Table of Contents
Part I Ordinary Differential Equations
1 Introduction to Differential Equations
1
2 First-Order Differential Equations
22
3 Higher-Order Differential Equations
99
4 The Laplace Transform
198
5 Series Solutions of Linear Differential Equations
252
6 Numerical Solutions of Ordinary Differential Equations
317
Part II Vectors, Matrices, and Vector Calculus
7 Vectors
339
8 Matrices
373
9 Vector Calculus
438
Part III Systems of Differential Equations
10 Systems of Linear Differential Equations
551
11 Systems of Nonlinear Differential Equations
604
Part IV Fourier Series and Partial Differential Equations
12 Orthogonal Functions and Fourier Series
634
13 Boundary-Value Problems in Rectangular Coordinates
680
14 Boundary-Value Problems in Other Coordinate Systems
755
15 Integral Transform Method
793
16 Numerical Solutions of Partial Differential Equations
832
Part V Complex Analysis
17 Functions of a Complex Variable
854
18 Integration in the Complex Plane
877
19 Series and Residues
896
20 Conformal Mappings
919
Appendices
Appendix II Gamma function
942
Projects
3.7 Road Mirages
944
3.10 The Ballistic Pendulum
946
8.1
Two-Ports in Electrical Circuits
947
8.2
Traffic Flow
948
8.15 Temperature Dependence of Resistivity
949
9.16 Minimal Surfaces
950
14.3 The Hydrogen Atom
952
15.4 The Uncertainity Inequality in Signal Processing
955
15.4 Fraunhofer Diffraction by a Circular Aperture
958
16.2 Instabilities of Numerical Methods
960
Part I
Ordinary Differential Equations
Introduction to
Differential Equations
1
EXERCISES 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.
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.
1
1.1
Definitions and Terminology
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
{x 5x = π/2 + nπ} or {x x =
π/10 + nπ/5}. From y = 25 sec2 5x we have
= 5 tan 5x is
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 or x = 2}. From y = 2x/(4 − x2 )2 we have
y = 2x
1
4 − x2
2
= 2xy.
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.
19. Writing ln(2X −1)−ln(X −1) = t and differentiating implicitly we obtain
X
2
dX
1 dX
−
=1
2X − 1 dt
X − 1 dt
2
1
dX
−
=1
2X − 1 X − 1 dt
4
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
-2
2
4
t
-2
-4
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.
2
1.1
Definitions and Terminology
20. Implicitly differentiating the solution, we obtain
y
dy
dy
−2x2
− 4xy + 2y
=0
dx
dx
−x2 dy − 2xy dx + y dy = 0
4
2
2xy dx + (x2 − y)dy = 0.
Using the quadratic formula to solve y 2 − 2x2 y − 1 = 0 for y, we get
√
√
y = 2x2 ± 4x4 + 4 /2 = x2 ± x4 + 1 . Thus, two explicit solutions
√
√
are y1 = x2 + x4 + 1 and y2 = x2 − x4 + 1 . Both solutions are defined
on (−∞, ∞). The graph of y1 (x) is solid and the graph of y2 is dashed.
-4
-2
2
4
x
-2
-4
21. Differentiating P = c1 et / (1 + c1 et ) we obtain
dP
(1 + c1 et ) c1 et − c1 et · c1 et
c1 et [(1 + c1 et ) − c1 et ]
=
=
2
dt
1 + c1 et
1 + c1 et
(1 + c1 et )
=
x
22. Differentiating y = e−x
2
c1 et
1 + c1 et
1−
c1 et
= P (1 − P ).
1 + c1 et
et dt + c1 e−x we obtain
2
2
0
x
y = e−x ex − 2xe−x
2
2
2
x
et dt − 2c1 xe−x = 1 − 2xe−x
2
2
2
0
et dt − 2c1 xe−x .
2
2
0
Substituting into the differential equation, we have
x
y + 2xy = 1 − 2xe−x
2
2
2
0
23. From y = c1 e2x + c2 xe2x we obtain
x
et dt − 2c1 xe−x + 2xe−x
2
et dt + 2c1 xe−x = 1.
2
2
0
dy
d2 y
= (2c1 + c2 )e2x + 2c2 xe2x and
= (4c1 + 4c2 )e2x + 4c2 xe2x , so that
dx
dx2
d2 y
dy
−4
+ 4y = (4c1 + 4c2 − 8c1 − 4c2 + 4c1 )e2x + (4c2 − 8c2 + 4c2 )xe2x = 0.
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
d3 y
d2 y
dy
+ y = (−6c1 + 4c1 + c1 + c1 )x−1 + (−c3 + 2c3 − c2 − c3 + c2 )x
+ 2x2 2 − x
3
dx
dx
dx
+ (−c3 + c3 )x ln x + (16 − 8 + 4)x2
= 12x2 .
