
.
Thus for λ < ω,
λ
−λt
2
2
2
2
cos ω − λ t − √
q(t) = E0 C 1 − e
sin ω − λ t .
ω 2 − λ2
For λ = ω, s2 + 2λ + ω 2 = (s + λ)2 and
1
E0
E0
=
L {q} =
2
L s(s + λ)
L
1/λ2
1/λ2
1/λ
−
−
s
s + λ (s + λ)2
E0
=
Lλ2
Thus for λ = ω,
q(t) = E0 C
1 − e−λt − λte−λt .
1
1
λ
−
−
s s + λ (s + λ)2
.
7.3
Operational Properties I
36. The differential equation is
R
1
dq
+ q = E0 e−kt , q(0) = 0.
dt
C
The Laplace transform of this equation is
RsL {q} +
1
1
L {q} = E0
.
C
s+k
Solving for L {q} we obtain
L {q} =
E0 /R
E0 C
=
.
(s + k)(RCs + 1)
(s + k)(s + 1/RC)
When 1/RC = k we have by partial fractions
E0
1
1
1
E0 1/(1/RC − k) 1/(1/RC − k)
−
=
−
.
L {q} =
R
s+k
s + 1/RC
R 1/RC − k s + k s + 1/RC
Thus
q(t) =
E0 C
e−kt − e−t/RC .
1 − kRC
When 1/RC = k we have
L {q} =
Thus
q(t) =
1
E0
.
R (s + k)2
E0 −kt E0 −t/RC
te
te
=
.
R
R
e−s
s2
e−2s
38. L e2−t U (t − 2) = L e−(t−2) U (t − 2) =
s+1
37. L {(t − 1)U (t − 1)} =
39. L {tU (t − 2)} = L {(t − 2)U (t − 2) + 2U (t − 2)} =
e−2s 2e−2s
+
s2
s
Alternatively, (16) of this section could be used:
−2s
L {tU (t − 2)} = e
−2s
L {t + 2} = e
1
2
+
2
s
s
40. L {(3t + 1)U (t − 1)} = 3L {(t − 1)U (t − 1)} + 4L {U (t − 1)} =
.
3e−s 4e−s
+
s2
s
Alternatively, (16) of this section could be used:
−s
L {(3t + 1)U (t − 1)} = e
−s
L {3t + 4} = e
3
4
+
2
s
s
.
451
452
CHAPTER 7
THE LAPLACE TRANSFORM
41. L {cos 2tU (t − π)} = L {cos 2(t − π)U (t − π)} =
se−πs
s2 + 4
Alternatively, (16) of this section could be used:
L {cos 2tU (t − π)} = e−πs L {cos 2(t + π)} = e−πs L {cos 2t} = e−πs
sin tU
s2
s
.
+4
π π
π se−πs/2
= L cos t −
U t−
= 2
2
2
2
s +1
Alternatively, (16) of this section could be used:
π s
π = e−πs/2 L sin t +
= e−πs/2 L {cos t} = e−πs/2 2
.
L sin t U t −
2
2
s +1
42. L
t−
43. L −1
e−2s
s3
44. L −1
(1 + e−2s )2
s+2
= L −1
1 2 −2s
· e
2 s3
1
= (t − 2)2 U (t − 2)
2
1
2e−2s
e−4s
+
+
s+2
s+2
s+2
= L −1
= e−2t + 2e−2(t−2) U (t − 2) + e−2(t−4) U (t − 4)
e−πs
s2 + 1
45. L −1
= sin (t − π)U (t − π) = − sin tU (t − π)
46. L −1
se−πs/2
s2 + 4
47. L −1
e−s
s(s + 1)
48. L −1
e−2s
s2 (s − 1)
= cos 2 t −
= L −1
π
U
2
t−
e−s
e−s
−
s
s+1
= L −1 −
π
2
= − cos 2tU
t−
π
2
= U (t − 1) − e−(t−1) U (t − 1)
e−2s e−2s
e−2s
− 2 +
s
s
s−1
= −U (t − 2) − (t − 2)U (t − 2) + et−2 U (t − 2)
49. (c)
50. (e)
51. (f )
55. L {2 − 4U (t − 3)} =
52. (b)
53. (a)
54. (d)
2 4 −3s
− e
s s
1 e−4s e−5s
−
+
s
s
s
2
57. L t U (t − 1) = L (t − 1)2 + 2t − 1 U (t − 1)
56. L {1 − U (t − 4) + U (t − 5)} =
=L
(t − 1) + 2(t − 1) − 1 U (t − 1) =
2
2
2
1
+
+
s3 s2
s
Alternatively, by (16) of this section,
−s
L {t U (t − 1)} = e
2
−s
L {t + 2t + 1} = e
2
2
2
1
+ 2+
3
s
s
s
.
e−s
7.3
58. L
3π
sin tU t −
2
=L
3π
3π
− cos t −
U t−
2
2
Operational Properties I
=−
59. L {t − tU (t − 2)} = L {t − (t − 2)U (t − 2) − 2U (t − 2)} =
1
e−2s 2e−2s
−
−
s2
s2
s
60. L {sin t − sin tU (t − 2π)} = L {sin t − sin (t − 2π)U (t − 2π)} =
61. L {f (t)} = L {U (t − a) − U (t − b)} =
e−2πs
1
−
s2 + 1 s2 + 1
e−as e−bs
−
s
s
62. L {f (t)} = L {U (t − 1) + U (t − 2) + U (t − 3) + · · · } =
=
se−3πs/2
s2 + 1
e−s e−2s e−3s
+
+
+ ···
s
s
s
1 e−s
s 1 − e−s
63. The Laplace transform of the differential equation is
5
sL {y} − y(0) + L {y} = e−s .
s
Solving for L {y} we obtain
L {y} =
1
1
5e−s
= 5e−s
−
.
s(s + 1)
s s+1
Thus
y = 5U (t − 1) − 5e−(t−1) U (t − 1).
64. The Laplace transform of the differential equation is
sL {y} − y(0) + L {y} =
1 2 −s
− e .
s s
Solving for L {y} we obtain
2e−s
1
1
1
1
−s 1
−
= −
− 2e
−
.
L {y} =
s(s + 1) s(s + 1)
s s+1
s s+1
Thus
y = 1 − e−t − 2 1 − e−(t−1) U (t − 1).
65. The Laplace transform of the differential equation is
sL {y} − y(0) + 2L {y} =
s+1
1
− e−s 2 .
2
s
s
Solving for L {y} we obtain
11 1 1
1 1
1 1
1 1
1
−s s + 1
−s 1 1
−e
=−
+
−e
+
+
−
L {y} = 2
.
2
2
2
s (s + 2)
s (s + 2)
4 s 2s
4 s+2
4 s 2s
4 s+2
Thus
1 −2t
1
1 1
1 1
+ (t − 1) − e−2(t−1) U (t − 1).
−
y =− + t+ e
4 2
4
4 2
4
453
454
CHAPTER 7
THE LAPLACE TRANSFORM
66. The Laplace transform of the differential equation is
s2 L {y} − sy(0) − y (0) + 4L {y} =
1 e−s
−
.
s
s
Solving for L {y} we obtain
1−s
11 1 s
1 2
1 s
1
−s
−s 1 1
L {y} =
−e
=
−
−
−e
−
.
2
2
2
2
s(s + 4)
s(s + 4)
4 s 4 s +4 2 s +4
4 s 4 s2 + 4
Thus
1
1 1
1 1
− cos 2(t − 1) U (t − 1).
y = − cos 2t − sin 2t −
4 4
2
4 4
67. The Laplace transform of the differential equation is
s2 L {y} − sy(0) − y (0) + 4L {y} = e−2πs
s2
1
.
+1
Solving for L {y} we obtain
1
1 2
s
−2πs 1
+e
−
.
L {y} = 2
2
s +4
3 s + 1 6 s2 + 4
Thus
y = cos 2t +
1
1
sin (t − 2π) − sin 2(t − 2π) U (t − 2π).
3
6
68. The Laplace transform of the differential equation is
s2 L {y} − sy(0) − y (0) − 5 [sL {y} − y(0)] + 6L {y} =
e−s
.
s
Solving for L {y} we obtain
1
1
+
s(s − 2)(s − 3) (s − 2)(s − 3)
1 1
1 1
1
1
−s 1 1
−
+
−
+
.
=e
6 s 2 s−2 3 s−3
s−2 s−3
L {y} = e−s
Thus
y=
1 1 2(t−1) 1 3(t−1)
U (t − 1) − e2t + e3t .
− e
+ e
6 2
3
69. The Laplace transform of the differential equation is
s2 L {y} − sy(0) − y (0) + L {y} =
Solving for L {y} we obtain
L {y} = e−πs
e−πs e−2πs
−
.
s
s
s
s
1
1
1
− 2
− e−2πs
−
+ 2
.
s s +1
s s2 + 1
s +1
Thus
y = [1 − cos (t − π)]U (t − π) − [1 − cos (t − 2π)]U (t − 2π) + sin t.
7.3
Operational Properties I
70. The Laplace transform of the differential equation is
s2 L {y} − sy(0) − y (0) + 4 [sL {y} − y(0)] + 3L {y} =
1 e−2s e−4s e−6s
−
−
+
.
s
s
s
s
Solving for L {y} we obtain
1 1
1 1
1 1
11 1 1
−2s 1 1
−
+
−e
−
+
L {y} =
3 s 2 s+1 6 s+3
3 s 2 s+1 6 s+3
1 1
1 1
1 1
1 1
−4s 1 1
−6s 1 1
−
+
+e
−
+
.
−e
3 s 2 s+1 6 s+3
3 s 2 s+1 6 s+3
Thus
1 1 −t 1 −3t
1 1 −(t−2) 1 −3(t−2)
y= − e + e
U (t − 2)
− e
−
+ e
3 2
6
3 2
6
1 1 −(t−4) 1 −3(t−4)
1 1 −(t−6) 1 −3(t−6)
−
− e
− e
+ e
+ e
U (t − 4) +
U (t − 6).
3 2
6
3 2
6
71. Recall from Section 3.8 that mx = −kx + f (t). Now m = W/g = 9.8/9.8 = 1 kg, and
9.8 = 0.6125k so that k = 16 N/m. Thus, the differential equation is x + 16x = f (t). The
initial conditions are x(0) = 0, x (0) = 0. Also, since
20t, 0 ≤ t < 5
f (t) =
0,
t≥5
and 20t = 20(t − 5) + 100 we can write
f (t) = 20t − 20tU (t − 5) = 20t − 20(t − 5)U (t − 5) − 100U (t − 5).
The Laplace transform of the differential equation is
s2 L {x} + 16L {x} =
20 20 −5s 100 −5s
e .
− 2e
−
s2
s
s
Solving for L {x} we obtain
20
20
100
− 2 2
e−5s −
e−5s
2
+ 16) s (s + 16)
s(s + 16)
4
s
5 1
5
25 1 25
−5s
·
·
· −
·
e−5s .
=
−
−
1−e
4 s2 16 s2 + 16
4 s
4 s2 + 16
L {x} =
s2 (s2
Thus
5
5
5
5
25 25
x(t) = t −
sin 4t − (t − 5) −
sin 4(t − 5) U (t − 5) −
−
cos 4(t − 5) U (t − 5)
4
16
4
16
4
4
5
5
5
25
5
sin 4t − tU (t − 5) +
sin 4(t − 5)U (t − 5) +
cos 4(t − 5)U (t − 5).
= t−
4
16
4
16
4
455
456
CHAPTER 7
THE LAPLACE TRANSFORM
72. Recall from Section 3.8 that mx = −kx + f (t). Now m = W/g = 32/32 = 1 slug, and
32 = 2k so that k = 16 lb/ft. Thus, the differential equation is x + 16x = f (t). The initial
conditions are x(0) = 0, x (0) = 0. Also, since
sin t, 0 ≤ t < 2π
f (t) =
0,
t ≥ 2π
and sin t = sin (t − 2π) we can write
f (t) = sin t − sin (t − 2π)U (t − 2π).
The Laplace transform of the differential equation is
s2 L {x} + 16L {x} =
s2
1
1
− 2
e−2πs .
+1 s +1
Solving for L {x} we obtain
1
1
− 2
e−2πs
2
+ 16) (s + 1) (s + 16) (s2 + 1)
1/15
−1/15
1/15
−1/15
+ 2
− 2
+ 2
e−2πs .
= 2
s + 16 s + 1
s + 16 s + 1
L {x} =
(s2
Thus
1
1
1
1
sin 4t +
sin t +
sin 4(t − 2π)U (t − 2π) −
sin (t − 2π)U (t − 2π)
60
15
60
15
⎧
1
1
⎪
⎨− sin 4t +
sin t, 0 ≤ t < 2π
60
15
=
⎪
⎩
0,
t ≥ 2π.
x(t) = −
73. The differential equation is
2.5
dq
+ 12.5q = 5U (t − 3).
dt
The Laplace transform of this equation is
sL {q} + 5L {q} =
2 −3s
e .
s
Solving for L {q} we obtain
2
L {q} =
e−3s =
s(s + 5)
Thus
2 1 2
1
· − ·
5 s 5 s+5
2
2
q(t) = U (t − 3) − e−5(t−3) U (t − 3).
5
5
e−3s .
7.3
Operational Properties I
74. The differential equation is
10
dq
+ 10q = 30et − 30et U (t − 1.5).
dt
The Laplace transform of this equation is
sL {q} − q0 + L {q} =
3e1.5 −1.5s
3
−
e
.
s − 1 s − 1.5
Solving for L {q} we obtain
3
1
3
1
2/5
1.5 −2/5
·
+ ·
− 3e
+
e−1.5s .
L {q} = q0 −
2
s+1 2 s−1
s + 1 s − 1.5
Thus
q(t) =
3
q0 −
2
3
6
e−t + et + e1.5 e−(t−1.5) − e1.5(t−1.5) U (t − 1.5).
2
5
75. (a) The differential equation is
di
3π
3π
+ 10i = sin t + cos t −
U t−
,
dt
2
2
i(0) = 0.
The Laplace transform of this equation is
sL {i} + 10L {i} =
se−3πs/2
1
+
.
s2 + 1
s2 + 1
Solving for L {i} we obtain
s
1
+ 2
e−3πs/2
+ 1)(s + 10) (s + 1)(s + 10)
1
s
10
1
−10
10s
1
1
−
+
+
+
+
e−3πs/2 .
=
101 s + 10 s2 + 1 s2 + 1
101 s + 10 s2 + 1 s2 + 1
L {i} =
(s2
Thus
i(t) =
(b) x
1 −10t
− cos t + 10 sin t
e
101
3π
3π
3π
1
−10e−10(t−3π/2) + 10 cos t −
+ sin t −
U t−
.
+
101
2
2
2
i
0.2
1
2
3
4
5
6
t
–0.2
The maximum value of i(t) is approximately 0.1 at t = 1.7, the minimum is approximately −0.1 at 4.7.
457
458
CHAPTER 7
THE LAPLACE TRANSFORM
76. (a) The differential equation is
50
1
dq
+
q = E0 [U (t − 1) − U (t − 3)],
dt
0.01
q(0) = 0
or
dq
+ 100q = E0 [U (t − 1) − U (t − 3)], q(0) = 0.
dt
The Laplace transform of this equation is
1 −s 1 −3s
.
50sL {q} + 100L {q} = E0
e − e
s
s
50
Solving for L {q} we obtain
E0
e−s
e−3s
E0 1 1
1
1
1 1
−s
L {q} =
−
=
−
e −
−
e−3s .
50 s(s + 2) s(s + 2)
50 2 s s + 2
2 s s+2
Thus
q(t) =
(b) x
E0 1 − e−2(t−1) U (t − 1) − 1 − e−2(t−3) U (t − 3) .
100
q
1
1
2
3
4
5
6
t
The maximum value of q(t) is approximately 1 at t = 3.
77. The differential equation is
d4 y
EI 4 = w0 [1 − U
dx
L
x−
].
2
Taking the Laplace transform of both sides and using y(0) = y (0) = 0 we obtain
s4 L {y} − sy (0) − y (0) =
w0 1
1 − e−Ls/2 .
EI s
Letting y (0) = c1 and y (0) = c2 we have
c2
w0 1
c1
+ 4+
1 − e−Ls/2
3
s
s
EI s5
1
1
1 w0
L 4
L
2
3
4
y(x) = c1 x + c2 x +
x − x−
.
U x−
2
6
24 EI
2
2
L {y} =
so that
To find c1 and c2 we compute
2 w
1
L
L
0
x2 − x −
y (x) = c1 + c2 x +
U x−
2 EI
2
2
and
L
L
w0
x− x−
U x−
.
y (x) = c2 +
EI
2
2
Then y (L) = y (L) = 0 yields the system
7.3
2 1 w0
L
2
=0
L −
c1 + c2 L +
2 EI
2
Operational Properties I
c1 + c2 L +
or
w0 L
c2 +
= 0.
EI 2
3 w0 L2
=0
8 EI
c2 +
1 w0 L
= 0.
2 EI
Solving for c1 and c2 we obtain c1 = w0 L2 /8EI and c2 = −w0 L/2EI. Thus
4 L
w0 1 2 2
1
1
1
L
L x − Lx3 + x4 −
x−
.
y(x) =
U x−
EI 16
12
24
24
2
2
78. The differential equation is
d4 y
L
2L
EI 4 = w0 U x −
−U x−
.
dx
3
3
Taking the Laplace transform of both sides and using y(0) = y (0) = 0 we obtain
s4 L {y} − sy (0) − y (0) =
w0 1 −Ls/3
− e−2Ls/3 .
e
EI s
Letting y (0) = c1 and y (0) = c2 we have
L {y} =
c2
w0 1
c1
e−Ls/3 − e−2Ls/3
+
+
s3 s4 EI s5
so that
1
1 w0
1
y(x) = c1 x2 + c2 x3 +
2
6
24 EI
L
x−
3
4
2L 4
L
2L
− x−
.
U x−
U x−
3
3
3
To find c1 and c2 we compute
1 w0
y (x) = c1 + c2 x +
2 EI
and
y (x) = c2 +
w0
EI
L
x−
3
2
2L 2
L
2L
− x−
U x−
U x−
3
3
3
L
L
2L
2L
x−
U x−
− x−
U x−
.
3
3
3
3
Then y (L) = y (L) = 0 yields the system
1 w0
c1 + c2 L +
2 EI
2L
3
2
2 L
−
=0
3
c2 +
w0 2L L
−
= 0.
EI 3
3
c1 + c2 L +
or
1 w0 L2
=0
6 EI
c2 +
1 w0 L
= 0.
3 EI
459
460
CHAPTER 7
THE LAPLACE TRANSFORM
Solving for c1 and c2 we obtain c1 = w0 L2 /6EI and c2 = −w0 L/3EI. Thus
1
w0 1 2 2
L x − Lx3
y(x) =
EI 12
18
L 4
2L 4
1
L
2L
x−
− x−
.
+
U x−
U x−
24
3
3
3
3
79. The differential equation is
L
L
2w0 L
d4 y
−x+ x−
U x−
.
EI 4 =
dx
L 2
2
2
Taking the Laplace transform of both sides and using y(0) = y (0) = 0 we obtain
1
2w0 L
1
4
− 2 + 2 e−Ls/2 .
s L {y} − sy (0) − y (0) =
EIL 2s s
s
Letting y (0) = c1 and y (0) = c2 we have
c2
2w0 L
1
1
c1
− 6 + 6 e−Ls/2
L {y} = 3 + 4 +
5
s
s
EIL 2s
s
s
so that
1
L 5
1
2w0 L 4
1 5
1
L
2
3
y(x) = c1 x + c2 x +
x −
x +
x−
U x−
2
6
EIL 48
120
120
2
2
5 1
w
5L
L
L
1
0
x4 − x5 + x −
.
U x−
= c1 x2 + c2 x3 +
2
6
60EIL 2
2
2
To find c1 and c2 we compute
w0
L 3
L
2
3
30Lx − 20x + 20 x −
U x−
y (x) = c1 + c2 x +
60EIL
2
2
2 w
L
L
0
60Lx − 60x2 + 60 x −
.
U x−
y (x) = c2 +
60EIL
2
2
and
Then y (L) = y (L) = 0 yields the system
c1 + c2 L +
w0
5
30L3 − 20L3 + L3 = 0
60EIL
2
w0
c2 +
[60L2 − 60L2 + 15L2 ] = 0.
60EIL
c1 + c2 L +
or
5w0 L2
=0
24EI
c2 +
w0 L
= 0.
4EI
7.3
Operational Properties I
Solving for c1 and c2 we obtain c1 = w0 L2 /24EI and c2 = −w0 L/4EI. Thus
5 w0 L 3
w0
L
L
5L 4
w0 L2 2
U x−
x −
x +
x − x5 + x −
.
y(x) =
48EI
24EI
60EIL 2
2
2
80. The differential equation is
L
d4 y
.
EI 4 = w0 1 − U x −
dx
2
Taking the Laplace transform of both sides and using y(0) = y (0) = 0 we obtain
s4 L {y} − sy (0) − y (0) =
w0 1
1 − e−Ls/2 .
EI s
Letting y (0) = c1 and y (0) = c2 we have
L {y} =
so that
c2
w0 1
c1
1 − e−Ls/2
+ 4+
3
s
s
EI s5
4 w
1
1
L
L
1
0
x4 − x −
.
U x−
y(x) = c1 x2 + c2 x3 +
2
6
24 EI
2
2
To find c1 and c2 we compute
1 w0
L 2
L
2
y (x) = c1 + c2 x +
x − x−
.
U x−
2 EI
2
2
Then y(L) = y (L) = 0 yields the system
4 1
1
1 w0
L
2
3
4
c1 L + c2 L +
L −
=0
2
6
24 EI
2
2 1 w0
L
2
= 0.
L −
c1 + c2 L +
2 EI
2
Solving for c1 and c2 we obtain c1 =
w0
y(x) =
EI
9
128
or
1
1
5w0
c1 L2 + c2 L3 +
L4 = 0
2
6
128EI
c1 + c2 L +
3w0 2
L = 0.
8EI
57
w0 L2 /EI and c2 = − 128
w0 L/EI. Thus
9 2 2
19
1
1
L x −
Lx3 + x4 −
256
256
24
24
L 4
L
x−
.
U x−
2
2
461
462
CHAPTER 7
THE LAPLACE TRANSFORM
81. (a) The temperature T of the cake inside the oven is modeled by
where Tm
dT
= k(T − Tm )
dt
is the ambient temperature of the oven. For 0 ≤ t ≤ 4, we have
Tm = 20 +
Hence for t ≥ 0,
Tm =
150 − 20
t = 20 + 32.5t.
4−0
70 + 32.5t,
0≤t<4
150,
t ≥ 4.
In terms of the unit step function,
Tm = (20 + 32.5t)[1 − U (t − 4)] + 150U (t − 4) = 20 + 32.5t + (130 − 32.5t)U (t − 4).
The initial-value problem is then
dT
= k[T − 20 − 32.5t − (130 − 32.5t)U (t − 4)],
dt
T (0) = 20.
(b) Let t(s) = L {T (t)}. Transforming the equation, using 130 − 32.5t = −32.5(t − 4) and
Theorem 7.3.2, gives
20 32.5 32.5 −4s
− 2 + 2 e
st(s) − 20 = k t(s) −
s
s
s
or
20k
32.5k
32.5k
20
−
− 2
+ 2
e−4s .
t(s) =
s − k s(s − k) s (s − k) s (s − k)
After using partial functions, the inverse transform is then
1
1 kt
1 k(t−4)
1
− 32.5
U (t − 4).
+t− e
+t−4− e
T (t) = 20 + 32.5
k
k
k
k
Of course, the obvious question is: What is k? If the cake is supposed to bake for, say,
20 minutes, then T (20) = 300. That is,
1 20k
1 16k
1
1
+ 20 − e
+ 16 − e
− 32.5
.
150 = 20 + 32.5
k
k
k
k
But this equation has no physically meaningful solution. This should be no surprise
since the model predicts the asymptotic behavior T (t) → 150 as t increases. Using
T (20) = 149 instead, we find, with the help of a CAS, that k ≈ −0.3.
82. In order to apply Theorem 7.3.2 we need the function to have the form f (t − a)U (t − a). To
accomplish this rewrite the functions given in the forms shown below.
(a) 2t + 1 = 2(t − 1 + 1) + 1 = 2(t − 1) + 3
(b) et = et−5+5 = e5 et−5
(c) cos t = − cos(t − π)
(d) t2 − 3t = (t − 2)2 + (t − 2) − 2
7.4 Operational Properties II
83. (a) From Theorem 7.3.1 we have L {tekti } = 1/(s − ki)2 . Then, using Euler’s formula,
L {tekti } = L {t cos kt + it sin kt} = L {t cos kt} + iL {t sin kt}
1
(s + ki)2
s2 − k 2
2ks
=
=
+i 2
.
2
2
2
2
2
2
2
(s − ki)
(s + k )
(s + k )
(s + k 2 )2
=
Equating real and imaginary parts we have
L {t cos kt} =
s2 − k 2
(s2 + k 2 )2
and L {t sin kt} =
(s2
2ks
.
+ k 2 )2
(b) The Laplace transform of the differential equation is
s2 L {x} + ω 2 L {x} =
s2
s
.
+ ω2
Solving for L {x} we obtain L {x} = s/(s2 + ω 2 )2 . Thus x = (1/2ω)t sin ωt.
7.4
Operational Properties II
−10t
1. L {te
d
}=−
ds
d
3. L {t cos 2t} = −
ds
d2
5. L {t sinh t} = 2
ds
2
1
s + 10
s
2
s +4
1
2
s −1
1
=
(s + 10)2
s2 − 4
=
2
(s2 + 4)
=
d3
2. L {t e } = (−1)
ds3
3 t
3
d
4. L {t sinh 3t} = −
ds
1
s−1
3
2
s −9
=
=
6
(s − 1)4
6s
(s2 − 9)2
6s2 + 2
(s2 − 1)3
2s s2 − 3
s
d2
d
1 − s2
2
=
=
6. L {t cos t} = 2
ds
s2 + 1
ds (s2 + 1)2
(s2 + 1)3
2t
6
12(s − 2)
d
=
7. L te sin 6t = −
ds (s − 2)2 + 36
[(s − 2)2 + 36]2
s+3
(s + 3)2 − 9
d
=
8. L te−3t cos 3t = −
2
ds (s + 3) + 9
[(s + 3)2 + 9]2
9. The Laplace transform of the differential equation is
sL {y} + L {y} =
(s2
2s
.
+ 1)2
Solving for L {y} we obtain
L {y} =
2s
1 1
1 s
1
1 1
s
−
+
+ 2
=−
+ 2
.
2
2
2
2
2
(s + 1)(s + 1)
2 s + 1 2 s + 1 2 s + 1 (s + 1)
(s + 1)2
463
464
CHAPTER 7
THE LAPLACE TRANSFORM
Thus
1
1
1
1
1
y(t) = − e−t − sin t + cos t + (sin t − t cos t) + t sin t
2
2
2
2
2
1
1
1
1
= − e−t + cos t − t cos t + t sin t.
2
2
2
2
10. The Laplace transform of the differential equation is
sL {y} − L {y} =
2(s − 1)
.
((s − 1)2 + 1)2
Solving for L {y} we obtain
L {y} =
2
.
((s − 1)2 + 1)2
Thus
y = et sin t − tet cos t.
11. The Laplace transform of the differential equation is
s2 L {y} − sy(0) − y (0) + 9L {y} =
s2
s
.
+9
Letting y(0) = 2 and y (0) = 5 and solving for L {y} we obtain
L {y} =
5
s
2s
2s3 + 5s2 + 19s − 45
+ 2
+ 2
= 2
.
2
2
(s + 9)
s + 9 s + 9 (s + 9)2
Thus
y = 2 cos 3t +
1
5
sin 3t + t sin 3t.
3
6
12. The Laplace transform of the differential equation is
s2 L {y} − sy(0) − y (0) + L {y} =
1
.
s2 + 1
Solving for L {y} we obtain
L {y} =
1
1
s
s3 − s2 + s
− 2
+ 2
= 2
.
2
2
(s + 1)
s + 1 s + 1 (s + 1)2
Thus
y = cos t − sin t +
1
1
sin t − t cos t
2
2
= cos t −
1
1
sin t − t cos t.
2
2
13. The Laplace transform of the differential equation is
s2 L {y} − sy(0) − y (0) + 16L {y} = L {cos 4t − cos 4tU (t − π)}
7.4 Operational Properties II
465
or by (16) of Section 7.3 in the text,
s
− e−πs L {cos 4(t + π)}
(s2 + 16)L {y} = 1 + 2
s + 16
s
s
s
=1+ 2
− e−πs L {cos 4t} = 1 + 2
− 2
e−πs .
s + 16
s + 16 s + 16
Thus
L {y} =
and
y=
s2
s
s
1
+ 2
− 2
e−πs
2
+ 16 (s + 16)
(s + 16)2
1
1
1
sin 4t + t sin 4t − (t − π) sin 4(t − π)U (t − π).
4
8
8
14. The Laplace transform of the differential equation is
s2 L {y} − sy(0) − y (0) + L {y} = L 1 − U
or
(s2 + 1)L {y} = s +
t−
π
+ sin tU
2
t−
π 2
1 1 −πs/2
π − e
+ e−πs/2 L sin t +
s s
2
=s+
1 1 −πs/2
+ e−πs/2 L {cos t}
− e
s s
=s+
1 1 −πs/2
s
− e
e−πs/2 .
+ 2
s s
s +1
Thus
1
1
s
s
+
−
e−πs/2 + 2
e−πs/2
2
2
+ 1 s(s + 1) s(s + 1)
(s + 1)2
1
s
1
s
s
s
+ − 2
−
− 2
= 2
e−πs/2
e−πs/2 + 2
s +1 s s +1
s s +1
(s + 1)2
1
s
1
s
− 2
e−πs/2 + 2
= −
e−πs/2
s
s s +1
(s + 1)2
L {y} =
and
s2
π y = 1 − 1 − cos t −
U
2
= 1 − (1 − sin t)U
15. x
t−
t−
π
1
π
π
+
t−
sin t −
U
2
2
2
2
π
1
π
−
t−
cos tU
2
2
2
y
16. x
4
0.5
2
2
3
4
5
6
π
.
2
3
4
π
2
y
1.0
1
t−
t−
f
1
0.5
2
1.0
4
2
5
6
7
8
9
f
466
CHAPTER 7
THE LAPLACE TRANSFORM
17. From (7) of Section 7.2 in the text along with Theorem 7.4.1,
L {ty } = −
d
d
dY
L {y } = − [s2 Y (s) − sy(0) − y (0)] = −s2
− 2sY + y(0),
ds
ds
ds
so that the transform of the given second-order differential equation is the linear first-order
differential equation in Y (s):
s2 Y + 3sY = −
4
s3
or
Y+
4
3
Y =− 5.
s
s
The solution of the latter equation is Y (s) = 4/s4 + c/s3 , so
y(t) = L −1 {Y (s)} =
2 3 c 2
t + t .
3
2
18. From Theorem 7.4.1 in the text
L {ty } = −
dY
d
d
L {y } = − [sY (s) − y(0)] = −s
−Y
ds
ds
ds
so that the transform of the given second-order differential equation is the linear first-order
differential equation in Y (s):
10
3
− 2s Y = − .
Y +
s
s
Using the integrating factor s3 e−s , the last equation yields
2
Y (s) =
5
c s2
+
e .
s3 s3
But if Y (s) is the Laplace transform of a piecewise-continuous function of exponential order,
we must have, in view of Theorem 7.2.3, lim Y (s) = 0. In order to obtain this condition we
s→∞
require c = 0. Hence
5
5
= t2 .
y(t) = L −1
s3
2
19. Identify f (τ ) = 4τ and g(t − τ ) = 3 (t − τ )2 . Therefore,
ˆ t
ˆ t
2
1 3 1 4
2
2
3
2 1 2
f ∗g =
(4τ ) 3 (t − τ ) dτ = 12
t τ − 2tτ + τ dτ = 12 t · t − 2t · t + t = t4
2
3
4
0
0
Then
24
4!
L {f ∗ g} = L t4 = 5 = 5
s
s
20. Identify f (τ ) = τ and g(t − τ ) = e−t+τ . Therefore,
ˆ t
ˆ t
f ∗g =
τ e−t+τ dτ = e−t
τ eτ dτ = e−t tet − et + 1 = t − 1 + e−t
0
Then
0
1
1
1
1
= 2
L {f ∗ g} = L t − 1 + e−t = 2 − +
s
s s+1
s (s + 1)
7.4 Operational Properties II
467
21. Identify f (τ ) = e−τ and g(t − τ ) = et−τ . Therefore,
ˆ
t
f ∗g =
−τ t−τ
e
e
ˆ
t
t
−2τ
dτ = e
e
0
0
Then
L {f ∗ g} = L
1 −2t 1
1
1
= − e−t + et
dτ = e · − e
+
2
2
2
2
t
1
1
− e−t + et
2
2
=−
1 1
1 1
1
+
= 2
2 s+1 2 s−1
s −1
22. Identify f (τ ) = cos 2τ and g(t − τ ) = et−τ . Therefore,
ˆ
f ∗g =
ˆ
t
t−τ
(cos 2τ ) e
t
t
dτ = e
0
−τ
e
0
1
1 2
cos 2τ dτ = e · − e−t cos 2t + + e−t sin 2t
5
5 5
t
2
1
1
= − cos 2t + sin 2t + et
5
5
5
Then
2
1
1
− cos 2t + sin 2t + et
5
5
5
L {f ∗ g} = L
=−
1 s
2 2
1 1
s
+
+
=
2
2
5 s +4 5 s +4 5 s−1
(s − 1) (s2 + 4)
24. L t2 ∗ tet =
1 3!s4 6
23. L 1 ∗ t3 =
s = s5
25. L e−t ∗ et cos t =
ˆ
27. L
t
eτ dτ
0
ˆ
28. L
t
0
29. L
t
ˆ
t
τ sin τ dτ
0
ˆ
31. L
t
2t
s−1
1
26.
L
e
∗
sin
t
=
(s + 1) [(s − 1)2 + 1]
(s − 2)(s2 + 1)
s
1
1
= 2
= L {cos t} =
s
s(s2 + 1)
s +1
e−τ cos τ dτ
0
30. L
τ et−τ dτ
1
s+1
s+1
1 =
= L e−t cos t =
2
2
s
s (s + 1) + 1
s (s + 2s + 2)
1
d
1
1 −2s
1
2
−
=−
= L {t sin t} =
=
2
2
s
s
ds s + 1
s (s2 + 1)
(s2 + 1)2
= L {t}L {et } =
0
ˆ
32. L
t
sin τ cos (t − τ ) dτ
1
− 1)
s2 (s
= L {sin t}L {cos t} =
0
ˆ
33. L
t
t
sin τ dτ
0
2
− 1)2
1
1
= L {et } =
s
s(s − 1)
cos τ dτ
ˆ
s3 (s
d
=− L
ds
ˆ
t
sin τ dτ
0
d
=−
ds
s
(s2
+ 1)2
1 1
s s2 + 1
=
3s2 + 1
s2 (s2 + 1)2
468
CHAPTER 7
ˆ
34. L
t
t
THE LAPLACE TRANSFORM
−τ
τe
d
=− L
ds
dτ
0
1
s(s − 1)
35. L −1
36. L
37. L
1
2
s (s − 1)
−1
1
3
s (s − 1)
−1
38. Using L −1
=L
−1
−1
1
(s − a)2
L
t
−τ
τe
d
=−
ds
dτ
0
ˆ
1/(s − 1)
s
= L −1
=L
ˆ
t
1
1
s (s + 1)2
=
3s + 1
+ 1)3
s2 (s
eτ dτ = et − 1
=
0
ˆ
1/s(s − 1)
s
t
=
(eτ − 1) dτ = et − t − 1
0
ˆ
1/s2 (s − 1)
s
t
=
0
1
(eτ − τ − 1) dτ = et − t2 − t − 1
2
= teat , (8) in the text gives
−1
ˆ
1
s(s − a)2
t
=
τ eaτ dτ =
0
1
(ateat − eat + 1).
a2
39. (a) The result in (4) in the text is L −1 {F (s)G(s)} = f ∗ g, so identify
F (s) =
2k 3
(s2 + k 2 )2
and
G(s) =
4s
.
+ k2
s2
Then
f (t) = sin kt − kt cos kt
and
g(t) = 4 cos kt
so
L
−1
8k 3 s
(s2 + k 2 )3
=L
−1
ˆ
{F (s)G(s)} = f ∗ g = 4
t
f (τ )g(t − τ ) dt
0
ˆ
=4
t
(sin kτ − kτ cos kτ ) cos k(t − τ ) dτ.
0
Using a CAS to evaluate the integral we get
L −1
8k 3 s
(s2 + k 2 )3
= t sin kt − kt2 cos kt.
(b) Observe from part (a) that
L {t(sin kt − kt cos kt)} =
8k 3 s
,
(s2 + k 2 )3
and from Theorem 7.4.1 that L {tf (t)} = −F (s). We saw in (5) in the text that
L {sin kt − kt cos kt} = 2k 3 /(s2 + k 2 )2 ,
so
L {t(sin kt − kt cos kt)} = −
2k 3
8k 3 s
d
=
.
ds (s2 + k 2 )2
(s2 + k 2 )3
7.4 Operational Properties II
40. The Laplace transform of the differential equation is
s2 L {y} + L {y} =
Thus
L {y} =
(s2
(s2
1
2s
+ 2
.
+ 1) (s + 1)2
2s
1
+ 2
2
+ 1)
(s + 1)3
y
50
5
10
–50
and, using Problem 39 with k = 1,
1
1
y = (sin t − t cos t) + (t sin t − t2 cos t).
2
4
41. The Laplace transform of the given equation is
L {f } + L {t}L {f } = L {t}.
Solving for L {f } we obtain L {f } =
s2
1
. Thus, f (t) = sin t.
+1
42. The Laplace transform of the given equation is
L {f } = L {2t} − 4L {sin t}L {f }.
Solving for L {f } we obtain
√
2s2 + 2
8
2 1
5
L {f } = 2 2
+ √ 2
=
.
2
s (s + 5)
5s
5 5 s +5
Thus
√
8
2
f (t) = t + √ sin 5 t.
5
5 5
43. The Laplace transform of the given equation is
L {f } = L tet + L {t}L {f }.
Solving for L {f } we obtain
L {f } =
Thus
469
s2
1
1 1
1 1
3
1
2
+
−
=
+
.
3
2
3
(s − 1) (s + 1)
8 s − 1 4 (s − 1)
4 (s − 1)
8 s+1
3
1
1
1
f (t) = et + tet + t2 et − e−t
8
4
4
8
44. The Laplace transform of the given equation is
L {f } + 2L {cos t}L {f } = 4L e−t + L {sin t}.
15
t
470
CHAPTER 7
THE LAPLACE TRANSFORM
Solving for L {f } we obtain
L {f } =
7
4
2
4s2 + s + 5
−
=
+4
.
3
2
(s + 1)
s + 1 (s + 1)
(s + 1)3
Thus
f (t) = 4e−t − 7te−t + 4t2 e−t .
45. The Laplace transform of the given equation is
L {f } + L {1}L {f } = L {1}.
1
. Thus, f (t) = e−t .
s+1
Solving for L {f } we obtain L {f } =
46. The Laplace transform of the given equation is
L {f } = L {cos t} + L e−t L {f }.
Solving for L {f } we obtain
L {f } =
s2
1
s
+ 2
.
+1 s +1
Thus
f (t) = cos t + sin t.
47. The Laplace transform of the given equation is
L {f } = L {1} + L {t} − L
=
8
3
ˆ
t
(t − τ )3 f (τ ) dτ
0
1
1
8
1
16
1
+ 2 + L {t3 }L {f } = + 2 + 4 L {f }.
s s
3
s s
s
Solving for L {f } we obtain
L {f } =
Thus
1 1
3 1
1 2
1 s
s2 (s + 1)
=
+
+
+
.
s4 − 16
8 s + 2 8 s − 2 4 s2 + 4 2 s2 + 4
1
3
1
1
f (t) = e−2t + e2t + sin 2t + cos 2t.
8
8
4
2
48. The Laplace transform of the given equation is
L {t} − 2L {f } = L et − e−t L {f }.
Solving for L {f } we obtain
L {f } =
Thus
1 1
1 3!
s2 − 1
=
−
.
4
2
2s
2s
12 s4
1
1
f (t) = t − t3 .
2
12
7.4 Operational Properties II
471
49. The Laplace transform of the given equation is
sL {y} − y(0) = L {1} − L {sin t} − L {1}L {y}.
Solving for L {y} we obtain
L {y} =
1
2s
1
s2 − s + 1
−
= 2
.
2
2
2
(s + 1)
s + 1 2 (s + 1)2
Thus
y = sin t −
1
t sin t.
2
50. The Laplace transform of the given equation is
sL {y} − y(0) + 6L {y} + 9L {1}L {y} = L {1}.
Solving for L {y} we obtain L {y} =
1
. Thus, y = te−3t .
(s + 3)2
51. The differential equation is
ˆ t
1
di
0.1 + 3i +
i(τ ) dτ = 100 [U (t − 1) − U (t − 2)]
dt
0.05 0
30
i
20
10
or
di
+ 30i + 200
dt
ˆ
t
i(τ ) dτ = 1000 [U (t − 1) − U (t − 2)] ,
0
where i(0) = 0. The Laplace transform of the differential
equation is
sL {i} − y(0) + 30L {i} +
0.5
1
1.5
2
2.5
–10
–20
–30
1000 −s
200
L {i} =
(e − e−2s ).
s
s
Solving for L {i} we obtain
1000e−s − 1000e−2s
=
L {i} =
s2 + 30s + 200
100
100
−
s + 10 s + 20
(e−s − e−2s ).
Thus
i(t) = 100 e−10(t−1) − e−20(t−1) U (t − 1) − 100 e−10(t−2) − e−20(t−2) U (t − 2).
3
t
472
CHAPTER 7
THE LAPLACE TRANSFORM
52. The differential equation is
ˆ t
1
di
i(τ ) dτ = 100 [t − (t − 1)U (t − 1)]
0.005 + i +
dt
0.02 0
i
2
1.5
or
di
+ 200i + 10,000
dt
ˆ
t
i(τ ) dτ = 20,000 [t − (t − 1)U (t − 1)] ,
0
1
0.5
where i(0) = 0. The Laplace transform of the differential
equation is
1
1 −s
10,000
L {i} = 20,000 2 − 2 e
.
sL {i} + 200L {i} +
s
s
s
0.5
1
1.5
2
t
Solving for L {i} we obtain
20,000
2
2
200
−s
L {i} =
(1 − e−s ).
−
−
(1 − e ) =
2
s(s + 100)
s s + 100 (s + 100)2
Thus
i(t) = 2−2e−100t −200te−100t −2U (t−1)+2e−100(t−1) U (t−1)+200(t−1)e−100(t−1) U (t−1).
53. L {f (t)} =
54. L {f (t)} =
1
1 − e−2as
1
1 − e−2as
ˆ
a
e−st dt −
ˆ
0
ˆ
a
2a
e−st dt =
a
e−st dt =
0
(1 − e−as )2
1 − e−as
=
s(1 − e−2as )
s(1 + e−as )
1
s(1 + e−as )
55. Using integration by parts,
ˆ
1
L {f (t)} =
1 − e−bs
56. L {f (t)} =
57. L {f (t)} =
58. L {f (t)} =
ˆ
1
1 − e−2s
1
1 − e−πs
1
0
ˆ
1
1 − e−2πs
π
0
ˆ
0
2
a
a −st
te dt =
b
s
(2 − t)e−st dt =
1
e−st sin t dt =
0
ˆ
te−st dt +
b
π
1
1
− bs
bs e − 1
.
1 − e−s
s2 (1 + e−s )
eπs/2 + e−πs/2
πs
1
1
·
coth
= 2
2
πs/2
−πs/2
s +1 e
s +1
2
−e
e−st sin t dt =
s2
1
1
·
+ 1 1 − e−πs
59. The differential equation is L di/dt + Ri = E(t), where i(0) = 0. The Laplace transform of
the equation is
LsL {i} + RL {i} = L {E(t)}.
From Problem 53 we have L {E(t)} = (1 − e−s )/s(1 + e−s ). Thus
(Ls + R)L {i} =
1 − e−s
s(1 + e−s )
7.4 Operational Properties II
and
1 − e−s
1 1 − e−s
1
1
=
−s
L s(s + R/L)(1 + e )
L s(s + R/L) 1 + e−s
1 1
1
−
(1 − e−s )(1 − e−s + e−2s − e−3s + e−4s − · · · )
=
R s s + R/L
1
1 1
−
(1 − 2e−s + 2e−2s − 2e−3s + 2e−4s − · · · ).
=
R s s + R/L
L {i} =
Therefore,
i(t) =
2
1
1 − e−Rt/L −
1 − e−R(t−1)/L U (t − 1)
R
R
+
=
2
2
1 − e−R(t−2)/L U (t − 2) −
1 − e−R(t−3)/L U (t − 3) + · · ·
R
R
2
1
1 − e−Rt/L +
R
R
∞
1 − e−R(t−n)/L U (t − n).
n=1
The graph of i(t) with L = 1 and R = 1 is shown below.
i
1
0.5
1
2
3
4
t
–0.5
–1
60. The differential equation is L di/dt + Ri = E(t), where i(0) = 0. The Laplace transform of
the equation is
LsL {i} + RL {i} = L {E(t)}.
From Problem 55 we have
1
L {E(t)} =
s
1
1
− s
s e −1
Thus
(Ls + R)L {i} =
=
1 1
1
.
−
2
s
s es − 1
1
1 1
−
2
s
s es − 1
and
1
1
1
1
1
−
2
s
L s (s + R/L) L s(s + R/L) e − 1
−s
1
1 1
1
L 1 L
1 1
+
−
−
e + e−2s + e−3s + · · · .
−
=
2
R s
R s R s + R/L
R s s + R/L
L {i} =
473
474
CHAPTER 7
THE LAPLACE TRANSFORM
Therefore
1
i(t) =
R
L
L −Rt/L
1
t− + e
−
1 − e−R(t−1)/L U (t − 1)
R R
R
1
1
1 − e−R(t−2)/L U (t − 2) −
1 − e−R(t−3)/L U (t − 3) − · · ·
R
R
∞
1
1
L
L
1 − e−R(t−n)/L U (t − n).
=
t − + e−Rt/L −
R
R R
R
−
n=1
The graph of i(t) with L = 1 and R = 1 is shown below.
i
1
0.5
1
2
4
3
t
–0.5
–1
61. The differential equation is x + 2x + 10x = 20f (t), where f (t) is the meander function in
Problem 53 with a = π. Using the initial conditions x(0) = x (0) = 0 and taking the Laplace
transform we obtain
(s2 + 2s + 10)L {x(t)} =
20
1
(1 − e−πs )
s
1 + e−πs
=
20
(1 − e−πs )(1 − e−πs + e−2πs − e−3πs + · · · )
s
=
20
(1 − 2e−πs + 2e−2πs − 2e−3πs + · · · )
s
∞
20 40
+
=
s
s
(−1)n e−nπs .
n=1
Then
L {x(t)} =
40
20
+
2
2
s(s + 2s + 10) s(s + 2s + 10)
2s + 4
2
+
= − 2
s s + 2s + 10
∞
n
(−1)
n=1
2 2(s + 1) + 2
+4
= −
s (s + 1)2 + 9
n
(−1)
n=1
(−1)n e−nπs
n=1
4s + 8
4
− 2
e−nπs
s s + 2s + 10
∞
∞
(s + 1) + 1 −nπs
1
−
e
s (s + 1)2 + 9
7.4 Operational Properties II
and
1 −t
x(t) = 2 1 − e cos 3t − e sin 3t
3
∞
1
(−1)n 1 − e−(t−nπ) cos 3(t − nπ) − e−(t−nπ) sin 3(t − nπ) U (t − nπ).
+4
3
−t
n=1
The graph of x(t) is shown below.
x
3
π
2π
t
–3
62. The differential equation is x + 2x + x = 5f (t), where f (t) is the square wave function with
a = π. Using the initial conditions x(0) = x (0) = 0 and taking the Laplace transform, we
obtain
(s2 + 2s + 1)L {x(t)} =
=
1
5
5
=
1 − e−πs + e−2πs − e−3πs + e−4πs − · · ·
−πs
s 1+e
s
5
s
∞
(−1)n e−nπs .
n=0
Then
5
L {x(t)} =
s(s + 1)2
∞
n −nπs
(−1) e
n=0
∞
n
=5
(−1)
n=0
1
1
1
−
−
s s + 1 (s + 1)2
and
∞
x(t) = 5
(−1)n (1 − e−(t−nπ) − (t − nπ)e−(t−nπ) )U (t − nπ).
n=0
The graph of x(t) is shown below.
e−nπs
475
476
CHAPTER 7
THE LAPLACE TRANSFORM
x
5
2π
4π
t
–5
1
63. f (t) = − L −1
t
=−
d
[ln(s − 3) − ln(s + 1)]
ds
1
= − L −1
t
1
1
−
s−3 s+1
1 3t
e − e−t
t
64. The transform of Bessel’s equation is
−
d 2
d
[s Y (s) − sy(0) − y (0)] + sY (s) − y(0) −
Y (s) = 0
ds
ds
or, after simplifying and using the initial condition, (s2 + 1)Y + sY = 0. This equation is
√
both separable and linear. Solving gives Y (s) = c/ s2 + 1 . Now Y (s) = L {J0 (t)}, where
J0 has a derivative that is continuous and of exponential order, implies by Problem 46 of
Exercises 7.2 that
s
=c
1 = J0 (0) = lim sY (s) = c lim √
s→∞
s→∞
s2 + k 2
so c = 1 and
1
1
Y (s) = √
or
L {J0 (t)} = √
.
s2 + 1
s2 + 1
65. (a) Using Theorem 7.4.1, the Laplace transform of the differential equation is
−
d
d 2
[sY − y(0)] + nY
s Y − sy(0) − y (0) + sY − y(0) +
ds
ds
d
d 2 =−
s Y + sY +
[sY ] + nY
ds
ds
dY
2 dY
− 2sY + sY + s
+ Y + nY
= −s
ds
ds
dY
2
= (s − s )
+ (1 + n − s)Y = 0.
ds
Separating variables, we find
1+n−s
dY
=
ds =
Y
s2 − s
n
1+n
−
s−1
s
ln Y = n ln (s − 1) − (1 + n) ln s + c
Y = c1
(s − 1)n
.
s1+n
ds
7.4 Operational Properties II
Since the differential equation is homogeneous, any constant multiple of a solution will
still be a solution, so for convenience we take c1 = 1. The following polynomials are
solutions of Laguerre’s differential equation:
n=0:
L0 (t) = L −1
1
s
n=1:
L1 (t) = L −1
s−1
s2
n=2:
L2 (t) = L −1
(s − 1)2
s3
= L −1
2
1
1
− 2+ 3
s s
s
n=3:
L3 (t) = L −1
(s − 1)3
s4
= L −1
3
1
3
1
− 2+ 3− 4
s s
s
s
n=4:
L4 (t) = L −1
(s − 1)4
s5
= L −1
4
1
6
4
1
− 2+ 3− 4+ 5
s s
s
s
s
=1
= L −1
1
1
−
s s2
=1−t
1
= 1 − 2t + t2
2
3
1
= 1 − 3t + t2 − t3
2
6
2
1
= 1 − 4t + 3t2 − t3 + t4 .
3
24
(b) Letting f (t) = tn e−t we note that f (k) (0) = 0 for k = 0, 1, 2, . . . , n − 1 and f (n) (0) = n!.
Now, by the first translation theorem,
1
1
et dn n −t
t (n)
(n)
L
{e
L
{f
=
t
e
f
(t)}
=
(t)}
L
n! dtn
n!
n!
s→s−1
1 n
=
s L {tn e−t } − sn−1 f (0) − sn−2 f (0) − · · · − f (n−1) (0)
n!
s→s−1
1 n
s L {tn e−t } s→s−1
n!
1 n
(s − 1)n
n!
s
=
=
= Y,
n!
(s + 1)n+1 s→s−1
sn+1
=
where Y = L {Ln (t)}. Thus
Ln (t) =
et dn n −t
(t e ),
n! dtn
n = 0, 1, 2, . . . .
2
66. Let L e−t = F (s) then
2
L y + L {y} = L e−t
s2 Y (s) + Y (s) = F (s)
Y (s) =
s2
1
· F (s)
+1
477
478
CHAPTER 7
THE LAPLACE TRANSFORM
Now take the inverse transform and use the inverse form of the convolution theorem to get
ˆ t
1
2
−1
−t2
y(t) = L
· F (s) = sin t ∗ e
=
sin τ e−(t−τ ) dτ
2
s +1
0
or
y(t) = L
−1
1
F (s) · 2
s +1
−t2
=e
ˆ
∗ sin t =
t
e−τ sin (t − τ ) dτ
2
0
67. Take the transform of both sides of the equation to get
ˆ t
f (τ )e−τ dτ
L {f (t)} = L et + L et ·
0
1
+ L f (t)e−t L {1} s→s−1
s−1
F (s) =
1
+ L f (t)e−t s→s−1 · [L {1}]s→s−1
s−1
1
F (s) = f rac1s − 1 + [F (s + 1)]s→s−1 ·
s s→s−1
F (s) =
1
1
+ F (s) ·
s−1
s−1
F (s) =
Solve the last equation for F (s) to get F (s) = 1/ (s − 2) therefore f (t) = e2t .
∞
(−1)k U (t − k) = U (t) − U (t − 1) + U (t − 2) + U (t − 3) + · · ·
68. (a) E(t) =
n=0
⎧
1, 0 ≤ t < 1
⎪
⎪
⎪
⎪
⎪
⎨0, 1 ≤ t < 2
=
⎪
1, 2 ≤ t < 3
⎪
⎪
⎪
⎪.
⎩
..
geometric series with r=−e−s
!
"#
$
1
1 − e−s + e−2s − e−3s + · · ·
=
s
1 1 −s 1 −2s 1 −3s
− e + e
− e
s s
s
s
1
1
1
=
.
=
−s
s 1 − (−e )
s (1 + e−s )
(b) L {E(t)} =
69. We know that
d
L {t sin 4t = −
ds
and so for s > 0,
ˆ
F (s) =
∞
4
s2 + 16
te−st sin 4t dt =
0
ˆ
Therefore
F (2) =
0
∞
te2t sin 4t dt =
=
8s
(s2
8s
(s2
8·2
(22
+ 16)2
+ 16)
2
+ 16)2
=
,
.
1
16
= .
400
25
7.4 Operational Properties II
70. (a) L
(b) L
ˆ
u ∞ π
s
a
1
a
arctan
du
=
a
·
= − arctan = arctan
2
2
u +a
a
a s
2
a
s
s
ˆ ∞
2
∞
2 (1 − cos kt)
2
2u
2
du
=
2
ln
u
−
ln
u
=
− 2
+
k
t
u u + k2
s
s
∞
∞
u2
= 2 ln u − ln u2 + k 2
= ln 2
u + k2 s
s
sin at
t
∞
=
= ln 1 − ln
s2
s2 + k 2
=
ln
s2 + k 2
s2
71. (a) Using the definition of the Laplace transform and integration by parts,
∞
ˆ ∞
ˆ ∞
−st
−st
e ln t dt = e (t ln t − t) + s
e−st (t ln t − t) dt
L {ln t} =
0
0
0
1
= sL {t ln t} − sL {t} = sL {t ln t} − .
s
(b) Letting Y (s) = L {ln t} we have
1
d
Y =s − Y −
ds
s
by Theorem 7.4.1. That is,
s
1
dY
+Y =−
ds
s
d
1
[sY ] = −
ds
s
sY = − ln s + c
Y =
(c) Now Y (1) = L {ln t}
ˆ
∞
=
s=1
c 1
− ln s, s > 0.
s s
e−t ln t dt = −γ. Thus
0
−γ = Y (1) = c − ln 1 = c
and
Y = L {ln t} = −
γ 1
− ln s.
s
s
72. The output for the first three lines of the program are
9y[t] + 6y [t] + y [t] == t sin[t]
2s
(1 + s2 )2
−11 − 4s − 22s2 − 4s3 − 11s4 − 2s5
Y →−
(1 + s2 )2 (9 + 6s + s2 )
1 − 2s + 9Y + s2 Y + 6(−2 + sY ) ==
479
480
CHAPTER 7
THE LAPLACE TRANSFORM
The fourth line is the same as the third line with Y → removed. The final line of output
shows a solution involving complex coefficients of eit and e−it . To get the solution in more
standard form write the last line as two lines:
euler={Eˆ(It)−>Cos[t] + I Sin[t], Eˆ(-It)−>Cos[t] - I Sin[t]}
InverseLaplaceTransform[Y, s, t]/.euler//Expand
We see that the solution is
487 247
1
y(t) =
+
t e−3t +
(13 cos t − 15t cos t − 9 sin t + 20t sin t) .
250
50
250
73. The solution is
1
1
y(t) = et − e−t/2 cos
6
6
√
1
15 t
− √ e−t/2 sin
2
2 15
√
15 t
.
2
74. The solution is
q(t) = 1 − cos t + (6 + 6 cos t)U (t − 3π) − (4 + 4 cos t)U (t − π).
q
5
π
3π
5π
–5
7.5
The Dirac Delta Function
1. The Laplace transform of the differential equation yields
L {y} =
1 −2s
e
s−3
so that
y = e3(t−2) U (t − 2).
2. The Laplace transform of the differential equation yields
L {y} =
2
e−s
+
s+1 s+1
so that
y = 2e−t + e−(t−1) U (t − 1).
t
7.5
The Dirac Delta Function
3. The Laplace transform of the differential equation yields
L {y} =
1 −2πs
1
+
e
s2 + 1
so that
y = sin t + sin tU (t − 2π).
4. The Laplace transform of the differential equation yields
L {y} =
so that
y=
4
1
e−2πs
2
4 s + 16
1
1
sin 4(t − 2π)U (t − 2π) = sin 4tU (t − 2π).
4
4
5. The Laplace transform of the differential equation yields
L {y} =
so that
s2
1
e−πs/2 + e−3πs/2
+1
3π
π
3π
U t−
t−
+ sin t −
2
2
2
3π
π
+ cos tU t −
.
t−
2
2
π
y = sin t −
U
2
= − cos tU
6. The Laplace transform of the differential equation yields
L {y} =
s2
1
s
+ 2
(e−2πs + e−4πs )
+1 s +1
so that
y = cos t + sin t[U (t − 2π) + U (t − 4π)].
7. The Laplace transform of the differential equation yields
1
11 1 1
−s
L {y} = 2
(1 + e ) =
−
(1 + e−s )
s + 2s
2 s 2 s+2
so that
1 1 −2(t−1)
1 1 −2t
− e
U (t − 1).
+
y= − e
2 2
2 2
8. The Laplace transform of the differential equation yields
1
3 1
31 1 1
1 1
1 1 −2s
s+1
−2s
+
e
−
−
−
e
=
+
L {y} = 2
2
s (s − 2) s(s − 2)
4 s−2 4 s 2 s
2 s−2 2 s
so that
1
1
3 2t 3 1
U (t − 2).
y = e − − t + e2(t−2) −
4
4 2
2
2
481
482
CHAPTER 7
THE LAPLACE TRANSFORM
9. The Laplace transform of the differential equation yields
L {y} =
1
e−2πs
(s + 2)2 + 1
so that
y = e−2(t−2π) sin tU (t − 2π).
10. The Laplace transform of the differential equation yields
L {y} =
1
e−s
(s + 1)2
so that
y = (t − 1)e−(t−1) U (t − 1).
11. The Laplace transform of the differential equation yields
L {y} =
=
e−πs + e−3πs
4+s
+
s2 + 4s + 13
s2 + 4s + 13
−πs
2
3
s+2
1
3
−3πs
e
+
+
+
e
3 (s + 2)2 + 32 (s + 2)2 + 32 3 (s + 2)2 + 32
so that
1
2
y = e−2t sin 3t + e−2t cos 3t + e−2(t−π) sin 3(t − π)U (t − π)
3
3
1
+ e−2(t−3π) sin 3(t − 3π)U (t − 3π).
3
12. The Laplace transform of the differential equation yields
L {y} =
e−2s + e−4s
1
+
(s − 1)2 (s − 6) (s − 1)(s − 6)
=−
−2s
1
1
1 1
1 1
1 1
1 1
−4s
−
+
−
+
+
+
e
e
25 s − 1 5 (s − 1)2 25 s − 6
5 s−1 5 s−6
so that
1
1 t 1 t
1 6t
1 t−2 1 6(t−2)
1
U (t − 2) + − et−4 + e6(t−4) U (t − 4).
y = − e − te + e + − e
+ e
25
5
25
5
5
5
5
13. The Laplace transform of the differential equation yields
∞
L {y } + L {y} =
L {δ (t − kπ)}
k=1
s2 + 1 Y (s) = 1 +
Y (s) =
k = 1∞ e−kπs
1
+
2
s +1
∞
k=1
e−kπs
s2 + 1
7.5
The Dirac Delta Function
483
so that
∞
sin (t − kπ)U (t − kπ)
y(t) = sin t +
k=1
∞
(−1)k U (t − kπ)
= sin t + sin t
sin (t − kπ) = (−1)k sin t
←−
k=1
= sin t − sin tU (t − π) + sin tU (t − 2π) − sin tU (t − 3π) + sin tU (t − 4π) − · · ·
⎧
⎪
sin t,
⎪
⎪
⎪
⎪
⎪
0,
⎪
⎪
⎪
⎪
⎪
⎪
sin t,
⎪
⎪
⎪
⎪
⎪
⎪
0,
⎪
⎪
⎨
= sin t,
⎪
⎪
⎪
⎪0,
⎪
⎪
⎪
⎪
⎪
sin t,
⎪
⎪
⎪
⎪
⎪
⎪
0,
⎪
⎪
⎪
⎪
⎪
⎩...
0≤t<π
π ≤ t < 2π
2π ≤ t < 3π
y
3π ≤ t < 4π
1
4π ≤ t < 5π
5π ≤ t < 6π
6π ≤ t < 7π
Π
2Π 3Π 4Π 5Π 6Π 7Π 8Π
t
7π ≤ t < 8π
The graph of y(t) given on the right is the half-wave rectification of sin t. Also, see Figure
7.4.11 in Exercises 7.4.
14. The Laplace transform of the differential equation yields
L {y } + L {y} =
∞
L {δ (t − 2kπ)}
k=1
2
s + 1 Y (s) = 1 +
∞
e−2kπs
k=1
1
+
Y (s) = 2
s +1
∞
k=1
e−2kπs
s2 + 1
so that
∞
sin (t − 2kπ)U (t − 2kπ)
y(t) = sin t +
k=1
∞
(−1)k U (t − 2kπ)
= sin t + sin t
←−
sin (t − 2kπ) = sin t
k=1
= sin t + sin tU (t − 2π) + sin tU (t − 4π) + sin tU (t − 6π) + sin tU (t − 8π) + · · ·
484
CHAPTER 7
THE LAPLACE TRANSFORM
y
⎧
sin t,
0≤t<π
⎪
⎪
⎪
⎪
⎪
⎪
2 sin t, 2π ≤ t < 4π
⎪
⎪
⎨
= 3 sin t, 4π ≤ t < 6π
⎪
⎪
⎪
⎪4 sin t, 6π ≤ t < 8π
⎪
⎪
⎪
⎪
⎩..
.
4
3
2
1
1
2
3
4
Π
2Π 3Π 4Π 5Π 6Π 7Π 8Π
t
The graph of y(t) is given on the right.
15. Take the transform of the equation y
(4)
L
w0
δ x−
and solve for Y (s) to get
=
EI
2
w0 (−L/2)s
e
EI
w0 (−L/2)s
e
s4 Y (s) − sy (0) − y (0) =
EI
s2 Y (s) − s3 y(0) − s2 y (0) − sy (0) − y (0) =
Y (s) =
1 1
w0 1 (−L/2)s
y (0) + 4 y (0) +
e
3
s
s
EI s4
L 3
1 1 w0
L
2
3
x−
.
The inverse transform is then y(x) = y (0)x + y (0)x +
U x−
2
6
6EI
2
2
Using the conditions y (L) = y (L) = 0 to find y (0) and y (0) we get
1
y(x) =
2
w0 L
2EI
1 w0
w0
x −
x3 +
6 EI
6EI
2
L
L 3
U x−
x−
2
2
We could also write the solution as
⎧
w0
L
w0 L
⎪
⎪
⎪
x2 −
x3 ,
0≤x<
⎪
⎨ 4EI
6EI
2
y(x) = ⎪
⎪
w0
w0
L
L 3
w0 L
⎪
2
3
⎪
≤x≤L
−
+
,
x
x
x
−
⎩
4EI
6EI
6EI
2
2
L 3
1
w0
L
1 y (0)x2 + y (0)x3 +
x−
. Using
U x−
2
6
6EI
2
2
the conditions y(L) = y (L) = 0 to find y (0) and y (0) we get the system of equations
⎧
w0 L3
⎪
2 3 ⎪
⎪
)y
(0)
+
(8L
)y
(0)
=
−
(24L
⎨
EI
⎪
⎪
w L2
⎪
⎩(8L)y (0) + (4L2 )y (0) = − 0
EI
16. From Problem 15 we have y(x) =
Solving the system leads to the solution
w0
L 3
1
w0
L
1 w0 L
2
3
x +
−
x +
x−
U x−
y(x) =
2 8EI
6
2EI
6EI
2
2
7.6
Systems of Linear Differential Equations
We could also write the solution as
⎧
w0
L
w0 L
⎪
⎪
⎪
x2 −
x3 ,
0≤x<
⎪
⎨ 16EI
12EI
2
y(x) = ⎪
⎪
w0
w0
L 3
L
w0 L
⎪
2
3
⎪
x
x
x
−
≤x≤L
−
+
,
⎩
16EI
12EI
6EI
2
2
17. You should disagree. Although formal manipulations of the Laplace transform lead to
y(t) = 13 e−t sin 3t in both cases, this function does not satisfy the initial condition y (0) = 0
of the second initial-value problem.
18. From (7) in (i) of the Remarks it follows with the identifications f (t) = et e−st = e(1−s)t and
t0 = 1 that
ˆ ∞
t
L e δ (t − 1) =
e(1−s)t δ (t − 1) dt = f (1) = e1−s = ee−s .
0
Using the initial conditions y(0) = 0 and y (0) = 2 we then have
L y + 4L y + 3L {y} = L et δ (t − 1)
(s + 1) (s + 3) Y (s) = 2 + ee−s
ee−s
2
+
(s + 1) (s + 3) (s + 1) (s + 3)
1
e
1
1
1
−
+
−
e−s
=
s+1 s+3 2 s+1 s+3
Y (s) =
y(t) = e−t − e−3t +
e −(t−1)
e
− e−3(t−1) U (t − 1) .
2
The graph of y(t) is given on the right.
y
0.5
1
7.6
Systems of Linear Differential Equations
1. Taking the Laplace transform of the system gives
sL {x} = −L {x} + L {y}
sL {y} − 1 = 2L {x}
t
485
486
CHAPTER 7
THE LAPLACE TRANSFORM
so that
L {x} =
and
L {y} =
Then
1
1 1
1 1
=
−
(s − 1)(s + 2)
3 s−1 3 s+2
1
2
2 1
1 1
+
=
+
.
s s(s − 1)(s + 2)
3 s−1 3 s+2
1
1
x = et − e−2t
3
3
and
2
1
y = et + e−2t .
3
3
2. Taking the Laplace transform of the system gives
1
s−1
1
sL {y} − 1 = 8L {x} − 2
s
sL {x} − 1 = 2L {y} +
so that
L {y} =
1 1
8 1
173 1
53 1
s3 + 7s2 − s + 1
=
−
+
−
2
s(s − 1)(s − 16)
16 s 15 s − 1
96 s − 4 160 s + 4
and
y=
Then
8
173 4t
53 −4t
1
− et +
e −
e .
16 15
96
160
1
1
1
173 4t
53 −4t
1
e +
e .
x = y + t = t − et +
8
8
8
15
192
320
3. Taking the Laplace transform of the system gives
sL {x} + 1 = L {x} − 2L {y}
sL {y} − 2 = 5L {x} − L {y}
so that
L {x} =
s
5 3
−s − 5
=− 2
−
s2 + 9
s + 9 3 s2 + 9
and
x = − cos 3t −
Then
5
sin 3t.
3
7
1
1
y = x − x = 2 cos 3t − sin 3t.
2
2
3
4. Taking the Laplace transform of the system gives
(s + 3)L {x} + sL {y} =
(s − 1)L {x} + (s − 1)L {y} =
1
s
1
s−1
7.6
so that
Systems of Linear Differential Equations
L {y} =
11 1 1
4
1
5s − 1
+
+
=−
3s(s − 1)2
3 s 3 s − 1 3 (s − 1)2
L {x} =
11 1 1
1
1
1 − 2s
−
−
=
.
2
3s(s − 1)
3 s 3 s − 1 3 (s − 1)2
and
Then
x=
1 1 t 1 t
− e − te
3 3
3
and
1 1
4
y = − + et + tet .
3 3
3
5. Taking the Laplace transform of the system gives
so that
L {x} =
and
L {y} =
Then
(2s − 2)L {x} + sL {y} =
1
s
(s − 3)L {x} + (s − 3)L {y} =
2
s
11 5 1
2
−s − 3
=−
+
−
s(s − 2)(s − 3)
2 s 2 s−2 s−3
11 5 1
8 1
3s − 1
=−
−
+
.
s(s − 2)(s − 3)
6 s 2 s−2 3 s−3
1 5
x = − + e2t − 2e3t
2 2
and
1 5
8
y = − − e2t + e3t .
6 2
3
6. Taking the Laplace transform of the system gives
(s + 1)L {x} − (s − 1)L {y} = −1
sL {x} + (s + 2)L {y} = 1
so that
L {y} =
and
s + 1/2
s + 1/2
√
=
s2 + s + 1
(s + 1/2)2 + ( 3/2)2
√
√
3/2
−3/2
√
=− 3
.
L {x} = 2
2
s +s+1
(s + 1/2) + ( 3/2)2
√
Then
−t/2
y=e
3
t
cos
2
and
√
√
−t/2
x = − 3e
sin
7. Taking the Laplace transform of the system gives
(s2 + 1)L {x} − L {y} = −2
−L {x} + (s2 + 1)L {y} = 1
so that
L {x} =
−2s2 − 1
1 1
3 1
=− 2 −
4
2
s + 2s
2s
2 s2 + 2
3
t.
2
487
488
CHAPTER 7
THE LAPLACE TRANSFORM
and
√
3
1
x = − t − √ sin 2 t.
2
2 2
Then
√
1
3
y = x + x = − t + √ sin 2 t.
2
2 2
8. Taking the Laplace transform of the system gives
(s + 1)L {x} + L {y} = 1
4L {x} − (s + 1)L {y} = 1
so that
L {x} =
and
L {y} =
Then
s2
s+2
s+1
1
2
=
+
2
2
+ 2s + 5
(s + 1) + 2
2 (s + 1)2 + 22
s+1
2
−s + 3
=−
+2
.
s2 + 2s + 5
(s + 1)2 + 22
(s + 1)2 + 22
1
x = e−t cos 2t + e−t sin 2t
2
and
y = −e−t cos 2t + 2e−t sin 2t.
9. Adding the equations and then subtracting them gives
1
d2 x
= t2 + 2t
dt2
2
d2 y
1
= t2 − 2t.
2
dt
2
Taking the Laplace transform of the system gives
1 4!
1 3!
1
+
L {x} = 8 +
5
s 24 s
3 s4
and
L {y} =
so that
x=8+
1 4 1 3
t + t
24
3
1 3!
1 4!
−
5
24 s
3 s4
and
y=
1 4 1 3
t − t .
24
3
10. Taking the Laplace transform of the system gives
(s − 4)L {x} + s3 L {y} =
s2
6
+1
(s + 2)L {x} − 2s3 L {y} = 0
so that
L {x} =
4 1
4 s
8 1
4
=
−
−
2
2
(s − 2)(s + 1)
5 s − 2 5 s + 1 5 s2 + 1
7.6
Systems of Linear Differential Equations
and
L {y} =
s3 (s
1
2
6 s
8 1
2s + 4
2
1 1
= − 2 −2 3 +
−
+
.
2
2
− 2)(s + 1)
s s
s
5 s − 2 5 s + 1 5 s2 + 1
Then
4
8
4
x = e2t − cos t − sin t
5
5
5
and
8
1
6
y = 1 − 2t − 2t2 + e2t − cos t + sin t.
5
5
5
11. Taking the Laplace transform of the system gives
s2 L {x} + 3(s + 1)L {y} = 2
s2 L {x} + 3L {y} =
so that
L {x} = −
Then
1
(s + 1)2
1
1
1 2
1
2s + 1
= + 2+
.
−
3
+ 1)
s s
2s
s+1
s3 (s
1
x = 1 + t + t2 − e−t
2
and
1
1
1
1
1
y = te−t − x = te−t + e−t − .
3
3
3
3
3
12. Taking the Laplace transform of the system gives
(s − 4)L {x} + 2L {y} =
−3L {x} + (s + 1)L {y} =
2e−s
s
1 e−s
+
2
s
so that
−1
2
+ e−s
(s − 1)(s − 2)
(s − 1)(s − 2)
1
1
1
1
−
+ e−s −2
+2
=
s−1 s−2
s−1
s−2
L {x} =
and
2/s + 1
s/2 − 2
+ e−s
(s − 1)(s − 2)
(s − 1)(s − 2)
1 1
1
1
1
3 1
−
+ e−s
−3
+2
.
=
2 s−1 2 s−2
s
s−1
s−2
L {y} =
Then
x = et − e2t + −2et−1 + 2e2(t−1) U (t − 1)
and
y=
3 t 1 2t e − e + 1 − 3et−1 + 2e2(t−1) U (t − 1).
2
2
489
490
CHAPTER 7
THE LAPLACE TRANSFORM
13. The system is
x1 = −3x1 + 2(x2 − x1 )
x2 = −2(x2 − x1 )
x1 (0) = 1
x1 (0) = 0
x2 (0) = 0.
x2 (0) = 1
Taking the Laplace transform of the system gives
(s2 + 5)L {x1 } − 2L {x2 } = 1
−2L {x1 } + (s2 + 2)L {x2 } = s
so that
and
√
s2 + 2s + 2
2 s
1 1
2 s
4
6
=
+
−
+ √
L {x1 } = 4
s + 7s2 + 6
5 s2 + 1 5 s2 + 1 5 s2 + 6 5 6 s2 + 6
√
s3 + 5s + 2
4 s
2 1
1 s
2
6
=
+
+
− √ 2
.
L {x2 } = 2
2
2
2
2
(s + 1)(s + 6)
5 s +1 5 s +1 5 s +6 5 6 s +6
Then
x1 =
√
√
1
2
2
4
cos t + sin t − cos 6 t + √ sin 6 t
5
5
5
5 6
x2 =
√
√
4
2
1
2
cos t + sin t + cos 6 t − √ sin 6 t.
5
5
5
5 6
and
14. In this system x1 and x2 represent displacements of masses m1 and m2 from their equilibrium
positions. Since the net forces acting on m1 and m2 are
−k1 x1 + k2 (x2 − x1 )
and
− k2 (x2 − x1 ) − k3 x2 ,
respectively, Newton’s second law of motion gives
m1 x1 = −k1 x1 + k2 (x2 − x1 )
m2 x2 = −k2 (x2 − x1 ) − k3 x2 .
Using k1 = k2 = k3 = 1, m1 = m2 = 1, x1 (0) = 0, x1 (0) = −1, x2 (0) = 0, and x2 (0) = 1, and
taking the Laplace transform of the system, we obtain
(2 + s2 )L {x1 } − L {x2 } = −1
L {x1 } − (2 + s2 )L {x2 } = −1
so that
1
+3
and
L {x2 } =
√
1
x1 = − √ sin 3 t
3
and
√
1
x2 = √ sin 3 t.
3
L {x1 } = −
Then
s2
s2
1
.
+3
7.6
Systems of Linear Differential Equations
15. (a) By Kirchhoff’s first law we have i1 = i2 + i3 . By Kirchhoff’s second law, on each loop
we have E(t) = Ri1 + L1 i2 and E(t) = Ri1 + L2 i3 or L1 i2 + Ri2 + Ri3 = E(t) and
L2 i3 + Ri2 + Ri3 = E(t).
(b) Taking the Laplace transform of the system
0.01i2 + 5i2 + 5i3 = 100
0.0125i3 + 5i2 + 5i3 = 100
gives
(s + 500)L {i2 } + 500L {i3 } =
10,000
s
400L {i2 } + (s + 400)L {i3 } =
8,000
s
so that
L {i3 } =
s2
8,000
80 1 80
1
=
−
.
+ 900s
9 s
9 s + 900
Then
i3 =
80 80 −900t
− e
9
9
and
i2 = 20 − 0.0025i3 − i3 =
100 100 −900t
−
e
.
9
9
(c) i1 = i2 + i3 = 20 − 20e−900t
16. (a) Taking the Laplace transform of the system
i2 + i3 + 10i2 = 120 − 120U (t − 2)
−10i2 + 5i3 + 5i3 = 0
gives
(s + 10)L {i2 } + sL {i3 } =
120 1 − e−2s
s
−10sL {i2 } + 5(s + 1)L {i3 } = 0
so that
60
12 120(s + 1)
48
−2s
−2s
L {i2 } =
−
+
=
1
−
e
1
−
e
(3s2 + 11s + 10)s
s + 5/3 s + 2
s
and
L {i3 } =
Then
and
240
240 240
−2s
−2s
1
−
e
−
=
.
1
−
e
3s2 + 11s + 10
s + 5/3 s + 2
i2 = 12 + 48e−5t/3 − 60e−2t − 12 + 48e−5(t−2)/3 − 60e−2(t−2) U (t − 2)
i3 = 240e−5t/3 − 240e−2t − 240e−5(t−2)/3 − 240e−2(t−2) U (t − 2).
491
492
CHAPTER 7
THE LAPLACE TRANSFORM
(b) i1 = i2 + i3 = 12 + 288e−5t/3 − 300e−2t − 12 + 288e−5(t−2)/3 − 300e−2(t−2) U (t − 2)
17. Taking the Laplace transform of the system
i2 + 11i2 + 6i3 = 50 sin t
i3 + 6i2 + 6i3 = 50 sin t
gives
(s + 11)L {i2 } + 6L {i3 } =
6L {i2 } + (s + 6)L {i3 } =
50
s2 + 1
50
+1
s2
so that
L {i2 } =
20 1
375
1
145 s
85
1
50s
=−
+
+
+
.
2
2
2
(s + 2)(s + 15)(s + 1)
13 s + 2 1469 s + 15 113 s + 1 113 s + 1
Then
i2 = −
and
i3 =
20 −2t
375 −15t 145
85
e
e
cos t +
sin t
+
+
13
1469
113
113
1
810
25
11
30
250 −15t 280
sin t − i2 − i2 = e−2t +
e
cos t +
sin t.
−
3
6
6
13
1469
113
113
18. Taking the Laplace transform of the system
0.5i1 + 50i2 = 60
0.005i2 + i2 − i1 = 0
gives
sL {i1 } + 100L {i2 } =
120
s
−200L {i1 } + (s + 200)L {i2 } = 0
so that
L {i2 } =
61 6
s + 100
100
24,000
6
=
−
−
.
s(s2 + 200s + 20,000)
5 s 5 (s + 100)2 + 1002 5 (s + 100)2 + 1002
Then
i2 =
6 6 −100t
6
− e
cos 100t − e−100t sin 100t
5 5
5
and
i1 = 0.005i2 + i2 =
6 6 −100t
− e
cos 100t.
5 5
7.6
Systems of Linear Differential Equations
19. Taking the Laplace transform of the system
2i1 + 50i2 = 60
0.005i2 + i2 − i1 = 0
gives
2sL {i1 } + 50L {i2 } =
60
s
−200L {i1 } + (s + 200)L {i2 } = 0
so that
L {i2 } =
s(s2
6,000
+ 200s + 5,000)
√
√
s + 100
50 2
6 2
61 6
√
√
−
−
.
=
5 s 5 (s + 100)2 − (50 2 )2
5 (s + 100)2 − (50 2 )2
Then
and
√
√
√
6 6 −100t
6 2 −100t
e
cosh 50 2 t −
sinh 50 2 t
i2 = − e
5 5
5
√
√
√
6 6 −100t
9 2 −100t
e
cosh 50 2 t −
sinh 50 2 t.
i1 = 0.005i2 + i2 = − e
5 5
10
20. (a) Using Kirchhoff’s first law we write i1 = i2 + i3 . Since i2 = dq/dt we have i1 − i3 = dq/dt.
Using Kirchhoff’s second law and summing the voltage drops across the shorter loop
gives
1
(1)
E(t) = iR1 + q,
C
so that
1
1
q.
E(t) −
i1 =
R1
R1 C
Then
1
1
dq
= i1 − i3 =
q − i3
E(t) −
dt
R1
R1 C
and
dq
1
+ q + R1 i3 = E(t).
dt
C
Summing the voltage drops across the longer loop gives
R1
E(t) = i1 R1 + L
di3
+ R2 i3 .
dt
Combining this with (1) we obtain
i1 R1 + L
or
L
di3
1
+ R2 i3 = i1 R1 + q
dt
C
1
di3
+ R2 i3 − q = 0.
dt
C
493
494
CHAPTER 7
THE LAPLACE TRANSFORM
(b) Using L = R1 = R2 = C = 1, E(t) = 50e−t U (t − 1) = 50e−1 e−(t−1) U (t − 1),
q(0) = i3 (0) = 0, and taking the Laplace transform of the system we obtain
(s + 1)L {q} + L {i3 } =
50e−1 −s
e
s+1
(s + 1)L {i3 } − L {q} = 0,
so that
L {q} =
50e−1 e−s
(s + 1)2 + 1
and
q(t) = 50e−1 e−(t−1) sin (t − 1)U (t − 1) = 50e−t sin (t − 1)U (t − 1).
21. (a) Taking the Laplace transform of the system
4θ1 + θ2 + 8θ1 = 0
θ1 + θ2 + 2θ2 = 0
gives
4 s2 + 2 L {θ1 } + s2 L {θ2 } = 3s
s2 L {θ1 } + s2 + 2 L {θ2 } = 0
so that
3s2 + 4 s2 + 4 L {θ2 } = −3s3
or
L {θ2 } =
Then
θ2 =
s
3 s
1
−
.
2 s2 + 4/3 2 s2 + 4
3
1
2
cos √ t − cos 2t
2
2
3
so that
θ1 =
θ1 = −θ2 − 2θ2
and
2
1
3
cos √ t + cos 2t.
4
4
3
(b) x
θ2
θ1
02
02
01
01
3π
6π
3π
t
–1
–1
–2
–2
6π
t
7.6
Systems of Linear Differential Equations
Mass m2 has extreme displacements of greater magnitude. Mass m1 first passes through
its equilibrium position at about t = 0.87, and mass m2 first passes through its equilibrium position at about t = 0.66. The motion of the pendulums is not periodic since
√
√
√
cos (2t/ 3 ) has period 3 π, cos 2t has period π, and the ratio of these periods is 3 ,
which is not a rational number.
(c) x
θ2
2
1
–1
–.5
.5
1
θ1
–1
–2
The Lissajous curve is plotted for 0 ≤ t ≤ 30.
(d)
t=0
t=3
t=6
t=1
t=4
t=7
t=2
t
θ1
θ2
t=5
1
2
3
4
5
6
7
8
9
10
–0.2111
–0.6585
0.4830
–0.1325
–0.4111
0.8327
0.0458
–0.9639
0.3534
0.4370
0.8263
0.6438
–1.9145
0.1715
1.6951
–0.8662
–0.3186
0.9452
–1.2741
–0.3502
t=8
t=9
(e) Using a CAS to solve θ1 (t) = θ2 (t) we see that θ1 = θ2 (so
that the double pendulum is straight out) when t is about 0.75
seconds.
t = 10
t = 0.75
495
496
CHAPTER 7
THE LAPLACE TRANSFORM
(f ) To make a movie of the pendulum it is necessary to locate the mass in the plane as a
function of time. Suppose that the upper arm is attached to the origin and that the
equilibrium position lies along the negative y-axis. Then mass m1 is at (x, (t), y1 (t)) and
mass m2 is at (x2 (t), y2 (t)), where
x1 (t) = 16 sin θ1 (t)
and
y1 (t) = −16 cos θ1 (t)
x2 (t) = x1 (t) + 16 sin θ2 (t)
and
y2 (t) = y1 (t) − 16 cos θ2 (t).
and
A reasonable movie can be constructed by letting t range from 0 to 10 in increments of
0.1 seconds.
Chapter 7 in Review
ˆ
1
1. L {f (t)} =
te−st dt +
ˆ
0
ˆ
(2 − t)e−st dt =
1
4
2. L {f (t)} =
∞
e−st dt =
2
1
2
− e−s
s2 s2
1 −2s
− e−4s
e
s
3. False; consider f (t) = t−1/2 .
4. False, since f (t) = (et )10 = e10t .
5. True, since lims→∞ F (s) = 1 = 0. (See Theorem 7.2.3 in the text.)
6. False; consider f (t) = 1 and g(t) = 1.
7. L e−7t =
1
s+7
8. L te−7t =
1
(s + 7)2
2
s2 + 4
2
10. L e−3t sin 2t =
(s + 3)2 + 4
2
4s
d
= 2
11. L {t sin 2t} = −
ds s2 + 4
(s + 4)2
9. L {sin 2t} =
12. L {sin 2tU (t − π)} = L {sin 2(t − π)U (t − π)} =
13. L −1
20
s6
14. L −1
1
3s − 1
= L −1
1 5!
6 s6
= L −1
1
= t5
6
1
1
3 s − 1/3
1
= et/3
3
s2
2
e−πs
+4
Chapter 7 in Review
15. L −1
1
(s − 5)3
16. L −1
1
2
s −5
17. L −1
18. L −1
L −1
19.
s2
1
= L −1
2
1
= t2 e5t
2
√
1
1 √
= − √ e− 5 t + √ e 5 t
2 5
2 5
1
1
1
1
√ + √
√
= L −1 − √
2 5 s+ 5 2 5 s− 5
s
− 10s + 29
1 −5s
e
s2
2
(s − 5)3
= L −1
2
s−5
5
+
2
2
(s − 5) + 2
2 (s − 5)2 + 22
5
= e5t cos 2t + e5t sin 2t
2
= (t − 5)U (t − 5)
s + π −s
e
s2 + π 2
= L −1
s
π
e−s + 2
e−s
s2 + π 2
s + π2
= cos π(t − 1)U (t − 1) + sin π(t − 1)U (t − 1)
20. L −1
L2 s2
1
+ n2 π 2
=
1 L −1
L
L2 nπ
s2
nπ/L
+ (n2 π 2 )/L2
=
1
nπ
sin
t
Lnπ
L
21. L e−5t exists for s > −5.
d
22. L te8t f (t) = − F (s − 8).
ds
23. L {eat f (t − k)U (t − k)} = e−ks L {ea(t+k) f (t)} = e−ks eak L {eat f (t)} = e−k(s−a) F (s − a)
ˆ t
1
F (s − a)
, whereas
eaτ f (τ ) dτ = L {eat f (t)} =
24. L
s
s
0
ˆ t
ˆ t
F
(s)
F (s − a)
.
L eat
f (τ ) dτ = L
f (τ ) dτ =
=
s s−a
0
0
s→s−a
s→s−a
25. f (t)U (t − t0 )
26. f (t) − f (t)U (t − t0 )
27. f (t − t0 )U (t − t0 )
28. f (t) − f (t)U (t − t0 ) + f (t)U (t − t1 )
29. f (t) = t − [(t − 1) + 1]U (t − 1) + U (t − 1) − U (t − 4) = t − (t − 1)U (t − 1) − U (t − 4)
1
1
1
L {f (t)} = 2 − 2 e−s − e−4s
s
s
s
t
1
1
1 −4(s−1)
e
L e f (t) =
−
e−(s−1) −
(s − 1)2 (s − 1)2
s−1
30. f (t) = sin tU (t − π) − sin tU (t − 3π) = − sin (t − π)U (t − π) + sin (t − 3π)U (t − 3π)
1
1
e−πs + 2
e−3πs
L {f (t)} = − 2
s +1
s +1
1
1
e−π(s−1) +
e−3π(s−1)
L et f (t) = −
2
(s − 1) + 1
(s − 1)2 + 1
497
498
CHAPTER 7
THE LAPLACE TRANSFORM
31. f (t) = 2 − 2U (t − 2) + [(t − 2) + 2]U (t − 2) = 2 + (t − 2)U (t − 2)
1
2
+ 2 e−2s
s s
t
1
2
+
L e f (t) =
e−2(s−1)
s − 1 (s − 1)2
L {f (t)} =
32. f (t) = t − tU (t − 1) + (2 − t)U (t − 1) − (2 − t)U (t − 2) = t − 2(t − 1)U (t − 1) + (t − 2)U (t − 2)
2
1
1
− 2 e−s + 2 e−2s
2
s
s
s
t
1
2
1
L e f (t) =
−
e−(s−1) +
e−2(s−1)
2
2
(s − 1)
(s − 1)
(s − 1)2
L {f (t)} =
33. The graph of
∞
(−1)k+1 U (t − k) = −1 + 2U (t − 1) − 2U (t − 2) + 2U (t − 3) − · · ·
f (t) = −1 + 2
k=1
is
y
1
1
2
3
4
5
6
t
1
One way of proceeding to find the Laplace transform is to take the transform term-by-term
of the series:
2
2
1 2
L {f (t)} = − + e−s − e−2s + e−3s − · · ·
s s
s
s
←−
geometric series
For s > 0,
1 2 −s
1 2
e−s
L {f (t)} = − +
e − e−2s + e−3s − · · · = − + ·
s s
s s 1 + e−s
=
e−s − 1
s (1 + e−s )
Alternatively, since f is a periodic functions it can also be defined by
f (t) =
−1, 0 ≤ t < 1
1,
1 ≤ t < 2,
where f (t + 2) = f (t).
Chapter 7 in Review
By Theorem 7.4.3 with p = 2 we get
ˆ 1
ˆ 2
1
−st
−st
(−1) e dt +
(1) e dt
L {f (t)} =
1 − e−2s
0
1
1 −s 1 1 −2s 1 −s
−1
1
e − − e
1 − 2e−s + e−2s
=
+ e
=
−2s
−2s
1−e
s
s s
s
s (1 − e )
− (1 − e−s )
=
s (1 − e−2s )
2
Using 1 − e−2s = (1 + e−s ) (1 − e−s ) and algebra the last expression is the same as
L {f (t)} =
e−s − 1
.
s (1 + e−s )
34. The graph of
∞
(2k + 1 − t) [U (t − 2k) − U (t − 2k − 1)]
f (t) =
k=0
= (1 − t) [U (t) − U (t − 1)] + (3 − t) [U (t − 2) − U (t − 3)] + · · ·
is
y
1
2
3
4
5
6
t
Since f is a periodic function it can also be defined by
1 − t, 0 ≤ t < 1
f (t) =
where f (t + 2) = f (t).
0,
1≤t<2
By Theorem 7.4.3 with p = 2 we get
ˆ 1
ˆ 2
1 −s
1
1
1
1
−st
−st
+
(1 − t) e dt +
0 · e dt =
e − 2
L {f (t)} =
1 − e−2s
1 − e−2s s s2
s
0
1
=
s − 1 + e−s
.
s2 (1 − e−2s )
35. Taking the Laplace transform of the differential equation we obtain
L {y} =
so that
2
1
5
+
2
(s − 1)
2 (s − 1)3
1
y = 5tet + t2 et .
2
499
500
CHAPTER 7
THE LAPLACE TRANSFORM
36. Taking the Laplace transform of the differential equation we obtain
L {y} =
=
(s −
1
− 8s + 20)
1)2 (s2
1
1
1
s−4
2
6
5
6
+
−
+
169 s − 1 13 (s − 1)2 169 (s − 4)2 + 22 338 (s − 4)2 + 22
so that
y=
1
6 4t
5 4t
6 t
e + tet −
e cos 2t +
e sin 2t.
169
13
169
338
37. Taking the Laplace transform of the given differential equation we obtain
L {y} =
1
2
s3 + 6s2 + 1
− 2
e−2s −
e−2s
2
s (s + 1)(s + 5) s (s + 1)(s + 5)
s(s + 1)(s + 5)
=−
6 1 1 1
3
1
13
1
· + · 2+ ·
−
·
25 s 5 s
2 s + 1 50 s + 5
6 1 1 1
1
1
1
1
− − · + · 2+ ·
−
·
e−2s
25 s 5 s
4 s + 1 100 s + 5
2 1 1
1
1
1
· − ·
+
·
e−2s
−
5 s 2 s + 1 10 s + 5
so that
y=−
1
3
13
4
1
6
+ t + e−t − e−5t − U (t − 2) − (t − 2)U (t − 2)
25 5
2
50
25
5
9 −5(t−2)
1
e
U (t − 2).
+ e−(t−2) U (t − 2) −
4
100
38. Taking the Laplace transform of the differential equation we obtain
L {y} =
s3 + 2
2 + 2s + s2 −s
−
e
s3 (s − 5)
s3 (s − 5)
2 1
2 1
1 2
127 1
37 1 12 1
1 2
37 1
−
− −
−
e−s
=−
−
+
−
+
125 s 25 s2 5 s3 125 s − 5
125 s 25 s2 5 s3 125 s − 5
so that
2
2
1 2 127 5t
37
12
1
37 5(t−1)
U (t − 1).
y=−
− t− t +
e − −
− (t − 1) − (t − 1)2 +
e
125 25
5
125
125 25
5
125
39. The function in Figure 7.R.10 is
f (t) =
⎧
0,
0≤t<1
⎪
⎪
⎪
⎪
⎨t − 1, 1 ≤ t < 2
⎪
3 − t,
⎪
⎪
⎪
⎩
0,
2≤t<3
t≥3
Chapter 7 in Review
or
f (t) = (t − 1)U (t − 1) − 2(t − 2)U (t − 2) + (t − 3)U (t − 3).
The transform of the differential equation is
sY (s) − 1 + 2Y (s) =
so
e−s 2e−2s e−3s
−
+ 2
s2
s2
s
1
1
2
1
+
e−s − 2
e−2s + 2
e−3s ,
s + 2 s2 (s + 2)
s (s + 2)
s (s + 2)
Y (s) =
and
−2t
y(t) = e
1
1 1
+ − + (t − 1) + e−2(t−1) U (t − 1)
4 2
4
1
1 1
− 2 − + (t − 2) + e−2(t−2) U (t − 2)
4 2
4
1 1
1
+ − + (t − 3) + e−2(t−3) U (t − 3).
4 2
4
40. The tranform of the differential equation is
∞
s2 Y (s) − 3 + 5sY (s) + 4Y (s) = 12
(−1)k
k=0
so
∞
s2 + 5s + 4 Y (s) = 3 + 12
(−1)k
k=0
and
3
+
Y (s) = 2
s + 5s + 4
Thus
1
1
−
+
Y (s) =
s+1 s+4
and
−t
y(t) = e
−4t
−e
∞
+
∞
(−1)k
k=0
k
(−1)
k=0
e−ks
s
12
e−ks .
s (s2 + 5s + 4)
∞
4
1
3
−
+
e−ks
s s+1 s−4
(−1)k 3 − 4e−(t−k) + e−4(t−k) U (t − k).
k=0
41. Taking the Laplace transform of the integral equation we obtain
L {y} =
so that
e−ks
s
1
1 2
1
+ 2+
s s
2 s3
1
y(t) = 1 + t + t2 .
2
501
502
CHAPTER 7
THE LAPLACE TRANSFORM
42. Taking the Laplace transform of the integral equation we obtain
(L {f })2 = 6 ·
6
s4
or L {f } = ±6 ·
1
s2
so that f (t) = ±6t.
43. Taking the Laplace transform of the system gives
sL {x} + L {y} =
1
+1
s2
4L {x} + sL {y} = 2
so that
L {x} =
Then
s2 − 2s + 1
11 1 1
9 1
=−
+
+
.
s(s − 2)(s + 2)
4 s 8 s−2 8 s+2
9
1 1
x = − + e2t + e−2t
4 8
8
9
1
and y = −x + t = e−2t − e2t + t.
4
4
44. Taking the Laplace transform of the system gives
s2 L {x} + s2 L {y} =
1
s−2
2sL {x} + s2 L {y} = −
so that
L {x} =
and
L {y} =
Then
x=
1
s−2
11 1 1
1
2
−
+
=
2
s(s − 2)
2 s 2 s − 2 (s − 2)2
1
31 1 1
3 1
−s − 2
−
−
=−
+
.
2
2
− 2)
4 s 2s
4 s − 2 (s − 2)2
s2 (s
1 1 2t
− e + te2t
2 2
45. The integral equation is
3
3 1
and y = − − t + e2t − te2t .
4 2
4
ˆ
t
10i + 2
i(τ ) dτ = 2t2 + 2t.
0
Taking the Laplace transform we obtain
s
2
s+2
9
2
45
9
2
9
4
= 2
=− + 2 +
=− + 2 +
.
+ 2
L {i} =
3
s
s
10s + 2
s (5s + 2)
s s
5s + 1
s s
s + 1/5
Thus
i(t) = −9 + 2t + 9e−t/5 .
Chapter 7 in Review
46. The differential equation is
dq
1 d2 q
+ 10 + 100q = 10 − 10U (t − 5).
2
2 dt
dt
Taking the Laplace transform we obtain
20
−5s
1
−
e
s(s2 + 20s + 200)
1 1
1
s + 10
1
10
=
−
−
1 − e−5s
2
2
2
2
10 s 10 (s + 10) + 10
10 (s + 10) + 10
L {q} =
so that
q(t) =
1
1
1
− e−10t cos 10t − e−10t sin 10t
10 10
10
1
1
1
− e−10(t−5) cos 10(t − 5) − e−10(t−5) sin 10(t − 5) U (t − 5).
−
10 10
10
47. Taking the Laplace transform of the given differential equation we obtain
c1 2!
1
1
c2 3!
5!
5! −sL/2
2w0 L 4!
+
· 5−
· 6+
· 6e
· 3+
·
L {y} =
EIL 48 s
120 s
120 s
2 s
6 s4
so that
1 5
1
L
2w0 L 4
L 5
c 1 2 c2 3
x −
x +
U x−
y=
x−
+ x + x
EIL 48
120
120
2
2
2
6
where y (0) = c1 and y (0) = c2 . Using y (L) = 0 and y (L) = 0 we find
c2 = −w0 L/4EI.
c1 = w0 L2 /24EI,
Hence
1 5 L 4 L2 3 L3 2 1
L
w0
L 5
− x + x −
x +
x +
.
U x−
y=
x−
12EIL
5
2
2
4
5
2
2
48. (a) In this case the boundary conditions are y(0) = 0, y (0) = 0, y(π) = 0, and y (π) = 0.
w 0
L y (4) + 4L {y} = L
EI
c
0
c
0
! "#1 $
! "# $ ! "#2 $
!"#$
w0
s4 Y (s) − s3 y(0) −s2 y (0) −s y (0) − y (0) +4Y (s) =
sEI
4
w0
s + 4 Y (s) = c1 s2 + c2 +
sEI
503
504
CHAPTER 7
THE LAPLACE TRANSFORM
Thus
Y (s) = c1
s4
w0 /EI
s2
1
+ c2 4
+
+4
s + 4 s (s4 + 4)
w0 /EI
s2
1
+ c2 4
+
s4 + 4
s + 4 s (s2 − 2s + 2) (s2 + 2s + 2)
w0 2
s−1
s+1
s2
1
+ c2 4
+
− 2
− 2
= c1 4
s +4
s + 4 8EI s s − 2s + 2 s + 2s + 2
c2 4
w0 2
s−1
s+1
c1 2s2
+
+
−
−
=
2
4
4
2 s +4
4 s + 4 8EI s (s − 1) + 1 (s + 1)2 + 1
= c1
Using the table in Appendix III the inverse transform is
c2
c1
(sin x cosh x + cos x sinh x) + (sin x cosh x − cos x sinh x)
2
4
w0
+
2 − ex cos x − e−x cos x
8EI
c2
w0
c1
(sin x cosh x + cos x sinh x) + (sin x cosh x − cos x sinh x) +
[1 − cos x cosh x]
=
2
4
4EI
y(x) =
The remaining conditions y(π) = 0 and y (0) = 0 then give
c1 =
w0
(1 + cosh π) csch π,
4EI
and c2 = −
w0
(1 + cosh π) csch π
2EI
Therefore,
y(x) =
w0
(1 + cosh π) csch π (sin x cosh x + cos x sinh x)
8EI
w0
w0
(1 + cosh π) csch π (sin x cosh x − cos x sinh x) +
[1 − cos x cosh x]
−
8EI
4EI
(b) In this case the boundary conditions are y(0) = y (0) = 0 and y(π) = y (π) = 0. If we
let c1 = y (0) and c2 = y (0) then
w0
π s4 L {y} − sy(0) − s2 y (0) − s3 y (0) − y (0) + 4L {y} =
L δ x−
EI
2
and
L {y} =
2s
c2
4
w0
4
c1
· 4
+
· 4
+
· 4
e−sπ/2 .
2 s +4
4 s + 4 4EI s + 4
From the table of transforms we get
y=
c2
c1
sin x sinh x + (sin x cosh x − cos x sinh x)
2
4
w0 π
π
π
π +
sin x −
cosh x −
− cos x −
sinh x −
U
4EI
2
2
2
2
Using y(π) = 0 and y (π) = 0 we find
c1 =
w0 sinh π2
,
EI sinh π
c2 = −
w0 cosh π2
.
EI sinh π
x−
π
2
Chapter 7 in Review
Hence
y=
w0 cosh π2
w0 sinh π2
sin x sinh x −
(sin x cosh x − cos x sinh x)
2EI sinh π
4EI sinh π
π
π
π
π w0 sin x −
cosh x −
− cos x −
sinh x −
U
+
4EI
2
2
2
2
x−
π
.
2
49. (a) With ω 2 = g/l and K = k/m the system of differential equations is
θ1 + ω 2 θ1 = −K(θ1 − θ2 )
θ2 + ω 2 θ2 = K(θ1 − θ2 ).
Denoting the Laplace transform of θ(t) by Θ(s) we have that the Laplace transform of
the system is
(s2 + ω 2 )Θ1 (s) = −KΘ1 (s) + KΘ2 (s) + sθ0
(s2 + ω 2 )Θ2 (s) = KΘ1 (s) − KΘ2 (s) + sψ0 .
If we add the two equations, we get
Θ1 (s) + Θ2 (s) = (θ0 + ψ0 )
s2
s
+ ω2
which implies
θ1 (t) + θ2 (t) = (θ0 + ψ0 ) cos ωt.
This enables us to solve for first, say, θ1 (t) and then find θ2 (t) from
θ2 (t) = −θ1 (t) + (θ0 + ψ0 ) cos ωt.
Now solving
(s2 + ω 2 + K)Θ1 (s) − KΘ2 (s) = sθ0
−kΘ1 (s) + (s2 + ω 2 + K)Θ2 (s) = sψ0
gives
[(s2 + ω 2 + K)2 − K 2 ]Θ1 (s) = s(s2 + ω 2 + K)θ0 + Ksψ0 .
Factoring the difference of two squares and using partial fractions we get
Θ1 (s) =
θ0 + ψ 0
s
s
s(s2 + ω 2 + K)θ0 + Ksψ0
θ0 − ψ 0
=
,
+
2
2
2
2
2
2
2
(s + ω )(s + ω + 2K)
2
s +ω
2
s + ω 2 + 2K
so
θ1 (t) =
θ0 − ψ 0
θ0 + ψ 0
cos ωt +
cos ω 2 + 2K t.
2
2
Then from θ2 (t) = −θ1 (t) + (θ0 + ψ0 ) cos ωt we get
θ2 (t) =
θ0 − ψ 0
θ0 + ψ 0
cos ωt −
cos ω 2 + 2K t.
2
2
505
506
CHAPTER 7
THE LAPLACE TRANSFORM
(b) With the initial conditions θ1 (0) = θ0 , θ1 (0) = 0, θ2 (0) = θ0 , θ2 (0) = 0 we have
θ1 (t) = θ0 cos ωt,
θ2 (t) = θ0 cos ωt.
Physically this means that both pendulums swing in the same direction as if they were
free since the spring exerts no influence on the motion (θ1 (t) and θ2 (t) are free of K).
With the initial conditions θ1 (0) = θ0 , θ1 (0) = 0, θ2 (0) = −θ0 , θ2 (0) = 0 we have
θ2 (t) = −θ0 cos ω 2 + 2K t.
θ1 (t) = θ0 cos ω 2 + 2K t,
Physically this means that both pendulums swing in the opposite directions, stretching
and compressing the spring. The amplitude of both displacements is |θ0 |. Moreover,
θ1 (t) = θ0 and θ2 (t) = −θ0 at precisely the same times. At these times the spring is
stretched to its maximum.
50. (a) We will find the first two times for which x (t) = 0 and then ontain the rest of the times
using periodicity of x (t). The solution of
x + ω 2 x = F,
is
x(t) =
x(0) = x0 ,
F
F
x0 − 2 cos ωt + 2
ω
ω
x (0) = 0
x (t) =
so
F
x0 − 2 (−ω sin ωt) .
ω
The latter equation is 0 when t = π/ω. The next initial-value problem is then
x + ω 2 x = −F,
x
π
ω
=
2F
− x0 ,
ω2
x
π
ω
= 0,
where π/ω = T /2. From the solution of this problem,
F
3F
x(t) = x0 − 2 cos ωt − 2 ,
ω
ω
we see from
3F
x (t) = x0 − 2 (−ω sin ωt) = 0
ω
that t = 2π/ω = T . Since x (t) has period T = 2π/ω we can see that x (t) = 0 at the
times
2π
2π 2pi
π
+
k
and
+
k,
for k = 0, 1, 2, . . .
ω
ω
ω
ω
which are
π 2π 3π
T
3T
0, , , , . . . or 0, , T,
,...
ω ω ω
2
2
(b) There is no motin unless the intial displacement is such that the force of the spring is
greater than the force due to friction. That is,
k|x0 | > fk
or
k
fk
|x0 | >
m
m
or ω 2 |x0 | > F.
Chapter 7 in Review
(c) From part (b), an intial displacement x(0) = x0 for which |x0 | > F/ω 2 will result in
motion. On the other hand, an initial displacement x0 for which |x0 | ≤ F/ω 2 will be
insufficient to overcome the force of friction and the system will be “dead”.
(d) The system can be described by the initial-value problem
x + ω 2 x = g(t),
x(0) = x0 ,
x (0) = 0,
where g is a version of the meander function shown in Figure 7.4.6 in the text. In this
case the amplitude of the function is F instead of 1, and the length of each line segment
is T /2 rather than a. Then
L x + ω 2 α {x} = α {g} ,
s2 X(s) − sx0 + ω 2 X(s) = G(s),
and
sx0
1
+
G(s).
s2 + w2 s2 + ω 2
Now, using Problem 49 in Section 7.4 with F instead of 1 and a = T /2, we have
F F 1 − e−sT /2
−sT /2
−sT
−3sT /2
=
+
2e
−
2e
+
·
·
·
.
L {g(t)} = G(s) =
1
−
2e
s 1 + e−sT /2
s
X(s) =
Then
1
F
s
−sT /2
−sT
−3sT /2
+
+
2e
−
2e
+
·
·
·
1
−
2e
s2 + ω 2
s s2 + ω 2
s
F 1
s
−sT /2
−sT
−3sT /2
−
= x0 2
+
+
2e
−
2e
+
·
·
·
1
−
2e
s + ω 2 ω 2 s s2 + ω 2
X(s) = x0
and
2F
F
T
T
T
x(t) = x0 cos ωt + (1 − cos ωt) − 2 U t −
− cos ω t −
U t−
ω
ω
2
2
2
+
or
2F
[U (t − T ) − cos ω (t − T )U (t − T )]
ω2
3T
2T
3T
2F
− cos ω t −
U t−
+ ···
− 2 U t−
ω
2
2
2
⎧
⎪
⎪
x0 cos ωt +
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
x0 cos ωt +
⎪
⎪
⎪
⎪
⎪
⎨
F
(1 − cos ωt) ,
ω2
T
F
2F
,
(1 − cos ωt) − 2 1 − cos ω t −
ω2
ω
2
T
F
2F
x(t) =
x0 cos ωt + 2 (1 − cos ωt) − 2 1 − cos ω t −
⎪
⎪
⎪
ω
ω
2
⎪
⎪
⎪
⎪
⎪
⎪
2F
⎪
⎪
+ 2 [1 − cos ω (t − T )] ,
⎪
⎪
ω
⎪
⎪
⎪
⎪
..
⎩
.
0 ≤ t < T /2
T /2 ≤ t < T
T ≤ t < 3T /2
507
508
CHAPTER 7
THE LAPLACE TRANSFORM
(e) The solutions from Problem 28 in Chapter 5 in Review are
x(t) =
4.5 cos t + 1, 0 ≤ t < π
2.5 cos t − 1,
π ≤ t < 2π.
On the interval [0, 2π) the solution is
⎧
F
⎪
⎪
0 ≤ t < T /2
⎪
⎨x0 cos ωt + ω 2 (1 − cos ωt),
x(t) =
⎪
T
F
2F
⎪
⎪
, T /2 ≤ t < T.
⎩x0 cos ωt + 2 (1 − cos ωt) − 2 1 − cos ω t −
ω
ω
2
We let m = 1, k = 1, fk = 1, and x0 = 5.5. Then, when T = 2π, the foregoing becomes
5.5 cos t + (1 − cos t),
0≤t<π
x(t) =
5.5 cos t + (1 − cos t) − 2[1 − cos (t − π)], π ≤ t < 2π.
Simplifying this we get
x(t) =
4.5 cos t + 1, 0 ≤ t < π
2.5 cos t − 1,
π ≤ t2π.
(f ) At t = T /2, x(T /2) = −x0 + 2F/ω 2 . At t = T ,
4F
x(T ) = x0 − 2 =
ω
2F
2F
T
2F
x0 − 2 − 2 = −x
− 2.
ω
ω
2
ω
At t = 3T /2, x(3T /2) = −x0 + 6F/ω 2 = −x(T ) + 2F/ω 2 . We see in general that each
successive oscillation is 2F/ω 2 shorter than the preceding one.
(g) The system will stay in motion until the oscillations bring the mass with the “dead zone”
at which time the motion will cease.
51. (a) Rewriting the system as
d2
=0
dt2
d2 y
= −g.
dt2
Then, taking the Laplace of each equation, we have
s2 X(s) − sx(0) − x (0) = 0
s2 Y (s) − sy(0) − y (0) =
g
.
s
Chapter 7 in Review
Using x(0) = 0, x (0) = v0 cos t, y(0) = 0, y (0) = v0 sin θ where v0 = |v0 |, we have
⎧
⎧
1
⎪
⎨X(s) = (v0 cos θ) 2
⎨s2 X(s) = v0 cos θ
s
g or
1
g
⎪
⎩s2 Y (s) = v0 sin θ −
⎩Y (s) = (v0 sin θ)
− 3
s
2
s
s
Then
⎧
⎨x(t) = (v0 cos θ) t
⎩y(t) = (v0 sin θ) t − 1 gt2
2
x
into the equation for y(t) yields
v0 cos θ
2
x
1
g
x
1
= −
x + tan θ x
+ (v0 sin θ)
y(x) = − g
2
v0 cos θ
v0 cos θ
2 v02 cos2 θ
(b) Substituting t =
The projectile hits the ground when y = 0. This occurs when x = 0, which is the initial
condition, or
0=
x=
−g
2v02 cos2 θ
x + tan θ
v 2 sin 2θ
(tan θ) 2v02 cos2 θ
= 0
g
g
which is the horizontal range.
(c) When 0 < θ < π/2 the complementary angle of θ is π/2 − θ. Substituting this into the
result of part (b) we have
v02 sin 2 π2 − θ
v 2 sin (π − 2θ)
v 2 (sin π cos 2θ − cos π sin 2θ)
π
−θ =
= 0
= 0
R
2
g
g
g
=
v02 sin 2θ
= R(θ).
g
(d) When g = 9.8, θ = 38◦ and v0 = 90 we have from part (b) that
R=
902
sin 76◦ ≈ 801 m.
9.8
Solving
x(t) = (90 cos 38◦ )t = 801 m
for t, we see that the projectile hits the ground after about 11.3 sec.
(e) For θ = 38◦ the curve is shown in blue, while
for θ = 52◦ the curve is shown in red.
509
510
CHAPTER 7
THE LAPLACE TRANSFORM
52. (a) Taking the Laplace transform of the first equation, we obtain
m s2 L {x(t)} − v0 cos θ = −βsL {x(t)} ,
so that
mv0 cos θ
v0 cos θ
=
L {x(t)} =
s (s + β/m)
β
1
1
−
s s + β/m
and
x(t) =
mv0 cos θ
1 − e−βt/m
β
Taking the Laplace transform of the second equation, we obtain
m s2 L {x(t)} − v0 cos θ = −βsL {x(t)} ,
so that
v0 sin θ
g
−
s (s + β/m) s2 (s2 + β/m)
1
mg 1
m2 g 1 m2 g
1
mv0 sin θ 1
−
−
+ 2
,
− 2
=
β
s s + β/m
β s2
β s
β s + β/m
L {x(t)} =
and
m2 g −βt/m
mg
mv0 sin θ m2 g
−βt/m
+ 2 1−e
t− 2 e
−
y(t) =
β
β
β
β
mg
m
mg
=
v0 sin θ +
1 − e−βt/m −
t.
β
β
β
(b) To find when the projectile hits the ground, we use a CAS to solve y(t) = 0. This gives
t0 = 10.23 sec. The range is then approximately x(t0 ) = 536.02 m.
(c) Solving y(t) = 0 when θ = 52◦ gives t1 = 12.79, so the range in this case is x(t1 ) =
418.8 m, which is considerably less than the range of 536.02 ft when θ = 38◦ .
Chapter 7 in Review
(d) For θ = 38◦ the curve is shown in blue, while
for θ = 52◦ the curve is shown in red. We see
that the larger angle of elevation results in a
smaller range for the projectile. Also, while
the curves look at first glance like parabolas,
a closer examination shows that they are not
parabolas (at least not with vertical axes).
511
200
100
0
0
200
400
600
Chapter 8
Systems of Linear First-Order Differential Equations
8.1
Preliminary Theory – Linear Systems
x
3
−5
1. Let X =
. Then X =
X.
y
4
8
4
−7
x
X.
2. Let X =
. Then X =
5
0
y
3. Let
4. Let
5. Let
6. Let
⎛ ⎞
x
⎜ ⎟
X = ⎝ y ⎠.
z
⎛ ⎞
x
⎜ ⎟
X = ⎝ y ⎠.
z
⎛ ⎞
x
⎜ ⎟
X = ⎝ y ⎠.
z
⎛ ⎞
x
⎜ ⎟
X = ⎝ y ⎠.
z
⎞
−3
4 −9
⎟
⎜
0⎠ X.
= ⎝ 6 −1
10
4
3
⎞
⎛
1 −1 0
⎟
⎜
0 2⎠ X.
=⎝ 1
−1
0 1
⎞
⎛
⎛
⎞ ⎛ ⎞ ⎛ ⎞
0
−1
1 −1
1
t
⎟
⎜
⎜
⎟ ⎜ ⎟ ⎜ ⎟
2
1 −1⎠ X + ⎝−3t ⎠ + ⎝ 0⎠ + ⎝ 0⎠.
= ⎝2
2
1
1
1
−t
t2
⎞
⎛ −t
⎛
⎞
e sin 2t
−3 4 0
⎟
⎜
⎜
⎟
= ⎝ 5 0 9⎠ X + ⎝4e−t cos 2t⎠.
0 1 6
−e−t
⎛
Then X
Then X
Then X
Then X
dy
= −x + 3y − et
dt
7.
dx
= 4x + 2y + et ;
dt
8.
dx
= 7x + 5y − 9z − 8e−2t ;
dt
dy
= 4x + y + z + 2e5t ;
dt
9.
dx
= x − y + 2z + e−t − 3t;
dt
dy
= 3x − 4y + z + 2e−t + t;
dt
10.
dx
= 3x − 7y + 4 sin t + (t − 4)e4t ;
dt
dz
= −2y + 3z + e5t − 3e−2t
dt
dz
= −2x + 5y + 6z + 2e−t − t
dt
dy
= x + y + 8 sin t + (2t + 1)e4t
dt
512
8.1
11. Since
X =
−5 −5t
e
−10
Preliminary Theory – Linear Systems
3 −4
−5 −5t
X=
e
4 −7
−10
and
3 −4
X.
X =
4 −7
we see that
12. Since
X =
5 cos t − 5 sin t t
e
2 cos t − 4 sin t
and
−2 5
5 cos t − 5 sin t t
X=
e
−2 4
2 cos t − 4 sin t
−2 5
X =
X.
−2 4
we see that
13. Since
X =
3
2
−3
e−3t/2
−1
1/4
X.
X =
1 −1
we see that
14. Since
and
5 t
4
e +
tet
−1
−4
X =
X =
⎛ ⎞
0
⎜ ⎟
X = ⎝0⎠
0
15. Since
2 1
X.
−1 0
⎛
⎞
⎛ ⎞
1
2
1
0
⎜
⎟
⎜ ⎟
0⎠ X = ⎝0⎠
⎝ 6 −1
−1 −2 −1
0
and
⎞
1
2
1
⎟
⎜
0⎠ X.
X = ⎝ 6 −1
−1 −2 −1
⎛
we see that
⎞
cos t
⎟
⎜
X = ⎝ 12 sin t − 12 cos t⎠
− cos t − sin t
we see that
2 1
5 t
4
X=
e +
tet
−1 0
−1
−4
and
we see that
16. Since
1
3
−1
4
2
X=
e−3t/2
1 −1
−3
⎞
⎛
⎞
cos t
1 0
1
⎟
⎜
⎜
⎟
0⎠ X = ⎝ 12 sin t − 12 cos t⎠
⎝ 1 1
−2 0 −1
− cos t − sin t
⎛
⎛
and
⎛
⎞
1 0
1
⎜
⎟
0⎠ X.
X = ⎝ 1 1
−2 0 −1
513
514
CHAPTER 8
SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
17. Yes, since W (X1 , X2 ) = −2e−8t = 0 the set X1 , X2 is linearly independent on the interval
(−∞, ∞).
18. Yes, since W (X1 , X2 ) = 8e2t = 0 the set X1 , X2 is linearly independent on the interval
(−∞, ∞).
19. No, since W (X1 , X2 , X3 ) = 0 the set X1 , X2 , X3 is linearly dependent on the interval
(−∞, ∞).
20. Yes, since W (X1 , X2 , X3 ) = −84e−t = 0 the set X1 , X2 , X3 is linearly independent on the
interval (−∞, ∞).
21. Since
Xp =
2
−1
and
we see that
Xp
22. Since
Xp
=
0
=
0
1 4
2
−7
2
Xp +
t+
=
3 2
−4
−18
−1
1 4
2
−7
Xp +
t+
.
3 2
−4
−18
and
2
1
−5
0
Xp +
=
1 −1
2
0
2
1
−5
Xp +
.
Xp =
1 −1
2
we see that
23. Since
2
1
et +
tet
Xp =
0
−1
and
we see that
Xp
2 1
1 t
2 t
1
Xp −
e =
e +
tet
3 4
7
0
−1
2 1
1 t
=
Xp −
e.
3 4
7
24. Since
⎞
3 cos 3t
⎟
⎜
0
Xp = ⎝
⎠
−3 sin 3t
we see that
⎞
⎞
⎛
⎛ ⎞
1 2 3
3 cos 3t
−1
⎟
⎟
⎜
⎜
⎜ ⎟
0
⎠
⎝−4 2 0⎠ Xp + ⎝ 4⎠ sin 3t = ⎝
−6 1 0
−3 sin 3t
3
⎛
⎛
and
⎞
⎛ ⎞
1 2 3
−1
⎟
⎜
⎜ ⎟
Xp = ⎝−4 2 0⎠ Xp + ⎝ 4⎠ sin 3t.
−6 1 0
3
⎛
8.2 Homogeneous Linear Systems
25. Let
⎞
6
⎜ ⎟
X1 = ⎝−1⎠ e−t ,
−5
⎞
−3
⎜ ⎟
X2 = ⎝ 1⎠ e−2t ,
1
⎛
Then
⎛ ⎞
2
⎜ ⎟ 3t
X3 = ⎝1⎠ e ,
1
⎛
⎛
⎞
−6
⎜ ⎟
X1 = ⎝ 1⎠ e−t = AX1 ,
5
⎛
⎞
6
⎜ ⎟
X2 = ⎝−2⎠ e−2t = AX2 ,
−2
⎞
⎛
0 6 0
⎟
⎜
and A = ⎝1 0 1⎠ .
1 1 0
⎛ ⎞
6
⎜ ⎟ 3t
X3 = ⎝3⎠ e = AX3 ,
3
and W (X1 , X2 , X3 ) = 20 = 0 so that X1 , X2 , and X3 form a fundamental set for X = AX
on the interval (−∞, ∞).
26. Let
X1 =
1
−1 −
√
2
√
e
2t
,
X2 =
1
−1 +
√
2
√
− 2t
e
,
1 2
−2
1
Xp =
t +
t+
,
0
4
0
−1 −1
A=
.
−1
1
and
Then
√
2
√ e 2 t = AX1 ,
=
−2 − 2
√ √
− 2
√ e− 2 t = AX2 ,
X2 =
−2 + 2
−2
−1
2
1
4
= AXp +
,
t+
t2 +
t+
Xp =
4
5
0
1
−6
√
X1
√
and W (X1 , X2 ) = 2 2 = 0 so that Xp is a particular solution and X1 and X2 form a
fundamental set on the interval (−∞, ∞).
8.2
Homogeneous Linear Systems
1. The system is
X =
1 2
X
4 3
and det(A − λI) = (λ − 5)(λ + 1) = 0. For λ1 = 5 we obtain
0
1 − 12
−4
2 0
−→
so that
4 −2 0
0
0 0
1
K1 =
.
2
515
516
CHAPTER 8
SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
For λ2 = −1 we obtain
2 2
4 4
1 1
0
−→
0
0 0
Then
0
0
so that
−1
K2 =
.
1
1 5t
−1 −t
e + c2
e .
X = c1
2
1
2. The system is
X =
2 2
X
1 3
and det(A − λI) = (λ − 1)(λ − 4) = 0. For λ1 = 1 we obtain
1 2
1 2
For λ2 = 4 we obtain
−2
2
1 −1
0
1 2
−→
0 0
0
0
0
Then
−→
0
0
−1 1
0 0
so that
K1 =
0
0
so that
−2
.
1
1
K2 =
.
1
−2 t
1 4t
e + c2
e .
X = c1
1
1
3. The system is
X =
−4 2
− 52
2
X
and det(A − λI) = (λ − 1)(λ + 3) = 0. For λ1 = 1 we obtain
−5 2
0
− 52
1
0
For λ2 = −3 we obtain
−1 2
0
− 52
Then
5
0
−5 2
−→
0 0
0
0
−1 2
−→
0 0
0
0
so that
so that
2 t
2 −3t
e + c2
e .
X = c1
5
1
2
K1 =
.
5
2
K2 =
.
1
8.2 Homogeneous Linear Systems
4. The system is
X =
2
3
4
−2
5
−2
X
and det(A − λI) = 12 (λ + 1)(2λ + 7) = 0. For λ1 = −7/2 we obtain
1 2 0
1 2 0
−2
−→
so
that
K
=
.
1
3
3
0
0
0
1
0
4
2
For λ2 = −1 we obtain
3
−2
3
4
2
−1
0
0
−→
Then
X = c1
5. The system is
−3 4
0 0
0
0
4
K2 =
.
3
so that
−2 −7t/2
4 −t
e
e .
+ c2
1
3
10
−5
X
X =
8 −12
and det(A − λI) = (λ − 8)(λ + 10) = 0. For λ1 = 8 we obtain
1 − 52
0
2 −5 0
5
−→
so that K1 =
.
8 −20 0
2
0
0 0
For λ2 = −10 we obtain
20 −5
8 −2
1 − 14
0
−→
0
0
0
0
0
so that
1
K2 =
.
4
5 8t
1 −10t
e + c2
e
.
X = c1
2
4
Then
6. The system is
X =
−6 2
X
−3 1
and det(A − λI) = λ(λ + 5) = 0. For λ1 = 0 we obtain
1 − 13
−6 2 0
0
−→
so that
−3 1 0
0
0 0
1
.
K1 =
3
For λ2 = −5 we obtain
−1 2
−3 6
2
K2 =
.
1
0
0
1 −2
−→
0
0
0
0
so that
517
518
CHAPTER 8
SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
Then
1
2 −5t
+ c2
e .
X = c1
3
1
7. The system is
⎞
1 1 −1
⎟
⎜
0⎠ X
X = ⎝0 2
0 1 −1
⎛
and det(A − λI) = (λ − 1)(2 − λ)(λ + 1) = 0. For λ1 = 1, λ2 = 2, and λ3 = −1 we obtain
⎛ ⎞
1
⎜ ⎟
K1 = ⎝0⎠ ,
0
so that
⎛ ⎞
2
⎜ ⎟
K2 = ⎝3⎠ ,
1
⎛ ⎞
1
⎜ ⎟
and K3 = ⎝0⎠ ,
2
⎛ ⎞
⎛ ⎞
⎛ ⎞
1
2
1
⎜ ⎟ t
⎜ ⎟ 2t
⎜ ⎟ −t
X = c1 ⎝0⎠ e + c2 ⎝3⎠ e + c3 ⎝0⎠ e .
0
1
2
8. The system is
⎛
⎞
2 −7 0
⎜
⎟
X = ⎝5 10 4⎠ X
0
5 2
and det(A − λI) = (2 − λ)(λ − 5)(λ − 7) = 0. For λ1 = 2, λ2 = 5, and λ3 = 7 we obtain
⎞
4
⎜ ⎟
K1 = ⎝ 0⎠ ,
−5
⎛
so that
⎛ ⎞
−7
⎜ ⎟
K2 = ⎝ 3⎠ ,
5
⎞
−7
⎜ ⎟
and K3 = ⎝ 5⎠ ,
5
⎛
⎞
⎛ ⎞
⎛ ⎞
4
−7
−7
⎜ ⎟ 2t
⎜ ⎟ 5t
⎜ ⎟ 7t
X = c1 ⎝ 0⎠ e + c2 ⎝ 3⎠ e + c3 ⎝ 5⎠ e .
−5
5
5
⎛
9. We have det(A − λI) = −(λ + 1)(λ − 3)(λ + 2) = 0. For λ1 = −1, λ2 = 3, and λ3 = −2 we
obtain
⎛ ⎞
⎛ ⎞
⎛ ⎞
−1
1
1
⎜ ⎟
⎜ ⎟
⎜ ⎟
K1 = ⎝ 0⎠ , K2 = ⎝4⎠ , and K3 = ⎝−1⎠ ,
1
3
3
so that
⎛ ⎞
⎛ ⎞
⎛ ⎞
−1
1
1
⎜ ⎟ −t
⎜ ⎟ 3t
⎜ ⎟ −2t
X = c1 ⎝ 0⎠ e + c2 ⎝4⎠ e + c3 ⎝−1⎠ e .
1
3
3
8.2 Homogeneous Linear Systems
10. We have det(A − λI) = −λ(λ − 1)(λ − 2) = 0. For λ1 = 0, λ2 = 1, and λ3 = 2 we obtain
⎛ ⎞
⎛ ⎞
⎛ ⎞
1
0
1
⎜ ⎟
⎜ ⎟
⎜ ⎟
K1 = ⎝ 0⎠ , K2 = ⎝1⎠ , and K3 = ⎝0⎠ ,
−1
0
1
so that
⎞
⎛ ⎞
⎛ ⎞
1
0
1
⎜ ⎟
⎜ ⎟ t
⎜ ⎟ 2t
X = c1 ⎝ 0⎠ + c2 ⎝1⎠ e + c3 ⎝0⎠ e .
−1
0
1
⎛
11. We have det(A − λI) = −(λ + 1)(λ + 1/2)(λ + 3/2) = 0. For λ1 = −1, λ2 = −1/2, and
λ3 = −3/2 we obtain
⎛ ⎞
⎛
⎞
⎛ ⎞
4
−12
4
⎜ ⎟
⎜
⎟
⎜ ⎟
K1 = ⎝ 0⎠ , K2 = ⎝ 6⎠ , and K3 = ⎝ 2⎠ ,
−1
5
−1
so that
⎞
⎞
⎛
⎛ ⎞
4
−12
4
⎟ −t/2
⎜ ⎟ −t
⎜
⎜ ⎟ −3t/2
+ c3 ⎝ 2⎠ e
.
X = c1 ⎝ 0⎠ e + c2 ⎝ 6⎠ e
−1
5
−1
⎛
12. We have det(A − λI) = (λ − 3)(λ + 5)(6 − λ) = 0. For λ1 = 3, λ2 = −5, and λ3 = 6 we obtain
⎛ ⎞
⎛ ⎞
⎛ ⎞
1
1
2
⎜ ⎟
⎜ ⎟
⎜ ⎟
K1 = ⎝1⎠ , K2 = ⎝−1⎠ , and K3 = ⎝−2⎠ ,
0
0
11
so that
⎛ ⎞
⎛ ⎞
⎛ ⎞
1
1
2
⎜ ⎟ 3t
⎜ ⎟ −5t
⎜ ⎟ 6t
+ c3 ⎝−2⎠ e .
X = c1 ⎝1⎠ e + c2 ⎝−1⎠ e
0
0
11
13. We have det(A − λI) = (λ + 1/2)(λ − 1/2) = 0. For λ1 = −1/2 and λ2 = 1/2 we obtain
0
1
K1 =
and K2 =
,
1
1
so that
If
0 −t/2
1 t/2
+ c2
e
e .
X = c1
1
1
3
X(0) =
5
then c1 = 2 and c2 = 3. The solutions is
1 t/2
0 −t/2
+3
e .
X=2
e
1
1
519
520
CHAPTER 8
SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
14. We have det(A − λI) = (2 − λ)(λ − 3)(λ + 1) = 0. For λ1 = 2, λ2 = 3, and λ3 = −1 we obtain
⎛ ⎞
⎛ ⎞
⎛ ⎞
5
2
−2
⎜ ⎟
⎜ ⎟
⎜ ⎟
K1 = ⎝−3⎠ , K2 = ⎝0⎠ , and K3 = ⎝ 0⎠ ,
2
1
1
⎞
⎛ ⎞
⎛ ⎞
5
2
−2
⎜ ⎟ 2t
⎜ ⎟ 3t
⎜ ⎟ −t
X = c1 ⎝−3⎠ e + c2 ⎝0⎠ e + c3 ⎝ 0⎠ e .
2
1
1
⎛
so that
⎛ ⎞
1
⎜ ⎟
X(0) = ⎝3⎠
0
If
then c1 = −1, c2 = 5/2, and c3 = −1/2. The solutions is
⎛ ⎞
⎛ ⎞
⎛ ⎞
2
−2
5
⎜ ⎟ 2t 5 ⎜ ⎟ 3t 1 ⎜ ⎟ −t
X = −1 ⎝−3⎠ e + ⎝0⎠ e − ⎝ 0⎠ e .
2
2
1
1
2
15. (a) From the discussion in Section 4.9 we get the system
⎧
3
1
dx1
⎪
3
⎪
⎨ dt = − 100 x1 + 100 x2
− 100
Thus
X =
1
⎪
⎪
⎩ dx2 = 1 x1 − 1 x2
50
dt
50
50
1 100
1
− 50
X
(b) The eigenvalues and eigenvectors of the coefficient matrix are found by solving
det(A − λI) = 0 to get
−1
1
1
1
so K1 =
so K2 =
and
λ2 = −
λ1 = −
25
100
1
2
The general solution is then
λ1 t
X(t) = c1 K1 e
λ2 t
+ c2 K 2 e
−1 −t/25
1 −t/100
e
e
= c1
+ c2
1
2
Using the initial conditions, we get
−c1 + c2
−1
1
20
+ c2
=
=
X(0) = c1
1
2
5
c1 + c2
Thus
35
X(t) = −
3
so
c1 = −
35
3
and c2 =
−1 −t/25 25 1 −t/100
e
e
+
3 2
1
Hence
x1 (t) =
35 −t/25 25 −t/100
e
+ e
3
3
and
x2 (t) = −
35 −t/25 50 −t/100
e
+ e
3
3
25
3
8.2 Homogeneous Linear Systems
(c)
(d) Set x1 (t) = x2 (t) and, using a scientific calculator, we find that t ≈ 34.3277 min.
16. (a) The system to solve is
⎧
1
dx1
⎪
⎪
= − x1
⎪
⎪
dt
50
⎪
⎪
⎪
⎨
1
2
dx2
= x1 − x2
⎪
dt
50
75
⎪
⎪
⎪
⎪
⎪
⎪
⎩ dx3 = 2 x2 − 1 x3
dt
75
25
Thus
⎛ 1
− 50
⎜
1
X = ⎜
⎝ 50
2
− 75
0
2
75
0
⎞
0
⎟
0⎟
⎠X
1
− 25
The eigenvalues and eigenvectors of the coefficient matrix are found by solving
det(A − λI) = 0 to get
λ1 = −
1
25
so
⎛ ⎞
0
⎜ ⎟
K1 = ⎝0⎠ ,
1
and
λ3 = −
λ2 = −
2
75
so
1
50
so
⎛ ⎞
0
⎜ ⎟
K3 = ⎝1⎠
2
⎛ ⎞
1
⎜ ⎟
K 2 = ⎝ 3⎠ ,
4
The general solution is then
⎛ ⎞
⎛ ⎞
⎛ ⎞
0
1
0
⎜ ⎟ −t/25
⎜ ⎟ −t/50
⎜ ⎟ −2t/75
X(t) = c1 ⎝0⎠ e
+ c2 ⎝3⎠ e
+ c3 ⎝1⎠ e
.
1
4
2
521
522
CHAPTER 8
SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
Using the initial conditions, we get
⎛ ⎞
⎛ ⎞
⎛ ⎞ ⎛
⎞ ⎛ ⎞
0
1
0
c2
15
⎜ ⎟
⎜ ⎟
⎜ ⎟ ⎜
⎟ ⎜ ⎟
X(0) = c1 ⎝0⎠ + c2 ⎝3⎠ + c3 ⎝1⎠ = ⎝ 3c2 + c3 ⎠ = ⎝10⎠
c1 + 4c2 + 2c3
1
4
2
5
c1 = 15, c2 = 15, and c3 = −35
⎛ ⎞
⎛ ⎞
⎛ ⎞
0
1
0
⎜ ⎟ −t/25
⎜ ⎟ −t/50
⎜ ⎟ −2t/75
X(t) = 15 ⎝0⎠ e
+ 15 ⎝3⎠ e
− 35 ⎝1⎠ e
1
4
2
Therefore,
x1 (t) = 15e−t/50
and x2 (t) = 45e−t/50 − 35e−2t/75
and x3 (t) = 15e−t/25 + 60e−t/50 − 70e−2t/75
(b) All three solutions tend to zero as t → ∞ which means eventually the tanks will contain
pure water.
⎞
⎞
⎞
⎛
⎛
⎛
0.382175
0.405188
−0.923562
⎟
⎟
⎟
⎜
⎜
⎜
17. X = c1 ⎝0.851161⎠ e8.58979t + c2 ⎝−0.676043⎠ e2.25684t + c3 ⎝−0.132174⎠ e−0.0466321t
0.359815
0.615458
0.35995
⎛
⎞
⎛
⎞
⎛
⎞
0.0312209
−0.280232
0.262219
⎜
⎟
⎜
⎟
⎜
⎟
⎜ 0.949058 ⎟
⎜−0.836611⎟
⎜−0.162664⎟
⎜
⎟ 5.05452t
⎜
⎟ 4.09561t
⎜
⎟ −2.92362t
⎟
⎟
⎟
18. X = c1 ⎜
+ c2 ⎜
+ c3 ⎜
⎜ 0.239535 ⎟ e
⎜−0.275304⎟ e
⎜−0.826218⎟ e
⎜
⎟
⎜
⎟
⎜
⎟
⎝ 0.195825 ⎠
⎝ 0.176045 ⎠
⎝−0.346439⎠
0.0508861
0.338775
0.31957
⎛
⎞
⎛
⎞
0.313235
−0.301294
⎜
⎟
⎜
⎟
⎜ 0.64181 ⎟
⎜ 0.466599 ⎟
⎜
⎟ 2.02882t
⎜
⎟
⎟e
⎜ 0.222136 ⎟ e−0.155338t
0.31754
+ c4 ⎜
+
c
5
⎜
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎝ 0.173787 ⎠
⎝ 0.0534311 ⎠
−0.599108
−0.799567
19. (a)
K2
K1
8.2 Homogeneous Linear Systems
(b) Letting c1 = 1 and c2 = 0 we get x = 5e8t , y = 2e8t . Eliminating the parameter we find
y = 25 x, x > 0. When c1 = −1 and c2 = 0 we find y = 25 x, x < 0. Letting c1 = 0 and
c2 = 1 we get x = e−10t , y = 4e−10t . Eliminating the parameter we find y = 4x, x > 0.
Letting c1 = 0 and c2 = −1 we find y = 4x, x < 0.
(c) The eigenvectors K1 = (5, 2) and K2 = (1, 4) are shown in the figure in part (A).
20. In Problem 2, letting c1 = 1 and c2 = 0 we get x = −2et ,
y = et . Eliminating the parameter we find y = − 12 x, x < 0.
When c1 = −1 and c2 = 0 we find y = − 12 x, x > 0. Letting
c1 = 0 and c2 = 1 we get x = e4t , y = e4t . Eliminating the
parameter we find y = x, x > 0. When c1 = 0 and c2 = −1
we find y = x, x < 0.
In Problem 4, letting c1 = 1 and c2 = 0 we get x = −2e−7t/2 ,
y = e−7t/2 . Eliminating the parameter we find y = − 12 x,
x < 0. When c1 = −1 and c2 = 0 we find y = − 12 x, x > 0.
Letting c1 = 0 and c2 = 1 we get x = 4e−t , y = 3e−t .
Eliminating the parameter we find y = 34 x, x > 0. When
c1 = 0 and c2 = −1 we find y = 34 x, x < 0.
21. We have det(A − λI) = λ2 = 0. For λ1 = 0 we obtain
1
.
K=
3
A solution of (A − λ1 I)P = K is
so that
1
P=
2
1
1
1
+ c2
t+
.
X = c1
3
3
2
22. We have det(A − λI) = (λ + 1)2 = 0. For λ1 = −1 we obtain
1
K=
.
1
A solution of (A − λ1 I)P = K is
P=
so that
0
1
5
0 −t
1 −t
1
−t
e + c2
te + 1 e
.
X = c1
1
1
5
K1
K2
K2
K1
523
524
CHAPTER 8
SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
23. We have det(A − λI) = (λ − 2)2 = 0. For λ1 = 2 we obtain
1
K=
.
1
A solution of (A − λ1 I)P = K is
P=
so that
X = c1
1
1
e2t + c2
1
−3
0
1
1
te2t +
1
−3
0
e2t .
24. We have det(A − λI) = (λ − 6)2 = 0. For λ1 = 6 we obtain
3
K=
.
2
A solution of (A − λ1 I)P = K is
P=
so that
1
2
0
1
3 6t
3
e + c2
te6t + 2 e6t .
X = c1
0
2
2
25. We have det(A − λI) = (1 − λ)(λ − 2)2 = 0. For λ1 = 1 we obtain
⎛ ⎞
1
⎜ ⎟
K1 = ⎝1⎠ .
1
For λ2 = 2 we obtain
Then
⎛ ⎞
1
⎜ ⎟
K 2 = ⎝ 0⎠
1
⎛ ⎞
1
⎜ ⎟
and K3 = ⎝1⎠ .
0
⎛ ⎞
⎛ ⎞
⎛ ⎞
1
1
1
⎜ ⎟ t
⎜ ⎟ 2t
⎜ ⎟ 2t
X = c1 ⎝1⎠ e + c2 ⎝0⎠ e + c3 ⎝1⎠ e .
1
1
0
8.2 Homogeneous Linear Systems
26. We have det(A − λI) = (λ − 8)(λ + 1)2 = 0. For λ1 = 8 we obtain
⎛ ⎞
2
⎜ ⎟
K1 = ⎝1⎠ .
2
For λ2 = −1 we obtain
⎛
⎞
0
⎜ ⎟
K2 = ⎝−2⎠
1
⎛
⎞
1
⎜ ⎟
and K3 = ⎝−2⎠ .
0
⎛ ⎞
⎛ ⎞
⎛ ⎞
2
0
1
⎜ ⎟ 8t
⎜ ⎟ −t
⎜ ⎟ −t
X = c1 ⎝1⎠ e + c2 ⎝−2⎠ e + c3 ⎝−2⎠ e .
2
1
0
Then
27. We have det(A − λI) = −λ(5 − λ)2 = 0. For λ1 = 0 we obtain
⎛ ⎞
−4
⎜ ⎟
K1 = ⎝−5⎠ .
2
For λ2 = 5 we obtain
A solution of (A − λ1 I)P = K is
⎛
⎞
−2
⎜ ⎟
K = ⎝ 0⎠ .
1
⎛5⎞
2
⎜ ⎟
1⎟
P=⎜
⎝2⎠
0
so that
⎛5⎞ ⎤
⎡⎛ ⎞
⎛ ⎞
⎛ ⎞
2
−4
−2
−2
⎢⎜ ⎟ 5t ⎜ ⎟ 5t ⎥
⎜ ⎟
⎜ ⎟ 5t
⎜1⎟ ⎥
X = c1 ⎝−5⎠ + c2 ⎝ 0⎠ e + c3 ⎢
⎣⎝ 0⎠ te + ⎝ 2 ⎠ e ⎦ .
2
1
1
0
28. We have det(A − λI) = (1 − λ)(λ − 2)2 = 0. For λ1 = 1 we obtain
⎛ ⎞
1
⎜ ⎟
K1 = ⎝0⎠ .
0
For λ2 = 2 we obtain
⎞
0
⎜ ⎟
K = ⎝−1⎠ .
1
⎛
525
526
CHAPTER 8
SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
A solution of (A − λ2 I)P = K is
so that
⎞
0
⎜ ⎟
P = ⎝−1⎠
0
⎛
⎛ ⎞ ⎤
⎛ ⎞
⎛ ⎞
⎡⎛ ⎞
0
1
0
0
⎜ ⎟ t
⎜ ⎟ 2t
⎢⎜ ⎟ 2t ⎜ ⎟ 2t ⎥
X = c1 ⎝0⎠ e + c2 ⎝−1⎠ e + c3 ⎣⎝−1⎠ te + ⎝−1⎠ e ⎦ .
0
0
1
1
29. We have det(A − λI) = −(λ − 1)3 = 0. For λ1 = 1 we obtain
⎛ ⎞
0
⎜ ⎟
K = ⎝1⎠ .
1
Solutions of (A − λ1 I)P = K and (A − λ1 I)Q = P are
⎛ ⎞
0
⎜ ⎟
P = ⎝1⎠
0
⎛1⎞
2
⎜ ⎟
and Q = ⎝ 0⎠
0
so that
⎛ ⎞ ⎤
⎛ ⎞
⎛1⎞ ⎤
⎛ ⎞
⎡⎛ ⎞
⎡⎛ ⎞
0
0
0
0
0
2
2
⎜ ⎟ ⎥
⎜ ⎟
⎜ ⎟ ⎥
⎜ ⎟
⎢⎜ ⎟
⎢⎜ ⎟ t
X = c1 ⎝1⎠ et + c2 ⎣⎝1⎠ tet + ⎝1⎠ et ⎦ + c3 ⎣⎝1⎠ et + ⎝1⎠ tet + ⎝ 0⎠ et ⎦ .
2
0
0
1
1
1
0
30. We have det(A − λI) = (λ − 4)3 = 0. For λ1 = 4 we obtain
⎛ ⎞
1
⎜ ⎟
K = ⎝0⎠ .
0
Solutions of (A − λ1 I)P = K and (A − λ1 I)Q = P are
⎛ ⎞
0
⎜ ⎟
P = ⎝1⎠
0
⎛ ⎞
0
⎜ ⎟
and Q = ⎝0⎠
1
so that
⎛ ⎞ ⎤
⎛ ⎞
⎛ ⎞ ⎤
⎛ ⎞
⎡⎛ ⎞
⎡⎛ ⎞
0
0
0
1
1
1
⎜ ⎟ 4t
⎢⎜ ⎟ 4t ⎜ ⎟ 4t ⎥
⎢⎜ ⎟ t2 4t ⎜ ⎟ 4t ⎜ ⎟ 4t ⎥
X = c1 ⎝0⎠ e + c2 ⎣⎝0⎠ te + ⎝1⎠ e ⎦ + c3 ⎣⎝0⎠ e + ⎝1⎠ te + ⎝0⎠ e ⎦ .
2
0
0
1
0
0
0
8.2 Homogeneous Linear Systems
31. We have det(A − λI) = (λ − 4)2 = 0. For λ1 = 4 we obtain
2
K=
.
1
A solution of (A − λ1 I)P = K is
so that
If
1
P=
1
2 4t
2
1 4t
4t
e + c2
te +
e .
X = c1
1
1
1
−1
X(0) =
6
then c1 = −7 and c2 = 13. The solutions is
2 4t
2t + 1 4t
X = −7
e + 13
e .
1
t+1
32. We have det(A − λI) = −(λ + 1)(λ − 1)2 = 0. For λ1 = −1 we obtain
⎛ ⎞
−1
⎜ ⎟
K1 = ⎝ 0⎠ .
1
For λ2 = 1 we obtain
so that
If
⎛ ⎞
1
⎜ ⎟
K2 = ⎝0⎠
1
⎛ ⎞
0
⎜ ⎟
and K3 = ⎝1⎠
0
⎛ ⎞
⎛ ⎞
⎛ ⎞
−1
1
0
⎜ ⎟ −t
⎜ ⎟ t
⎜ ⎟ t
X = c1 ⎝ 0⎠ e + c2 ⎝0⎠ e + c3 ⎝1⎠ e .
1
1
0
⎛ ⎞
1
⎜ ⎟
X(0) = ⎝2⎠
5
then c1 = 2, c2 = 3, and c3 = 2. The solution is
⎛ ⎞
⎛ ⎞
⎛ ⎞
1
0
−1
⎜ ⎟ t
⎜ ⎟ t
⎜ ⎟ −t
X = 2 ⎝ 0⎠ e + 3 ⎝0⎠ e + 2 ⎝1⎠ e .
1
0
1
527
528
CHAPTER 8
SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
33. In this case det(A − λI) = (2 − λ)5 , and λ1 = 2 is an eigenvalue of multiplicity 5. Linearly
independent eigenvectors are
⎛ ⎞
⎛ ⎞
⎛ ⎞
1
0
0
⎜ ⎟
⎜ ⎟
⎜ ⎟
⎜0⎟
⎜0⎟
⎜0⎟
⎜ ⎟
⎜ ⎟
⎜ ⎟
⎜
⎜
⎟
⎟
⎟
K2 = ⎜1⎟ ,
and
K3 = ⎜
K1 = ⎜0⎟ ,
⎜0⎟ .
⎜ ⎟
⎜ ⎟
⎜ ⎟
⎝0⎠
⎝0⎠
⎝1⎠
0
0
0
34. In Problem 22 letting c1 = 1 and c2 = 0 we get x = et , y = et . Eliminating the parameter we
find y = x, x > 0. When c1 = −1 and c2 = 0 we find y = x, x < 0.
In Problem 23 letting c1 = 1 and c2 = 0 we get x = e2t , y = e2t . Eliminating the parameter
we find y = x, x > 0. When c1 = −1 and c2 = 0 we find y = x, x < 0.
Phase portarit for Problem 20
Phase portarit for Problem 21
In Problems 35-46 the form of the answer will vary according to the choice of eigenvector. For
1
the solution has the form
example, in Problem 35, if K1 is chosen to be
2−i
X = c1
cos t 4t
sin t 4t
e + c2
e .
2 cos t + sin t
2 sin t − cos t
35. We have det(A − λI) = λ2 − 8λ + 17 = 0. For λ1 = 4 + i we obtain
2+i
K1 =
5
so that
Then
2 + i (4+i)t
2 cos t − sin t 4t
cos t + 2 sin t 4t
X1 =
e
=
e +i
e .
5
5 cos t
5 sin t
2 cos t − sin t 4t
cos t + 2 sin t 4t
e + c2
e .
X = c1
5 cos t
5 sin t
8.2 Homogeneous Linear Systems
36. We have det(A − λI) = λ2 + 1 = 0. For λ1 = i we obtain
−1 − i
K1 =
2
so that
X1 =
−1 − i it
sin t − cos t
− cos t − sin t
e =
+i
.
2
2 cos t
2 sin t
Then
X = c1
sin t − cos t
2 cos t
+ c2
− cos t − sin t
.
2 sin t
37. We have det(A − λI) = λ2 − 8λ + 17 = 0. For λ1 = 4 + i we obtain
−1 − i
K1 =
2
so that
−1 − i (4+i)t
sin t − cos t 4t
− sin t − cos t 4t
e
=
e +i
e .
X1 =
2
2 cos t
2 sin t
Then
X = c1
sin t − cos t 4t
− sin t − cos t 4t
e + c2
e .
2 cos t
2 sin t
38. We have det(A − λI) = λ2 − 10λ + 34 = 0. For λ1 = 5 + 3i we obtain
1 − 3i
K1 =
2
so that
X1 =
1 − 3i (5+3i)t
cos 3t + 3 sin 3t 5t
sin 3t − 3 cos 3t 5t
e
=
e +i
e .
2
2 cos 3t
2 sin 3t
cos 3t + 3 sin 3t 5t
sin 3t − 3 cos 3t 5t
e + c2
e .
X = c1
2 cos 3t
2 sin 3t
Then
39. We have det(A − λI) = λ2 + 9 = 0. For λ1 = 3i we obtain
4 + 3i
K1 =
5
so that
X1 =
4 + 3i 3it
4 cos 3t − 3 sin 3t
4 sin 3t + 3 cos 3t
e =
+i
.
5
5 cos 3t
5 sin 3t
Then
X = c1
4 cos 3t − 3 sin 3t
4 sin 3t + 3 cos 3t
+ c2
.
5 cos 3t
5 sin 3t
529
530
CHAPTER 8
SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
40. We have det(A − λI) = λ2 + 2λ + 5 = 0. For λ1 = −1 + 2i we obtain
2 + 2i
K1 =
1
so that
2 + 2i (−1+2i)t
2 cos 2t − 2 sin 2t −t
2 cos 2t + 2 sin 2t −t
e
=
e +i
e .
X1 =
1
cos 2t
sin 2t
2 cos 2t − 2 sin 2t −t
2 cos 2t + 2 sin 2t −t
e + c2
e .
X = c1
cos 2t
sin 2t
41. We have det(A − λI) = −λ λ2 + 1 = 0. For λ1 = 0 we obtain
⎛ ⎞
1
⎜ ⎟
K1 = ⎝0⎠ .
0
Then
For λ2 = i we obtain
⎛ ⎞
−i
⎜ ⎟
K2 = ⎝ i⎠
1
so that
⎛
⎞
⎛
⎞
⎛
⎞
−i
sin t
− cos t
⎜ ⎟
⎜
⎟
⎜
⎟
X2 = ⎝ i⎠ eit = ⎝− sin t⎠ + i ⎝ cos t⎠ .
1
cos t
sin t
Then
⎞
⎞
⎛ ⎞
⎛
⎛
1
sin t
− cos t
⎟
⎟
⎜ ⎟
⎜
⎜
X = c1 ⎝0⎠ + c2 ⎝− sin t⎠ + c3 ⎝ cos t⎠ .
0
cos t
sin t
42. We have det(A − λI) = −(λ + 3)(λ2 − 2λ + 5) = 0. For λ1 = −3 we obtain
⎛ ⎞
0
⎜ ⎟
K1 = ⎝−2⎠ .
1
For λ2 = 1 + 2i we obtain
so that
⎞
⎛
−2 − i
⎟
⎜
K2 = ⎝ −3i ⎠
2
⎞
⎞
⎛
−2 cos 2t + sin 2t
− cos 2t − 2 sin 2t
⎟ t
⎟ t
⎜
⎜
3 sin 2t
−3 cos 2t
X2 = ⎝
⎠e + i⎝
⎠e .
2 cos 2t
2 sin 2t
⎛
8.2 Homogeneous Linear Systems
Then
⎞
⎞
⎞
⎛
⎛
0
−2 cos 2t + sin 2t
− cos 2t − 2 sin 2t
⎟ t
⎟ t
⎜ ⎟
⎜
⎜
3 sin 2t
−3 cos 2t
X = c1 ⎝−2⎠ e−3t + c2 ⎝
⎠ e + c3 ⎝
⎠e .
1
2 cos 2t
2 sin 2t
⎛
43. We have det(A − λI) = (1 − λ)(λ2 − 2λ + 2) = 0. For λ1 = 1 we obtain
⎛ ⎞
0
⎜ ⎟
K1 = ⎝2⎠ .
1
For λ2 = 1 + i we obtain
⎛ ⎞
1
⎜ ⎟
K2 = ⎝ i⎠
i
so that
⎛ ⎞
⎛
⎞
⎛
⎞
1
cos t
sin t
⎜ ⎟
⎜
⎟
⎜
⎟
X2 = ⎝ i⎠ e(1+i)t = ⎝− sin t⎠ et + i ⎝cos t⎠ et .
i
− sin t
cos t
Then
⎛ ⎞
⎛
⎞
⎛
⎞
0
cos t
sin t
⎜ ⎟
⎜
⎟
⎜
⎟
X = c1 ⎝2⎠ et + c2 ⎝− sin t⎠ et + c3 ⎝cos t⎠ et .
1
− sin t
cos t
44. We have det(A − λI) = −(λ − 6)(λ2 − 8λ + 20) = 0. For λ1 = 6 we obtain
⎛ ⎞
0
⎜ ⎟
K1 = ⎝1⎠ .
0
For λ2 = 4 + 2i we obtain
⎛ ⎞
−i
⎜ ⎟
K2 = ⎝ 0⎠
2
so that
⎞
⎞
⎛ ⎞
⎛
⎛
−i
sin 2t
− cos 2t
⎟
⎟
⎜ ⎟
⎜
⎜
X2 = ⎝ 0⎠ e(4+2i)t = ⎝ 0 ⎠ e4t + i ⎝ 0 ⎠ e4t .
2
2 cos 2t
2 sin 2t
Then
⎛ ⎞
⎛
⎞
⎛
⎞
0
sin 2t
− cos 2t
⎜ ⎟
⎜
⎟
⎜
⎟
X = c1 ⎝1⎠ e6t + c2 ⎝ 0 ⎠ e4t + c3 ⎝ 0 ⎠ e4t .
0
2 cos 2t
2 sin 2t
531
532
CHAPTER 8
SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
45. We have det(A − λI) = (2 − λ)(λ2 + 4λ + 13) = 0. For λ1 = 2 we obtain
⎞
28
⎜ ⎟
K1 = ⎝−5⎠ .
25
⎛
For λ2 = −2 + 3i we obtain
⎞
4 + 3i
⎟
⎜
K2 = ⎝ −5 ⎠
0
⎛
so that
⎛
⎞
⎛
⎞
⎛
⎞
4 + 3i
4 cos 3t − 3 sin 3t
4 sin 3t + 3 cos 3t
⎜
⎟
⎜
⎟ −2t
⎜
⎟ −2t
−5 cos 3t
−5 sin 3t
X2 = ⎝ −5 ⎠ e(−2+3i)t = ⎝
+ i⎝
⎠e
⎠e .
0
0
0
Then
⎞
⎞
⎞
⎛
⎛
28
4 cos 3t − 3 sin 3t
4 sin 3t + 3 cos 3t
⎟ −2t
⎟ −2t
⎜ ⎟
⎜
⎜
−5 cos 3t
−5 sin 3t
+ c3 ⎝
X = c1 ⎝−5⎠ e2t + c2 ⎝
⎠e
⎠e .
25
0
0
⎛
46. We have det(A − λI) = −(λ + 2)(λ2 + 4) = 0. For λ1 = −2 we obtain
⎛
⎞
0
⎜ ⎟
K1 = ⎝−1⎠ .
1
For λ2 = 2i we obtain
⎞
⎛
−2 − 2i
⎟
⎜
K2 = ⎝ 1 ⎠
1
so that
⎛
⎞
⎛
⎞
⎛
⎞
−2 − 2i
−2 cos 2t + 2 sin 2t
−2 cos 2t − 2 sin 2t
⎜
⎟
⎜
⎟
⎜
⎟
cos 2t
sin 2t
X2 = ⎝ 1 ⎠ e2it = ⎝
⎠ + i⎝
⎠.
1
cos 2t
sin 2t
Then
⎛
⎞
⎛
⎞
⎛
⎞
0
−2 cos 2t + 2 sin 2t
−2 cos 2t − 2 sin 2t
⎜ ⎟
⎜
⎟
⎜
⎟
cos 2t
sin 2t
X = c1 ⎝−1⎠ e−2t + c2 ⎝
⎠ + c3 ⎝
⎠.
1
cos 2t
sin 2t
8.2 Homogeneous Linear Systems
47. We have det(A − λI) = (1 − λ)(λ2 + 25) = 0. For λ1 = 1 we obtain
⎛ ⎞
25
⎜ ⎟
K1 = ⎝−7⎠ .
6
For λ2 = 5i we obtain
⎞
1 + 5i
⎟
⎜
K2 = ⎝ 1 ⎠
1
⎛
⎛
⎞
⎛
⎞
⎛
⎞
1 + 5i
cos 5t − 5 sin 5t
sin 5t + 5 cos 5t
⎜
⎟
⎜
⎟
⎜
⎟
cos 5t
sin 5t
X2 = ⎝ 1 ⎠ e5it = ⎝
⎠ + i⎝
⎠.
1
cos 5t
sin 5t
so that
⎛
⎞
⎛
⎞
⎛
⎞
25
cos 5t − 5 sin 5t
sin 5t + 5 cos 5t
⎜ ⎟
⎜
⎟
⎜
⎟
cos 5t
sin 5t
X = c1 ⎝−7⎠ et + c2 ⎝
⎠ + c3 ⎝
⎠.
6
cos 5t
sin 5t
Then
⎞
4
⎜ ⎟
X(0) = ⎝ 6⎠
−7
⎛
If
then c1 = c2 = −1 and c3 = 6. The solution is
⎛
⎞
⎛ ⎞
⎛
⎞
cos 5t − 5 sin 5t
25
sin 5t + 5 cos 5t
⎜
⎟
⎜ ⎟
⎜
⎟
cos 5t
sin 5t
X = −1 ⎝−7⎠ et − ⎝
⎠ + 6⎝
⎠.
cos 5t
6
sin 5t
48. We have det(A − λI) = λ2 − 10λ + 29 = 0. For λ1 = 5 + 2i we obtain
1
K1 =
1 − 2i
so that
X1 =
1
cos 2t
sin 2t
(5+2i)t
5t
e
=
e +i
e5t .
1 − 2i
cos 2t + 2 sin 2t
sin 2t − 2 cos 2t
and
X = c1
If X(0) =
cos 2t
sin
2t
e5t + c2
e5t .
cos 2t + 2 sin 2t
sin 2t − 2 cos 2t
−2
, then c1 = −2 and c2 = −5. The solution is
8
sin 2t
cos 2t
5t
e5t .
X = −2
e −5
sin 2t − 2 cos 2t
cos 2t + 2 sin 2t
533
534
CHAPTER 8
SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
49. (a) The system to solve is
⎧
1
1
dx
⎪
⎪ 1 = − x1 + x3
⎪
⎪
dt
20
10
⎪
⎪
⎪
⎨
1
1
dx2
= x1 − x2
⎪
dt
20
20
⎪
⎪
⎪
⎪
⎪
⎪
⎩ dx3 = 1 x2 − 1 x3
dt
20
10
⎛
Thus
⎜
X = ⎜
⎝
1
− 20
0
1
20
1
− 20
0
1
20
1 ⎞
10
⎟
0⎟
⎠X
1
− 10
(b) The eigenvalues and eigenvectors of the coefficient matrix are found by solving
det(A − λI) = 0 to get
λ1 = 0 so
⎛ ⎞
2
⎜ ⎟
K1 = ⎝2⎠ ,
1
and
⎛
⎞
−1 − i
1
1
⎜
⎟
λ2 = − + i so K2 = ⎝ i ⎠ ,
10 20
1
⎞
⎛
−1 + i
1
1
⎟
⎜
λ3 = − − i so K3 = ⎝ −i ⎠
10 20
1
The general solution is then
⎛ ⎞
⎛
⎛
1
1 ⎞
1
1 ⎞
t + sin 20
t
t − sin 20
t
2
− cos 20
− cos 20
⎟
⎟ −t/10
⎜ ⎟
⎜
⎜
1
1
⎟ e−t/10 + c3 ⎜
⎟e
⎟ + c2 ⎜
t
t
−
sin
cos
2
X(t) = c1 ⎜
20
20
⎠
⎠
⎝ ⎠
⎝
⎝
1
t
cos 20
1
1
t
sin 20
Using the initial conditions, we get
⎛ ⎞
⎛ ⎞
⎛ ⎞ ⎛ ⎞
30
2
−1
−1
⎜ ⎟
⎜ ⎟
⎜ ⎟ ⎜ ⎟
X(0) = c1 ⎝2⎠ + c2 ⎝ 0 ⎠ + c3 ⎝ 1 ⎠ = ⎝20⎠
5
1
1
0
c1 = 11, c2 = −6, and c3 = −2
⎛
⎛ ⎞
⎛
1
1 ⎞
1
1 ⎞
− cos 20
t + sin 20
t
t − sin 20
t
2
− cos 20
⎜
⎟
⎟ −t/10
⎜ ⎟
⎜
1
1
⎟ e−t/10 − 2 ⎜
⎟e
⎟ − 6⎜
t
t
−
sin
cos
2
X(t) = 11 ⎜
20
20
⎝
⎠
⎠
⎝ ⎠
⎝
1
1
t
cos 20
1
t
sin 20
50. (a) From Problem 49,
⎛
⎛ ⎞
⎛
1
1 ⎞
1
1 ⎞
t + sin 20
t
t − sin 20
t
cos 20
2
− cos 20
⎜
⎜ ⎟
⎟
⎟ −t/10
⎜
1
1
⎟ e−t/10 − 2 ⎜
⎟e
⎟ − 6⎜
t
t
−
sin
−
cos
2
X(t) = 11 ⎜
20
20
⎝
⎝ ⎠
⎠
⎠
⎝
1
1
t
cos 20
1
t
sin 20
8.2 Homogeneous Linear Systems
The sum of the three individual solutions is
1
1
1
1
−t/10
−t/10
+
+ −2 cos t + 2 sin t e
22 + 6 cos t − 6 sin t e
20
20
20
20
1
1
−t/10
−t/10
22 + 6 sin t e
+
+ 2 cos t e
20
20
1
1
−t/10
−t/10
= 55
11 + −6 cos t e
+ −2 sin t e
20
20
This shows that the total amount of salt in the system remains constant over time. The
fact that the sum x1 (t) + x2 (t) + x3 (t) is a constant can also be seen from the fact that
d
dx1 dx2 dx3
+
+
=
(x1 + x2 + x3 ) = 0 .
dt
dt
dt
dt
⎞
⎛ ⎞
22
x1 (t)
⎟
⎜ ⎟
⎜
(b) From the last three equation in part (b) of Problem 49 we see that ⎝x2 (t)⎠ → ⎝22⎠
11
x3 (t)
⎛
as t → ∞. These are steady state solutions and represent the fact that regardless of the
initial amount of salt in each tank (see note below) the salt attains a uniform distribution
in the system over time, that is, 22/55 or 40% of salt in each of the tanks A and B and
11/55 or 20% in tank C.
(Note: Regardless of intial conditions note that from the equations in part (b) of Problem
49,
⎞
⎛ ⎞
⎛
2c3
x1 (t)
⎟
⎜ ⎟
⎜
⎝x2 (t)⎠ → ⎝2c3 ⎠ as t → ∞
c3
x3 (t)
Moreover, x1 (t) + x2 (t) + x3 (t) = 5c3 . So over time there is 2c3 /5c3 = 0.4 or 40% in tank
A, 2c + 3/5c3 = 0.4 or 40% in tank B, and c3 /5c3 = 0.2 or 20% in tank C.)
51.
Phase portrait for Problem 38
Phase portrait for Problem 39
52. Letting x1 = y1 , x1 = y2 , x2 = y3 , and x2 = y4 we have
y2 = x1 = −10x1 + 4x2 = −10y1 + 4y3
y4 = x2 = 4x1 − 4x2 = 4y1 − 4y3 .
Phase portrait for Problem 40
535
536
CHAPTER 8
SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
The corresponding linear system is
y1 = y2
y2 = −10y1 + 4y3
y3 = y4
y4 = 4y1 − 4y3
or
⎛
0
⎜
−10
Y = ⎜
⎝ 0
4
1
0
0
4
0
0
0 −4
⎞
0
0⎟
⎟ Y.
1⎠
0
√
√
Using a CAS, we find eigenvalues ± 2i and ±2 3i with corresponding eigenvectors
⎞ ⎛
⎞
⎛ √
⎞
⎛ √
0
∓ 2 i/4
∓ 2/4
⎟ ⎜
⎟
⎜
⎟
⎜
⎜ 1/2 ⎟ ⎜1/2⎟
⎜ 0 ⎟
⎟ ⎜
⎟
⎜
⎟
⎜
⎟=⎜
⎟
⎜ √
⎟ + i⎜ √
⎜∓ 2 i/2⎟ ⎜ 0 ⎟
⎜∓ 2/2⎟
⎠ ⎝
⎠
⎝
⎠
⎝
1
1
0
and
⎞ ⎛ ⎞
⎛ √
⎛ √ ⎞
0
± 3 i/3
± 3/3
⎟ ⎜ ⎟
⎟
⎜
⎜
⎜ −2 ⎟ ⎜−2⎟
⎜ 0 ⎟
⎟ ⎜ ⎟
⎟
⎜
⎜
⎟ = ⎜ ⎟ + i⎜ √
⎟.
⎜ √
⎜∓ 3 i/6⎟ ⎜ 0⎟
⎜∓ 3/6⎟
⎠ ⎝ ⎠
⎠
⎝
⎝
1
1
0
8.2 Homogeneous Linear Systems
Thus
⎡⎛
⎛ √ ⎞
⎤
− 2/4
⎢⎜
⎜
⎥
⎟
⎟
⎢⎜1/2⎟
⎜ 0 ⎟
√ ⎥
√
⎢⎜
⎜
⎥
⎟
⎟
Y(t) = c1 ⎢⎜
⎟ sin 2 t⎥
⎟ cos 2 t − ⎜ √
⎢⎜ 0 ⎟
⎜− 2/2⎟
⎥
⎣⎝
⎝
⎦
⎠
⎠
1
0
⎡⎛ √
⎛
⎤
⎞
⎞
− 2/4
0
⎢⎜
⎜
⎥
⎟
⎟
⎢⎜ 0 ⎟
⎜1/2⎟
√
√ ⎥
⎢⎜
⎜
⎥
⎟
⎟
+ c2 ⎢⎜ √
⎟ cos 2 t + ⎜
⎟ sin 2 t⎥
⎢⎜− 2/2⎟
⎜ 0 ⎟
⎥
⎣⎝
⎝
⎦
⎠
⎠
1
0
⎞
⎛√
⎤
⎡⎛ ⎞
3/3
0
⎟
⎜
⎥
⎢⎜ ⎟
⎜ 0 ⎟
⎥
⎢⎜−2⎟
√
√
⎟
⎜
⎥
⎢⎜ ⎟
+ c3 ⎢⎜ ⎟ cos 2 3 t − ⎜ √ ⎟ sin 2 3 t⎥
⎜− 3/6⎟
⎥
⎢⎜ 0⎟
⎠
⎝
⎦
⎣⎝ ⎠
1
0
⎞
⎛ ⎞
⎤
⎡⎛ √
3/3
0
⎟
⎜ ⎟
⎥
⎢⎜
⎜−2⎟
⎢⎜ 0 ⎟
√
√ ⎥
⎟
⎜ ⎟
⎥
⎢⎜
+ c4 ⎢⎜ √ ⎟ cos 2 3 t + ⎜ ⎟ sin 2 3 t⎥ .
⎜0⎟
⎥
⎢⎜− 3/6⎟
⎠
⎝ ⎠
⎦
⎣⎝
1
0
0
⎞
The initial conditions y1 (0) = 0, y2 (0) = 1, y3 (0) = 0, and y4 (0) = −1 imply c1 = − 25 , c2 = 0,
c3 = − 35 , and c4 = 0. Thus,
√
√
2
sin 2 t +
x1 (t) = y1 (t) = −
10
√
√
2
sin 2 t −
x2 (t) = y3 (t) = −
5
√
√
3
sin 2 3 t
5
√
√
3
sin 2 3 t.
10
53. (a) From det(A − λI) = λ(λ − 2) = 0 we get λ1 = 0 and λ2 = 2. For λ1 = 0 we obtain
1 1 0
1 1 0
−1
−→
so that K1 =
.
1 1 0
0 0 0
1
For λ2 = 2 we obtain
−1
1
1 −1
Then
0
0
−→
−1 1
0 0
0
0
so that
−1
1 2t
+ c2
e .
X = c1
1
1
1
K2 =
.
1
537
538
CHAPTER 8
SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
The line y = −x is not a trajectory of the system.
Trajectories are x = −c1 + c2 e2t , y = c1 + c2 e2t or
y = x + 2c1 . This is a family of lines perpendicular to
the line y = −x. All of the constant solutions of the
system do, however, lie on the line y = −x.
(b) From det(A − λI) = λ2 = 0 we get λ1 = 0 and
−1
K=
.
1
A solution of (A − λ1 I)P = K is
so that
−1
P=
0
−1
−1
−1
.
+ c2
t+
X = c1
0
1
1
All trajectories are parallel to y = −x, but y = −x is
not a trajectory. There are constant solutions of the
system, however, that do lie on the line y = −x.
54. The system of differential equations is
x1 = 2x1 + x2
x2 = 2x2
x3 = 2x3
x4 = 2x4 + x5
x5 = 2x5 .
We see immediately that x2 = c2 e2t , x3 = c3 e2t , and x5 = c5 e2t . Then
x1 = 2x1 + c2 e2t
so
x1 = c2 te2t + c1 e2t ,
x4 = 2x4 + c5 e2t
so
x4 = c5 te2t + c4 e2t .
and
8.2 Homogeneous Linear Systems
The general solution of the system is
⎛ ⎞ ⎤
⎞
⎛ ⎞
⎡⎛ ⎞
0
1
c2 te2t + c1 e2t
1
⎜ ⎟ ⎥
⎟
⎜ ⎟
⎢⎜ ⎟
⎜
2t
c2 e
⎜ 1⎟ ⎥
⎟
⎜0⎟
⎢⎜0⎟
⎜
⎟
⎜ ⎟ 2t
⎢⎜ ⎟ 2t ⎜ ⎟ 2t ⎥
⎜
2t
⎜ ⎟ ⎥
⎟
⎜
⎟
⎢
⎜
⎜
⎟
c3 e
X=⎜
⎟ = c1 ⎜0⎟ e + c2 ⎢⎜0⎟ te + ⎜0⎟ e ⎥
⎜ ⎟ ⎥
⎟
⎜ ⎟
⎢⎜ ⎟
⎜
⎝ 0⎠ ⎦
⎝0⎠
⎣⎝0⎠
⎝c5 te2t + c4 e2t ⎠
2t
0
0
c5 e
0
⎛ ⎞
⎛ ⎞
⎡⎛ ⎞
⎛ ⎞ ⎤
0
0
0
0
⎜ ⎟
⎜ ⎟
⎢⎜ ⎟
⎜ ⎟ ⎥
⎜0⎟
⎜0⎟
⎢⎜0⎟
⎜0⎟ ⎥
⎜ ⎟ 2t
⎜ ⎟ 2t
⎢⎜ ⎟ 2t ⎜ ⎟ 2t ⎥
⎜
⎟
⎜
⎟
⎢
⎟ ⎥
⎜
⎟
+ c3 ⎜1⎟ e + c4 ⎜0⎟ e + c5 ⎢⎜0⎟ te + ⎜
⎜0⎟ e ⎥
⎜ ⎟
⎜ ⎟
⎢⎜ ⎟
⎜ ⎟ ⎥
⎝0⎠
⎝1⎠
⎣⎝1⎠
⎝0⎠ ⎦
0
0
1
0
⎛
⎛ ⎞ ⎤
⎛ ⎞ ⎤
⎡
0
0
⎜ ⎟ ⎥
⎜ ⎟ ⎥
⎢
⎢
⎜1⎟ ⎥
⎜0⎟ ⎥
⎢
⎢
⎜ ⎟ 2t ⎥
⎜ ⎟ 2t ⎥
⎢
⎢
2t
2t
2t
2t
2t
⎜
⎟
⎟ ⎥
⎢
⎥
⎢
= c1 K1 e + c2 ⎢K1 te + ⎜0⎟ e ⎥ + c3 K2 e + c4 K3 e + c5 ⎢K3 te + ⎜
⎜0⎟ e ⎥ .