25. From y =
−x2 ,
2
x ,
x<0
x≥0
we obtain y =
−2x,
2x,
x<0
so that xy − 2y = 0.
x≥0
3
1.1
Definitions and Terminology
26. 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
x→0−
x→0+
exist at x = 0.
27. (a) From y = emx we obtain y = memx . Then y + 2y = 0 implies
memx + 2emx = (m + 2)emx = 0.
Since emx > 0 for all x, m = −2. Thus y = e−2x is a solution.
(b) From y = emx we obtain y = memx and y = m2 emx . Then y − 5y + 6y = 0 implies
m2 emx − 5memx + 6emx = (m − 2)(m − 3)emx = 0.
Since emx > 0 for all x, m = 2 and m = 3. Thus y = e2x and y = e3x are solutions.
28. (a) From y = xm we obtain y = mxm−1 and y = m(m − 1)xm−2 . Then xy + 2y = 0 implies
xm(m − 1)xm−2 + 2mxm−1 = [m(m − 1) + 2m]xm−1 = (m2 + m)xm−1
= m(m + 1)xm−1 = 0.
Since xm−1 > 0 for x > 0, m = 0 and m = −1. Thus y = 1 and y = x−1 are solutions.
(b) From y = xm we obtain y = mxm−1 and y = m(m − 1)xm−2 . Then x2 y − 7xy + 15y = 0 implies
x2 m(m − 1)xm−2 − 7xmxm−1 + 15xm = [m(m − 1) − 7m + 15]xm
= (m2 − 8m + 15)xm = (m − 3)(m − 5)xm = 0.
Since xm > 0 for x > 0, m = 3 and m = 5. Thus y = x3 and y = x5 are solutions.
In Problems 29–32, we substitute y = c into the differential equations and use y = 0 and y = 0
29. Solving 5c = 10 we see that y = 2 is a constant solution.
30. Solving c2 + 2c − 3 = (c + 3)(c − 1) = 0 we see that y = −3 and y = 1 are constant solutions.
31. Since 1/(c − 1) = 0 has no solutions, the differential equation has no constant solutions.
32. Solving 6c = 10 we see that y = 5/3 is a constant solution.
33. 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
34. From x = cos 2t + sin 2t + 15 et and y = − cos 2t − sin 2t − 15 et we obtain
and
dx
1
= −2 sin 2t + 2 cos 2t + et
dt
5
and
dy
1
= 2 sin 2t − 2 cos 2t − et
dt
5
1
d2 x
= −4 cos 2t − 4 sin 2t + et
2
dt
5
and
d2 y
1
= 4 cos 2t + 4 sin 2t − et .
2
dt
5
Then
and
1
1
d2 x
4y + et = 4(− cos 2t − sin 2t − et ) + et = −4 cos 2t − 4 sin 2t + et = 2
5
5
dt
4
1.1
Definitions and Terminology
1
1
d2 y
4x − et = 4(cos 2t + sin 2t + et ) − et = 4 cos 2t + 4 sin 2t − et = 2 .
5
5
dt
35. (y )2 + 1 = 0 has no real solutions because (y )2 + 1 is positive for all functions y = φ(x).
36. 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.
37. 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.
38. 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.
√
39. 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 d
=
= cos xx=π = cos π = −1,
(sin x)
dx dx
x=π
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.
40. 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 =
particular solution is y =
15
13
sin t −
10
13
15
13
and B = − 10
13 . A
cos t.
41. 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).
42. 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.
43. 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 )
44. A tangent line will be vertical where y is undefined, or in this case, where x(2y 3 − x3 ) = 0. This gives x = 0
and 2y 3 = x3 . Substituting y 3 = x3 /2 into x3 + y 3 = 3xy we get
5
1.1
Definitions and Terminology
1
1
x3 + x3 = 3x
x
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 42 were close.
√
√
45. 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.
46. 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).
47. For the first-order differential equation integrate f (x). For the second-order differential equation integrate twice.
In the latter case we get y = ( f (x)dx)dx + c1 x + c2 .
48. Solving for y using the quadratic formula we obtain the two differential equations
1
1
y =
and y =
2 + 2 1 + 3x6
2 − 2 1 + 3x6 ,
x
x
so the differential equation cannot be put in the form dy/dx = f (x, y).
49. 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.
50. Differentiating we get y = c1 + 3c2 x2 and y = 6c2 x. Then c2 = y /6x
y
xy y= y −
x+
x3 = xy −
2
6x
and c1 = y − xy /2, so
1 2 x y
3
and the differential equation is x2 y − 3xy + 3y = 0.
51. (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
2
dy
dy
(b) lim
= lim e−x = 0 and lim
= lim e−x = 0. Since dy/dx approaches 0 as x approaches −∞
x→−∞ dx
x→−∞
x→∞ dx
x→∞
and ∞, the solution curve has horizontal asymptotes to the left and to the right.