⎜ ⎟ ⎥
⎜ ⎟ ⎥
⎢
⎢
⎝0⎠ ⎦
⎝0⎠ ⎦
⎣
⎣
0
1
⎡
There are three solutions of the form X = Ke2t , where K is an eigenvector, and two solutions
of the form X = Kte2t + Pe2t . See (12) in the text. From (13) and (14) in the text
(A − 2I)K1 = 0
and
(A − 2I)P1 = K1 .
This implies
⎛
0
⎜
⎜0
⎜
⎜0
⎜
⎜
⎝0
0
1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
⎞⎛ ⎞ ⎛ ⎞
1
0
p1
⎟⎜ ⎟ ⎜ ⎟
0⎟ ⎜p2 ⎟ ⎜0⎟
⎟⎜ ⎟ ⎜ ⎟
⎜ ⎟ ⎜ ⎟
0⎟
⎟ ⎜p3 ⎟ = ⎜0⎟ ,
⎟⎜ ⎟ ⎜ ⎟
1⎠ ⎝p4 ⎠ ⎝0⎠
0
p5
0
so p2 = 1 and p5 = 0, while p1 , p3 , and p4 are arbitrary. Choosing p1 = p3 = p4 = 0 we have
⎛ ⎞
0
⎜ ⎟
⎜1⎟
⎜ ⎟
⎟
P1 = ⎜
⎜0⎟ .
⎜ ⎟
⎝0⎠
0
539
540
CHAPTER 8
SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
Therefore a solution is
Repeating for K3 we find
so another solution is
⎛ ⎞
⎛ ⎞
0
1
⎜ ⎟
⎜ ⎟
⎜1⎟
⎜0⎟
⎜ ⎟ 2t ⎜ ⎟ 2t
⎜ ⎟
⎟
X=⎜
⎜0⎟ te + ⎜0⎟ e .
⎜ ⎟
⎜ ⎟
⎝0⎠
⎝0⎠
0
0
⎛ ⎞
0
⎜ ⎟
⎜0⎟
⎜ ⎟
⎟
P2 = ⎜
⎜0⎟ ,
⎜ ⎟
⎝0⎠
1
⎛ ⎞
⎛ ⎞
0
0
⎜ ⎟
⎜ ⎟
⎜0⎟
⎜0⎟
⎜ ⎟ 2t ⎜ ⎟ 2t
⎜ ⎟
⎟
X=⎜
⎜0⎟ te + ⎜0⎟ e .
⎜ ⎟
⎜ ⎟
⎝0⎠
⎝1⎠
1
0
55. From x = 2 cos 2t − 2 sin 2t, y = − cos 2t we find x + 2y = −2 sin 2t. Then
(x + 2y)2 = 4 sin2 2t = 4(1 − cos2 2t) = 4 − 4 cos2 2t = 4 − 4y 2
and
x2 + 4xy + 4y 2 = 4 − 4y 2
or
x2 + 4xy + 8y 2 = 4.
This is a rotated conic section and, from the discriminant b2 − 4ac = 16 − 32 < 0, we see that
the curve is an ellipse.
56. Suppose the eigenvalues are α ± iβ, β > 0. In Problem 38 the eigenvalues are 5 ± 3i, in
Problem 39 they are ±3i, and in Problem 40 they are −1 ± 2i. From Problem 51 we deduce
that the phase portrait will consist of a family of closed curves when α = 0 and spirals when
α = 0. The origin will be a repellor when α > 0, and an attractor when α < 0.
8.3
Nonhomogeneous Linear Systems
1. Solving
2 − λ
3 det(A − λI) = = λ2 − 1 = (λ − 1)(λ + 1) = 0
−1 −2 − λ
we obtain eigenvalues λ1 = −1 and λ2 = 1. Corresponding eigenvectors are
−1
−3
and K2 =
.
K1 =
1
1
8.3
Thus
X c = c1
Substituting
Nonhomogeneous Linear Systems
−1 −t
−3 t
e + c2
e.
1
1
a1
Xp =
b1
into the system yields
2a1 + 3b1 = 7
−a1 − 2b1 = −5,
from which we obtain a1 = −1 and b1 = 3. Then
−1
−1 −t
−3 t
X(t) = c1
.
e + c2
e +
3
1
1
2. Solving
5 − λ
9 det(A − λI) = = λ2 − 16λ + 64 = (λ − 8)2 = 0
−1 11 − λ
we obtain the eigenvalue λ = 8. A corresponding eigenvector is
3
K=
.
1
Solving (A − 8I)P = K we obtain
Thus
Substituting
2
P=
.
1
3 8t
3
2 8t
e + c2
te8t +
e .
X c = c1
1
1
1
a1
Xp =
b1
into the system yields
5a1 + 9b1 = −2
−a1 + 11b1 = −6,
from which we obtain a1 = 1/2 and b1 = −1/2. Then
1
3 8t
3
2 8t
8t
2
e + c2
te +
+
.
e
X(t) = c1
− 12
1
1
1
541
542
CHAPTER 8
3. Solving
SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
1 − λ
3
det(A − λI) = = λ2 − 2λ − 8 = (λ − 4)(λ + 2) = 0
3
1 − λ
we obtain eigenvalues λ1 = −2 and λ2 = 4. Corresponding eigenvectors are
1
1
and K2 =
.
K1 =
−1
1
Thus
X c = c1
Substituting
1 −2t
1 4t
e
e .
+ c2
−1
1
a1
a3 2
a2
t +
t+
Xp =
b3
b2
b1
into the system yields
a3 + 3b3 = 2
a2 + 3b2 = 2a3
a1 + 3b1 = a2
3a3 + b3 = 0
3a2 + b2 + 1 = 2b3
3a1 + b1 + 5 = b2
from which we obtain a3 = −1/4, b3 = 3/4, a2 = 1/4, b2 = −1/4, a1 = −2, and b1 = 3/4.
Then
1
1
−4
1
1
−2
4
−2t
4t
2
X(t) = c1
+ c2
e
e +
t +
t+
.
3
1
3
−
−1
1
4
4
4
4. Solving
1 − λ −4 det(A − λI) = = λ2 − 2λ + 17 = 0
4
1 − λ
we obtain eigenvalues λ1 = 1 + 4i and λ2 = 1 − 4i. Corresponding eigenvectors are
i
−i
and K2 =
.
K1 =
1
1
Thus
0
−1
−1
0
t
cos 4t +
sin 4t e + c2
cos 4t −
sin 4t et
X c = c1
1
0
0
1
− sin 4t t
− cos 4t t
e + c2
e.
= c1
cos 4t
− sin 4t
Substituting
a2
a1 6t
a3
Xp =
t+
+
e
b3
b2
b1
8.3
Nonhomogeneous Linear Systems
into the system yields
a3 − 4b3 = −4
a2 − 4b2 = a3
−5a1 − 4b1 = −9
4a3 + b3 = 1
4a2 + b2 = b3
4a1 − 5b1 = −1
from which we obtain a3 = 0, b3 = 1, a2 = 4/17, b2 = 1/17, a1 = 1, and b1 = 1. Then
X(t) = c1
5. Solving
− sin 4t
cos 4t
et + c2
− cos 4t
− sin 4t
et +
4
0
t+
1
17
1
17
1
+
e6t .
1
1 4 − λ
3 det(A − λI) = = λ2 − 10λ + 21 = (λ − 3)(λ − 7) = 0
9
6 − λ
we obtain the eigenvalues λ1 = 3 and λ2 = 7. Corresponding eigenvectors are
1
1
and K2 =
.
K1 =
−3
9
Thus
1 3t
1 7t
e + c2
e .
−3
9
X c = c1
Substituting
a1 t
e
Xp =
b1
into the system yields
1
3a1 + b1 = 3
3
9a1 + 5b1 = −10
from which we obtain a1 = 55/36 and b1 = −19/4. Then
X(t) = c1
6. Solving
1
−3
3t
e + c2
1
e +
9
55 36
et .
19
−4
7t
−1 − λ
5 det(A − λI) = = λ2 + 4 = 0
−1
1 − λ
we obtain the eigenvalues λ1 = 2i and λ2 = −2i. Corresponding eigenvectors are
5
5
and K2 =
.
K1 =
1 + 2i
1 − 2i
543
544
CHAPTER 8
SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
Thus
X c = c1
Substituting
5 cos 2t
cos 2t − 2 sin 2t
+ c2
5 sin 2t
.
2 cos 2t + sin 2t
a1
a2
cos t +
sin t
Xp =
b2
b1
into the system yields
−a2 + 5b2 − a1 = 0
−a2 + b2 − b1 − 2 = 0
−a1 + 5b1 + a2 + 1 = 0
−a1 + b1 + b2 = 0
from which we obtain a2 = −3, b2 = −2/3, a1 = −1/3, and b1 = 1/3. Then
X(t) = c1
7. Solving
5 cos 2t
cos 2t − 2 sin 2t
+ c2
5 sin 2t
2 cos 2t + sin 2t
+
−3
− 23
cos t +
− 13
1
3
sin t.
1 − λ
1
1 2−λ
3 = (1 − λ)(2 − λ)(5 − λ) = 0
det(A − λI) = 0
0
0
5 − λ
we obtain the eigenvalues λ1 = 1, λ2 = 2, and λ3 = 5. Corresponding eigenvectors are
⎛ ⎞
⎛ ⎞
⎛ ⎞
1
1
1
⎜ ⎟
⎜ ⎟
⎜ ⎟
K1 = ⎝0⎠ , K2 = ⎝1⎠ and K3 = ⎝2⎠ .
0
0
2
Thus
Substituting
⎛ ⎞
⎛ ⎞
⎛ ⎞
1
1
1
⎜ ⎟ t
⎜ ⎟ 2t
⎜ ⎟ 5t
Xc = C1 ⎝0⎠ e + C2 ⎝1⎠ e + C3 ⎝2⎠ e .
0
0
2
⎛ ⎞
a1
⎜ ⎟ 4t
Xp = ⎝ b1 ⎠ e
c1
into the system yields
−3a1 + b1 + c1 = −1
−2b1 + 3c1 = 1
c1 = −2
8.3
Nonhomogeneous Linear Systems
from which we obtain c1 = −2, b1 = −7/2, and a1 = −3/2. Then
⎛ 3⎞
⎛ ⎞
⎛ ⎞
⎛ ⎞
−2
1
1
1
⎜
⎟
⎜ ⎟
⎜ ⎟
⎜ ⎟
7 ⎟ 4t
X(t) = C1 ⎝0⎠ et + C2 ⎝1⎠ e2t + C3 ⎝2⎠ e5t + ⎜
⎝− 2 ⎠ e .
0
0
2
−2
8. Solving
−λ
0
5
det(A − λI) = 0 5 − λ 0 = −(λ − 5)2 (λ + 5) = 0
5
0
−λ
we obtain the eigenvalues λ1 = 5, λ2 = 5, and λ3 = −5. Corresponding eigenvectors are
⎛ ⎞
⎛ ⎞
⎛ ⎞
1
1
1
⎜ ⎟
⎜ ⎟
⎜ ⎟
K1 = ⎝0⎠ , K2 = ⎝1⎠ and K3 = ⎝ 0⎠ .
0
1
−1
Thus
⎛ ⎞
⎛ ⎞
⎛ ⎞
1
1
1
⎜ ⎟ 5t
⎜ ⎟ 5t
⎜ ⎟ −5t
Xc = C1 ⎝0⎠ e + C2 ⎝1⎠ e + C3 ⎝ 0⎠ e .
1
1
−1
Substituting
⎛ ⎞
a1
⎜ ⎟
Xp = ⎝ b1 ⎠
c1
into the system yields
5c1 = −5
5b1 = 10
5a1 = −40
from which we obtain c1 = −1, b1 = 2, and a1 = −8. Then
⎛ ⎞
⎛ ⎞
⎛ ⎞
⎛ ⎞
−8
1
1
1
⎜ ⎟ 5t
⎜ ⎟ 5t
⎜ ⎟ −5t ⎜ ⎟
+ ⎝ 2⎠ .
X(t) = C1 ⎝0⎠ e + C2 ⎝1⎠ e + C3 ⎝ 0⎠ e
−1
1
1
−1
9. First solve the associated homogeneous system
−1 −2
X =
X
3
4
The eigenvalues and eigenvectors of the coefficient matrix are found by solving det(A−λI) = 0
to get
1
2
and
λ2 = 2 so K2 =
λ1 = 1 so K1 =
−1
−3
545
546
CHAPTER 8
SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
The complementary solution is then
λ1 t
Xc (t) = c1 K1 e
λ2 t
+ c2 K 2 e
= c1
1 t
2 2t
e + c2
e
−1
−3
a1
and force it into the original system to get
Based on the form of F(t), guess Xp =
b1
−9
. The general solution is then
Xp =
6
1 t
2 2t
−9
e + c2
e +
X = X c + X p = c1
−1
−3
6
Next use the initial condition to solve for c1 and c2 :
1
2
−9
−4
X(0) = c1
+ c2
+
=
−1
−3
6
5
c1 = 13 and c2 = −4
2 2t
−9
1 t
e +
e −4
X = 13
−3
6
−1
10. First solve the associated homogeneous system
1 −1
X =
X
1
3
The eigenvalues of the coefficient matrix are found by solving det(A − λI) = 0 to get λ1 =
λ2 = 2 so we get
−1
−1 2t
Thus X1 =
e
K1 =
1
1
−1
1 2t
A second solution is found to be X2 =
te2t +
e , and so we have
1
0
−1 2t
−1
1 2t
e + c2
te2t +
e
Xc (t) = c1
1
1
0
a2 t + a1
and force it into the original system to
Based on the form of F(t), guess Xp =
b2 t + b1
−t − 1
. The general solution is then
get Xp =
0
−1 2t
−1
−t
−
1
1
e + c2
te2t +
e2t +
X(t) = c1
1
1
0
0
8.3
Nonhomogeneous Linear Systems
Next use the initial condition to solve for c1 and c2 :
−1
1
−1
3
+ c2
+
=
X(0) = c1
1
0
0
2
c1 = 2 and c2 = 6
−1 2t
−t − 1
−1 2t
1 2t
X(t) = 2
e +6
+
e +
e
1
0
1
0
11. (a) From the discussion in Section 4.9 we get the system
⎧
3
dx1
1
⎪
3
⎪
⎨ dt = − 100 x1 + 100 x2
− 100
Thus X =
1
⎪ dx2
1
1
⎪
⎩
50
= x1 − x2 + 1
dt
50
25
(b) First solve the associated homogeneous system
3
− 100
X =
1
50
1 100
1
− 25
1 100
1
− 25
0
X+
1
X
The eigenvalues and eigenvectors of the coefficient matrix are found by solving
det(A − λI) = 0 to get
−1
1
1
1
so K1 =
so K2 =
λ1 = −
and
λ2 = −
20
50
2
1
The complementary solution is then
Xc (t) = c1 K1 eλ1 t + c2 K2 eλ2 t = c1
−1 −t/20
1 −t/50
e
e
+ c2
2
1
a1
and force it into the original system to get
Based on the form of F(t), guess Xp =
b1
10
. The general solution is then
Xp =
30
−1 −t/20
1 −t/50
10
e
e
X = X c + X p = c1
+ c2
+
2
1
30
Next use the initial condition to solve for c1 and c2 :
−1
1
10
60
+ c2
+
=
X(0) = c1
2
1
30
10
70
80
and c2 =
c1 = −
3
3
10
70 −1 −t/20 80 1 −t/50
+
+
e
e
X=−
3
3 1
30
2
547
548
CHAPTER 8
SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
10
(c) The solution X(t) →
as t → ∞. Over a long period of time the total amount of
30
salt in the system of tanks approaches 40 lb.
(d)
i2
so that
12. (a) Let X =
i3
60
−2 −2
X+
X =
60
−2 −5
2 −t
1 −6t
e + c2
e .
−1
2
and
X c = c1
a1
30
then Xp =
so that
If Xp =
b1
0
X = c1
2 −t
1 −6t
30
e + c2
e
+
.
−1
2
0
0
For X(0) =
we find c1 = −12 and c2 = −6.
0
(b) i1 (t) = i2 (t) + i3 (t) = −12e−t − 18e−6t + 30.
13. From
X =
we obtain
Then
3 −3
4
X+
2 −2
−1
1
3 t
X c = c1
+ c2
e.
1
2
1 3et
Φ=
1 2et
and Φ−1 =
−2
3
−t
−e−t
e
8.3
so that
ˆ
U=
Φ−1 F dt =
ˆ and
Xp = ΦU =
Nonhomogeneous Linear Systems
−11t
−11
dt =
−5e−t
5e−t
−11
−15
t+
.
−11
−10
The solution is
1
3 t
11
15
+ c2
e −
t−
X = X c + X p = c1
1
2
11
10
2
−1
0
X+
t
X =
3 −2
4
14. From
1 t
1 −t
e + c2
e .
X c = c1
1
3
we obtain
Then
et e−t
Φ=
et 3e−t
so that
ˆ
U=
Φ−1 F dt =
3
and Φ−1 =
ˆ −2te−t
2tet
−t
2e
− 12 e−t
− 12 et
1 t
2e
dt =
2te−t + 2e−t
2tet − 2et
4
0
t+
.
Xp = ΦU =
8
−4
and
The solution is
1 t
1 −t
4
0
e + c2
e +
t−
X = X c + X p = c1
1
3
8
4
15. From
X =
3 −5
3
4
−1
X+
1 t/2
e
−1
10 3t/2
2 t/2
e
e .
+ c2
X c = c1
3
1
we obtain
Then
Φ=
10e3t/2 2et/2
3e3t/2
et/2
−1
and Φ
=
1 −3t/2
4e
− 12 e−3t/2
− 34 e−t/2
5 −t/2
2e
549
550
CHAPTER 8
SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
so that
ˆ
U=
Φ−1 F dt =
and
Xp = ΦU =
The solution is
X = X c + X p = c1
ˆ 3 e−t 4
− 13
2
− 13
4
− 13
4
dt =
t/2
te
+
3 −t −4e
− 13
4 t
− 15
2
et/2 .
− 94
13 15 10 3t/2
2 t/2
2
2
t/2
e
e −
te −
et/2
+ c2
13
9
3
1
4
16. From
X =
2 −1
sin 2t
X+
4
2
2 cos 2t
− sin 2t 2t
cos 2t 2t
e + c2
e .
X c = c1
2 cos 2t
2 sin 2t
we obtain
Then
4
Φ=
−e2t sin 2t e2t cos 2t
2e2t cos 2t 2e2t sin 2t
so that
ˆ
U=
−1
and Φ
Φ−1 F dt =
=
−e−2t sin 2t
1
2
e−2t cos 2t
e−2t cos 2t
1
2
e−2t sin 2t
ˆ cos 4t
dt =
Xp = ΦU =
The solution is
X = X c + X p = c1
sin 4t
− 14 sin 2t sin 4t − 14 cos 2t cos 4t
1
2
− 14 cos 4t
sin 4t
and
1
4
cos 2t sin 4t − sin 2t cos 4t
1
2
e2t .
1
− 4 sin 2t sin 4t − 14 cos 2t cos 4t
− sin 2t 2t
cos 2t 2t
e2t
e + c2
e +
1
1
2 cos 2t
2 sin 2t
cos 2t sin 4t − sin 2t cos 4t
2
17. From
X =
we obtain
Then
2
1 t
0 2
e
X+
−1
−1 3
2 t
1 2t
e + c2
e .
X c = c1
1
1
2et e2t
Φ=
et e2t
and Φ−1 =
e−t −e−t
−e−2t 2e−2t
8.3
so that
ˆ
U=
Φ−1 F dt =
ˆ 2
−3e−t
Nonhomogeneous Linear Systems
dt =
2t
3e−t
4
3 t
Xp = ΦU =
tet +
e.
2
3
and
The solution is
2 t
1 2t
4
3 t
e + c2
e +
tet +
e
X = X c + X p = c1
1
1
2
3
0 2
2
X + −3t
−1 3
e
18. From
X =
2 t
1 2t
e + c2
e .
X c = c1
1
1
we obtain
2et e2t
Φ=
et e2t
Then
so that
ˆ
U=
Φ−1 F dt =
−1
and Φ
=
ˆ 2e−t − e−4t
−2e−2t + 2e−5t
and
Xp = ΦU =
e−t −e−t
−e−2t 2e−2t
1 −3t
10 e
dt =
−2e−t + 14 e−4t
e−2t − 25 e−5t
−3
.
3 −3t
e
−1
− 20
The solution is
1 −3t
−3
2 t
1 2t
10 e
X = X c + X p = c1
e + c2
e +
1
1
− 3 e−3t − 1
20
19. From
1
8
12
X =
X+
t
1 −1
12
we obtain
Then
4 3t
−2 −3t
e + c2
e .
X c = c1
1
1
4e3t −2e−3t
Φ=
e3t
e−3t
−3t
6e
1 −3t 3e
− 16 e3t
2 3t
3e
1
−1
and Φ
=
551
552
CHAPTER 8
SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
so that
ˆ
U=
Φ−1 F dt =
ˆ 6te−3t
6te3t
and
Xp = ΦU =
dt =
−2te−3t − 23 e−3t
2te3t − 23 e3t
4
−3
−12
.
t+
0
−4
3
The solution is
4
4 3t
−2 −3t
−12
3
e + c2
e
+
t−
X = X c + X p = c1
4
1
1
0
3
20. From
X =
1
8
e−t
X+
tet
1 −1
4 3t
−2 −3t
e + c2
e .
X c = c1
1
1
we obtain
Then
4e3t −2e−3t
Φ=
e3t
e−3t
so that
ˆ
Φ−1 F dt =
U=
and Φ−1 =
ˆ 1 e−4t + 1 te−2t 6
3
− 16 e2t + 23 te4t
and
Xp = ΦU =
The solution is
−3t
6e
1 −3t 3e
− 16 e3t
2 3t
3e
1
dt =
1 −4t
e
− 16 te−2t −
− 24
1 2t
− 12
e + 16 te4t −
−tet − 14 et
− 18 e−t − 18 et
1 −2t 12 e
1 4t
24 e
.
t 1 t te + 4 e
4 3t
−2 −3t
e + c2
e
−
X = X c + X p = c1
1 −t
1
1
e + 1 et
8
21. From
X =
8
2 −t
3
2
e
X+
1
−2 −1
we obtain
X c = c1
Then
Φ=
et
−et
0
1 t
1
e + c2
tet + 1 et .
−1
−1
2
tet
1 t
2e
− tet
and Φ−1
e−t − 2te−t −2te−t
=
2e−t
2e−t
8.3
so that
ˆ
U=
ˆ Φ−1 F dt =
2e−2t − 6te−2t
6e−2t
and
Xp = ΦU =
1
2
Nonhomogeneous Linear Systems
dt =
1 −2t
2e
+ 3te−2t
−3e−2t
−2
e−t .
The solution is
1
t
1 t
2
t
e + c2 1
e +
e−t
−1
−2
2 −t
X = X c + X p = c1
22. From
X =
3
2
1
X+
−2 −1
1
0
1 t
1
t
e + c2
te + 1 et .
−1
−1
2
we obtain
X c = c1
Then
Φ=
et
tet
−et
so that
ˆ
U=
1 t
2e
and Φ
− tet
Φ−1 F dt =
ˆ e−t − 4te−t
2e−t
dt =
3e−t + 4te−t
−2e−t
and
Xp = ΦU =
The solution is
X = X c + X p = c1
23. From
X =
X c = c1
3
.
−5
t
1 t
3
et +
e + c2 1
−1
−5
2 −t
0 −1
sec t
X+
1
0
0
we obtain
Then
e−t − 2te−t −2te−t
=
2e−t
2e−t
−1
cos t
sin t
+ c2
.
sin t
− cos t
cos t
sin t
Φ=
sin t − cos t
and Φ−1 =
cos t
sin t
sin t − cos t
553
554
CHAPTER 8
SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
so that
ˆ
U=
ˆ Φ−1 F dt =
and
Xp = ΦU =
1
t
dt =
tan t
− ln | cos t|
t cos t − sin t ln | cos t|
.
t sin t + cos t ln | cos t|
The solution is
X = X c + X p = c1
cos t
sin t
+ c2
sin t
− cos t
+
t cos t − sin t ln | cos t|
t sin t + cos t ln | cos t|
1 −1
3 t
X+
e
X =
1
1
3
24. From
we obtain
X c = c1
− sin t t
cos t t
e + c2
e.
cos t
sin t
− sin t cos t t
Φ=
e
cos t sin t
Then
so that
ˆ
U=
Φ−1 F dt =
ˆ −1
and Φ
−3 sin t + 3 cos t
3 cos t + 3 sin t
and
Xp = ΦU =
=
− sin t cos t −t
e
cos t sin t
dt =
3 cos t + 3 sin t
3 sin t − 3 cos t
−3 t
e.
3
The solution is
X = X c + X p = c1
25. From
1
−1
cos
t
X+
et
X =
1
1
sin t
we obtain
X c = c1
Then
− sin t t
cos t t
−3 t
e + c2
e +
e
cos t
sin t
3
− sin t t
cos t t
e + c2
e.
cos t
sin t
− sin t cos t t
Φ=
e
cos t sin t
and Φ−1 =
− sin t cos t −t
e
cos t sin t
8.3
so that
ˆ
U=
ˆ 0
0
dt =
Φ−1 F dt =
1
t
and
Xp = ΦU =
The solution is
Nonhomogeneous Linear Systems
X = X c + X p = c1
cos t
tet .
sin t
− sin t t
cos t t
cos t
e + c2
e +
tet
cos t
sin t
sin t
2
−2
1 1 −2t
e
X+
X =
8 −6
3 t
26. From
we obtain
1
1 −2t
1
2
−2t
−2t
e
te
e
.
X c = c1
+ c2
+
1
2
2
2
Then
1
t+
Φ=
2 2t +
so that
ˆ
U=
−1
Φ
1
2
e−2t
1
2
−1
and Φ
=
−4t − 1 2t + 1 2t
e
4
−2
ˆ 2 + 2/t
2t + 2 ln t
dt =
F dt =
−2/t
−2 ln t
and
Xp = ΦU =
2t + ln t − 2t ln t
e−2t .
4t + 3 ln t − 4t ln t
The solution is
1
1 −2t
2t
+
ln
t
−
2t
ln
t
2 +t
e−2t +
e
+ c2
e−2t
X = X c + X p = c1
1
2
4t
+
3
ln
t
−
4t
ln
t
+ 2t
2
27. From
X =
0 1
0
X+
−1 0
sec t tan t
we obtain
X c = c1
Then
Φ=
cos t
− sin t
cos t sin t
t
− sin t cos t
+ c2
sin t
.
cos t
and Φ−1 =
cos t − sin t
sin t
cos t
555
556
CHAPTER 8
SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
so that
ˆ
ˆ Φ−1 F dt =
U=
− tan2 t
tan t
dt =
t − tan t
− ln | cos t|
and
cos t
− sin t
sin t
t+
−
ln | cos t|.
− sin t
sin t tan t
cos t
Xp = ΦU =
The solution is
X = X c + X p = c1
cos t
− sin t
+ c2
sin t
cos t
− sin t
sin t
+
t+
−
ln | cos t|
cos t
− sin t
sin t tan t
cos t
28. From
0 1
1
X+
−1 0
cot t
X =
we obtain
cos t
− sin t
X c = c1
Then
cos t sin t
− sin t cos t
Φ=
so that
ˆ
U=
Φ−1 F dt =
+ c2
sin t
.
cos t
−1
ˆ and Φ
=
0
csc t
dt =
cos t − sin t
sin t
cos t
0
ln | csc t − cot t|
and
sin t ln | csc t − cot t|
.
cos t ln | csc t − cot t|
Xp = ΦU =
The solution is
X = X c + X p = c1
cos t
− sin t
29. From
X =
we obtain
X c = c1
Then
+ c2
csc t t
X+
e
sec t
1
1 2
− 12
sin t
sin t ln | csc t − cot t|
+
cos t
cos t ln | csc t − cot t|
2 sin t t
2 cos t t
e + c2
e.
cos t
− sin t
2 sin t 2 cos t t
Φ=
e
cos t − sin t
1
−1
and Φ
=
2
sin t
cos t
1
2
cos t − sin t
e−t
8.3
so that
ˆ
U=
Φ−1 F dt =
and
Xp = ΦU =
3 sin t
3
2
ˆ The solution is
X = X c + X p = c1
te +
X =
we obtain
X c = c1
Φ=
ˆ
U=
Φ−1 F dt =
ˆ cos t − sin t
cos t
Xp = ΦU =
ln | sin t| + ln | cos t|
e ln | sin t| +
t
+ c2
2 cos t t
e ln | cos t|.
− sin t
cos t + sin t
.
sin t
and Φ
2 cos t + sin t − sec t
2 sin t − cos t
and
1 −2
tan t
X+
1 −1
1
cos t − sin t cos t + sin t
cos t
sin t
so that
1
2
− 12 sin t
3
2t
3 sin t
2 sin t t
2 cos t t
tet
e + c2
e + 3
cos t
− sin t
cos
t
2
cos t t
2 cos t t
e ln | sin t| +
+
e ln | cos t|
− sin t
− 12 sin t
30. From
Then
cos t
dt =
cot t − tan t
t
cos t
3
2
1
2
Nonhomogeneous Linear Systems
−1
=
− sin t cos t + sin t
cos t sin t − cos t
2 sin t − cos t − ln | sec t + tan t|
dt =
−2 cos t − sin t
−3 − cos t ln |sec t + tan t| + sin t ln |sec t + tan t|
.
−1 − cos t ln |sec t + tan t|
The solution is
cos t − sin t
cos t + sin t
+ c2
X = X c + X p = c1
cos t
sin t
−3 − cos t ln |sec t + tan t| + sin t ln |sec t + tan t|
+
−1 − cos t ln |sec t + tan t|
31. From
⎞
⎛ t⎞
⎛
e
1 1 0
⎟
⎜ 2t ⎟
⎜
X = ⎝1 1 0⎠ X + ⎝ e ⎠
0 0 3
te3t
557
558
CHAPTER 8
SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
⎞
⎛ ⎞
⎛ ⎞
1
1
0
⎜ ⎟
⎜ ⎟ 2t
⎜ ⎟ 3t
Xc = c1 ⎝−1⎠ + c2 ⎝1⎠ e + c3 ⎝0⎠ e .
0
0
1
⎛
we obtain
Then
⎛
⎛
⎞
1 e2t 0
⎜
⎟
Φ = ⎝−1 e2t 0 ⎠
0
0 e3t
⎜
1 −2t
and Φ−1 = ⎜
⎝2e
0
so that
ˆ
U=
Φ−1 F dt =
ˆ
and
− 12
1
2
⎛1
t
2e
− 12 e2t
⎜
⎜ 1 e−t +
⎝2
t
⎛
1
2
⎞
1 −2t
2e
0
⎛
1 t
2e
0
⎞
⎟
0 ⎟
⎠
e−3t
− 14 e2t
⎞
⎟
⎜
⎟
⎟ dt = ⎜− 1 e−t + 1 t⎟
2 ⎠
⎠
⎝ 2
1 2
2t
− 14 e2t + 12 te2t
⎞
⎜ t 1 2t 1 2t ⎟
⎟
Xp = ΦU = ⎜
⎝−e + 4 e + 2 te ⎠ .
1 2 3t
2t e
The solution is
⎛
⎞
⎞
⎛ ⎞
⎛ ⎞
− 14 e2t + 12 te2t
1
1
0
⎜ t 1 2t 1 2t ⎟
⎜ ⎟
⎜ ⎟
⎜ ⎟
⎟
X = Xc + Xp = c1 ⎝−1⎠ + c2 ⎝1⎠ e2t + c3 ⎝0⎠ e3t + ⎜
⎝−e + 4 e + 2 te ⎠
0
0
1
1 2 3t
2t e
⎛
⎞
⎛ ⎞
0
3 −1 −1
⎟
⎜ ⎟
⎜
1 −1⎠ X + ⎝ t ⎠
X = ⎝1
2et
1 −1
1
⎛
32. From
⎛ ⎞
⎛ ⎞
⎛ ⎞
1
1
1
⎜ ⎟ t
⎜ ⎟ 2t
⎜ ⎟ 2t
Xc = c1 ⎝1⎠ e + c2 ⎝1⎠ e + c3 ⎝0⎠ e .
1
0
1
we obtain
Then
⎛
⎞
et e2t e2t
⎜
⎟
Φ = ⎝et e2t 0 ⎠
et 0 e2t
so that
ˆ
U=
Φ−1 F dt =
ˆ
⎞
−e−t
e−t
e−t
⎟
⎜
0
−e−2t ⎠
= ⎝ e−2t
e−2t −e−2t
0
⎛
and Φ−1
⎛ −t
⎞
⎛
⎞
−te−t − e−t + 2t
te + 2
⎜
⎟
⎜
⎟
2e−t
⎝ −2e−t ⎠ dt = ⎝
⎠
−2t
1
1
−2t
−2t
−te
+ 4e
2 te
8.3
⎛
and
⎞
⎛
Nonhomogeneous Linear Systems
⎞
⎛ ⎞
⎛ ⎞
2
2
⎜ ⎟
⎜ ⎟ ⎜ ⎟ t ⎜ ⎟ t
⎟
⎟
⎜
⎜
Xp = ΦU = ⎝ −1⎠ t + ⎝ −1⎠ + ⎝2⎠ e + ⎝2⎠ te .
2
0
−1
−3
− 12
− 34
2
4
The solution is
⎛ 1⎞
⎛ 3⎞ ⎛ ⎞
⎛ ⎞
⎛ ⎞
⎛ ⎞
⎛ ⎞
−2
−4
2
1
1
1
2
⎟
⎜ ⎟ ⎜ ⎟ t ⎜ ⎟ t
⎜ ⎟ t
⎜ ⎟ 2t
⎜ ⎟ 2t ⎜
⎜
⎟
⎜
⎟
X = Xc + Xp = c1 ⎝1⎠ e + c2 ⎝1⎠ e + c3 ⎝0⎠ e + ⎝ −1⎠ t + ⎝ −1⎠ + ⎝2⎠ e + ⎝2⎠ te
2
1
0
1
0
−1
−3
2
33. From
X =
we obtain
and
3 −1
4e2t
X+
4e4t
−1
3
−e4t e2t
Φ=
,
e4t e2t
−1
Φ
=
− 12 e−4t
1 −2t
2e
1 −4t 2e
1 −2t
2e
,
0
e−2t + 2t − 1
Φ F ds = Φ ·
+Φ·
X = ΦΦ (0)X(0) + Φ
e2t + 2t − 1
1
0
−1 2t
−2
2 4t
2
2t
4t
e +
te +
e .
=
te +
1
2
0
2
ˆ
−1
we obtain
−1
1 1+t
Φ=
,
1
t
and
X = ΦΦ−1 (1)X(1) + Φ
t
1
−1
1/t
X =
X+
1 −1
1/t
34. From
ˆ
1
t
Φ−1
−t 1 + t
=
,
1
−1
1
1
3
ln
t
−4
ln t.
+
t−
=
+Φ·
Φ−1 F ds = Φ ·
1
4
3
0
3
i1
35. Let X =
so that
i2
X =
and
4
−11
3
100 sin t
X+
3 −3
0
1 −2t
3 −12t
e
e
+ c2
.
X C = c1
3
−1
559
560
CHAPTER 8
SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
Then
e−2t 3e−12t
3e−2t −e−12t
Φ=
ˆ
U=
−1
Φ
ˆ F dt =
Φ−1 =
,
10e2t sin t
30e12t sin t
and
Xp = ΦU =
3 2t
0e
3 12t
10 e
1 12t
− 10
e
dt =
1 2t
10 e
,
2e2t (2 sin t − cos t)
6 12t
(12 sin t
29 e
− cos t)
,
332
29
sin t −
76
29
cos t
276
29
sin t −
168
29
cos t
1 −2t
3 −12t
e
e
+ c2
+ Xp .
X = c1
3
−1
so that
0
If X(0) =
then c1 = 2 and c2 =
0
6
29 .
The solutions is
3 −12t
1 −2t
6
4 19
4 83
e
cos t +
sin t.
+
−
X=2
e
29 −1
29 42
29 69
3
36. Write the differential equation as a system
y = v
v = −Qy − P v + f
or
y
0
1
y
0
=
+
.
v
−Q −P
v
f
From (9) in the text of this section, a particular solution is then Xp = Φ(x)
where
u1
y1 y2
and
Xp =
.
Φ(x) =
y1 y2
u2
Then
1
Φ−1 (x) =
y1 y2 − y2 y1
so
ˆ
Xp =
1
W
y2 −y2
−y1
y1
y2 −y2
−y1
y1
´
Φ−1 (x)F(x) dx
,
0
dx
f
and W = y1 y2 − y2 y1 . Thus
ˆ
u1 =
−y2 f (x)
dx
W
ˆ
and
u2 =
y1 f (x)
dx,
W
which are the antiderivative forms of the equations in (5) of Section 4.6 in the text.
8.3
Nonhomogeneous Linear Systems
37. (a) The eigenvalues are 0, 1, 3, and 4, with corresponding eigenvectors
⎛ ⎞
−6
⎜−4⎟
⎜ ⎟
⎜ ⎟,
⎝ 1⎠
⎛ ⎞
2
⎜1⎟
⎜ ⎟
⎜ ⎟,
⎝0⎠
⎛ ⎞
3
⎜1⎟
⎜ ⎟
⎜ ⎟,
⎝2⎠
2
0
1
⎛
⎛
⎞
−6 2et 3e3t −e4t
⎜−4 et e3t
e4t ⎟
⎜
⎟
(b) Φ = ⎜
⎟,
3t
⎝1
0 2e
0 ⎠
0
2
0
e3t
Φ−1
⎛
2
3
and
0
0
⎜
⎜ 1 e−t
⎜ 3
=⎜
⎜
⎜ 0
⎝
− 13 e−4t
− 13 e2t
⎞
−1
⎜ 1⎟
⎜ ⎟
⎜ ⎟.
⎝ 0⎠
⎛
− 13
0
1
3
2
3
e−t
−2e−t
0
2
3
e−4t
⎞
⎜
⎟
⎜ 1 e−2t + 8 e−t − 2et + 1 t⎟
⎜
3
3
3 ⎟
⎟,
(c) Φ−1 (t)F(t) = ⎜
⎜
⎟
⎜
− 13 e−3t + 23 e−t ⎟
⎝
⎠
1 −4t
1
2 −5t
−3t
+3e
− 3 te
3e
⎛
ˆ
⎜
⎜
⎜
−1
Φ (t)F(t) dt = ⎜
⎜
⎜
⎝
ˆ
Xp (t) = Φ(t)
⎛
⎞
− 16 e2t + 23 t
− 16 e−2t − 83 e−t − 2et + 16 t2
2 −5t
− 15
e
−
1
9
e−3t − 23 e−t
1
12
e−4t +
1
27
e−3t + 19 te−3t
⎟
⎟
⎟
⎟,
⎟
⎟
⎠
Φ−1 (t)F(t) dt
−5e2t − 15 e−t −
⎜
⎜−2e2t −
⎜
=⎜
⎜
⎜
⎝
3
10
1
27
e−t +
et − 19 tet + 13 t2 et − 4t −
1 t
27 e
− 32
e2t
+ 19 tet + 16 t2 et − 83 t −
2
3
+ t+
−e2t + 43 t −
⎛
2
9
1
9
⎞
−6c1 + 2c2 et + 3c3 e3t − c4 e4t
⎜ −4c + c et + c e3t + c e4t ⎟
1
2
3
4
⎜
⎟
Xc (t) = Φ(t)C = ⎜
⎟,
3t
⎝
⎠
c1 + 2c3 e
3t
2c1 + c3 e
59
12
⎞
⎟
95 ⎟
36 ⎟
⎟,
⎟
⎟
⎠
e−3t
0
2
3
⎞
⎟
e−t ⎟
⎟
⎟
⎟
− 13 e−3t ⎟
⎠
1 −4t
3e
8
3
561
562
CHAPTER 8
SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
⎛
⎞
−6c1 + 2c2 et + 3c3 e3t − c4 e4t
ˆ
⎜ −4c + c et + c e3t + c e4t ⎟
1
2
3
4
⎜
⎟
X(t) = Φ(t)C + Φ(t) Φ−1 (t)F(t) dt = ⎜
⎟
⎝
⎠
c1 + 2c3 e3t
3t
2c1 + c3 e
⎞
⎛
1 t
e − 19 tet + 13 t2 et − 4t − 59
−5e2t − 15 e−t − 27
12
⎟
⎜
⎜−2e2t − 3 e−t + 1 et + 1 tet + 1 t2 et − 8 t − 95 ⎟
⎜
10
27
9
6
3
36 ⎟
⎟
+⎜
⎟
⎜
⎟
⎜
− 32 e2t + 23 t + 29
⎠
⎝
4
1
2t
−e + 3 t − 9
⎛ ⎞
⎛ ⎞
⎛ ⎞
⎛ ⎞
−6
2
3
−1
⎜−4⎟
⎜1⎟
⎜1⎟
⎜ 1⎟
⎜ ⎟
⎜ ⎟
⎜ ⎟
⎜ ⎟
(d) X(t) = c1 ⎜ ⎟ + c2 ⎜ ⎟ et + c3 ⎜ ⎟ e3t + c4 ⎜ ⎟ e4t
⎝ 1⎠
⎝0⎠
⎝2⎠
⎝ 0⎠
2
0
⎛
1
0
−5e2t − 15 e−t −
⎜
⎜−2e2t −
⎜
+⎜
⎜
⎜
⎝
3
10
1
27
e−t +
et − 19 tet + 13 t2 et − 4t −
1 t
27 e
− 32
e2t
+ 19 tet + 16 t2 et − 83 t −
2
3
+ t+
−e2t + 43 t −
8.4
Matrix Exponential
1. For A =
1 0
we have
0 2
1
0
1
A2 =
0 2
0
1
A3 = AA2 =
0
1
4
3
A = AA =
0
0
1 0
=
,
2
0 4
1 0
1 0
0
,
=
0 8
0 4
2
1 0
1 0
0
,
=
0 16
0 8
2
and so on. In general
Ak =
1 0
0 2k
for
k = 1, 2, 3, . . . .
2
9
1
9
59
12
⎞
⎟
95 ⎟
36 ⎟
⎟
⎟
⎟
⎠
8.4 Matrix Exponential
Thus
A
A2 2 A3 3
eAt = I + t +
t +
t + ···
1!
2!
3!
1 0
1 1 0 2
1 1 0 3
1 1 0
t+
t +
t + ···
=
+
1! 0 2
2! 0 4
3! 0 8
0 1
⎞
⎛
t2 t3
0⎟
0
et
⎜1 + t + 2! + 3! + · · ·
=⎝
⎠=
(2t)2 (2t)3
0 e2t
0 1+t+
+
+ ···
2!
3!
and
e−At =
2. For A =
e−t
0
.
0 e−2t
0 1
we have
1 0
0 1
0
A =
1 0
1
0
3
2
A = AA =
1
2
1
1
=
0
0
1
0
I=
0
1
0
=I
1
1
=A
0
A4 = (A2 )2 = I
A5 = AA4 = AI = A,
and so on. In general,
k
A, k = 1, 3, 5, . . .
A =
I,
k = 2, 4, 6, . . . .
Thus
A2 2 A3 3
A
t+
t +
t + ···
1!
2!
3!
1
1
= I + At + It2 + At3 + · · ·
2!
3!
1 4
1 3
1 5
1 2
= I 1 + t + t + ··· + A t + t + t + ···
2!
4!
3!
5!
cosh t sinh t
= I cosh t + A sinh t =
sinh t cosh t
eAt = I +
and
−At
e
=
cosh (−t) sinh (−t)
sinh (−t) cosh (−t)
=
cosh t − sinh t
.
− sinh t
cosh t
563
564
CHAPTER 8
SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
⎞
1
1
1
⎟
⎜
1
1⎠
A=⎝ 1
−2 −2 −2
⎛
3. For
we have
⎞
⎞ ⎛
⎞⎛
0 0 0
1
1
1
1
1
1
⎟
⎟ ⎜
⎟⎜
⎜
1
1⎠ = ⎝0 0 0⎠ .
1
1⎠ ⎝ 1
A2 = ⎝ 1
0 0 0
−2 −2 −2
−2 −2 −2
⎛
Thus, A3 = A4 = A5 = · · · = 0 and
⎞
⎞ ⎛
⎞ ⎛
⎛
t+1
t
t
t
t
t
1 0 0
⎟
⎟ ⎜
⎟ ⎜
⎜
t+1
t
t
t⎠ = ⎝ t
eAt = I + At = ⎝0 1 0⎠ + ⎝ t
⎠.
−2t −2t −2t + 1
−2t −2t −2t
0 0 1
⎛
⎞
0 0 0
⎜
⎟
A = ⎝3 0 0⎠
5 1 0
4. For
we have
⎛
⎞⎛
0 0 0
0
⎜
⎟⎜
2
A = ⎝3 0 0⎠ ⎝3
5 1 0
5
⎛
0 0
⎜
3
2
A = AA = ⎝3 0
5 1
⎞ ⎛
⎞
0 0
0 0 0
⎟ ⎜
⎟
0 0⎠ = ⎝0 0 0⎠
1 0
3 0 0
⎞⎛
⎞ ⎛
⎞
0
0 0 0
0 0 0
⎟⎜
⎟ ⎜
⎟
0⎠ ⎝0 0 0⎠ = ⎝0 0 0⎠ .
0
3 0 0
0 0 0
Thus, A4 = A5 = A6 = · · · = 0 and
1
eAt = I + At + A2 t2
2
⎞ ⎛
⎞
⎛
⎞ ⎛
⎞ ⎛
1
0 0
0 0 0
1 0 0
0 0 0
⎟ ⎜
⎜
⎟ ⎜
⎟ ⎜
1 0⎟
= ⎝0 1 0⎠ + ⎝3t 0 0⎠ + ⎝ 0 0 0⎠ = ⎝ 3t
⎠.
3
3
2 0 0
2
0 0 1
5t t 0
2t
2 t + 5t t 1
5. Using the result of Problem 1,
X=
0
c1
et
0
et
+ c2 2t .
= c1
2t
c2
0
0 e
e
6. Using the result of Problem 2,
cosh t
sinh t
cosh t sinh t
c1
= c1
+ c2
.
X=
c2
sinh t
cosh t
sinh t cosh t
8.4 Matrix Exponential
7. Using the result of Problem 3,
⎛
⎞⎛ ⎞
⎛
⎞
⎛
⎞
⎛
⎞
t+1
t
t
t+1
t
t
c1
⎜
⎟⎜ ⎟
⎜
⎟
⎜
⎟
⎜
⎟
t+1
t
t
X=⎝ t
⎠ ⎝c2 ⎠ = c1 ⎝ t ⎠ + c2 ⎝t + 1⎠ + c3 ⎝
⎠.
c3
−2t −2t −2t + 1
−2t
−2t
−2t + 1
8. Using the result of Problem 4,
⎞⎛ ⎞
⎞
⎛
⎛ ⎞
⎛ ⎞
1
0 0
c1
0
0
⎟⎜ ⎟
⎜ 3t ⎟
⎜ ⎟
⎜ ⎟
⎜
1
0
X=⎝
⎠ ⎝c2 ⎠ = c1 ⎝
⎠ + c2 ⎝1⎠ + c3 ⎝0⎠ .
3 2
3
2
t
1
c3
2 t + 5t t 1
2 t + 5t
⎛
1
3t
9. To solve
1
0
3
X =
X+
0 2
−1
3
, and use the results of Problem 1 and Equation (6) in the
−1
we identify t0 = 0, F(t) =
text.
ˆ
X(t) = eAt C + eAt
t
e−As F(s) ds
t0
ˆ t e−s
0
0
0
c1
et
3
et
+
ds
=
2t
2t
−2s
c2
0 e
0 e
0 e
−1
0
ˆ t
0
et
3e−s
c1 et
+
ds
=
0 e2t
c2 e2t
−e−2s
0
=
c1 et
c2 e2t
0
−3e−t + 3
0
et
c1 et
+
=
1 −2t
0 e2t
c2 e2t
− 12
2e
−3 + 3et
−3
1 t
0 2t
c1 et
+ 1 1
= c3
e + c4
e +
.
=
2t
1
2t
c2 e
0
1
2 − 2e
2
t
−3e−s 0
et
+
1 −2s 0 e2t
2e
10. To solve
X =
we identify t0 = 0, F(t) =
text.
t
e4t
1 0
t
X + 4t
0 2
e
, and use the results of Problem 1 and Equation (6) in the
565
566
CHAPTER 8
SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
ˆ
At
At
t
X(t) = e C + e
e−As F(s) ds
t0
ˆ t e−s
0
0
0
c1
et
s
et
+
ds
=
0 e2t
0 e2t
0 e−2s
c2
e4s
0
ˆ t se−s
0
et
c1 et
+
ds
=
0 e2t
c2 e2t
e2s
0
t
−se−s − e−s 0
et
c1 et
+
=
1 2s
0 e2t
c2 e2t
2e
0
−te−t − e−t + 1
0
et
c1 et
+
=
1 2t
1
c2 e2t
0 e2t
2e − 2
−t − 1 + et
−t − 1
1 t
0 2t
c1 et
e + c4
e +
+ 1
= c3
.
=
1 2t
1 4t
4t
0
1
c2 e2t
2e − 2e
2e
11. To solve
X =
0 1
1
X+
1 0
1
1
, and use the results of Problem 2 and Equation (6) in the
we identify t0 = 0, F(t) =
1
8.4 Matrix Exponential
text.
ˆ
At
At
t
X(t) = e C + e
e−As F(s) ds
t0
ˆ t
cosh t sinh t
c1
cosh t sinh t
cosh s − sinh s
1
=
+
ds
c2
sinh t cosh t
sinh t cosh t
− sinh s
cosh s
1
0
ˆ t
cosh t sinh t
cosh s − sinh s
c1 cosh t + c2 sinh t
+
ds
=
sinh t cosh t
c1 sinh t + c2 cosh t
− sinh s + cosh s
0
t
c1 cosh t + c2 sinh t
cosh t sinh t
sinh s − cosh s =
+
c1 sinh t + c2 cosh t
sinh t cosh t
− cosh s + sinh s 0
cosh t sinh t
sinh t − cosh t + 1
c1 cosh t + c2 sinh t
+
=
c1 sinh t + c2 cosh t
sinh t cosh t
− cosh t + sinh t + 1
sinh2 t − cosh2 t + cosh t + sinh t
c1 cosh t + c2 sinh t
+
=
c1 sinh t + c2 cosh t
sinh2 t − cosh2 t + sinh t + cosh t
cosh t
sinh t
cosh t
sinh t
1
+ c2
+
+
−
= c1
sinh t
cosh t
sinh t
cosh t
1
1
cosh t
sinh t
.
+ c4
−
= c3
1
sinh t
cosh t
12. To solve
X =
cosh t
0 1
X+
sinh t
1 0
cosh t
we identify t0 = 0, F(t) =
, and use the results of Problem 2 and Equation (6) in
sinh t
the text.
567
568
CHAPTER 8
SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
ˆ
t
X(t) = eAt C + eAt
e−As F(s) ds
t0
ˆ t
cosh t sinh t
cosh s − sinh s
cosh s
cosh t sinh t
c1
+
ds
=
c2
sinh t cosh t
− sinh s
cosh s
sinh s
sinh t cosh t
0
ˆ t 1
cosh t sinh t
c1 cosh t + c2 sinh t
+
=
ds
c1 sinh t + c2 cosh t
sinh t cosh t
0
0
t
cosh t sinh t
s c1 cosh t + c2 sinh t
+
=
sinh t cosh t
0 c1 sinh t + c2 cosh t
0
cosh t sinh t
t
c1 cosh t + c2 sinh t
+
=
c1 sinh t + c2 cosh t
sinh t cosh t
0
t cosh t
cosh t
sinh t
cosh t
c1 cosh t + c2 sinh t
+
= c1
+ c2
+t
.
=
t sinh t
c1 sinh t + c2 cosh t
sinh t
cosh t
sinh t
13. We have
⎛ ⎞
⎛ ⎞
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
c1
1
0
0
1
⎜ ⎟
⎜ ⎟
⎜ ⎟ ⎜ ⎟ ⎜ ⎟
X(0) = c1 ⎝0⎠ + c2 ⎝1⎠ + c3 ⎝0⎠ = ⎝c2 ⎠ = ⎝−4⎠ .
0
0
1
6
c3
Thus, the solution of the initial-value problem is
⎞
⎞
⎛
⎞
⎛
⎛
t
t
t+1
⎟
⎟
⎜
⎟
⎜
⎜
t
X = ⎝ t ⎠ − 4 ⎝t + 1⎠ + 6 ⎝
⎠.
−2t + 1
−2t
−2t
14. We have
−3
c3 − 3
1
0
4
X(0) = c3
+ c4
+
=
=
.
1
1
0
1
3
c4 + 2
2
Thus, c3 = 7 and c4 =
5
2
, so
−3
1 t 5 0 2t
.
e +
X=7
e +
1
2 1
0
2
s − 4 −3
15. From sI − A =
we find
4
s+4
⎛
(sI − A)−1
1/2
3/2
⎜s − 2 − s + 2
⎜
=⎜
⎝ −1
1
+
s−2 s+2
⎞
3/4
3/4
−
s − 2 s + 2⎟
⎟
⎟
−1/2
3/2 ⎠
+
s−2 s+2
8.4 Matrix Exponential
and
3
2t
2e
eAt =
− 12 e−2t
3 2t
4e
− 34 e−2t
−e2t + e−2t − 12 e2t + 32 e−2t
.
The general solution of the system is then
3
c1
X = eAt C =
c2
−e2t + e−2t − 12 e2t + 32 e−2t
3
1
3
3
−4
−2
4
2
2t
−2t
2t
e + c1
e
e + c2
e−2t
+ c2
= c1
1
3
−1
1
−2
2
3 2t
1 −2t
1
1
1
3
e + − c1 − c 2
e
c1 + c2
=
2
4
2
4
−2
−2
3 2t
1 −2t
e + c4
e .
= c3
−2
−2
2t
2e
− 12 e−2t
3 2t
4e
− 34 e−2t
s−4
2
16. From sI − A =
we find
−1 s − 1
⎛
(sI − A)−1
1
2
⎜s − 3 − s − 2
⎜
=⎜
⎝ 1
1
−
s−3 s−2
and
eAt =
⎞
2
2
+
s − 3 s − 2⎟
⎟
⎟
−1
2 ⎠
+
s−3 s−2
−
2e3t − e2t −2e3t + 2e2t
e3t − e2t −e3t + 2e2t
.
The general solution of the system is then
c1
2e3t − e2t −2e3t + 2e2t
X=e C=
3t
2t
3t
2t
e −e
−e + 2e
c2
2 3t
−1 2t
−2 3t
2 2t
e + c1
e + c2
e + c2
e
= c1
1
−1
−1
2
2 3t
1 2t
= (c1 − c2 )
e + (−c1 + 2c2 )
e
1
1
2 3t
1 2t
e + c4
e .
= c3
1
1
At
569
570
CHAPTER 8
SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
s−5
9
17. From sI − A =
we find
−1 s + 1
⎛
1
3
+
⎜ s − 2 (s − 2)2
⎜
=⎜
⎝
1
(s − 2)2
(sI − A)−1
and
At
e
9
−
(s − 2)2
3
1
−
s − 2 (s − 2)2
⎞
⎟
⎟
⎟
⎠
−9te2t
e2t + 3te2t
.
=
te2t
e2t − 3te2t
The general solution of the system is then
2t + 3te2t
2t
−9te
c1
e
X = eAt C =
2t
2t
2t
te
e − 3te
c2
1 2t
3
0 2t
−9
2t
e + c1
te + c2
e + c2
te2t
= c1
0
1
1
−3
1 + 3t 2t
−9t
= c1
e + c2
e2t .
t
1 − 3t
18. From sI − A =
s −1
we find
2 s+2
⎛
(sI − A)−1
and
eAt =
s+1+1
⎜ (s + 1)2 + 1
⎜
=⎜
⎝
−2
(s + 1)2 + 1
⎞
1
(s + 1)2 + 1 ⎟
⎟
⎟
s+1−1 ⎠
(s + 1)2 + 1
e−t sin t
e−t cos t + e−t sin t
.
−2e−t sin t
e−t cos t − e−t sin t
The general solution of the system is then
−t cos t + e−t sin t
−t sin t
e
c1
e
X = eAt C =
−t
−t
−t
−2e sin t
e cos t − e sin t
c2
1 −t
1 −t
0 −t
1 −t
e cos t + c1
e sin t + c2
e cos t + c2
e sin t
= c1
0
−2
1
−1
cos t + sin t −t
sin t
e + c2
e−t .
= c1
−2 sin t
cos t − sin t
8.4 Matrix Exponential
19. Solving
2 − λ
1 det(A − λI) = = λ2 − 8λ + 15 = (λ − 3)(λ − 5) = 0
−3 6 − λ
we find eigenvalues λ1 = 3 and λ2 = 5. Corresponding eigenvectors are
1
K1 =
1
Then
1
and K2 =
.
3
1 1
P=
,
1 3
P
−1
=
3/2 −1/2
,
−1/2
1/2
so that
PDP
20. Solving
−1
=
and D =
3 0
,
0 5
2 1
.
−3 6
2 − λ
1 det(A − λI) = = λ2 − 4λ + 3 = (λ − 1)(λ − 3) = 0
2 − λ
1
we find eigenvalues λ1 = 1 and λ2 = 3. Corresponding eigenvectors are
−1
K1 =
1
Then
−1 1
P=
,
1 1
P−1
1
and K2 =
.
1
−1/2 1/2
=
,
1/2 1/2
so that
PDP
−1
and D =
1 0
,
0 3
2 1
=
.
1 2
21. From equation (3) in the text
etA = etPDP
−1
1 2
1
t (PDP−1 )2 + t3 (PDP−1 )3 + · · ·
2!
3!
1
1
2
3
= P I + tD + (tD) + (tD) + · · · P−1 = PetD P−1 .
2!
3!
= I + t(PDP−1 ) +
571
572
CHAPTER 8
SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
22. From Equation (3) in the text
eDt
⎛
⎞
⎛ 2
⎞ ⎛
1 0 ··· 0
0
λ1
λ1 0 · · ·
⎜0 1 · · · 0⎟ ⎜ 0 λ · · ·
⎟
⎜
0⎟ 1 ⎜ 0
2
⎜
⎟ ⎜
2⎜
⎟
⎟+⎜
=⎜
..⎟ ⎜ ..
.. . .
..⎟ + 2! t ⎜ ..
⎜ .. .. . .
. .⎠ ⎝ .
.
⎝. .
⎝ .
.
.⎠
0 0 ··· 1
0 0 · · · λn
0
⎛ 3
λ1
⎜ 0
1 ⎜
+ t3 ⎜
.
3! ⎜
⎝ ..
0
⎛
0
1 + λ1 t + 2!1 (λ1 t)2 + · · ·
⎜
1
0
1 + λ2 t + 2! (λ2 t)2 + · · ·
⎜
⎜
=⎜
..
..
⎝
.
.
0
0
⎞
⎛ λ1 t
0 ···
0
e
⎜ 0 eλ2 t · · ·
0⎟
⎟
⎜
⎟
⎜
=⎜ .
.
.
.. . . .
..⎟
⎠
⎝ ..
0
0
λ22
..
.
0
···
···
..
.
0
λ32
..
.
0
···
···
..
.
⎞
0
0⎟
⎟
⎟
..⎟
.⎠
· · · λ2n
⎞
0
0⎟
⎟
⎟
..⎟ + · · ·
.⎠
· · · λ3n
⎞
0
⎟
0
⎟
⎟
..
⎟
⎠
.
1
2
· · · 1 + λn t + 2! (λn t) + · · ·
···
···
..
.
0 · · · eλn t
23. From Problems 19, 21, and 22 and Equation (1) in the text
At
Dt
X = e C = Pe
P
−1
3
0
1 1
e3t
2
C=
0 e5t
1 3
−1
2
3
=
c1
1
c2
− 12
3t
2e
− 12 e5t
3 3t
2e
− 32 e5t
2
− 12 e3t + 12 e5t
c1
.
1 3t
3 5t
c2
−2e + 2e
24. From Problems 20, 21, and 22 and Equation (1) in the text
At
Dt
X = e C = Pe
P
−1
1
−2
0
−1 1
et
C=
3t
1
0 e
1 1
=
2
1 t
2e
1 c1
2
1
2
c2
c1
.
1 t
1 3t
c2
2e + 2e
+ 12 e3t − 12 et + 12 e3t
− 12 et + 12 e3t
25. If det(sI − A) = 0, then s is an eigenvalue of A. Thus sI − A has an inverse if s is not an
eigenvalue of A. For the purposes of the discussion in this section, we take s to be larger
than the largest eigenvalue of A. Under this condition sI − A has an inverse.
8.4 Matrix Exponential
26. Since A3 = 0, A is nilpotent. Since
eAt = I + At + A2
t2
tk
+ · · · + Ak
+ ··· ,
2!
k!
if A is nilpotent and Am = 0, then Ak = 0 for k ≥ m and
eAt = I + At + A2
t2
tm−1
+ · · · + Am−1
.
2!
(m − 1)!
In this problem A3 = 0, so
eAt
⎞
⎞
⎛
⎞ ⎛
⎛
−1
0
1
−1
1
1
1
0
0
t2 ⎜
⎟ t2
⎟
⎜
⎟ ⎜
= ⎝0 1 0⎠ + ⎝−1 0 1⎠ t + ⎝ 0 0 0⎠
= I + At + A2
2
2
−1 0 1
−1 1 1
0 0 1
⎞
⎛
t + t2 /2
1 − t − t2 /2 t
⎟
⎜
−t
1
t
=⎝
⎠
2
2
−t − t /2 t 1 + t + t /2
and the solution of X = AX is
⎞
⎛ ⎞ ⎛
c1
c1 (1 − t − t2 /2) + c2 t + c3 (t + t2 /2)
⎟
⎜ ⎟ ⎜
−c1 t + c2 + c3 t
X(t) = eAt C = eAt ⎝c2 ⎠ = ⎝
⎠.
c1 (−t − t2 /2) + c2 t + c3 (1 + t + t2 /2)
c3
27. (a) The following commands can be used in Mathematica:
A = {{4, 2},{3, 3}};
c = {c1, c2};
m = MatrixExp[A t];
sol = Expand[m.c]
Collect[sol, {c1, c2}]//MatrixForm
The output gives
2
+ c2 − et +
x(t) = c1
5
3
3 t
3
e +
y(t) = c1 − et + e6t + c2
5
5
5
2 t 3 6t
e + e
5
5
2 6t
e
5
2 6t
e
.
5
The eigenvalues are 1 and 6 with corresponding eigenvectors
1
−2
,
and
1
3
so the solution of the system is
−2 t
1 6t
X(t) = b1
e + b2
e
3
1
573
574
CHAPTER 8
SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
or
x(t) = −2b1 et + b2 e6t
y(t) = 3b1 et + b2 e6t .
If we replace b1 with − 15 c1 + 15 c2 and b2 with 35 c1 + 25 c2 , we obtain the solution found
using the matrix exponential.
(b) x(t) = c1 e−2t cos t − (c1 + c2 )e−2t sin t
y(t) = c2 e−2t cos t + (2c1 + c2 )e−2t sin t
28. x(t) = c1 (3e−2t − 2e−t ) + c3 (−6e−2t + 6e−t )
y(t) = c2 (4e−2t − 3e−t ) + c4 (4e−2t − 4e−t )
z(t) = c1 (e−2t − e−t ) + c3 (−2e−2t + 3e−t )
w(t) = c2 (−3e−2t + 3e−t ) + c4 (−3e−2t + 4e−t )
Chapter 8 in Review
4
1. If X = k
, then X = 0 and
5
4
8
24
8
0
−
=k
−
=
.
5
1
3
1
0
1
4
k
2 −1
We see that k =
1
3
.
2. Solving for c1 and c2 we find c1 = − 34 and c2 =
3. Since
1
4
.
⎛ ⎞
⎞⎛ ⎞ ⎛ ⎞
3
12
3
4
6
6
⎜ ⎟
⎟⎜ ⎟ ⎜ ⎟
⎜
3
2⎠ ⎝ 1⎠ = ⎝ 4⎠ = 4 ⎝ 1⎠ ,
⎝ 1
−1
−4
−1
−1 −4 −3
⎛
we see that λ = 4 is an eigenvalue with eigenvector K3 . The corresponding solution is
X3 = K3 e4t .
1
. The general
4. The other eigenvalue is λ2 = 1 − 2i with corresponding eigenvector K2 =
−i
solution is
cos 2t t
sin 2t t
X(t) = c1
e + c2
e.
− sin 2t
cos 2t
Chapter 8 in Review
1
. A solution to (A − λI)P = K is
−1
5. We have det(A − λI) = (λ − 1)2 = 0 and K =
0
P=
so that
1
1 t
1
0 t
t
e + c2
te +
e .
X = c1
−1
−1
1
6. We have det(A − λI) = (λ + 6)(λ + 2) = 0 so that
1 −6t
1 −2t
e
e .
X = c1
+ c2
−1
1
1
and
7. We have det(A − λI) = λ2 − 2λ + 5 = 0. For λ = 1 + 2i we obtain K1 =
i
1 (1+2i)t
cos 2t t
sin 2t t
e
=
e +i
e.
X1 =
i
− sin 2t
cos 2t
Then
X = c1
cos 2t t
sin 2t t
e + c2
e.
− sin 2t
cos 2t
8. We have det(A − λI) = λ2 − 2λ + 2 = 0. For λ = 1 + i we obtain K1 =
3−i
and
2
3 − i (1+i)t
3 cos t + sin t t
− cos t + 3 sin t t
e
=
e +i
e.
X1 =
2
2 cos t
2 sin t
Then
3 cos t + sin t t
− cos t + 3 sin t t
e + c2
e.
X = c1
2 cos t
2 sin t
9. We have det(A − λI) = −(λ − 2)(λ − 4)(λ + 3) = 0 so that
⎞
⎛ ⎞
⎛ ⎞
⎛
−2
0
7
⎟
⎜ ⎟
⎜ ⎟
⎜
X = c1 ⎝ 3⎠ e2t + c2 ⎝1⎠ e4t + c3 ⎝ 12⎠ e−3t .
1
1
−16
√
10. We have det(A−λI) = −(λ+2)(λ2 −2λ+3) = 0. The eigenvalues are λ1 = −2, λ2 = 1+ 2i,
√
and λ2 = 1 − 2i, with eigenvectors
⎛
⎛
⎞
⎞
⎛ ⎞
1
1
−7
⎜1√ ⎟
⎜ 1√ ⎟
⎜ ⎟
⎜
⎟
⎟
K1 = ⎝ 5⎠ , K2 = ⎜
⎝ 2 2 i⎠ , and K3 = ⎝− 2 2 i⎠ .
4
1
1
575
576
CHAPTER 8
SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
Thus
⎛
⎤
⎡⎛ ⎞
⎞
⎞
0
−7
1
√
√ ⎥ t
⎜1√ ⎟
⎢⎜ ⎟
⎜ ⎟
⎜
⎥
⎟
X = c1 ⎝ 5⎠ e−2t + c2 ⎢
⎣⎝0⎠ cos 2 t − ⎝ 2 2⎠ sin 2 t⎦ e
4
1
0
⎛
⎡⎛
⎤
⎞
⎛ ⎞
0
1
√ ⎥ t
√
⎢⎜ 1 √ ⎟
⎟
⎜ 2⎟ cos 2t + ⎜
0
sin
+ c3 ⎢
2 t⎥
⎠
⎝
⎣⎝ 2 ⎠
⎦e
1
0
⎞
⎞
⎛
⎛
√
√
⎛ ⎞
cos 2t
sin 2 t
−7
⎜ 1√
⎜1√
√ ⎟ t
√ ⎟ t
⎜ ⎟
⎟
⎟
⎜
= c1 ⎝ 5⎠ e−2t + c2 ⎜
⎝− 2 2 sin 2t⎠ e + c3 ⎝ 2 2 cos 2 t⎠ e .
√
√
4
cos 2 t
sin 2 t
1 2t
4 4t
e + c2
e .
X c = c1
0
1
11. We have
Then
Φ=
and
ˆ
U=
−1
Φ
e2t 4e4t
0
e4t
,
Φ−1
e−2t −4e−2t
,
=
0
e−4t
ˆ −2t
2e
15e−2t + 32te−2t
− 64te−2t
F dt =
dt =
,
16te−4t
−e−4t − 4te−4t
11 + 16t
.
Xp = ΦU =
−1 − 4t
so that
The solution is
1 2t
4 4t
11 + 16t
e + c2
e +
.
X = X c + X p = c1
0
1
−1 − 4t
12. We have
X c = c1
2 cos t
2
sin
t
et + c2
et .
− sin t
cos t
Φ=
and
ˆ
U=
2 cos t 2 sin t t
e,
− sin t cos t
1
Then
Φ−1 =
2 cos t − sin t
1
2
sin t
cos t
e−t ,
ˆ cos
t
−
sec
t
sin
t
−
ln
|
sec
t
+
tan
t|
dt =
,
Φ−1 F dt =
sin t
− cos t
Chapter 8 in Review
so that
Xp = ΦU =
−2 cos t ln | sec t + tan t|
et .
−1 + sin t ln | sec t + tan t|
The solution is
−2
cos
t
ln
|
sec
t
+
tan
t|
2 cos t
2
sin
t
et .
et + c2
et +
−1 + sin t ln | sec t + tan t|
− sin t
cos t
X = X c + X p = c1
13. We have
X c = c1
Then
Φ=
cos t + sin t
2 cos t
cos t + sin t sin t − cos t
,
2 cos t
2 sin t
+ c2
sin t − cos t
.
2 sin t
Φ−1 =
sin t
− cos t
1
2
1
2
cos t − 12 sin t
cos t + 12 sin t
,
and
sin t − 12 cos t + 12 csc t
dt
U = Φ F dt =
sin t − 12 cos t + 12 csc t
1
− 2 cos t − 12 sin t + 12 ln | csc t − cot t|
,
=
1
1
1
cos
t
−
sin
t
+
ln
|
csc
t
−
cot
t|
2
2
2
ˆ
ˆ 1
2
− 12
−1
so that
Xp = ΦU =
−1
−1
sin t
ln | csc t − cot t|.
sin t + cos t
+
The solution is
X = Xc + Xp
cos t + sin t
sin t − cos t
−1
sin t
+ c2
+
+
ln | csc t − cot t|.
= c1
2 cos t
2 sin t
−1
sin t + cos t
14. We have
X c = c1
Then
Φ=
1 2t
1
1 2t
e + c2
te2t +
e .
−1
−1
0
e2t te2t + e2t
−e2t
−te2t
and
ˆ
U=
,
Φ−1
−te−2t −te−2t − e−2t
,
=
e−2t
e−2t
ˆ 1 2
t
−
t
t
−
1
,
Φ−1 F dt =
dt = 2
−t
−1
577
578
CHAPTER 8
SYSTEMS OF LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
so that
Xp = ΦU =
The solution is
X = X c + X p = c1
− 12
1
2
−2
t2 e2t +
te2t .
1
1
−2
1 2t
1 + t 2t
−2
2 2t
e + c2
e +
t e +
te2t .
1
−1
−t
1
2
15. (a) Letting
⎛ ⎞
k1
⎜ ⎟
K = ⎝k2 ⎠
k3
we note that (A − 2I)K = 0 implies that 3k1 + 3k2 + 3k3 = 0, so k1 = −(k2 + k3 ).
Choosing k2 = 0, k3 = 1 and then k2 = 1, k3 = 0 we get
⎛ ⎞
⎛ ⎞
−1
−1
⎜ ⎟
⎜ ⎟
K1 = ⎝ 0⎠ and K2 = ⎝ 1⎠ ,
1
0
respectively. Thus,
⎛ ⎞
−1
⎜ ⎟ 2t
X1 = ⎝ 0⎠ e
1
⎛ ⎞
−1
⎜ ⎟ 2t
and X2 = ⎝ 1⎠ e
0
are two solutions.
(b) From det(A − λI) = λ2 (3 − λ) = 0 we see that λ1 = 3, and λ2 = 0 is an eigenvalue of
multiplicity two. Letting
⎛ ⎞
k1
⎜ ⎟
K = ⎝k2 ⎠ ,
k3
as in part (A), we note that (A − 0I)K = AK = 0 implies that k1 + k2 + k3 = 0, so
k1 = −(k2 + k3 ). Choosing k2 = 0, k3 = 1, and then k2 = 1, k3 = 0 we get
⎛ ⎞
⎛ ⎞
−1
−1
⎜ ⎟
⎜ ⎟
K2 = ⎝ 0⎠ and K3 = ⎝ 1⎠ ,
1
0
respectively. Since an eigenvector corresponding to λ1 = 3 is
⎛ ⎞
1
⎜ ⎟
K1 = ⎝1⎠ ,
1
the general solution of the system is
⎛ ⎞
⎛ ⎞
⎛ ⎞
1
−1
−1
⎜ ⎟ 3t
⎜ ⎟
⎜ ⎟
X = c1 ⎝1⎠ e + c2 ⎝ 0⎠ + c3 ⎝ 1⎠ .
1
1
0
Chapter 8 in Review
c1 t
e we have X = X = IX.
16. For X =
c2
579
Chapter 9
Numerical Solutions of Ordinary Differential Equations
9.1
1. x
Euler Methods and Error Analysis
h = 0.1
xn
1.00
1.10
1.20
1.30
1.40
1.50
3. x
h = 0.1
xn
0.00
0.10
0.20
0.30
0.40
0.50
yn
5.0000
3.9900
3.2546
2.7236
2.3451
2.0801
yn
0.0000
0.1005
0.2030
0.3098
0.4234
0.5470
h = 0.05
xn
1.00
1.05
1.10
1.15
1.20
1.25
1.30
1.35
1.40
1.45
1.50
h = 0.05
xn
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
yn
2. x
5.0000
4.4475
3.9763
3.5751
3.2342
2.9452
2.7009
2.4952
2.3226
2.1786
2.0592
yn
h = 0.1
xn
0.00
0.10
0.20
0.30
0.40
0.50
4. x
0.0000
0.0501
0.1004
0.1512
0.2028
0.2554
0.3095
0.3652
0.4230
0.4832
0.5465
h = 0.1
xn
0.00
0.10
0.20
0.30
0.40
0.50
580
yn
2.0000
1.6600
1.4172
1.2541
1.1564
1.1122
yn
1.0000
1.1110
1.2515
1.4361
1.6880
2.0488
h = 0.05
xn
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
h = 0.05
xn
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
yn
2.0000
1.8150
1.6571
1.5237
1.4124
1.3212
1.2482
1.1916
1.1499
1.1217
1.1056
yn
1.0000
1.0526
1.1113
1.1775
1.2526
1.3388
1.4387
1.5556
1.6939
1.8598
2.0619
9.1
5. x
h = 0.1
xn
0.00
0.10
0.20
0.30
0.40
0.50
7. x
h = 0.1
xn
0.00
0.10
0.20
0.30
0.40
0.50
9. x
h = 0.1
xn
1.00
1.10
1.20
1.30
1.40
1.50
yn
0.0000
0.0952
0.1822
0.2622
0.3363
0.4053
yn
0.5000
0.5215
0.5362
0.5449
0.5490
0.5503
yn
1.0000
1.0095
1.0404
1.0967
1.1866
1.3260
h = 0.05
xn
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
h = 0.05
xn
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
h = 0.05
xn
1.00
1.05
1.10
1.15
1.20
1.25
1.30
1.35
1.40
1.45
1.50
yn
6. x
0.0000
0.0488
0.0953
0.1397
0.1823
0.2231
0.2623
0.3001
0.3364
0.3715
0.4054
yn
0.00
0.10
0.20
0.30
0.40
0.50
8. x
0.5000
0.5116
0.5214
0.5294
0.5359
0.5408
0.5444
0.5469
0.5484
0.5492
0.5495
yn
1.0000
1.0024
1.0100
1.0228
1.0414
1.0663
1.0984
1.1389
1.1895
1.2526
1.3315
h = 0.1
xn
h = 0.1
xn
0.00
0.10
0.20
0.30
0.40
0.50
10. x
h = 0.1
xn
0.00
0.10
0.20
0.30
0.40
0.50
Euler Methods and Error Analysis
yn
0.0000
0.0050
0.0200
0.0451
0.0805
0.1266
yn
1.0000
1.1079
1.2337
1.3806
1.5529
1.7557
yn
0.5000
0.5250
0.5498
0.5744
0.5986
0.6224
h = 0.05
xn
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
h = 0.05
xn
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
h = 0.05
xn
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
yn
0.0000
0.0013
0.0050
0.0113
0.0200
0.0313
0.0451
0.0615
0.0805
0.1022
0.1266
yn
1.0000
1.0519
1.1079
1.1684
1.2337
1.3043
1.3807
1.4634
1.5530
1.6503
1.7560
yn
0.5000
0.5125
0.5250
0.5374
0.5498
0.5622
0.5744
0.5866
0.5987
0.6106
0.6224
581
582
CHAPTER 9
NUMERICAL SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS
11. To obtain the analytic solution use the substitution u = x + y − 1. The resulting differential
equation in u(x) will be separable.
h = 0.05
h = 0.1
xn
yn
Actual
Value
xn
yn
Actual
Value
0.00
1.10
0.20
0.30
0.40
0.50
2.0000
2.1220
2.3049
2.5858
3.0378
3.8254
2.0000
2.1230
2.3085
2.5958
3.0650
3.9082
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
2.0000
2.0553
2.1228
2.2056
2.3075
2.4342
2.5931
2.7953
3.0574
3.4057
3.8840
2.0000
2.0554
2.1230
2.2061
2.3085
2.4358
2.5958
2.7997
3.0650
3.4189
3.9082
12. (a) x
(b) x
y
20
15
10
5
1.1
1.2
1.3
1.4
xn
Euler
Imp. Euler
1.00
1.10
1.20
1.30
1.40
1.0000
1.2000
1.4938
1.9711
2.9060
1.0000
1.2000
1.4938
1.9711
2.9060
x
13. (a) Using Euler’s method we obtain y(0.1) ≈ y1 = 1.2.
(b) Using y = 4e2x we see that the local truncation error is
y (c)
h2
(0.1)2
= 4e2c
= 0.02e2c .
2
2
Since e2x is an increasing function, e2c ≤ e2(0.1) = e0.2 for 0 ≤ c ≤ 0.1. Thus an upper
bound for the local truncation error is 0.02e0.2 = 0.0244.
(c) Since y(0.1) = e0.2 = 1.2214, the actual error is y(0.1) − y1 = 0.0214, which is less than
0.0244.
(d) Using Euler’s method with h = 0.05 we obtain y(0.1) ≈ y2 = 1.21.
(e) The error in (d) is 1.2214 − 1.21 = 0.0114. With global truncation error O(h), when the
step size is halved we expect the error for h = 0.05 to be one-half the error when h = 0.1.
Comparing 0.0114 with 0.0214 we see that this is the case.
9.1
Euler Methods and Error Analysis
14. (a) Using the improved Euler’s method we obtain y(0.1) ≈ y1 = 1.22.
(b) Using y = 8e2x we see that the local truncation error is
y (c)
h3
(0.1)3
= 8e2c
= 0.001333e2c .
6
6
Since e2x is an increasing function, e2c ≤ e2(0.1) = e0.2 for 0 ≤ c ≤ 0.1. Thus an upper
bound for the local truncation error is 0.001333e0.2 = 0.001628.
(c) Since y(0.1) = e0.2 = 1.221403, the actual error is y(0.1) − y1 = 0.001403 which is less
than 0.001628.
(d) Using the improved Euler’s method with h = 0.05 we obtain y(0.1) ≈ y2 = 1.221025.
(e) The error in (d) is 1.221403 − 1.221025 = 0.000378. With global truncation error O(h2 ),
when the step size is halved we expect the error for h = 0.05 to be one-fourth the error
for h = 0.1. Comparing 0.000378 with 0.001403 we see that this is the case.
15. (a) Using Euler’s method we obtain y(0.1) ≈ y1 = 0.8.
(b) Using y = 5e−2x we see that the local truncation error is
5e−2c
(0.1)2
= 0.025e−2c .
2
Since e−2x is a decreasing function, e−2c ≤ e0 = 1 for 0 ≤ c ≤ 0.1. Thus an upper bound
for the local truncation error is 0.025(1) = 0.025.
(c) Since y(0.1) = 0.8234, the actual error is y(0.1) − y1 = 0.0234, which is less than 0.025.
(d) Using Euler’s method with h = 0.05 we obtain y(0.1) ≈ y2 = 0.8125.
(e) The error in (d) is 0.8234 − 0.8125 = 0.0109. With global truncation error O(h), when
the step size is halved we expect the error for h = 0.05 to be one-half the error when
h = 0.1. Comparing 0.0109 with 0.0234 we see that this is the case.
16. (a) Using the improved Euler’s method we obtain y(0.1) ≈ y1 = 0.825.
(b) Using y = −10e−2x we see that the local truncation error is
10e−2c
(0.1)3
= 0.001667e−2c .
6
Since e−2x is a decreasing function, e−2c ≤ e0 = 1 for 0 ≤ c ≤ 0.1. Thus an upper bound
for the local truncation error is 0.001667(1) = 0.001667.
583
584
CHAPTER 9
NUMERICAL SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS
(c) Since y(0.1) = 0.823413, the actual error is y(0.1) − y1 = 0.001587, which is less than
0.001667.
(d) Using the improved Euler’s method with h = 0.05 we obtain y(0.1) ≈ y2 = 0.823781.
(e) The error in (d) is |0.823413 − 0.8237181| = 0.000305. With global truncation error
O(h2 ), when the step size is halved we expect the error for h = 0.05 to be one-fourth
the error when h = 0.1. Comparing 0.000305 with 0.001587 we see that this is the case.
17. (a) Using y = 38e−3(x−1) we see that the local truncation error is
y (c)
h2
h2
= 38e−3(c−1)
= 19h2 e−3(c−1) .
2
2
(b) Since e−3(x−1) is a decreasing function for 1 ≤ x ≤ 1.5, e−3(c−1) ≤ e−3(1−1) = 1 for
1 ≤ c ≤ 1.5 and
h2
≤ 19(0.1)2 (1) = 0.19.
y (c)
2
(c) Using Euler’s method with h = 0.1 we obtain y(1.5) ≈ 1.8207. With h = 0.05 we obtain
y(1.5) ≈ 1.9424.
(d) Since y(1.5) = 2.0532, the error for h = 0.1 is E0.1 = 0.2325, while the error for h = 0.05
is E0.05 = 0.1109. With global truncation error O(h) we expect E0.1 /E0.05 ≈ 2. We
actually have E0.1 /E0.05 = 2.10.
18. (a) Using y = −114e−3(x−1) we see that the local truncation error is
3
3
y (c) h = 114e−3(x−1) h = 19h3 e−3(c−1) .
6
6
(b) Since e−3(x−1) is a decreasing function for 1 ≤ x ≤ 1.5, e−3(c−1) ≤ e−3(1−1) = 1 for
1 ≤ c ≤ 1.5 and
3
h
y (c) ≤ 19(0.1)3 (1) = 0.019.
6
(c) Using the improved Euler’s method with h = 0.1 we obtain y(1.5) ≈ 2.080108. With
h = 0.05 we obtain y(1.5) ≈ 2.059166.
(d) Since y(1.5) = 2.053216, the error for h = 0.1 is E0.1 = 0.026892, while the error
for h = 0.05 is E0.05 = 0.005950. With global truncation error O(h2 ) we expect
E0.1 /E0.05 ≈ 4. We actually have E0.1 /E0.05 = 4.52.
9.2 Runge–Kutta Methods
19. (a) Using y = −1/(x + 1)2 we see that the local truncation error is
2
h2
1
y (c) h =
.
2
(c + 1)2 2
(b) Since 1/(x + 1)2 is a decreasing function for 0 ≤ x ≤ 0.5, 1/(c + 1)2 ≤ 1/(0 + 1)2 = 1 for
0 ≤ c ≤ 0.5 and
2
2
y (c) h ≤ (1) (0.1) = 0.005.
2
2
(c) Using Euler’s method with h = 0.1 we obtain y(0.5) ≈ 0.4198. With h = 0.05 we obtain
y(0.5) ≈ 0.4124.
(d) Since y(0.5) = 0.4055, the error for h = 0.1 is E0.1 = 0.0143, while the error for h = 0.05
is E0.05 = 0.0069. With global truncation error O(h) we expect E0.1 /E0.05 ≈ 2. We
actually have E0.1 /E0.05 = 2.06.
20. (a) Using y = 2/(x + 1)3 we see that the local truncation error is
y (c)
h3
h3
1
=
.
3
6
(c + 1) 3
(b) Since 1/(x + 1)3 is a decreasing function for 0 ≤ x ≤ 0.5, 1/(c + 1)3 ≤ 1/(0 + 1)3 = 1 for
0 ≤ c ≤ 0.5 and
h3
(0.1)3
y (c)
≤ (1)
= 0.000333.
6
3
(c) Using the improved Euler’s method with h = 0.1 we obtain y(0.5) ≈ 0.405281. With
h = 0.05 we obtain y(0.5) ≈ 0.405419.
(d) Since y(0.5) = 0.405465, the error for h = 0.1 is E0.1 = 0.000184, while the error for
h = 0.05 is E0.05 = 0.000046. With global truncation error O(h2 ) we expect
E0.1 /E0.05 ≈ 4. We actually have E0.1 /E0.05 = 3.98.
∗
depends on yn and is used to determine yn+1 , all of the yn∗ cannot be computed
21. Because yn+1
at one time independently of the corresponding yn values. For example, the computation of
y4∗ involves the value of y3 .
9.2
1. x
Runge–Kutta Methods
xn
yn
Actual
Value
0.00
1.10
0.20
2.0000
2.1230
2.3085
2.0000
2.1230
2.3085
0.30
0.40
0.50
2.5958
3.0649
3.9078
2.5958
3.0650
3.9082
585
586
CHAPTER 9
NUMERICAL SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS
2. In this problem we use h = 0.1. Substituting w2 =
into the equations in (4) in the text, we obtain
w 1 = 1 − w2 =
3
4
1
1
1
2
2
α=
= and β =
= .
4
2w2
3
2w2
3
The resulting second-order Runge-Kutta method is
1
3
h
yn+1 = yn + h
k1 + k2 = yn + (k1 + 3k2 )
4
4
4
xn
Second–Order
Runge–Kutta
Improved
Euler
0.00
1.10
2.0000
2.1213
2.0000
2.1220
0.20
0.30
0.40
2.3030
2.5814
3.0277
2.3049
2.5858
3.0378
0.50
3.8002
3.8254
where
k1 = f (xn , yn ) and k2 = f
2
2
xn + h, yn + hk1 .
3
3
The table compares the values obtained using this second-order Runge-Kutta method with
the values obtained using the improved Euler’s method.
3. x
5. x
7. x
9. x
4. x
xn
yn
5.0000
3.9724
3.2284
2.6945
0.00
0.10
0.20
2.0000
1.6562
1.4110
0.30
1.2465
1.40
2.3163
1.50
2.0533
0.40
0.50
1.1480
1.1037
xn
yn
xn
yn
0.00
0.0000
0.00
1.0000
0.10
0.20
0.1003
0.2027
0.10
0.20
1.1115
1.2530
0.30
0.40
0.50
0.3093
0.4228
0.5463
0.30
0.40
0.50
1.4397
1.6961
2.0670
xn
yn
xn
yn
0.00
0.10
0.20
0.30
0.40
0.50
0.0000
0.0953
0.1823
0.2624
0.3365
0.4055
0.00
0.10
0.20
0.30
0.40
0.50
0.0000
0.0050
0.0200
0.0451
0.0805
0.1266
xn
yn
xn
yn
0.00
0.10
0.20
0.30
0.40
0.50
0.5000
0.5213
0.5358
0.5443
0.5482
0.5493
0.00
0.10
0.20
0.30
0.40
0.50
1.0000
1.1079
1.2337
1.3807
1.5531
1.7561
xn
yn
1.00
1.10
1.20
1.30
6. x
8. x
10. x
9.2 Runge–Kutta Methods
11. x
12. x
xn
yn
1.0000
1.0101
0.00
0.10
0.5000
0.5250
1.20
1.30
1.40
1.0417
1.0989
1.1905
0.20
0.30
0.40
0.5498
0.5744
0.5987
1.50
1.3333
0.50
0.6225
xn
yn
1.00
1.10
13. (a) Write the equation in the form
dv
k
= g − v 2 = f (t, v).
dt
m
tn
vn
0.0
1.0
0.0
24.948
33.086
34.672
34.945
34.991
2.0
3.0
(b) x
4.0
5.0
(c) Separating variables and using partial fractions we have
⎞
⎛
1
1
1 ⎝
⎠ dv = dt
+
√
2 g √g − k v √ g + k v
m
m
and
1
√
k
2 g m
√
ln g +
√
k v − ln g −
m k v
m = t + c.
Since v(0) = 0 we find c = 0. Solving for v we obtain
4 kg
t
m
mg e
−1
v(t) =
4 kg
k
t
m
e
+1
and v(5) ≈ 35. Alternatively, the solution can be expressed as
v(t) =
14. (a) x
mg
tanh
k
4 kg
t.
m
t (days)
1
2
3
4
5
A (observed)
2.78
13.53
36.30
47.50
49.40
A (approximated)
1.93
12.50
36.46
47.23
49.00
587
588
CHAPTER 9
NUMERICAL SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS
A(t)
50
(b) From the graph we estimate A(1) ≈ 1.68,
A(2) ≈ 13.2, A(3) ≈ 36.8, A(4) ≈ 46.9, and
A(5) ≈ 48.9.
40
30
20
10
0
1
2
(c) Let α = 2.128 and β = 0.0432. Separating variables we obtain
dA
= dt
A(α − βA)
β
1 1
+
dA = dt
α A α − βA
1
[ln A − ln (α − βA)] = t + c
α
ln
A
= α(t + c)
α − βA
A
= eα(t+c)
α − βA
A = αeα(t+c) − βAeα(t+c)
1 + βeα(t+c) A = αeα(t+c) .
Thus
A(t) =
α
α
αeα(t+c)
=
=
.
−αc
α(t+c)
−α(t+c)
β
+
e
e−αt
1 + βe
β+e
From A(0) = 0.24 we obtain
0.24 =
α
β + e−αc
so that e−αc = α/0.24 − β ≈ 8.8235 and
A(t) ≈
2.128
0.0432 + 8.8235e−2.128t .
+
t (days)
1
2
3
4
5
A (observed)
2.78
13.53
36.30
47.50
49.40
A (actual)
1.95
12.64
36.63
49.02
49.02
3
4
5
t
9.2 Runge–Kutta Methods
15. (a) x x
n
h = 0.05
h = 0.1
1.00
1.05
1.10
1.15
1.0000
1.1112
1.2511
1.4348
1.0000
1.20
1.25
1.30
1.35
1.6934
2.1047
2.9560
7.8981
1.6934
1.40
1.0608 ×1015
903.0282
1.2511
(b) x
589
y
20
15
10
5
2.9425
1.1
1.2
1.3
1.4
x
16. (a) Using the RK4 method we obtain y(0.1) ≈ y1 = 1.2214.
(b) Using y (5) (x) = 32e2x we see that the local truncation error is
y (5) (c)
h5
(0.1)5
= 32e2c
= 0.000002667e2c .
120
120
Since e2x is an increasing function, e2c ≤ e2(0.1) = e0.2 for 0 ≤ c ≤ 0.1. Thus an upper
bound for the local truncation error is 0.000002667e0.2 = 0.000003257.
(c) Since y(0.1) = e0.2 = 1.221402758, the actual error is y(0.1) − y1 = 0.000002758 which is
less than 0.000003257.
(d) Using the RK4 formula with h = 0.05 we obtain y(0.1) ≈ y2 = 1.221402571.
(e) The error in (d) is 1.221402758 − 1.221402571 = 0.000000187. With global truncation
error O(h4 ), when the step size is halved we expect the error for h = 0.05 to be onesixteenth the error for h = 0.1. Comparing 0.000000187 with 0.000002758 we see that
this is the case.
17. (a) Using the RK4 method we obtain y(0.1) ≈ y1 = 0.823416667.
(b) Using y (5) (x) = −40e−2x we see that the local truncation error is
40e−2c
(0.1)5
= 0.000003333.
120
Since e−2x is a decreasing function, e−2c ≤ e0 = 1 for 0 ≤ c ≤ 0.1. Thus an upper bound
for the local truncation error is 0.000003333(1) = 0.000003333.
(c) Since y(0.1) = 0.823413441, the actual error is |y(0.1) − y1 | = 0.000003225, which is less
than 0.000003333.
(d) Using the RK4 method with h = 0.05 we obtain y(0.1) ≈ y2 = 0.823413627.
590
CHAPTER 9
NUMERICAL SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS
(e) The error in (d) is |0.823413441 − 0.823413627| = 0.000000185. With global truncation
error O(h4 ), when the step size is halved we expect the error for h = 0.05 to be onesixteenth the error when h = 0.1. Comparing 0.000000185 with 0.000003225 we see that
this is the case.
18. (a) Using y (5) = −1026e−3(x−1) we see that the local truncation error is
5 (5)
y (c) h = 8.55h5 e−3(c−1) .
120 (b) Since e−3(x−1) is a decreasing function for 1 ≤ x ≤ 1.5, e−3(c−1) ≤ e−3(1−1) = 1 for
1 ≤ c ≤ 1.5 and
y (5) (c)
h5
≤ 8.55(0.1)5 (1) = 0.0000855.
120
(c) Using the RK4 method with h = 0.1 we obtain y(1.5) ≈ 2.053338827. With h = 0.05 we
obtain y(1.5) ≈ 2.053222989.
19. (a) Using y (5) = 24/(x + 1)5 we see that the local truncation error is
y (5) (c)
1
h5
h5
=
.
120
(c + 1)5 5
(b) Since 1/(x + 1)5 is a decreasing function for 0 ≤ x ≤ 0.5, 1/(c + 1)5 ≤ 1/(0 + 1)5 = 1 for
0 ≤ c ≤ 0.5 and
y (5) (c)
h5
(0.1)5
≤ (1)
= 0.000002.
5
5
(c) Using the RK4 method with h = 0.1 we obtain y(0.5) ≈ 0.405465168. With h = 0.05 we
obtain y(0.5) ≈ 0.405465111.
20. Each step of Euler’s method requires only 1 function evaluation, while each step of the
improved Euler’s method requires 2 function evaluations – once at (xn , yn ) and again at
∗ ). The second-order Runge-Kutta methods require 2 function evaluations per step,
(xn+1 , yn+1
while the RK4 method requires 4 function evaluations per step. To compare the methods we
approximate the solution of y = (x + y − 1)2 , y(0) = 2, at x = 0.2 using h = 0.1 for the
Runge-Kutta method, h = 0.05 for the improved Euler’s method, and h = 0.025 for Euler’s
method. For each method a total of 8 function evaluations is required. By comparing with
the exact solution we see that the RK4 method appears to still give the most accurate result.
9.2 Runge–Kutta Methods
xn
Euler
h = 0.025
Imp. Euler
h = 0.05
RK4
h = 0.1
Actual
0.000
2.0000
2.0000
2.0000
2.0000
0.025
0.050
2.0250
2.0526
2.0553
0.075
0.100
0.125
2.0830
2.1165
2.1535
2.1228
0.150
2.1943
2.2056
0.175
0.200
2.2395
2.2895
2.3075
591
2.0263
2.0554
2.1230
2.0875
2.1230
2.1624
2.2061
2.3085
2.2546
2.3085
21. (a) For y + y = 10 sin 3x an integrating factor is ex so that
y
d x
[e y] = 10ex sin 3x
dx
5
ex y = ex sin 3x − 3ex cos 3x + c
y = sin 3x − 3 cos 3x + ce−x .
2
When x = 0, y = 0, so 0 = −3 + c and c = 3. The solution is
x
y = sin 3x − 3 cos 3x + 3e−x .
−5
Using Newton’s method we find that x = 1.53235 is the only positive
root in [0, 2].
(b) Using the RK4 method with h = 0.1 we obtain the table of values shown. These values are
used to obtain an interpolating function in Mathematica. The graph of the interpolating
function is shown. Using Mathematica’s root finding capability we see that the only
positive root in [0, 2] is x = 1.53236.
xn
yn
xn
yn
0.0
0.1
0.0000
0.1440
1.0
1.1
4.2147
3.8033
0.2
0.3
0.5448
1.1409
1.2
1.3
3.1513
2.3076
0.4
0.5
0.6
1.8559
2.6049
3.3019
1.4
1.5
1.6
1.3390
0.3243
– 0.6530
0.7
3.8675
1.7
– 1.5117
0.8
0.9
1.0
4.2356
4.3593
4.2147
1.8
1.9
2.0
– 2.1809
– 2.6061
– 2.7539
y
5
2
−5
x
592
CHAPTER 9
9.3
NUMERICAL SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS
Multistep Methods
In the tables in this section “ABM” stands for Adams-Bashforth-Moulton.
1. Writing the differential equation in the form y − y = x − 1 we see that an integrating factor
´
is e− dx = e−x , so that
d −x e y = (x − 1)e−x
dx
and
y = ex (−xe−x + c) = −x + cex .
From y(0) = 1 we find c = 1, so the solution of the initial-value problem is y = −x + ex .
Actual values of the analytic solution above are compared with the approximated values in
the table.
xn
yn
Actual
0.0
0.2
0.4
0.6
0.8
1.00000000
1.02140000
1.09181796
1.22210646
1.42552788
1.00000000
1.02140276
1.09182470
1.22211880
1.42554093
init. cond.
RK4
RK4
RK4
ABM
2. The following program is written in Mathematica. It uses the Adams-Bashforth-Moulton
method to approximate the solution of the initial-value problem y = x + y − 1, y(0) = 1, on
the interval [0, 1].
Clear[f, x, y, h, a, b, y0];
f[x , y ]:= x + y - 1;
h = 0.2;
a = 0; y0 = 1; b = 1;
f[x, y]
Clear[k1, k2, k3, k4, x, y, u, v]
x = u[0] = a;
y = v[0] = y0;
n = 0;
While[x < a + 3h,
x
(* define the differential equation *)
(* set the step size *)
(* set the initial condition and the interval *)
(* display the DE *)
x
x
x
x
(* use RK4 to compute the first 3 values after y(0) *)
9.3
n = n + 1;
k1 = f[x, y];
k2 = f[x + h/2, y + h k1/2];
k3 = f[x + h/2, y + h k2/2];
k4 = f[x + h, y + h k3];
x = x + h;
y = y + (h/6)(k1 + 2k2 + 2k3 + k4);
u[n] = x;
v[n] = y];
Multistep Methods
x
While[x ≤ b,
(* use Adams-Bashforth-Moulton *)
p3 = f[u[n - 3], v[n - 3]];
x
p2 = f[u[n - 2], v[n - 2]];
x
p1 = f[u[n - 1], v[n - 1]];
x
p0 = f[u[n], v[n]];
x
pred = y + (h/24)(55p0 - 59p1 + 37p2 - 9p3);
(* predictor *)
x = x + h;
x
p4 = f[x, pred];
x
y = y + (h/24)(9p4 + 19p0 - 5p1 + p2);
(* corrector *)
n = n + 1;
x
u[n] = x;
x
v[n] = y]
x
TableForm[Prepend[Table[{u[n], v[n]}, {n, 0, (b-a)/h}], {"x(n)", "y(n)"}]];
(* display the table *)
3. The first predictor is y4∗ = 0.73318477.
xn
yn
0.0
1.00000000
0.73280000
0.2
0.4
0.6
0.8
init. cond.
0.64608032
RK4
RK4
0.65851653
0.72319464
RK4
ABM
4. The first predictor is y4∗ = 1.21092217.
xn
yn
0.0
0.2
0.4
0.6
2.00000000
1.41120000
1.14830848
1.10390600
init. cond.
RK4
RK4
RK4
0.8
1.20486982
ABM
593
594
CHAPTER 9
NUMERICAL SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS
5. The first predictor for h = 0.2 is y4∗ = 1.02343488.
xn
h = 0.2
0.0
0.00000000
init. cond.
0.2
0.3
0.20270741
RK4
0.4
0.5
0.42278899
RK4
0.6
0.68413340
0.7
0.8
0.9
1.0
h = 0.1
0.00000000
init. cond.
0.10033459
RK4
0.20270988
0.30933604
RK4
RK4
0.42279808
0.54631491
ABM
ABM
RK4
0.68416105
ABM
ABM
0.84233188
1..02971420
1..26028800
ABM
1.02969040
1.55685960
ABM
1.55762558
ABM
0.1
ABM
ABM
6. The first predictor for h = 0.2 is y4∗ = 3.34828434.
xn
h = 0.2
0.0
1.00000000
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.44139950
h = 0.1
init. cond.
RK4
1.97190167
RK4
2.60280694
RK4
3.34860927
ABM
4.22797875
ABM
1.00000000
1.21017082
1.44140511
1.69487942
1.97191536
2.27400341
2.60283209
2.96031780
3.34863769
3.77026548
4.22801028
init. cond.
RK4
RK4
RK4
ABM
ABM
ABM
ABM
ABM
ABM
ABM
9.4 Higher-Order Equations and Systems
7. The first predictor for h = 0.2 is y4∗ = 0.13618654.
xn
h = 0.2
0.0
0.00000000
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
h = 0.1
0.00262739
init. cond.
0.00000000
init. cond.
RK4
0.00033209
0.00262486
RK4
RK4
RK4
ABM
ABM
ABM
ABM
ABM
ABM
ABM
0.02005764
RK4
0.06296284
RK4
0.13598600
ABM
0.00868768
0.02004821
0.03787884
0.06294717
0.09563116
0.13596515
ABM
0.18370712
0.23841344
0.23854783
8. The first predictor for h = 0.2 is y4∗ = 2.61796154.
xn
h = 0.2
0.0
1.00000000
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
9.4
h = 0.1
init. cond.
1.00000000
init. cond.
1.23369623
RK4
1.10793839
1.23369772
1.38068454
RK4
RK4
RK4
1.55308554
RK4
1.99610329
RK4
2.62136177
ABM
3.52079042
ABM
1.55309381
1.75610064
1.99612995
2.28119129
2.62131818
3.02914333
3.52065536
ABM
ABM
ABM
ABM
ABM
ABM
ABM
Higher-Order Equations and Systems
1. The substitution y = u leads to the iteration formulas
yn+1 = yn + hun ,
un+1 = un + h(4un − 4yn ).
The initial conditions are y0 = −2 and u0 = 1. Then
y1 = y0 + 0.1u0 = −2 + 0.1(1) = −1.9
u1 = u0 + 0.1(4u0 − 4y0 ) = 1 + 0.1(4 + 8) = 2.2
y2 = y1 + 0.1u1 = −1.9 + 0.1(2.2) = −1.68.
595
596
CHAPTER 9
NUMERICAL SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS
The general solution of the differential equation is y = c1 e2x + c2 xe2x . From the initial
conditions we find c1 = −2 and c2 = 5. Thus y = −2e2x + 5xe2x and y(0.2) ≈ 1.4918.
2. The substitution y = u leads to the iteration formulas
yn+1 = yn + hun ,
un+1 = un + h
2
2
u n − 2 yn .
x
x
The initial conditions are y0 = 4 and u0 = 9. Then
y1 = y0 + 0.1u0 = 4 + 0.1(9) = 4.9
2
2
u1 = u0 + 0.1
u0 − y0 = 9 + 0.1[2(9) − 2(4)] = 10
1
1
y2 = y1 + 0.1u1 = 4.9 + 0.1(10) = 5.9.
The general solution of the Cauchy-Euler differential equation is y = c1 x + c2 x2 . From the
initial conditions we find c1 = −1 and c2 = 5. Thus y = −x + 5x2 and y(1.2) = 6.
3. The substitution y = u leads to
the system
y = u,
u = 4u − 4y.
Using formula (4), we obtain the
table shown.
4. The substitution y = u leads to
the system
y = u,
u =
2
2
u − 2 y.
x
x
Using formula (4), we obtain the
table shown.
5. The substitution y = u leads to
the system
y = u,
u = 2u − 2y + et cos t.
Using formula (4), we obtain the
table shown.
xn
h = 0.2
yn
h = 0.2
un
h = 0.1
yn
h = 0.1
un
0.0
–2.0000
1.0000
–2.0000
1.0000
–1.4928
4.4731
–1.8321
–1.4919
2.4427
4.4753
xn
h = 0.2
yn
h = 0.2
un
h = 0.1
yn
h = 0.1
un
1.0
4.0000
9.0000
4.0000
9.0000
1.1
1.2
6.0001
11.0002
4.9500
6.0000
10.0000
11.0000
h = 0.2
yn
h = 0.2
un
h = 0.1
yn
h = 0.1
un
1.0000
2.0000
1.4640
2.6594
1.0000
1.2155
1.4640
2.0000
2.3150
2.6594
0.1
0.2
xn
0.0
0.1
0.2
9.4 Higher-Order Equations and Systems
597
6. Using h = 0.1, the RK4 method for a system, and a numerical solver, we obtain
i1
i2
tn
h = 0.2
i1n
h = 0.2
i3n
7
6
7
6
0.0
0.1
0.2
0.3
0.4
0.5
0.0000
2.5000
2.8125
2.0703
0.6104
–1.5619
0.0000
3.7500
5.7813
7.4023
9.1919
11.4877
5
5
4
4
3
2
3
2
1
1
1 2 3 4 5
7.
tn
0.0
0.1
0.2
h = 0.2
xn
h = 0.2
yn
h = 0.1
xn
h = 0.1
yn
6.0000
2.0000
8.3055
3.4199
6.0000
7.0731
8.3055
2.0000
2.6524
3.4199
t
1 2 3 4 5
t
x,y
20
x(t)
15
10
y(t)
5
0.5
8.
tn
0.0
0.1
0.2
h = 0.2
xn
h = 0.2
yn
h = 0.1
xn
h = 0.1
yn
1.0000
1.0000
1.0000
1.4006
2.0845
1.0000
1.8963
3.3502
2.0785
3.3382
1
1.5
2
1
1.5
2
t
x,y
50
40
y(t)
30
20
10
x(t)
0.5
t
598
CHAPTER 9
9.
NUMERICAL SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS
h = 0.2
xn
h = 0.2
yn
h = 0.1
xn
h = 0.1
yn
0.0
0.1
–3.0000
5.0000
–3.0000
–3.4790
5.0000
4.6707
0.2
–3.9123
4.2857
–3.9123
4.2857
tn
x,y
30
25
20
x(t)
15
10
y(t)
5
5
15
10
25
20
30
t
−5
10.
tn
0.0
0.1
0.2
h = 0.2
xn
h = 0.2
yn
h = 0.1
xn
h = 0.1
yn
0.5000
0.2000
2.1589
2.3279
0.5000
1.0207
2.1904
0.2000
1.0115
2.3592
x,y
2
x(t)
1.5
1
y(t)
0.5
0.1
11. Solving for x and y we obtain the system
0.2
x,y
x = −2x + y + 5t
1
2
3
y = 2x + y − 2t
−5
tn
0.0
0.1
0.2
t
0.3
h = 0.2
xn
h = 0.2
yn
h = 0.1
xn
h = 0.1
yn
1.0000
–2.0000
0.4179
–2.1824
1.0000
0.6594
0.4173
–2.0000
–2.0476
–2.1821
x(t)
−10
−15
y(t)
−20
4t
9.5 Second-Order Boundary-Value Problems
12. Solving for x and y we obtain the system
599
x,y
1
x = y − 3t2 + 2t − 5
2
60
40
1
y = − y + 3t2 + 2t + 5.
2
y(t)
20
tn
h = 0.2
xn
h = 0.2
yn
h = 0.1
xn
0.0
3.0000
–1.0000
3.0000
–1.0000
−20
0.0933
2.4727
1.9867
–0.4527
0.0933
−40
0.1
0.2
1.9867
h = 0.1
yn
2
4
6
8
t
x(t)
−60
9.5
Second-Order Boundary-Value Problems
1. We identify P (x) = 0, Q(x) = 9, f (x) = 0, and h = (2 − 0)/4 = 0.5. Then the finite difference
equation is
yi+1 + 0.25yi + yi−1 = 0.
The solution of the corresponding linear system gives
x
y
0.0
4.0000
0.5
–5.6774
1.0
–2.5807
1.5
–6.3226
2.0
1.0000
2. We identify P (x) = 0, Q(x) = −1, f (x) = x2 , and h = (1 − 0)/4 = 0.25. Then the finite
difference equation is
yi+1 − 2.0625yi + yi−1 = 0.0625x2i .
The solution of the corresponding linear system gives
x
y
0.00
0.0000
0.25
–0.0172
0.50
–0.0316
0.75
–0.0324
1.00
0.0000
3. We identify P (x) = 2, Q(x) = 1, f (x) = 5x, and h = (1 − 0)/5 = 0.2. Then the finite
difference equation is
1.2yi+1 − 1.96yi + 0.8yi−1 = 0.04(5xi ).
The solution of the corresponding linear system gives
x
y
0.0
0.0000
0.2
–0.2259
0.4
–0.3356
0.6
–0.3308
0.8
–0.2167
1.0
0.0000
600
CHAPTER 9
NUMERICAL SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS
4. We identify P (x) = −10, Q(x) = 25, f (x) = 1, and h = (1 − 0)/5 = 0.2. Then the finite
difference equation is
−yi + 2yi−1 = 0.04.
The solution of the corresponding linear system gives
x
y
0.00
1.0000
0.2
1.9600
0.4
3.8800
0.6
7.7200
0.8
15.4000
1.0
0.0000
5. We identify P (x) = −4, Q(x) = 4, f (x) = (1 + x)e2x , and h = (1 − 0)/6 = 0.1667. Then the
finite difference equation is
0.6667yi+1 − 1.8889yi + 1.3333yi−1 = 0.2778(1 + xi )e2xi .
The solution of the corresponding linear system gives
x
y
0.0000
3.0000
0.1667
3.3751
0.3333
3.6306
0.5000
3.6448
0.6667
3.2355
0.8333
2.1411
1.0000
0.0000
√
6. We identify P (x) = 5, Q(x) = 0, f (x) = 4 x , and h = (2 − 1)/6 = 0.1667. Then the finite
difference equation is
√
1.4167yi+1 − 2yi + 0.5833yi−1 = 0.2778(4 xi ).
The solution of the corresponding linear system gives
x
y
1.0000
1.0000
1.1667
–0.5918
1.3333
–1.1626
1.5000
–1.3070
1.6667
–1.2704
1.8333
–1.1541
2.0000
–1.0000
7. We identify P (x) = 3/x, Q(x) = 3/x2 , f (x) = 0, and h = (2 − 1)/8 = 0.125. Then the finite
difference equation is
0.1875
0.0469
0.1875
yi+1 + −2 +
yi + 1 −
yi−1 = 0.
1+
xi
xi
x2i
The solution of the corresponding linear system gives
x
y
1.000
5.0000
1.125
3.8842
1.250
2.9640
1.375
2.2064
1.500
1.5826
1.625
1.0681
1.750
0.6430
1.875
0.2913
2.000
0.0000
8. We identify P (x) = −1/x, Q(x) = x−2 , f (x) = ln x/x2 , and h = (2 − 1)/8 = 0.125. Then the
finite difference equation is
0.0625
0.0156
0.0625
yi+1 + −2 +
yi + 1 +
yi−1 = 0.0156 ln xi .
1−
xi
xi
x2i
The solution of the corresponding linear system gives
x
y
1.000
0.0000
1.125
– 0.1988
1.250
– 0.4168
1.375
– 0.6510
1.500
–0.8992
1.625
–1.1594
1.750
– 1.4304
1.875
–1.7109
2.000
– 2.0000
9.5 Second-Order Boundary-Value Problems
9. We identify P (x) = 1 − x, Q(x) = x, f (x) = x, and h = (1 − 0)/10 = 0.1. Then the finite
difference equation is
[1 + 0.05(1 − xi )]yi+1 + [−2 + 0.01xi ]yi + [1 − 0.05(1 − xi )]yi−1 = 0.01xi .
The solution of the corresponding linear system gives
x
y
0.0
0.0000
0.1
0.2660
0.2
0.5097
0.3
0.7357
0.5
1.1465
0.4
0.9471
0.7
1.5149
0.6
1.3353
0.8
1.6855
0.9
1.8474
1.0
2.0000
10. We identify P (x) = x, Q(x) = 1, f (x) = x, and h = (1 − 0)/10 = 0.1. Then the finite
difference equation is
(1 + 0.05xi )yi+1 − 1.99yi + (1 − 0.05xi )yi−1 = 0.01xi .
The solution of the corresponding linear system gives
x
y
0.0
1.0000
0.1
0.8929
0.2
0.7789
0.3
0.6615
0.4
0.5440
0.5
0.4296
0.7
0.2225
0.6
0.3216
0.8
0.1347
0.9
0.0601
1.0
0.0000
11. We identify P (x) = 0, Q(x) = −4, f (x) = 0, and h = (1 − 0)/8 = 0.125. Then the finite
difference equation is
yi+1 − 2.0625yi + yi−1 = 0.
The solution of the corresponding linear system gives
x
y
0.000
0.0000
0.125
0.3492
0.250
0.7202
0.375
1.1363
0.500
1.6233
0.625
2.2118
0.750
2.9386
0.875
3.8490
1.000
5.0000
12. We identify P (r) = 2/r, Q(r) = 0, f (r) = 0, and h = (4 − 1)/6 = 0.5. Then the finite
difference equation is
0.5
0.5
ui+1 − 2ui + 1 −
ui−1 = 0.
1+
ri
ri
The solution of the corresponding linear system gives
r
u
1.0
50.0000
1.5
72.2222
2.0
83.3333
2.5
90.0000
3.0
94.4444
3.5
97.6190
4.0
100.0000
601
602
CHAPTER 9
NUMERICAL SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS
13. (a) The difference equation
1+
h
h
Pi yi+1 + (−2 + h2 Qi )yi + 1 − Pi yi−1 = h2 fi
2
2
is the same as equation (8) in the text. The equations are the same because the derivation
was based only on the differential equation, not the boundary conditions. If we allow i
to range from 0 to n − 1 we obtain n equations in the n + 1 unknowns y−1 , y0 , y1 , . . . ,
yn−1 . Since yn is one of the given boundary conditions, it is not an unknown.
(b) Identifying y0 = y(0), y−1 = y(0 − h), and y1 = y(0 + h) we have from equation (5) in
the text
1
[y1 − y−1 ] = y (0) = 1 or y1 − y−1 = 2h.
2h
The difference equation corresponding to i = 0,
h
h
2
1 + P0 y1 + (−2 + h Q0 )y0 + 1 − P0 y−1 = h2 f0
2
2
becomes, with y−1 = y1 − 2h,
h
h
2
1 + P0 y1 + (−2 + h Q0 )y0 + 1 − P0 (y1 − 2h) = h2 f0
2
2
or
2y1 + (−2 + h2 Q0 )y0 = h2 f0 + 2h − P0 .
Alternatively, we may simply add the equation y1 − y−1 = 2h to the list of n difference
equations obtaining n + 1 equations in the n + 1 unknowns y−1 , y0 , y1 , . . . , yn−1 .
(c) Using n = 5 we obtain
x
y
0.0
–2.2755
0.2
–2.0755
0.4
–1.8589
0.6
–1.6126
0.8
–1.3275
1.0
–1.0000
14. Using h = 0.1 and, after shooting a few times, y (0) = 0.43535 we obtain the following table
with the RK4 method.
x
y
0.0
1.00000
0.1
1.04561
0.2
1.09492
0.3
1.14714
0.4
1.20131
0.5
1.25633
0.6
1.31096
0.7
1.36392
0.8
1.41388
0.9
1.45962
1.0
1.50003
Chapter 9 in Review
Chapter 9 in Review
1. x
xn
Euler
h = 0.1
Euler
h = 0.05
Imp. Euler
h = 0.1
Imp. Euler
h = 0.05
RK4
h = 0.1
RK4
h = 0.05
1.00
2.0000
2.0000
2.0000
2.0000
2.0000
2.0000
2.1386
2.0693
2.1469
2.2328
2.1549
2.0735
2.1554
2.2459
2.1556
2.0736
2.1556
2.2462
1.05
1.10
1.15
1.20
1.25
2.3097
2.3272
2.4299
2.3439
2.3450
2.4527
2.3454
2.3454
2.4532
1.30
1.35
1.40
1.45
1.50
2.5136
2.5672
3.1157
2.5689
2.6937
2.8269
2.9686
3.1187
2.5695
3.0201
2.5409
2.6604
2.7883
2.9245
3.0690
3.1197
2.5695
2.6944
2.8278
2.9696
3.1197
xn
Euler
h = 0.1
Euler
h = 0.05
Imp. Euler
h = 0.1
Imp. Euler
h = 0.05
RK4
h = 0.1
RK4
h = 0.05
0.00
0.05
0.0000
0.0000
0.0500
0.0000
0.0000
0.0501
0.0000
0.0000
0.0500
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.1000
0.1001
0.1506
0.2017
0.1005
0.1004
0.1512
0.2027
0.1003
0.1003
0.1511
0.2026
0.2026
0.2551
0.3087
0.3087
0.3637
0.4201
0.4781
2. x
2.7504
0.2010
0.4207
0.5327
0.5382
0.4782
0.5378
0.5376
0.5376
Euler
h = 0.1
Euler
h = 0.05
Imp. Euler
h = 0.1
Imp. Euler
h = 0.05
RK4
h = 0.1
RK4
h = 0.05
0.5000
0.5000
0.5000
0.5000
0.5000
0.5000
0.6000
0.5500
0.6024
0.6048
0.5512
0.6049
0.6049
0.5512
0.6049
0.3049
0.4135
0.5279
xn
0.50
0.55
0.60
0.65
0.70
0.75
0.80
0.85
0.90
0.95
1.00
0.2030
2.8278
0.2552
0.3088
0.3638
0.4202
0.45
0.50
3. x
2.8246
0.7090
0.8283
0.9559
1.0921
0.2537
0.3067
0.3610
0.4167
0.4739
0.6573
0.7144
0.7739
0.8356
0.8996
0.9657
1.0340
1.1044
0.3092
0.7191
0.8427
0.9752
1.1163
0.6609
0.7193
0.7800
0.8430
0.9082
0.9755
1.0451
1.1168
0.4201
0.7194
0.8431
0.9757
1.1169
0.6610
0.7194
0.7801
0.8431
0.9083
0.9757
1.0452
1.1169
603
604
CHAPTER 9
4. x
NUMERICAL SOLUTIONS OF ORDINARY DIFFERENTIAL EQUATIONS
xn
Euler
h = 0.1
Euler
h = 0.05
Imp. Euler
h = 0.1
Imp. Euler
h = 0.05
RK4
h = 0.1
RK4
h = 0.05
1.00
1.05
1.10
1.15
1.20
1.0000
1.0000
1.1000
1.2183
1.3595
1.5300
1.0000
1.0000
1.1091
1.2405
1.4010
1.6001
1.0000
1.0000
1.1095
1.2415
1.4029
1.6036
1.2000
1.4760
1.25
1.30
1.35
1.40
1.45
1.50
1.2380
1.5910
1.7389
1.8710
1.9988
2.3284
2.7567
3.3296
4.1253
2.4643
3.4165
1.2415
1.6036
1.8523
2.1524
3.1458
5.2510
2.1799
2.6197
3.2360
4.1528
5.6404
1.8586
2.1909
3.2745
5.8338
2.1911
2.6401
3.2755
4.2363
5.8446
5. Using
yn+1 = yn + hun ,
y0 = 3
un+1 = un + h(2xn + 1)yn ,
u0 = 1
we obtain (when h = 0.2) y1 = y(0.2) = y0 + hu0 = 3 + (0.2)1 = 3.2. When h = 0.1 we have
y1 = y0 + 0.1u0 = 3 + (0.1)1 = 3.1
u1 = u0 + 0.1(2x0 + 1)y0 = 1 + 0.1(1)3 = 1.3
y2 = y1 + 0.1u1 = 3.1 + 0.1(1.3) = 3.23.
6. The first predictor is y3∗ = 1.14822731.
xn
yn
0.0
0.1
2.00000000
1.65620000
init. cond.
RK4
0.2
0.3
0.4
1.41097281
1.24645047
1.14796764
RK4
RK4
ABM
7. Using x0 = 1, y0 = 2, and h = 0.1 we have
x1 = x0 + h(x0 + y0 ) = 1 + 0.1(1 + 2) = 1.3
y1 = y0 + h(x0 − y0 ) = 2 + 0.1(1 − 2) = 1.9
and
x2 = x1 + h(x1 + y1 ) = 1.3 + 0.1(1.3 + 1.9) = 1.62
y2 = y1 + h(x1 − y1 ) = 1.9 + 0.1(1.3 − 1.9) = 1.84.
Thus, x(0.2) ≈ 1.62 and y(0.2) ≈ 1.84.
Chapter 9 in Review
8. We identify P (x) = 0, Q(x) = 6.55(1 + x), f (x) = 1, and h = (1 − 0)/10 = 0.1. Then the
finite difference equation is
yi+1 + [−2 + 0.0655(1 + xi )]yi + yi−1 = 0.001
or
yi+1 + (0.0655xi − 1.9345)yi + yi−1 = 0.001.
The solution of the corresponding linear system gives
x
y
0.0
0.0000
0.1
4.1987
0.2
8.1049
0.3
11.3840
0.4
13.7038
0.5
14.7770
0.7
12.5396
0.6
14.4083
0.8
9.2847
0.9
4.9450
1.0
0.0000
605
Chapter 10
Systems of Nonlinear First-Order Differential
Equations
10.1
Autonomous Systems
1. The corresponding plane autonomous system is
x = y,
y = −9 sin x.
If (x, y) is a critical point, y = 0 and −9 sin x = 0. Therefore x = ±nπ and so the critical
points are (±nπ, 0) for n = 0, 1, 2, . . . .
2. The corresponding plane autonomous system is
x = y,
y = −2x − y 2 .
If (x, y) is a critical point, then y = 0 and so −2x − y 2 = −2x = 0. Therefore (0, 0) is the
sole critical point.
3. The corresponding plane autonomous system is
x = y,
y = x2 − y(1 − x3 ).
If (x, y) is a critical point, y = 0 and so x2 − y(1 − x3 ) = x2 = 0. Therefore (0, 0) is the sole
critical point.
4. The corresponding plane autonomous system is
x = y,
y = −4
x
− 2y.
1 + x2
If (x, y) is a critical point, y = 0 and so −4x/(1 + x2 ) − 2(0) = 0. Therefore x = 0 and so
(0, 0) is the sole critical point.
606
10.1
Autonomous Systems
5. The corresponding plane autonomous system is
x = y,
y = −x + x3 .
If (x, y) is a critical point, y = 0 and −x + x3 = 0. Hence x(−1 + x2 ) = 0 and so x = 0,
1/ , − 1/ . The critical points are (0, 0), ( 1/ , 0) and (− 1/ , 0).
6. The corresponding plane autonomous system is
x = y,
y = −x + x|x|.
If (x, y) is a critical point, y = 0 and −x + x|x| = x(−1 + |x|) = 0. Hence x = 0, 1/, −1/.
The critical points are (0, 0), (1/, 0) and (−1/, 0).
7. From x+xy = 0 we have x(1+y) = 0. Therefore x = 0 or y = −1. If x = 0, then, substituting
into −y − xy = 0, we obtain y = 0. Likewise, if y = −1, 1 + x = 0 or x = −1. We can
conclude that (0, 0) and (−1, −1) are critical points of the system.
8. From y 2 − x = 0 we have x = y 2 . Substituting into x2 − y = 0, we obtain y 4 − y = 0 or
y(y 3 − 1) = 0. It follows that y = 0, 1 and so (0, 0) and (1, 1) are the critical points of the
system.
9. From x − y = 0 we have y = x. Substituting into 3x2 − 4y = 0 we obtain
3x2 − 4x = x(3x − 4) = 0. It follows that (0, 0) and (4/3, 4/3) are the critical points of the
system.
10. From x3 − y = 0 we have y = x3 . Substituting into x − y 3 = 0 we obtain x − x9 = 0 or
x(1 − x8 ). Therefore x = 0, 1, −1 and so the critical points of the system are (0, 0), (1, 1),
and (−1, −1).
11. From x(10 − x − 12 y) = 0 we obtain x = 0 or x + 12 y = 10. Likewise y(16 − y − x) = 0 implies
that y = 0 or x + y = 16. We therefore have four cases. If x = 0, y = 0 or y = 16. If
x + 12 y = 10, we can conclude that y(− 12 y + 6) = 0 and so y = 0, 12. Therefore the critical
points of the system are (0, 0), (0, 16), (10, 0), and (4, 12).
12. Adding the two equations we obtain 10 − 15y/(y + 5) = 0. It follows that y = 10, and from
−2x + y + 10 = 0 we can conclude that x = 10. Therefore (10, 10) is the sole critical point of
the system.
13. From x2 ey = 0 we have x = 0. Since ex − 1 = e0 − 1 = 0, the second equation is satisfied for
an arbitrary value of y. Therefore any point of the form (0, y) is a critical point.
14. From sin y = 0 we have y = ±nπ. From ex−y = 1, we can conclude that x − y = 0 or x = y.
The critical points of the system are therefore (±nπ, ±nπ) for n = 0, 1, 2, . . . .
607
608
CHAPTER 10
SYSTEMS OF NONLINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
15. From x(1 − x2 − 3y 2 ) = 0 we have x = 0 or x2 + 3y 2 = 1. If x = 0, then substituting into
y(3 − x2 − 3y 2 ) gives y(3 − 3y 2 ) = 0. Therefore y = 0, 1, −1. Likewise x2 = 1 − 3y 2 yields
2y = 0 so that y = 0 and x2 = 1 − 3(0)2 = 1. The critical points of the system are therefore
(0, 0), (0, 1), (0, −1), (1, 0), and (−1, 0).
16. From −x(4 − y 2 ) = 0 we obtain x = 0, y = 2, or y = −2. If x = 0, then substituting into
4y(1 − x2 ) yields y = 0. Likewise y = 2 gives 8(1 − x2 ) = 0 or x = 1, −1. Finally y = −2
yields −8(1 − x2 ) = 0 or x = 1, −1. The critical points of the system are therefore (0, 0),
(1, 2), (−1, 2), (1, −2), and (−1, −2).
17. (a) From Exercises 8.2, Problem 1, x = c1 e5t − c2 e−t and y = 2c1 e5t + c2 e−t .
(b) From X(0) = (−2, 2) it follows that c1 = 0 and c2 = 2. Therefore x = −2e−t and
y = 2e−t .
y
(c)
(–2, 2)
2
–2
x
2
–2
18. (a) From Exercises 8.2, Problem 6, x = c1 + 2c2 e−5t and y = 3c1 + c2 e−5t , which is not
periodic.
(b) From X(0) = (3, 4) it follows that c1 = c2 = 1. Therefore x = 1 + 2e−5t and y = 3 + e−5t
gives y = 12 (x − 1) + 3.
y
(c)
4
(3, 4)
2
–4
y = 3x
–2
2
4
x
–2
–4
19. (a) From Exercises 8.2, Problem 34, x = c1 (4 cos 3t − 3 sin 3t) + c2 (4 sin 3t + 3 cos 3t) and
y = c1 (5 cos 3t) + c2 (5 sin 3t). All solutions are periodic with p = 2π/3.
10.1
Autonomous Systems
(b) From X(0) = (4, 5) it follows that c1 = 1 and c2 = 0. Therefore x = 4 cos 3t − 3 sin 3t
and y = 5 cos 3t.
y
(c)
6
(4, 5)
4
2
–6
–4
–2
2
4
6
x
–2
–4
–6
20. (a) From Exercises 8.2, Problem 35, x = c1 (sin t − cos t) + c2 (− cos t − sin t) and
y = 2c1 cos t + 2c2 sin t. All solutions are periodic with p = 2π.
(b) From X(0) = (−2, 2) it follows that c1 = c2 = 1.
y = 2 cos t + 2 sin t.
Therefore x = −2 cos t and
y
(c)
4
(–2, 2)
–4
2
–2
2
4
x
–2
–4
21. (a) From Exercises 8.2, Problem 38, x = c1 (sin t − cos t)e4t + c2 (− sin t − cos t)e4t and y =
2c1 (cos t) e4t +2c2 (sin t) e4t . Because of the presence of e4t , there are no periodic solutions.
(b) From X(0) = (−1, 2) it follows that c1 = 1 and c2 = 0. Therefore x = (sin t − cos t)e4t
and y = 2(cos t) e4t .
609
610
CHAPTER 10
SYSTEMS OF NONLINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
y
(c)
5
(–1, 2)
–5
5
x
–5
22. (a) From Exercises 10.2, Problem 40, x = c1 e−t (2 cos 2t − 2 sin 2t) + c2 e−t (2 cos 2t + 2 sin 2t)
and y = c1 e−t cos 2t + c2 e−t sin 2t. Because of the presence of e−t , there are no periodic
solutions.
(b) From X(0) = (2, 1) it follows that c1 = 1 and c2 = 0. Therefore x = e−t (2 cos 2t−2 sin 2t)
and y = e−t cos 2t.
y
(c)
2
(2, 1)
–2
2
x
–2
23. Switching to polar coordinates,
1
dy
dr
dx
1
=
+y
x
= (−xy − x2 r4 + xy − y 2 r4 ) = −r5
dt
r
dt
dt
r
dx
1
dy
1
dθ
= 2 −y
+x
= 2 (y 2 + xyr4 + x2 − xyr4 ) = 1 .
dt
r
dt
dt
r
dr
= −r5 we obtain
dt
1/4
1
and θ = t + c2 .
r=
4t + c1
If we use separation of variables on
Since X(0) = (4, 0), r = 4 and θ = 0 when t = 0. It follows that c2 = 0 and c1 =
final solution can be written as
4
r= √
,
θ=t
4
1024t + 1
and so the solution spirals toward the origin as t increases.
1
256
. The
10.1
Autonomous Systems
611
24. Switching to polar coordinates,
1
dx
dy
1
dr
=
x
+y
= (xy + x2 r2 − xy + y 2 r2 ) = r3
dt
r
dt
dt
r
1
dy
1
dx
dθ
= 2 −y
+x
= 2 (−y 2 − xyr2 − x2 + xyr2 ) = −1 .
dt
r
dt
dt
r
If we use separation of variables, it follows that
r=√
1
−2t + c1
and θ = −t + c2 .
Since X(0) = (4, 0), r = 4 and θ = 0 when t = 0. It follows that c2 = 0 and c1 =
final solution can be written as
r=√
Note that r → ∞ as t →
4
,
1 − 32t
1
32 . Because 0 ≤ t ≤
1
16 .
The
θ = −t.
1
32 ,
the curve is not a spiral.
25. Switching to polar coordinates,
dr
1
dx
dy
1
=
x
+y
= [−xy + x2 (1 − r2 ) + xy + y 2 (1 − r2 )] = r(1 − r2 )
dt
r
dt
dt
r
1
dy
1
dx
dθ
= 2 −y
+x
= 2 [y 2 − xy(1 − r2 ) + x2 + xy(1 − r2 )] = 1.
dt
r
dt
dt
r
Now dr/dt = r − r3 or (dr/dt) − r = −r3 is a Bernoulli differential equation. Following the
procedure in Section 2.5 of the text, we let w = r−2 so that w = −2r−3 (dr/dt). Therefore
w + 2w = 2, a linear first order differential equation. It follows that w = 1 + c1 e−2t and so
r2 = 1/(1 + c1 e−2t ). The general solution can be written as
r=√
1
,
1 + c1 e−2t
θ = t + c2 .
If X(0) = (1, 0), r = 1 and θ = 0 when t = 0. Therefore c1 = 0 = c2 and so x = r cos t = cos t
and y = r sin t = sin t. This solution generates the circle r = 1. If X(0) = (2, 0), r = 2 and
θ = 0 when t = 0. Therefore c1 = −3/4, c2 = 0 and so
1
,
r=
1 − 34 e−2t
θ = t.
This solution spirals toward the circle r = 1 as t increases.
26. Switching to polar coordinates,
dr
1
dx
dy
1
x2
y2
=
x
+y
=
xy − (4 − r2 ) − xy − (4 − r2 ) = r2 − 4
dt
r
dt
dt
r
r
r
1
dy
1
dx
xy
xy
dθ
= 2 −y
+x
= 2 −y 2 +
(4 − r2 ) − x2 −
(4 − r2 ) = −1.
dt
r
dt
dt
r
r
r
612
CHAPTER 10
SYSTEMS OF NONLINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
From Example 3, Section 2.2,
r=2
1 + c1 e4t
1 − c1 e4t
and θ = −t + c2 .
If X(0) = (1, 0), r = 1 and θ = 0 when t = 0. It follows that c2 = 0 and c1 = − 13 . Therefore
r=2
1 − 13 e4t
1 + 13 e4t
and θ = −t.
Note that r = 0 when e4t = 3 or t = (ln 3)/4 and r → −2 as t → ∞. The solution therefore
approaches the circle r = 2. If X(0) = (2, 0), it follows that c1 = c2 = 0. Therefore r = 2 and
θ = −t so that the solution generates the circle r = 2 traversed in the clockwise direction.
Note also that the original system is not defined at (0, 0) but the corresponding polar system
is defined for r = 0. If the Runge-Kutta method is applied to the original system, the solution
corresponding to X(0) = (1, 0) will stall at the origin.
27. The system has no critical points, so there are no periodic solutions.
28. From x(6y − 1) = 0 and y(2 − 8x) = 0 we see that (0, 0) and
(1/4, 1/6) are critical points. From the graph we see that there
are periodic solutions around (1/4, 1/6).
29. The only critical point is (0, 0). There appears to be a single
periodic solution around (0, 0).
30. The system has no critical points, so there are no periodic solutions.
10.2
Stability of Linear Systems
1. (a) If X(0) = X0 lies on the line y = 2x, then X(t) approaches (0, 0) along this line. For
all other initial conditions, X(t) approaches (0, 0) from the direction determined by the
line y = −x/2.
10.2
Stability of Linear Systems
y
(b)
1
–1
x
1
y = – x/2
–1
2. (a) If X(0) = X0 lies on the line y = −x, then X(t) becomes unbounded along this line.
For all other initial conditions, X(t) becomes unbounded and y = −3x/2 serves as an
asymptote.
y
4
(b)
2
(1, 1)
–4
–2
4 x
2
y = – 3x/2
–2
–4
3. (a) All solutions are unstable spirals which become unbounded as t increases.
y
(b)
5
(1, 1)
–5
5
–5
x
613
614
CHAPTER 10
SYSTEMS OF NONLINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
4. (a) All solutions are spirals which approach the origin.
y
(b)
(1, 1)
1
–1
1
x
–1
5. (a) All solutions approach (0, 0) from the direction specified by the line y = x.
y
(b)
1
–1
1
x
–1
6. (a) All solutions become unbounded and y = x/2 serves as the asymptote.
y
4
(b)
3
2
1
–4 –3 –2 –1
1
–1
–2
y = x/2
–3
–4
2
3
4 x
10.2
Stability of Linear Systems
7. (a) If X(0) = X0 lies on the line y = 3x, then X(t) approaches (0, 0) along this line. For all
other initial conditions, X(t) becomes unbounded and y = x serves as the asymptote.
y
(b)
3
2
1
–3 –2 –1
1
2
3
x
–1
–2
–3
8. (a) The solutions are ellipses which encircle the origin.
y
3
(b)
2
1
–3
–2
–1
(1, 1)
1
2
3 x
–1
–2
–3
9. Since Δ = −41 < 0, we can conclude from Figure 10.2.12 that (0, 0) is a saddle point.
10. Since Δ = 29 and τ = −12, τ 2 − 4Δ > 0 and so from Figure 10.2.12, (0, 0) is a stable node.
11. Since Δ = −19 < 0, we can conclude from Figure 10.2.12 that (0, 0) is a saddle point.
12. Since Δ = 1 and τ = −1, τ 2 − 4Δ = −3 and so from Figure 10.2.12, (0, 0) is a stable spiral
point.
13. Since Δ = 1 and τ = −2, τ 2 − 4Δ = 0 and so from Figure 10.2.12, (0, 0) is a degenerate
stable node.
14. Since Δ = 1 and τ = 2, τ 2 −4Δ = 0 and so from Figure 10.2.12, (0, 0) is a degenerate unstable
node.
15. Since Δ = 0.01 and τ = −0.03, τ 2 − 4Δ < 0 and so from Figure 10.2.12, (0, 0) is a stable
spiral point.
16. Since Δ = 0.0016 and τ = 0.08, τ 2 − 4Δ = 0 and so from Figure 10.2.12, (0, 0) is a degenerate
unstable node.
615
616
CHAPTER 10
SYSTEMS OF NONLINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
17. Δ = 1 − μ2 , τ = 0, and so we need Δ = 1 − μ2 > 0 for (0, 0) to be a center. Therefore |μ| < 1.
18. Note that Δ = 1 and τ = μ. Therefore we need both τ = μ < 0 and τ 2 − 4Δ = μ2 − 4 < 0
for (0, 0) to be a stable spiral point. These two conditions can be written as −2 < μ < 0.
19. Note that Δ = μ + 1 and τ = μ + 1 and so τ 2 − 4Δ = (μ + 1)2 − 4(μ + 1) = (μ + 1)(μ − 3).
It follows that τ 2 − 4Δ < 0 if and only if −1 < μ < 3. We can conclude that (0, 0) will be a
saddle point when μ < −1. Likewise (0, 0) will be an unstable spiral point when τ = μ+1 > 0
and τ 2 − 4Δ < 0. This condition reduces to −1 < μ < 3.
20. τ = 2α, Δ = α2 + β 2 > 0, and τ 2 − 4Δ = −4β < 0. If α < 0, (0, 0) is a stable spiral point. If
α > 0, (0, 0) is an unstable spiral point. Therefore (0, 0) cannot be a node or saddle point.
21. AX1 +F = 0 implies that AX1 = −F or X1 = −A−1 F. Since Xp (t) = −A−1 F is a particular
solution, it follows from Theorem 10.1.6 that X(t) = Xc (t) + X1 is the general solution to
X = AX + F. If τ < 0 and Δ > 0 then Xc (t) approaches (0, 0) by Theorem 10.2.1(a). It
follows that X(t) approaches X1 as t → ∞.
22. If bc < 1, Δ = adx̂ŷ(1 − bc) > 0 and τ 2 − 4Δ = (ax̂ − dŷ)2 + 4abcdx̂ŷ > 0. Therefore (0, 0) is
a stable node.
23. (a) The critical point is X1 = (−3, 4).
(b) From the graph, X1 appears to be an unstable node or a
saddle point.
(c) Since Δ = −1, (0, 0) is a saddle point.
24. (a) The critical point is X1 = (−1, −2).
(b) From the graph, X1 appears to be a stable node or a degenerate stable node.
(c) Since τ = −16, Δ = 64, and τ 2 − 4Δ = 0, (0, 0) is a degenerate stable node.
25. (a) The critical point is X1 = (0.5, 2).
(b) From the graph, X1 appears to be an unstable spiral point.
(c) Since τ = 0.2, Δ = 0.03, and τ 2 − 4Δ = −0.08, (0, 0) is an
unstable spiral point.
10.3
Linearization and Local Stability
26. (a) The critical point is X1 = (1, 1).
(b) From the graph, X1 appears to be a center.
(c) Since τ = 0 and Δ = 1, (0, 0) is a center.
10.3
Linearization and Local Stability
1. Switching to polar coordinates,
1
dx
dy
1
dr
1
=
x
+y
= (αx2 − βxy + xy 2 + βxy + αy 2 − xy 2 ) = αr2 = αr.
dt
r
dt
dt
r
r
Therefore r = ceαt and so r → 0 if and only if α < 0.
2. The differential equation dr/dt = αr(5 − r) is a logistic differential equation. [See Section
3.2, (4) and (5).] It follows that
r=
5
1 + c1 e−5αt
and θ = −t + c2 .
If α > 0, r → 5 as t → +∞ and so the critical point (0, 0) is unstable. If α < 0, r → 0 as
t → +∞ and so (0, 0) is asymptotically stable.
3. The critical points are x = 0 and x = n + 1. Since g (x) = k(n + 1) − 2kx, g (0) = k(n + 1) > 0
and g (n + 1) = −k(n + 1) < 0. Therefore x = 0 is unstable while x = n + 1 is asymptotically
stable. See Theorem 10.2.
4. Note that x = k is the only critical point since ln(x/k) is not defined at x = 0. Since
g (x) = −k − k ln (x/k), g (k) = −k < 0. Therefore x = k is an asymptotically stable critical
point by Theorem 10.2.
5. The only critical point is T = T0 . Since g (T ) = k, g (T0 ) = k > 0. Therefore T = T0 is
unstable by Theorem 10.2.
6. The only critical point is v = mg/k. Now g(v) = g − (k/m)v and so g (v) = −k/m < 0.
Therefore v = mg/k is an asymptotically stable critical point by Theorem 10.2.
7. Critical points occur at x = α, β. Since g (x) = k(−α − β + 2x), g (α) = k(α − β) and
g (β) = k(β − α). Since α > β, g (α) > 0 and so x = α is unstable. Likewise x = β is
asymptotically stable.
8. Critical points occur at x = α, β, γ. Since
g (x) = k(α − x)(−β − γ − 2x) + k(β − x)(γ − x)(−1),
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CHAPTER 10
SYSTEMS OF NONLINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
g (α) = −k(β − α)(γ − α) < 0 since α > β > γ. Therefore x = α is asymptotically stable.
Similarly g (β) > 0 and g (γ) < 0. Therefore x = β is unstable while x = γ is asymptotically
stable.
9. Critical points occur at P = a/b, c but not at P = 0. Since g (P ) = (a − bP ) + (P − c)(−b),
g (a/b) = (a/b − c)(−b) = −a + bc
and g (c) = a − bc.
Since a < bc, −a + bc > 0 and a − bc < 0. Therefore P = a/b is unstable while P = c is
asymptotically stable.
10. Since A > 0, the only critical point is A = K 2 . Since
g (A) = 12 kKA−1/2 − k, g (K 2 ) = −k/2 < 0. Therefore A = K 2 is asymptotically stable.
11. The sole critical point is (1/2, 1) and
g (X) =
−2y
−2x
.
2y 2x − 1
Computing g ((1/2, 1)) we find that τ = −2 and Δ = 2 so that τ 2 − 4Δ = −4 < 0. Therefore
(1/2, 1) is a stable spiral point.
12. Critical points are (1, 0) and (−1, 0), and
g (X) =
2x −2y
.
0
2
At X = (1, 0), τ = 4, Δ = 4, and so τ 2 − 4Δ = 0. We can conclude that (1, 0) is unstable but
we are unable to classify this critical point any further. At X = (−1, 0), Δ = −4 < 0 and so
(−1, 0) is a saddle point.
13. y = 2xy − y = y(2x − 1). Therefore if (x, y) is a critical point, either x = 1/2 or y = 0. The
case x = 1/2 and y − x2 + 2 = 0 implies that (x, y) = (1/2, −7/4). The case y = 0 leads to
√
√
the critical points ( 2 , 0) and (− 2 , 0). We next use the Jacobian matrix
−2x
1
2y 2x − 1
√
√
to classify these three critical points. For X = ( 2 , 0) or (− 2 , 0), τ = −1 and Δ < 0.
Therefore both critical points are saddle points. For X = (1/2, −7/4), τ = −1, Δ = 7/2 and
so τ 2 − 4Δ = −13 < 0. Therefore (1/2, −7/4) is a stable spiral point.
g (X) =
14. y = −y + xy = y(−1 + x). Therefore if (x, y) is a critical point, either y = 0 or x = 1. The
case y = 0 and 2x − y 2 = 0 implies that (x, y) = (0, 0). The case x = 1 leads to the critical
√
√
points (1, 2 ) and (1, − 2 ). We next use the Jacobian matrix
g (X) =
2 −2y
y x−1
10.3
Linearization and Local Stability
to classify these critical points. For X = (0, 0), Δ = −2 < 0 and so (0, 0) is a saddle point.
√
√
√
For either (1, 2 ) or (1, − 2 ), τ = 2, Δ = 4, and so τ 2 − 4Δ = −12. Therefore (1, 2 ) and
√
(1, − 2 ) are unstable spiral points.
15. Since x2 − y 2 = 0, y 2 = x2 and so x2 − 3x + 2 = (x − 1)(x − 2) = 0. It follows that the critical
points are (1, 1), (1, −1), (2, 2), and (2, −2). We next use the Jacobian
g (X) =
−3
2y
2x −2y
to classify these four critical points. For X = (1, 1), τ = −5, Δ = 2, and so τ 2 − 4Δ = 17 > 0.
Therefore (1, 1) is a stable node. For X = (1, −1), Δ = −2 < 0 and so (1, −1) is a saddle point.
For X = (2, 2), Δ = −4 < 0 and so we have another saddle point. Finally, if X = (2, −2),
τ = 1, Δ = 4, and so τ 2 − 4Δ = −15 < 0. Therefore (2, −2) is an unstable spiral point.
16. From y 2 − x2 = 0, y = x or y = −x. The case y = x leads to (4, 4) and (−1, 1) but the case
y = −x leads to x2 − 3x + 4 = 0 which has no real solutions. Therefore (4, 4) and (−1, 1) are
the only critical points. We next use the Jacobian matrix
g (X) =
y x−3
−2x
2y
to classify these two critical points. For X = (4, 4), τ = 12, Δ = 40, and so τ 2 − 4Δ < 0.
Therefore (4, 4) is an unstable spiral point. For X = (−1, 1), τ = −3, Δ = 10, and so
x2 − 4Δ < 0. It follows that (−1, −1) is a stable spiral point.
17. Since x = −2xy = 0, either x = 0 or y = 0. If x = 0, y(1 − y 2 ) = 0 and so (0, 0), (0, 1), and
(0, −1) are critical points. The case y = 0 leads to x = 0. We next use the Jacobian matrix
−2y
−2x
g (X) =
−1 + y 1 + x − 3y 2
to classify these three critical points. For X = (0, 0), τ = 1 and Δ = 0 and so the test is
inconclusive. For X = (0, 1), τ = −4, Δ = 4 and so τ 2 − 4Δ = 0. We can conclude that
(0, 1) is a stable critical point but we are unable to classify this critical point further in this
borderline case. For X = (0, −1), Δ = −4 < 0 and so (0, −1) is a saddle point.
18. We found that (0, 0), (0, 1), (0, −1), (1, 0) and (−1, 0) were the critical points in Problem 15,
Section 10.1. The Jacobian is
g (X) =
−6xy
1 − 3x2 − 3y 2
.
2
−2xy 3 − x − 9y 2
For X = (0, 0), τ = 4, Δ = 3 and so τ 2 − 4Δ = 4 > 0. Therefore (0, 0) is an unstable node.
Both (0, 1) and (0, −1) give τ = −8, Δ = 12, and τ 2 − 4Δ = 16 > 0. These two critical points
are therefore stable nodes. For X = (1, 0) or (−1, 0), Δ = −4 < 0 and so saddle points occur.
619
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CHAPTER 10
SYSTEMS OF NONLINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
19. We found the critical points (0, 0), (10, 0), (0, 16) and (4, 12) in Problem 11, Section 10.1.
Since the Jacobian is
− 12 x
10 − 2x − 12 y
g (X) =
−y 16 − 2y − x
we can classify the critical points as follows:
X
(0, 0)
(10, 0)
(0, 16)
(4, 12)
τ
Δ
26
−4
−14
−16
160
−60
−32
24
τ 2 − 4Δ
36
−
−
160
Conclusion
unstable node
saddle point
saddle point
stable node
20. We found the sole critical point (10, 10) in Problem 12, Section 10.1. The Jacobian is
⎛
⎞
−2
1
g (X) = ⎝
15 ⎠ ,
2 −1 −
(y + 5)2
g ((10, 10)) has trace τ = −46/15, Δ = 2/15, and τ 2 − 4Δ > 0. Therefore (0, 0) is a stable
node.
21. The corresponding plane autonomous system is
θ = y,
1
y = (cos θ − ) sin θ.
2
Since |θ| < π, it follows that critical points are (0, 0), (π/3, 0) and (−π/3, 0). The Jacobian
matrix is
0
1
g (X) =
1
cos 2θ − 2 cos θ 0
and so at (0, 0), τ = 0 and Δ = −1/2. Therefore (0, 0) is a saddle point. For X = (±π/3, 0),
τ = 0 and Δ = 3/4. It is not possible to classify either critical point in this borderline case.
22. The corresponding plane autonomous system is
1
− 3y 2 y − x2 .
x = y, y = −x +
2
If (x, y) is a critical point, y = 0 and so −x − x2 = −x(1 + x) = 0. Therefore (0, 0) and (−1, 0)
are the only two critical points. We next use the Jacobian matrix
g (X) =
0
−1 − 2x
1
1
2
− 9y 2
to classify these critical points. For X = (0, 0), τ = 1/2, Δ = 1, and τ 2 − 4Δ < 0. Therefore
(0, 0) is an unstable spiral point. For X = (−1, 0), τ = 1/2, Δ = −1 and so (−1, 0) is a saddle
point.
10.3
Linearization and Local Stability
23. The corresponding plane autonomous system is
x = y,
y = x2 − y(1 − x3 )
and the only critical point is (0, 0). Since the Jacobian matrix is
0
g (X) =
2x +
1
3x2 y
x3
−1
,
τ = −1 and Δ = 0, and we are unable to classify the critical point in this borderline case.
24. The corresponding plane autonomous system is
4x
− 2y
1 + x2
and the only critical point is (0, 0). Since the Jacobian matrix is
⎛
⎞
0
1
⎜
⎟
g (X) = ⎝
⎠,
1 − x2
−2
−4
2
2
(1 + x )
x = y,
y = −
τ = −2, Δ = 4, τ 2 − 4Δ = −12, and so (0, 0) is a stable spiral point.
25. In Problem 5, Section 10.1, we showed that (0, 0), ( 1/ , 0) and (− 1/ , 0) are the critical
points. We will use the Jacobian matrix
g (X) =
0
1
2
−1 + 3x 0
to classify these three critical points. For X = (0, 0), τ = 0 and Δ = 1 and we are unable to
classify this critical point. For (± 1/ , 0), τ = 0 and Δ = −2 and so both of these critical
points are saddle points.
26. In Problem 6, Section 10.1, we showed that (0, 0), (1/, 0), and (−1/, 0) are the critical
points. Since Dx x|x| = 2|x|, the Jacobian matrix is
g (X) =
0
1
.
2|x| − 1 0
For X = (0, 0), τ = 0, Δ = 1 and we are unable to classify this critical point. For (±1/, 0),
τ = 0, Δ = −1, and so both of these critical points are saddle points.
27. The corresponding plane autonomous system is
x = y,
and the Jacobian matrix is
⎛
y = −
(β + α2 y 2 )x
1 + α2 x2
0
⎜
g (X) = ⎝ (β + αy 2 )(α2 x2 − 1)
(1 + α2 x2 )2
1
⎞
⎟
−2α2 yx ⎠.
1 + α2 x2
For X = (0, 0), τ = 0 and Δ = β. Since β < 0, we can conclude that (0, 0) is a saddle point.
621
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SYSTEMS OF NONLINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
28. From x = −αx + xy = x(−α + y) = 0, either x = 0 or y = α. If x = 0, then 1 − βy = 0 and
so y = 1/β. The case y = α implies that 1 − βα − x2 = 0 or x2 = 1 − αβ. Since αβ > 1, this
equation has no real solutions. It follows that (0, 1/β) is the unique critical point. Since the
Jacobian matrix is
−α + y x
,
g (X) =
−2x
−β
τ = −α − β +
1 − αβ
1
= −β +
< 0 and Δ = αβ − 1 > 0. Therefore (0, 1/β) is a stable
β
β
critical point.
29. (a) The graphs of −x + y − x3 = 0 and −x − y + y 2 = 0 are shown
in the figure. The Jacobian matrix is
g (X) =
−1 −
−1
3x2
y
x = y2 – y
1
.
−1 + 2y
For X = (0, 0), τ = −2, Δ = 2, τ 2 − 4Δ = −4, and so (0, 0)
is a stable spiral point.
x1
x
y = x3 + x
(b) For X1 , Δ = −6.07 < 0 and so a saddle point occurs at X1 .
30. (a) The corresponding plane autonomous system is
1 3
x = y, y = y − y − x
3
and so the only critical point is (0, 0). Since the Jacobian matrix is
g (X) =
0
1
,
−1 (1 − y 2 )
τ = , Δ = 1, and so τ 2 − 4Δ = 2 − 4 at the critical point (0, 0).
(b) When τ = > 0, (0, 0) is an unstable critical point.
(c) When < 0 and τ 2 − 4Δ = 2 − 4 < 0, (0, 0) is a stable spiral point. These two
requirements can be written as −2 < < 0.
(d) When = 0, x + x = 0 and so x = c1 cos t + c2 sin t. Therefore all solutions are periodic
(with period 2π) and so (0, 0) is a center.
31. The differential equation dy/dx = y /x = −2x3 /y can be solved by separating variables. It
follows that y 2 + x4 = c. If X(0) = (x0 , 0) where x0 > 0, then c = x40 so that y 2 = x40 − x4 .
Therefore if −x0 < x < x0 , y 2 > 0 and so there are two values of y corresponding to each
value of x. Therefore the solution X(t) with X(0) = (x0 , 0) is periodic and so (0, 0) is a
center.
10.3
Linearization and Local Stability
623
32. The differential equation dy/dx = y /x = (x2 − 2x)/y can be solved by separating variables.
It follows that y 2 /2 = (x3 /3) − x2 + c and since X(0) = (x(0), x (0)) = (1, 0), c = 23 .
Therefore
x3 − 3x2 + 2
(x − 1)(x2 − 2x − 2)
y2
=
=
.
2
3
3
√
But (x − 1)(x2 − 2x − 2) > 0 for 1 − 3 < x < 1 and so each x in this interval has 2
corresponding values of y. therefore X(t) is a periodic solution.
33. (a) x = 2xy = 0 implies that either x = 0 or y = 0. If x = 0, then from 1 − x2 + y 2 = 0,
y 2 = −1 and there are no real solutions. If y = 0, 1 − x2 = 0 and so (1, 0) and (−1, 0)
are critical points. The Jacobian matrix is
g (X) =
2y 2x
−2x 2y
and so τ = 0 and Δ = 4 at either X = (1, 0) or (−1, 0). We obtain no information about
these critical points in this borderline case.
(b) The differential equation is
y
2
1− +
dy
= =
dx
x
2xy
or
2xy
y 2 /x,
x2
y2
1
–2
dy
= 1 − x2 + y 2 .
dx
2
–1
(1/x2 ) − 1
Letting μ =
it follows that dμ/dx =
and
2
so μ = −(1/x)−x+2c. Therefore y /x = −(1/x)−x+2c
which can be put in the form (x − c)2 + y 2 = c2 − 1. The
solution curves are shown and so both (1, 0) and (−1, 0)
are centers.
–2
624
CHAPTER 10
SYSTEMS OF NONLINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
34. (a) The differential equation is dy/dx = y /x = (−x − y 2 )/y = −(x/y) − y and so
dy/dx + y = −xy −1 .
(b) Let w = y 1−n = y 2 . It follows that dw/dx + 2w = −2x, a linear first order dif
ferential equation whose solution is y 2 = w = ce−2x + 12 − x . Since x(0) = 12 and
y(0) = x (0) = 0, 0 = c and so y 2 = 12 − x, a parabola with vertex at (1/2, 0). Therefore
the solution X(t) with X(0) = (1/2, 0) is not periodic.
35. The differential equation is dy/dx = y /x = (x3 − x)/y and so y 2 /2 = x4 /4 − x2 /2 + c or
y 2 = x4 /2 − x2 + c1 . Since x(0) = 0 and y(0) = x (0) = v0 , it follows that c1 = v02 and so
1
(x2 − 1)2 + 2v02 − 1
.
y 2 = x4 − x2 + v02 =
2
2
The x-intercepts on this graph satisfy
x = 1 ± 1 − 2v02
2
√
and so we must require that 1 − 2v02 ≥ 0 (or |v0 | ≤ 12 2 ) for real solutions to exist. If
x20 = 1 − 1 − 2v02 and −x0 < x < x0 , then (x2 − 1)2 + 2v02 − 1 > 0 and so there are
two corresponding values of y. Therefore X(t) with X(0) = (0, v0 ) is periodic provided that
√
|v0 | ≤ 12 2 .
36. The corresponding plane autonomous system is
x = y,
y = x2 − x + 1
and so the critical points must satisfy y = 0 and
x=
Therefore we must require that ≤
matrix
1
4
1±
√
1 − 4
.
2
for real solutions to exist. We will use the Jacobian
g (X) =
0 1
2x − 1 0
√
√
to attempt to classify ((1± 1 − 4 )/2, 0) when ≤ 1/4. Note that τ = 0 and Δ = ∓ 1 − 4 .
√
For X = ((1 + 1 − 4 )/2, 0) and < 1/4, Δ < 0 and so a saddle point occurs. For
√
X = ((1 − 1 − 4 )/2, 0), Δ ≥ 0 and we are not able to classify this critical point using
linearization.
10.3
Linearization and Local Stability
37. The corresponding plane autonomous system is
x = y,
y = −
β
α
R
x − x3 − y
L
L
L
where x = q and y = q . If X = (x, y) is a critical point, y = 0 and
−αx − βx3 = −x(α + βx2 ) = 0. If β > 0, α + βx2 = 0 has no real solutions and so (0, 0) is
the only critical point. Since
⎞
⎛
0
1
⎠
g (X) = ⎝ −α − 3βx2
R ,
−
L
L
τ = −R/L < 0 and Δ = α/L > 0. Therefore (0, 0) is a stable critical point. If β < 0,
(0, 0) and (±x̂, 0), where x̂2 = −α/β are critical points. At X(±x̂, 0), τ = −R/L < 0 and
Δ = −2α/L < 0. Therefore both critical points are saddles.
38. If we let dx/dt = y, then dy/dt = −x3 − x. From this we obtain the first-order differential
equation
dy/dt
x3 + x
dy
=
=−
.
dx
dx/dt
y
Separating variables and integrating we obtain
ˆ
ˆ
y dy = − (x3 + x) dx
and
1
1
1 2
y = − x4 − x2 + c1 .
2
4
2
Completing the square we can write the solution as y 2 = − 12 (x2 + 1)2 + c2 . If X(0) = (x0 , 0),
then c2 = 12 (x20 + 1)2 and so
1
1
x4 + 2x20 + 1 − x4 − 2x2 − 1
y 2 = − (x2 + 1)2 + (x20 + 1)2 = 0
2
2
2
=
(x2 + x2 + 2)(x20 − x2 )
(x20 + x2 )(x20 − x2 ) + 2(x20 − x2 )
= 0
.
2
2
Note that y = 0 when x = −x0 . In addition, the right-hand side is positive for −x0 < x < x0 ,
and so there are two corresponding values of y for each x between −x0 and x0 . The solution
X = X(t) that satisfies X(0) = (x0 , 0) is therefore periodic, and so (0, 0) is a center.
39. (a) Letting x = θ and y = x we obtain the system x = y and y = 1/2 − sin x. Since
sin π/6 = sin 5π/6 = 1/2 we see that (π/6, 0) and (5π/6, 0) are critical points of the
system.
(b) The Jacobian matrix is
g (X) =
0
1
− cos x 0
625
626
CHAPTER 10
SYSTEMS OF NONLINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
and so
A1 = g = ((π/6, 0)) =
0
1
√
− 3/2 0
and A2 = g = ((5π/6, 0)) =
√
0
1
3/2 0
.
Since det A1 > 0 and the trace of A1 is 0, no conclusion can be drawn regarding the
critical point (π/6, 0). Since det A2 < 0, we see that (5π/6, 0) is a saddle point.
(c) From the system in part (A) we obtain the first-order differential equation
1/2 − sin x
dy
=
.
dx
y
Separating variables and integrating we obtain
ˆ
ˆ 1
− sin x dx
y dy =
2
and
1 2 1
y = x + cos x + c1
2
2
or
y 2 = x + 2 cos x + c2 .
For x0 near π/6, if X(0) = (x0 , 0) then c2 = −x0 − 2 cos x0 and
y 2 = x + 2 cos x − x0 − 2 cos x0 . Thus, there are two values of y for each x in a sufficiently
small interval around π/6. Therefore (π/6, 0) is a center.
40. (a) Writing the system as x = x(x3 − 2y 3 ) and y = y(2x3 − y 3 ) we see that (0, 0) is a critical
point. Setting x3 − 2y 3 = 0 we have x3 = 2y 3 and 2x3 − y 3 = 4y 3 − y 3 = 3y 3 . Thus,
(0, 0) is the only critical point of the system.
(b) From the system we obtain the first-order differential equation
2x3 y − y 4
dy
= 4
dx
x − 2xy 3
or
(2x3 y − y 4 ) dx + (2xy 3 − x4 ) dy = 0
which is homogeneous. If we let y = ux it follows that
(2x4 u − x4 u4 ) dx + (2x4 u3 − x4 )(u dx + x du) = 0
x4 u(1 + u3 ) dx + x5 (2u3 − 1) du = 0
2u3 − 1
1
dx +
du = 0
x
u(u3 + 1)
1
1
2u − 1
1
dx +
− +
du = 0.
x
u + 1 u u2 − u + 1
10.4
Autonomous Systems as Mathematical Models
Integrating gives
ln |x| + ln |u + 1| − ln |u| + ln |u2 − u + 1| = c1
or
u+1
(u2 − u + 1) = c2
x
u
2
y+x
y
y
x
− + 1 = c2
y
x2 x
(xy + x2 )(y 2 − xy + x2 ) = c2 x2 y
xy 3 + x4 = c2 x2 y
x3 + y 2 = 3c3 xy.
(c) We see from the graph that (0, 0) is unstable. It is not
possible to classify the critical point as a node, saddle,
center, or spiral point.
10.4
Autonomous Systems as Mathematical Models
1. We are given that x(0) = θ(0) = π/3 and y(0) = θ (0) = w0 . Since y 2 = (2g/l) cos x + c,
w02 = (2g/l) cos (π/3) + c = g/l + c and so c = w02 − g/l. Therefore
l 2
2g
1
2
w
y =
cos x − +
l
2 2g 0
and the x-intercepts occur where cos x = 1/2 − (l/2g)w02 and so 1/2 − (l/2g)w02 must be
greater than −1 for solutions to exist. This condition is equivalent to |w0 | < 3g/l .
2. (a) Since y 2 = (2g/l) cos x + c, x(0) = θ(0) = θ0 and y(0) = θ (0) = 0, c = −(2g/l) cos θ0 and
so y 2 = 2g(cos θ − cos θ0 )/l. When θ = −θ0 , y 2 = 2g[cos (−θ0 ) − cos θ0 ]/l = 0. Therefore
y = dθ/dt = 0 when θ = −θ0 .
(b) Since y = dθ/dt and θ is decreasing between the time when θ = θ0 , t = 0, and θ = −θ0 ,
that is, t = T ,
2g dθ
=−
cos θ − cos θ0 .
dt
l
Therefore
1
l
dt
√
=−
dθ
2g cos θ − cos θ0
627
628
CHAPTER 10
SYSTEMS OF NONLINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
and so
T =−
l
2g
ˆ
θ=−θ0
θ=θ0
1
√
dθ =
cos θ − cos θ0
l
2g
ˆ
θ0
−θ0
√
1
dθ.
cos θ − cos θ0
3. The corresponding plane autonomous system is
x = y,
and
∂
∂x
f (x)
β
−g
− y
2
1 + [f (x)]
m
y = −g
= −g
f (x)
β
− y
2
1 + [f (x)]
m
(1 + [f (x)]2 )f (x) − f (x)2f (x)f (x)
.
(1 + [f (x)]2 )2
If X1 = (x1 , y1 ) is a critical point, y1 = 0 and f (x1 ) = 0. The Jacobian at this critical point
is therefore
⎞
⎛
0
1
⎟
⎜
g (X1 ) = ⎝
β ⎠.
−gf (x1 ) −
m
4. When β = 0 the Jacobian matrix is
0
−gf (x1 )
1
0
which has complex eigenvalues λ = ± gf (x1 ) i. The approximating linear system with
x (0) = 0 has solution
x(t) = x(0) cos gf (x1 ) t
and period 2π/ gf (x1 ) . Therefore p ≈ 2π/ gf (x1 ) for the actual solution.
5. (a) If f (x) = x2 /2, f (x) = x and so
x 1
y
dy
= = −g
.
dx
x
1 + x2 y
We can separate variables to show that y 2 = −g ln(1 + x2 ) + c. But x(0) = x0 and
y(0) = x (0) = v0 . Therefore c = v02 + g ln(1 + x20 ) and so
2
y =
Now
v02
− g ln
2
1 + x2
1 + x20
v02
− g ln
1 + x2
.
1 + x20
2
≥ 0 if and only if x2 ≤ ev0 /g (1 + x20 ) − 1.
Therefore, if |x| ≤ [ev0 /g (1 + x20 ) − 1]1/2 , there are two values of y for a given value of x
and so the solution is periodic.
10.4
Autonomous Systems as Mathematical Models
(b) Since z = x2 /2, the maximum height occurs at the largest value of x on the cycle. From
2
(A), xmax = [ev0 /g (1 + x20 ) − 1]1/2 and so
zmax =
x2max
1 2
= [ev0 /g (1 + x20 ) − 1].
2
2
6. (a) If f (x) = cosh x, f (x) = sinh x and [f (x)]2 + 1 = sinh2 x + 1 = cosh2 x. Therefore
y
sinh x 1
dy
= = −g
.
dx
x
cosh2 x y
We can separate variables to show that y 2 = 2g/ cosh x + c. But x(0) = x0 and y(0) =
x (0) = v0 . Therefore c = v02 − (2g/ cosh x0 ) and so
y2 =
Now
2g
2g
−
+ v02 .
cosh x cosh x0
2g
2g
−
+ v02 ≥ 0
cosh x cosh x0
if and only if
cosh x ≤
2g cosh x0
2g − v02 cosh x0
and the solution to this inequality is an interval [−a, a]. Therefore each x in (−a, a) has
two corresponding values of y and so the solution is periodic.
(b) Since z = cosh x, the maximum height occurs at the largest value of x on the cycle. From
(a), xmax = a where cosh a = 2g cosh x0 /(2g − v02 cosh x0 ). Therefore
zmax =
2g cosh x0
.
2g − v02 cosh x0
7. If xm < x1 < xn , then F (x1 ) > F (xm ) = F (xn ). Letting x = x1 ,
G(y) =
F (xm )G(a/b)
c0
=
< G(a/b).
F (x1 )
F (x1 )
Therefore from Property (ii ) on page 418 in this section of the text, G(y) = c0 /F (x1 ) has
two solutions y1 and y2 that satisfy y1 < a/b < y2 .
8. From Property (i ) on page 418 in this section of the text, when y = a/b, xn is taken on at
some time t. From Property (iii ), if x > xn there is no corresponding value of y. Therefore
the maximum number of predators is xn and xn occurs when y = a/b.
9. (a) In the Lotka–Volterra Model the average number of predators is d/c and the average
number of prey is a/b. But
x = −ax + bxy − 1 x = −(a + 1 )x + bxy
y = −cxy + dy − 2 y = −cxy + (d − 2 )y
and so the new critical point in the first quadrant is (d/c − 2 /c, a/b + 1 /b).
629
630
CHAPTER 10
SYSTEMS OF NONLINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
(b) The average number of predators d/c − 2 /c has decreased while the average number of
prey a/b + 1 /b has increased. The fishery science model is consistent with Volterra’s
principle.
x, y
10. (a) Solving
x(t)
10
x(−0.1 + 0.02y) = 0
y(t)
5
y(0.2 − 0.025x) = 0
20
40
60
80
100 t
in the first quadrant we obtain the
critical point (8, 5). The graphs are
plotted using x(0) = 7 and y(0) = 4.
(b) The graph in part (A) was obtained using NDSolve in Mathematica. We see that
the period is around 40. Since x(0) = 7, we use the FindRoot equation solver in
Mathematica to approximate the solution of x(t) = 7 for t near 40. From this we see
that the period is more closely approximated by t = 44.65.
11. Solving
x(20 − 0.4x − 0.3y) = 0
y(10 − 0.1y − 0.3x) = 0
we see that critical points are (0, 0), (0, 100), (50, 0), and (20, 40). The Jacobian matrix is
g (X) =
0.08(20 − 0.8x − 0.3y)
−0.024x
−0.018y
0.06(10 − 0.2y − 0.3x)
and so
A1 = g ((0, 0)) =
A3 = g ((50, 0)) =
1.6 0
0 0.6
−1.6 −1.2
0
−0.3
A2 = g ((0, 100)) =
−0.8
0
−1.8 −0.6
A4 = g ((20, 40)) =
−0.64 −0.48
.
−0.72 −0.24
Since det(A1 ) = Δ1 = 0.96 > 0, τ = 2.2 > 0, and τ12 − 4Δ1 = 1 > 0, we see that (0, 0) is
an unstable node. Since det(A2 ) = Δ2 = 0.48 > 0, τ = −1.4 < 0, and τ22 − 4Δ2 = 0.04 > 0,
we see that (0, 100) is a stable node. Since det(A3 ) = Δ3 = 0.48 > 0, τ = −1.9 < 0, and
τ32 − 4Δ3 = 1.69 > 0, we see that (50, 0) is a stable node. Since det(A4 ) = −0.192 < 0 we see
that (20, 40) is a saddle point.
12. Δ = r1 r2 , τ = r1 + r2 and τ 2 − 4Δ = (r1 + r2 )2 − 4r1 r2 = (r1 − r2 )2 . Therefore when r1 = r2 ,
(0, 0) is an unstable node.
10.4
Autonomous Systems as Mathematical Models
13. For X = (K1 , 0), τ = −r1 + r2 [1 − (K1 α21 /K2 )] and Δ = −r1 r2 [1 − (K1 α21 /K2 )]. If we let
c = 1 − K1 α21 /K2 , τ 2 − 4Δ = (cr2 + r1 )2 > 0. Now if k1 > K2 /α21 , c < 0 and so τ < 0,
Δ > 0. Therefore (K1 , 0) is a stable node. If K1 < K2 /α21 , c > 0 and so Δ < 0. In this case
(K1 , 0) is a saddle point.
14. (x̂, ŷ) is a stable node if and only if K1 /α12 > K2 and K2 /α21 > K1 . [See Figure 10.4.11(a)
in the text.] From Problem 12, (0.0) is an unstable node and from Problem 13, since
K1 < K2 /α21 , (K1 , 0) is a saddle point. Finally, when K2 < K1 /α12 , (0, K2 ) is a saddle
point. This is Problem 12 with the roles of 1 and 2 interchanged. Therefore (0, 0), (K1 , 0),
and (0, K2 ) are unstable.
15. K1 /α12 < K2 < K1 α21 and so α12 α21 > 1. Therefore Δ = (1 − α12 α21 )x̂ŷ r1 r2 /K1 K2 < 0
and so (x̂, ŷ) is a saddle point.
16. (a) The corresponding plane autonomous system is
x = y,
y =
β
−g
sin x −
y
l
ml
and so critical points must satisfy both y = 0 and sin x = 0. Therefore (±nπ, 0) are
critical points.
(b) The Jacobian matrix
⎛
0
1
⎞
⎝ g
β ⎠
− cos x −
l
ml
has trace τ = −β/ml and determinant Δ = g/l > 0 at (0, 0). Therefore
τ 2 − 4Δ =
β 2 − 4glm2
β2
g
=
−
4
.
m2 l2
l
m2 l2
We can conclude that (0, 0) is a stable spiral point provided
√
β 2 − 4glm2 < 0 or β < 2m gl .
17. (a) The corresponding plane autonomous system is
x = y,
y = −
k
β
y|y| − x
m
m
and so a critical point must satisfy both y = 0 and x = 0. Therefore (0, 0) is the unique
critical point.
(b) The Jacobian matrix is
⎛
0
1
⎞
⎟
⎠
β
k
− 2|y|
−
m
m
and so τ = 0 and Δ = k/m > 0. Therefore (0, 0) is a center, stable spiral point, or an
unstable spiral point. Physical considerations suggest that (0, 0) must be asymptotically
stable and so (0, 0) must be a stable spiral point.
⎜
⎝
631
632
CHAPTER 10
SYSTEMS OF NONLINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
18. (a) The magnitude of the frictional force between the bead and the wire is μ(mg cos θ) for
some μ > 0. The component of this frictional force in the x-direction is
(μmg cos θ) cos θ = μmg cos2 θ.
But
1
cos θ = 1 + [f (x)]2
and so
μmg cos2 θ =
μmg
.
1 + [f (x)]2
It follows from Newton’s Second Law that
mx = −mg
f (x)
μ
− βx + mg
2
1 + [f (x)]
1 + [f (x)]2
and so
x = g
μ − f (x)
β
− x .
2
1 + [f (x)]
m
(b) A critical point (x, y) must satisfy y = 0 and f (x) = μ. Therefore critical points occur
at (x1 , 0) where f (x1 ) = μ. The Jacobian matrix of the plane autonomous system is
⎛
⎞
0
1
⎜
⎟
g (X) = ⎝ (1 + [f (x)]2 )(−f (x)) − (μ − f (x))2f (x)f (x)
β⎠
−
g
(1 + [f (x)]2 )2
m
and so at a critical point X1 ,
⎛
0
⎜
g (X) = ⎝ −gf (x )
1
1 + μ2
1
⎞
⎟
β ⎠.
−
m
Therefore τ = −β/m < 0 and Δ = gf (x1 )/(1 + μ2 ). When f (x1 ) < 0, Δ < 0 and so
a saddle point occurs. When f (x1 ) > 0 and
τ 2 − 4Δ =
β2
f (x1 )
−
4g
< 0,
m2
1 + μ2
(x1 , 0) is a stable spiral point. This condition can also be written as
β 2 < 4gm2
f (x1 )
.
1 + μ2
19. We have dy/dx = y /x = −f (x)/y and so, using separation of variables,
ˆ x
y2
=−
f (μ) dμ + c
or
y 2 + 2F (x) = c.
2
0
We can conclude that for a given value of x there are at most two corresponding values of y.
If (0, 0) were a stable spiral point there would exist an x with more than two corresponding
values of y. Note that the condition f (0) = 0 is required for (0, 0) to be a critical point of the
corresponding plane autonomous system x = y, y = −f (x).
10.4
Autonomous Systems as Mathematical Models
20. (a) x = x(−a + by) = 0 implies that x = 0 or y = a/b. If x = 0, then, from
−cxy +
r
y(K − y) = 0,
K
y = 0 or K. Therefore (0, 0) and (0, K) are critical points. If ŷ = a/b, then
ŷ −cx +
r
(K − ŷ) = 0.
K
The corresponding value of x, x = x̂, therefore satisfies the equation cx̂ = r(K − ŷ)/K.
(b) The Jacobian matrix is
⎛
g (X) = ⎝
−a + by
−cy
bx
−cx +
⎞
⎠
r
(K − 2y)
K
and so at X1 = (0, 0), Δ = −ar < 0. For X1 = (0, K), Δ = n(Kb − a) = −rb(K − a/b).
Since we are given that K > a/b, Δ < 0 in this case. Therefore (0, 0) and (0, K) are
each saddle points. For X1 = (x̂, ŷ) where ŷ = a/b and cx̂ = r(K − ŷ)/K, we can write
the Jacobian matrix as
⎞
⎛
0
bx̂
g ((x̂, ŷ)) = ⎝
r ⎠
−cŷ − ŷ
K
and so τ = −rŷ/K < 0 and Δ = bcx̂ŷ > 0. Therefore (x̂, ŷ) is a stable critical point and
so it is either a stable node (perhaps degenerate) or a stable spiral point.
(c) Write
τ 2 − 4Δ =
2
2
r2 2
r
r
r
ŷ
−
4bcx̂ŷ
=
ŷ
ŷ
−
4bcx̂
=
ŷ
ŷ − 4b (K − ŷ)
K2
K2
K2
K
using
cx̂ =
r
r
r
(K − ŷ) =
ŷ
+ 4b ŷ − 4bK .
K
K
K
Therefore τ 2 − 4Δ < 0 if and only if
ŷ <
4bK
4bK 2
=
.
+ 4b
r + 4bK
r
K
Note that
4bK
4bK 2
=
·K ≈K
r + 4bK
r + 4bK
where K is large, and ŷ = a/b < K. Therefore τ 2 − 4Δ < 0 when K is large and a stable
spiral point will result.
633
634
CHAPTER 10
SYSTEMS OF NONLINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
21. The equation
y
x−x=x
x =α
1+y
αy
−1
1+y
=0
implies that x = 0 or y = 1/(α − 1). When α > 0, ŷ = 1/(α − 1) > 0. If x = 0, then from
the differential equation for y , y = β. On the other hand, if ŷ = 1/(α − 1), ŷ/(1 + ŷ) = 1/α
and so x̂/α − 1/(α − 1) + β = 0. It follows that
α
1
=
[(α − 1)β − 1]
x̂ = α β −
α−1
α−1
and if β(α − 1) > 1, x̂ > 0. Therefore (x̂, ŷ) is the unique critical point in the first quadrant.
The Jacobian matrix is
⎞
⎛
αx
y
−1
α
2
(1 + y) ⎟
⎜ y+1
⎟
g (X) = ⎜
⎠
⎝
−x
y
−
−
1
1+y
(1 + y)2
and for X = (x̂, ŷ), the Jacobian can be written in the form
⎞
⎛
(α − 1)2
x̂ ⎟
⎜ 0
α
⎟
⎜
g ((x̂, ŷ)) = ⎜
⎟.
⎠
⎝ 1
(α − 1)2
−
−
−
1
α
α2
It follows that
τ =−
(α − 1)2
x̂ + 1 < 0,
α2
Δ=
(α − 1)2
x̂
α2
and so τ = −(Δ + 1). Therefore τ 2 − 4Δ = (Δ + 1)2 − 4Δ = (Δ − 1)2 > 0. Therefore (x̂, ŷ)
is a stable node.
22. Letting y = x we obtain the plane autonomous system
x = y
y = −8x + 6x3 − x5 .
Solving x5 −6x3 +8x = x(x2 −4)(x2 −2) = 0 we see that crit√
√
ical points are (0, 0), (0, −2), (0, 2), (0, − 2 ), and (0, 2 ).
The Jacobian matrix is
g (X) =
−8 +
18x2
0 1
− 5x4 0
and we see that det(g (X)) = 5x4 − 18x2 + 8 and the trace of g (X) is 0. Since
√
√
det(g ((± 2 , 0))) = −8 < 0, (± 2 , 0) are saddle points. For the other critical points the
determinant is positive and linearization discloses no information. The graph of the phase
plane suggests that (0, 0) and (±2, 0) are centers.
Chapter 10 in Review
Chapter 10 in Review
1. True
2. True
3. A center or a saddle point
4. Complex with negative real parts
5. False; there are initial conditions for which lim X(t) = (0, 0).
t→∞
6. True
7. False; this is a borderline case. See Figure 10.3.7 in the text.
8. False; see Figure 10.4.2 in the text.
9. The system is linear and we identify Δ = −α and τ = α + 1. Since a critical point will be
a center when Δ > 0 and τ = 0 we see that for α = −1 critical points will be centered and
solutions will be periodic. Not also that when α = −1 the system is
x = −x − 2y
y = x + y
which does have an isolated critical point at (0, 0).
10. We identify g(x) = sin x in Theorem 10.3.1. Then x1 = nπ is a critical point for n an integer
and g (nπ) = cos nπ < 0 when n is an odd integer. Thus, nπ is an asymptotically stable
critical point when n is an odd integer.
11. Switching to polar coordinates,
1
dx
dy
1
dr
=
x
+y
= (−xy − x2 r3 + xy − y 2 r3 ) = −r4
dt
r
dt
dt
r
dx
dθ
1
dy
1
= 2 −y
+x
= 2 (y 2 + xyr3 + x2 − xyr3 ) = 1.
dt
r
dt
dt
r
1
and θ = t+c2 . Since X(0) = (1, 0),
3t + c1
r = 1 and θ = 0. It follows that c1 = 1, c2 = 0, and so
Using separation of variables it follows that r = √
3
r= √
3
1
,
3t + 1
θ = t.
As t → ∞, r → 0 and the solution spirals toward the origin.
635
636
CHAPTER 10
SYSTEMS OF NONLINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
12. (a) If X(0) = X0 lies on the line y = −2x, then X(t) approaches (0, 0) along this line. For
all other initial conditions, X(t) approaches (0, 0) from the direction determined by the
line y = x.
(b) If X(0) = X0 lies on the line y = −x, then X(t) approaches (0, 0) along this line. For all
other initial conditions, X(t) becomes unbounded and y = 2x serves as an asymptote.
13. (a) τ = 0, Δ = 11 > 0 and so (0, 0) is a center.
(b) τ = −2, Δ = 1, τ 2 − 4Δ = 0 and so (0, 0) is a degenerate stable node.
14. From x = x(1 + y − 3x) = 0, either x = 0 or 1 + y − 3x = 0. If x = 0, then, from
y(4 − 2x − y) = 0 we obtain y(4 − y) = 0. It follows that (0, 0) and (0, 4) are critical points.
If 1 + y − 3x = 0, then y(5 − 5x) = 0. Therefore (1/3, 0) and (1, 2) are the remaining critical
points. We will use the Jacobian matrix
g (X) =
1 + y − 6x
x
−2y
4 − 2x − 2y
to classify these four critical points. The results are as follows:
X
τ
Δ
τ 2 − 4Δ
(0, 0)
5
4
9
unstable node
(0, 4)
1 3, 0
−
−20
−
saddle point
−
− 10
3
−
saddle point
(1, 2)
−5
10
−15
Conclusion
stable spiral point
15. From x = r cos θ, y = r sin θ we have
dθ dr
dx
= −r sin θ
+
cos θ
dt
dt
dt
dy
dθ dr
= r cos θ
+
sin θ.
dt
dt
dt
Then r = αr, θ = 1 gives
dx
= −r sin θ + αr cos θ
dt
dy
= r cos θ + αr sin θ.
dt
We see that r = 0, which corresponds to X = (0, 0), is a critical point. Solving r = αr we
have r = c1 eαt . Thus, when α < 0, limt→∞ r(t) = 0 and (0, 0) is a stable critical point. When
α = 0, r = 0 and r = c1 . In this case (0, 0) is a center, which is stable. Therefore, (0, 0) is a
stable critical point for the system when α ≤ 0.
Chapter 10 in Review
16. The corresponding plane autonomous system is x = y, y = μ(1−x2 )y−x and so the Jacobian
at the critical point (0, 0) is
0 1
.
g ((0, 0)) =
−1 μ
Therefore τ = μ, Δ = 1 and τ 2 − 4Δ = μ2 − 4. Now μ2 − 4 < 0 if and only if −2 < μ < 2.
We may therefore conclude that (0, 0) is a stable node for μ < −2, a stable spiral point for
−2 < μ < 0, an unstable spiral point for 0 < μ < 2, and an unstable node for μ > 2.
In the case that μ = 2, (0, 0) is unstable but is a borderline case that may be an unstable
spiral, unstable node, or degenerate unstable node. Similarly, in the case that μ = −2, (0, 0)
is but is a borderline case that may be a stable spiral, a stable node, or a degenerate stable
node.
17. Critical points occur at x = ±1. Since
1
g (x) = − e−x/2 (x2 − 4x − 1),
2
g (1) > 0 and g (−1) < 0. Therefore x = 1 is unstable and x = −1 is asymptotically stable.
y
−2x y 2 + 1
dy
= =
. We may separate variables to show that y 2 + 1 = −x2 + c. But
18.
dx
x
y
x(0) = x0 and y(0) = x (0) = 0. It follows that c = 1 + x20 so that
y 2 = (1 + x20 − x2 )2 − 1.
Note that 1+x20 −x2 > 1 for −x0 < x < x0 and y = 0 for x = ±x0 . Each x with −x0 < x < x0
has two corresponding values of y and so the solution X(t) with X(0) = (x0 , 0) is periodic.
19. The corresponding plane autonomous system is
x = y,
y = −
k
β
y − (s + x)3 + g
m
m
⎛
and so the Jacobian is
0
1
⎞
⎟
⎜
g (X) = ⎝ 3k
β ⎠.
2
− (s + x) −
m
m
For X = (0, 0), τ = −
3k 2
β
< 0, Δ =
s > 0.
m
m
Therefore
τ 2 − 4Δ =
β2
12k 2
1
s = 2 (β 2 − 12kms2 ).
−
2
m
m
m
Therefore (0, 0) is a stable node if β 2 > 12kms2 and a stable spiral point provided
β 2 < 12kms2 , where ks3 = mg.
637
638
CHAPTER 10
SYSTEMS OF NONLINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
20. (a) If (x, y) is a critical point, y = 0 and so sin x(ω 2 cos x − g/l) = 0. Either sin x = 0 (in
which case x = 0) of cos x = g/ω 2 l. But if ω 2 < g/l, g/ω 2 l > 1 and so the latter
equation has no real solutions. Therefore (0, 0) is the only critical point if ω 2 < g/l. The
Jacobian matrix is
⎞
⎛
0
1
⎟
⎜
g (X) = ⎝
β ⎠
g
2
ω cos 2x − cos x −
l
ml
and so τ = −β/ml < 0 and Δ = g/l − ω 2 > 0 for X = (0, 0). It follows that (0, 0)
is asymptotically stable and so after a small displacement, the pendulum will return to
θ = 0, θ = 0.
(b) If ω 2 > g/l, cos x = g/ω 2 l will have two solutions x = ±x̂ that satisfy −π < x < π.
Therefore (±x̂, 0) are two additional critical points. If X1 = (0, 0), Δ = g/l − ω 2 < 0
and so (0, 0) is a saddle point. If X1 = (±x̂, 0), τ = −β/ml < 0 and
g2
g2
g2
g
2
2
Δ = cos x̂ − ω cos 2x̂ = 2 2 − ω 2 4 2 − 1 = ω 2 − 2 2 > 0.
l
ω l
ω l
ω l
Therefore (x̂, 0) and (−x̂, 0) are each stable. When θ(0) = θ0 , θ (0) = 0 and θ0 is small
we expect the pendulum to reach one of these two stable equilibrium positions.
Chapter 11
Fourier Series
11.1
Orthogonal Functions
ˆ
1.
2
1
xx2 dx = x4 = 0
4 −2
2
−2
1
1
1 6 1 4 x (x + 1) dx = x + x = 0
6 4 −1
ˆ
2.
1
3
2
−1
ˆ
2
3.
ex (xe−x − e−x ) dx =
0
ˆ
0
ˆ
5.
(x − 1) dx =
2
1 2
x −x =0
2
0
π
1
cos x sin2 x dx = sin3 x = 0
3
0
π/2
1
x cos 2x dx =
2
−π/2
ˆ
2
0
π
4.
ˆ
−1
5π/4
6.
ex sin x dx =
π/4
π/2
1
cos 2x + x sin 2x =0
2
−π/2
5π/4
1 x
1 x
e sin x − e cos x =0
2
2
π/4
7. For m = n
ˆ
0
π/2
ˆ
1 π/2
sin(2n + 1)x sin(2m + 1)x dx =
(cos 2(n − m)x − cos 2(n + m + 1)x) dx
2 0
π/2
π/2
1
sin 2(n − m)x
sin 2(n + m + 1)x
=
−
= 0.
4(n − m)
4(n + m + 1)
0
639
0
640
CHAPTER 11
FOURIER SERIES
For m = n
ˆ
π/2
π/2 ˆ
sin2 (2n + 1)x dx =
0
0
1 1
− cos 2(2n + 1)x dx
2 2
π/2
π/2
1
π
1 sin 2(2n + 1)x
−
=
= x
2 4(2n + 1)
4
0
0
so that
1√
π.
2
sin (2n + 1)x =
8. For m = n
ˆ
π/2
0
ˆ
1 π/2
cos (2n + 1)x cos (2m + 1)x dx =
(cos 2(n − m)x + cos 2(n + m + 1)x) dx
2 0
π/2
π/2
1
1
sin 2(n − m)x
sin 2(n + m + 1)x
=
+
= 0.
4(n − m)
4(n + m + 1)
0
0
For m = n
ˆ
π/2
π/2 ˆ
cos2 (2n + 1)x dx =
0
0
1 1
+ cos 2(2n + 1)x dx
2 2
π/2
π/2
1 1
π
= x
sin 2(2n + 1)x
+
=
2 4(2n + 1)
4
0
0
so that
cos (2n + 1)x =
1√
π.
2
9. For m = n
ˆ
ˆ π
1 π
sin nx sin mx dx =
(cos(n − m)x − cos(n + m)x) dx
2 0
0
π
π
1
1
sin (n − m)x −
sin (n + m)x = 0.
=
2(n − m)
2(n + m)
0
0
For m = n
ˆ
0
π
ˆ
π
sin2 nx dx =
0
π
π
1 1 1
1
π
− cos 2nx dx = x −
sin 2nx =
2 2
2
4n
2
0
so that
sin nx =
π
.
2
0
11.1
Orthogonal Functions
10. For m = n
ˆ
p
0
p
(n − m)π
(n + m)π
cos
x − cos
x dx
p
p
0
p
p
(n − m)π (n + m)π p
p
sin
x −
sin
x = 0.
=
2(n − m)π
p
2(n + m)π
p
mπ
1
nπ
x sin
x dx =
sin
p
p
2
ˆ
0
0
For m = n
ˆ
p
0
nπ
sin
x dx =
p
p
ˆ
2
0
p
p
1 1
p
p
2nπ
1 2nπ − cos
x dx = x −
sin
x =
2 2
p
2 4nπ
p 2
0
so that
0
sin nπ x = p .
p 2
11. For m = n
ˆ
p
cos
0
p
(n − m)π
(n + m)π
x + cos
x dx
p
p
0
p
p
(n − m)π (n + m)π p
p
sin
x +
sin
x = 0.
=
2(n − m)π
p
2(n + m)π
p
nπ
mπ
1
x cos
x dx =
p
p
2
ˆ
cos
0
0
For m = n
ˆ
0
Also
p
nπ
x dx =
cos2
p
ˆ
p
0
ˆ
0
p
p
p
2nπ
1 2nπ p
p
1 1
+ cos
x dx = x +
sin
x = .
2 2
p
2 4nπ
p 2
0
p
p
nπ nπ
x dx =
sin
x = 0 and
1 · cos
p
nπ
p 0
so that
1 =
√
p and
0
ˆ
p
12 dx = p
0
cos nπ x = p .
p 2
12. For m = n, we use Problems 11 and 10:
ˆ p
nπ
nπ
mπ
mπ
cos
cos
x cos
x dx = 2
x cos
x dx = 0
p
p
p
p
−p
0
ˆ p
ˆ p
nπ
mπ
nπ
mπ
x sin
x dx = 2
x sin
x dx = 0.
sin
sin
p
p
p
p
−p
0
ˆ
p
641
642
CHAPTER 11
Also
ˆ
FOURIER SERIES
p
mπ
1
nπ
x cos
x dx =
sin
p
p
2
−p
(n − m)π
(n + m)π
sin
x + sin
x dx = 0,
p
p
−p
p
ˆ p
p
nπ nπ
x dx =
sin
x = 0,
1 · cos
p
nπ
p −p
ˆ
p
−p
p
ˆ p
nπ
p
nπ x dx = −
cos
x = 0,
1 · sin
p
nπ
p −p
−p
and
ˆ
p
nπ
nπ
x cos
x dx =
sin
p
p
−p
For m = n
ˆ
p
2nπ
−
2nπ 1
sin
x dx = 4nπ cos
x = 0.
2
p
p
p p
−p
−p
2nπ
1 1
+ cos
x dx = p,
2
p
−p
−p 2
ˆ p
ˆ p
2nπ
1 1
2 nπ
x dx =
− cos
x dx = p,
sin
p
2
p
−p
−p 2
ˆ
p
cos2
nπ
x dx =
p
ˆ
ˆ
and
p
p
12 dx = 2p
−p
so that
1 = 2p ,
13. Since
ˆ
∞
∞
−x2
e
−∞
e−x · 1 · 2x dx = 0,
2
−∞
ˆ
√
nπ
cos
x
= p,
p ˆ
∞
· 1 · (4x − 2) dx = 2
2
ˆ
and
←− integrand is an odd function
e−x (4x2 − 2) dx
2
0
∞
−x2
x 2xe
=4
√
nπ
sin
x
= p.
p 0
←− integrand is an even function
ˆ
dx − 4
∞
e−x dx
2
0
∞ ˆ
∞
2
2
e−x dx
= 4 −xe−x +
0
0
−x2
= −4xe
ˆ
−4
∞
e−x dx
2
0
∞
ˆ ∞
ˆ ∞
2
−x2
e
dx − 4
e−x dx = 0
+4
0
0
0
and
ˆ
∞
−∞
e−x · 2x · (4x2 − 2) dx = 0,
2
the functions are orthogonal.
←− integrand is an odd function
11.1
14. Since
ˆ
∞
0
ˆ
∞
e−x · 1 ·
0
Orthogonal Functions
∞ ˆ
∞
e−x · 1(1 − x) dx = (x − 1)e−x −
e−x dx = 0,
0
0
∞ ˆ
∞
1 2
1 2 −x x − 2x + 1 dx = 2x − 1 − x e +
e−x (x − 2) dx
2
2
0
0
∞ ˆ
∞
e−x dx = 0,
= 1 + (2 − x)e−x +
0
0
and
ˆ
∞
−x
e
· (1 − x)
0
ˆ ∞
1 3 5 2
1 2
−x
x − 2x + 1 dx =
− x + x − 3x + 1 dx
e
2
2
2
0
∞ ˆ ∞
3 2
5 2
−x 1 3
−x
− x + 5x − 3 dx
x − x + 3x − 1 +
e
=e
2
2
2
0
0
∞ ˆ ∞
3
x2 − 5x + 3 +
e−x (5 − 3x) dx
= 1 + e−x
2
0
0
∞
ˆ ∞
−x
= 1 − 3 + e (3x − 5) − 3
e−x dx = 0,
0
0
the functions are orthogonal.
´b
´b
15. By orthogonality a φ0 (x)φn (x)dx = 0 for n = 1, 2, 3, . . . ; that is, a φn (x) dx = 0 for n = 1,
2, 3, . . . .
16. Using the facts that φ0 and φ1 are orthogonal to φn for n > 1, we have
ˆ b
ˆ b
ˆ b
(αx + β)φn (x) dx = α
xφn (x) dx + β
1 · φn (x) dx
a
a
ˆ
a
ˆ
b
b
φ1 (x)φn (x) dx + β
=α
a
φ0 (x)φn (x) dx
a
=α·0+β·0=0
for n = 2, 3, 4, . . . .
17. Using the fact that φn and φm are orthogonal for n = m we have
ˆ b
ˆ b
2
2
φ2m (x) + 2φm (x)φn (x) + φ2n (x) dx
[φm (x) + φn (x)] dx =
φm (x) + φn (x) =
a
ˆ
=
a
a
b
ˆ
φ2m (x) dx + 2
ˆ
b
φm (x)φn (x) dx +
a
= φm (x)2 + φn (x)2 .
a
b
φ2n (x) dx
643
644
CHAPTER 11
FOURIER SERIES
18. Setting
ˆ
0=
−2
and
ˆ
2
f3 (x)f1 (x) dx =
ˆ
−2
−2
ˆ
2
0=
2
f3 (x)f2 (x) dx =
16 64
+ c2
x2 + c1 x3 + c2 x4 dx =
3
5
3
64
x + c1 x4 + c2 x5 dx = c1
5
−2
2
we obtain c1 = 0 and c2 = −5/12.
19. The fundamental period is 2π/2π = 1.
20. The fundamental period is 2π/(4/L) = 12 πL.
21. The fundamental period of sin x + sin 2x is 2π.
22. The fundamental period of sin 2x + cos 4x is 2π/2 = π.
23. The fundamental period of sin 3x + cos 2x is 2π since the smallest integer multiples of 2π/3
and 2π/2 = π that are equal are 3 and 2, respectively.
24. The fundamental period of sin2 πx is 1 since
sin π(x + 1) = sin (πx + π) = sin πx cos π + cos πx sin π = − sin πx
sin2 π(x + 1) = (− sin πx)2 = sin2 πx.
Alternatively, we can also use the trigonometric identity sin2 πx = 12 (1 − cos 2πx).
25. (a) For m and n positive integers, we have orthogonality on the interval [−π, π] because
ˆ
ˆ π
1 π
sin nx sin mx dx =
[cos (m − n)x − cos (m + n)x] dx
2 −π
−π
1 sin (m − n)x sin (m + n)x π
−
= 0, m = n
=
2
m−n
m+n
−π
(b) The function f (x) = 1 is continuous on [−π, π] but is orthogonal to every function in the
orthogonal set:
ˆ π
cos π − cos (−π)
= 0, n = 1, 2, 3, . . .
1 · sin nx dx =
n
π
Therefore the set {sin nx}, n = 1, 2, 3, . . . is not complete on the interval [−π, π].
26. (a) Following the pattern established by φ1 (x) and φ2 (x) we have
φ3 (x) = f3 (x) −
(f3 , φ0 )
(f3 , φ1 )
(f3 , φ2 )
φ0 (x) −
φ1 (x) −
φ2 (x).
(φ0 , φ0 )
(φ1 , φ1 )
(φ2 , φ2 )
11.1
Orthogonal Functions
645
(b) To show mutual orthogonality we compute (φ0 , φ1 ), (φ0 , φ2 ), and (φ1 , φ2 ) using properties
(i), (ii), (iii), and (iv) from this section in the text.
(f1 , φ0 )
(f1 , φ0 )
φ0 = (φ0 , f1 ) −
(φ0 , φ0 ) = (φ0 , f1 ) − (f1 , φ0 ) = 0
(φ0 , φ1 ) = φ0 , f1 −
(φ0 , φ0 )
(φ0 , φ0 )
(f2 , φ0 )
(f2 , φ1 )
(f2 , φ0 )
(f2 , φ1 )
φ0 −
φ1 = (φ0 , f2 ) −
(φ0 , φ0 ) −
(φ0 , φ1 )
(φ0 , φ2 ) = φ0 , f2 −
(φ0 , φ0 )
(φ1 , φ1 )
(φ0 , φ0 )
(φ1 , φ1 )
= (φ0 , f2 ) − (f2 , φ0 ) − 0 = 0
(f2 , φ0 )
(f2 , φ1 )
(f2 , φ0 )
(f2 , φ1 )
φ0 −
φ1 = (φ1 , f2 ) −
(φ1 , φ0 ) −
(φ1 , φ1 )
(φ1 , φ2 ) = φ1 , f2 −
(φ0 , φ0 )
(φ1 , φ1 )
(φ0 , φ0 )
(φ1 , φ1 )
= (φ1 , f2 ) − 0 − (f2 , φ1 ) = 0.
27. First we identify f0 (x) = 1, f1 (x) = x, f2 (x) = x2 , and f3 (x) = x3 . Then, we use the formulas
from Problem 26. First, we have φ0 (x) = f0 (x) = 1. Then
ˆ 1
ˆ 1
(f1 , φ0 ) = (x, 1) =
x dx = 0 and (φ0 , φ0 ) =
1 dx = 2,
−1
so
φ1 (x) = f1 (x) −
Next
ˆ
1
−1
(f1 , φ0 )
0
φ0 (x) = x − (1) = x.
(φ0 , φ0 )
2
2
x dx = , (f2 , φ1 ) = (x2 , x) =
(f2 , φ0 ) = (x , 1) =
3
−1
ˆ 1
2
and (φ1 , φ1 ) =
x2 dx = ,
3
−1
2
ˆ
1
2
x3 dx = 0,
−1
so
φ2 (x) = f2 (x) −
0
(f2 , φ0 )
(f2 , φ1 )
2/3
1
φ0 (x) −
φ1 (x) = x2 −
(1) − (x) = x2 − .
(φ0 , φ0 )
(φ1 , φ1 )
2
2
3
Finally,
ˆ
(f3 , φ0 ) = (x3 , 1) =
1
−1
and
ˆ
x3 dx = 0,
1
x ,x −
3
3
(f3 , φ2 ) =
2
(f3 , φ1 ) = (x3 , x) =
1
2
x4 dx = ,
5
−1
1 3
5
x − x
dx = 0,
=
3
−1
ˆ
1
so
φ3 (x) = f3 (x) −
(f3 , φ0 )
(f3 , φ1 )
(f3 , φ2 )
φ0 (x) −
φ1 (x) −
φ2 (x)
(φ0 , φ0 )
(φ1 , φ1 )
(φ2 , φ2 )
= x3 − 0 −
2/5
3
(x) − 0 = x3 − x.
2/3
5
646
CHAPTER 11
FOURIER SERIES
28. Let Pn (x) denote the Legendre polynomials given in Section 6.4. The entries in the orthogonal set found by the Gram-Schmidt process in Problem 27 are proportional to the Legenre
polynomials.
φ0 = P0 = 1
φ1 = P1 = x
1 2
3
φ2 = P2 =
3x − 1
2
2
1 3
5
φ3 = P3 =
5x − 3x
2
2
11.2
1. a0 =
an =
Fourier Series
1
π
1
π
1
bn =
π
f (x) =
ˆ
π
f (x) dx =
−π
ˆ
π
f (x) cos
ˆ
−π
1
π
ˆ
π
1 dx = 1
0
1
nπ
x dx =
f (x) sin
π
π
−π
1 1
+
2 π
Converges to
1
2
0
π
1
1
(1 − cos nπ) =
[1 − (−1)n ]
nπ
nπ
sin nx dx =
0
1 − (−1)n
sin nx
n
at x = 0.
ˆ
ˆ
1 0
1 π
f (x) dx =
(−1) dx +
2 dx = 1
π −π
π 0
−π
ˆ
ˆ
ˆ
1 π
1 0
1 π
f (x) cos nx dx =
(− cos nx) dx +
2 cos nx dx = 0
an =
π −π
π −π
π 0
ˆ
ˆ
ˆ
1 π
1 0
1 π
3
[1 − (−1)n ]
f (x) sin nx dx =
(− sin nx) dx +
2 sin nx dx =
bn =
π −π
π −π
π 0
nπ
1
2. a0 =
π
ˆ
n=1
π
cos nx dx = 0
ˆ
π
∞
ˆ
1
nπ
x dx =
pi
π
π
∞
1 3 1 − (−1)n
sin nx
f (x) = +
2 π
n
n=1
Converges to
ˆ
3. a0 =
at x = 0.
ˆ
1
−1
an =
ˆ
0
f (x) dx =
ˆ
1 dx +
−1
1
0
ˆ
−1
ˆ
0
−1
1
x cos nπx dx =
0
ˆ
0
f (x) sin nπx dx =
−1
3
2
cos nπx dx +
ˆ
1
1
x dx =
f (x) cos nπx dx =
ˆ
bn =
1
2
1
sin nπx dx +
−1
0
1
n2 π 2
x sin nπx dx = −
1
nπ
[(−1)n − 1]
11.2
f (x) =
Fourier Series
∞ 3 (−1)n − 1
1
+
sin
nπx
cos
nπx
−
4
n2 π 2
nπ
n=1
Converges to 12 at x = 0.
ˆ 1
ˆ 1
1
f (x) dx =
x dx =
4. a0 =
2
−1
0
ˆ 1
ˆ 1
an =
f (x) cos nπx dx =
x cos nπx dx =
ˆ
bn =
−1
ˆ
1
0
1
f (x) sin nπx dx =
−1
f (x) =
1
+
4
x sin nπx dx =
0
∞ n=1
1
n2 π 2
[(−1)n − 1]
(−1)n+1
nπ
(−1)n − 1
(−1)n+1
sin nπx
cos nπx +
2
2
n π
nπ
f is continous on the interval
ˆ
ˆ
1 π
1 π 2
1
f (x) dx =
x dx = π 2
5. a0 =
π −π
π 0
3
ˆ π
ˆ π
1
1
1
an =
f (x) cos nx dx =
x2 cos nx dx =
π −π
π 0
π
2(−1)n
n2
ˆ π
1
1
bn =
x2 sin nx dx =
π 0
π
π
ˆ
x2
2 π
sin nx −
x sin nx dx
π
n 0
0
=
π
ˆ
x2
2 π
− cos nx +
x cos nx dx
n
n 0
0
2
π
(−1)n+1 + 3 [(−1)n − 1]
n
n π
∞
π
2[(−1)n − 1]
π 2 2(−1)n
n+1
+
(−1)
sin nx
cos nx +
+
f (x) =
6
n2
n
n3 π
=
n=1
f is continous on the interval
ˆ
ˆ
ˆ
1 π
1 0 2
1 π 2
5
f (x) dx =
π dx +
π − x2 dx = π 2
6. a0 =
π −π
π −π
π 0
3
ˆ
ˆ
ˆ
1 π
1 0 2
1 π 2
π − x2 cos nx dx
an =
f (x) cos nx dx =
π cos nx dx +
π −π
π −π
π 0
π
ˆ
2 π
2
1 π 2 − x2
sin nx +
x sin nx dx = 2 (−1)n+1
=
π
n
n 0
n
0
ˆ π
ˆ
ˆ
1
1 0 2
1 π 2
π − x2 sin nx dx
bn =
f (x) sin nx dx =
π sin nx dx +
π −π
π −π
π 0
π
ˆ
1 x2 − π 2
2 π
π
2
π
n
cos nx −
x cos nx dx = (−1)n + 3 [1 − (−1)n ]
= [(−1) − 1] +
n
π
n
n 0
n
n π
0
647
648
CHAPTER 11
f (x) =
FOURIER SERIES
∞ 5π 2 2
π
2[1 − (−1)n ]
n+1
n
+
(−1)
sin
nx
(−1)
cos
nx
+
+
6
n2
n
n3 π
n=1
f is continous on the interval
ˆ
ˆ
1 π
1 π
f (x) dx =
(x + π) dx = 2π
7. a0 =
π −π
π −π
ˆ
ˆ
1 π
1 π
an =
f (x) cos nx dx =
(x + π) cos nx dx = 0
π −π
π −π
ˆ
1 π
2
f (x) sin nx dx = (−1)n+1
bn =
π −π
n
f (x) = π +
∞
2
(−1)n+1 sin nx
n
n=1
f is continous on the interval
ˆ
ˆ
1 π
1 π
f (x) dx =
(3 − 2x) dx = 6
8. a0 =
π −π
π −π
ˆ
ˆ
1 π
1 π
an =
f (x) cos nx dx =
(3 − 2x) cos nx dx = 0
π −π
π −π
ˆ
1 π
4
(3 − 2x) sin nx dx = (−1)n
bn =
π −π
n
f (x) = 3 + 4
∞
(−1)n
n=1
n
sin nx
f is continous on the interval
ˆ
ˆ
1 π
1 π
2
9. a0 =
f (x) dx =
sin x dx =
π −π
π 0
π
ˆ π
ˆ π
ˆ π
1
1
1
f (x) cos nx dx =
sin x cos nx dx =
(sin (1 + n)x + sin (1 − n)x) dx
an =
π −π
π 0
2π 0
1 + (−1)n
forn = 2, 3, 4, . . .
π(1 − n2 )
ˆ π
1
a1 =
sin 2x dx = 0
2π 0
ˆ
ˆ
1 π
1 π
bn =
f (x) sin nx dx =
sin x sin nx dx
π −π
π 0
ˆ π
1
(cos (1 − n)x − cos (1 + n)x) dx = 0 forn = 2, 3, 4, . . .
=
2π 0
ˆ π
1
1
b1 =
(1 − cos 2x) dx =
2π 0
2
=
∞
1 + (−1)n
1 1
cos nx
f (x) = + sin x +
π 2
π(1 − n2 )
n=2
f is continous on the interval
11.2
10. a0 =
an =
=
ˆ
2
π
f (x) dx =
−π/2
2
π
1
π
2
bn =
π
1
=
π
π/2
ˆ
2
π
ˆ
π/2
cos x dx =
0
π/2
f (x) cos 2nx dx =
−π/2
ˆ
π/2
2
π
ˆ
2
π
π/2
cos x cos 2nx dx
0
(cos (2n − 1)x + cos (2n + 1)x) dx =
0
ˆ
π/2
2
f (x) sin 2nx dx =
π
−π/2
ˆ
π/2
Fourier Series
ˆ
2(−1)n+1
π(4n2 − 1)
π/2
cos x sin 2nx dx
0
(sin (2n − 1)x + sin (2n + 1)x) dx =
0
4n
π(4n2 − 1)
∞ 4n
1 2(−1)n+1
cos 2nx +
sin 2nx
f (x) = +
π
π(4n2 − 1)
π(4n2 − 1)
n=1
Converges to
1
2
at x = 0.
ˆ 0
ˆ 1
1
1
f (x) dx =
−2 dx +
1 dx = −
2
2
−2
−1
0
ˆ 0
ˆ
ˆ 1
1 2
nπ
1
nπ
nπ
1
nπ
x dx =
x dx +
x dx = −
sin
f (x) cos
(−2) cos
cos
an =
2 −2
2
2
2
2
nπ
2
−1
0
ˆ 0
ˆ
ˆ 1
1
3 1 2
nπ
nπ
nπ
nπ
x dx =
x dx +
x dx =
f (x) sin
(−2) sin
sin
bn =
1 − cos
2 −2
2
2
2
2
nπ
2
−1
0
1
11. a0 =
2
ˆ
2
∞ 1
nπ
nπ
3 nπ
nπ
1 −
sin
cos
x+
1 − cos
sin
x
f (x) = − +
4
nπ
2
2
nπ
2
2
n=1
Converges to −1 at x = −1, − 12 at x = 0, and
12. a0 =
1
2
ˆ
2
f (x) dx =
−2
1
2
ˆ
ˆ
1
x dx +
0
2
1 dx
1
=
at x = 1.
3
4
ˆ 2
1
1
2 nπ
nπ
nπ
nπ
x dx =
x dx +
x dx = 2 2 cos
−1
f (x) cos
x cos
cos
2
2
2
2
n π
2
−2
0
1
ˆ 1
ˆ 2
ˆ
1
1 2
nπ
nπ
nπ
x dx =
x dx +
x dx
bn =
f (x) sin
x sin
sin
2 −2
2
2
2
2
0
1
nπ nπ
2 +
(−1)n+1
= 2 2 sin
n π
2
2
∞ nπ
nπ nπ
nπ
nπ
2 2 3 n+1
cos
sin
− 1 cos
x + 2 2 sin
+
(−1)
x
f (x) = +
8
n2 π 2
2
2
n π
2
2
2
1
an =
2
ˆ
ˆ
1
2
2
n=1
f is continous on the interval
649
650
CHAPTER 11
FOURIER SERIES
ˆ 0
ˆ 5
1
9
f (x) dx =
1 dx +
(1 + x) dx =
5
2
−5
−5
0
ˆ 0
ˆ 5
ˆ 5
1
nπ
1
nπ
nπ
x dx =
x dx +
x dx
f (x) cos
cos
(1 + x) cos
an =
5 −5
5
5
5
5
−5
0
1
13. a0 =
5
=
ˆ
5
5
n2 π 2
[(−1)n − 1]
ˆ 0
ˆ 5
1
5
nπ
nπ
nπ
x dx =
x dx +
x dx =
(−1)n+1
f (x) sin
sin
(1 + x) cos
5
5
5
5
nπ
−5
−5
0
∞ 5
5
9 nπ
nπ
n
n+1
x+
(−1)
x
f (x) = +
[(−1) − 1] cos
sin
4
n2 π 2
5
nπ
5
1
bn =
5
ˆ
5
n=1
f is continous on the interval
ˆ 0
ˆ
ˆ 2
1 2
1
f (x) dx =
(2 + x) dx +
2 dx = 3
14. a0 =
2 −2
2
−2
0
ˆ 0
ˆ
ˆ 2
1
2
1 2
nπ
nπ
nπ
x dx =
x dx +
x dx = 2 2 [1 −
f (x) cos
(2 + x) cos
2 cos
an =
2 −2
2
2
2
2
n
π
−2
0
n
(−1) ]
ˆ 0
ˆ 2
ˆ
1
2
1 2
nπ
nπ
nπ
x dx =
x dx +
x dx =
(−1)n+1
f (x) sin
(2 + x) sin
2 sin
bn =
2 −2
2
2
2
2
nπ
−2
0
∞
2
2
3 nπ
nπ
n
n+1
x+
(−1)
x
f (x) = +
[1 − (−1) ] cos
sin
2
n2 π 2
2
nπ
2
n=1
f is continous on the interval
ˆ
ˆ
1 π
1 π x
1
f (x) dx =
e dx = (eπ − e−π )
15. a0 =
π −π
π −π
π
ˆ π
1
(−1)n (eπ − e−π )
an =
f (x) cos nx dx =
π −π
π(1 + n2 )
ˆ π
ˆ π
1
1
(−1)n n(e−π − eπ )
f (x) sin nx dx =
ex sin nx dx =
bn =
π −π
π −π
π(1 + n2 )
∞ (−1)n n(e−π − eπ )
eπ − e−π (−1)n (eπ − e−π )
+
cos nx +
sin nx
f (x) =
2π
π(1 + n2 )
π(1 + n2 )
n=1
f is continous on the interval
ˆ
ˆ
1 π
1 π x
1
f (x) dx =
(e − 1) dx = (eπ − π − 1)
16. a0 =
π −π
π 0
π
ˆ π
ˆ π
1
1
[eπ (−1)n − 1]
f (x) cos nx dx =
(ex − 1) cos nx dx =
an =
π −π
π 0
π(1 + n2 )
ˆ
ˆ
1 π
1 π x
f (x) sin nx dx =
(e − 1) sin nx dx
bn =
π −π
π 0
1
1 neπ (−1)n+1
n
(−1)n
−
=
+
+
π
1 + n2
1 + n2
n
n
11.2
f (x) =
Fourier Series
∞ eπ − π − 1 eπ (−1)n − 1
+
cos nx
2π
π(1 + n2 )
n=1
+
n
(−1)n − 1
eπ (−1)n+1 + 1 +
2
1+n
n
sin nx
f is continous on the interval
y
17.
1
3 Π
2 Π
Π
Π
2Π
3Π
x
y
18.
2
6
4
2
2
x
4
19. The function in Problem 5 is discontinuous at x = π, so the corresponding Fourier series
converges to π 2 /2 at x = π. That is,
∞ π 2 2(−1)n
π
2[(−1)n − 1]
π2
n+1
=
+
(−1)
sin nπ
cos nπ +
+
2
6
n2
n
n3 π
n=1
∞
∞
1
π2 2
π2
1
π 2 2(−1)n
n
+
+
+ 2 1 + 2 + 2 + ···
(−1) =
=
=
6
n2
6
n2
6
2
3
n=1
and
n=1
1
π2
=
6
2
π2 π2
−
2
6
=1+
1
1
+ 2 + ··· .
2
2
3
At x = 0 the series converges to 0 and
∞
π 2 2(−1)n
1
π2
1
1
0=
+
+ 2 −1 + 2 − 2 + 2 − · · ·
=
6
n2
6
2
3
4
n=1
so
π2
1
1
1
= 1 − 2 + 2 − 2 + ··· .
12
2
3
4
20. From Problem 19
π2
1
=
8
2
π2 π2
+
6
12
1
=
2
2
2
2 + 2 + 2 + ···
3
5
=1+
1
1
+ 2 + ··· .
2
3
5
21. The function in Problem 7 is continuous at x = π/2 so
π
3π
=f
2
2
∞
1 1 1
nπ
2
n+1
(−1)
= π + 2 1 − + − + ···
sin
=π+
n
2
3 5 7
n=1
651
652
CHAPTER 11
FOURIER SERIES
and
1 1 1
π
= 1 − + − + ··· .
4
3 5 7
22. The function in Problem 9 is continuous at x = π/2 so
1=f
1=
π
∞
nπ
1 1 1 + (−1)n
cos
= + +
2
π 2
π(1 − n )
2
2
n=2
1 1
2
2
2
+ +
−
+
− ···
π 2 3π 3 · 5π 5 · 7π
and
2
2
π 2
+ −
+
− ···
2 3 3·5 5·7
π =1+
or
1
1
1
1
π
= +
−
+
− ··· .
4
2 1·3 3·5 5·7
23. (a) Letting c0 = a0 /2, cn = (an − ibn )/2, and c−n = (an + ibn )/2 we have
∞ nπ
nπ
a0 an cos
+
x + bn sin
x
f (x) =
2
p
p
n=1
= c0 +
∞
einπx/p + e−inπx/p
ieinπx/p − ie−inπx/p
− bn
2
2
an
n=1
= c0 +
∞ inπx/p
cn e
i(−n)πx/p
+ c−n e
∞
=
cn einπx/p .
n=−∞
n=1
(b) From part (a) we have
1
1
cn = (an − ibn ) =
2
2p
ˆ p
nπ
nπ
1
x − i sin
x dx =
f (x) cos
f (x)e−inπx/p dx
p
p
2p −p
−p
ˆ
and
1
1
c−n = (an + ibn ) =
2
2p
p
ˆ p
nπ
1
nπ
x + i sin
x dx =
f (x) cos
f (x)einπx/p dx
p
p
2p
−p
−p
ˆ
p
for n = 1, 2, 3, . . .. Thus, for n = ±1, ±2, ±3, . . .,
ˆ p
1
cn =
f (x)e−inπx/p dx.
2p −p
When n = 0 the above formula gives
c0 =
1
2p
ˆ
p
f (x) dx,
−p
which is a0 /2 where a0 is (9) in the text. Therefore
ˆ p
1
f (x)e−inπx/p dx, n = 0, ±1, ±2, . . . .
cn =
2p −p
11.3
Fourier Cosine and Sine Series
24. Identifying f (x) = e−x and p = π, we have
1
cn =
2π
ˆ
π
1
e e dx =
2π
−π
π
1
=−
e−(in+1)x 2(in + 1)π
−x inx
ˆ
π
e−(in+1)x dx
−π
π
=−
1
e−(in+1)π − e(in+1)π
2(in + 1)π
=
e(in+1)π − e−(in+1)π
2(in + 1)π
=
eπ (cos nπ + i sin nπ) − e−π (cos nπ − i sin nπ)
2(in + 1)π
=
(eπ − e−π ) cos nπ
(eπ − e−π ) (−1)n
=
.
2(in + 1)π
2(in + 1)π
Thus
f (x) =
∞
(−1)n
n=−∞
11.3
eπ − e−π inx
e .
2(in + 1)π
Fourier Cosine and Sine Series
1. Since f (−x) = sin (−3x) = − sin 3x = −f (x), f (x) is an odd function.
2. Since f (−x) = −x cos (−x) = −x cos x = −f (x), f (x) is an odd function.
3. Since f (−x) = (−x)2 − x = x2 − x, f (x) is neither even nor odd.
4. Since f (−x) = (−x)3 + 4x = −(x3 − 4x) = −f (x), f (x) is an odd function.
5. Since f (−x) = e|−x| = e|x| = f (x), f (x) is an even function.
6. Since f (−x) = e−x − ex = −f (x), f (x) is an odd function.
7. For 0 < x < 1, f (−x) = (−x)2 = x2 = −f (x), f (x) is an odd function.
8. For 0 ≤ x < 2, f (−x) = −x + 5 = f (x), f (x) is an even function.
9. Since f (x) is not defined for x < 0, it is neither even nor odd.
10. Since f (−x) = (−x)5 = x5 = f (x), f (x) is an even function.
653
654
CHAPTER 11
FOURIER SERIES
11. Since f (x) is an odd function, we have
ˆ
2 π
2
bn =
[1 − (−1)n ]
1 · sin nx dx =
π 0
nπ
Thus
f (x) =
∞
2
[1 − (−1)n ] sin nx.
nπ
n=1
12. Since f (x) is an even function, we expand in a cosine series:
ˆ 2
1 dx = 1
a0 =
1
ˆ
an =
2
cos
1
Thus
2
nπ
nπ
x dx = −
sin
.
2
nπ
2
∞
f (x) =
nπ
nπ
1 −2
+
sin
cos
x.
2
nπ
2
2
n=1
13. Since f (x) is an even function, we expand in a cosine series:
ˆ
2 π
x dx = π
a0 =
π 0
ˆ
2 π
2
an =
x cos nx dx = 2 [(−1)n − 1].
π 0
n π
Thus
∞
f (x) =
π 2
+
[(−1)n − 1] cos nx.
2
n2 π
n=1
14. Since f (x) is an odd function, we expand in a sine series:
ˆ
2 π
2
bn =
x sin nx dx = (−1)n+1 .
π 0
n
Thus
f (x) =
∞
2
(−1)n+1 sin nx.
n
n=1
15. Since f (x) is an even function, we expand in a cosine series:
ˆ 1
2
x2 dx =
a0 = 2
3
0
⎛
⎞
1
ˆ 1
ˆ 1
2
x
2
4
sin nπx −
x2 cos nπx dx = 2 ⎝
x sin nπx dx⎠ = 2 2 (−1)n .
an = 2
nπ
nπ
n
π
0
0
0
Thus
∞
f (x) =
1 4
+
(−1)n cos nπx.
3
n2 π 2
n=1
11.3
Fourier Cosine and Sine Series
16. Since f (x) is an odd function, we expand in a sine series:
⎛
⎞
1
ˆ 1
ˆ 1
2
2
x
cos nπx +
x2 sin nπx dx = 2 ⎝−
x cos nπx dx⎠
bn = 2
nπ
nπ 0
0
0
=
4
2(−1)n+1
+ 3 3 [(−1)n − 1].
nπ
n π
Thus
f (x) =
∞ 2(−1)n+1
nπ
n=1
4
n
+ 3 3 [(−1) − 1] sin nπx.
n π
17. Since f (x) is an even function, we expand in a cosine series:
ˆ
2 π 2
4
(π − x2 ) dx = π 2
a0 =
π 0
3
π
ˆ
ˆ
2 − x2
π
2 π 2
2
2 π
sin nx +
an =
(π − x2 ) cos nx dx =
x sin nx dx
π 0
π
n
n 0
0
Thus
=
4
(−1)n+1 .
n2
∞
4
2
(−1)n+1 cos nx dx.
f (x) = π 2 +
3
n2
n=1
18. Since f (x) is an odd function, we expand in a sine series:
π
ˆ
ˆ
x3
2 π 3
2
3 π 2
− cos nx +
x sin nx dx =
x cos nx dx
bn =
π 0
π
n
n 0
=
2π 2
12
(−1)n+1 − 2
n
n π
12
2π 2
(−1)n+1 − 2
=
n
n π
0
ˆ
π
x sin nx dx
0
π
ˆ
x
1 π
− cos nx +
cos nx dx
n
n 0
Thus
f (x) =
0
∞ 2π 2
n=1
n
n+1
(−1)
=
2π 2
12
(−1)n+1 + 3 (−1)n .
n
n
12
n
+ 3 (−1) sin nx.
n
19. Since f (x) is an odd function, we expand in a sine series:
ˆ
2 π
2(π + 1)
2
(−1)n+1 +
.
bn =
(x + 1) sin nx dx =
π 0
nπ
nπ
Thus
f (x) =
∞ 2(π + 1)
n=1
nπ
n+1
(−1)
2
+
nπ
sin nx.
655
656
CHAPTER 11
FOURIER SERIES
20. Since f (x) is an odd function, we expand in a sine series:
ˆ 1
ˆ 1
ˆ
bn = 2
(x − 1) sin nπx dx = 2
x sin nπx dx −
0
=2
0
1
n2 π 2
x
1
cos nπx +
cos nπx
nπ
nπ
sin nπx −
Thus
1
sin nπx dx
0
1
=−
0
2
.
nπ
∞
2
sin nπx.
f (x) = −
nπ
n=1
21. Since f (x) is an even function, we expand in a cosine series:
ˆ
ˆ
1
a0 =
2
x dx +
0
ˆ
an =
1
1
x cos
0
Thus
3
2
1 dx =
ˆ
nπ
x dx +
2
2
cos
1
nπ
4 nπ
x dx = 2 2 cos
−1 .
2
n π
2
∞
nπ
nπ
3 4 cos
− 1 cos
x.
f (x) = +
2
2
4
n π
2
2
n=1
22. Since f (x) is an odd function, we expand in a sine series:
bn =
1
π
Thus
ˆ
π
x sin
0
n
x dx +
2
ˆ
2π
π sin
π
n
4
nπ
2
x dx = 2 sin
+ (−1)n+1 .
2
n π
2
n
∞ nπ
2
4
n
n+1
sin
+ (−1)
sin x.
f (x) =
2
n π
2
n
2
n=1
23. Since f (x) is an even function, we expand in a cosine series:
ˆ
2 π
4
sin x dx =
a0 =
π 0
π
ˆ π
ˆ
2
1 π
sin x cos nx dx =
(sin (1 + n) x + sin (1 − n) x) dx
an =
π 0
π 0
2
(1 + (−1)n )
π(1 − n2 )
ˆ
1 π
sin 2x dx = 0.
a1 =
π 0
for n = 2, 3, 4, . . .
=
Thus
∞
f (x) =
2 2[1 + (−1)n ]
+
cos nx.
π
π(1 − n2 )
n=2
11.3
Fourier Cosine and Sine Series
24. Since f (x) is an even function, we expand in a cosine series. [See the solution of Problem 10
in Exercises 11.2 for the computation of the integrals.]
ˆ π/2
2
4
cos x dx =
a0 =
π/2 0
π
ˆ π/2
2
nπ
4(−1)n+1
an =
x dx =
cos x cos
π/2 0
π/2
π (4n2 − 1)
Thus
∞
2 4(−1)n+1
cos 2nx.
f (x) = +
π
π (4n2 − 1)
n=1
ˆ
1/2
25. a0 = 2
1 dx = 1
0
ˆ
1/2
nπ
2
sin
nπ
2
0
ˆ 1/2
nπ
2 1 − cos
1 · sin nπx dx =
bn = 2
nπ
2
0
an = 2
1 · cos nπx dx =
f (x) =
nπ
1 2
+
sin
cos nπx
2
nπ
2
∞
n=1
∞
f (x) =
n=1
ˆ
2 nπ
1 − cos
sin nπx
nπ
2
1
26. a0 = 2
1 dx = 1
1/2
ˆ
1
an = 2
1 · cos nπx dx = −
1/2
ˆ
nπ
2
sin
nπ
2
2 nπ
+ (−1)n+1
cos
nπ
2
1/2
∞ 2
nπ
1 −
sin
cos nπx
f (x) = +
2
nπ
2
1
1 · sin nπx dx =
bn = 2
n=1
f (x) =
∞
n=1
4
27. a0 =
π
4
an =
π
bn =
4
π
ˆ
2 nπ
+ (−1)n+1 sin nπx
cos
nπ
2
π/2
cos x dx =
0
ˆ
ˆ
π/2
0
4
π
2
cos x cos 2nx dx =
π
π/2
cos x sin 2nx dx =
0
2
π
ˆ
ˆ
π/2
[cos (2n + 1)x + cos (2n − 1)x] dx =
0
0
π/2
[sin (2n + 1)x + sin (2n − 1)x] dx =
4(−1)n
π(1 − 4n2 )
8n
π(4n2 − 1)
657
658
CHAPTER 11
FOURIER SERIES
∞
f (x) =
2 4(−1)n
+
cos 2nx
π
π(1 − 4n2 )
n=1
f (x) =
∞
8n
sin 2nx
π(4n2 − 1)
n=1
28. a0 =
2
π
2
an =
π
ˆ
π
sin x dx =
0
ˆ
π
4
π
ˆ
1
sin x cos nx dx =
π
0
π
[sin (n + 1)x − sin (n − 1)x] dx
0
2[(−1)n + 1]
for n = 2, 3, 4, . . .
π(1 − n2 )
ˆ
ˆ
2 π
1 π
bn =
sin x sin nx dx =
[cos (n − 1)x − cos (n + 1)x] dx = 0
π 0
π 0
ˆ
1 π
a1 =
sin 2x dx = 0
π 0
ˆ
2 π 2
sin x dx = 1
b1 =
π 0
=
for n = 2, 3, 4, . . .
f (x) = sin x
f (x) =
∞
2
2 (−1)n + 1
+
cos nx
π π
1 − n2
n=2
ˆ
2
29. a0 =
π
ˆ
π/2
π
(π − x) dx
x dx +
0
π/2
ˆ
2
an =
π
ˆ
π/2
π
x cos nx dx +
0
ˆ
2
bn =
π
=
π
2
(π − x) cos nx dx
=
π/2
ˆ
π/2
π
x sin nx dx +
0
(π − x) sin nx dx
=
π/2
nπ
2 2 cos
+ (−1)n+1 − 1
2
n π
2
nπ
4
sin
n2 π
2
∞
nπ
π 2 2 cos
+ (−1)n+1 − 1 cos nx
f (x) = +
4
n2 π
2
n=1
f (x) =
∞
n=1
1
30. a0 =
π
an =
bn =
1
π
1
π
ˆ
4
nπ
sin
sin nx
2
n π
2
2π
(x − π) dx =
π
ˆ
ˆ
2π
(x − π) cos
n
4 nπ
x dx = 2
(−1)n − cos
2
n π
2
(x − π) sin
n
2
4
nπ
x dx = (−1)n+1 − 2 sin
2
n
n π
2
π
2π
π
π
2
11.3
Fourier Cosine and Sine Series
∞
π 4 nπ
n
+
(−1)n − cos
cos x
4
n2 π
2
2
f (x) =
n=1
∞ f (x) =
n=1
ˆ
nπ
2
4
(−1)n+1 − 2 sin
n
n π
2
ˆ
1
31. a0 =
2
x dx +
1 dx =
0
1
ˆ
sin
n
x
2
3
2
4 nπ
nπ
x dx = 2 2 cos
−1
2
n
π
2
0
ˆ 2
ˆ 1
4
2
nπ
nπ
nπ
bn =
x dx +
x dx = 2 2 sin
+
(−1)n+1
x sin
1 · sin
2
2
n
π
2
nπ
0
1
1
an =
x cos
∞
nπ
nπ
3 4 − 1 cos
x
cos
f (x) = +
2
2
4
n π
2
2
n=1
∞ f (x) =
n=1
ˆ
2
4
nπ
nπ
n+1
+
(−1)
x
sin
sin
2
2
n π
2
nπ
2
ˆ
1
32. a0 =
0
1
0
1
1 · sin
bn =
0
f (x) =
f (x) =
3
+
4
ˆ
1
33. a0 = 2
n=1
1
an = 2
0
2
n2 π 2
ˆ
bn = 2
0
1
ˆ
2
3
2
(2 − x) cos
1
2
(2 − x) sin
1
4 nπ
nπ
x dx = 2 2 cos
+ (−1)n+1
2
n π
2
2
4
nπ
nπ
x dx =
+ 2 2 sin
2
nπ n π
2
nπ
nπ
4 + (−1)n+1 cos
x
cos
n2 π 2
2
2
4
2
nπ
+ 2 2 sin
nπ n π
2
(x2 + x) dx =
0
ˆ
nπ
x dx +
2
∞
∞ n=1
=
ˆ
nπ
x dx +
1 · cos
2
an =
ˆ
(2 − x) dx =
1
ˆ
=
2
1 dx +
sin
nπ
x
2
5
3
1
ˆ 1
2(x2 + x)
2
sin nπx −
(x + x) cos nπx dx =
(2x + 1) sin nπx dx
nπ
nπ 0
2
0
[3(−1)n − 1]
1
ˆ 1
2 + x)
2(x
2
2
cos nπx +
(x + x) sin nπx dx = −
(2x + 1) cos nπx dx
nπ
nπ 0
4
4
(−1)n+1 + 3 3 [(−1)n − 1]
nπ
n π
0
659
660
CHAPTER 11
FOURIER SERIES
∞
5 2
+
[3(−1)n − 1] cos nπx
6
n2 π 2
f (x) =
n=1
∞ f (x) =
n=1
ˆ
2
34. a0 =
4
4
(−1)n+1 + 3 3 [(−1)n − 1] sin nπx
nπ
n π
(2x − x2 ) dx =
0
ˆ
2
(2x − x2 ) cos
8
nπ
x dx = 2 2 [(−1)n+1 − 1]
2
n π
(2x − x2 ) sin
16
nπ
x dx = 3 3 [1 − (−1)n ]
2
n π
an =
ˆ
0
2
bn =
4
3
0
∞
nπ
2 8
+
x
[(−1)n+1 − 1] cos
2
2
3
n π
2
f (x) =
n=1
∞
nπ
16
x
[1 − (−1)n ] sin
3
3
n π
2
f (x) =
n=1
35. a0 =
an =
bn =
ˆ
1
π
2π
0
1
π
1
π
ˆ
2π
8
x2 dx = π 2
3
x2 cos nx dx =
0
ˆ
2π
x2 sin nx dx = −
0
∞
4
f (x) = π 2 +
3
36. a0 =
2
π
2
π
4π
n
4
4π
sin nx
cos nx −
n2
n
π
x dx = π
0
2
an =
π
bn =
n=1
ˆ
4
n2
ˆ
π
x cos 2nx dx = 0
0
ˆ π
x sin 2nx dx = −
0
1
n
∞
f (x) =
π 1
−
sin 2nx
2
n
n=1
ˆ
1
37. a0 = 2
(x + 1) dx = 3
0
ˆ
1
an = 2
(x + 1) cos 2nπx dx = 0
ˆ
0
bn = 2
0
1
(x + 1) sin 2nπx dx = −
1
nπ
11.3
Fourier Cosine and Sine Series
∞
f (x) =
3 1
−
sin 2nπx
2
nπ
n=1
38. a0 =
2
2
2
an =
2
2
bn =
2
ˆ
2
(2 − x) dx = 2
0
ˆ
2
(2 − x) cos nπx dx = 0
0
ˆ
2
(2 − x) sin nπx dx =
0
f (x) = 1 +
2
nπ
∞
2
sin nπx
nπ
n=1
39. The periodic extensions for the cosine, sine, and Fourier series are shown below:
–3L
–2L
–L
L
2L
3L
–3L
–2L
–L
L
2L
3L
–3L
–2L
–L
L
2L
3L
661
662
CHAPTER 11
FOURIER SERIES
40. The periodic extensions for the cosine, sine, and Fourier series are shown below:
–3L
–2L
–L
L
2L
3L
–3L
–2L
–L
L
2L
3L
–3L
–2L
–L
L
2L
3L
41. The periodic extensions for the cosine, sine, and Fourier series are shown below:
–3L
–2L
–L
L
2L
3L
–3L
–2L
–L
L
2L
3L
–3L
–2L
–L
L
2L
3L
11.3
Fourier Cosine and Sine Series
42. The periodic extensions for the cosine, sine, and Fourier series are shown below:
–3L
–2L
–L
L
–3L
–2L
–L
L
2L
3L
–3L
–2L
–L
L
2L
3L
43. We have
2
bn =
π
ˆ
π
5 sin nt dt =
0
so that
f (t) =
Substituting the assumption xp (t) =
+ 10xp =
∞
3L
10
[1 − (−1)n ]
nπ
∞
10[1 − (−1)n ]
nπ
n=1
xp
2L
∞
n=1 Bn sin nt
Bn (10 − n ) sin nt =
2
n=1
sin nt.
into the differential equation then gives
∞
10[1 − (−1)n ]
n=1
nπ
and so Bn = 10[1 − (−1)n ]/nπ(10 − n2 ). Thus
∞
10 1 − (−1)n
sin nt.
xp (t) =
π
n(10 − n2 )
n=1
44. We have
2
bn =
π
so that
ˆ
1
(1 − t) sin nπt dt =
0
∞
2
sin nπt.
f (t) =
nπ
n=1
2
nπ
sin nt
663
664
CHAPTER 11
FOURIER SERIES
∞
Substituting the assumption xp (t) =
xp + 10xp =
∞
n=1 Bn sin nπt
into the differential equation then gives
Bn (10 − n2 π 2 ) sin nπt =
n=1
and so Bn = 2/nπ(10 −
n2 π 2 ).
∞
2
sin nπt
nπ
n=1
Thus
∞
2
1
sin nπt.
π
n(10 − n2 π 2 )
xp (t) =
n=1
45. We have
ˆ
2
a0 =
π
an =
π
0
ˆ
2
π
π
4
(2πt − t2 ) dt = π 2
3
(2πt − t2 ) cos nt dt = −
0
so that
4
n2
∞
f (t) =
2π 2 4
−
cos nt.
3
n2
n=1
Substituting the assumption
∞
A0 +
An cos nt
xp (t) =
2
n=1
into the differential equation then gives
∞
∞
1 2
2π 2 4
1 x + 12xp = 6A0 +
−
An − n + 12 cos nt =
cos nt
4 p
4
3
n2
n=1
n=1
and A0 = π 2 /9, An = 16/n2 (n2 − 48). Thus
∞
π2
1
+ 16
cos nt.
xp (t) =
18
n2 (n2 − 48)
n=1
46. We have
2
a0 =
1/2
an =
2
1/2
ˆ
1/2
t dt =
0
ˆ
1
2
1/2
t cos 2nπt dt =
0
so that
1
n2 π 2
[(−1)n − 1]
∞
f (t) =
1 (−1)n − 1
+
cos 2nπt.
4
n2 π 2
n=1
Substituting the assumption
∞
A0 +
An cos 2nπt
xp (t) =
2
n=1
11.3
Fourier Cosine and Sine Series
into the differential equation then gives
∞
∞
n=1
n=1
1 (−1)n − 1
1 xp + 12xp = 6A0 +
An (12 − n2 π 2 ) cos 2nπt = +
cos 2nπt
4
4
n2 π 2
and A0 = 1/24, An = [(−1)n − 1]/n2 π 2 (12 − n2 π 2 ). Thus
∞
1 (−1)n − 1
1
+
cos 2nπt.
xp (t) =
48 π 2
n2 (12 − n2 π 2 )
n=1
47. (a) The general solution is x(t) = c1 cos
√
10 t + c2 sin
√
10 t + xp (t), where
∞
xp (t) =
10 1 − (−1)n
sin nt.
π
n(10 − n2 )
n=1
The initial condition x(0) = 0 implies c1 + xp (0) = 0. Since xp (0) = 0, we have c1 = 0
√
√
√
and x(t) = c2 sin 10 t + xp (t). Then x (t) = c2 10 cos 10 t + xp (t) and x (0) = 0
implies
∞
√
10 1 − (−1)n
cos 0 = 0.
c2 10 +
π
10 − n2
n=1
Thus
√
∞
10 1 − (−1)n
c2 = −
π
10 − n2
n=1
and
∞
√
1
10 1 − (−1)n 1
sin nt − √ sin 10 t .
x(t) =
π
10 − n2 n
10
n=1
(b) The graph is plotted using eight nonzero terms in the series expansion of x(t).
x
4
2
20
40
60
80
–2
–4
√
√
48. (a) The general solution is x(t) = c1 cos 4 3 t + c2 sin 4 3t + xp (t), where
∞
xp (t) =
π2
1
+ 16
cos nt.
18
n2 (n2 − 48)
n=1
The initial condition x(0) = 0 implies c1 + xp (0) = 1 or
∞
π2
1
− 16
.
c1 = 1 − xp (0) = 1 −
2
2
18
n (n − 48)
n=1
t
665
666
CHAPTER 11
FOURIER SERIES
√
√
√
√
Now x (t) = −4 3c1 sin 4 3 t + 4 3c2 cos 4 3 t + xp (t), so x (0) = 0 implies
√
4 3c2 + xp (0) = 0. Since xp (0) = 0, we have c2 = 0 and
∞
1
π2
− 16
1−
2
2
18
n (n − 48)
x(t) =
n=1
=
∞
π2
1
cos 4 3 t +
+ 16
cos nt
2
2
18
n (n − 48)
√
n=1
∞
√
√ π2
π2
1
+ 1−
cos 4 3 t + 16
3t .
cos
nt
−
cos
4
18
18
n2 (n2 − 48)
n=1
(b) The graph is plotted using five nonzero terms in the series expansion of x(t).
x
1.5
1
0.5
2
4
6
8
10
12
t
14
–0.5
–1
49. (a) We have
2
bn =
L
ˆ
L
0
nπ
2w0
w0 x
sin
x dx =
(−1)n+1
L
L
nπ
so that
w(x) =
∞
2w0
n=1
nπ
(−1)n+1 sin
∞
(b) If we assume yp (x) =
n=1 Bn sin (nπx/L)
yp(4)
=
then
∞
n4 π 4
n=1
L4
nπ
x.
L
Bn sin
nπ
x
L
(4)
and so the differential equation EIyp = w(x) gives
Bn =
Thus
yp (x) =
2w0 (−1)n+1 L4
.
EIn5 π 5
∞
2w0 L4 (−1)n+1
nπ
x.
sin
EIπ 5
n5
L
n=1
50. We have
bn =
so that
2
L
ˆ
2L/3
w0 sin
L/3
2w0
nπ
x dx =
L
nπ
nπ
2nπ
cos
− cos
3
3
∞
nπ
2nπ
nπ
2w0
cos
− cos
sin
x.
w(x) =
nπ
3
3
L
n=1
11.3
If we assume yp (x) =
∞
n=1 Bn sin (nπx/L)
yp(4) (x) =
Fourier Cosine and Sine Series
then
∞
n4 π 4
L4
n=1
Bn sin
nπ
x
L
(4)
and so the differential equation EIyp (x) = w(x) gives
2nπ
cos nπ
3 − cos 3
.
EIn5 π 5
Bn = 2w0 L4
Thus
yp (x) =
∞
2nπ
2w0 L4 cos nπ
nπ
3 − cos 3
x.
sin
5
5
EIπ
n
L
n=1
51. We note that w(x)is 2π-periodic and even. With p = π we find the cosine expansion of
w0 , 0 < x < π/2
f (x) =
0,
π/2 < x < π
We have
2
a0 =
π
an =
Thus,
2
π
ˆ
π
0
ˆ
2
f (x) dx =
π
ˆ
π/2
w0 dx = w0
0
π
f (x) cos nx dx =
0
2
π
ˆ
π/2
w0 cos nx dx =
0
2w0
nπ
sin
.
nπ
2
∞
nπ
w0 2w0 1
+
sin
cos nx.
w(x) =
2
π
n
2
n=1
Now we assume a particular solution of the form yp (x) = A0 /2 + ∞
n=1 An cos nx. Then
∞
(4)
4
yp (x) = n=1 An n cos nx and substituting into the differential equation, we obtain
∞
EIyp(4) (x) + kyp (x) =
kA0 +
An (EIn4 + k) cos nx
2
n=1
∞
w0 2w0 1
nπ
=
+
sin
cos nx.
2
π
n
2
n=1
Thus
A0 =
w0
k
and
yp (x) =
and
An =
2w0 sin (nπ/2)
,
π n(EIn4 + k)
∞
w0 2w0 sin (nπ/2)
+
cos nx.
2k
π
n(EIn4 + k)
n=1
667
668
CHAPTER 11
FOURIER SERIES
52. (a) If f and g are even and h(x) = f (x)g(x) then
h(−x) = f (−x)g(−x) = f (x)g(x) = h(x),
and h is even.
(c) If f is even and g is odd and h(x) = f (x)g(x) then
h(−x) = f (−x)g(−x) = f (x)[−g(x)] = −h(x),
and h is odd.
(d) Let h(x) = f (x) ± g(x) where f and g are even. Then
h(−x) = f (−x) ± g(−x) = f (x) ± g(x) = h(x),
and h is even.
(e) If f and g are odd and h(x) = f (x) ± g(x) then
h(−x) = f (−x) ± g(−x) = −f (x) ± [−g(x)] = −[f (x) ± g(x)] = −h(x)
and h is odd.
(f ) If f is even then
ˆ
ˆ
a
ˆ
0
f (x) dx =
−a
a
f (x) dx +
ˆ
ˆ
a
f (−u)(−du) +
a
ˆ
0
ˆ
a
f (u) du +
←− f is an even function
f (x) dx
0
ˆ
a
=
du = −dx
0
0
=
←− u = −x,
f (x) dx
−a
a
f (x) dx = 2
f (x) dx.
0
0
(g) If f is odd then
ˆ
ˆ
a
−a
ˆ
0
f (x) dx =
a
f (x) dx +
−a
ˆ
f (x) dx
ˆ
0
a
f (−u)(−du) +
=
ˆ
a
f (x) dxf
←− f is an odd function
a
[−f (u)] du +
0
du = −dx
0
ˆ
a
=
←− u = −x,
0
f (x) dx = 0.
0
53. If f (x) is even then f (−x) = f (x). If f (x) is odd then f (−x) = −f (x). Thus, if f (x) is both
even and odd f (x) = f (−x) = −f (x), and f (x) = 0.
11.4
Sturm–Liouville Problem
54. For EIy (4) + ky = 0 the roots of the auxiliary equation are m1 = α + αi, m2 = α − αi,
√
m3 = −α + αi, and m4 = −α − αi, where α = (k/EI)1/4 / 2 . Thus
yc = eαx (c1 cos αx + c2 sin αx) + e−αx (c3 cos αx + c4 sin αx).
We expect y(x) to be bounded as x → ∞, so we must have c1 = c2 = 0. We also expect y(x)
to be bounded as x → −∞, so we must have c3 = c4 = 0. Thus, yc = 0 and the solution of
the differential equation in Problem 51 is yp (x).
11.4
Sturm–Liouville Problem
1. For λ ≤ 0 the only solution of the boundary-value problem is y = 0. For λ = α2 > 0 we have
y = c1 cos αx + c2 sin αx.
Now
y (x) = −c1 α sin αx + c2 α cos αx
and y (0) = 0 implies c2 = 0, so
y(1) + y (1) = c1 (cos α − α sin α) = 0 or
cot α = α.
The eigenvalues are λn = αn2 where α1 , α2 , α3 , . . . are the consecutive positive solutions
of cot α = α. The corresponding eigenfunctions are cos αn x for n = 1, 2, 3, . . . . Using a
CAS we find that the first four eigenvalues are approximately 0.7402, 11.7349, 41.4388, and
90.8082 with corresponding approximate eigenfunctions cos 0.8603x, cos 3.4256x, cos 6.4373x,
and cos 9.5293x.
2. For λ < 0 the only solution of the boundary-value problem is y = 0. For λ = 0 we have
y = c1 x + c2 . Now y = c1 and the boundary conditions both imply c1 + c2 = 0. Thus, λ = 0
is an eigenvalue with corresponding eigenfunction y0 = x − 1.
For λ = α2 > 0 we have
y = c1 cos αx + c2 sin αx
and
y (x) = −c1 α sin αx + c2 α cos αx.
The boundary conditions imply
c 1 + c2 α = 0
c1 cos α + c2 sin α = 0
which gives
−c2 α cos α + c2 sin α = 0
or
tan α = α.
669
670
CHAPTER 11
FOURIER SERIES
The eigenvalues are λn = αn2 where α1 , α2 , α3 , . . . are the consecutive positive solutions
of tan α = α. The corresponding eigenfunctions are α cos αx − sin αx (obtained by taking
c2 = −1 in the first equation of the system.) Using a CAS we find that the first four positive
eigenvalues are 20.1907, 59.6795, 118.9000, and 197.858 with corresponding eigenfunctions
4.4934 cos 4.4934x − sin 4.4934x, 7.7253 cos 7.7253x − sin 7.7253x,
10.9041 cos 10.9041x − sin 10.9041x, and 14.0662 cos 14.0662x − sin 14.0662x.
3. For λ = 0 the solution of y = 0 is y = c1 x + c2 . The condition y (0) = 0 implies c1 = 0, so
λ = 0 is an eigenvalue with corresponding eigenfunction 1.
For λ = −α2 < 0 we have y = c1 cosh αx + c2 sinh αx and y = c1 α sinh αx + c2 α cosh αx.
The condition y (0) = 0 implies c2 = 0 and so y = c1 cosh αx. Now the condition y (L) = 0
implies c1 = 0. Thus y = 0 and there are no negative eigenvalues.
For λ = α2 > 0 we have y = c1 cos αx + c2 sin αx and y = −c1 α sin αx + c2 α cos αx. The
condition y (0) = 0 implies c2 = 0 and so y = c1 cos αx. Now the condition y (L) = 0 implies
−c1 α sin αL = 0. For c1 = 0 this condition will hold when αL = nπ or λ = α2 = n2 π 2 /L2 ,
where n = 1, 2, 3, . . . . These are the positive eigenvalues with corresponding eigenfunctions
cos(nπx/L), n = 1, 2, 3, . . . .
4. For λ = −α2 < 0 we have
y = c1 cosh αx + c2 sinh αx
y = c1 α sinh αx + c2 α cosh αx.
Using the fact that cosh x is an even function and sinh x is odd we have
y(−L) = c1 cosh (−αL) + c2 sinh (−αL)
= c1 cosh αL − c2 sinh αL
and
y (−L) = c1 α sinh (−αL) + c2 α cosh (−αL)
= −c1 α sinh αL + c2 α cosh αL.
The boundary conditions imply
c1 cosh αL − c2 sinh αL = c1 cosh αL + c2 sinh αL
or
2c2 sinh αL = 0
and
−c1 α sinh αL + c2 α cosh αL = c1 α sinh αL + c2 α cosh αL
or
2c1 α sinh αL = 0.
11.4
Sturm–Liouville Problem
Since αL = 0, c1 = c2 = 0 and the only solution of the boundary-value problem in this case
is y = 0.
For λ = 0 we have
y = c1 x + c 2
y = c1 .
From y(−L) = y(L) we obtain
−c1 L + c2 = c1 L + c2 .
Then c1 = 0 and y = 1 is an eigenfunction corresponding to the eigenvalue λ = 0.
For λ = α2 > 0 we have
y = c1 cos αx + c2 sin αx
y = −c1 α sin αx + c2 α cos αx.
The first boundary condition implies
c1 cos αL − c2 sin αL = c1 cos αL + c2 sin αL
or
2c2 sin αL = 0.
Thus, if c1 = 0 and c2 = 0,
αL = nπ
or λ = α2 =
n2 π 2
, n = 1, 2, 3, . . . .
L2
The corresponding eigenfunctions are sin (nπx/L), for n = 1, 2, 3, . . . . Similarly, the second
boundary condition implies
2c1 α sin αL = 0.
If c1 = 0 and c2 = 0,
n2 π 2
, n = 1, 2, 3, . . . ,
L2
and the corresponding eigenfunctions are cos (nπx/L), for n = 1, 2, 3, . . . .
αL = nπ
or λ = α2 =
5. The eigenfunctions are cos αn x where cot αn = αn . Thus
ˆ 1
ˆ
1 1
2
2
cos αn x =
cos αn x dx =
(1 + cos 2αn x) dx
2 0
0
1
1
1
1
1
x+
1+
sin 2αn x =
sin 2αn
=
2
2αn
2
2αn
1
1+
=
2
1
1+
=
2
0
1
1
1
1+
(2 sin αn cos αn ) =
sin αn cot αn sin αn
2αn
2
αn
1
1
1 + sin2 αn .
(sin αn ) αn (sin αn ) =
αn
2
671
672
CHAPTER 11
FOURIER SERIES
6. The eigenfunctions are sin αn x where tan αn = −αn . Thus
ˆ 1
ˆ
1 1
sin αn x2 =
sin2 αn x dx =
(1 − cos 2αn x) dx
2 0
0
1
1
1
1
1
x−
1−
sin 2αn x =
sin 2αn
=
2
2αn
2
2αn
1
=
1−
2
1
=
1−
2
0
1
1
1
(2 sin αn cos αn ) =
tan αn cos αn cos αn
1−
2αn
2
αn
1
1 2
−αn cos αn =
1 + cos2 αn .
αn
2
7. (a) If λ ≤ 0 the initial conditions imply y = 0. For λ = α2 > 0 the general solution of the
Cauchy-Euler differential equation is y = c1 cos (α ln x) + c2 sin (α ln x). The condition
y(1) = 0 implies c1 = 0, so that y = c2 sin (α ln x). The condition y(5) = 0 implies
α ln 5 = nπ, n = 1, 2, 3, . . . . Thus, the eigenvalues are n2 π 2 /(ln 5)2 for n = 1, 2, 3, . . . ,
with corresponding eigenfunctions sin [(nπ/ ln 5) ln x].
(b) The self-adjoint form is
λ
d
[xy ] + y = 0.
dx
x
(c) An orthogonality relation is
ˆ
1
5
mπ
nπ
1
sin
ln x sin
ln x dx = 0,
x
ln 5
ln 5
m = n.
√
8. (a) The roots of the auxiliary equation m2 + m + λ = 0 are 12 (−1 ± 1 − 4λ ). When λ = 0
the general solution of the differential equation is c1 + c2 e−x . The boundary conditions
imply c1 + c2 = 0 and c1 + c2 e−2 = 0. Since the determinant of the coefficients is not
0, the only solution of this homogeneous system is c1 = c2 = 0, in which case y = 0.
When λ = 14 , the general solution of the differential equation is c1 e−x/2 + c2 xe−x/2 . The
boundary conditions imply c1 = 0 and c1 + 2c2 = 0, so c1 = c2 = 0 and y = 0. Similarly,
if 0 < λ < 14 , the general solution is
1
y = c1 e 2 (−1+
√
1−4λ )x
1
+ c2 e 2 (−1−
√
1−4λ )x
.
In this case the boundary conditions again imply c1 = c2 = 0, and so y = 0. Now, for
λ > 14 , the general solution of the differential equation is
y = c1 e−x/2 cos
√
√
4λ − 1 x + c2 e−x/2 sin 4λ − 1 x.
√
The condition y(0) = 0 implies c1 = 0 so y = c2 e−x/2 sin 4λ − 1 x. From
√
y(2) = c2 e−1 sin 2 4λ − 1 = 0
11.4
Sturm–Liouville Problem
√
we see that the eigenvalues are determined by 2 4λ − 1 = nπ for n = 1, 2, 3, . . . . Thus,
the eigenvalues are n2 π 2 /42 + 1/4 for n = 1, 2, 3, . . . , with corresponding eigenfunctions
e−x/2 sin(nπx/2).
(b) The self-adjoint form is
d x [e y ] + λex y = 0.
dx
(c) An orthogonality relation is
ˆ
2
x
e
0
−x/2
e
ˆ 2
nπ
mπ
nπ
mπ
−x/2
x e
x dx =
x cos
x dx = 0.
sin
cos
sin
2
2
2
2
0
1
9. We divide by lead coefficient of the differential equation to obtain the form y + −1 +
y +
x
n
y = 0. The integrating factor is then
x
´
e
(−1+1/x) dx
= e−x+ln x = e−x eln x = xe−x .
Thus, the differential equation is
xe−x y + (1 − x)e−x y + ne−x y = 0
and the self-adjoint form is
d
xe−x y + ne−x y = 0.
dx
Identifying the weight function p(x) = e−x and noting that since r(x) = xe−x , r(0) = 0 and
limx→∞ r(x) = 0, we have the orthogonality relation
ˆ ∞
e−x Lm (x)Ln (x) dx = 0, m = n.
0
´
10. To obtain the self-adjoint form we note that an integrating factor is e
the differential equation is
−2x dx
= e−x . Thus,
e−x y − 2xe−x y + 2ne−x y = 0
2
and the self-adjoint form is
2
2
d −x2 2
e y + 2ne−x y = 0.
dx
Identifying the weight function p(x) = e−x and noting that since r(x) = e−x ,
lim r(x) = lim r(x) = 0, we have the orthogonality relation
2
x→−∞
x→∞
ˆ
∞
−∞
e−x Hm (x)Hn (x) dx = 0, m = n.
2
2
2
673
674
CHAPTER 11
FOURIER SERIES
11. (a) The differential equation is
(1 + x2 )y + 2xy +
λ
y = 0.
1 + x2
Letting x = tan θ we have θ = tan−1 x and
dy dθ
1 dy
dy
=
=
dx
dθ dx
1 + x2 dθ
2
1 dy
1
2x
d y dθ
d
dy
d2 y
=
−
=
2
2
2
2
2
2
dx
dx 1 + x dθ
1+x
dθ dx
(1 + x ) dθ
=
2x
d2 y
dy
1
.
−
2
2
2
2
2
(1 + x ) dθ
(1 + x ) dθ
The differential equation can then be written in terms of y(θ) as
(1 + x2 )
1
2x
1 dy
λ
d2 y
dy
+
2x
+
−
y
(1 + x2 )2 dθ2
(1 + x2 )2 dθ
1 + x2 dθ
1 + x2
=
or
1 d2 y
λ
+
y=0
2
2
1 + x dθ
1 + x2
d2 y
+ λy = 0.
dθ2
The boundary conditions become y(0) = y(π/4) = 0. For λ ≤ 0 the only solution of
the boundary-value problem is y = 0. For λ = α2 > 0 the general solution of the
differential equation is y = c1 cos αθ + c2 sin αθ. The condition y(0) = 0 implies c1 = 0
so y = c2 sin αθ. Now the condition y(π/4) = 0 implies c2 sin απ/4 = 0. For c2 = 0 this
condition will hold when απ/4 = nπ or λ = α2 = 16n2 , where n = 1, 2, 3, . . . . These are
the eigenvalues with corresponding eigenfunctions sin 4nθ = sin(4n tan−1 x), for n = 1,
2, 3, . . . .
(b) An orthogonality relation is
ˆ
0
1
x2
1
sin (4m tan−1 x) sin (4n tan−1 x) dx = 0,
+1
m = n.
12. (a) Letting λ = α2 the differential equation becomes x2 y + xy + (α2 x2 − 1)y = 0. This is
the parametric Bessel equation with ν = 1. The general solution is
y = c1 J1 (αx) + c2 Y1 (αx).
Since Y is unbounded at 0 we must have c2 = 0, so that y = c1 J1 (αx). The condition
J1 (3α) = 0 defines the eigenvalues λn = αn2 for n = 1, 2, 3, . . . . The corresponding
eigenfunctions are J1 (αn x).
11.5
Bessel and Legendre Series
(b) Using a CAS or Table 5.1 in the text to solve J1 (3α) = 0 we find 3α1 = 3.8317, 3α2 =
7.0156, 3α3 = 10.1735, and 3α4 = 13.3237. The corresponding eigenvalues are λ1 =
α12 = 1.6313, λ2 = α22 = 5.4687, λ3 = α32 = 11.4999, and λ4 = α42 = 19.7245.
13. When λ = 0 the differential equation is r(x)y + r (x)y = 0. By inspection we see that y = 1
is a solution of the boundary-value problem. Thus, λ = 0 is an eigenvalue.
14. (a) An orthogonality relation is
ˆ
1
cos xm x cos xn x dx = 0
0
where xm = xn are positive solutions of cot x = x.
(b) Referring to Problem 1 we use a CAS to compute
ˆ 1
(cos 0.8603x)(cos 3.4256x) dx = −1.8771 × 10−6 ≈ 0.
0
15. (a) An orthogonality relation is
ˆ 1
(xm cos xm x − sin xm x)(xn cos xn x − sin xn x) dx = 0
0
where xm = xn are positive solutions of tan x = x.
(b) Referring to Problem 2 we use a CAS to compute
ˆ 1
(4.4934 cos 4.4934x − sin 4.4934x)(7.7253 cos 7.7253x − sin 7.7253x) dx = −2.5650 × 10−4
0
≈ 0.
11.5
Bessel and Legendre Series
1. Identifying b = 3, we have α1 = 1.2772, α2 = 2.3385, α3 = 3.3912, and α4 = 4.4412.
2. By (6) in the text J0 (2α) = −J1 (2α). Thus, J0 (2α) = 0 is equivalent to J1 (2α) = 0. Then
α1 = 1.9159, α2 = 3.5078, α3 = 5.0867, and α4 = 6.6618.
3. The boundary condition indicates that we use (15) and (16) in the text. With b = 2 we obtain
ˆ 2
2
xJ0 (αi x) dx
t = αi x, dt = αi dx
ci =
4J12 (2αi ) 0
ˆ 2αi
1
1
·
=
tJ0 (t) dt
2J12 (2αi ) αi2 0
ˆ 2αi
d
1
[tJ1 (t)] dt
[From (5) in the text]
=
2
2
dt
2αi J1 (2αi ) 0
2αi
1
1
tJ
=
.
(t)
=
1
2
2
αi J1 (2αi )
2αi J1 (2αi )
0
675
676
CHAPTER 11
FOURIER SERIES
Thus
f (x) =
∞
i=1
1
J0 (αi x).
αi J1 (2αi )
4. The boundary condition indicates that we use (19) and (20) in the text. With b = 2 we obtain
2
c1 =
4
ci =
ˆ
0
2
2
2 x2 x dx =
= 1,
4 2
2
2
4J0 (2αi )
0
ˆ
2
xJ0 (αi x) dx
t = αi x,
dt = αi dx
0
ˆ 2αi
1
1
·
=
tJ0 (t) dt
2J02 (2αi ) αi2 0
ˆ 2αi
d
1
[tJ1 (t)] dt
=
2
2
dt
2αi J0 (2αi ) 0
2αi
1
J1 (2αi )
tJ
.
=
(t)
=
1
2
2
2αi J0 (2αi )
αi J02 (2αi )
[From (5) in the text]
0
Now since J0 (2αi ) = 0 is equivalent to J1 (2αi ) = 0 we conclude ci = 0 for i = 2, 3, 4, . . . .
Thus the expansion of f on 0 < x < 2 consists of a series with one nontrivial term:
f (x) = c1 = 1.
5. The boundary condition indicates that we use (17) and (18) in the text. With b = 2 and
h = 1 we obtain
ci =
2αi2
(4αi2 + 1)J02 (2αi )
ˆ
2
xJ0 (αi x) dx
t = αi x,
dt = αi dx
0
ˆ 2αi
1
2αi2
·
=
tJ0 (t) dt
(4αi2 + 1)J02 (2αi ) αi2 0
ˆ 2αi
d
2
[tJ1 (t)] dt
[From (5) in the text]
=
2
2
dt
(4αi + 1)J0 (2αi ) 0
2αi
2
4αi J1 (2αi )
tJ
.
=
(t)
=
1
2
2
(4αi + 1)J0 (2αi )
(4αi2 + 1)J02 (2αi )
0
Thus
f (x) = 4
∞
αi J1 (2αi )
J0 (αi x).
+ 1)J02 (2αi )
(4αi2
i=1
11.5
Bessel and Legendre Series
6. Writing the boundary condition in the form
2J0 (2α) + 2αJ0 (2α) = 0
we identify b = 2 and h = 2. Using (17) and (18) in the text we obtain
2αi2
ci =
(4αi2 + 4)J02 (2αi )
ˆ
2
xJ0 (αi x) dx
t = αi x,
dt = αi dx
0
ˆ 2αi
1
αi2
·
=
tJ0 (t) dt
2(αi2 + 1)J02 (2αi ) αi2 0
ˆ 2αi
1
d
=
[tJ1 (t)] dt
[From (5) in the text]
2
2
dt
2(αi + 1)J0 (2αi ) 0
2αi
1
αi J1 (2αi )
tJ1 (t)
.
=
= 2
2(αi2 + 1)J02 (2αi )
(αi + 1)J02 (2αi )
0
Thus
f (x) =
∞
i=1
αi J1 (2αi )
J0 (αi x).
(αi2 + 1)J02 (2αi )
7. The boundary condition indicates that we use (17) and (18) in the text. With n = 1, b = 4,
and h = 3 we obtain
2αi2
ci =
(16αi2 − 1 + 9)J12 (4αi )
ˆ
4
xJ1 (αi x)5x dx
t = αi x,
dt = αi dx
0
ˆ 4αi
1
5αi2
·
t2 J1 (t) dt
4(2αi2 + 1)J12 (4αi ) αi3 0
ˆ 4αi
d 2
5
[t J2 (t)] dt
[From (5) in the text]
=
2
2
dt
4αi (2αi + 1)J1 (4αi ) 0
4αi
5
20αi J2 (4αi )
2
t
.
=
J
(t)
=
2
4αi (2αi2 + 1)J12 (4αi )
(2αi2 + 1)J12 (4αi )
=
0
Thus
f (x) = 20
∞
αi J2 (4αi )
J1 (αi x).
+ 1)J12 (4αi )
(2αi2
i=1
677
678
CHAPTER 11
FOURIER SERIES
8. The boundary condition indicates that we use (15) and (16) in the text. With n = 2 and
b = 1 we obtain
ˆ 1
2
xJ2 (αi x)x2 dx
t = αi x, dt = αi dx
c1 = 2
J3 (αi ) 0
ˆ αi
1
2
·
= 2
t3 J2 (t) dt
J3 (αi ) αi4 0
ˆ αi
2
d 3
[t J3 (t)] dt
= 4 2
[From (5) in the text]
αi J3 (αi ) 0 dt
αi
2
2
t3 J3 (t) =
.
= 4 2
αi J3 (αi )
αi J3 (αi )
0
Thus
f (x) = 2
∞
i=1
1
J2 (αi x).
αi J3 (αi )
9. The boundary condition indicates that we use (19) and (20) in the text. With b = 3 we obtain
3
ˆ
2 3 2
2 x4 9
xx dx =
c1 =
= ,
9 0
9 4
2
ci =
2
9J02 (3αi )
0
ˆ
3
xJ0 (αi x)x2 dx
t = αi x,
dt = αi dx
0
1
2
· 4
=
2
9J0 (3αi ) αi
2
=
4
9αi J02 (3αi )
ˆ
ˆ
0
3αi
t3 J0 (t) dt
0
3αi
t2
d
[tJ1 (t)] dt
dt
⎛
u = t2
du = 2t dt
⎞
d
dv = dt
[tJ1 (t)] dt
v = tJ1 (t)
3αi
ˆ 3αi
2
3
⎝
t J1 (t) − 2
=
t2 J1 (t) dt⎠ .
9αi4 J02 (3αi )
0
0
With n = 0 in equation (6) in the text we have J0 (x) = −J1 (x), so the boundary condition
J0 (3αi ) = 0 implies J1 (3αi ) = 0. Then
⎛
3αi ⎞
ˆ 3αi
2
2
d 2
⎝−2t2 J2 (t) ⎠
−2
t
J
(t)
dt
=
ci =
2
dt
9αi4 J02 (3αi )
9αi4 J02 (3αi )
0
0
=
−4J2 (3αi )
2
.
−18αi2 J2 (3αi ) = 2 2
9αi4 J02 (3αi )
αi J0 (3αi )
Thus
∞
f (x) =
J2 (3αi )
9
−4
2 J 2 (3α ) J0 (αi x).
2
α
i
0
i
i=1
11.5
Bessel and Legendre Series
10. The boundary condition indicates that we use (15) and (16) in the text. With b = 1 it follows
that
ˆ 1
2
x 1 − x2 J0 (αi x) dx
ci = 2
J1 (αi ) 0
ˆ 1
ˆ 1
2
3
xJ0 (αi x) dx −
x J0 (αi x) dx
t = αi x, dt = αi dx
= 2
J1 (αi ) 0
0
ˆ αi
ˆ αi
1
2
1
3
= 2
tJ0 (t) dt − 4
t J0 (t) dt
J1 (αi ) αi2 0
αi 0
ˆ αi
ˆ αi
1
1
u = t2
d
2
2 d
[tJ
[tJ
(t)]
dt
−
t
(t)]
dt
1
1
2
2
4
dt
du = 2t dt
J1 (αi ) αi 0 dt
αi 0
αi
αi
ˆ αi
1
2
1
= 2
tJ1 (t) − 4 t3 J1 (t) − 2
t2 J1 (t) dt
J1 (αi ) αi2
αi
0
=
0
d
dv = dt
[tJ1 (t)] dt
v = tJ1 (t)
0
αi ˆ αi
J1 (αi ) J1 (αi )
2 2
2
2
d 2
2
−
+ 4
t
J
(t)
= 2
t J2 (t) dt = 2
2
αi
αi
J1 (αi )
αi 0 dt
J1 (αi ) αi4
0
=
4J2 (αi )
.
αi2 J12 (αi )
Thus
f (x) = 4
11. (a)
∞
J2 (αi )
2 J 2 (α ) J0 (αi x).
α
i
1
i
i=1
y
4
2
5
10
15
20
25
30
x
−2
−4
(b) Using FindRoot in Mathematica we find the roots x1 = 2.9496, x2 = 5.8411,
x3 = 8.8727, x4 = 11.9561, and x5 = 15.0624.
(c) Dividing the roots in part (b) by 4 we find the eigenvalues α1 = 0.7374, α2 = 1.4603,
α3 = 2.2182, α4 = 2.9890, and α5 = 3.7656.
(d) The next five eigenvalues are α6 = 4.5451, α7 = 5.3263, α8 = 6.1085, α9 = 6.8915, and
α10 = 7.6749.
679
680
CHAPTER 11
FOURIER SERIES
12. (a) From Problem 7, the coefficients of the Fourier-Bessel series are
20αi J2 (4αi )
.
(2αi2 + 1)J12 (4αi )
ci =
Using a CAS we find c1 = 26.7896, c2 = −12.4624, c3 = 7.1404, c4 = −4.68705, and
c5 = 3.35619.
(b)
S1
20
S2
20
S3
20
S4
20
S5
20
15
15
15
15
15
10
10
10
10
10
5
5
5
5
5
1
(c)
2
3
4
5
x
1
S10
20
2
3
4
5
x
1
2
3
4
5
x
1
2
3
4
5
x
1
2
3
4
5
x
S10
20
15
15
10
10
5
5
10
2
1
3
4
x
20
30
40
x
50
−5
−10
13. Since f is expanded as a series of Bessel functions, J1 (αi x) and J1 is an odd function, the
series should represent an odd function.
y
14. (a) Since J0 is an even function, a series expansion of a
function defined on (0, 2) would converge to the even
extension of the function on (−2, 0).
2
1.5
1
0.5
−2
−1
2
1
(b) In Section 6.3 we saw that J2 (x) = 2J2 (x)/x − J3 (x). Since J2 is
even and J3 is odd we see that
20
J2 (−x) = 2J2 (−x)/(−x) − J3 (−x)
15
x
y
= −2J2 (x)/x + J3 (x) = −J2 (x),
10
J2
so that
is an odd function. Now, if f (x) = 3J2 (x) +
we see that
2xJ2 (x),
5
f (−x) = 3J2 (−x) − 2xJ2 (−x)
= 3J2 (x) + 2xJ2 (x) = f (x),
−4 −2
2
4
x
so that f is an even function. Thus, a series expansion of a function defined on (0, 4)
would converge to the even extension of the function on (−4, 0).
11.5
15. We compute
ˆ
1 1
c0 =
xP0 (x) dx =
2 0
ˆ
3 1
c1 =
xP1 (x) dx =
2 0
ˆ
5 1
xP2 (x) dx =
c2 =
2 0
7
c3 =
2
9
2
c4 =
ˆ
13
2
c6 =
0
ˆ
11
2
c5 =
1
1
2
3
2
5
2
ˆ
1
1
4
x dx =
0
ˆ
1
x2 dx =
0
ˆ
1
0
7
xP3 (x) dx =
2
0
9
2
1
xP5 (x) dx =
0
1
xP6 (x) dx =
0
0.5
1
2
−1
0.5
−0.5
1
x
5
1
(3x3 − x) dx =
2
16
xP4 (x) dx =
ˆ
681
S5
1
1
ˆ
Bessel and Legendre Series
ˆ
1
1
(5x4 − 3x2 ) dx = 0
2
1
3
1
(35x5 − 30x3 + 3x) dx = −
8
32
0
ˆ
0
11
2
13
2
ˆ
1
1
(63x6 − 70x4 + 15x2 ) dx = 0
8
1
13
1
(231x7 − 315x5 + 105x3 − 5x) dx =
.
16
256
0
ˆ
0
Thus
1
5
3
13
1
P6 (x) + · · · .
f (x) = P0 (x) + P1 (x) + P2 (x) − P4 (x) +
4
2
16
32
256
The figure above is the graph of S5 (x) = 14 P0 (x) + 12 P1 (x) +
16. We compute
ˆ
1 1 x
c0 =
e P0 (x) dx =
2 −1
ˆ
3 1 x
e P1 (x) dx =
c1 =
2 −1
ˆ
5 1 x
c2 =
e P2 (x) dx =
2 −1
5
16 P2 (x)
−
3
32 P4 (x)
+
S5
1
2
3
2
5
2
ˆ
ˆ
ˆ
3
1
1
ex dx = (e − e−1 )
2
−1
1
2
xex dx = 3e−1
1
−1
1
−1
1
(3x2 ex − ex ) dx
2
−1
−0.5
5
= (e − 7e−1 )
2
ˆ
7 1
e P3 (x) dx =
2 −1
−1
ˆ
ˆ
9 1 x
9 1
e P4 (x) dx =
c4 =
2 −1
2 −1
7
c3 =
2
ˆ
1
x
13
256 P6 (x).
1
7
(5x3 ex − 3xex ) dx = (−5e + 37e−1 )
2
2
9
1
(35x4 ex − 30x2 ex + 3ex ) dx = (36e − 266e−1 ).
8
2
0.5
1
x
682
CHAPTER 11
FOURIER SERIES
Thus
5
1
f (x) = (e − e−1 )P0 (x) + 3e−1 P1 (x) + (e − 7e−1 )P2 (x)
2
2
9
7
+ (−5e + 37e−1 )P3 (x) + (36e − 266e−1 )P4 (x) + · · · .
2
2
The figure above is the graph of S5 (x).
17. Using cos2 θ = 12 (cos 2θ + 1) we have
1
3
1
3
1
3
1
1
P2 (cos θ) = (3 cos2 θ − 1) = cos2 θ − = (cos 2θ + 1) − = cos 2θ + = (3 cos 2θ + 1).
2
2
2
4
2
4
4
4
18. From Problem 17 we have
1
P2 (cos θ) = (3 cos 2θ + 1)
4
4
1
cos 2θ = P2 (cos θ) − .
3
3
or
Then, using P0 (cos θ) = 1,
F (θ) = 1 − cos 2θ = 1 −
1
4
P2 (cos θ) −
3
3
4 4
4
4
− P2 (cos θ) = P0 (cos θ) − P2 (cos θ).
3 3
3
3
=
19. If f is an even function on (−1, 1) then
ˆ
ˆ 1
f (x)P2n (x) dx = 2
−1
1
f (x)P2n (x) dx
0
and
ˆ
1
−1
f (x)P2n+1 (x) dx = 0.
Thus
c2n =
2(2n) + 1
2
ˆ
1
−1
f (x)P2n (x) dx =
4n + 1
2
ˆ
2
1
f (x)P2n (x) dx
0
ˆ
1
f (x)P2n (x) dx,
= (4n + 1)
0
c2n+1 = 0, and
f (x) =
∞
c2n P2n (x).
n=0
20. If f is an odd function on (−1, 1) then
ˆ 1
−1
and
ˆ
f (x)P2n (x) dx = 0
ˆ
1
−1
1
f (x)P2n+1 (x) dx = 2
f (x)P2n+1 (x) dx.
0
11.5
Bessel and Legendre Series
683
Thus
c2n+1
2(2n + 1) + 1
=
2
ˆ
1
4n + 3
f (x)P2n+1 (x) dx =
2
−1
ˆ
2
1
f (x)P2n+1 (x) dx
0
ˆ
1
f (x)P2n+1 (x) dx,
= (4n + 3)
0
c2n = 0, and
f (x) =
∞
c2n+1 P2n+1 (x).
n=0
21. From (26) in Problem 19 in the text we find
ˆ 1
ˆ 1
1
xP0 (x) dx =
x dx =
c0 =
2
0
0
ˆ 1
ˆ 1
5
1
c2 = 5
(3x3 − x) dx =
xP2 (x) dx = 5
2
8
0
0
ˆ 1
ˆ 1
3
1
c4 = 9
(35x5 − 30x3 + 3x) dx = −
xP4 (x) dx = 9
8
16
0
0
and
ˆ
ˆ
1
c6 = 13
1
xP6 (x) dx = 13
0
0
S4
1
0.5
−1
−0.5
0.5
1
x
13
1
(231x7 − 315x5 + 105x3 − 5x) dx =
.
16
128
Hence, from (25) in the text,
1
5
3
13
f (x) = P0 (x) + P2 (x) − P4 (x) +
P6 + · · · .
2
8
16
128
On the interval −1 < x < 1 this series represents the function f (x) = |x|.
22. From (28) in Problem 20 in the text we find
ˆ 1
ˆ 1
3
P1 (x) dx = 3
x dx = ,
c1 = 3
2
0
0
ˆ 1
ˆ 1
7
1 3
c3 = 7
P3 (x) dx = 7
5x − 3x dx = − ,
8
0
0 2
ˆ 1
ˆ 1
11
1
c5 = 11
P5 (x) dx = 11
63x5 − 70x3 + 15x dx =
16
0
0 8
and
S4
1
0.5
−1
−0.5
0.5
−0.5
−1
ˆ
c7 = 15
ˆ
1
1
P7 (x) dx = 15
0
0
75
1 .
429x7 − 693x5 + 315x3 − 35x dx = −
16
128
Hence, from (27) in the text,
7
11
75
3
P7 (x) + · · · .
f (x) = P1 (x) − P3 (x) + P5 (x) −
2
8
16
128
1
x
684
CHAPTER 11
FOURIER SERIES
On the interval −1 < x < 1 this series represents the odd function
f (x) =
−1, −1 < x < 0
1,
0 < x < 1.
23. Since there is a Legendre polynomial of any specified degree, every polynomial can be represented as a finite linear combination of Legendre polynomials.
24. We want to express both x2 and x3 as linear combinations of P0 (x) = 1, P1 (x) = x,
P2 (x) = 12 (3x2 − 1), and P3 (x) = 12 (5x3 − 3x). Setting
x2 = c0 P0 (x) + c1 P1 (x) + c2 P2 (x) = c0 + c1 x + c2
1
3
3
(3x2 − 1) = c0 − c2 + c1 x + c2 x2 ,
2
2
2
we obtain the system
c0 −
1
c2 = 0
2
c1 = 0
3
c2 = 1.
2
The solution is c0 = 13 , c1 = 0, c2 = 23 . Thus, x2 =
1
3
P0 (x) + 23 P2 (x). Setting
3
x = c0 P0 (x) + c1 P1 (x) + c2 P2 (x) + c3 P3 (x) = c0 + c1 x + c2
1
1
2
3
(3x − 1) + c3 (5x − 3x)
2
2
3
1
3
5
= c0 − c2 + c1 − c3 x + c2 x2 + c3 x3 ,
2
2
2
2
we obtain the system
c0 −
1
c2 = 0
2
c1 −
3
c3 = 0
2
3
c2 = 0
2
5
c3 = 1.
2
The solution is c0 = 0, c1 = 35 , c2 = 0, c3 = 25 . Thus x3 =
3
5
P1 (x) + 25 P3 (x).
Chapter 11 in Review
Chapter 11 in Review
1. True, since
´π
−π
(x2 − 1)x5 dx = 0
2. Even, since if f and g are odd then h(−x) = f (−x)g(−x) = −f (x)[−g(x)] = f (x)g(x) = h(x)
3. cosine, since f is even
4. True
5. False; the Sturm-Liouville problem,
d
[r(x)y ] + λp(x)y = 0,
dx
on the interval [a, b], has eigenvalue λ = 0.
y (a) = 0,
y (b) = 0,
6. Periodically extending the function we see that at x = −1 the function converges to 12 (−1 +
0) = − 12 ; at x = 0 it converges to 12 (0 + 1) = 12 , and at x = 1 it converges to 12 (−1 + 0) = − 12 .
7. The Fourier series will converge to 1, the cosine series to 1, and the sine series to 0 at x = 0.
Respectively, this is because the rule (x2 + 1) defining f (x) determines a continuous function
on (−3, 3), the even extension of f to (−3, 0) is continuous at 0, and the odd extension of f
to (−3, 0) approaches −1 as x approaches 0 from the left.
8. cos 5x, since the general solution is y = c1 cos αx + c2 sin αx and y (0) = 0 implies c2 = 0.
9. Since the coefficient of y in the differential equation is n2 , the weight function is the integrating
factor
√
´
1
1
1
1
1 − x2
1 ´ (b/a) dx
2)
− x 2 dx
ln
(1−x
1−x
2
e
=
e
=
e
=
=√
2
2
2
a(x)
1−x
1−x
1−x
1 − x2
on the interval [−1, 1]. The orthogonality relation is
ˆ 1
1
√
Tm (x)Tn (x) dx = 0,
1 − x2
−1
m = n.
10. Since Pn (x) is orthogonal to P0 (x) = 1 for n > 0,
ˆ 1
ˆ 1
Pn (x) dx =
P0 (x)Pn (x) dx = 0.
−1
−1
11. We know from a half-angle formula in trigonometry that cos2 x =
cosine series.
1
2
+
1
2
cos 2x, which is a
12. (a) For m = n
ˆ ˆ L
n−m
(2n + 1)π
(2m + 1)π
1 L
n+m+1
cos
sin
x sin
x dx =
πx − cos
πx dx
2L
2L
2 0
L
L
0
= 0.
685
686
CHAPTER 11
FOURIER SERIES
(b) From
ˆ
0
L
(2n + 1)π
x dx =
sin
2L
L
ˆ
2
we see that
0
1 1
(2n + 1)π
L
− cos
x dx =
2 2
L
2
sin (2n + 1)π
2L
13. Since
ˆ
a0 =
L
x
= 2 .
0
(−2x) dx = 1,
−1
ˆ
an =
0
(−2x)cos nπx dx =
−1
and
ˆ
bn =
2
[(−1)n − 1],
n2 π 2
0
(−2x) sin nπx dx =
−1
2
(−1)n
nπ
for n = 1, 2, 3, . . . we have
∞
1 f (x) = +
2
n=1
2
2
n
n
(−1) sin nπx .
[(−1) − 1] cos nπx +
n2 π 2
nπ
14. Since
ˆ
1
2
(2x2 − 1) dx = − ,
3
−1
ˆ 1
an =
(2x2 − 1) cos nπx dx =
a0 =
−1
and
ˆ
bn =
1
−1
8
n2 π 2
(−1)n ,
(2x2 − 1) sin nπx dx = 0
for n = 1, 2, 3, . . . we have
∞
1 8
(−1)n cos nπx.
f (x) = − +
3
n2 π 2
n=1
15. (a) Since
2
a0 =
1
and
an =
2
1
ˆ
1
ˆ
1
e−x dx = 2(1 − e−1 )
0
e−x cos nπx dx =
−1
2
[1 − (−1)n e−1 ],
1 + n2 π 2
for n = 1, 2, 3, . . . , we have the cosine series
f (x) = 1 − e−1 + 2
∞
1 − (−1)n e−1
n=1
1 + n2 π 2
cos nπx.
Chapter 11 in Review
(b) Since
2
bn =
1
ˆ
1
e−x sin nπx dx =
0
687
2nπ
[1 − (−1)n e−1 ],
1 + n2 π 2
for n = 1, 2, 3, . . . , we have the sine series
f (x) = 2π
∞
n[1 − (−1)n e−1 ]
n=1
16.
3
2
1 + n2 π 2
f
3
f
3
2
2
1
1
1
1
2
3
x
3
2
2
1
1
1
1
2
2
3
3
f (x) = |x| − x, −1 < x < 1
3
sin nπx.
f
3
2
2
1
1
1
2
3
x
3
2
1
1
1
1
2
2
3
3
f (x) = e−|x|
3
x
f (x) = 2x2 − 1, −1 < x < 1
f
3
1
2
⎧
⎪
−e−x , 0 < x < 1
⎪
⎨
x=0
f (x) = 0,
⎪
⎪
⎩−ex ,
−1 < x < 0
2
3
x
688
CHAPTER 11
FOURIER SERIES
17. The cosine series of f in Problem 15 converges to F (x) on the interval −1 < x < 1 since F is
the even extension of f to the interval.
18. Expanding in a full Fourier series we have
1
a0 =
2
ˆ
ˆ
2
4
x dx +
0
2
(−1)n − 1
nπx
dx = 2
2 cos
2
n2 π 2
0
2
ˆ 2
ˆ 4
1
nπx
nπx
2
dx +
dx = −
bn =
x sin
2 sin
2
2
2
nπ
0
2
1
an =
2
ˆ
2 dx = 3
2
nπx
dx +
x cos
2
so
∞
3
f (x) = + 2
2
n=1
ˆ
4
1
nπx
(−1)n − 1
nπx
−
sin
.
cos
n2 π 2
2
nπ
2
19. For λ = α2 > 0 a general solution of the given differential equation is
y = c1 cos (3α ln x) + c2 sin (3α ln x)
and
y = −
3c1 α
3c2 α
sin (3α ln x) +
cos (3α ln x).
x
x
Since ln 1 = 0, the boundary condition y (1) = 0 implies c2 = 0. Therefore
y = c1 cos (3α ln x).
Using ln e = 1 we find that y(e) = 0 implies c1 cos 3α = 0 or 3α = (2n − 1)π/2, for n = 1,
2, 3, . . . . The eigenvalues are λ = α2 = (2n − 1)2 π 2 /36 with corresponding eigenfunctions
cos [(2n − 1)π(ln x)/2] for n = 1, 2, 3, . . . .
20. To obtain the self-adjoint form of the differential equation in Problem 19 we note that an in´
tegrating factor is (1/x2 )e dx/x = 1/x. Thus the weight function is 1/x and an orthogonality
relation is
ˆ e
2n − 1
2m − 1
1
cos
π ln x cos
π ln x dx = 0, m = n.
2
2
1 x
Chapter 11 in Review
21. The boundary condition indicates that we use (15) and (16) of Section 11.5 in the text. With
b = 4 we obtain
ˆ 4
2
ci =
xJ0 (αi x)f (x) dx
16J12 (4αi ) 0
ˆ 2
1
xJ0 (αi x) dx
t = αi x, dt = αi dx
=
8J12 (4αi ) 0
ˆ 2αi
1
1
·
=
tJ0 (t) dt
8J12 (4αi ) αi2 0
ˆ 2αi
d
1
[tJ1 (t)] dt
[From (5) in 11.5 in the text]
=
2
dt
8J1 (4αi ) 0
2αi
J1 (2αi )
1
tJ
.
(t)
=
=
1
2
8J1 (4αi )
4αi J12 (4αi )
0
Thus
∞
f (x) =
1 J1 (2αi )
2 (4α ) J0 (αi x.
4
α
J
i
i
1
i=1
22. Since f (x) = x4 is a polynomial in x, an expansion of f in Legendre polynomials in x must
terminate with the term having the same degree as f . Using the fact that x4 P1 (x) and
x4 P3 (x) are odd functions, we see immediately that c1 = c3 = 0. Now
1
c0 =
2
c2 =
c4 =
Thus
5
2
9
2
ˆ
ˆ
ˆ
1
1
x P0 (x) dx =
2
−1
1
−1
1
−1
ˆ
4
x4 P2 (x) dx =
x4 P4 (x) dx =
5
2
9
2
ˆ
ˆ
1
−1
1
−1
1
−1
x4 dx =
1
5
4
1
(3x6 − x4 ) dx =
2
7
8
1
(35x8 − 30x6 + 3x4 ) dx =
.
8
35
4
8
1
f (x) = P0 (x) + P2 (x) + P4 (x).
5
7
35
23. (a)
fe (x) + fo (x) =
2f (x)
f (x) + f (−x) + f (x) − f (−x)
=
= f (x)
2
2
(b)
fe (−x) =
f (−x) + f (−(−x))
f (−x) + f (x)
=
= fe (x)
2
2
fo (−x) =
f (−x) − f (x)
f (−x) − f (−(−x))
=
= −fo (x)
2
2
689
690
CHAPTER 11
FOURIER SERIES
24.
fe (x) =
f (x) + f (−x)
ex + e−x
=
= cosh x
2
2
fo (x) =
ex − e−x
f (x) − f (−x)
=
= sinh x
2
2
25. If we let u = x + 2p then dx = du and we get
ˆ a
ˆ a+2p
f (x) dx =
f (u − 2p) du ←− since f is 2p − periodic, f (u − 2p) = f (u)
0
2p
ˆ
a+2p
f (u) du
=
2p
ˆ
a+2p
f (x) dx
=
2p
Therefore we get
ˆ
ˆ
a+2p
a
ˆ
2p
f (x) dx =
a+2p
f (x) dx +
f (x) dx
a
ˆ
2p
ˆ
2p
f (x) dx +
=
f (x) dx
a
ˆ
0
ˆ
a
f (x) dx +
=
0
ˆ
2p
0
2p
f (x) dx
a
f (x) dx
=
a
Chapter 12
Boundary-Value Problems in Rectangular Coordinates
12.1
Separable Partial Differential Equations
1. Substituting u(x, y) = X(x)Y (y) into the partial differential equation yields X Y = XY .
Separating variables and using the separation constant −λ, where λ = 0, we obtain
Y
X
=
= −λ.
X
Y
When λ = 0
X + λX = 0
and
Y + λY = 0
X = c1 e−λx
and
Y = c2 e−λy .
so that
A particular product solution of the partial differential equation is
u = XY = c3 e−λ(x+y) ,
λ = 0.
When λ = 0 the differential equations become X = 0 and Y = 0, so in this case X = c4 ,
Y = c5 , and u = XY = c6 .
2. Substituting u(x, y) = X(x)Y (y) into the partial differential equation yields X Y +3XY = 0.
Separating variables and using the separation constant −λ we obtain
Y
X
=
= −λ.
−3X
Y
When λ = 0
X − 3λX = 0
and
Y + λY = 0
so that
X = c1 e3λx
and
Y = c2 e−λy .
A particular product solution of the partial differential equation is
u = XY = c3 eλ(3x−y) .
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CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
When λ = 0 the differential equations become X = 0 and Y = 0, so in this case X = c4 ,
Y = c5 , and u = XY = c6 .
3. Substituting u(x, y) = X(x)Y (y) into the partial differential equation yields
X Y + XY = XY . Separating variables and using the separation constant −λ we obtain
Y −Y
X
=
= −λ.
X
Y
Then
X + λX = 0
and
Y − (1 + λ)Y = 0
so that
X = c1 e−λx
and
Y = c2 e(1+λ)y .
A particular product solution of the partial differential equation is
u = XY = c3 ey+λ(y−x) .
4. Substituting u(x, y) = X(x)Y (y) into the partial differential equation yields
X Y = XY + XY . Separating variables and using the separation constant −λ we obtain
Y +Y
X
=
= −λ.
X
Y
Then
X + λX = 0
and
y + (1 + λ)Y = 0
so that
X = c1 e−λx
Y = c2 e−(1+λ)y = 0.
and
A particular product solution of the partial differential equation is
u = XY = c3 e−y−λ(x+y) .
5. Substituting u(x, y) = X(x)Y (y) into the partial differential equation yields xX Y = yXY .
Separating variables and using the separation constant −λ we obtain
yY xX =
= −λ.
X
Y
When λ = 0
xX + λX = 0
and
yY + λY = 0
X = c1 x−λ
and
Y = c2 y −λ .
so that
A particular product solution of the partial differential equation is
u = XY = c3 (xy)−λ .
When λ = 0 the differential equations become X = 0 and Y = 0, so in this case X = c4 ,
Y = c5 , and u = XY = c6 .
12.1
Separable Partial Differential Equations
6. Substituting u(x, y) = X(x)Y (y) into the partial differential equation yields
yX Y + xXY = 0. Separating variables and using the separation constant −λ we obtain
X
Y
=−
= −λ.
xX
yY
When λ = 0
X + λxX = 0
and
Y − λyY = 0
so that
X = c1 eλx
2 /2
and
Y = c2 e−λy
2 /2
.
A particular product solution of the partial differential equation is
u = XY = c3 eλ(x
2 −y 2 )/2
.
When λ = 0 the differential equations become X = 0 and Y = 0, so in this case X = c4 ,
Y = c5 , and u = XY = c6 .
7. Substituting u(x, y) = X(x)Y (y) into the partial differential equation yields
X Y + X Y + XY = 0 , which is not separable.
8. Substituting u(x, y) = X(x)Y (y) into the partial differential equation yields yX Y +XY = 0.
Separating variables and using the separation constant −λ we obtain
Y
X
= − = −λ.
X
yY
When λ = 0
X + λX = 0
and
λyY − Y = 0
X = c1 e−λx
and
Y = c2 y 1/λ .
so that
A particular product solution of the partial differential equation is
u = XY = c3 e−λx y 1/λ .
In this case λ = 0 yields no solution.
9. Substituting u(x, t) = X(x)T (t) into the partial differential equation yields
kX T − XT = XT . Separating variables and using the separation constant −λ we obtain
T
kX − X
=
= −λ.
X
T
Then
X +
λ−1
X=0
k
and
T + λT = 0.
693
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CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
The second differential equation implies T (t) = c1 e−λt . For the first differential equation we
consider three cases:
I. If (λ − 1)/k = 0 then λ = 1, X = 0, and X(x) = c2 x + c3 , so
u = XT = e−t (A1 x + A2 ).
II. If (λ − 1)/k = −α2 < 0, then λ = 1 − kα2 , X − α2 X = 0, and
X(x) = c4 cosh αx + c5 sinh αx, so
u = XT = (A3 cosh αx + A4 sinh αx)e−(1−kα
2 )t
.
III. If (λ−1)/k = α2 > 0, then λ = 1+kα2 , X +α2 X = 0, and X(x) = c6 cos αx+c7 sin αx,
so
2
u = XT = (A5 cos αx + A6 sin αx)e−(1+λα )t .
10. Substituting u(x, t) = X(x)T (t) into the partial differential equation yields kX T = XT .
Separating variables and using the separation constant −λ we obtain
T
X =
= −λ.
X
kT
Then
X + λX = 0
and
T + λkT = 0.
The second differential equation implies T (t) = c1 e−λkt . For the first differential equation we
consider three cases:
I. If λ = 0 then X = 0 and X(x) = c2 x + c3 , so
u = XT = A1 x + A2 .
II. If λ = −α2 < 0, then X − α2 X = 0, and X(x) = c4 cosh αx + c5 sinh αx, so
2
u = XT = (A3 cosh αx + A4 sinh αx)ekα t .
III. If λ = α2 > 0, then X + α2 X = 0, and X(x) = c6 cos αx + c7 sin αx, so
u = XT = (A5 cos αx + A6 sin αx)e−kα t .
2
11. Substituting u(x, t) = X(x)T (t) into the partial differential equation yields a2 X T = XT .
Separating variables and using the separation constant −λ we obtain
T X = 2 = −λ.
X
a T
Then
X + λX = 0
and
T + a2 λT = 0.
12.1
Separable Partial Differential Equations
We consider three cases:
I. If λ = 0 then X = 0 and X(x) = c1 x + c2 . Also, T = 0 and T (t) = c3 t + c4 , so
u = XT = (c1 x + c2 )(c3 t + c4 ).
II. If λ = −α2 < 0, then X − α2 X = 0, and X(x) = c5 cosh αx + c6 sinh αx. Also,
T − α2 a2 T = 0 and T (t) = c7 cosh αat + c8 sinh αat, so
u = XT = (c5 cosh αx + c6 sinh αx)(c7 cosh αat + c8 sinh αat).
III. If λ = α2 > 0, then X + α2 X = 0, and X(x) = c9 cos αx + c10 sin αx. Also,
T + α2 a2 T = 0 and T (t) = c11 cos αat + c12 sin αat, so
u = XT = (c9 cos αx + c10 sin αx)(c11 cos αat + c12 sin αat).
12. Substituting u(x, t) = X(x)T (t) into the partial differential equation yields
a2 X T = XT + 2kXT . Separating variables and using the separation constant −λ we
obtain
T + 2kT X =
= −λ.
X
a2 T
Then
X + λX = 0
and
T + 2kT + a2 λT = 0.
We consider three cases:
I. If λ = 0 then X = 0 and X(x) = c1 x + c2 . Also, T + 2kT = 0 and T (t) = c3 + c4 e−2kt ,
so
u = XT = (c1 x + c2 )(c3 + c4 e−2kt ).
II. If λ = −α2 < 0, then X − α2 X = 0, and X(x) = c5 cosh αx + c6 sinh αx. The auxiliary
2 + 2km − α2 a2 = 0. Solving for m we obtain
equation of T + 2kT − α2 a2 T = 0 is m
√
√
√
2
2 2
2
2 2
m = −k ± k 2 + α2 a2 , so T (t) = c7 e(−k+ k +α a )t + c8 e(−k− k +α a )t . Then
√
√
2
2 2
2
2 2
u = XT = (c5 cosh αx + c6 sinh αx) c7 e(−k+ k +α a )t + c8 e(−k− k +α a )t .
III. If λ = α2 > 0, then X + α2 X = 0, and X(x) = c9 cos αx + c10 sin αx. The auxiliary
equation of T + 2kT + α2 a2 T = 0 is m2 + 2km + α2 a2 = 0. Solving for m we obtain
√
m = −k ± k 2 − α2 a2 . We consider three possibilities for the discriminant k 2 − α2 a2 :
(i) If k 2 − α2 a2 = 0 then T (t) = c11 e−kt + c12 te−kt and
u = XT = (c9 cos αx + c10 sin αx)(c11 e−kt + c12 te−kt ).
From k 2 − α2 a2 = 0 we have α = k/a so the solution can be written
u = XT = (c9 cos kx/a + c10 sin kx/a)(c11 e−kt + c12 te−kt ).
695
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CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
√
√
(ii) If k 2 − α2 a2 < 0 then T (t) = e−kt c13 cos α2 a2 − k 2 t + c14 sin α2 a2 − k 2 t and
u = XT = (c9 cos αx+c10 sin αx)e−kt c13 cos α2 a2 − k 2 t + c14 sin α2 a2 − k 2 t .
√
√
(iii) If k 2 − α2 a2 > 0 then T (t) = c15 e(−k+ k −α a )t + c16 e(−k− k −α a )t and
√
√
2
2 2
2
2 2
u = XT = (c9 cos αx + c10 sin αx) c15 e(−k+ k −α a )t + c16 e(−k− k −α a )t .
2
2 2
2
2 2
13. Substituting u(x, y) = X(x)Y (y) into the partial differential equation yields X Y +XY = 0.
Separating variables and using the separation constant −λ we obtain
−
Y X =
= −λ.
X
Y
Then
X − λX = 0
and
Y + λY = 0.
We consider three cases:
I. If λ = 0 then X = 0 and X(x) = c1 x + c2 . Also, Y = 0 and Y (y) = c3 y + c4 so
u = XY = (c1 x + c2 )(c3 x + c4 ).
II. If λ = −α2 < 0 then X + α2 X = 0 and X(x) = c5 cos αx + c6 sin αx. Also, Y − α2 Y = 0
and Y (y) = c7 cosh αy + c8 sinh αy so
u = XY = (c5 cos αx + c6 sin αx)(c7 cosh αy + c8 sinh αy).
III. If λ = α2 > 0 then X − α2 X = 0 and X(x) = c9 cosh αx + c10 sinh αx. Also,
Y + α2 Y = 0 and Y (y) = c11 cos αy + c12 sin αy so
u = XY = (c9 cosh αx + c10 sinh αx)(c11 cos αy + c12 sin αy).
14. Substituting u(x, y) = X(x)Y (y) into the partial differential equation yields
x2 X Y + XY = 0. Separating variables and using the separation constant −λ we obtain
−
Y x2 X =
= −λ.
X
Y
Then
x2 X − λX = 0
and
Y + λY = 0.
We consider three cases:
I. If λ = 0 then x2 X = 0 and X(x) = c1 x + c2 . Also, Y = 0 and Y (y) = c3 y + c4 so
u = XY = (c1 x + c2 )(c3 y + c4 ).
12.1
Separable Partial Differential Equations
II. If λ = −α2 < 0 then x2 X + α2 X = 0 and Y − α2 Y = 0. The solution of the second
differential equation is Y (y) = c5 cosh αy + c6 sinh αy. The first equation is Cauchy-Euler
√
with auxiliary equation m2 − m + α2 = 0. Solving for m we obtain m = 12 ± 12 1 − 4α2 . We
consider three possibilities for the discriminant 1 − 4α2 .
(i) If 1 − 4α2 = 0 then X(x) = c7 x1/2 + c8 x1/2 ln x and
u = XY = x1/2 (c7 + c8 ln x)(c5 cosh αy + c6 sinh αy).
(ii) If 1 − 4α2 < 0 then
1/2
√
c9 cos
X(x) = x
√
4α2 − 1
4α2 − 1
ln x + c10 sin
ln x
2
2
and
√
u = XY = x1/2 c9 cos
4α2 − 1
ln x
2
√
+c10 sin
4α2 − 1
ln x
(c5 cosh αy + c6 sinh αy).
2
√
√
2
2
(iii) If 1 − 4α2 > 0 then X(x) = x1/2 c11 x 1−4α /2 + c12 x− 1−4α /2 and
1/2
u = XY = x
√
c11 x
1−4α2 /2
√
− 1−4α2 /2
+ c12 x
(c5 cosh αy + c6 sinh αy).
III. If λ = α2 > 0 then x2 X − α2 X = 0 and Y + α2 Y = 0. The solution of the second
differential equation is Y (y) = c13 cos αy + c14 sin αy. The first equation is Cauchy-Euler
√
with auxiliary equation m2 − m − α2 = 0. Solving for m we obtain m = 12 ± 12 1 + 4α2 .
In this case the
is always
positive
discriminant
so the solution of the differential equation is
√
√
2 /2
2 /2
1/2
1+4α
−
1+4α
c15 x
and
+ c16 x
X(x) = x
√
√
2
2
u = XY = x1/2 c15 x 1+4α /2 + c16 x− 1+4α /2 (c13 cos αy + c14 sin αy).
15. Substituting u(x, y) = X(x)Y (y) into the partial differential equation yields X Y + XY =
XY . Separating variables and using the separation constant −λ we obtain
Y − Y X =
= −λ.
X
Y
Then
X + λX = 0
We consider three cases:
and
Y − (1 + λ)Y = 0.
697
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CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
I. If λ = 0 then X = 0 and X(x) = c1 x+c2 . Also Y −Y = 0 and Y (y) = c3 cosh y+c4 sinh y
so
u = XY = (c1 x + c2 )(c3 cosh y + c4 sinh y).
II. If λ = −α2 < 0 then X − α2 X = 0 and Y + (α2 − 1)Y = 0. The solution of the first
differential equation is X(x) = c5 cosh αx + c6 sinh αx. The solution of the second differential
equation depends on the nature of α2 − 1. We consider three cases:
(i) If α2 − 1 = 0, or α2 = 1, then Y (y) = c7 y + c8 and
u = XY = (c5 cosh αx + c6 sinh αx)(c7 y + c8 ).
√
√
(ii) If α2 − 1 < 0, or 0 < α2 < 1, then Y (y) = c9 cosh 1 − α2 y + c10 sinh 1 − α2 y and
u = XY = (c5 cosh αx + c6 sinh αx) c9 cosh 1 − α2 y + c10 sinh 1 − α2 y .
√
√
(iii) If α2 − 1 > 0, or α2 > 1, then Y (y) = c11 cos α2 − 1 y + c12 sin α2 − 1 y and
u = XY = (c5 cosh αx + c6 sinh αx) c11 cos α2 − 1 y + c12 sin α2 − 1 y .
III. If λ = α2 > 0, then X + α2 X = 0 and X(x) = c13 cos αx + c14 sin αx. Also,
√
√
Y − (1 + α2 )Y = 0 and Y (y) = c15 cosh 1 + α2 y + c16 sinh 1 + α2 y so
u = XY = (c13 cos αx + c14 sin αx) c15 cosh 1 + α2 y + c16 sinh 1 + α2 y .
16. Substituting u(x, t) = X(x)T (t) into the partial differential equation yields a2 X T −g = XT ,
which is not separable.
17. Identifying A = B = C = 1, we compute B 2 − 4AC = −3 < 0. The equation is elliptic.
18. Identifying A = 3, B = 5, and C = 1, we compute B 2 − 4AC = 13 > 0. The equation is
hyperbolic.
19. Identifying A = 1, B = 6, and C = 9, we compute B 2 − 4AC = 0. The equation is parabolic.
20. Identifying A = 1, B = −1, and C = −3, we compute B 2 − 4AC = 13 > 0. The equation is
hyperbolic.
21. Identifying A = 1, B = −9, and C = 0, we compute B 2 − 4AC = 81 > 0. The equation is
hyperbolic.
22. Identifying A = 0, B = 1, and C = 0, we compute B 2 − 4AC = 1 > 0. The equation is
hyperbolic.
23. Identifying A = 1, B = 2, and C = 1, we compute B 2 − 4AC = 0. The equation is parabolic.
12.1
Separable Partial Differential Equations
24. Identifying A = 1, B = 0, and C = 1, we compute B 2 − 4AC = −4 < 0. The equation is
elliptic.
25. Identifying A = a2 , B = 0, and C = −1, we compute B 2 − 4AC = 4a2 > 0. The equation is
hyperbolic.
26. Identifying A = k > 0, B = 0, and C = 0, we compute B 2 − 4AC = 0. The equation is
parabolic.
27. Substituting u(r, t) = R(r)T (t) into the partial differential equation yields
1
k R T + R T
r
= RT .
Separating variables and using the separation constant −λ we obtain
rR + R
T
=
= −λ.
rR
kT
Then
rR + R + λrR = 0
and
T + λkT = 0.
Letting λ = α2 and writing the first equation as r2 R + rR = α2 r2 R = 0 we see that it is a
parametric Bessel equation of order 0. As discussed in Chapter 5 of the text, it has solution
2
R(r) = c1 J0 (αr) + c2 Y0 (αr). Since a solution of T + α2 kT is T (t) = e−kα t , we see that a
solution of the partial differential equation is
u = RT = e−kα t [c1 J0 (αr) + c2 Y0 (αr)].
2
28. Substituting u(r, θ) = R(r)Θ(θ) into the partial differential equation yields
R Θ +
1 1
R Θ + 2 RΘ = 0.
r
r
Separating variables and using the separation constant −λ we obtain
r2 R + rR
Θ
=−
= −λ.
R
Θ
Then
r2 R + rR + λR = 0
and
Θ − λΘ = 0.
Letting λ = −α2 we have the Cauchy-Euler equation r2 R + rR − α2 R = 0 whose solution
is R(r) = c3 rα + c4 r−α . Since the solution of Θ + α2 Θ = 0 is Θ(θ) = c1 cos αθ + c2 sin αθ we
see that a solution of the partial differential equation is
u = RΘ = (c1 cos αθ + c2 sin αθ)(c3 rα + c4 r−α ).
699
700
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
29. For u = A1 + B1 x we compute ∂ 2 u/∂x2 = 0 = ∂u/∂y. Then ∂ 2 u/∂x2 = 4 ∂u/∂y.
For u = A2 eα
2y
cos 2αx + B2 eα
2y
sin 2αx we compute
∂u
2
2
= 2αA2 eα y sinh 2αx + 2αB2 eα y cosh 2αx
∂x
∂2u
2
2
= 4α2 A2 eα y cosh 2αx + 4α2 B2 eα y sinh 2αx
2
∂x
and
∂u
2
2
= α2 A2 eα y cosh 2αx + α2 B2 eα y sinh 2αx.
∂y
Then ∂ 2 u/∂x2 = 4 ∂u/∂y.
For u = A3 e−α
2y
cosh 2αx + B3 e−α
2y
sinh 2αx we compute
∂u
2
2
= −2αA3 e−α y sin 2αx + 2αB3 e−α y cos 2αx
∂x
∂2u
2
2
= −4α2 A3 e−α y cos 2αx − 4α2 B3 e−α y sin 2αx
2
∂x
and
∂u
2
2
= −α2 A3 e−α y cos 2αx − α2 B3 e−α y sin 2αx.
∂y
Then ∂ 2 u/∂x2 = 4 ∂u/∂y.
30. We identify A = xy +1, B = x+2y, and C = 1. Then B 2 −4AC = x2 +4y 2 −4. The equation
x2 + 4y 2 = 4 defines an ellipse. The partial differential equation is hyperbolic outside the
ellipse, parabolic on the ellipse, and elliptic inside the ellipse.
31. Assuming u(x, y) = X(x)Y (y) and substituting into ∂ 2 u/∂x2 − u = 0 we get X Y − XY = 0
or Y (X − X) = 0. This implies X(x) = c1 ex or X(x) = c2 e−x . For these choices of X, Y
can be any function of y. Two solutions of the partial differential equation are then
u1 (x, y) = A(y)ex
and u2 (x, y) = B(y)e−x .
Since the partial differential equation is linear and homogeneous the superposition principle
indicates that another solution is
u(x, y) = u1 (x, y) + u2 (x, y) = A(y)ex + B(y)e−x .
32. Assuming u(x, y) = X(x)Y (y) and substituting into ∂ 2 u/∂x∂y + ∂u/∂x = 0 we get
X Y + X Y = 0 or X (Y + Y ) = 0. This implies Y (y) = c1 e−y . For this choice of Y , X can
be any function of x. A solution of the partial differential equation is then u(x, y) = A(x)e−y .
In addition, noting that the partial differential equation can be written
∂ ∂u
+ u = 0,
∂x ∂y
12.2
Classical PDEs and Boundary-Value Problems
any function u2 (x, y) = B(y) will satisfy the partial differential equation since, in this case,
∂u2 /∂y + u2 = B (y) + B(y) and the x-partial of B (y) + B(y) is 0. Thus, using the superposition principle, a solution of the partial differential equation is
u(x, y) = u1 (x, y) + u2 (x, y) = A(x)e−y + B(y).
12.2
1. k
Classical PDEs and Boundary-Value Problems
∂2u
∂u
,
=
2
∂x
∂t
u(0, t) = 0,
0 < x < L, t > 0
∂u
∂x
u(x, 0) = f (x),
= 0,
t>0
x=L
0<x<L
∂2u
∂u
, 0 < x < L, t > 0
=
∂x2
∂t
u(0, t) = u0 , u(L, t) = u1 , t > 0
2. k
u(x, 0) = 0,
3. k
0<x<L
∂2u
∂u
,
=
2
∂x
∂t
u(0, t) = 100,
u(x, 0) = f (x),
4. k
0 < x < L, t > 0
∂u
∂x
= −hu(L, t),
x=L
0<x<L
∂2u
∂u
,
+ h(u − 50) =
2
∂x
∂t
∂u
∂x
= 0,
x=0
u(x, 0) = 100,
t>0
∂u
∂x
= 0,
0 < x < L, t > 0
t>0
x=L
0<x<L
∂2u
∂u
, 0 < x < L, t > 0, h a constant
− hu =
2
∂x
∂t
πt
u(0, t) = sin
, u(L, t) = 0, t > 0
L
u(x, 0) = f (x), 0 < x < L
5. k
6. k
∂2u
∂u
,
+ h(u − 50) =
∂x2
∂t
∂u
∂x
= 0,
x=0
u(x, 0) = 100,
∂u
∂x
= 0,
x=L
0<x<L
0 < x < L, t > 0
t>0
701
702
CHAPTER 12
7. a2
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
∂2u
∂2u
=
,
∂x2
∂t2
u(0, t) = 0,
0 < x < L, t > 0
u(L, t) = 0,
u(x, 0) = x(L − x),
8. a2
∂2u
∂2u
= 2 ,
2
∂x
∂t
u(0, t) = 0,
∂u
∂t
t>0
t=0
∂u
∂t
∂u
∂t
∂u
∂y
0<x<L
0 < x < L, t > 0
= 0,
t>0
0<x<L
t=0
0 < x < L, t > 0, A a constant
u(L, t) = 0,
∂2u ∂2u
+ 2 = 0,
∂x2
∂y
∂u
∂x
πx
,
L
u(L, t) = sin πt,
u(x, 0) = 0,
12.
t=0
= sin
∂2u
∂2u
+
Ax
=
,
∂x2
∂t2
u(0, t) = 0,
0<x<L
0 < x < L, t > 0
u(x, 0) = f (x),
11.
= 0,
∂u
∂2u
∂2u
−
2β
,
=
∂x2
∂t
∂t2
u(0, t) = 0,
10. a2
∂u
∂t
u(L, t) = 0,
u(x, 0) = 0,
9. a2
t>0
t>0
= 0,
0<x<L
t=0
0 < x < 4, 0 < y < 2
= 0,
u(4, y) = f (y),
= 0,
u(x, 2) = 0,
0<y<2
x=0
0<x<4
y=0
∂2u ∂2u
+ 2 = 0,
∂x2
∂y
u(0, y) = e−y ,
u(x, 0) = f (x),
0 < x < π, y > 0
u(π, y) =
100, 0 < y ≤ 1
0<x<π
0,
y>1
12.3
12.3
Heat Equation
1. Using u = XT and −λ as a separation constant we obtain
X + λX = 0,
X(0) = 0,
X(L) = 0,
and
T + kλT = 0.
This leads to
X = c1 sin
nπ
x
L
for n = 1, 2, 3, . . . so that
u=
∞
and
T = c2 e−kn
2 π 2 t/L2
An e−kn
2 π 2 t/L2
sin
n=1
Imposing
u(x, 0) =
∞
An sin
n=1
gives
2
An =
L
ˆ
L/2
sin
0
nπ
x.
L
nπ
x
L
nπ
2 nπ x dx =
1 − cos
L
nπ
2
for n = 1, 2, 3, . . . so that
u(x, t) =
∞
nπ
2 1 − cos nπ
2 n −kn2 π 2 t/L2
x.
e
sin
π
L
n=1
2. Using u = XT and −λ as a separation constant we obtain
X + λX = 0,
X(0) = 0,
X(L) = 0,
and
T + kλT = 0.
This leads to
X = c1 sin
nπ
x
L
for n = 1, 2, 3, . . . so that
u=
∞
n=1
and
An e−kn
T = c2 e−kn
2 π 2 t/L2
2 π 2 t/L2
sin
nπ
x.
L
Heat Equation
703
704
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
Imposing
∞
u(x, 0) =
nπ
x
L
An sin
n=1
gives
2
An =
L
for n = 1, 2, 3, . . . so that
ˆ
L
x(L − x) sin
0
u(x, t) =
4L2
nπ
x dx = 3 3 [1 − (−1)n ]
L
n π
∞
nπ
4L2 1 − (−1)n −kn2 π2 t/L2
x.
e
sin
3
3
π
n
L
n=1
3. Using u = XT and −λ as a separation constant we obtain
X + λX = 0,
X (0) = 0,
X (L) = 0,
and
T + kλT = 0.
This leads to
nπ
2 2
2
x
and
T = c2 e−kn π t/L
L
for n = 0, 1, 2, . . . (λ = 0 is an eigenvalue in this case) so that
X = c1 cos
u=
∞
An e−kn
2 π 2 t/L2
cos
n=0
Imposing
u(x, 0) = f (x) = A0 +
∞
nπ
x.
L
An cos
n=1
nπ
x
L
gives
1
u(x, t) =
L
ˆ
L
0
∞
2
f (x) dx +
L
n=1
ˆ
L
f (x) cos
0
nπ
nπ
2 2
2
x dx e−kn π t/L cos
x.
L
L
4. If L = 2 and f (x) is x for 0 < x < 1 and f (x) is 0 for 1 < x < 2 then
u(x, t) =
∞
nπ
1 1
nπ
nπ
1
2 2
+4
sin
+ 2 2 cos
− 1 e−kn π t/4 cos
x.
4
2nπ
2
n π
2
2
n=1
5. Using u = XT and −λ as a separation constant leads to
X + λX = 0,
X (0) = 0,
X (L) = 0,
12.3
Heat Equation
and
T + (h + kλ)T = 0.
Then
nπ
2 2
2
x
and
T = c2 e−ht−kn π t/L
L
for n = 0, 1, 2, . . . (λ = 0 is an eigenvalue in this case) so that
X = c1 cos
u = A0 e−ht + e−ht
∞
An e−kn
2 π 2 t/L2
cos
n=1
Imposing
u(x, 0) = f (x) =
∞
An cos
n=0
nπ
x.
L
nπ
x
L
gives
e−ht
u(x, t) =
L
ˆ
L
0
ˆ L
∞
2e−ht nπ
nπ
2 2
2
x dx e−kn π t/L cos
x.
f (x) dx +
f (x) cos
L
L
L
0
n=1
6. In Problem 5 we instead find that X(0) = 0 and X(L) = 0 so that
X = c1 sin
and
nπ
x
L
ˆ L
∞
nπ
nπ
2e−ht 2 2
2
x dx e−kn π t/L sin
x.
f (x) sin
u=
L
L
L
0
n=1
7. Using −λ as the separation constant implies X + λX = 0 and T + kλT = 0. The boundary
conditions are then X(−L) = X(L) and X (−L) = X (L).
For λ = 0, X(x) = c1 + c2 x. The condition X(−L) = X(L) implies c2 = 0. Therefore an
eigenfunction is X(x) = c1 = 0. The boundary condition X (−L) = X (L) is automatically
satisfied.
For λ = −α2 < 0, X(x) = c3 cosh αx + c4 sinh αx. From the condition X(−L) = X(L) we
obtain
c3 cosh αL − c4 sinh αL = c3 cosh αL + c4 sinh αL
so 2c4 sinh αL = 0.
This implies c4 = 0 and so X(x) = c3 cosh αx. The boundary condition X (−L) = X (L)
implies
−c3 α sinh αL = c3 sinh αL so 2αc3 sinh αL = 0.
Therefore c3 = 0 and X(x) = 0.
For λ = α2 > 0, X(x) = c5 cos αx + c6 sin αx. The boundary condition X(−L) = X(L)
implies
c5 cos αL − c6 sin αL = c5 cos αL + c6 sin αL so 2c6 sin αL = 0.
705
706
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
If c6 = 0, then α = nπ/L for n = 1, 2, . . . . The boundary condition X (−L) = X (L) implies
−c5 α sin αL + c6 α cos αL = αc5 sin αL + c6 α cos αL
so 2αc5 sin αL = 0.
Then, for c5 = 0,
sin αL = 0 and α = nπ/L, n = 1, 2, . . . .
Thus the coefficients c5 and c6 are arbitrary but nonzero. Therefore the eigenvalues are
λn = (nπ/L)2 , n = 0, 1, 2, . . . , and the corresponding eigenfunctions are
1, cos
nπ
nπ
x, sin
x for
L
L
n = 0, 1, 2, . . . .
Forming product solutions with
T (t) =
c7 ,
λ=0
−k(nπ/L)2 t
c7 e
,
λ > 0,
relabeling constants, and summing gives
u(x, t) = A0 +
∞
e−k(nπ/L)
2t
An cos
k=1
nπ
nπ x + Bn sin
x .
L
L
When t = 0 we get the full Fourier series of f on (−L, L),
f (x) = A0 +
∞ An cos
k=1
nπ
nπ x + Bn sin
x .
L
L
The coefficients are then A0 = 12 a0 , An = an , Bn = bn , or
1
A0 =
2L
ˆ
L
f (x) dx,
−L
1
An =
L
ˆ
L
nπ
dx,
f (x) cos
L
−L
1
Bn =
L
ˆ
L
f (x) sin
−L
nπ
dx.
L
8. In this case we have from (13) of this section in the text that
un (x, 0) = An sin
nπx
5πx
= f (x) = 10 sin
,
L
L
so we can take n = 5 and A5 = 10. All other values of An are 0. Therefore, we can take the
solution of the boundary-value problem to be
u(x, t) = 10e−k(25π
2 /L2 )t
sin
5πx
.
L
12.3
Heat Equation
u
9.
100
x = 3π/4
80
x = 5π/6
x = 11π/12
60
x=π
40
20
1
2
3
4
5
6
t
10. (a) The solution is
u(x, t) =
∞
An e−kn
2 π 2 t/1002
sin
n=1
nπ
x,
100
where
2
An =
100
ˆ
50
0
nπ
x dx +
0.8x sin
100
ˆ
100
0.8(100 − x) sin
50
320
nπ
nπ
x dx = 2 2 sin
.
100
n π
2
Thus,
∞
nπ
nπ −kn2 π2 t/1002
320 1 x.
sin
sin
u(x, t) = 2
e
2
π
n
2
100
n=1
(b) Since An = 0 for n even, the first five nonzero terms correspond to n = 1, 3, 5, 7, 9. In
this case sin (nπ/2) = sin (2p − 1)/2 = (−1)p+1 for p = 1, 2, 3, 4, 5, and
∞
u(x, t) =
320 (−1)p+1 (−1.6352(2p−1)2 π2 /1002 )t
(2p − 1)π
x.
e
sin
π2
(2p − 1)2
100
p=1
40
30
20 u
10
0
20
40
60
x
80
100
50
100
t
150
0
200
707
708
CHAPTER 12
12.4
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
Wave Equation
For Problems 1–10, recall that the solution to the wave equation is given by
u(x, t) =
∞ An cos
n=1
ˆ
L
nπ
2
x dx and Bn =
L
nπa
0
∂u
.
Here, f (x) = u(x, 0) and g(x) =
∂t
where An =
2
L
nπa
nπa nπ
t + Bn sin
t sin
x
L
L
L
ˆ
L
f (x) sin
g(x) sin
0
nπ
x dx.
L
t=0
1. By the discussion in Section 12.4, pages 474 and 475,
∞ nπa
nπa nπ
An cos
t + Bn sin
t sin
x
u(x, t) =
L
L
L
n=1
The coefficients are given by
ˆ
ˆ
2 L
nπ
2 L
x dx =
f (x) sin
An =
L 0
L
L 0
ˆ L
2
nπ
x dx = 0
Bn =
g(x) sin
nπa 0
L
1
nπ
L2 [1 − (−1)n ]
x(L − x) sin
x dx =
4
L
n3 π 3
Therefore the solution to the problem is
u(x, t) =
∞
n=1
L2 [1 − (−1)n ]
nπa
nπ
t sin
x
· cos
3
3
n π
L
L
2. By the discussion in Section 12.4, pages 474 and 475,
u(x, t) =
∞ nπa
nπa nπ
An cos
t + Bn sin
t sin
x
L
L
L
n=1
The coefficients are given by
ˆ
2 L
nπ
x dx = 0
f (x) sin
An =
L 0
L
ˆ L
ˆ L
2
4L3 [1 − (−1)n ]
2
nπ
nπ
x dx =
x dx =
g(x) sin
x(L − x) sin
Bn =
nπa 0
L
nπa 0
L
an4 π 4
Therefore the solution to the problem is
u(x, t) =
∞
n=1
nπ
4L3 [1 − (−1)n ]
nπa
t sin
x
· sin
4
4
an π
L
L
12.4
Wave Equation
3. By the discussion in Section 12.4, pages 474 and 475,
u(x, t) =
∞
(An cos nat + Bn sin nat) sin nx
n=1
The coefficients are given by
ˆ
2 π
f (x) sin nx dx = 0
An =
π 0
ˆ π
ˆ π
2
1
g(x) sin nx dx =
sin x sin nx dx = 0 for n = 2, 3, 4, . . .
Bn =
nπa 0
nπa 0
ˆ π
2
1
sin x sin x dx =
B1 =
πa 0
a
Therefore the solution to the problem is
u(x, t) = B1 sin at sin x =
1
sin at sin x
a
4. By the discussion in Section 12.4, pages 474 and 475,
u(x, t) =
∞
(An cos nat + Bn sin nat) sin nx
n=1
The coefficients are given by
ˆ
ˆ
2 π
2 π1 2
2(−1)n+1
x π − x2 sin nx dx =
An =
f (x) sin nx dx =
π 0
π 0 6
n3
ˆ π
2
Bn =
g(x) sin nx dx = 0
nπa 0
Therefore the solution to the problem is
u(x, t) =
∞
n=1
2(−1)n+1
· cos nat sin nx
n3
5. By the discussion in Section 12.4, pages 474 and 475,
u(x, t) =
∞
(An cos nπat + Bn sin nπat) sin nπx
n=1
The coefficients are given by
ˆ
ˆ
2 1
2 1
4 [1 − (−1)n ]
An =
f (x) sin nπx dx =
x(1 − x) sin nπx dx =
1 0
1 0
n3 π 3
ˆ 1
ˆ 1
2
2
4 [1 − (−1)n ]
Bn =
g(x) sin nπx dx =
x(1 − x) sin nπx dx =
nπa 0
nπa 0
an4 π 4
Therefore the solution to the problem is
u(x, t) =
∞
n=1
4 [1 − (−1)n ]
4 [1 − (−1)n ]
cos
nπat
+
sin nπat sin nπx
n3 π 3
an4 π 4
709
710
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
6. By the discussion in Section 12.4, pages 474 and 475,
u(x, t) =
∞
(An cos nπat + Bn sin nπat) sin nπx
n=1
The coefficients are
ˆ
ˆ
2 1
2 1
f (x) sin nπx dx =
0.01 sin 3πx sin nπx dx = 0 for n = 1, 2, 4, 5, 6, . . .
An =
1 0
1 0
ˆ
2 1
0.01 sin 3πx sin 3πx dx = 0.01
A3 =
1 0
ˆ 1
2
g(x) sin nπx dx = 0
Bn =
nπa 0
Therefore the solution to the problem is
u(x, t) = (A3 cos 3πat) sin 3πx = 0.01 cos 3πat sin 3πx
For Problems 7–10, we have g(x) = 0 because the string is released from the rest, so Bn = 0. So,
our general solution is of the form
u(x, t) =
∞
n=1
2
where An =
L
ˆ
L
f (x) sin
0
An cos
nπx
nπat
sin
L
L
nπ
xdx as before.
L
7. By the discussion in Section 12.4, pages 474 and 475,
u(x, t) =
∞ nπ
nπa
nπa t + Bn sin
t sin
x
An cos
L
L
L
n=1
The coefficients are
ˆ
ˆ
ˆ L
L/2
2 L
nπ
2
nπ
x
nπ
2hx
x dx =
sin
x dx +
sin
x dx
f (x) sin
2h 1 −
An =
L 0
L
L 0
L
L
L
L
L/2
8h sin nπ
2
=
n2 π 2
ˆ L
2
nπ
x dx = 0
Bn =
g(x) sin
nπa 0
L
Therefore the solution to the problem is
∞
8h sin nπ
nπ
nπa
2
t sin
x
u(x, t) =
cos
2
2
n π
L
L
n=1
12.4
Wave Equation
8. By the discussion in Section 12.4, pages 474 and 475,
∞ nπ
nπa
nπa An cos
t + Bn sin
t sin
x
u(x, t) =
L
L
L
n=1
The coefficients are
ˆ
ˆ
ˆ L
L/3
2
nπ
2 L
nπ
3hx
3h x
nπ
An =
x dx =
sin
x dx +
x dx
f (x) sin
1−
sin
L 0
L
L 0
L
L
L
L
L/3 2
9h sin nπ
3
=
n2 π 2
ˆ L
2
nπ
x dx = 0
Bn =
g(x) sin
nπa 0
L
Therefore the solution to the problem is
∞
9h sin nπ
nπ
nπa
3
t sin
x
cos
u(x, t) =
2
2
n π
L
L
n=1
9. By the discussion in Section 12.4, pages 474 and 475,
∞ nπa
nπa nπ
An cos
u(x, t) =
t + Bn sin
t sin
x
L
L
L
n=1
The coefficients are
ˆ
2 L
nπ
x dx
f (x) sin
An =
L 0
L
ˆ
ˆ 2L/3
ˆ L
L/3
nπ
nπ
x
nπ
3hx
2
sin
x dx +
x dx +
sin
x dx
h sin
3h 1 −
=
L 0
L
L
L
L
L
L/3
2L/3
+ sin nπ
6h sin 2nπ
3
3
=
n2 π 2
ˆ L
2
nπ
x dx = 0
Bn =
g(x) sin
nπa 0
L
Therefore the solution to the problem is
nπ ∞
+
sin
6h sin 2nπ
nπ
nπa
3
3
t sin
x
cos
u(x, t) =
2
2
n π
L
L
n=1
Using the trigonometric identity
2nπ
nπ nπ
nπ
nπ
sin
= sin nπ −
= sin nπ cos
− cos nπ sin
= −(−1)n sin
3
3
3
3
3
we have
∞
nπa
nπ
nπ
6h 1 − (−1)n
cos
t sin
x.
sin
u(x, t) = 2
2
π
n
3
L
L
n=1
711
712
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
10. By the discussion in Section 12.4, pages 474 and 475,
u(x, t) =
∞ nπ
nπa
nπa An cos
t + Bn sin
t sin
x
L
L
L
n=1
The coefficients are
2
Bn =
nπa
and
An =
Using sin
2
L
ˆ
L
f (x) sin
0
ˆ
L
g(x) sin
0
nπ
x dx = 0
L
18h
nπ
x dx = 2 2
L
n π
sin
2nπ
nπ
− sin
3
3
.
2nπ
nπ
= −(−1)n
we have
3
3
∞
18h 1 + (−1)n
nπ
nπa
nπ
u(x, t) = 2
cos
t sin
x
sin
2
π
n
3
L
L
n=1
11. By the discussion in Section 12.4, pages 474 and 475, we get
⎧ ⎪
X + λX = 0,
⎪
⎨
X (0) = 0,
⎪
⎪
⎩X (L) = 0,
For λ = 0,
⎧ ⎪
X = 0,
⎪
⎨
X (0) = 0,
⎪
⎪
⎩X (L) = 0,
Thus X = c1 + c2 x and therefore X = c1 . Moreover {T = 0 implies T = c3 + c4 t.
Therefore we have u0 (x, t) = c1 (c3 + c4 t) = A0 + B0 t. Using the given conditions and the
results from Problem 3 in Section 12.5, the rest of the eigenvalues are λn = n2 π 2 /L2 with
corresponding eigenfunctions Xn = cos nπ
L x, n = 1, 2, 3, . . . therefore
nπa
nπa
T = c3 cos L t + c4 sin L t and we now have
u(x, t) = (A0 + B0 t) +
∞ An cos
n=1
nπa
nπa nπ
t + Bn sin
t cos
x
L
L
L
To complete the problem we need only to find the coefficients. At t = 0 we have
u(x, 0) = x = A0 +
∞
n=1
where
1
A0 =
L
ˆ
0
L
L
x dx =
2
An cos
nπ
x
L
2
and An =
L
←− A Fourier cosine series for x
ˆ
L
x cos
0
nπ
2L [(−1)n − 1]
x dx =
L
n2 π 2
12.4
Similarly at t = 0,
ut (x, 0) = 0 = B0 +
∞ Bn ·
n=1
Wave Equation
nπ
nπa cos
x
L
L
and so we get Bn = 0 for n = 0, 1, 2, 3, . . . therefore the solution to the problem is
∞
L u(x, t) = +
2
n=1
2L [(−1)n − 1]
nπ
nπa
t cos
x
cos
2
2
n π
L
L
12. By the discussion in Section 12.4, pages 474 and 475, we get
⎧ ⎪
X + λX = 0,
⎪
⎨
X (0) = 0,
⎪
⎪
⎩X (L) = 0.
For λ = 0,
⎧ ⎪
X = 0,
⎪
⎨
X (0) = 0,
⎪
⎪
⎩X (L) = 0.
Thus X = c1 + c2 x and therefore X = c1 . Moreover, T = 0 implies T = c3 + c4 t.
Therefore we have u0 (x, t) = c1 (c3 + c4 t) = A0 + B0 t. Using the given conditions and the
results from Problem 3 in Section 12.5, the rest of the eigenvalues are λn = n2 π 2 /L2 with
corresponding eigenfunctions Xn = cos nπ
L x, n = 1, 2, 3, . . . therefore
nπa
nπa
T = c3 cos L t + c4 sin L t and we now have
u(x, t) = (A0 + B0 t) +
∞ An cos
n=1
nπ
nπa
nπa t + Bn sin
t cos
x
L
L
L
To complete the problem we need only to find the coefficients. At t = 0 we have
u(x, 0) = f (x) = A0 +
∞
An cos
n=1
where
1
A0 =
L
ˆ
nπ
x
L
L
f (x) dx
and
0
←− A Fourier cosine series for f (x)
2
An =
L
ˆ
L
f (x) cos
0
nπ
x dx
L
Similarly at t = 0,
ut (x, 0) = g(x) = B0 +
∞ Bn ·
n=1
nπa nπ
cos
x
L
L
and so we get
B0 =
1
L
ˆ
L
g(x) dx
0
and
Bn =
2
nπa
ˆ
L
g(x) cos
0
nπ
x dx
L
713
714
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
13. From Figure 12.4.6 we expect u(L/2, t) = 0 for t ≥ 0. To prove this we have from Problem
10,
∞
u(L/2, t) =
nπa
nπ
18h 1 + (−1)n
nπ
cos
t sin
sin
π2
n2
3
L
2
n=1
⎧
2m+1
=1−1=0
⎨0, n = 1, 3, 5, . . . , 2m + 1, . . . , we have 1 + (−1)
=
⎩0, n = 2, 4, 6, . . . , 2m, . . . , we have sin nπ = sin mπ = 0.
2
14. For Problem 1,
∞ u1 (x, t) =
n=1
nπa nπ
An cos
t sin
x,
L
L
2
An =
L
where
ˆ
L
f (x) sin
0
nπ
x dx
L
For Problem 2,
u2 (x, t) =
∞ n=1
nπ
nπa t sin
x
Bn sin
L
L
where
2
Bn =
nπa
ˆ
L
g(x) sin
0
nπ
x dx
L
By superposition,
u(x, t) = u1 (x, t) + u2 (x, t) =
∞ An cos
n=1
=
∞ n=1
An cos
∞ nπ
nπ
nπa nπa t sin
x+
t sin
x
Bn sin
L
L
L
L
n=1
nπ
nπa
nπa , t + Bn sin
t sin
x
L
L
L
15. Using u = XT and −λ as a separation constant we obtain
X + λX = 0,
X(0) = 0,
X(π) = 0,
and
T + 2βT + λT = 0,
T (0) = 0.
Solving the differential equations we get
X = c1 sin nx + c2 cos nx
and
T = e−βt c3 cos n2 − β 2 t + c4 sin n2 − β 2 t
The boundary conditions on X imply c2 = 0 so
and
T = e−βt c3 cos n2 − β 2 t + c4 sin n2 − β 2 t
X = c1 sin nx
12.4
and
u=
∞
Wave Equation
e−βt An cos n2 − β 2 t + Bn sin n2 − β 2 t sin nx.
n=1
Imposing
u(x, 0) = f (x) =
∞
An sin nx
n=1
and
ut (x, 0) = 0 =
∞ Bn
n2 − β 2 − βAn sin nx
n=1
gives
−βt
u(x, t) = e
∞
An
n=1
β
sin n2 − β 2 t sin nx,
cos n2 − β 2 t + n2 − β 2
where
2
An =
π
ˆ
π
f (x) sin nx dx.
0
16. Using u = XT and −λ = as a separation constant leads to X + α2 X = 0, X(0) = 0,
√
X(π) = 0 and T + (1 + α2 )T = 0, T (0) = 0. Then X = c2 sin nx and T = c3 cos n2 + 1 t
for n = 1, 2, 3, . . . so that
u=
∞
Bn cos
n2 + 1 t sin nx.
n1
Imposing u(x, 0) =
∞
Bn sin nx gives
n=1
Bn =
=
ˆ
ˆ
2 π/2
2 π
4
nπ
x sin nx dx +
(π − x) sin nx dx =
sin
π 0
π π/2
πn2
2
⎧
n even,
⎪
⎨0,
⎪
⎩ 4 (−1)(n+3)/2 ,
πn2
n = 2k − 1, k = 1, 2, 3, . . .
Thus with n = 2k − 1,
∞
4 sin nπ
2
u(x, t) =
cos
n2 + 1 t sin nx
π
n2
n=1
=
∞
4 (−1)k+1
cos (2k − 1)2 + 1 t sin (2k − 1) x.
2
π
(2k − 1)
k=1
715
716
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
17. Separating variables in the partial differential equation and using the separation constant
−λ = α4 gives
T X (4)
= − 2 = α4
X
a T
so that
X (4) − α4 X = 0
T + a2 α4 T = 0
and
X = c1 cosh αx + c2 sinh αx + c3 cos αx + c4 sin αx
T = c5 cos aα2 t + c6 sin aα2 t.
The boundary conditions translate into X(0) = X(L) = 0 and X (0) = X (L) = 0. From
X(0) = X (0) = 0 we find c1 = c3 = 0. From
X(L) = c2 sinh αL + c4 sin αL = 0
X (L) = α2 c2 sinh αL − α2 c4 sin αL = 0
we see by subtraction that c2 = 0 and c4 sin αL = 0. This equation yields the eigenvalues
α = nπL for n = 1, 2, 3, . . . . The corresponding eigenfunctions are
nπ
x.
X = c4 sin
L
Thus
u(x, t) =
∞
An cos
n=1
n2 π 2
n2 π 2
nπ
x.
at
+
B
sin
at sin
n
2
2
L
L
L
From
u(x, 0) = f (x) =
∞
An sin
n=1
we obtain
2
An =
L
From
ˆ
L
f (x) sin
0
nπ
x
L
nπ
x dx.
L
∞
n2 π 2
n2 π 2
nπ
n2 π 2 a
n2 π 2 a
∂u −An
=
x
sin
at
+
B
cos
at sin
n
2
2
2
2
∂t
L
L
L
L
L
n=1
and
∂u
∂t
we obtain
= g(x) =
n=1
t=0
2
n2 π 2 a
=
Bn
2
L
L
and
Bn =
∞
2L
n2 π 2 a
ˆ
Bn
n2 π 2 a
nπ
x
sin
2
L
L
L
g(x) sin
0
ˆ
L
g(x) sin
0
nπ
x dx
L
nπ
x dx.
L
12.4
Wave Equation
18. (a) Write the differential equation in X from Problem 23 as X (4) − α4 X = 0 where the
eigenvalues are λ = α2 . Then
X = c1 cosh αx + c2 sinh αx + c3 cos αx + c4 sin αx
and using X(0) = 0 and X (0) = 0 we find c3 = −c1 and c4 = −c2 . The conditions
X(L) = 0 and X (L) = 0 yield the system of equations
c1 (cosh αL − cos αL) + c2 (sinh αL − sin αL) = 0
c1 (α sinh αL + α sin αL) + c2 (α cosh αL − α cos αL) = 0.
In order for this system to have nontrivial solutions the determinant of the coefficients
must be zero. That is,
α(cosh αL − cos αL)2 − α(sinh2 αL − sin2 αL) = 0.
Since α = 0 leads to X = 0, λ = α2 = 02 = 0 is not an eigenvalue. Then, dividing the
above equation by α, we have
(cosh αL − cos αL)2 − (sinh2 αL − sin2 αL)
= cosh2 αL − 2 cosh αL cos αL + cos2 αL − sinh2 αL + sin2 αL
= −2 cosh αL cos αL + 2 = 0
or cosh αL cos αL = 1. Letting x = αL we see that the eigenvalues are λn = αn2 = x2n /L2
where xn , n = 1, 2, 3, . . . , are the positive roots of the equation cosh x cos x = 1.
(b) The equation cosh x cos x = 1 is the same as cos x = sech x.
The figure indicates that the equation has an infinite number of roots.
1
0.5
cos x
sech x
2
4
6
8
–0.5
–1
(c) Using a CAS we find the first four positive roots of cosh x cos x = 1 to be x1 = 4.7300,
x2 = 7.8532, x3 = 10.9956, and x4 = 14.1372. Thus the first four eigenvalues are
λ1 = x21 /L = 22.3733/L, λ2 = x22 /L = 61.6728/L, λ3 = x23 /L = 120.9034/L, and
λ4 = x24 /L = 199.8594/L.
19. From (8) in the text we have
u(x, t) =
∞ An cos
n=1
nπ
nπa
nπa t + Bn sin
t sin
x.
L
L
L
Since ut (x, 0) = g(x) = 0 we have Bn = 0 and
u(x, t) =
∞
n=1
=
nπ
nπ
nπa nπa nπa
1 nπ
t sin
x=
x+
t + sin
x−
t
An cos
An sin
L
L
2
L
L
L
L
∞
n=1
∞
nπ
nπ
1
(x + at) + sin
(x − at) .
An sin
2
L
L
n=1
717
718
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
From
u(x, 0) = f (x) =
∞
An sin
n=1
we identify
∞
f (x + at) =
An sin
nπ
(x + at)
L
An sin
nπ
(x − at),
L
n=1
and
f (x − at) =
∞
n=1
so that
nπ
x
L
1
u(x, t) = [f (x + at) + f (x − at)].
2
20. (a) We note that ξx = ηx = 1, ξt = a, and ηt = −a. Then
∂u ∂ξ
∂u ∂η
∂u
=
+
= uξ + uη
∂x
∂ξ ∂x ∂η ∂x
and
∂uξ ∂η ∂uη ∂ξ
∂uξ ∂ξ
∂uη ∂η
∂
∂2u
(uξ + uη ) =
+
+
+
=
2
∂x
∂x
∂ξ ∂x
∂η ∂x
∂ξ ∂x
∂η ∂x
= uξξ + 2uξη + uηη .
Similarly
∂2u
= a2 (uξξ − 2uξη + uηη ).
∂t2
Thus
a2
∂2u
∂2u
=
∂x2
∂t2
∂2u
= 0.
∂ξ∂η
becomes
(b) Integrating
∂2u
∂
=
uξ = 0
∂ξ∂η
∂η
we obtain
ˆ
∂
uξ dη =
∂η
ˆ
0 dη
uξ = f (ξ).
Integrating this result with respect to ξ we obtain
ˆ
ˆ
∂u
dξ = f (ξ) dξ
∂ξ
u = F (ξ) + G(η).
12.4
Wave Equation
Since ξ = x + at and η = x − at, we then have
u = F (ξ) + G(η) = F (x + at) + G(x − at).
Next, we have
u(x, t) = F (x + at) + G(x − at)
u(x, 0) = F (x) + G(x) = f (x)
ut (x, 0) = aF (x) − aG (x) = g(x)
Integrating the last equation with respect to x gives
ˆ
1 x
F (x) − G(x) =
g(s) ds + c1 .
a x0
Substituting G(x) = f (x) − F (x) we obtain
1
1
F (x) = f (x) +
2
2a
where c = c1 /2. Thus
G(x) =
1
1
f (x) −
2
2a
ˆ
x
g(s) ds + c
x0
ˆ
x
g(s) ds − c.
x0
(c) From the expressions for F and G,
1
1
F (x + at) = f (x + at) +
2
2a
1
1
G(x − at) = f (x − at) −
2
2a
ˆ
x+at
g(s) ds + c
x0
ˆ
x−at
g(s) ds − c.
x0
Thus,
1
1
u(x, t) = F (x + at) + G(x − at) = [f (x + at) + f (x − at)] +
2
2a
ˆ
Here we have used −
ˆ
x−at
1
1
21. u(x, t) = [sin (x + at) + sin (x − at)] +
2
2a
1
1
[sin (x + at) + sin (x − at)] +
2
2a
= sin x cos at +
g(s) ds.
x−at
g(s) ds.
x−at
ˆ
x+at
ds
x−at
1
1
= [sin x cos at + cos x sin at + sin x cos at − cos x sin at] + s
2
2a
22. u(x, t) =
x+at
x0
g(s) ds =
x0
ˆ
ˆ
x+at
= sin x cos at + t
x−at
x+at
cos s ds
x−at
1
1
[sin (x + at) − sin (x − at)] = sin x cos at + cos x sin at
2a
a
719
720
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
1
23. u(x, t) = 0 +
2a
ˆ
x+at
sin 2s ds =
x−at
1 − cos (2x + 2at) + cos (2x − 2at)
2a
2
=
1
[− cos 2x cos 2at + sin 2x sin 2at + cos 2x cos 2at + sin 2x sin 2at]
4a
=
1
sin 2x sin 2at
2a
24. u(x, t) =
1
1 −(x+at)2
2
2
2 2
2
2 2
e
e−(x +2axt+a t ) + e−(x −2axt+a t )
+ e−(x−at) =
2
2
= e−(x
2 +a2 t2 )
e−2axt + e2axt
2
2 2
= e−(x +a t ) cosh 2axt
2
25. (a) x
(b) x
u
26. (a)
1
x
–6
6
(b) Since g(x) = 0, d’Alembert’s solution with a = 1 is
1
u(x, t) = [f (x + t) + f (x − t)].
2
Sample plots are shown below.
12.4
1
Wave Equation
1
t = 0.4
–6
–4
–2
0
t = 0.6
4
2
6
–6
–4
–2
1
0
2
–4
–2
0
2
4
6
–6
–4
–2
0
2
–2
0
2
4
6
–6
–4
–2
0
2
–2
4
6
4
6
1
t = 4.0
–4
6
t = 3.0
1
–6
4
1
t = 2.0
–4
6
t = 1.0
1
–6
4
1
t = 0.8
–6
721
0
t = 5.0
4
2
6
–6
–4
–2
0
2
(c) The single peaked wave disolves into two peaks moving outward.
27. (a) With a = 1, d’Alembert’s solution is
1
u(x, t) =
2
ˆ
x+t
g(s) ds
x−t
where
g(s) =
1, |s| ≤ 0.1
0, |s| > 0.1.
Sample plots are shown below.
t=0.0
t=0.2
t=0.4
t=0.6
722
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
t=0.8
t=1.0
t=2.0
t=3.0
t=5.0
t=4.0
(b) Some frames of the movie are shown in part (a), The string has a roughly rectangular
shape with the base on the x-axis increasing in length.
28. (a)-(b) With the given parameters, the solution is
u(x, t) =
∞
nπ
8 1
cos nt sin nx.
sin
2
2
π
n
2
n=1
For n even, sin (nπ/2) = 0, so the first six nonzero terms correspond to n = 1, 3, 5, 7,
9, 11. In this case sin (nπ/2) = sin (2p − 1)/2 = (−1)p+1 for p = 1, 2, 3, 4, 5, 6, and
u(x, t) =
∞
8 (−1)p+1
cos (2p − 1)t sin (2p − 1)x.
π2
(2p − 1)2
p=1
Frames of the movie corresponding to t = 0.5, 1, 1.5, and 2 are shown.
u
1
u
1
0.5
0.5
0.5
−0.5
1
1.5
2
2.5
3
x
1.5
2
2.5
3
0.5
1
1.5
2
2.5
3
x
u
1
u
1
0.5
0.5
0.5
−1
1
−1
−1
−0.5
0.5
−0.5
1
1.5
2
2.5
3
x
−0.5
−1
x
12.5
12.5
Laplace’s Equation
Laplace’s Equation
1. Using u = XY and −λ as a separation constant we obtain
X + λX = 0,
X(0) = 0,
X(a) = 0,
and
Y − λY = 0,
Y (0) = 0.
With λ = α2 > 0 the solutions of the differential equations are
X = c1 cos αx + c2 sin αx
and
Y = c3 cosh αy + c4 sinh αy
and
Y = c4 sinh
The boundary and initial conditions imply
X = c2 sin
nπ
x
a
for n = 1, 2, 3, . . . so that
u=
∞
An sin
n=1
Imposing
u(x, b) = f (x) =
nπ
nπ
x sinh
y.
a
a
∞
An sinh
n=1
gives
2
nπb
=
An sinh
a
a
so that
u(x, y) =
∞
ˆ
f (x) sin
0
An sin
nπb
2
An = csch
a
a
nπb
nπ
sin
x
a
a
a
n=1
where
nπ
y
a
ˆ
nπ
x dx
a
nπ
nπ
x sinh
y
a
a
a
f (x) sin
0
nπ
x dx.
a
2. Using u = XY and −λ as a separation constant we obtain
X + λX = 0,
X(0) = 0,
X(a) = 0,
723
724
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
and
Y − λY = 0,
Y (0) = 0.
With λ = α2 > 0 the solutions of the differential equations are
X = c1 cos αx + c2 sin αx
and
Y = c3 cosh αy + c4 sinh αy
and
Y = c3 cosh
The boundary and initial conditions imply
X = c2 sin
nπ
x
a
for n = 1, 2, 3, . . . so that
∞
u=
An sin
n=1
Imposing
u(x, b) = f (x) =
nπ
nπ
x cosh
y.
a
a
∞
An cosh
n=1
gives
An cosh
nπb
2
=
a
a
so that
u(x, y) =
∞
ˆ
An =
f (x) sin
0
An sin
nπb
2
sech
a
a
nπ
nπb
sin
x
a
a
a
n=1
where
nπ
y
a
ˆ
nπ
x dx
a
nπ
nπ
x cosh
y
a
a
a
f (x) sin
0
nπ
x dx.
a
3. Using u = XY and −λ as a separation constant we obtain
X + λX = 0,
X(0) = 0,
X(a) = 0,
and
Y − λY = 0,
Y (b) = 0.
With λ = α2 > 0 the solutions of the differential equations are
X = c1 cos αx + c2 sin αx
and
Y = c3 cosh αy + c4 sinh αy
12.5
Laplace’s Equation
The boundary and initial conditions imply
X = c2 sin
nπ
x
a
and
Y = c2 cosh
cosh nπb
nπ
nπ
a
y − c2
y
sinh
nπb
a
a
sinh a
for n = 1, 2, 3, . . . so that
u=
∞
An
n=1
cosh nπb
nπ
nπ
nπ
a
y−
y sin
x.
cosh
sinh
nπb
a
a
a
sinh a
Imposing
u(x, 0) = f (x) =
∞
An sin
n=1
gives
2
An =
a
ˆ
f (x) sin
0
so that
∞
2
u(x, y) =
a
ˆ
n=1
a
0
a
nπ
x dx
f (x) sin
a
nπ
x
a
nπ
x dx
a
cosh nπb
nπ
nπ
nπ
a
cosh
y−
y sin
x.
sinh
nπb
a
a
a
sinh a
4. Using u = XY and −λ as a separation constant we obtain
X + λX = 0,
X (0) = 0,
X (a) = 0,
and
Y − λY = 0,
Y (b) = 0.
With λ = α2 > 0 the solutions of the differential equations are
X = c1 cos αx + c2 sin αx
and
Y = c3 cosh αy + c4 sinh αy
The boundary and initial conditions imply
X = c1 cos
nπ
x
a
and
Y = c3 cosh
cosh nπb
nπ
nπ
a
y − c3
y
sinh
nπb
a
a
sinh a
for n = 1, 2, 3, . . . . Since λ = 0 is an eigenvalue for both differential equations with corresponding eigenfunctions 1 and y − b, respectively we have
∞
cosh nπb
nπ
nπ
nπ
a
x cosh
y−
y .
An cos
sinh
u = A0 (y − b) +
nπb
a
a
a
sinh
a
n=1
725
726
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
Imposing
u(x, 0) = x = −A0 b +
∞
An cos
n=1
gives
1
−A0 b =
a
and
2
An =
a
ˆ
a
x cos
0
ˆ
a
0
nπ
x
a
1
x dx = a
2
2a
nπ
x dx = 2 2 [(−1)n − 1]
a
n π
so that
∞
cosh nπb
2a (−1)n − 1
nπ
nπ
nπ
a
a
x cosh
y−
y .
cos
sinh
u(x, y) = (b − y) + 2
nπb
2b
π
n2
a
a
a
sinh
a
n=1
5. Using u = XY and −λ as a separation constant we obtain
X + λX = 0,
X (0) = 0,
X (a) = 0,
and
Y − λY = 0,
Y (b) = 0.
With λ = −α2 < 0 the solutions of the differential equations are
X = c1 cosh αx + c2 sinh αx
and
Y = c3 cos αy + c4 sin αy
for n = 1, 2, 3 . . . . The boundary and initial conditions imply
X = c2 sinh nπx
and
Y = c3 cos nπy
for n = 1, 2, 3, . . . . Since λ = 0 is an eigenvalue for the differential equation in X with
corresponding eigenfunction x we have
u = A0 x +
∞
An sinh nπx cos nπy.
n=1
Imposing
u(1, y) = 1 − y = A0 +
∞
An sinh nπ cos nπy
n=1
gives
ˆ
1
A0 =
0
(1 − y) dy
12.5
and
ˆ
1
An sinh nπ = 2
(1 − y) cos nπy dy =
0
Laplace’s Equation
2[1 − (−1)n ]
2π2
for n = 1, 2, 3, . . . so that
∞
2 1 − (−1)n
1
sinh nπx cos nπy.
u(x, y) = x + 2
2
π
n2 sinh nπ
n=1
6. Using u = XY and −λ as a separation constant we obtain
X + λX = 0,
X (1) = 0
and
Y − λY = 0,
Y (0) = 0,
Y (π) = 0.
With λ = α2 < 0 the solutions of the differential equations are
X = c1 cosh αx + c2 sinh αx
and
Y = c3 cos αy + c4 sin αy
The boundary and initial conditions imply
X = c1 cosh nx − c1
sinh n
sinh nx
cosh n
and
Y = c3 cos ny
for n = 1, 2, 3, . . . . Since λ = 0 is an eigenvalue for both differential equations with corresponding eigenfunctions 1 and 1 we have
u = A0 +
∞
An cosh nx −
n=1
sinh n
sinh nx cos ny.
cosh n
Imposing
u(0, y) = g(y) = A0 +
∞
An cos ny
n=1
gives
1
A0 =
π
ˆ
π
g(y) dy
and
0
2
An =
π
ˆ
π
g(y) cos ny dy
0
for n = 1, 2, 3, . . . so that
u(x, y) =
1
π
ˆ
π
g(y) dy +
0
∞
n=1
2
π
ˆ
π
g(y) cos ny dy
0
cosh nx −
sinh n
sinh nx cos ny.
cosh n
727
728
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
7. Using u = XY and −λ as a separation constant we obtain
X + λX = 0,
X (0) = X(0)
and
Y − λY = 0,
Y (0) = 0,
Y (π) = 0.
With λ = α2 < 0 the solutions of the differential equations are
X = c1 cosh αx + c2 sinh αx
and
Y = c3 cos αy + c4 sin αy
The boundary and initial conditions imply
Y = c4 sin ny
and
X = c2 (n cosh nx + sinh nx)
for n = 1, 2, 3, . . . so that
u=
∞
An (n cosh nx + sinh nx) sin ny.
n=1
Imposing
u(π, y) = 1 =
∞
An (n cosh nπ + sinh nπ) sin ny
n=1
gives
An (n cosh nπ + sinh nπ) =
2
π
ˆ
π
sin ny dy =
0
2[1 − (−1)n ]
nπ
for n = 1, 2, 3, . . . so that
u(x, y) =
∞
2 1 − (−1)n n cosh nx + sinh nx
sin ny.
π
n
n cosh nπ + sinh nπ
n=1
8. Using u = XY and −λ as a separation constant we obtain
X + λX = 0,
X(0) = 0,
X(1) = 0,
12.5
Laplace’s Equation
and
Y − λY = 0,
Y (0) = Y (0).
With λ = α2 > 0 the solutions of the differential equations are
X = c1 cos αx + c2 sin αx
and
Y = c3 cosh αy + c4 sinh αy
The boundary and initial conditions imply
X = c2 sin nπx
and
Y = c4 (nπ cosh nπy + sinh nπy)
for n = 1, 2, 3, . . . so that
u=
∞
An (nπ cosh nπy + sinh nπy) sin nπx.
n=1
Imposing
u(x, 1) = f (x) =
∞
An (nπ cosh nπ + sinh nπ) sin nπx
n=1
gives
2
An (nπ cosh nπ + sinh nπ) =
π
for n = 1, 2, 3, . . . so that
u(x, y) =
∞
ˆ
π
f (x) sin nπx dx
0
An (nπ cosh nπy + sinh nπy) sin nπx
n=1
where
2
An =
nπ cosh nπ + π sinh nπ
ˆ
1
f (x) sin nπx dx.
0
9. This boundary-value problem has the form of Problem 1 from the text of this section, with
a = b = 1, f (x) = 100, and g(x) = 200. The solution, then, is
u(x, y) =
∞
(An cosh nπy + Bn sinh nπy) sin nπx,
n=1
where
ˆ
1
An = 2
100 sin nπx dx = 200
0
and
1
2
Bn =
sinh nπ
=
ˆ
200 sin nπx dx − An cosh nπ
0
1
400
sinh nπ
= 200
1
1 − (−1)n
nπ
1 − (−1)n
nπ
− 200
1 − (−1)n
nπ
1 − (−1)n
[2 csch nπ − coth nπ].
nπ
cosh nπ
729
730
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
10. This boundary-value problem has the form of Problem 2 from the text of this section, with
a = 1 and b = 1. Thus, the solution has the form
u(x, y) =
∞
(An cosh nπx + Bn sinh nπx) sin nπy.
n=1
The boundary condition u(0, y) = 10y implies
10y =
∞
An sin nπy
n=1
and
2
An =
1
ˆ
1
10y sin nπy dy =
0
20
(−1)n+1 .
nπ
The boundary condition ux (1, y) = −1 implies
−1 =
∞
(nπAn sinh nπ + nπBn cosh nπ) sin nπy
n=1
and
nπAn sinh nπ + nπBn cosh nπ =
2
1
An sinh nπ + Bn cosh nπ = −
Bn =
ˆ
1
(− sin nπy) dy
0
2
[1 − (−1)n ]
nπ
2
20
[(−1)n − 1] sech nπ −
(−1)n+1 tanh nπ.
nπ
nπ
11. Using u = XY and −λ as a separation constant we obtain
X + λX = 0,
X(0) = 0,
X(π) = 0,
and
Y − λY = 0.
With λ = α2 > 0 the solutions of the differential equations are
X = c1 cos αx + c2 sin αx
and
Y = c3 eαy + c4 e−αy
Then the boundedness of u as y → ∞ implies c3 = 0, so Y = c4 e−ny . The boundary conditions
at x = 0 and x = π imply c1 = 0 so X = c2 sin nx for n = 1, 2, 3, . . . and
u=
∞
n=1
An e−ny sin nx.
12.5
Imposing
u(x, 0) = f (x) =
∞
Laplace’s Equation
An sin nx
n=1
gives
An =
so that
u(x, y) =
∞
n=1
2
π
ˆ
2
π
π
f (x) sin nx dx
0
ˆ
π
f (x) sin nx dx e−ny sin nx.
0
12. Using u = XY and −λ as a separation constant we obtain
X + λX = 0,
X (0) = 0,
X (π) = 0,
and
Y − λY = 0.
With λ = α2 > 0 the solutions of the differential equations are
X = c1 cos αx + c2 sin αx
Y = c3 eαy + c4 e−αy
and
The boundary conditions at x = 0 and x = π imply c2 = 0 so X = c1 cos nx for n = 1, 2, 3, . . . .
Now the boundedness of u as y → ∞ implies c3 = 0, so Y = c4 e−ny . In this problem λ = 0
is also an eigenvalue with corresponding eigenfunction 1 so that
u = A0 +
∞
An e−ny cos nx.
n=1
Imposing
u(x, 0) = f (x) = A0 +
∞
An cos nx
n=1
gives
1
A0 =
π
so that
1
u(x, y) =
π
ˆ
π
f (x) dx
2
An =
π
and
0
ˆ
π
f (x) dx +
0
∞
n=1
2
π
ˆ
π
0
ˆ
π
f (x) cos nx dx
0
f (x) cos nx dx e−ny cos nx.
731
732
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
13. Since the boundary conditions at y = 0 and y = b are functions of x we choose to separate
Laplace’s equation as
X Y =−
= −λ
X
Y
so that
X + λX = 0
Y − λY = 0.
Then with λ = α2 we have
X(x) = c1 cos αx + c2 sin αx
Y (y) = c3 cosh αy + c4 sinh αy.
Now X(0) = 0 gives c1 = 0 and X(a) = 0 implies sin αa = 0 or α = nπ/a for n = 1, 2, 3, . . ..
Thus
nπ
nπ
nπ y + Bn sinh
y sin
x
un (x, y) = XY = An cosh
a
a
a
and
∞ nπ
nπ
nπ y + Bn sinh
y sin
x.
(1)
u(x, y) =
An cosh
a
a
a
n=1
At y = 0 we then have
∞
f (x) =
An sin
n=1
and consequently
2
An =
a
At y = b,
g(y) =
∞ ˆ
f (x) sin
0
An cosh
n=1
a
nπ
x
a
nπ
x dx.
a
(2)
nπ
nπ
nπ b + Bn sinh
a sin
x
a
b
a
indicates that the entire expression in the parentheses is given by
ˆ
nπ
nπ
2 a
nπ
An cosh
b + Bn sinh
b=
x dx.
g(x) sin
a
a
a 0
a
We can now solve for Bn :
Bn sinh
2
nπ
b=
a
a
Bn =
ˆ
a
nπ
nπ
x dx − An cosh
b
a
a
ˆ
2 a
nπ
nπ
x dx − An cosh
b .
g(x) sin
a 0
a
a
g(x) sin
0
1
sinh nπ
a b
(3)
A solution to the given boundary-value problem consists of the series (1) with coefficients An
and Bn given in (2) and (3), respectively.
12.5
Laplace’s Equation
733
14. Since the boundary conditions at x = 0 and x = a are functions of y we choose to separate
Laplace’s equation as
X Y =−
= −λ
X
Y
so that
X + λX = 0
Y − λY = 0.
Then with λ = −α2 we have
X(x) = c1 cosh αx + c2 sinh αx
Y (y) = c3 cos αy + c4 sin αy.
Now Y (0) = 0 gives c3 = 0 and Y (b) = 0 implies sin αb = 0 or α = nπ/b for n = 1, 2, 3, . . . .
Thus
nπ
nπ
nπ x + Bn sinh
x sin
y
un (x, y) = XY = An cosh
b
b
b
and
∞ nπ
nπ
nπ x + Bn sinh
x sin
y.
(4)
u(x, y) =
An cosh
b
b
b
n=1
At x = 0 we then have
F (y) =
∞
An sin
n=1
and consequently
2
An =
b
At x = a,
G(y) =
∞ ˆ
F (y) sin
0
An cosh
n=1
b
nπ
y
b
nπ
y dy.
b
(5)
nπ
nπ nπ
a + Bn sinh
a sin
y
b
b
b
indicates that the entire expression in the parentheses is given by
ˆ
2 b
nπ
nπ
nπ
a + Bn sinh
a=
y dy.
G(y) sin
An cosh
b
b
b 0
b
We can now solve for Bn :
Bn sinh
2
nπ
a=
b
b
Bn =
ˆ
b
nπ
nπ
y dy − An cosh
a
b
b
ˆ
2 b
nπ
nπ
y dy − An cosh
a .
G(y) sin
b 0
b
b
G(y) sin
0
1
sinh nπ
b a
(6)
A solution to the given boundary-value problem consists of the series (4) with coefficients An
and Bn given in (5) and (6), respectively.
734
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
15. Referring to the discussion in this section of the text we identify a = b = π, f (x) = 0,
g(x) = 1, F (y) = 1, and G(y) = 1. Then An = 0 and
u1 (x, y) =
∞
Bn sinh ny sin nx
n=1
where
Bn =
2
π sinh nπ
ˆ
π
sin nx dx =
0
2[1 − (−1)n ]
.
nπ sinh nπ
Next
u2 (x, y) =
∞
(An cosh nx + Bn sinh nx) sin ny
n=1
where
An =
2
π
ˆ
π
sin ny dy =
0
2[1 − (−1)n ]
nπ
and
1
Bn =
sinh nπ
2
π
ˆ
π
sin ny dy − An cosh nπ
0
2[1 − (−1)n ] 2[1 − (−1)n ]
−
cosh nπ
nπ
nπ
=
1
sinh nπ
=
2[1 − (−1)n ]
(1 − cosh nπ).
nπ sinh nπ
Now
An cosh nx + Bn sinh nx =
sinh nx
2[1 − (−1)n ]
cosh nx +
(1 − cosh nπ)
nπ
sinh nπ
=
2[1 − (−1)n ]
[cosh nx sinh nπ + sinh nx − sinh nx cosh nπ]
nπ sinh nπ
=
2[1 − (−1)n ]
[sinh nx + sinh n(π − x)]
nπ sinh nπ
and
u(x, y) = u1 + u2 =
∞
2 1 − (−1)n
sinh ny sin nx
π
n sinh nπ
n=1
+
∞
2 [1 − (−1)n ][sinh nx + sinh n(π − x)]
sin ny.
π
n sinh nπ
n=1
12.5
Laplace’s Equation
16. Referring to the discussion in this section of the text we identify a = b = 2, f (x) = 0,
x,
0<x<1
g(x) =
2 − x, 1 < x < 2,
F (y) = 0, and G(y) = y(2 − y). Then An = 0 and
u1 (x, y) =
∞
Bn sinh
n=1
nπ
nπ
y sin
x
2
2
where
1
Bn =
sinh nπ
=
=
ˆ
g(x) sin
0
ˆ
1
sinh nπ
8 sin
2
1
x sin
0
nπ
2
n2 π 2 sinh nπ
nπ
x dx
2
nπ
x dx +
2
ˆ
2
(2 − x) sin
1
nπ
x dx
2
.
Next, since An = 0 in u2 , we have
u2 (x, y) =
∞
Bn sinh
n=1
where
1
Bn =
sinh nπ
ˆ
2
y(2 − y) sin
0
nπ
nπ
x sin
2
2
nπ
16[1 − (−1)n ]
y dy = 3 3
.
2
n π sinh nπ
Thus
∞
8 sin nπ
nπ
nπ
2
u(x, y) = u1 + u2 = 2
sinh
y sin
x
2
π
n sinh nπ
2
2
n=1
+
∞
16 [1 − (−1)n ]
nπ
nπ
sinh
x sin
y.
π3
n3 sinh nπ
2
2
n=1
17. (a)
200
u
100
3
2
0
0
y
1
1
x
2
3
0
735
736
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
(b) The maximum value occurs at (π/2, π) and is f (π/2) = 25π 2 .
(c) The coefficients are
2
An = csch nπ
π
=
ˆ
π
100x(π − x) sin nx dx
0
400
200 csch nπ 200
(1 − (−1)n ) = 3 [1 − (−1)n ] csch nπ.
3
π
n
n π
See part (a) for the graph.
y
18. From the figure showing the boundary conditions we
see that the maximum value of the temperature is 1
at (1, 2) and (2, 1).
3
2.5
2
u=x
u=2−x
u=1
u = y (2 − y)
1.5 u = 0
1 u=0
u=1
u = y (2 − y)
0.5 u = 0
u=0 u=0 u=0
0.5
1
1.5
x
2
2.5
3
19. Assuming u(x, y) = X(x)Y (y) and substituting into the partial differential equationwe get
X Y + XY = 0. Separating variables and using λ = α2 we get
X − α2 X = 0,
X (0) = 0,
which implies X(x) = c3 cosh αx. From
Y + α2 Y = 0,
Y (0) = 0,
Y (b) = 0
we get Y (y) = c1 cos αy + c2 sin αy and eigenvalues λn = n2 π 2 /b2 , n = 1, 2, 3, . . . . The
corresponding eigenfunctions are Y (y) = c1 cos(nπy/b). For λ = 0 the boundary conditions
applied to X(x) = c3 + c4 x and Y (y) = c1 + c2 y imply X = c3 and Y = c1 . Forming products
and using the superposition principle then gives
u(x, y) = A0 +
∞
An cosh
n=1
nπ
nπ
x cos
y.
b
b
The remaining boundary condition, ux (a) = g(y) implies
∂u
g(y) =
∂x
=
x=a
∞
n=1
An
nπ
nπ
nπ
sinh
x cos
y,
b
b
b
12.5
Laplace’s Equation
and so A0 remains arbitrary. In order that the series expression for g(y) be a cosine series,
the constant term in the series, a0 /2, must be 0. Thus, from Section 11.3 in the text,
ˆ b
ˆ
2 b
g(y) dy = 0 so
g(y) dy = 0.
a0 =
b 0
0
Also,
nπ
2
nπ
An
sinh
a=
b
b
b
and
An =
2
nπ sinh nπa/b
The solution is then
u(x, y) = A0 +
∞
ˆ
b
g(y) cos
nπ
y dy
b
g(y) cos
nπ
y dy.
b
0
ˆ
b
0
An cosh
n=1
nπ
nπ
x cos
y,
b
b
where the An are defined above and A0 is arbitrary. In general, Neumann problems do not
have unique solutions.
´b
For a physical interpretation of the compatibility condition 0 g(y)dy = 0 see the texts Elementary Partial Differential Equations by Paul Berg and James McGregor (Holden-Day) and
Partial Differential Equations of Mathematical Physics by Tyn Myint-U (North Holland).
20. Separating variables in the boundary-value problem leads to
Y + λY = 0,
Y (0) = 0, Y (π) = 0
X − λX = 0.
The boundary conditions yield
λ0 = 0
so
Y = c1
and X = c3 + c4 x.
Also
λn = n2 , n > 0 so
Y (y) = c5 cos ny
and X(x) = c7 cosh nx + c8 sinh nx.
Thus product solutions are
u(x, y) = A0 + B0 x +
∞
(An cosh nx + Bn sinh nx) cos ny.
n=1
The last part of this problem involves matching coefficients.
At x = 0,
u0 cos y = A0 +
∞
n=1
An cos ny
implies A0 = 0, A1 = u0 , and An = 0, for n ≥ 2.
737
738
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
Therefore
u(x, y) = B0 x + u0 cosh x cos y + B1 sinh x cos y + B2 sinh 2x cos 2y +
∞
Bn sinh nx cos ny.
n=3
At x = 1,
u0 (1 + cos 2y) = B0 + u0 cosh 1 cos y + B1 sinh 1 cos y + B2 sinh 2 cos 2y +
∞
Bn sinh n cos ny,
n=3
so
u0 − u0 cosh 1 cos y + u0 cos 2y = B0 + B1 sinh 1 cos y + B2 sinh 2 cos 2y +
∞
Bn sinh n cos ny.
n=3
Then
B0 = u0 , B1 = −u0
u0
cosh 1
, B2 =
, and Bn = 0 for n ≥ 3.
sinh 1
sinh 2
Therefore
u(x, y) = u0 x + u0 cosh x cos y − u0
= u0 x + u0
sinh 1 cosh x − cosh 1 sinh x
sinh 1
or
u(x, y) = u0 x + u0
1
21. (a)
u0
cosh 1
sinh x cos y +
sinh 2x cos 2y
sinh 1
sinh 2
140
110
0.6
80
30
0.4
30
60
60
80
0.2
0
0
0.2
0.4
0.6
0.8
1
(b)
200
u
150
100
1
0.8
0.6
y
0.4
50
0
0
0.2
0.4
x
u0
sinh 2x cos 2y
sinh 2
u0
sinh (1 − x)
cos y +
sinh 2x cos 2y.
sinh 1
sinh 2
170
0.8
cos y +
0.6
0.2
0.8
10
12.6
1
22.
Nonhomogeneous Boundary-Value Problems
0
0.8
2
0.6
1
0.5
0.4
0.2
0.2
−0.05
0
0.2
0
12.6
0.4
0.6
0.8
1
Nonhomogeneous Boundary-Value Problems
1. Using v(x, t) = u(x, t) − 100 we wish to solve kvxx = vt subject to v(0, t) = 0, v(1, t) = 0, and
v(x, 0) = −100. Let v = XT and use −λ as a separation constant so that
X + λX = 0,
X(0) = 0,
X(1) = 0,
and
T + λkT = 0.
This leads to
X = c2 sin (nπx)
T = c3 e−kn
2 π2 t
and
for n = 1, 2, 3, . . . so that
v(x, t) =
∞
An e−kn
2 π2 t
sin nπx.
n=1
Imposing
v(x, 0) = −100 =
∞
An sin nπx
n=1
gives
ˆ
An = 2
1
(−100) sin nπx dx =
0
so that
∞
u(x, t) = v(x, t) + 100 = 100 +
−200
[1 − (−1)n ]
nπ
200 (−1)n − 1 −kn2 π2 t
e
sin nπx.
π
n
n=1
739
740
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
2. Letting u(x, t) = v(x, t) + ψ(x) and proceeding as in Example 1 in the text we find
ψ(x) = u0 − u0 x. Then v(x, t) = u(x, t) + u0 x − u0 and we wish to solve kvxx = vt
subject to v(0, t) = 0, v(1, t) = 0, and v(x, 0) = f (x) + u0 x − u0 . Let v = XT and use
−λ as a separation constant so that
X + λX = 0,
X(0) = 0,
X(1) = 0,
and
T + λkT = 0.
Then
X = c2 sin nπx
for n = 1, 2, 3, . . . so that
v=
∞
T = c3 e−kn
2 π2 t
and
An e−kn
2 π2 t
sin nπx.
n=1
Imposing
v(x, 0) = f (x) + u0 x − u0 =
∞
An sin nπx
n=1
gives
ˆ
An = 2
1
(f (x) + u0 x − u0 ) sin nπx dx
0
so that
u(x, t) = v(x, t) + u0 − u0 x = u0 − u0 x +
∞
An e−kn
2 π2 t
sin nπx.
n=1
3. If we let u(x, t) = v(x, t) + ψ(x), then we obtain as in Example 1 in the text
kψ + r = 0
or
ψ(x) = −
r 2
x + c1 x + c2 .
2k
The boundary conditions become
u(0, t) = v(0, t) + ψ(0) = u0
u(1, t) = v(1, t) + ψ(1) = u0 .
12.6
Nonhomogeneous Boundary-Value Problems
741
Letting ψ(0) = ψ(1) = u0 we obtain homogeneous boundary conditions in v:
v(0, t) = 0
and
v(1, t) = 0.
Now ψ(0) = ψ(1) = u0 implies c2 = u0 and c1 = r/2k. Thus
ψ(x) = −
r 2
r
r
x + x + u0 = u0 − x(x − 1).
2k
2k
2k
To determine v(x, t) we solve
k
∂v
∂2v
,
=
2
∂x
dt
0 < x < 1, t > 0
v(0, t) = 0,
v(x, 0) =
v(1, t) = 0,
r
x(x − 1) − u0 .
2k
Separating variables, we find
v(x, t) =
∞
An e−kn
2 π2 t
sin nπx,
n=1
where
ˆ
1
An = 2
0
u
r
r 0
x(x − 1) − u0 sin nπx dx = 2
+ 3 3 [(−1)n − 1].
2k
nπ kn π
Hence, a solution of the original problem is
∞
u(x, t) = ψ(x) + v(x, t) = u0 −
r
2 2
x(x − 1) +
An e−kn π t sin nπx,
2k
n=1
where An is defined in (1).
4. If we let u(x, t) = v(x, t) + ψ(x), then we obtain as in Example 1 in the text
kψ + r = 0.
Integrating gives
ψ(x) = −
r 2
x + c1 x + c2 .
2k
The boundary conditions become
u(0, t) = v(0, t) + ψ(0) = u0
u(1, t) = v(1, t) + ψ(1) = u1 .
Letting ψ(0) = u0 and ψ(1) = u1 we obtain homogeneous boundary conditions in v:
v(0, t) = 0
and
v(1, t) = 0.
(1)
742
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
Now ψ(0) = u0 and ψ(1) = u1 imply c2 = u0 and c1 = u1 − u0 + r/2k. Thus
r r
ψ(x) = − x2 + u1 − u0 +
x + u0 .
2k
2k
To determine v(x, t) we solve
∂v
∂2v
=
,
2
∂x
dt
k
0 < x < 1, t > 0
v(0, t) = 0,
v(1, t) = 0,
v(x, 0) = f (x) − ψ(x).
Separating variables, we find
v(x, t) =
∞
An e−kn
2 π2 t
sin nπx,
n=1
where
ˆ
1
An = 2
[f (x) − ψ(x)] sin nπx dx.
(2)
0
Hence, a solution of the original problem is
r r 2 2 2
x + u1 − u0 +
x + u0 +
An e−kn π t sin nπx,
2k
2k
∞
u(x, t) = ψ(x) + v(x, t) = −
n=1
where An is defined in (2).
5. Substituting u(x, t) = v(x, t) + ψ(x) into the partial differential equation gives
k
∂2v
∂v
+ kψ + Ae−βx =
.
2
∂x
∂t
This equation will be homogeneous provided ψ satisfies
kψ + Ae−βx = 0.
The solution of this differential equation is obtained by successive integrations:
ψ(x) = −
A −βx
+ c1 x + c 2 .
e
β2k
From ψ(0) = 0 and ψ(1) = 0 we find
c1 =
A −β
(e − 1)
β2k
and
c2 =
A
.
β2k
Hence
ψ(x) = −
A −β
A
A A −βx
−βx
−β
e
(e
=
1
−
e
+
−
1)x
+
+
(e
−
1)x
.
β2k
β2k
β2k
β2k
12.6
Nonhomogeneous Boundary-Value Problems
Now the new problem is
k
∂v
∂2v
=
∂x2
∂t
, 0 < x < 1,
t > 0,
v(1, t) = 0,
t > 0,
v(0, t) = 0,
v(x, 0) = f (x) − ψ(x),
0 < x < 1.
Identifying this as the heat equation solved in Section 12.3 in the text with L = 1 we obtain
∞
v(x, t) =
An e−kn
2 π2 t
sin nπx
n=1
where
ˆ
An = 2
1
[f (x) − ψ(x)] sin nπx dx.
0
Thus
u(x, t) =
∞
A 2 2
−βx
−β
+
(e
−
1)x
+
An e−kn π t sin nπx.
1
−
e
2
β k
n=1
6. Substituting u(x, t) = v(x, t) + ψ(x) into the partial differential equation gives
k
∂v
∂2v
.
+ kψ − hv − hψ =
2
∂x
∂t
This equation will be homogeneous provided ψ satisfies
kψ − hψ = 0.
Since k and h are positive, the general solution of this latter linear second-order equation is
h
h
x + c2 sinh
x.
ψ(x) = c1 cosh
k
k
From ψ(0) = 0 and ψ(π) = u0 we find c1 = 0 and c2 = u0 / sinh
h/k π. Hence
h/k x
ψ(x) = u0
.
sinh h/k π
sinh
Now the new problem is
k
∂v
∂2v
,
− hv =
2
∂x
∂t
v(0, t) = 0,
0 < x < π, t > 0
v(π, t) = 0,
v(x, 0) = −ψ(x),
t>0
0 < x < π.
743
744
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
If we let v = XT then
T + hT
X =
= −λ.
X
kT
With λ = α2 > 0, the separated differential equations
X + α2 X = 0
T + h + kα2 T = 0.
and
have the respective solutions
X(x) = c3 cos αx + c4 sin αx
2
T (t) = c5 e−(h+kα )t .
From X(0) = 0 we get c3 = 0 and from X(π) = 0 we find α = n for n = 1, 2, 3, . . . .
Consequently, it follows that
v(x, t) =
∞
2
An e−(h+kn )t sin nx
n=1
where
2
An = −
π
ˆ
π
ψ(x) sin nx dx.
0
Hence a solution of the original problem is
∞
sinh h/k x
2
u(x, t) = u0
+ e−ht
An e−kn t sin nx
sinh h/k π
n=1
where
2
An = −
π
ˆ
π
0
h/k x
sin nx dx.
u0
sinh h/k π
sinh
Using the exponential definition of the hyperbolic sine and integration by parts we find
An =
2u0 nk(−1)n
.
π (h + kn2 )
7. Substituting u(x, t) = v(x, t) + ψ(x) into the partial differential equation gives
k
∂v
∂2v
.
+ kψ − hv − hψ + hu0 =
2
∂x
∂t
This equation will be homogeneous provided ψ satisfies
kψ − hψ + hu0 = 0
or
kψ − hψ = −hu0 .
This non-homogeneous, linear, second-order, differential equation has solution
h
h
x + c2 sinh
x + u0 ,
ψ(x) = c1 cosh
k
k
12.6
Nonhomogeneous Boundary-Value Problems
where we assume h > 0 and k > 0. From ψ(0) = u0 and ψ(1) = 0 we find c1 = 0 and
c2 = −u0 / sinh h/k . Thus, the steady-state solution is
⎛
⎞
h
sinh
u0
h
k x
sinh
⎠.
ψ(x) = −
x + u0 = u0 ⎝1 −
k
h
sinh
sinh h
k
k
8. The partial differential equation is
k
∂2u
∂u
.
− hu =
∂x2
∂t
Substituting u(x, t) = v(x, t) + ψ(x) gives
k
∂v
∂2v
.
+ kψ − hv − hψ =
2
∂x
∂t
This equation will be homogeneous provided ψ satisfies
kψ − hψ = 0.
Assuming h > 0 and k > 0, we have
√
√
ψ = c1 e h/k x + c2 e− h/k x ,
where we have used the exponential form of the solution since the rod is infinite. Now, in
order that the steady-state temperature ψ(x) be bounded as x → ∞, we require c1 = 0. Then
√
ψ(x) = c2 e− h/k x
and ψ(0) = u0 implies c2 = u0 . Thus
ψ(x) = u0 e−
√
h/k x
.
9. Substituting u(x, t) = v(x, t) + ψ(x) into the partial differential equation gives
a2
∂2v
∂2v
2 +
a
ψ
+
Ax
=
.
∂x2
∂t2
This equation will be homogeneous provided ψ satisfies
a2 ψ + Ax = 0.
The solution of this differential equation is
ψ(x) = −
A 3
x + c1 x + c 2 .
6a2
From ψ(0) = 0 we obtain c2 = 0, and from ψ(1) = 0 we obtain c1 = A/6a2 . Hence
ψ(x) =
A
(x − x3 ).
6a2
745
746
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
Now the new problem is
a2
∂2v
∂2v
=
∂x2
∂t2
v(0, t) = 0,
v(1, t) = 0,
v(x, 0) = −ψ(x),
t > 0,
vt (x, 0) = 0,
0 < x < 1.
Identifying this as the wave equation solved in Section 12.4 in the text with L = 1, f (x) =
−ψ(x), and g(x) = 0 we obtain
v(x, t) =
∞
An cos nπat sin nπx
n=1
where
ˆ
1
An = 2
0
Thus
A
[−ψ(x)] sin nπx dx = 2
3a
ˆ
1
(x3 − x) sin nπx dx =
0
2A(−1)n
.
a2 π 3 n3
∞
2A (−1)n
A
3
cos nπat sin nπx.
u(x, t) = 2 (x − x ) + 2 3
6a
a π
n3
n=1
10. We solve
a2
∂2u
∂2u
−
g
=
,
∂x2
∂t2
u(0, t) = 0,
u(1, t) = 0,
∂u
∂t
u(x, 0) = 0,
0 < x < 1, t > 0
= 0,
t>0
0 < x < 1.
t=0
The partial differential equation is nonhomogeneous. The substitution u(x, t) = v(x, t)+ψ(x)
yields a homogeneous partial differential equation provided ψ satisfies
a2 ψ − g = 0.
By integrating twice we find
ψ(x) =
g 2
x + c1 x + c 2 .
2a2
The imposed conditions ψ(0) = 0 and ψ(1) = 0 then lead to c2 = 0 and c1 = −g/2a2 . Hence
ψ(x) =
g 2
x −x .
2
2a
12.6
Nonhomogeneous Boundary-Value Problems
The new problem is now
∂2v
∂2v
=
,
∂x2
∂t2
a2
v(0, t) = 0,
v(x, 0) =
0 < x < 1, t > 0
v(1, t) = 0
g x − x2 ,
2
2a
∂v
∂t
= 0.
t=0
Substituting v = XT we find in the usual manner
X + α2 X = 0
T + a2 α2 T = 0
with solutions
X(x) = c3 cos αx + c4 sin αx
T (t) = c5 cos aαt + c6 sin aαt.
The conditions X(0) = 0 and X(1) = 0 imply in turn that c3 = 0 and α = nπ for
n = 1, 2, 3, . . . . The condition T (0) = 0 implies c6 = 0. Hence, by the superposition
principle
∞
An cos (anπt) sin (nπx).
v(x, t) =
n=1
At t = 0,
∞
g 2
An sin (nπx)
x
−
x
=
2a2
n=1
and so
An =
g
a2
ˆ
1
x − x2 sin (nπx) dx =
0
2g
a2 n3 π 3
[1 − (−1)n ] .
Thus the solution to the original problem is
u(x, t) = ψ(x) + v(x, t) =
∞
2g 1 − (−1)n
g 2
−
x
+
cos (anπt) sin (nπx).
x
2a2
a2 π 3
n3
n=1
11. Substituting u(x, y) = v(x, y) + ψ(y) into Laplace’s equation we obtain
∂2v
∂2v
+
+ ψ (y) = 0.
∂x2 ∂y 2
This equation will be homogeneous provided ψ satisfies ψ(y) = c1 y + c2 . Considering
u(x, 0) = v(x, 0) + ψ(0) = u1
u(x, 1) = v(x, 1) + ψ(1) = u0
u(0, y) = v(0, y) + ψ(y) = 0
747
748
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
we require that ψ(0) = u1 , ψ1 = u0 and v(0, y) = −ψ(y). Then c1 = u0 − u1 and c2 = u1 .
The new boundary-value problem is
∂2v
∂2v
+
=0
∂x2 ∂y 2
v(x, 0) = 0,
v(x, 1) = 0,
v(0, y) = −ψ(y),
0 < y < 1,
where v(x, y) is bounded at x → ∞. This problem is similar to Problem 11 in Section 12.5.
The solution is
ˆ 1
∞
2
[−ψ(y) sin nπy] dy e−nπx sin nπy
v(x, y) =
n=1
=2
∞
n=1
0
ˆ
(u1 − u0 )
1
ˆ
y sin nπy dy − u1
0
1
sin nπy dy e−nπx sin nπy
0
∞
2 u0 (−1)n − u1 −nπx
=
e
sin nπy.
π
n
n=1
Thus
u(x, y) = v(x, y) + ψ(y) = (u0 − u1 )y + u1 +
∞
2 u0 (−1)n − u1 −nπx
e
sin nπy.
π
n
n=1
12. Substituting u(x, y) = v(x, y) + ψ(x) into Poisson’s equation we obtain
∂2v
∂2v
+
ψ
(x)
+
h
+
= 0.
∂x2
∂y 2
The equation will be homogeneous provided ψ satisfies ψ (x) + h = 0 or
ψ(x) = − h2 x2 + c1 x + c2 . From ψ(0) = 0 we obtain c2 = 0. From ψ(π) = 1 we obtain
c1 =
Then
ψ(x) =
1 hπ
+
.
π
2
1 hπ
+
π
2
x−
h 2
x .
2
The new boundary-value problem is
∂2v
∂2v
+ 2 =0
2
∂x
∂y
v(0, y) = 0,
v(π, y) = 0,
v(x, 0) = −ψ(x),
0 < x < π.
12.6
Nonhomogeneous Boundary-Value Problems
This is Problem 11 in Section 12.5. The solution is
v(x, y) =
∞
An e−ny sin nx
n=1
where
An =
2
π
ˆ
π
2(−1)n
n
1 hπ
+
π
2
1 hπ
+
π
2
h
x − x2 +
An e−ny sin nx.
2
[−ψ(x) sin nx] dx =
0
Thus
u(x, y) = v(x, y) + ψ(x) =
− h(−1)n
π
2
+ 2
n n
.
∞
n=1
13. From (13) and (14) we have
ψ(x, t) = sin t + x[0 − sin t] = (1 − x) sin t,
Then the substitution
u(x, t) = v(x, t) + ψ(x, t) = v(x, t) + (1 − x) sin t
leads to the boundary-value problem
∂2v
∂v
,
+ (x − 1) cos t =
2
∂x
∂t
v(0, t) = 0,
0 < x < 1,
v(1, t) = 0,
v(x, 0) = 0,
t>0
t>0
0 < x < 1.
The eigenvalues and eigenfunctions of the Sturm-Liouville problem
X + λX = 0,
X(0) = 0,
X(1) = 0
are λn = αn2 = n2 π 2 and sin nπx, n = 1, 2, 3, . . . . With G(x, t) = (x − 1) cos t we assume for
fixed t that v and G can be written as Fourier sine series:
v(x, t) =
∞
vn (t) sin nπx
and
G(x, t) =
n=1
∞
Gn (t) sin nπx.
n=1
By treating t as a parameter, the coefficients Gn can be computed:
ˆ 1
ˆ
2 1
2
cos t.
(x − 1) cos t sin nπx dx = 2 cos t
(x − 1) sin nπx dx = −
Gn (t) =
1 0
nπ
0
Hence
(x − 1) cos t =
∞
−2 cos t
n=1
nπ
sin nπx.
749
750
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
Now, using the series representation for v(x, t), we have
∞
∞
∂2v
=
vn (t)(−n2 π 2 ) sin nπx and
∂x2
∂v =
vn (t) sin nπx.
∂t
n=1
n=1
Writing the partial differential equation as vt − vxx = (x − 1) cos t and using the above results
we have
∞
∞
−2 cos t
2 2
sin nπx.
[vn (t) + n π vn (t)] sin nπx =
nπ
n=1
n=1
Equating coefficients we get
vn (t) + n2 π 2 vn (t) = −
2 cos t
.
nπ
For each n this is a linear first-order differential equation whose general solution is
vn (t) = −
Thus
v(x, t) =
∞
2 n2 π 2 cos t + sin t
2 2
+ Cn e−n π t .
4
4
nπ
n π +1
−
n=1
2n2 π 2 cos t + 2 sin t
2 2
+ Cn e−n π t sin nπx.
4
4
nπ(n π + 1)
The initial condition v(x, 0) = 0 implies
∞
n=1
−
2nπ
+ Cn sin nπx = 0
+1
n4 π 4
so that Cn = 2nπ/(n4 π 4 + 1). Therefore
v(x, t) =
∞
−
n=1
2n2 π 2 cos t + 2 sin t
2nπ
2 2
+ 4 4
e−n π t sin nπx
4
4
nπ(n π + 1)
n π +1
2 2
∞
2 n2 π 2 e−n π t − n2 π 2 cos t − sin t
=
sin nπx
π
n(n4 π 4 + 1)
n=1
and
2 2
∞
2 n2 π 2 e−n π t − n2 π 2 cos t − sin t
sin nπx.
u(x, t) = v(x, t) + ψ(x, t) = (1 − x) sin t +
π
n(n4 π 4 + 1)
n=1
14. From (13) and (14) we have
ψ(x, t) = u0 (t) +
X
[u1 (t) − u0 (t)] = x + (1 − x)t2 ,
L
Then the substitution
u(x, t) = v(x, t) + ψ(x, t) = v(x, t) + x + (1 − x)t2
12.6
Nonhomogeneous Boundary-Value Problems
leads to the boundary-value problem
∂2v
∂v
,
+ 5tx =
2
∂x
∂t
v(0, t) = 0,
0 < x < 1,
v(1, t) = 0,
v(x, 0) = x2 − x,
t>0
t>0
0 < x < 1.
The eigenvalues and eigenfunctions of the Sturm-Liouville problem
X + λX = 0,
X(0) = 0,
X(1) = 0
are λn = αn2 = n2 π 2 and sin nπx, n = 1, 2, 3, . . . . With G(x, t) = 5tx we assume for fixed t
that v and G can be written as Fourier sine series:
v(x, t) =
∞
vn (t) sin nπx
and
∞
G(x, t) =
n=1
Gn (t) sin nπx.
n=1
By treating t as a parameter, the coefficients Gn can be computed:
ˆ
ˆ 1
2 1
10t
(−1)n+1 .
5tx sin nπx dx = 10t
x sin nπx dx =
Gn (t) =
1 0
nπ
0
Hence
5tx =
∞
(−1)n+1
n=1
10t
sin nπx.
nπ
Now, using the series representation for v(x, t), we have
∞
∂2v
=
vn (t)(−n2 π 2 ) sin nπx and
∂x2
n=1
∞
∂v =
vn (t) sin nπx.
∂t
n=1
Writing the partial differential equation as vt − vxx = 5tx and using the above results we have
∞
[vn (t)
2 2
+ n π vn (t)] sin nπx =
n=1
∞
(−1)n+1
n=1
10t
sin nπx.
nπ
Equating coefficients we get
vn (t) + n2 π 2 vn (t) = (−1)n+1
10t
.
nπ
For each n this is a linear first-order differential equation whose general solution is
vn (t) = 10(−1)n+1
Thus
v(x, t) =
∞
n=1
10(−1)n+1
n2 π 2 t − 1
2 2
+ Cn e−n π t .
n5 π 5
n2 π 2 t − 1
2 2
+ Cn e−n π t sin nπx.
5
5
n π
751
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CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
The initial condition v(x, 0) = x2 − x implies
∞
10(−1)n+1
n=1
−1
+ Cn sin nπx = x2 − x.
n5 π 5
Thinking of x2 − x as a Fourier sine series with coefficients
ˆ 1
(x2 − x) sin nπx dx = [4(−1)n − 4]/n3 π 3
2
0
we equate coefficients to obtain
10(−1)n
4(−1)n − 4
+
C
=
n
n5 π 5
n3 π 3
so
Cn =
4(−1)n − 4 10(−1)n
−
.
n3 π 3
n5 π 5
Therefore
v(x, t) =
∞
10(−1)n+1
n=1
n2 π 2 t − 1
+
n5 π 5
4(−1)n − 4 10(−1)n
−
n3 π 3
n5 π 5
e−n
2 π2 t
sin nπx
and
u(x, t) = v(x, t) + ψ(x, t)
= x + (1 − x)t +
2
∞
10(−1)n+1
n=1
n2 π 2 t − 1
+
n5 π 5
4(−1)n − 4 10(−1)n
−
n3 π 3
n5 π 5
15. From (13) and (14) we have
ψ(x, t) = u0 (t) +
X
[u1 (t) − u0 (t)] = x sin t,
L
Then the substitution
u(x, t) = v(x, t) + ψ(x, t) = x sin t
leads to the boundary-value problem
∂2v
∂2v
+
x
sin
t
=
,
∂x2
∂t2
v(0, t) = 0,
v(x, 0) = 0,
∂v
∂t
0 < x < 1,
v(1, t) = 0,
= −x,
t>0
t>0
0 < x < 1.
t=0
The eigenvalues and eigenfunctions of the Sturm-Liouville problem
X + λX = 0,
X(0) = 0,
X(1) = 0
e−n
2 π2 t
sin nπx.
12.6
Nonhomogeneous Boundary-Value Problems
are λn = αn2 = n2 π 2 and sin nπx, n = 1, 2, 3, . . . . With G(x, t) = x sin t we assume for fixed
t that v and G can be written as Fourier sine series:
v(x, t) =
∞
vn (t) sin nπx
and
G(x, t) =
n=1
∞
Gn (t) sin nπx.
n=1
By treating t as a parameter, the coefficients Gn can be computed:
ˆ
2
Gn (t) =
1
1
0
2
x sin t sin nπx dx = sin t
1
Hence
x sin t =
∞
n=1
−2
ˆ
1
x sin nπx dx = −2
0
(−1)n sin t
.
nπ
(−1)n sin t
sin nπx.
nπ
Now, using the series representation for v(x, t), we have
∞
∞
∂2v
=
vn (t)(−n2 π 2 ) sin nπx and
2
∂x
∂v =
vn (t) sin nπx.
∂t
n=1
n=1
Writing the partial differential equation as vtt − vxx = x sin t and using the above results we
have
∞
∞
(−1)n sin t
2 2
sin nπx.
[vn (t) + n π vn (t)] sin nπx =
−2
nπ
n=1
n=1
Equating coefficients we get
vn (t) + n2 π 2 vn (t) = −2
(−1)n sin t
.
nπ
For each n the general solution is
vn (t) = An cos nπt + Bn sin nπt − 2
Thus
v(x, t) =
∞
An cos nπt + Bn sin nπt − 2
n=1
(−1)n
sin t
nπ (n2 π 2 − 1)
(−1)n
sin t sin nπx.
nπ (n2 π 2 − 1)
The initial condition v(x, 0) = 0 implies
∞
An sin nπx = 0.
n=1
or An = 0 for n = 1, 2, 3, . . . . So
v(x, t) =
∞
n=1
Bn sin nπt − 2
(−1)n
sin t sin nπx.
nπ (n2 π 2 − 1)
753
754
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
and
∂v
∂t
= −x =
∞
nπBn − 2
n=1
t=0
(−1)n
sin nπx.
nπ (n2 π 2 − 1)
Thinking of −x as a Fourier sine series with coefficients
ˆ 1
(−1)n
x sin nπx dx = 2
−2
nπ
0
we equate coefficients to obtain
nπBn − 2
so
Bn = 2
(−1)n
(−1)n
=
2
nπ (n2 π 2 − 1)
nπ
(−1)n
(−1)n
+2
.
nπ
nπ (n2 π 2 − 1)
Therefore
v(x, t) =
∞
2
n=1
(−1)n
(−1)n
+2
nπ
nπ (n2 π 2 − 1)
sin nπt − 2
(−1)n
sin t sin nπx
nπ (n2 π 2 − 1)
and
u(x, t) = v(x, t) + ψ(x, t)
= x sin t + 2
∞
n=1
(−1)n
(−1)n
+
nπ
nπ (n2 π 2 − 1)
sin nπt −
(−1)n
sin t sin nπx.
nπ (n2 π 2 − 1)
16. From (13) and (14) we have
ψ(x, t) = u0 (t) +
X
[u1 (t) − u0 (t)] = 1 − e−t ,
L
Then the substitution
u(x, t) = v(x, t) + ψ(x, t) = v(x, t) + 1 − e−t
leads to the boundary-value problem
∂v
∂2v
,
− e−t =
2
∂x
∂t
v(0, t) = 0,
0 < x < 1,
v(1, t) = 0,
v(x, 0) = 0,
t>0
t>0
0 < x < 1.
The eigenvalues and eigenfunctions of the Sturm-Liouville problem
X + λX = 0,
X(0) = 0,
X(1) = 0
12.6
Nonhomogeneous Boundary-Value Problems
are λn = αn2 = n2 π 2 and sin nπx, n = 1, 2, 3, . . . . With G(x, t) = −e−t we assume for fixed
t that v and G can be written as Fourier sine series:
v(x, t) =
∞
vn (t) sin nπx
and
G(x, t) =
n=1
∞
Gn (t) sin nπx.
n=1
By treating t as a parameter, the coefficients Gn can be computed:
ˆ 1
2
(−1)n − 1
.
sin nπx dx = 2e−t
Gn (t) = e−t
1
nπ
0
Hence
−t
−e
−t
= 2e
∞
(−1)n − 1
n=1
nπ
sin nπx.
Now, using the series representation for v(x, t), we have
∞
∂2v
=
vn (t)(−n2 π 2 ) sin nπx and
∂x2
n=1
∞
∂v =
vn (t) sin nπx.
∂t
n=1
Writing the partial differential equation as vt − vxx = −e−t and using the above results we
have
∞
∞
(−1)n − 1
sin nπx.
[vn (t) + n2 π 2 vn (t)] sin nπx = 2e−t
nπ
n=1
n=1
Equating coefficients we get
vn (t) + n2 π 2 vn (t) = 2
(−1)n − 1 −t
e .
nπ
For each n this is a linear first-order differential equation whose general solution is
vn (t) = 2
Thus
v(x, t) =
∞
2
n=1
(−1)n − 1
2 2
e−t + Cn e−n π t .
2
2
nπ (n π − 1)
(−1)n − 1
2 2
e−t + Cn e−n π t sin nπx.
2
2
nπ (n π − 1)
The initial condition v(x, 0) = 0 implies
Cn = −2
Therefore
v(x, t) = 2
∞
n=1
(−1)n − 1
.
nπ (n2 π 2 − 1)
(−1)n − 1 −t
−n2 π 2 t
sin nπx
−
e
e
nπ (n2 π 2 − 1)
and
u(x, t) = v(x, t) + ψ(x, t) = 1 − e−t +
∞
2 (−1)n − 1 −t
−n2 π 2 t
sin nπx.
e
−
e
π
n (n2 π 2 − 1)
n=1
755
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CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
17. Identifying k = 1 and L = π we see that the eigenfunctions of X + λX = 0, X(0) = 0,
∞
un (t) sin nx, the formal
X(π) = 0 are sin nx, n = 1, 2, 3, . . . . Assuming that u(x, t) =
n=1
partial derivatives of u are
∞
∞
∂2u =
un (t)(−n2 ) sin nx
∂x2
∂u =
un (t) sin nx.
∂t
and
n=1
∞
Assuming that xe−3t =
n=1
Fn (t) sin nx we have
n=1
Fn (t) =
2
π
ˆ
π
xe−3t sin nx dx =
0
Then
xe−3t =
2e−3t
π
ut − uxx =
∞
π
x sin nx dx =
0
∞
2e−3t (−1)n+1
n=1
and
ˆ
n
2e−3t (−1)n+1
.
n
sin nx
[un (t) + n2 un (t)] sin nx = xe−3t =
n=1
∞
2e−3t (−1)n+1
n=1
n
sin nx.
Equating coefficients we obtain
un (t) + n2 un (t) =
2e−3t (−1)n+1
.
n
This is a linear first-order differential equation whose solution is
un (t) =
Thus
2(−1)n+1 −3t
2
e
+ Cn e−n t .
2
n(n − 3)
∞
∞
2(−1)n+1 −3t
2
e sin nx +
Cn e−n t sin nx
u(x, t) =
n(n2 − 3)
n=1
n=1
and u(x, 0) = 0 implies
∞
∞
2(−1)n+1
sin
nx
+
Cn sin nx = 0
n(n2 − 3)
n=1
n=1
so that Cn = 2(−1)n /n(n2 − 3). Therefore
u(x, t) = 2
∞
∞
(−1)n+1 −3t
(−1)n −n2 t
e
e
sin
nx
+
2
sin nx.
n(n2 − 3)
n(n2 − 3)
n=1
n=1
12.6
Nonhomogeneous Boundary-Value Problems
18. Identifying k = 1 and L = π we see that the eigenfunctions of X + λX = 0, X(0) = 0,
∞
1
un (t) cos nx,
X (π) = 0 are 1, cos nx, n = 1, 2, 3, . . . . Assuming that u(x, t) = u0 (t) +
2
n=1
the formal partial derivatives of u are
∞
∞
∂2u =
un (t)(−n2 ) cos nx
∂x2
∂u
1
= u0 +
un (t) cos nx.
∂t
2
and
n=1
n=1
∞
1
Fn (t) cos nx we have
Assuming that xe−3t = F0 (t) +
2
n=1
2e−3t
F0 (t) =
π
and
Fn (t) =
2e−3t
π
Then
xe−3t =
ˆ
ˆ
π
x dx = πe−3t
0
π
x cos nx dx =
0
2e−3t [(−1)n − 1]
.
πn2
∞
π −3t 2e−3t [(−1)n − 1]
e
+
cos nx
2
πn2
n=1
and
∞
ut − uxx
1
= u0 (t) +
[un (t) + n2 un (t)] cos nx
2
n=1
= xe−3t =
∞
π −3t 2e−3t [(−1)n − 1]
e
+
cos nx.
2
πn2
n=1
Equating coefficients, we obtain
u0 (t) = πe−3t
un (t) + n2 un (t) =
and
2e−3t [(−1)n − 1]
cos nx.
πn2
The first equation yields u0 (t) = −(π/3)e−3t + C0 and the second equation, which is a linear
first-order differential equation, yields
un (t) =
2[(−1)n − 1] −3t
2
e
+ Cn e−n t .
2
2
πn (n − 3)
Thus
∞
∞
n=1
n=1
2[(−1)n − 1]
π
2
−3t
e
u(x, t) = − e−3t + C0 +
cos
nx
+
Cn e−n t cos nx
3
πn2 (n2 − 3)
and u(x, 0) = 0 implies
∞
∞
n=1
n=1
2[(−1)n − 1]
π
cos
nx
+
Cn cos nx = 0
− + C0 +
3
πn2 (n2 − 3)
757
758
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
so that C0 = π/3 and Cn = 2[(−1)n − 1]/πn2 (n2 − 3). Therefore
∞
∞
π
2 (−1)n − 1 −3t
2 1 − (−1)n −n2 t
−3t
u(x, t) = (1 − e ) +
e cos nx +
e
cos nx.
3
π
n2 (n2 − 3)
π
n2 (n2 − 3)
n=1
n=1
19. Identifying k = 1 and L = 1 we see that the eigenfunctions of X + λX = 0, X(0) = 0,
∞
un (t) sin nπx, the formal
X(1) = 0 are sin nπx, n = 1, 2, 3, . . . . Assuming that u(x, t) =
n=1
partial derivatives of u are
∞
∞
∂2u =
un (t)(−n2 π 2 ) sin nπx
∂x2
∂u =
un (t) sin nπx.
∂t
and
n=1
Assuming that −1 + x − x cos t =
2
Fn (t) =
1
ˆ
1
n=1
∞
n=1 Fn (t) sin nπx
we have
(−1 + x − x cos t) sin nπx dx =
0
Then
−1 + x − x cos t =
2[−1 + (−1)n cos t]
.
nπ
∞
2 −1 + (−1)n cos t
sin nπx
π
n
n=1
and
ut − uxx =
∞
[un (t) + n2 π 2 un (t)] sin nπx
n=1
= −1 + x − x cos t =
∞
2 −1 + (−1)n cos t
sin nπx.
π
n
n=1
Equating coefficients we obtain
un (t) + n2 π 2 un (t) =
2[−1 + (−1)n cos t]
.
nπ
This is a linear first-order differential equation whose solution is
un (t) =
1
n2 π 2 cos t + sin t
2
2 2
− 2 2 + (−1)n
+ Cn e−n π t .
4
4
nπ
n π
n π +1
Thus
u(x, t) =
∞
∞
1
2
n2 π 2 cos t + sin t
2 2
− 2 2 + (−1)n
sin
nπx
+
Cn e−n π t sin nπx
4
4
nπ
n π
n π +1
n=1
n=1
and u(x, 0) = x(1 − x) implies
∞
1
2
n2 π 2
− 2 2 + (−1)n 4 4
+ Cn sin nπx = x(1 − x).
nπ
n π
n π +1
n=1
12.6
Nonhomogeneous Boundary-Value Problems
Hence
ˆ
1
2
n2 π 2
2
− 2 2 + (−1)n 4 4
+ Cn =
nπ
n π
n π +1
1
and
Cn =
1
x(1 − x) sin nπx dx = 2
0
1 − (−1)n
n3 π 3
4 − 2(−1)n
2nπ
.
− (−1)n 4 4
3
3
n π
n π +1
Therefore
u(x, t) =
∞
1
2
n2 π 2 cos t + sin t
− 2 2 + (−1)n
sin nπx
nπ
n π
n4 π 4 + 1
n=1
∞
4 − 2(−1)n
2nπ
2 2
+
e−n π t sin nπx.
− (−1)n 4 4
3
3
n π
n π +1
n=1
20. Identifying k = 1 and L = π we see that the eigenfunctions of X + λX = 0, X(0) = 0,
∞
un (t) sin nx, the formal
X(π) = 0 are sin nx, n = 1, 2, 3, . . . . Assuming that u(x, t) =
n=1
partial derivatives of u are
∞
∂2u =
un (t)(−n2 ) sin nx
∂x2
∞
∂ 2 u =
un (t) sin nx.
∂t2
and
n=1
Then
utt − uxx =
∞
n=1
[un (t) + n2 un (t)] sin nx = cos t sin x.
n=1
Equating coefficients, we obtain u1 (t) + u1 (t) cos t and un (t) + n2 un (t) = 0 for n = 2, 3, 4, . . ..
Solving the first differential equation we obtain u1 (t) = A1 cos t + B1 sin t + 12 t sin t. From the
second differential equation we obtain un (t) = An cos nt + Bn sin nt for n = 2, 3, 4, . . .. Thus
∞
u(x, t) =
1
A1 cos t + B1 sin t + t sin t sin x +
(An cos nt + Bn sin nt) sin nx.
2
n=2
From
u(x, 0) = A1 sin x +
∞
An sin nx = 0
n=2
we see that An = 0 for n = 1, 2, 3, . . . . Thus
∞
u(x, t) =
1
B1 sin t + t sin t sin x +
Bn sin nt sin nx
2
n=2
and
∂u
=
∂t
∞
1
1
B1 cos t + t cos t + sin t sin x +
nBn cos nt sin nx,
2
2
n=2
so
∂u
∂t
= B1 sin x +
t=0
∞
nBn sin nx = 0.
n=2
We see that Bn = 0 for all n so u(x, t) = 12 t sin t sin x.
759
760
CHAPTER 12
12.7
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
Orthogonal Series Expansions
1. Referring to Example 1 in the text we have
X(x) = c1 cos αx + c2 sin αx
and
T (t) = c3 e−kα t .
2
From X (0) = 0 (since the left end of the rod is insulated), we find c2 = 0.
X(x) = c1 cos αx and the other boundary condition X (1) = −hX(1) implies
−α sin α + h cos α = 0
or
cot α =
Then
α
.
h
Denoting the consecutive positive roots of this latter equation by αn for n = 1, 2, 3, . . . , we
have
∞
2
An e−kαn t cos αn x.
u(x, t) =
n=1
From the initial condition u(x, 0) = 1 we obtain
1=
∞
An cos αn x
n=1
and
ˆ
1
cos αn x dx
An =
ˆ 01
=
1
2
2
cos αn x dx
sin αn /αn
1+
1
2αn
=
sin 2αn
αn 1 +
2 sin αn
1
αn
sin αn cos αn
0
=
αn 1 +
2h sin αn
=
.
αn [h + sin2 αn ]
sin αn (αn sin αn )
2 sin αn
1
hαn
The solution is
u(x, t) = 2h
∞
sin αn
2
e−kαn t cos αn x.
2
α (h + sin αn )
n=1 n
2. Substituting u(x, t) = v(x, t) + ψ(x) into the partial differential equation gives
k
∂2v
∂v
.
+ kψ =
∂x2
∂t
This equation will be homogeneous if ψ (x) = 0 or ψ(x) = c1 x + c2 . The boundary condition
u(0, t) = 0 implies ψ(0) = 0 which implies c2 = 0. Thus ψ(x) = c1 x. Using the second
boundary condition we obtain
−
∂v
+ ψ
∂x
= −h[v(1, t) + ψ(1) − u0 ],
x=1
12.7
Orthogonal Series Expansions
which will be homogeneous when
−ψ (1) = −hψ(1) + hu0 .
Since ψ(1) = ψ (1) = c1 we have −c1 = −hc1 + hu0 and c1 = hu0 /(h − 1). Thus
ψ(x) =
hu0
x.
h−1
The new boundary-value problem is
k
∂v
∂2v
,
=
∂x2
∂t
v(0, t) = 0,
∂v
∂x
0 < x < 1,
t>0
= −hv(1, t),
h > 0,
hu0
x,
h−1
0 < x < 1.
t>0
x=1
v(x, 0) = f (x) −
Referring to Example 1 in the text we see that
v(x, t) =
∞
An e−kαn t sin αn x
2
n=1
and
∞
hu0
2
x+
u(x, t) = v(x, t) + ψ(x) =
An e−kαn t sin αn x
h−1
n=1
where
∞
f (x) −
hu0
x=
An sin αn x
h−1
n=1
and αn is a solution of αn cos αn = −h sin αn . The coefficients are
ˆ 1
ˆ 1
[f (x) − hu0 x/(h − 1)] sin αn x dx
[f (x) − hu0 x/(h − 1)] sin αn x dx
0
0
=
An =
ˆ 1
1
1
2
sin
2α
1
−
n
2
2αn
sin αn x dx
0
ˆ
1
2
[f (x) − hu0 x/(h − 1)] sin αn x dx
0
=
=
ˆ
1−
2
´1
0
1
αn
sin αn cos αn
1
2
=
[f (x) − hu0 x/(h − 1)] sin αn x dx
0
1−
[f (x) − hu0 x/(h − 1)] sin αn x dx
2h
=
1
h + cos2 αn
1 − hαn (−αn cos αn ) cos αn
3. Separating variables in Laplace’s equation gives
X + α2 X = 0
Y − α2 Y = 0
1
hαn (h sin αn ) cos αn
ˆ
0
1
f (x) −
hu0
x sin αn x dx.
h−1
761
762
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
and
X(x) = c1 cos αx + c2 sin αx
Y (y) = c3 cosh αy + c4 sinh αy.
From u(0, y) = 0 we obtain X(0) = 0 and c1 = 0. From ux (a, y) = −hu(a, y) we obtain
X (a) = −hX(a) and
α cos αa = −h sin αa
tan αa = −
or
α
.
h
Let αn , where n = 1, 2, 3, . . . , be the consecutive positive roots of this equation. From
u(x, 0) = 0 we obtain Y (0) = 0 and c3 = 0. Thus
u(x, y) =
∞
An sinh αn y sin αn x.
n=1
Now
f (x) =
∞
An sinh αn b sin αn x
n=1
ˆ
and
An sinh αn b =
Since
ˆ
a
sin2 αn x dx =
0
a
0´
f (x) sin αn x dx
a
2
0 sin αn x dx
.
1
1
1
1
a−
a−
sin 2αn a =
sin αn a cos αn a
2
2αn
2
αn
=
1
1
a−
(h sin αn a) cos αn a
2
hαn
=
1
1 1
a−
ah + cos2 αn a ,
(−αn cos αn a) cos αn a =
2
hαn
2h
we have
An =
2h
sinh αn b[ah + cos2 αn a]
ˆ
a
f (x) sin αn x dx.
0
4. Letting u(x, y) = X(x)Y (y) and separating variables gives
X Y + XY = 0.
The boundary conditions
∂u
∂y
=0
y=0
and
∂u
∂y
= −hu(x, 1)
y=1
12.7
Orthogonal Series Expansions
correspond to
X(x)Y (0) = 0
X(x)Y (1) = −hX(x)Y (1)
and
or
Y (0) = 0
Y (1) = −hY (1).
and
Since these homogeneous boundary conditions are in terms of Y , we separate the differential
equation as
X Y =−
= α2 .
X
Y
Then
Y + α2 Y = 0
and
X − α2 X = 0
have solutions
Y (y) = c1 cos αy + c2 sin αy
and
X(x) = c3 e−αx + c4 eαx .
We use exponential functions in the solution of X(x) since the interval over which X is defined
is infinite. (See the informal rule given in Section 11.5 of the text that discusses when to use
the exponential form and when to use the hyperbolic form of the solution of y − α2 y = 0.)
Now, Y (0) = 0 implies c2 = 0, so Y (y) = c1 cos αy. Since Y (y) = −c1 α sin αy, the boundary
condition Y (1) = −hY (1) implies
−c1 α sin α = −hc1 cos α
or
cot α =
α
.
h
Consideration of the graphs of f (α) = cot α and g(α) = α/h show that cos α = αh has an
infinite number of roots. The consecutive positive roots αn for n = 1, 2, 3, . . . , are the
eigenvalues of the problem. The corresponding eigenfunctions are Yn (y) = c1 cos αn y. The
condition lim u(x, y) = 0 is equivalent to lim X(x) = 0. Thus c4 = 0 and X(x) = c3 e−αx .
x→∞
x→∞
Therefore
un (x, y) = Xn (x)Yn (x) = An e−αn x cos αn y
and by the superposition principle
u(x, y) =
∞
An e−αn x cos αn y.
n=1
[It is easily shown that there are no eigenvalues corresponding to α = 0.]
condition u(0, y) = u0 implies
∞
An cos αn y.
u0 =
n=1
Finally, the
763
764
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
This is not a Fourier cosine series since the coefficients αn of y are not integer multiples of
π/p, where p = 1 in this problem. The functions cos αn y are however orthogonal since they
are eigenfunctions of the Sturm-Lionville problem
Y + α2 Y = 0,
Y (0) = 0
Y (1) + hY (1) = 0,
with weight function p(x) = 1. Thus we find
ˆ 1
u0 cos αn y dy
0
.
An = ˆ 1
2
cos αn y dy
0
Now
ˆ
0
and
ˆ
1
1
u0
u0 cos αn y dy =
sin αn y
αn
1
2
cos2 αn y dy =
0
ˆ
1
=
0
1
(1 + cos 2αn y) dy =
0
u0
sin αn
αn
1
1
sin 2αn y
y+
2
2αn
1
0
1
1
1
1
1+
1+
sin 2αn =
sin αn cos αn .
2
2αn
2
αn
=
Since cot α = α/h,
cos α
sin α
=
α
h
and
ˆ
1
cos2 αn y dy =
0
Then
u0
αn
1
2 1+
An =
and
u(x, y) = 2hu0
sin αn
1
h
2
sin αn
∞
n=1
αn
1
sin2 αn
1+
2
h
=
.
2hu0 sin αn
αn h + sin2 αn
sin αn
e−αn x cos αn y
2
h + sin αn
where αn for n = 1, 2, 3, . . . are the consecutive positive roots of cot α = α/h.
5. The boundary-value problem is
k
∂u
∂2u
,
=
2
∂x
∂t
u(0, t) = 0,
0 < x < L,
∂u
∂x
= 0,
t > 0,
t > 0,
x=L
u(x, 0) = f (x),
0 < x < L.
12.7
Orthogonal Series Expansions
Separation of variables leads to
X + α2 X = 0
T + kα2 T = 0
and
X(x) = c1 cos αx + c2 sin αx
T (t) = c3 e−kα t .
2
From X(0) = 0 we find c1 = 0. From X (L) = 0 we obtain cos αL = 0 and
π(2n − 1)
, n = 1, 2, 3, . . . .
2L
α=
Thus
u(x, t) =
∞
An e−k(2n−1)
2 π 2 t/4L2
sin
n=1
where
ˆ
L
f (x) sin
An =
0
ˆ
L
2
sin
0
2n − 1
2L
2n − 1
ˆ
πx dx
2 L
2L
=
f (x) sin
L 0
2n − 1
πx dx
2L
πx
2n − 1
2L
πx dx.
6. Substituting u(x, t) = v(x, t) + ψ(x) into the partial differential equation gives
a2
∂2v
∂2v
+
ψ
(x)
=
.
∂x2
∂t2
This equation will be homogeneous if ψ (x) = 0 or ψ(x) = c1 x + c2 . The boundary condition
u(0, t) = 0 implies ψ(0) = 0 which implies c2 = 0. Thus ψ(x) = c1 x. Using the second
boundary condition, we obtain
E
∂v
+ ψ
∂x
= F0 ,
x=L
which will be homogeneous when
Eψ (L) = F0 .
Since ψ (x) = c1 we conclude that c1 = F0 /E and
ψ(x) =
F0
x.
E
765
766
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
The new boundary-value problem is
a2
∂2v
∂2v
=
,
∂x2
∂t2
∂v
∂x
v(0, t) = 0,
v(x, 0) = −
0 < x < L,
= 0,
t > 0,
= 0,
0 < x < L.
x=L
∂v
∂t
F0
x,
E
t>0
t=0
Referring to Example 2 in the text we see that
v(x, t) =
∞
2n − 1
2L
An cos a
n=1
where
∞
F0
An sin
− x=
E
n=1
and
2n − 1
2L
πt sin
2n − 1
2L
πx
πx
ˆ L
2n − 1
πx dx
−F0
x sin
8F0 L(−1)n
2L
0
=
.
An =
ˆ L
Eπ 2 (2n − 1)2
2 2n − 1
πx dx
E
sin
2L
0
Thus
∞
u(x, t) = v(x, t) + ψ(x) =
F0
8F0 L (−1)n
x+
cos a
E
Eπ 2
(2n − 1)2
n=1
2n − 1
2L
πt sin
7. Separation of variables leads to
Y + α2 Y = 0
X − α2 X = 0
and
Y (y) = c1 cos αy + c2 sin αy
X(x) = c3 cosh αx + c4 sinh αx.
From Y (0) = 0 we find c1 = 0. From Y (1) = 0 we obtain cos α = 0 and
α=
π(2n − 1)
, n = 1, 2, 3, . . . .
2
Thus
Y (y) = c2 sin
2n − 1
2
πy.
2n − 1
2L
πx.
12.7
Orthogonal Series Expansions
From X (0) = 0 we find c4 = 0. Then
u(x, y) =
∞
2n − 1
2
An cosh
n=1
where
u0 = u(1, y) =
∞
An cosh
n=1
and
ˆ
An cosh
2n − 1
2
2n − 1
2
1
u0 sin
π = ˆ0
1
sin2
0
2n − 1
2
πx sin
π sin
πy
2n − 1
2
πy
2n − 1
πy dy
4u0
2
.
=
(2n − 1)π
2n − 1
πy dy
2
Thus
∞
4u0 1
2n−1 cosh
u(x, y) =
π
(2n
−
1)
cosh
π
2
n=1
2n − 1
2
πx sin
2n − 1
2
πy.
8. Using u = XT and separation of variables leads to
X + λX = 0
T + (1 + λ)T = 0
with
X (0) = 0 and X(1) = 0
This is a regular Sturm-Liouville problem. For λ = 0 we have X = 0 or X = c1 + c2 x. From
X (0) = 0 we find c2 = 0, so X = c1 . From X(1) = 0 we find c1 = 0. Thus X = 0.
For λ = α2 the general solution is X = c1 cos aαx + c2 sin αx. From X (0) = 0 we find c2 = 0,
2n − 1
π for n = 1, 2, 3, . . . .
so X = c1 cos αx. From X(1) = 0 we find cos α = 0 or α =
2
Therefore
2 2
2n − 1
X = c1 cos
πx
and
T = c3 e−[(2n−1) π t/4+1]
2
Thus
u(x, t) = e−t
∞
n=1
where
ˆ
An =
0
1
An e−(2n−1)
2 π 2 t/4
cos
2n − 1
2
πx.
2n − 1
−16 cos nπ
1 − x2 cos
πx dx
−32(−1)n
2
(2n − 1)3 π 3
=
=
.
ˆ 1
1/2
(2n − 1)3 π 3
2 2n − 1
πx dx
cos
2
0
767
768
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
9. The boundary-value problem is
k
∂u
∂x
∂2u
∂u
,
=
2
∂x
∂t
∂u
∂x
= hu(0, t),
x=0
0 < x < 1,
t>0
= −hu(1, t),
h > 0,
t > 0,
x=1
u(x, 0) = f (x),
0 < x < 1.
Referring to Example 1 in the text we have
X(x) = c1 cos αx + c2 sin αx
and
T (t) = c3 e−kα t .
2
Applying the boundary conditions, we obtain
X (0) = hX(0)
and X (1) = −hX(1)
or
αc2 = hc1
−αc1 sin α + αc2 cos α = −hc1 cos α − hc2 sin α.
Choosing c1 = α and c2 = h (to satisfy the first equation above) we obtain
−α2 sin α + hα cos α = −hα cos α − h2 sin α
2hα cos α = (α2 − h2 ) sin α.
The eigenvalues αn are the consecutive positive roots of
tan α =
Then
u(x, t) =
∞
2hα
.
α2 − h2
An e−kαn t (αn cos αn x + h sin αn x)
2
n=1
where
f (x) = u(x, 0) =
∞
An (αn cos αn x + h sin αn x)
n=1
and
ˆ
1
f (x)(αn cos αn x + h sin αn x) dx
An =
0
ˆ
1
(αn cos αn x + h sin αn x)2 dx
0
2
= 2
αn + 2h + h2
ˆ
1
f (x)(αn cos αn x + h sin αn x) dx.
0
[Note: the evaluation and simplification of the integral in the denominator requires the use
of the relationship (α2 − h2 ) sin α = 2hα cos α.]
12.7
Orthogonal Series Expansions
10. The eigenfunctions of the associated homogeneous boundary-value problem are sin αn x,
n = 1, 2, 3, . . . , where the αn are the consecutive positive roots of tan α = −α. We
assume that
u(x, t) =
∞
un (t) sin αn x
xe−2t =
and
n=1
∞
Fn (t) sin αn x.
n=1
Then
−2t
ˆ
1
x sin αn x dx
e
ˆ
Fn (t) =
0
1
.
2
sin αn x dx
0
Since αn cos αn = − sin αn and
ˆ 1
1
1
1−
sin2 αn x dx =
sin 2αn ,
2
2αn
0
we have
−2t
ˆ
1
e
sin αn − αn cos αn
αn2
x sin αn x dx = e−2t
0
ˆ
1
0
=
2 sin αn −2t
e
αn2
1
sin2 αn x dx = [1 + cos2 αn ]
2
and so
Fn (t) =
4 sin αn
e−2t .
αn2 (1 + cos2 αn )
Substituting the assumptions into ut − kuxx = xe−2t and equating coefficients leads to the
linear first-order differential equation
un (t) + kαn2 u(t) =
4 sin αn
e−2t
αn2 (1 + cos2 αn )
whose solution is
un (t) =
From
u(x, t) =
4 sin αn
2
e−2t + Cn e−kαn t .
αn2 (1 + cos2 αn )(kαn2 − 2)
∞
n=1
αn2 (1
4 sin αn
2
e−2t + Cn e−kαn t sin αn x
2
2
+ cos αn )(kαn − 2)
and the initial condition u(x, 0) = 0 we see
Cn = −
αn2 (1
4 sin αn
.
+ cos2 αn )(kαn2 − 2)
The formal solution of the original problem is then
u(x, t) =
∞
α2 (1
n=1 n
4 sin αn
2
(e−2t − e−kαn t ) sin αn x.
+ cos2 αn )(kαn2 − 2)
769
770
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
11. Using u = XT and separation constant −λ = α4 we find
X (4) − α4 X = 0
and
X(x) = c1 cos αx + c2 sin αx + c3 cosh αx + c4 sinh αx.
Since u = XT the boundary conditions become
X(0) = 0,
X (0) = 0,
X (1) = 0,
X (1) = 0.
Now X(0) = 0 implies c1 + c3 = 0, while X (0) = 0 implies c2 + c4 = 0. Thus
X(x) = c1 cos αx + c2 sin αx − c1 cosh αx − c2 sinh αx.
The boundary condition X (1) = 0 implies
−c1 cos α − c2 sin α − c1 cosh α − c2 sinh α = 0
while the boundary condition X (1) = 0 implies
c1 sin α − c2 cos α − c1 sinh α − c2 cosh α = 0.
We then have the system of two equations in two unknowns
(cos α + cosh α)c1 + (sin α + sinh α)c2 = 0
(sin α − sinh α)c1 − (cos α + cosh α)c2 = 0.
This homogeneous system will have nontrivial solutions for c1 and c2 provided
cos α + cosh α
sin α + sinh α
=0
sin α − sinh α − cos α − cosh α
or
−2 − 2 cos α cosh α = 0.
Thus, the eigenvalues are determined by the equation cos α cosh α = −1.
50
Using a computer to graph cosh α and
−1/ cos α = − sec α we see that the first two positive
eigenvalues occur near 1.9 and 4.7. Applying Newton’s
method with these initial values we find that the eigenvalues are α1 = 1.8751 and α2 = 4.6941.
40
30
20
10
0
12. (a) In this case the boundary conditions are
u(0, t) = 0,
u(1, t) = 0,
∂u
∂x
∂u
∂x
=0
x=0
= 0.
x=1
1
2
3
4
α
5
12.7
Orthogonal Series Expansions
771
Separating variables leads to
X(x) = c1 cos αx + c2 sin αx + c3 cosh αx + c4 sinh αx
subject to
X(0) = 0,
X (0) = 0,
X(1) = 0,
X (1) = 0.
and
Now X(0) = 0 implies c1 + c3 = 0 while X (0) = 0 implies c2 + c4 = 0. Thus
X(x) = c1 cos αx + c2 sin αx − c1 cosh αx − c2 sinh αx.
The boundary condition X(1) = 0 implies
c1 cos α + c2 sin α − c1 cosh α − c2 sinh α = 0
while the boundary condition X (1) = 0 implies
−c1 sin α + c2 cos α − c1 sinh α − c2 cosh α = 0.
We then have the system of two equations in two unknowns
(cos α − cosh α)c1 + (sin α − sinh α)c2 = 0
−(sin α + sinh α)c1 + (cos α − cosh α)c2 = 0.
This homogeneous system will have nontrivial solutions for c1 and c2 provided
cos α − cosh α sin α − sinh α
=0
− sin α − sinh α cos α − cosh α
or
2 − 2 cos α cosh α = 0.
Thus, the eigenvalues are determined by the equation cos α cosh α = 1.
(b) Using a computer to graph cosh α and 1/ cos α = sec α
we see that the first two positive eigenvalues occur
near the vertical asymptotes of sec α, at 3π/2 and
5π/2. Applying Newton’s method with these initial
values we find that the eigenvalues are α1 = 4.7300
and α2 = 7.8532.
100
80
60
40
20
α
0
2
4
6
8
10
772
CHAPTER 12
12.8
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
Higher Dimensional Problems
1. This boundary-value problem was solved in Example 1 in the text. Identifying b = c = π and
f (x, y) = u0 we have
u(x, y, t) =
∞ ∞
Amn e−k(m
2 +n2 )t
sin mx sin ny
m=1 n=1
where
Amn
4
= 2
π
=
ˆ
π
ˆ
0
π
0
4u0
u0 sin mx sin ny dx dy = 2
π
ˆ
ˆ
π
π
sin mx dx
0
sin ny dy
0
4u0
[1 − (−1)m ][1 − (−1)n ].
mnπ 2
2. As shown in Example 1 in the text, separation of variables leads to
X(x) = c1 cos αx + c2 sin αx
Y (y) = c3 cos βy + c4 sin βy
and
T (t) + c5 e−k(α
2 +β 2 )t
.
The boundary conditions
!
ux (0, y, t) = 0,
ux (1, y, t) = 0
uy (x, 0, t) = 0,
uy (x, 1, t) = 0
imply
X (0) = 0,
X (1) = 0
Y (0) = 0,
Y (1) = 0
Applying these conditions to
X (x) = −αc1 sin αx + αc2 cos αx
and
Y (y) = −βc3 sin βy + βc4 cos βy
gives c2 = c4 = 0 and sin α = sin β = 0. Then
α = mπ, m = 0, 1, 2, . . .
and
β = nπ, n = 0, 1, 2, . . . .
By the superposition principle
u(x, y, t) = A00 +
∞
−km2 π 2 t
Am0 e
cos mπx +
m=1
+
∞ ∞
m=1 n=1
∞
A0n e−kn
2 π2 t
n=1
Amn e−k(m
2 +n2 )π 2 t
cos mπx cos nπy.
cos nπy
12.8
Higher Dimensional Problems
773
We now compute the coefficients of the double cosine series: Identifying b = c = 1 and
f (x, y) = xy we have
ˆ
ˆ
1ˆ 1
A00 =
xy dx dy =
0
0
ˆ
0
1
1
1
1 2
x y dy =
2
2
0
ˆ
1ˆ 1
Am0 = 2
1
xy cos mπx dx dy = 2
0
ˆ
ˆ
0
0
1
y dy =
0
1
,
4
1
1
(cos mπx + mπx sin mπx) y dy
2
m π2
0
cos mπ − 1
(−1)m − 1
cos mπ − 1
y
dy
=
=
,
m2 π 2
m2 π 2
m2 π 2
0
ˆ 1ˆ 1
(−1)n − 1
=2
xy cos nπy dx dy =
,
n2 π 2
0
0
1
=2
A0n
and
ˆ
Amn = 4
0
(−1)m − 1
m2 π 2
ˆ
1
xy cos mπx cos nπy dx dy = 4
0
=4
ˆ
1ˆ 1
x cos mπx dx
0
(−1)n − 1
n2 π 2
1
y cos nπy dy
0
.
In Problems 3 and 4 we need to solve the partial differential equation
a2
∂2u ∂2u
+ 2
∂x2
∂y
=
∂2u
.
∂t2
To separate this equation we try u(x, y, t) = X(x)Y (y)T (t):
a2 (X Y T + XY T ) = XY T Y T X =−
+ 2 = −α2 .
X
Y
a T
Then
X + α2 X = 0
(1)
T Y = 2 + α2 = −β 2
Y
a T
Y + β 2 Y = 0
T + a2 α2 + β 2 T = 0.
(2)
(3)
774
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
The general solutions of equations (1), (2), and (3) are, respectively,
X(x) = c1 cos αx + c2 sin αx
Y (y) = c3 cos βy + c4 sin βy
T (t) = c5 cos a α2 + β 2 t + c6 sin a α2 + β 2 t.
3. The conditions X(0) = 0 and Y (0) = 0 give c1 = 0 and c3 = 0. The conditions X(π) = 0 and
Y (π) = 0 yield two sets of eigenvalues:
α = m, m = 1, 2, 3, . . .
and
β = n, n = 1, 2, 3, . . . .
A product solution of the partial differential equation that satisfies the boundary conditions
is
umn (x, y, t) = Amn cos a m2 + n2 t + Bmn sin a m2 + n2 t sin mx sin ny.
To satisfy the initial conditions we use the superposition principle:
u(x, y, t) =
∞ ∞ 2
2
2
2
Amn cos a m + n t + Bmn sin a m + n t sin mx sin ny.
m=1 n=1
The initial condition ut (x, y, 0) = 0 implies Bmn = 0 and
u(x, y, t) =
∞ ∞
Amn cos a m2 + n2 t sin mx sin ny.
m=1 n=1
At t = 0 we have
xy(x − π)(y − π) =
∞ ∞
Amn sin mx sin ny.
m=1 n=1
Using (12) and (13) in the text, it follows that
ˆ πˆ π
4
xy(x − π)(y − π) sin mx sin ny dx dy
Amn = 2
π 0 0
ˆ π
ˆ π
4
x(x − π) sin mx dx
y(y − π) sin ny dy
= 2
π 0
0
=
16
[(−1)m − 1][(−1)n − 1].
m3 n3 π 2
4. The conditions X(0) = 0 and Y (0) = 0 give c1 = 0 and c3 = 0. The conditions X(b) = 0 and
Y (c) = 0 yield two sets of eigenvalues
α = mπ/b, m = 1, 2, 3, . . .
and
β = nπ/c, n = 1, 2, 3, . . . .
A product solution of the partial differential equation that satisfies the boundary conditions
is
nπ mπ x sin
y ,
umn (x, y, t) = (Amn cos aωmn t + Bmn sin aωmn t) sin
b
c
12.8
where ωmn =
principle:
Higher Dimensional Problems
775
(mπ/b)2 + (nπ/c)2 . To satisfy the initial conditions we use the superposition
u(x, y, t) =
∞ ∞
(Amn cos aωmn t + Bmn sin aωmn t) sin
m=1 n=1
At t = 0 we have
f (x, y) =
∞ ∞
Amn sin
m=1 n=1
and
g(x, y) =
∞ ∞
nπ mπ x sin
y .
b
c
mπ nπ x sin
y
b
c
Bmn aωmn sin
m=1 n=1
nπ mπ x sin
y .
b
c
Using (12) and (13) in the text, it follows that
Amn
4
=
bc
Bmn =
ˆ cˆ
b
f (x, y) sin
0
0
4
abc ωmn
ˆ cˆ
mπ x sin (nπc y) dx dy
b
b
g(x, y) sin
0
0
nπ mπ x sin
y dx dy.
b
c
In Problems 5 and 6 we try u(x, y, z) = X(x)Y (y)Z(z) to separate Laplace’s equation in three
dimensions:
X Y Z + XY Z + XY Z = 0
Y Z X =−
−
= −α2 .
X
Y
Z
Then
X + α2 X = 0
(4)
Z Y =−
+ α2 = −β 2
Y
Z
Y + β 2 Y = 0
(5)
Z − (α2 + β 2 )Z = 0.
(6)
The general solutions of equations (4), (5), and (6) are, respectively
X(x) = c1 cos αx + c2 sin αx
Y (y) = c3 cos βy + c4 sin βy
Z(z) = c5 cosh α2 + β 2 z + c6 sinh α2 + β 2 z.
776
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
5. The boundary and initial conditions are
u(0, y, z) = 0,
u(a, y, z) = 0
u(x, 0, z) = 0,
u(x, b, z) = 0
u(x, y, 0) = 0,
u(x, y, c) = f (x, y).
The conditions X(0) = Y (0) = Z(0) = 0 give c1 = c3 = c5 = 0. The conditions X(a) = 0
and Y (b) = 0 yield two sets of eigenvalues:
α=
mπ
, m = 1, 2, 3, . . .
a
and
β=
nπ
, n = 1, 2, 3, . . . .
b
By the superposition principle
u(x, y, t) =
∞ ∞
Amn sinh ωmn z sin
m=1 n=1
where
2
ωmn
=
and
Amn
4
=
ab sinh ωmn c
m2 π 2 n2 π 2
+ 2
a2
b
ˆ bˆ
a
f (x, y) sin
0
mπ
nπ
x sin
y
a
b
0
nπ
mπ
x sin
y dx dy.
a
b
6. The boundary and initial conditions are
u(0, y, z) = 0,
u(a, y, z) = 0,
u(x, 0, z) = 0,
u(x, b, z) = 0,
u(x, y, 0) = f (x, y),
u(x, y, c) = 0
The conditions X(0) = Y (0) = 0 give c1 = c3 = 0. The conditions X(a) = Y (b) = 0 yield
two sets of eigenvalues:
α=
mπ
, m = 1, 2, 3, . . .
a
Let
2
ωmn
=
and
β=
nπ
, n = 1, 2, 3, . . . .
b
m2 π 2 n2 π 2
+ 2 .
a2
b
Then the boundary condition Z(c) = 0 gives
c5 cosh c ωmn + c6 sinh c ωmn = 0
from which we obtain
Z(z) = c5 cosh wmn z −
=
cosh c ωmn
sinh ωz
sinh c ωmn
c5
(sinh c ωmn cosh ωmn z − cosh c ωmn sinh ωmn z) = cmn sinh ωmn (c − z).
sinh c ωmn
Chapter 12 in Review
By the superposition principle
u(x, y, t) =
∞ ∞
Amn sinh ωmn (c − z) sin
m=1 n=1
where
Amn =
4
ab sinh c ωmn
ˆ bˆ
a
f (x, y) sin
0
0
nπ
mπ
x sin
y
a
b
mπ
nπ
x sin
y dx dy.
a
b
Chapter 12 in Review
1. Letting u(x, y) = X(x) + Y (y) we have X Y = XY and
X
= Y Y = −λ.
X
If λ = 0 then X = 0 and X(x) = c1 . Also Y (y) = 0 so u = 0.
If λ = 0 then X + λX = 0 and Y + (1/λ)Y = 0. Thus X(x) = c1 e−λx and Y (y) = c2 e−y/λ so
u(x, y) = Ae(−λx−y/λ) .
2. Letting u = XY we have X Y +XY +2X Y +2XY = 0 so that (X +2X )Y +X(Y +2Y ) =
0. Separating variables and using the separation constant −λ we obtain
Y + 2Y X + 2X =
= −λ
−X
Y
so that
X + 2X − λX = 0
Y + 2Y + λY = 0.
and
The corresponding auxiliary equations are m2 + 2m − λ = 0 and m2 + 2m + λ with solutions
√
√
m = −1 ± 1 + λ and m = −1 ± 1 − λ, respectively. We consider five cases:
√
I. λ = −1: In this case X = c1 e−x + c2 xe−x and Y = c3 e(−1+ 2)y + c4 e(−1−
√
√
u = (c1 ex + c2 xe−x ) c3 e(−1+ 2)y + c4 e(−1− 2)y .
√
√
2)y
so that
√
II. λ = 1: In this case X = c5 e(−1+ 2)x + c6 e(−1− 2)y and Y = c7 e−y + c8 ye−y so that
√
√
u = c5 e(−1+ 2)x + c6 e(−1− 2)x c7 e−y + c8 ye−y .
III. −1 < λ < 1: Here both 1 + λ and 1 − λ are positive so
√
√
√
√
u = c9 e(−1+ 1+λ)x + c10 e(−1− 1+λ)x c11 e(−1+ 1−λ)y + c12 e(−1− 1−λ)y .
777
778
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
IV. λ < −1: Here 1 + λ < 0 and 1 − λ > 0 so
√
√
√
√
u = e−x c13 cos −1 − λ x + c14 sin −1 − λ x + c15 e(−1+ 1−λ)y + c16 e(−1− 1−λ)y .
V. λ > 1: Here 1 + λ > 0 and 1 − λ < 0 so
√
√
√
√
u = c17 e(−1+ 1+λ)x + c18 e(−1− 1+λ)x + e−x c19 cos λ − 1 y + c20 sin λ − 1 y .
We see from the above that it is not possible to choose λ so that both X and Y are oscillatory.
3. Substituting u(x, t) = v(x, t) + ψ(x) into the partial differential equation we obtain
k
∂v
∂2v
.
+ kψ (x) =
2
∂x
∂t
This equation will be homogeneous provided ψ satisfies
kψ = 0
or
ψ = c1 x + c2 .
Considering
u(0, t) = v(0, t) + ψ(0) = u0
we set ψ(0) = u0 so that ψ(x) = c1 x + u0 . Now
−
∂u
∂x
=−
x=π
∂v
∂x
− ψ (x) = v(π, t) + ψ(π) − u1
x=π
is equivalent to
∂v
∂x
+ v(π, t) = u1 − ψ (x) − ψ(π) = u1 − c1 − (c1 π + u0 ),
x=π
which will be homogeneous when
u 1 − c1 − c1 π − u 0 = 0
or
c1 =
u1 − u0
.
1+π
The steady-state solution is
ψ(x) =
u1 − u0
1+π
x + u0 .
4. The solution of the problem represents the heat of a thin rod of length π. The left boundary
x = 0 is kept at constant temperature u0 for t > 0. Heat is lost from the right end of the
rod by being in contact with a medium that is held at constant temperature u1 . Moreover,
intially the entire rod is at temperature zero.
Chapter 12 in Review
5. The boundary-value problem is
a2
∂2u
∂2u
=
,
∂x2
∂t2
u (0, t) = 0,
0 < x < 1,
t > 0,
u = (1, t) = 0,
t > 0,
∂u
∂t
u (x, 0) = 0,
= g(x),
0 < x < 1.
t=0
From Section 12.4 in the text we see that An = 0,
2
Bn =
nπa
¨1
2
g (x) sin nπ dx =
nπa
0
2h
=
nπa
3/4
¨
h sin nπx dx
1/4
3/4
1
−
cos nπx
nπ
and
u (x, t) =
=
1/4
∞
2h
n2 π 2 a
cos
nπ
3nπ
− cos
4
4
Bn sin nπat sin nπx.
n=1
6. The boundary-value problem is
∂2u
∂2u
2
+
x
=
,
∂x2
∂t2
0 < x < 1,
t > 0,
u(0, t) = 1,
u(1, t) = 0,
t > 0,
u(x, 0) = f (x),
ut (x, 0) = 0,
0 < x < 1.
Substituting u(x, t) = v(x, t) + ψ(x) into the partial differential equation gives
∂2v
∂2v
2
+
ψ
(x)
+
x
=
.
∂x2
∂t2
This equation will be homogeneous provided ψ (x) + x2 = 0 or
ψ(x) = −
1 4
x + c1 x + c 2 .
12
From ψ(0) = 1 and ψ(1) = 0 we obtain c1 = −11/12 and c2 = 1. The new problem is
∂2v
∂2v
= 2 ,
2
∂x
∂t
v(0, t) = 0,
0 < x < 1,
t > 0,
v(1, t) = 0,
t > 0,
v(x, 0) = f (x) − ψ(x),
vt (x, 0) = 0,
0 < x < 1.
779
780
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
From Section 12.4 in the text we see that Bn = 0,
ˆ
1
An = 2
ˆ
1
[f (x) − ψ(x)] sin nπx dx = 2
0
f (x) +
0
and
∞
v(x, t) =
1 4 11
x + x − 1 sin nπx dx,
12
12
An cos nπt sin nπx.
n=1
Thus
∞
u(x, t) = v(x, t) + ψ(x) = −
1 4 11
x − x+1+
An cos nπt sin nπx.
12
12
n=1
7. Using u = XY and −λ as a separation constant leads to
X − λX = 0,
X (0) = 0,
and
Y + λY = 0,
Y (0) = 0,
Y (π) = 0.
This leads to
Y = c4 sin ny
for n = 1, 2, 3, ... so that
u=
and
∞
X = c2 sinh nx
An sinh nx sin ny.
n=1
Imposing
u (π, y) = 50 =
∞
An sinh nπ sin ny
n=1
gives
An =
so that
100 1 − (−1)n
nπ sinh nπ
∞
100 1 − (−1)n
sinh nx sin ny.
u (x, y) =
π
n sinh nπ
n=1
8. Using u = XY and −λ as a separation constant leads to
X − λX = 0,
Chapter 12 in Review
and
Y + λY = 0,
Y (0) = 0,
Y (π) = 0.
This leads to
and X = c2 e−nx
Y = c3 cos ny
for n = 1, 2, 3, .... In this problem we also have λ = 0 is an eigenvalue with corresponding
eigenfunctions 1 and 1. Thus
u = A0 +
∞
An e−nx cos ny.
n=1
Imposing
u (0, y) = 50 = A0 +
∞
An cos ny
n=1
gives
A0 =
and
2
An
π
1
π
ˆ
ˆ
π
50 dy = 50
0
π
50 cos ny dy = 0
0
for n = 1, 2, 3, ... so that
u (x, y) = 50.
9. Using u = XY and −λ as a separation constant leads to
X − λX = 0,
and
Y + λY = 0,
Y (0) = 0,
Y (π) = 0.
Then
X = c1 enx + c2 e−nx
and
Y = c3 cos ny + c4 sin ny
for n = 1, 2, 3, . . . . Since u must be bounded as x → ∞, we define c1 = 0. Also Y (0) = 0
implies c3 = 0 so
∞
u=
An e−nx sin ny.
n=1
781
782
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
Imposing
u(0, y) = 50 =
∞
An sin ny
n=1
gives
2
π
An =
ˆ
π
100
[1 − (−1)n ]
nπ
50 sin ny dy =
0
so that
u(x, y) =
∞
100
n=1
nπ
[1 − (−1)n ]e−nx sin ny.
10. The boundary-value problem is
∂2u
∂u
=
,
2
∂x
∂t
−L < x < L,
t > 0,
u (−L, t) = 0,
u (L, t) = 0,
t > 0,
k
u (x, 0) = u0 ,
−L < x < L.
Referring to Section 12.3 in the text we have
X (x) = c1 cos ax + c2 sin ax
and
T (t) = c3 e−ka t .
2
Using the boundary conditions u (−L, 0) = X (−L) T (0) = 0 and u (L, 0) = X (L) T (0) = 0
we obtain X (−L) = 0 and X (L) = 0. Thus
c1 cos (−αL) + c2 sin (−αL) = 0
c1 cos αL + c2 sin αL = 0
or
c1 cos αL − c2 sin αL = 0
c1 cos αL + c2 sin αL = 0.
Adding, we find cos αL = 0 which gives the eigenvalues
α=
Thus
u (x, t) =
∞
2n − 1
π,
2L
n = 1, 2, 3, . . .
An e−[(2n−1)π/2L]
2
kt
n=1
From
u (x, 0) = u0 =
∞
n=1
An cos
cos
2n − 1
2L
2n − 1
2L
πx
πx.
Chapter 12 in Review
we find
ˆ
2n − 1
πx dx
u0 (−1)n+1 2L/π (2n − 1)
4u0 (−1)n+1
2L
=
=
.
L/2
π (2n − 1)
2n − 1
πx dx
2L
L
u0 cos
2
0
ˆ
An =
2
L
cos2
0
11. (a) The coefficients of the series
u (x, 0) =
∞
Bn sin nx
n=1
are
¨π
Bn = f rac2π
1
=
π
2
sin x sin nx dx =
π
0
¨π
0
sin (1 − n) x
1−n
For n = 1,
2
B1 =
π
π
0
¨π
sin (1 + n) x
−
1+n
1
sin x dx =
π
π
= 0 for n = 1.
0
¨π
(1 − cos 2x) dx = 1.
2
0
1
[cos (1 − n) x − cos (1 + n) x] dx
2
0
Thus
u (x, t) =
∞
Bn e−n t sin nx
2
n=1
reduces to u (x, t) = e−t sin x for n = 1.
(b) This is like part (a), but in this case, for n = 3 and n = 5,
2
Bn =
π
¨π
(100 sin 3x − 30 sin 5x) sin nx dx = 0;
0
while B3 = 100 and B5 = −30. Therefore
u (x, t) = 100e−9t sin 3x − 30e−25t sin 5x.
12. Substituting u(x, t) = v(x, t) + ψ(x) into the partial differential equation results in
ψ = − sin x and ψ(x) = c1 x+c2 +sin x. The boundary conditions ψ(0) = 400 and ψ(π) = 200
imply c1 = −200/π and c2 = 400 so
ψ(x) = −
200
x + 400 + sin x.
π
783
784
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
Solving
∂v
∂2v
,
=
2
∂x
∂t
0 < x < π,
t>0
v(0, t) = 0,
v(π, t) = 0,
t>0
v(x, 0) = 400 + sin x − −
200
x + 400 + sin x
π
=
200
x,
π
0<x<π
using separation of variables with separation constant −λ, where λ = α2 , gives
X + α2 X = 0
T + α2 T = 0.
and
Using X(0) = 0 and X(π) = 0 we determine α2 = n2 , X(x) = c2 sin nx, and T (t) = c3 e−n t .
Then
∞
2
An e−n t sin nx
v(x, t) =
2
n=1
and
∞
v(x, 0) =
200
x=
An sin nx
π
n=1
so
An =
Thus
400
π2
ˆ
π
400
(−1)n+1 .
nπ
x sin nx dx =
0
∞
u(x, t) = −
200
400 (−1)n+1 −n2 t
x + 400 + sin x +
e
sin nx.
π
π
n
n=1
13. Using u = XT and −λ, where λ = α2 , as a separation constant we find
X + 2X + α2 X = 0 and T + 2T + 1 + α2 T = 0.
Thus for α > 1
X = c1 e−x cos
α2 − 1 x + c2 e−x sin α2 − 1 x
T = c3 e−t cos αt + c4 e−t sin αt.
For 0 ≤ α ≤ 1 we only obtain X = 0. Now the boundary conditions X (0) = 0 and X (π) = 0
√
give, in turn, c1 = 0 and α2 − 1π = nπ or α2 = n2 = n2 +1, n = 1, 2, 3, .... The corresponding
solutions are X = c2 e−x sin nx. The initial condition T (0) = 0 implies c3 = αc4 and so
n2 + 1 cos n2 + 1 t + sin n2 + 1 t .
T = c4 e−t
Using u = XT and the superposition principle, a formal series solution is
u (x, t) = e−(x+t)
∞
n=1
An
n2 + 1 cos n2 + 1 t + sin n2 + 1 t sin nx.
Chapter 12 in Review
14. Letting c = XT and separating variables we obtain
kX − hX T
=
X
T
or
X − aX T
=
= −λ
X
kT
where a = h/k. Setting λ = α2 leads to the separated differential equations
X − aX + α2 X = 0
T + kα2 T = 0.
and
The solution of the second equation is
T (t) = c3 e−kα t .
2
√
For the first equation we have m = 12 a ± a2 − 4α2 , and we consider three cases using the
boundary conditions X (0) = X (1) = 0:
a2 > 4α2 The solution is X = c1 em1 x + c2 em2 x , where the boundary conditions imply
c1 = c2 = 0, so X = 0. (Note in this case that if α = 0, the solution is X = c1 + c2 eax
and the boundary conditions again imply c1 = c2 = 0, so X = 0.)
a2 = 4α2 The solution is X = c1 em1 x + c2 xem1 x , where the boundary conditions imply
c1 = c2 = 0, so X = 0.
a2 < 4α2 The solution is
√
ax/2
X (x) = c1 e
cos
4α2 − a2
x + c2 eax/2 sin
2
√
4α2 − a2
x.
2
From X (0) = 0 we see that c1 = 0. From X (1) = 0 we find
1 2
4α − a2 = nπ
2
or
α2 =
1 2 2
4n π + a2 .
4
Thus
X (x) = c2 eax/2 sin nπx,
and
c (x, t) =
∞
An eax/2 e−k(4n
2 π 2 +a2
)t/4 sin nπx.
n=1
The initial condition c (x, 0) = c0 implies
c0 =
∞
An eax/2 sin nπx.
n=1
From the self-adjoint form
d −ax e X + α2 e−ax X = 0
dx
785
786
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
the eigenfunctions are orthogonal on [0, 1] with weight function e−ax . That is
¨1
e−ax eax/2 sin nπx eax/2 sin mπx dx = 0,
n = m.
0
Multiplying (1) by e−ax eax/2 sin mπx and integrating we obtain
¨1
−ax ax/2
c0 e
e
sin mπx, dx =
∞
¨1
An
n=1
0
ˆ
1
e−ax/2 sin nπx dx = An
c0
ˆ
1
An = 2c0
0
ˆ
0
and
0
−ax/2
e
0
e−ax eax/2 (sin mπx) eax/2 sin nπx dx
1
1
sin2 nπx dx = An
2
8nπc0 ea/2 − (−1)n
4c0 2ea/2 nπ − 2nπ (−1)n
=
.
sin nπx dx =
ea/2 (a2 + 4n2 π 2 )
ea/2 (a2 + 4n2 π 2 )
15. By the discussion in Section 12.6, we assume a solution in the form u(x, t) = v(x, t) + ψ(x).
Force this into the differential equation to get ψ (x) = 0 together with the conditions ψ(0) = 0
and ψ (1) + ψ(1) = u1 .
Integrating twice and imposing the two conditions leads to ψ(x) = 12 (u1 − u0 )x + u0 . Now
u(x, t) = v(x, t) + ψ(x)
u(0, t) = v(0, t) + ψ(0) = u0
v(0, t) +
1
(u1 − u0 )(0) + u0 = u0
2
v(0, t) = 0
Similarly we find that
ux (x, t) = vx (x, t) + ψ (x)
ux (1, t) = vx (1, t) + ψ (1) = u1 − u(1, t)
1
(u1 − u0 ) = u1 − [v(1, t) + ψ(1)]
2
1
vx (1, t) = u1 − [v(1, t) + ψ(1)] − (u1 − u0 )
2
vx (1, t) = −v(1, t)
vx (1, t) +
So now we solve
∂v
∂2v
,
=
2
∂x
∂t
v(0, t) = 0,
∂v
∂x
0 < x < 1, t > 0
= −v(1, t), t > 0
x=1
1
v(x, 0) = (u0 − u1 )x, 0 < x < 1
2
Chapter 12 in Review
Using the results of Example 1 from Section 12.7, v(x, t) =
∞
An e−αn t sin αn x and the
2
n=1
coefficients are given by
ˆ 1
sin αn − αn cos αn
1
(u0 − u1 )
x sin αn x dx
1
αn2
2
An =
= (u0 − u1 ) 1
ˆ 1 0
2
2
2 (1 + cos αn )
sin2 αn x dx
0
= (u0 − u1 )
sin αn − αn cos αn
.
αn2 (1 + cos2 αn )
Since tan αn = −αn we have sin αn = −αn cos αn and so
An = 2(u1 − u0 )
∞
n=1
and
cos αn
2
e−αn t sin (αn x)
2
αn (1 + cos αn )
∞
1
cos αn
2
u(x, t) = u0 + (u1 − u0 )x + 2(u1 − u0 )
e−αn t sin (αn x)
2
αn (1 + cos2 αn )
n=1
16. This is similar Problem 13 in Exercises 12.5 with a = b = π and f (x) = g(x) = x2 − πx. Thus
u(x, y) =
∞
(An cosh ny + Bn sinh ny) sin nx.
n=1
At y = 0 we then have
f (x) =
∞
n=1
and consequently
An =
2
π
ˆ
At y = π,
g(x) =
π
0
∞
An sin
nπ
x
a
4 [(−1)n − 1]
x2 − πx sin nx dx =
πn3
(An cosh nπ + Bn sinh nπ) sin nx
n=1
indicates that the entire expression in the parentheses is given by
ˆ
2 π 2
An cosh nπ + Bn sinh nπ =
x − πx sin nx dx.
π 0
We can now solve for Bn :
2
1
Bn =
sinh nπ π
=
4 [(−1)n
ˆ
π
0
− 1]
πn3 sinh nπ
2
cosh nπ
x − πx sin nx dx −
An
sinh nπ
(1 − cosh nπ) = An
1 − cosh nπ
sinh nπ
787
788
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
Then
u(x, y) =
∞
4 [(−1)n − 1]
n=1
=
cosh ny +
πn3
∞
4 [(−1)n − 1]
π
n3
1 − cosh nπ
sinh ny sin nx
sinh nπ
sinh n(π − y) + sinh ny
sinh nπ
n=1
sin nx.
17. Subtituting u(x, y) = v(x, y) + ψ(x) into the boundary-value problem yields
∂2v
∂2v
+
= 0.
∂x2 ∂y 2
The condition ψ (x) = −2 has the general solution ψ(x) = −x2 + c1 x + c2 . The boundary
conditions x = 0 and x = π will be homogeneous if we choose ψ(0) = 0. The condition
ψ(π) = 0 gives c2 = 0 and c1 = π. Thus
ψ(π) = −x2 + πx.
The boundary-value problem in v is then
∂2v
∂2v
+
= 0,
∂x2 ∂y 2
v(0, y) = 0,
v(x, 0) = −ψ(x),
0 < x < π,
v(π, y) = 0,
0<y<π
0<y<π
v(x, π) = −ψ(x),
0 < y < π.
From Problem 16,
v(x, y) =
∞
(An cosh ny + Bn sinh ny) sin nx
n=1
=
∞
4 [(−1)n − 1]
π
n3
n=1
sinh n(π − y) + sinh ny
sinh nπ
sin nx.
Therefore
∞
4 [(−1)n − 1]
u(x, y) = ψ(x) + v(x, y) = −x + πx +
π
n3
2
n=1
sinh n(π − y) + sinh ny
sinh nπ
sin nx.
18. Using the subtitution u(x, t) = v(x, t) + ψ(x) into the boundary-value problem yields
∂v
∂2v
.
+ ψ + e−x =
2
∂x
∂t
Now u(0, t) = v(0, t) + ψ(0) = 0 and ux (π, t) = vx (π, t) + ψ (π) = 0. The condition ψ(0) = 0
gives −1 + c1 = 0 or c1 = 1. The condition ψ (π) = 0 gives e−π + c2 = 0 or c2 = −e−π . Thus,
ψ(x) = −e−x + 1 − e−π x.
Chapter 12 in Review
Now the homogeneous boundary-value problem for v(x, t) is
∂v
∂2v
,
=
2
∂x
∂t
0 < x < π,
∂v
∂x
v(0, t) = 0,
−x
t>0
,
t>0
x=π
− 1 + e−π x,
v(x, 0) = f (x) + e
0 < x < π.
Using v = XT and separation of variables leads to
X + λX = 0
T + λT = 0
with
X(0) = 0 and X (π) = 0
This is a regular Sturm-Liouville problem. For λ = α2 the general solution is
X = c3 cos αx + c4 sin αx. From X(0) = 0 we find c3 = 0, so X = c4 sin αx. From X (π) = 0
2n − 1
π for n = 1, 2, 3, . . . . Therefore
we find cos απ = 0 or απ =
2
2n − 1
2
X = c1 sin
Thus
v(x, t) =
∞
x
An e−(2n−1)
2 t/4
n=1
When t = 0,
−x
f (x) + e
−π
−1+e
T = c5 e−(2n−1)
and
x=
∞
sin
2n − 1
2
An sin
n=1
where
ˆ
An =
π
0
2
=
π
ˆ
x.
2n − 1
2
2n − 1
f (x) + e−x − 1 + e−π x sin
2
ˆ π
2n − 1
x
sin2
2
0
π
f (x) + e−x − 1 + e−π x sin
0
2 t/4
2n − 1
2
x
x dx
.
x dx.
The solutions is then
−x
u(x, t) = −e
−π
+1−e
x+
∞
n=1
where the coefficients An are defined above.
An e−(2n−1)
2 t/4
sin
2n − 1
2
x
789
790
CHAPTER 12
BOUNDARY-VALUE PROBLEMS IN RECTANGULAR COORDINATES
19. For w(x, y) = C sin
πx
πy
sin
,
a
b
π 4
πy
πx
∂4w
sin
=
C
sin
4
∂x
a
a
b
π 2 π 2
∂4w
πx
πy
sin
=
C
sin
2
2
∂x ∂y
a
b
a
b
π 4
πy
πx
∂4w
sin
=
C
sin
4
∂y
b
a
b
Thus
π 4
π 2 π 2 π 4
πx
∂4w
∂4w
πy
∂4w
sin
sin
+
2
+
=
C
+
2
+
4
2
2
4
∂x
∂x ∂y
∂y
a
a
b
b
a
b
=
πx
πy
q0
sin
sin
D
a
b
Hence
q0 /D
C=
1
1
+ 2
2
a
b
π4
Therefore
w(x, y) =
2
=
q0 a4 b4
π 4 D (a2 + b2 )2
q0 a4 b4
π 4 D (a2 +
b2 )2
sin
πy
πx
sin
.
a
b
20. Since sin 0 = 0 and sin π = 0, the first four boundary conditions are obvious:
w(0, y) = 0,
w(a, y) = 0,
w(x, 0) = 0
w(x, b) = 0.
Also, from
q0 a2 b4
πx
πy
∂2w
=
sin
sin
2
2
∂x
a
b
π 4 D (a2 + b2 )
q0 a4 b2
πx
πy
∂2w
sin
=
sin
2
2
4
2
2
∂x
a
b
π D (a + b )
and sin 0 = 0 and sin π = 0 we see that
∂2w
∂x2
= 0,
x=0
∂2w
∂x2
= 0,
x=a
∂2w
∂y 2
= 0,
y=0
∂2w
∂y 2
= 0.
y=b
Chapter 13
Boundary-Value Problems in Other Coordinate
Systems
13.1
Polar Coordinates
1. We have
ˆ π
1
u0
u0 dθ =
A0 =
2π 0
2
ˆ π
1
u0 cos nθ dθ = 0
An =
π 0
ˆ
1 π
u0
[1 − (−1)n ]
u0 sin nθ dθ =
Bn =
π 0
nπ
and so
u(r, θ) =
∞
u0 u0 1 − (−1)n n
+
r sin nθ.
2
π
n
n=1
2. We have
ˆ π
ˆ 2π
1
1
θ dθ +
(π − θ) dθ = 0
2π 0
2π π
ˆ
ˆ
1 π
1 2π
2
θ cos nθ dθ +
(π − θ) cos nθ dθ = 2 [(−1)n − 1]
An =
π 0
π π
n π
ˆ
ˆ
1 π
1 2π
1
Bn =
θ sin nθ dθ +
(π − θ) sin nθ dθ = [1 − (−1)n ]
π 0
π π
n
A0 =
and so
u(r, θ) =
∞
n=1
r
n
1 − (−1)n
2 [(−1)n − 1]
cos nθ +
sin nθ .
n2 π
n
791
792
CHAPTER 13
BOUNDARY-VALUE PROBLEMS IN OTHER COORDINATE SYSTEMS
3. We have
ˆ 2π
1
2π 2
(2πθ − θ2 ) dθ =
A0 =
2π 0
3
ˆ 2π
1
4
An =
(2πθ − θ2 ) cos nθ dθ = − 2
π 0
n
ˆ
1 2π
Bn =
(2πθ − θ2 ) sin nθ dθ = 0
π 0
and so
∞
u(r, θ) =
rn
2π 2
−4
cos nθ.
3
n2
n=1
4. We have
ˆ 2π
1
A0 =
θ dθ = π
2π 0
ˆ
1 2π
θ cos nθ dθ = 0
An =
π 0
ˆ
1 2π
2
θ sin nθ dθ = −
Bn =
π 0
n
and so
u(r, θ) = π − 2
∞
rn
n=1
n
sin nθ.
5. Proceeding in the usual way by letting u(r, θ) = R(r)Θ(θ) and substituting it into the PDE
we get
r2 R + rR − λR = 0 and
Θ + λΘ = 0
As in Example 1 we have R(r) = c3 rn + c4 r−n so that we must take c3 = 0 in order for the
solution to remain bounded as r → ∞. We therefore have
u(r, θ) = A0 +
∞
r−n (An cos nθ + Bn sin nθ)
n=1
The condition at r = c leads to the coefficients given by
1
A0 =
2π
cn
An =
π
Bn =
cn
π
ˆ
2π
f (θ) dθ
ˆ
0
π
f (θ) cos nθ dθ
ˆ
0
π
f (θ) sin nθ dθ
0
13.1
Polar Coordinates
6. The nature of the given conditions suggest we let u(r, θ) = v(r, θ) + ψ(θ). From this we get
ψ (θ) = 0 so that ψ(θ) = c1 θ + c2 . Now ψ(0) = 0 gives c2 = 0 and ψ(π) = u0 gives c1 = u0 /π.
Therefore ψ(θ) = (u0 /π) θ. Next we have
u(1, θ) = v(1, θ) + ψ(θ)
u0 = v(1, θ) +
From
v(r, θ) =
∞
u0
θ
π
An rn sin nθ
or v(1, θ) = u0 −
and
v(1, θ) =
n=1
u0
θ
π
∞
An sin nθ
n=1
We get the coefficients
2
An =
π
ˆ
0
π
u0 −
2u0
u0 θ sin nθ dθ =
π
nπ
The final solution is therefore
∞
u0
θ+
u(r, θ) =
π
n=1
2u0
nπ
rn sin nθ
7. Proceeding in the usual way by letting u(r, θ) = R(r)Θ(θ) and substituting it into the PDE
we get
r2 R + rR − λR = 0 and Θ + λΘ = 0
Using the results from Example 1 from Section 11.5, the eigenvalues and eigenfunctions for
the boundary-value problem Θ + λΘ = 0, Θ (0) = Θ (π) = 0 are λn = n2 for n = 0, 1, 2, . . .
and Θ = c1 cos (nθ). Now for n = 0 the equation r2 R + rR − λR = 0 has solution R(r) =
c3 + c4 ln r which, in order to remain bounded as r → 0, requires that c4 = 0. When
n = 1, 2, 3, . . ., the equation r2 R + rR − λR = 0 has solution R(r) = c3 rn + c4 r−n which
again requires c4 = 0. Product solutions are therefore
u0 (r, θ) = R0 (r)Θ0 (θ) = c3 c1 = A0 for n = 0
un (r, θ) = Rn (r)Θn (θ) = (c3 rn )(c1 cos nθ) = An rn cos nθ for n = 1, 2, 3, . . .
By the superposition principle
u(r, θ) = A0 +
∞
An rn cos nθ
n=1
From the given condition at r = 2, we obtain the coefficients
ˆ
1 2 π/2
u0
A0 = ·
u0 dθ =
2 π 0
2
ˆ
1 2 π/2
1 2u0 sin
u0 cos nθ dθ = n ·
An = n ·
2 π 0
2
nπ
nπ
2
793
794
CHAPTER 13
BOUNDARY-VALUE PROBLEMS IN OTHER COORDINATE SYSTEMS
The final solution is therefore
∞
u0 u(r, θ) =
+
2
n=1
1 2u0 sin
·
2n
nπ
nπ
2
∞
u0 2u0 r cos nθ =
+
2
π
n
sin
n=1
nπ
2
n
r n
2
cos nθ
8. We solve
∂ 2 u 1 ∂u
1 ∂2u
+
= 0,
+
∂r2
r ∂r
r2 ∂θ2
0<θ<
π
,
2
u(c, θ) = f (θ),
u(r, 0) = 0,
0 < r < c,
0<θ<
u(r, π/2) = 0,
π
,
2
0 < r < c.
Proceeding as in Example 1 in the text we obtain the separated differential equations
r2 R + rR − λR = 0
Θ + λΘ = 0.
Taking λ = α2 the solutions are
Θ(θ) = c1 cos αθ + c2 sin αθ
R(r) = c3 rα + c4 r−α .
Since we want R(r) to be bounded as r → 0 we require c4 = 0. Applying the boundary
conditions Θ(0) = 0 and Θ(π/2) = 0 we find that c1 = 0 and α = 2n for n = 1, 2, 3, . . . .
Therefore
∞
An r2n sin 2nθ.
u(r, θ) =
n=1
From
u(c, θ) = f (θ) =
∞
An c2n sin 2nθ
n=1
we find
4
An = 2n
πc
ˆ
π/2
f (θ) sin 2nθ dθ.
0
9. Referring to the solution of Problem 6 above we have
Θ(θ) = c1 cos αθ + c2 sin αθ
R(r) = c3 rα .
Applying the boundary conditions Θ (0) = 0 and Θ (π/2) = 0 we find that c2 = 0 and α = 2n
for n = 0, 1, 2, . . . .Therefore
u(r, θ) = A0 +
∞
n=1
An r2n cos 2nθ.
13.1
⎧
⎨1, 0 < θ < π/4
From
u(c, θ) =
= A0 +
⎩0,
π/4 < θ < π/2
we find
1
π/2
A0 =
and
2
c An =
π/2
ˆ
2n
Thus
u(r, θ) =
ˆ
∞
Polar Coordinates
An c2n cos 2nθ
n=1
π/4
dθ =
0
1
2
π/4
2
nπ
sin
.
nπ
2
cos 2nθ dθ =
0
∞
nπ r 2n
1 21
+
sin
cos 2nθ.
2 π
n
2 c
n=1
10. We solve
1 ∂2u
∂2u 1
∂u∂r
+
+
= 0,
∂r2
r
r2 ∂θ2
u(r, 0) = 0,
0 < θ < π/4,
r>0
r>0
u(r, π/4) = 30,
r > 0.
Proceeding as in Example 1 in the text we find the separated ordinary differential equations
to be
r2 R + rR − λR = 0
Θ + λΘ = 0.
With λ = α2 > 0 the corresponding general solutions are
R(r) = c1 rα + c2 r−α
Θ(θ) = c3 cos αθ + c4 sin αθ.
The condition Θ(0) = 0 implies c3 = 0 so that Θ = c4 sin αθ. Now, in order that the
temperature be bounded as r → ∞ we define c1 = 0. Similarly, in order that the temperature
be bounded as r → 0 we are forced to define c2 = 0. Thus R(r) = 0 and so no nontrivial
solution exists for λ > 0. For λ = 0 the separated differential equations are
r2 R + rR = 0
and
Θ = 0.
Solutions of these latter equations are
R(r) = c1 + c2 ln r
and
Θ(θ) = c3 θ + c4 .
795
796
CHAPTER 13
BOUNDARY-VALUE PROBLEMS IN OTHER COORDINATE SYSTEMS
Θ(0) = 0 still implies c4 = 0, whereas boundedness as r → 0 demands c2 = 0. Thus, a
product solution is
u = c1 c3 θ = Aθ.
From u(r, π/4) = 30 we obtain A = 120/π. Thus, a solution to the problem is
u(r, θ) =
120
θ.
π
11. Proceeding as in Example 1 in the text and again using the periodicity of u(r, θ), we have
Θ(θ) = c1 cos αθ + c2 sin αθ
where α = n for n = 0, 1, 2, . . . . Then
R(r) = c3 rn + c4 r−n .
[We do not have c4 = 0 in this case since 0 < a ≤ r.] Since u(b, θ) = 0 we have
∞
r u(r, θ) = A0 ln +
b
n=1
From
n b
r n
[An cos nθ + Bn sin nθ] .
−
r
b
∞
a u(a, θ) = f (θ) = A0 ln +
b
n=1
n b
a n
[An cos nθ + Bn sin nθ]
−
a
b
we find
ˆ 2π
a
1
f (θ) dθ,
A0 ln =
b
2π 0
n ˆ
a n
1 2π
b
An =
−
f (θ) cos nθ dθ,
a
b
π 0
and
n ˆ
a n
1 2π
b
Bn =
−
f (θ) sin nθ dθ.
a
b
π 0
12. Substituting u(r, θ) = v(r, θ) + ψ(r) into the partial differential equation we obtain
1 ∂v
1 ∂2v
∂2v
+
ψ
+
ψ
(r)
+
(r)
+
= 0.
∂r2
r ∂r
r2 ∂θ2
This equation will be homogeneous provided
1
ψ (r) + ψ (r) = 0
r
or
r2 ψ (r) + rψ (r) = 0.
The general solution of this Cauchy-Euler differential equation is
ψ(r) = c1 + c2 ln r.
13.1
Polar Coordinates
From
u0 = u(a, θ) = v(a, θ) + ψ(a)
and
u1 = u(b, θ) = v(b, θ) + ψ(b)
we see that in order for the boundary values v(a, θ) and v(b, θ) to be 0 we need ψ(a) = u0
and ψ(b) = u1 . From this we have
ψ(a) = c1 + c2 ln a = u0
ψ(b) = c1 + c2 ln b = u1 .
Solving for c1 and c2 we obtain
c1 =
Then
ψ(r) =
u1 ln a − u0 ln b
ln(a/b)
and
c2 =
u0 − u1
.
ln(a/b)
u0 ln(r/b) − u1 ln(r/a)
u1 ln a − u0 ln b u0 − u1
+
ln r =
.
ln(a/b)
ln(a/b)
ln(a/b)
From Problem 11 with f (θ) = 0 we see that the solution of
1 ∂2v
∂ 2 v 1 ∂v
+
+
= 0,
∂r2
r ∂r r2 ∂θ2
v(a, θ) = 0,
0 < θ < 2π,
a < r < b,
v(b, θ) = 0,
0 < θ < 2π
is v(r, θ) = 0. Thus the steady-state temperature of the ring is
u(r, θ) = v(r, θ) + ψ(r) =
u0 ln(r/b) − u1 ln(r/a)
.
ln(a/b)
13. Solutions of the separated equations are
Θ(θ) = c1 ,
n=0
Θ(θ) = c1 cos nθ + c2 sin nθ,
R(r) = c3 rn + c4 r−n ,
n = 1, 2, . . . R(r)
= c)3 + c2 ln r,
n=0
n = 1, 2, . . .
Thus
u(r, θ) = A0 + B0 ln r +
∞
An rn + Bn r−n cos nθ + Cn rn + Dn r−n sin nθ .
n=1
When r = 1,
ˆ 2π
1
75 sin θ dθ = 0 ← ln 1 = 0
A0 + B0 ln 1 =
2π 0
ˆ
1 2π
75 sin θ cos nθ dθ = 0, n = 1, 2, . . .
An + Bn =
π 0
ˆ
0,
n>0
1 2π
75 sin θ sin nθ dθ =
Cn + Dn =
, n = 1, 2, . . . ,
π 0
75, n = 1
797
798
CHAPTER 13
BOUNDARY-VALUE PROBLEMS IN OTHER COORDINATE SYSTEMS
so
A0 = 0,
A1 + B1 = 0,
C1 + D1 = 75,
and
An + Bn = 0,
Cn + Dn = 0,
When r = 2
A0 + B0 ln 2 =
An 2n + Bn 2−n
−n
n
Cn 2 + Dn 2
1
2π
1
=
π
1
=
π
ˆ
n > 1.
2π
60 cos θ dθ = 0
0
ˆ
2π
60 cos θ cos nθ dθ =
0
ˆ
0,
n>1
60,
n=1
∞
60 cos θ sin nθ dθ = 0,
n = 1, 2, . . . ,
0
so
B0 = 0,
for
2A1 +
1
B1 = 60,
2
2C1 +
1
D1 = 0,
2
and
An 2n + Bn 2−n = 0,
Cn 2n + Dn 2−n = 0,
for
n > 1.
Whe have A0 = 0 and B0 = 0, and solving the nonhomogeneous systems for n = 1,
A1 + B1 = 0
C1 + D1 = 75
1
1
B1 = 60
2C1 + D1 = 0
2
2
yields A1 = 40, B1 = −40, C1 = −25, and D1 = 100. Finally, solving the homogeneous
systems
2A1 +
An + Bn = 0
Cn + Dn = 0
An 2n + Bn 2−n = 0
Cn 2n + Dn 2−n = 0
gives An = Bn = Cn = Dn = 0 for n > 1. The solution is then
u(r, θ) = A1 r + B1 r−1 cos θ + C1 r + D1 r−1 sin θ
= 4 − r − 40r−1 cos θ + −25r + 100r−1 sin θ
4
1
cos θ − 25 r −
sin θ.
= 40 r −
r
r
14. We solve
1 ∂2u
∂ 2 u 1 ∂u
+
+
= 0,
∂r2
r ∂r
r2 ∂θ2
u(a, θ) = θ(π − θ),
u(r, 0) = 0,
0 < θ < π,
u(b, θ) = 0,
u(r, π) = 0,
a < r < b,
0 < θ < π,
a < r < b.
13.1
Polar Coordinates
Proceeding as in Example 1 in the text we obtain the separated differential equations
r2 R + rR − λR = 0
Θ + λΘ = 0.
Taking λ = α2 the solutions are
Θ(θ) = c1 cos αθ + c2 sin αθ
R(r) = c3 rα + c4 r−α .
Applying the boundary conditions Θ(0) = 0 and Θ(π) = 0 we find that c1 = 0 and α = n for
n = 1, 2, 3, . . . . The boundary condition R(b) = 0 gives
c3 bn + c4 b−n = 0
c4 = −c3 b2n .
and
2n
r − b2n
b2n
n
R(r) = c3 r − n = c3
r
rn
Then
and
u(r, θ) =
∞
An
n=1
From
u(a, θ) = θ(π − θ) =
r2n − b2n
rn
∞
n=1
we find
An
Thus
a2n − b2n
an
2
=
π
ˆ
π
An
sin nθ.
a2n − b2n
an
(θπ − θ2 ) sin nθ dθ =
0
sin nθ
4
[1 − (−1)n ].
n3 π
∞
4 1 − (−1)n r2n − b2n a n
sin nθ.
u(r, θ) =
π
n3
a2n − b2n r
n=1
15. The homogeneous boundary conditions Θ(θ) = 0 and Θ(π) = 0 imply that λ = 0 is not an
eigenvalue, but, imply for Θ(θ) = c1 cos λθ + c2 sin λθ, that c1 = 0 and λn = n2 , n = 1, 2, . . ..
Then Θ(θ) = c2 sin nθ, n = 1, 2, . . .. Applying R(1) = 0 to R(r) = c3 rn +c4 r−n gives c4 = −c3
∞
n
−n
so R(r) = c3 (r − r ). Thus u(r, θ) =
An (rn − r−n ) sin nθ and the boundary condition
n=1
u(2, θ) = u0 =
∞
An (2n − 2−n ) sin nθ
n=1
implies
−n
An 2 − 2
n
2u0
=
π
ˆ
π
sin nθ dθ =
0
2u0 1 − (−1)n
π
n
or An =
2u0 1 − (−1)n
.
π n (2n − 2−n )
799
800
CHAPTER 13
BOUNDARY-VALUE PROBLEMS IN OTHER COORDINATE SYSTEMS
Hence
u(r, θ) =
∞
2u0 1 − (−1)n rn − r−n
sin nθ .
π
n
2n − 2−n
n=1
16. Separating variables in the partial differential equation urr + (1/r)ur + (1/r2 )uθθ = 0 and
using λ = −α2 as the separation constant leads to
Θ − α2 Θ = 0
r2 R + rR + α2 R = 0,
and
Solving these two equations we obtain
Θ(θ) = c1 cosh αθ + c2 sinh αθ
and
R(r) = c3 cos (α ln r) + c4 sin (α ln r).
From the boundary condition u(1, θ) = R(1)Θ(θ) = 0 we see that
R(1) = c3 cos (α · 0) + c4 sin (α · 0) = c3 = 0. Similarly, the boundary condition
u(2, θ) = R(2)Θ(θ) = 0 means that R(2) = c4 sin (α ln 2) = 0. Since sin (α ln 2) = 0 when
α ln 2 = nπ, we have the eigenvalues λn = (nπ/ ln 2)2 for n = 1, 2, . . . . The corresponding
eigenfunctions are
nπ
ln r .
R(r) = c4 sin
ln 2
From the boundary condition u(r, 0) = 0 we have Θ(0) = c1 = 0 and so
Θ(θ) = c2 sinh (nπθ/ ln 2). Therefore
u(r, θ) =
∞
An sinh
n=1
nπ nπ
θ sin
ln r .
ln 2
ln 2
When θ = π, u(r, π) = r so
r=
∞
n=1
An sinh
nπ
π sin
ln r ,
ln 2
ln 2
nπ
which is an orthogonal function expansion. Using the idea of Problem 7 in Section 11.4 and
the information in Section 12.7,
ˆ 2
nπ
1
ln r dr
r · sin
nπ r
ln 2
An sinh
π = ˆ1 2
.
ln 2
1
2 nπ
sin
ln r dr
ln 2
1 r
Now, with the substitutions t = ln r (so dt = dr/r), r = et , and αn = nπ/ ln 2, a CAS gives,
after simplifying,
2πn [1 − 2(−1)n ]
.
An sinh (αn π) =
(nπ)2 + (ln 2)2
Therefore, the solution is
u(r, θ) = 2π
∞
n [1 − 2(−1)n ] sinh (αn θ)
sin (αn ln r).
(nπ)2 + (ln 2)2 sinh (αn π)
n=1
13.1
Polar Coordinates
17. The boundary-value problem is
1 ∂2u
∂ 2 u 1 ∂u
+
+
= 0,
∂r2
r ∂r
r2 ∂θ2
∂u ∂u = 0,
∂θ ∂θ θ=0
1 < r < 2,
= 0,
0 < θ < π/2
1<r<2
θ=π/2
u(1, θ) = 0,
u(2, θ) = f (θ),
0 < θ < π/2
Using u = R(r)Θ(θ) and Θ (0) = 0, Θ (π/2) = 0, R(1) = 0 we get Θ(θ) = c1 and R(r) = c4 ln r
for λ = α2 = 0 and for λ = −α2 we have Θ(θ) = c1 2nθ and R(r) = c3 r2n − r−2n . Hence
u(r, θ) = A0 ln r +
∞
An r2n − r−2n cos 2nθ.
n=1
For r = 2 we have
f (θ) = A0 ln 2 +
∞
An 22n − 2−2n cos 2nθ
n=1
where
a0
A0 ln 2 =
2
2
gives A0 =
π ln 2
ˆ
π/2
f (θ) dθ
0
and
2n
An 2
−2n
−2
= an
4
An =
2n
π (2 − 2−2n )
gives
ˆ
π/2
f (θ) cos 2nθ dθ.
0
18. Separating variables we get Θ(θ) = c1 cos αθ + c2 sin αθ, so Θ(0) = 0 and c1 = 0. Now
Θ(θ) = c2 sin αθ and Θ(π/4) = 0 implies sin(απ/4) = 0 or α = 4n. Then λ = (4n)2 and
Θ(θ) = c2 sin 4nθ. Now R(r) = c3 r4n + c4 r−4n , so R(a) = 0 implies c3 a4n + c4 a−4n = 0 and
c4 = −a4n /a−4n . Thus
(r/a)4n − (a/r)4n
a4n
R(r) = c3
u(r, θ) =
∞
An
n=1
u(b, θ) = 100 =
∞
(r/a)4n − (a/r)4n
sin 4nθ
a4n
An
n=1
8
(b/a)4n − (a/b)4n
An =
4n
a
π
and
∞
u(r, θ) =
ˆ
(b/a)4n − (a/b)4n
sin 4nθ
a4n
π/4
100 sin 4nθ dθ =
0
800 1 − (−1)n
π
4n
200 (r/a)4n − (a/r)4n 1 − (−1)n
sin 4nθ.
π
(b/a)4n − (a/b)4n
n
n=1
801
802
CHAPTER 13
BOUNDARY-VALUE PROBLEMS IN OTHER COORDINATE SYSTEMS
19. Let u1 be the solution of the boundary-value problem
1 ∂ 2 u1
∂ 2 2 1 ∂u1
+ 2
∂r +
= 0,
u1
r ∂r
r ∂θ2
u1 (a, θ) = f (θ),
u1 (b, θ) = 0,
0 < θ < 2π,
a<r<b
0 < θ < 2π
0 < θ < 2π
and let u2 be the solution to the boundary-value problem
1 ∂ 2 u2
∂ 2 u2 1 ∂u2
+
+
= 0,
∂r2
r ∂r
r2 ∂θ2
u2 (a, θ) = 0,
u2 (b, θ) = g(θ),
0 < θ < 2π,
a<r<b
0 < θ < 2π
0 < θ < 2π
Each of these problems can be solved using the methods shown in Problem 9 of this section.
Now if u(r, θ) = u1 (r, θ) + u2 (r, θ), then
u(a, θ) = u1 (a, θ) + u2 (a, θ) = f (θ)
u(b, θ) = u1 (b, θ) + u2 (b, θ) = g(θ)
and u(r, θ) will be the steady-state temperature of the circular ring with boundary conditions
u(a, θ) = f (θ) and u(b, θ) = g(θ).
20. Referring to Problem 19 above we solve the boundary-value problems for u1 (r, θ) and u2 (r, θ).
Using the answer to Problem 11 we find A0 = −100/ ln 2, A1 = 100/3, An = 0, n > 1, and
Bn = 0 for all n. Then
u1 (r, θ) = −100
ln r 100 −1
+
r − r cos θ .
ln 2
3
Using the answer to Problem 10 we find
ln r
ln 2r
= 200 1 +
u2 (r, θ) = 200
ln 2
ln 2
so
u(r, θ) = u1 (r, θ) + u2 (r, θ) = 200 + 100
ln r 100 −1
+
r − r cos θ .
ln 2
3
21. Using the same reasoning as in Example 1 in the text we obtain
u(r, θ) = A0 +
∞
rn (An cos nθ + Bn sin nθ).
n=1
The boundary condition at r = c implies
f (θ) =
∞
n=1
ncn−1 (An cos nθ + Bn sin nθ).
13.1
Polar Coordinates
Since this condition does not determine A0 , it is an arbitrary constant. However, to be a full
Fourier series on [0, 2π] we must require that f (θ) satisfy the condition A0 = a0 /2 = 0 or
´ 2π
0 f (θ) dθ = 0. If this integral were not 0, then the series for f (θ) would contain a nonzero
constant, which it obviously does not. With this as a necessary compatibility condition we
can then make the identifications
ncn−1 An = an
or
An =
1
ncn−1 π
ˆ
and ncn−1 Bn = bn
2π
f (θ) cos nθ dθ
and
1
Bn =
ncn−1 π
0
ˆ
2π
f (θ) sin nθ dθ.
0
22. Rather than emplyoing the method of separation of variables as in Example 1, we simply show
that the function satifies the partial differential equation and the given boundary condition.
First we note that the boundary condition is satisfied:
trig identity for sin 3θ
3
1
3
1
u(1, θ) = sin θ − sin 3θ = sin θ −
3 sin θ − 4 sin3 θ
4
4
4
4
=
3
3
sin θ − sin θ + sin3 θ = sin3 θ
4
4
Next
3
3
∂u
= sin θ − r2 sin 3θ
∂r
4
4
∂2u
3
= − r sin 3θ
2
∂r
2
∂2u
9
3
= − r sin θ + r3 sin 3θ
∂θ2
4
4
Therefore
1 ∂2u
3
3
3
3
9
∂ 2 u 1 ∂u
+
sin θ − r sin 3θ −
sin θ + r sin 3θ = 0
+
= − r sin 3θ +
2
2
2
∂r
r ∂r
r ∂θ
2
4r
4
4r
4
23. (a) From Problem 1 in this section, with u0 = 100,
∞
u(r, θ) = 50 +
100 1 − (−1)n n
r sin nθ.
π
n
n=1
803
804
CHAPTER 13
(b) x
BOUNDARY-VALUE PROBLEMS IN OTHER COORDINATE SYSTEMS
u
100
r = 0.9
80
60
r = 0.1
40
r = 0.3
r = 0.5
20
r = 0.7
1
2
3
4
5
6
θ
(c) We could use S5 from part (b) of this problem to compute the approximations, but in a
CAS it is just as easy to compute the sum with a much larger number of terms, thereby
getting greater accuracy. In this case we use partial sums including the term with r99
to find
u(0.9, 1.3) ≈ 96.5268
u(0.9, 2π − 1.3) ≈ 3.4731
u(0.7, 2) ≈ 87.871
u(0.7, 2π − 2) ≈ 12.129
u(0.5, 3.5) ≈ 36.0744
u(0.5, 2π − 3.5) ≈ 63.9256
u(0.3, 4) ≈ 35.2674
u(0.3, 2π − 4) ≈ 64.7326
u(0.1, 5.5) ≈ 45.4934
u(0.1, 2π − 5.5) ≈ 54.5066
(d) At the center of the plate u(0, 0) = 50. From the graphs in part (b) we observe that
the solution curves are symmetric about the point (π, 50). In part (c) we observe that
the horizontal pairs add up to 100, and hence average 50. This is consistent with the
observation about part (b), so it is appropriate to say the average temperature in the
plate is 50◦ .
13.2
13.2
Polar and Cylindrical Coordinates
Polar and Cylindrical Coordinates
1. Referring to the solution of Example 1 in the text we have
R(r) = c1 J0 (αn r)
and
T (t) = c3 cos aαn t + c4 sin aαn t
where the αn are the positive roots of J0 (αc) = 0. Now, the initial condition
u(r, 0) = R(r)T (0) = 0 implies T (0) = 0 and so c3 = 0. Thus
u(r, t) =
∞
∞
An sin aαn tJ0 (αn r)
∂u =
aαn An cos aαn tJ0 (αn r).
∂t
and
n=1
n=1
∂u ∂t From
we find
2
aαn An = 2 2
c J1 (αn c)
2
= 2 2
c J1 (αn c)
ˆ
∞
=1=
c
rJ0 (αn r) dr
ˆ
ˆ
aαn An J0 (αn r)
n=1
t=0
x = αn r, dx = αn dr
0
αn c
0
1
xJ0 (x) dx
αn2
αn c
1 d
[xJ1 (x)] dx
see (4) of Section 11.5 in text
αn2 dx
0
αn c
2
2
xJ1 (x)
.
= 2 2 2
=
cαn J1 (αn c)
c αn J1 (αn c)
=
2
2
2
c J1 (αn c)
0
Then
2
An =
and
acαn2 J1 (αn c)
∞
u(r, t) =
2 J0 (αn r)
sin aαn t.
ac
αn2 J1 (αn c)
n=1
2. From Example 1 in the text we have Bn = 0 because ∂u/∂t = 0 and
2
An = 2
J1 (αn )
ˆ
1
r(1 − r2 )J0 (αn r) dr.
0
From Problem 10, Exercises 11.6 we obtained An =
u(r, t) = 4
4J2 (αn )
. Thus
αn2 J12 (αn )
∞
J2 (αn )
2 (α ) cos aαn tJ0 (αn r).
J
n
1
n=1
805
806
CHAPTER 13
BOUNDARY-VALUE PROBLEMS IN OTHER COORDINATE SYSTEMS
3. Referring to Example 2 in the text we have
R(r) = c1 J0 (αr) + c2 Y0 (αr)
Z(z) = c3 cosh αz + c4 sinh αz
where c2 = 0 and J0 (2α) = 0 defines the positive eigenvalues λn = αn2 . From Z(4) = 0 we
obtain
cosh 4αn
or
c4 = −c3
.
c3 cosh 4αn + c4 sinh 4αn = 0
sinh 4αn
Then
cosh 4αn
sinh 4αn cosh αn z − cosh 4αn sinh αn z
sinh αn z = c3
Z(z) = c3 cosh αn z −
sinh 4αn
sinh 4αn
= c3
sinh αn (4 − z)
sinh 4αn
and
u(r, z) =
∞
An
n=1
sinh αn (4 − z)
J0 (αn r).
sinh 4αn
From
u(r, 0) = u0 =
∞
An J0 (αn r)
n=1
we obtain
2u0
An =
2
4J1 (2αn )
ˆ
2
rJ0 (αn r) dr =
0
u0
.
αn J1 (2αn )
Thus the temperature in the cylinder is
u(r, z) = u0
∞
sinh αn (4 − z)J0 (αn r)
n=1
αn sinh (4αn )J1 (2αn )
.
4. (a) The boundary condition ur (2, z) = 0 implies R (2) = 0 or J0 (2α) = 0. Thus α = 0 is
also an eigenvalue and the separated equations are in this case rR + R = 0 and z = 0.
The solutions of these equations are then
R(r) = c1 + c2 ln r,
Z(z) = c3 z + c4 .
Now Z(0) = 0 yields c4 = 0 and the implicit condition that the temperature is bounded
as r → 0 demands that we define c2 = 0. Thus we have
u(r, z) = A1 z +
∞
An sinh αn zJ0 (αn r).
n=2
At z = 4 we obtain
f (r) = 4A1 +
∞
n=2
An sinh 4αn J0 (αn r).
(1)
13.2
Polar and Cylindrical Coordinates
807
Thus from (17) and (18) of Section 11.5 in the text we can write with b = 2,
1
A1 =
8
An =
ˆ
2
rf (r) dr
(2)
0
1
2 sinh 4αn J02 (2αn )
ˆ
2
rf (r)J0 (αn r) dr
(3)
0
A solution of the problem consists of the series (1) with coefficients A1 and An defined
in (2) and (3), respectively.
(b) When f (r) = u0 we get A1 = u0 /4 and
An =
u0 J1 (2αn )
=0
αn sinh 4αn J02 (2αn )
since J0 (2α) = 0 is equivalent to J1 (2α) = 0. A solution of the problem is then
u0
z.
u(r, z) =
4
5. Letting the separation constant be λ = α2 and referring to Example 2 in Section 13.2 in the
text we have
R(r) = c1 J0 (αr) + c2 Y0 (αr)
Z(z) = c3 cosh αz + c4 sinh αz
where c2 = 0 and the positive eigenvalues λn are determined by J0 (2α) = 0. From Z (0) = 0
we obtain c4 = 0. Then
∞
An cosh αn zJ0 (αn r).
u(r, z) =
n=1
From
u(r, 4) = 50 =
∞
An cosh 4αn J0 (αn r)
n=1
we obtain (as in Example 1 of Section 13.1)
2(50)
An cosh 4αn =
4J12 (2αn )
ˆ
2
rJ0 (αn r) dr =
0
50
.
αn J1 (2αn )
Thus the temperature in the cylinder is
u(r, z) = 50
∞
cosh (αn z)J0 (αn r)
.
αn cosh (4αn )J1 (2αn )
n=1
808
CHAPTER 13
BOUNDARY-VALUE PROBLEMS IN OTHER COORDINATE SYSTEMS
6. The boundary-value problem in this case is
∂ 2 u 1 ∂u ∂ 2 u
+ 2 = 0,
+
∂r2
r ∂r
∂z
u(2, z) = 50,
∂u = 0,
∂z z=0
0 < r < 2,
0<z<4
∂u = 0,
∂z 0<z<4
0 < r < 2.
z=4
We have u(r, z) = v(r, z) + 50, so
∂ 2 v 1 ∂v ∂ 2 v
+
+
= 0,
∂r2
r ∂r ∂z 2
0 < r < 2,
v(2, z) = 0, 0 < z < 4
∂v ∂v = 0,
= 0,
∂z ∂z z=0
0<z<4
0 < r < 2,
z=4
which implies that v(r, z) = 0 and so u(r, z) = 50.
7. Using λ as the separation constant the separated equations are
rR + R − rλR = 0 and Z + λZ = 0.
The boundary conditions are Z (0) = 0 and Z (1) = 0.
If λ = 0 the solutions of the ordinary differential equations are
R = c1 + c2 ln r
and Z = c3 + c4 z.
Since Z (0) = 0, c4 = 0. Therefore Z = c3 , which satisfies Z (1) = 0. Boundedness at r = 0
implies c2 = 0. Therefore λ = 0 is an eigenvalue with eigenfunction R = c1 = 0.
If λ = −α2 < 0, the solutions of the ordinary differential equations are
R = c1 J0 (αr) + c2 Y0 (αr)
and Z = c3 cosh αz + c4 sinh αz.
Since Z (0) = 0, c4 = 0. Therefore Z = c3 cosh αz. Then Z (1) = 0, so c4 α sinh α = 0 and
c4 = 0. Thus Z = 0, and therefore u = 0.
If λ = α2 > 0, the solutions of the ordinary differential equations are
R = c1 I0 (αr) + c2 K0 (αr)
and Z = c3 cos αz + c4 sin αz.
Since Z (0) = 0, c4 = 0 and so Z = c3 cos αz. Now Z (1) = 0 so c4 α sin α = 0 and α =
nπ, n = 1, 2, 3, . . . . The eigenvalues are λn = n2 π 2 and corresponding eigenfunctions are
Z = c3 cos nπz. Now, the usual requirement that u be bounded at r = 0 implies c2 = 0.
13.2
Polar and Cylindrical Coordinates
(See Figure 5.3.4 in the text.) Therefore R = c1 I0 (αr) or R = c1 I0 (nπr). The superposition
principle then yields
∞
An I0 (nπr) cos nπz.
u(r, z) = A0 +
n=1
At r = 1
u(1, z) = z = A0 +
∞
An I0 (nπ) cos nπz
n=1
so
ˆ
1
1 2 1
1
z dz =
A0 = a0 = ·
2
2 1 0
2
ˆ
2 1
(−1)n − 1
an I0 (nπ) =
z cos nπz dz = 2
1 0
n2 π 2
and
An = 2
(−1)n − 1
.
n2 π 2 I0 (nπ)
where we note that I0 (nπ) has no real zeros. Therefore
∞
u(r, z) =
(−1)n − 1
1
+2
I0 (nπr) cos nπz.
2
n2 π 2 I0 (nπ)
n=1
8. Using λ as the separation constant the separated equations are
rR + R − rλR = 0 and Z + λZ = 0.
The boundary conditions are Z (0) = 0 and Z(1) = 0.
If λ = 0 the solutions of the ordinary differential equations are
R = c1 + c2 ln r
and Z = c3 + c4 z.
Since Z (0) = 0, c4 = 0. Therefore Z = c3 and Z(1) = 0 so Z(1) = c3 = 0. The product
solution is u = R(r)Z(z) = 0. Thus λ = 0 is not an eigenvalue.
If λ = −α2 < 0, the solutions of the ordinary differential equations are
R = c1 J0 (αr) + c2 Y0 (αr)
and Z = c3 cosh αz + c4 sinh αz.
Since Z (0) = 0, c4 = 0. Therefore Z = c3 cosh αz and Z(1) = 0 so c4 cosh α = 0, which
implies c4 = 0. Thus Z = 0 and therefore u = 0.
If λ = α2 > 0, the solutions of the ordinary differential equations are
R = c1 I0 (αr) + c2 K0 (αr)
and Z = c3 cos αz + c4 sin αz.
809
810
CHAPTER 13
BOUNDARY-VALUE PROBLEMS IN OTHER COORDINATE SYSTEMS
Since Z (0) = 0, c4 = 0 and so Z = c3 cos αz. Now Z(1) = 0 so c4 cos α = 0 and α =
(2n − 1)π/2, n = 1, 2, 3, . . . . The eigenvalues are λn = (2n − 1)2 π 2 /4 and corresponding
eigenfunctions are Z = c3 cos (2n − 1)πz/2. Now, the usual implicit requirement that u be
bounded at r = 0 implies c2 = 0. (See Figure 5.3.4 in the text.) Therefore R = c1 I0 (αr) or
R = c1 I0 ((2n − 1)πr/2). The superposition principle then yields
∞
2n − 1
2n − 1
u(r, z) =
An I0
πr cos
πz.
2
2
n=1
At r = 1
u(1, z) = z =
∞
An I0
n=1
2n − 1
2n − 1
π cos
πz,
2
2
which is not a Fourier series. Thus
ˆ 1
2n − 1
πz dz
z cos
2n − 1
−8 − 4(2n − 1)π(−1)n
2
π = ˆ0 1
.
An I0
=
2
(2n − 1)2 π 2
2 2n − 1
πz dz
cos
2
0
Therefore
u(r, z) = 4
∞
2n − 1
2n − 1
(2n − 1)π(−1)n+1 − 2
I
πr
cos
πz.
0
2n−1
2π2I
2
2
(2n
−
1)
π
0
2
n=1
9. Letting u(r, t) = R(r)T (t) and separating variables we obtain
R + 1r R
T
=
= −λ
R
kT
and
1
R + R + λR = 0,
r
T + λkT = 0.
From the last equation we find T (t) = e−λkt . If λ < 0, T (t) increases without bound as
t → ∞. Thus we assume λ = α2 > 0. Now
1
R + R + α2 R = 0
r
is a parametric Bessel equation with solution
R(r) = c1 J0 (αr) + c2 Y0 (αr).
Since Y0 is unbounded as r → 0 we take c2 = 0. Then R(r) = c1 J0 (αr) and the boundary
condition u(c, t) = R(c)T (t) = 0 implies J0 (αc) = 0. This latter equation defines the positive
eigenvalues λn = αn2 . Thus
u(r, t) =
∞
An J0 (αn r)e−αn kt .
2
n=1
From
u(r, 0) = f (r) =
∞
An J0 (αn r)
n=1
we find
2
An = 2 2
c J1 (αn c)
ˆ
c
rJ0 (αn r)f (r) dr, n = 1, 2, 3, . . . .
0
13.2
Polar and Cylindrical Coordinates
10. If the edge r = c is insulated we have the boundary condition ur (c, t) = 0. Referring to the
solution of Problem 9 above we have
R (c) = αc1 J0 (αc) = 0
which defines an eigenvalue λ = α2 = 0 and positive eigenvalues λn = αn2 . Thus
u(r, t) = A0 +
∞
An J0 (αn r)e−αn kt .
2
n=1
From
u(r, 0) = f (r) = A0 +
∞
An J0 (αn r)
n=1
we find
2
A0 = 2
c
An =
ˆ
c
rf (r) dr
0
2
2
2
c J0 (αn c)
ˆ
c
rJ0 (αn r)f (r) dr.
0
11. Referring to Problem 9 we have T (t) = e−λkt and R(r) = c1 J0 (αr). The boundary condition
hu(1, t)+ur (1, t) = 0 implies hJ0 (α)+αJ0 (α) = 0 which defines positive eigenvalues λn = αn2 .
Now
∞
2
An J0 (αn r)e−αn kt
u(r, t) =
n=1
where
An =
2αn2
(αn2 + h2 )J02 (αn )
ˆ
1
rJ0 (αn r)f (r) dr.
0
12. We solve
∂ 2 u 1 ∂u ∂ 2 u
+ 2 = 0, 0 < r < 1,
+
∂r2
r ∂r
∂z
∂u = −hu(1, z), z > 0
∂r z>0
r=1
u(r, 0) = u0 ,
assuming u = RZ we get
0 < r < 1.
R + 1r R
Z =−
= −λ
R
Z
and so
rR + R + λ2 rR = 0
and
Z − λZ = 0.
Letting λ = α2 we then have
R(r) = c1 J0 (αr) + c2 Y0 (αr)
and
Z(z) = c3 e−αz + c4 eαz .
811
812
CHAPTER 13
BOUNDARY-VALUE PROBLEMS IN OTHER COORDINATE SYSTEMS
We use the exponential form of the solution of Z − α2 Z = 0 since the domain of the variable
z is a semi-infinite interval. As usual we define c2 = 0 since the temperature is surely bounded
as r → 0. Hence R(r) = c1 J0 (αr). Now the boundary-condition ur (1, z) + hu(1, z) = 0 is
equivalent to
(4)
αJ0 (α) + hJ0 (α) = 0.
The eigenvalues αn are the positive roots of (4) above. Finally, we must now define c4 = 0
since the temperature is also expected to be bounded as z → ∞. A product solution of the
partial differential equation that satisfies the first boundary condition is given by
un (r, z) = An e−αn z J0 (αn r).
Therefore
u(r, z) =
∞
An e−αn z J0 (αn r)
n=1
is another formal solution. At z = 0 we have u0 = An J0 (αn r). In view of (4) above we use
equations (17) and (18) of Section 11.5 in the text with the identification b = 1:
ˆ 1
2αn2
rJ0 (αn r)u0 dr
An = 2
(αn + h2 ) J02 (αn ) 0
αn
2αn2 u0
2αn u0 J1 (αn )
.
(5)
= 2
tJ
(t)
= 2
1
2
2
2
(αn + h ) J0 (αn )αn
(αn + h2 ) J02 (αn )
0
Since J0 = −J1 [see equation (6) of Section 11.5 in the text] it follows from (4) above that
αn J1 (αn ) = hJ0 (αn ). Thus (5) from this section of the manual simplifies to
An =
(αn2
2u0 h
.
+ h2 ) J0 (αn )
A solution to the boundary-value problem is then
u(r, z) = 2u0 h
∞
n=1
e−αn z
J0 (αn r).
(αn2 + h2 ) J0 (αn )
13. Substituting u(r, t) = v(r, t) + ψ(r) into the partial differential equation gives
1
∂v
∂ 2 v 1 ∂v
+ ψ + ψ =
.
+
2
∂r
r ∂r
r
∂t
This equation will be homogeneous provided ψ + 1r ψ = 0 or ψ(r) = c1 ln r + c2 . Since ln r is
unbounded as r → 0 we take c1 = 0. Then ψ(r) = c2 and using u(2, t) = v(2, t) + ψ(2) = 100
we set c2 = ψ(2) = 100. Therefore ψ(r) = 100. Referring to Problem 5 above, the solution of
the boundary-value problem
∂v
∂ 2 v 1 ∂v
=
,
+
∂r2
r ∂r
∂t
v(2, t) = 0,
0 < r < 2, t > 0,
t > 0,
v(r, 0) = u(r, 0) − ψ(r)
13.2
is
v(r, t) =
∞
Polar and Cylindrical Coordinates
An J0 (αn r)e−αn t
2
n=1
where
An =
=
=
=
=
=
ˆ 2
2
rJ0 (αn r)[u(r, 0) − ψ(r)] dr
22 J12 (2αn ) 0
ˆ 1
ˆ 2
1
rJ0 (αn r)[200 − 100] dr +
rJ0 (αn r)[100 − 100] dr
2J12 (2αn ) 0
1
ˆ 1
50
rJ0 (αn r) dr
x = αn r, dx = αn dr
J12 (2αn ) 0
ˆ αn
50
1
xJ0 (x) dx
2
J1 (2αn ) 0 αn2
ˆ αn
d
50
[xJ1 (x)] dx
see (5) of Section 11.5 in text
2
2
αn J1 (2αn ) 0 dx
αn
50
50J1 (αn )
(xJ
.
(x))
=
1
2
2
αn J1 (2αn )
αn J12 (2αn )
0
Thus
u(r, t) = v(r, t) + ψ(r) = 100 + 50
∞
J1 (αn )J0 (αn r)
n=1
αn J12 (2αn )
e−αn t .
2
rψ + ψ = −βr. The general solution of this
14. Letting u(r, t) = v(r, t) + ψ(r) we obtain
nonhomogeneous Cauchy-Euler equation is found with the aid of variation of parameters:
ψ = c1 + c2 ln r − βr2 /4. In order that this solution be bounded as r → 0 we define c2 = 0.
Using ψ(1) = 0 then gives c1 = β/4 and so ψ(r) = β(1 − r2 )/4. Using v = RT we find that a
solution of
∂v
∂ 2 v 1 ∂v
=
,
+
∂r2
r ∂r
∂t
0 < r < 1, t > 0
v(1, t) = 0,
v(r, 0) = −ψ(r),
is
v(r, t) =
∞
t>0
0<r<1
An e−αn t J0 (αn r)
2
n=1
where
β
2
An = −
2
4 J1 (αn )
ˆ
1
r(1 − r2 )J0 (αn r) dr
0
and the αn are defined by J0 (α) = 0. From the result of Problem 10, Exercises 11.5 (see also
Problem 2 of this exercise set) we get
An = −
βJ2 (αn )
.
αn2 J12 (αn )
813
814
CHAPTER 13
BOUNDARY-VALUE PROBLEMS IN OTHER COORDINATE SYSTEMS
Thus from u = v + ψ(r) it follows that
∞
J2 (αn )
β
2
(1 − r2 ) − β
e−αn t J0 (αn r).
2 J 2 (α )
4
α
n
n=1 n 1
u(r, t) =
15. (a) Writing the partial differential equation in the form
2
∂ u ∂u
∂2u
g x 2+
= 2
∂x
∂x
∂t
and separating variables we obtain
xX + X T =
= −λ.
X
gT
Letting λ = α2 we obtain
xX + X + α2 X = 0
and
T + gα2 T = 0.
Letting x = τ 2 /4 in the first equation we obtain dx/dτ = τ /2 or dτ /dx = 2τ . Then
dX dτ
2 dX
dX
=
=
dx
dτ dx
τ dτ
and
2 dX
2 d dX
dX d 2
=
+
τ dτ
τ dx dτ
dτ dx τ
dX d 2 dτ
4 d2 X
4 dX
2 d dX dτ
+
= 2
.
− 3
=
2
τ dτ dτ dx
dτ dτ τ dx
τ dτ
τ dτ
d2 X
d
=
2
dx
dx
Thus
xX + X + α2 X =
τ2
4
4 d2 X
4 dX
− 3
τ 2 dτ 2
τ dτ
+
2 dX
d2 X 1 dX
+ α2 X =
+ α2 X = 0.
+
τ dτ
dτ 2
τ dτ
This is a parametric Bessel equation with solution
X(τ ) = c1 J0 (ατ ) + c2 Y0 (ατ ).
(b) To insure a finite solution at x = 0 (and thus τ = 0) we set c2 = 0. The condition
√
u(L, t) = X(L)T (t) = 0 implies X = X √ = c1 J0 (2α L ) = 0, which defines
x=L
τ =2 L
positive eigenvalues λn = αn2 . The solution of T + gα2 T = 0 is
√
√
T (t) = c3 cos (αn g t) + c4 sin (αn g t).
The boundary condition ut (x, 0) = X(x)T (0) = 0 implies c4 = 0. Thus
u(τ, t) =
∞
n=1
√
An cos (αn g t)J0 (αn τ ).
13.2
From
u(τ, 0) = f (τ 2 /4) =
∞
Polar and Cylindrical Coordinates
An J0 (αn τ )
n=1
we find
1
√
=
2
2LJ1 (2αn L )
2
√
=
2
LJ1 (2αn L )
√
ˆ
√
2 L
ˆ
2
√
An = √
(2 L )2 J12 (2αn L )
τ J0 (αn τ )f (τ 2 /4) dτ
v = τ /2, dv = dτ /2
0
L
2vJ0 (2αn v)f (v 2 )2 dv
0
√
ˆ
L
vJ0 (2αn v)f (v 2 ) dv.
0
The solution of the boundary-value problem is
u(x, t) =
∞
√
√
An cos (αn g t)J0 (2αn x ).
n=1
16. The boundary-value problem is
1
urr + ur = ut ,
r
0 < r < 1,
ur (1, t) = 1,
u(r, 0) = 0,
t>0
t>0
0 < r < 1.
(a) When u(r, t) = v(r, t) + Bt in the above boundary-value problem we have urr = vrr ,
ur = vr , ut = vt + B, ur (1, t) = vr (1, t), and u(r, 0) = v(r, 0) + 0. Thus, in terms of v
the boundary-value problem becomes
1
vrr + vr = vt + B,
r
vr (1, t) = 1,
v(r, 0) = 0,
t>0
0 < r < 1,
t>0
←− boundary value (r = 1)
0 < r < 1.
←− initial value (t = 0)
(b) The boundary-value problem in part(a) is still not homogeneous so we let
v(r, t) = w(r, t) + ψ(r) and try to determine a homogeneous boundary-value problem in
w. This leads to ψ (r)+(1/r)ψ (r) = B which can be written as r2 ψ (r)+rψ (r) = Br2 .
This is a nonhomogeneous Cauchy-Euler differential equation and can be solved using
variation of parameters. It’s solution is
ψ(r) =
Br2
+ c1 ln r + c2 .
4
Since we implicitly assume that lim ψ(r) is bounded, we conclude that c1 = 0. The
r→0
boundary condition at r = 1 is then
1 = vr (1, t) = wr (1, t) + ψ (1) = wr (1, t) +
B
,
2
815
816
CHAPTER 13
BOUNDARY-VALUE PROBLEMS IN OTHER COORDINATE SYSTEMS
so setting B = 2 forces wr (1, t) = 0. The initial condition becomes w(r, 0) + ψ(r) = 0.
In terms of w the boundary-value problem is
wrr +
1
wr = wt ,
r
wr (1, t) = 0,
w(r, 0) = −ψ(r) = −
0 < r < 1,
t>0
←− boundary value (r = 1)
t>0
r2
− c2 ,
2
0 < r < 1.
←− initial value (t = 0)
From Problem 10 (which refers to Problem 9) with k = 1, c = 1, and f (r) = −(r2 /2)−c2 ,
the solution for this boundary-value problem is
w(r, t) = A0 +
∞
An J0 (αn r)e−αn t
2
n=1
where
1
−r
A0 = 2
dr = − c2 +
,
4
0
2
ˆ 1
2
r
+ c2 dr
−rJ0 (αn r)
An = 2
2
J1 (αn ) 0
ˆ 1
ˆ 1
2c2
1
r3 J0 (αn r) dr − 2
rJ0 (αn r) dr.
=− 2
J1 (αn ) 0
J1 (αn ) 0
ˆ
1
r2
+ c2
2
Using the substitution t = αn r and the recursive properties of Jn , we find that
ˆ
1
r3 J0 (αn r) dr =
0
Since
ˆ
1
(2αn J2 (αn ) − J3 (αn )) .
αn3
1
rJ0 (αn r) dr =
0
J1 (αn )
,
αn
the coefficients are
An = −
1
αn3 J12 (αn )
[2αn J2 (αn ) − J3 (αn )] −
2c2
.
αn J1 (αn )
(c) The final solution is
u(r, t) = v(r, t) + 2t = w(r, t) +
where w(r, t) is given in part (b).
r2
+ c2 + 2t,
2
13.2
Polar and Cylindrical Coordinates
17. (a) First we see that
R Θ +
1 1
R Θ + 2 RΘ
T r
r
= 2 = −λ.
RΘ
a T
This gives T + a2 λT = 0 and from
1 R + λR
Θ
r
= −ν
=
−R/r2
Θ
R +
we get Θ + νΘ = 0 and r2 R + rR + (λr2 − ν)R = 0.
(b) With λ = α2 and ν = β 2 the general solutions of the differential equations in part (a)
are
T = c1 cos aαt + c2 sin aαt
Θ = c3 cos βθ + c4 sin βθ
R = c5 Jβ (αr) + c6 Yβ (αr).
(c) Implicitly we expect u(r, θ, t) = u(r, θ + 2π, t) and so Θ must be 2π-periodic. Therefore
β = n, n = 0, 1, 2, . . . . The corresponding eigenfunctions are 1, cos θ, cos 2θ, . . . , sin θ,
sin 2θ, . . . . Arguing that u(r, θ, t) is bounded as r → 0 we then define c6 = 0 and so
R = c3 Jn (αr). But R(c) = 0 gives Jn (αc) = 0; this equation defines the eigenvalues
λn = αn2 . For each n, αni = xni /c, i = 1, 2, 3, . . . , where xni are positive roots of
Jn (ac) = 0. The corresponding eigenfunctions are Jn (λni r) = 0.
(d) u(r, θ, t) =
n
(A0i cos aα0i t + B0i sin aα0i t)J0 (α0i r)
i=1
+
∞ ∞ (Ani cos aαni t + Bni sin aαni t) cos nθ
n=1 i=1
+ (Cni cos aαni t + Dni sin aαni t) sin nθ Jn (αni r)
18. (a) The boundary-value problem is
2
1 ∂u
∂2u
2 ∂ u
+
,
=
a
∂r2
r ∂r
∂t2
u(r, 0) = 0,
0 < r < 1, t > 0
u(1, t) = 0, t > 0
−v0 , 0 ≤ r < b
∂u =
,
∂t 0,
b≤r<1
t=0
0 < r < 1,
and the solution is
u(r, t) =
∞
n=1
(An cos aαn t + Bn sin aαn t)J0 (αn r),
817
818
CHAPTER 13
BOUNDARY-VALUE PROBLEMS IN OTHER COORDINATE SYSTEMS
where the eigenvalues λn = αn2 are defined by J0 (α) = 0 and An = 0 since f (r) = 0.
The coefficients Bn are given by
Bn =
2
aαn J12 (αn )
ˆ
b
rJ0 (αn r)g(r) dr = −
0
2v0
=−
aαn J12 (αn )
ˆ
αn b
2v0
aαn J12 (αn )
ˆ
b
rJ0 (αn r) dr
1
2v0
x
J0 (x)
dx = − 3 2
αn
αn
aαn J1 (αn )
0
let x = αn r
0
ˆ
αn b
xJ0 (x) dx
0
αn b
2v0
2v0 b J1 (αn b)
2v0
(xJ1 (x))
(αn bJ1 (αn b)) = − 2 2
.
=− 3
=− 3 2
aαn J1 (αn )
aαn J1 (αn )
aαn J1 (αn )
0
Thus,
∞
u(r, t) =
−2v0 b 1 J1 (αn b)
sin (aαn t)J0 (αn r).
a
αn2 J12 (αn )
n=1
(b) The standing wave un (r, t) is given by un (r, t) = Bn sin (aαn t)J0 (αn r), which has frequency fn = aαn /2π, where αn is the nth positive zero of J0 (x). The fundamental
frequency is f1 = aα1 /2π. The next two frequencies are
f2 =
α2 aα1 5.520
aα2
=
f1 = 2.295f1
=
2π
α1 2π
2.405
f3 =
α3 aα1 8.654
aα3
=
f1 = 3.598f1 .
=
2π
α1 2π
2.405
and
(c) With a = 1, b =
1
4
, and v0 = 1, the solution becomes
∞
u(r, t) = −
1 1 J1 (αn /4)
sin (αn t)J0 (αn r).
2
α2 J12 (αn )
n=1 n
The graphs of S5 (r, t) for t = 1, 2, 3, 4, 5, 6 are shown below.
u
u
0.2
0.1
−1
−0.5 −0.1
−0.2
0.5
1
r
−1
−0.5
−0.5
−0.1
−0.2
0.2
0.1
0.5
−0.1
−0.2
1
r
−1
−0.5 −0.1
−0.2
1
r
−1
−0.5
−0.1
−0.2
1
r
0.2
0.1
0.2
0.1
0.5
0.5
u
u
u
0.2
0.1
−1
u
0.2
0.1
0.5
1
r
−1
−0.5
−0.1
−0.2
0.5
1
r
13.2
Polar and Cylindrical Coordinates
(d) Three frames from the movie are shown.
19. (a) With c = 10 in Example 1 in the text the eigenvalues are λn = αn2 = x2n /100 where xn
is a positive root of J0 (x) = 0. From a CAS we find that x1 = 2.4048, x2 = 5.5201,
and x3 = 8.6537, so that the first three eigenvalues are λ1 = 0.0578, λ2 = 0.3047, and
λ3 = 0.7489. The corresponding coefficients are
ˆ 10
2
rJ0 (x1 r/10)(1 − r/10) dr = 0.7845,
A1 =
100J12 (x1 ) 0
ˆ 10
2
rJ0 (x2 r/10)(1 − r/10) dr = 0.0687,
A2 =
100J12 (x2 ) 0
and
2
A3 =
100J12 (x3 )
ˆ
10
rJ0 (x3 r/10)(1 − r/10) dr = 0.0531.
0
Since g(r) = 0, Bn = 0, n = 1, 2, 3, . . . , and the third partial sum of the series solution
is
S3 (r, t) =
∞
An cos (xn t/10)J0 (xn r/10)
n=1
= 0.7845 cos (0.2405t)J0 (0.2405r) + 0.0687 cos (0.5520t)J0 (0.5520r)
+ 0.0531 cos (0.8654t)J0 (0.8654r).
(b)
1 u
t=0
0.5
t=4
t = 20
4
2
6
8
10
r
t = 10
−0.5
t = 12
−1
20. Because of the nonhomogeneous boundary condition u(c, t) = 200 we use the substitution
u(r, t) = v(r, t) + ψ(r). This gives
2
∂v
1 ∂ v 1 ∂v
+ψ + ψ =
.
+
k
∂r2
r ∂r
r
∂t
819
820
CHAPTER 13
BOUNDARY-VALUE PROBLEMS IN OTHER COORDINATE SYSTEMS
This equation will be homogeneous provided ψ +(1/r)ψ = 0 or ψ(r) = c1 ln r +c2 .