2
(c) To test concavity we consider the second derivative
2
2
d2 y
d
d dy
=
e−x = −2xe−x .
=
dx2
dx dx
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, ∞).
6
1.1
(d)
Definitions and Terminology
y
x
52. (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.
53. (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.
(c) Using implicit differentiation we compute
d2 y
= y(−by ) + y (a − by) = y (a − 2by).
dx2
Solving d2 y/dx2 = 0 we obtain 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=aêb
y=0
x
54. (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.
7
1.1
Definitions and Terminology
(d)
y
x
55. 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.
56. 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.
EXERCISES 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 (−∞, ∞).
8
1.2 Initial-Value Problems
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 .
7. From the initial conditions we obtain the system
c1 = −1
c2 = 8.
The solution of the initial-value problem is x = − cos t + 8 sin t.
8. From the initial conditions we obtain the system
c2 = 0
−c1 = 1.
The solution of the initial-value problem is x = − cos t.
9. From the initial conditions we obtain
Solving, we find c1 =
√
√
3
1
1
c1 + c2 =
2
2
2
√
1
3
− c1 +
c2 = 0.
2
2
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 .
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 + c2 = 1
c1 − c2 = 2.
Solving, we find c1 =
3
2
and c2 = − 12 . The solution of the initial-value problem is y = 32 ex − 12 e−x .
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
y=
13. From the initial conditions we obtain
1 x 1 2 −x
1
1
e − e e = ex − e2−x .
2
2
2
2
e−1 c1 + ec2 = 5
e−1 c1 − ec2 = −5.
9
1.2
Initial-Value Problems
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.)
∂f
2
= y −1/3 . Thus, the differential equation will have a unique solution in any
∂y
3
rectangular region of the plane where y = 0.
√
18. For f (x, y) = xy we have ∂f /∂y = 12 x/y . Thus, the differential equation will have a unique solution in any
17. For f (x, y) = y 2/3 we have
region where x > 0 and y > 0 or where x < 0 and y < 0.
19. For f (x, y) =
where x = 0.
y
∂f
1
we have
= . Thus, the differential equation will have a unique solution in any region
x
∂y
x
20. For f (x, y) = x + y we have
∂f
= 1. Thus, the differential equation will have a unique solution in the entire
∂y
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.
x2
∂f
−3x2 y 2
we
have
=
2 . Thus, the differential equation will have a unique solution in
1 + y3
∂y
(1 + y 3 )
any region where y = −1.
22. For f (x, y) =
y2
2x2 y
∂f
=
we
have
2 . Thus, the differential equation will have a unique solution in
x2 + y 2
∂y
(x2 + y 2 )
any region not containing (0, 0).
23. For f (x, y) =
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.
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.
In Problems 25–28, we identify f (x, y) =
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.
10
1.2 Initial-Value Problems
(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
30. (a) Since
tan(x + c) = sec2 (x + c) = 1 + tan2 (x + c), we see that y = tan(x + c) satisfies the differential
dx
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).
d
1
1
1
31. (a) Since
= y 2 , we see that y = −
−
=
is a solution of the differential equation.
dx
x+c
(x + c)2
x+c
(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, ∞).
32. (a) Solving y(0) = −1/c = y0 we obtain c = −1/y0 and
y=−
1
y0
=
,
−1/y0 + x
1 − y0 x
y0 = 0.
Since we must have −1/y0 + x = 0, the largest interval of existence (which must contain 0) is either
(−∞, 1/y0 ) when y0 > 0 or (1/y0 , ∞) when y0 < 0.
(b) By inspection we see that y = 0 is a solution on (−∞, ∞).
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 = φ1 (x) = 3(x2 − 1) ,
y = φ2 (x) = − 3(x2 − 1) ,
y = φ3 (x) = 3(x2 − 1) ,
y = φ4 (x) = − 3(x2 − 1) ,
y
4
1 < x < ∞,
2
1 < x < ∞,
−∞ < x < −1,
-4
-2
−∞ < x < −1.
2
4
x
2
4
x
-2
-4
(c) Only y = φ3 (x) satisfies y(−2) = 3.
y
34. (a) Setting x = 2 and y = −4 in 3x2 − y 2 = c we get 12 − 16 = −4 = c, so the
explicit solution is
y = − 3x2 + 4 , −∞ < x < ∞.
(b) Setting c = 0 we have y =
√
√
3x and y = − 3x, both defined on
4
2
-4
-2
-2
(−∞, ∞).
-4
11
1.2
Initial-Value Problems
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).
39. Integrating y = 8e2x + 6x we obtain
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.
40. Integrating y = 12x − 2 we obtain
y =
(12x − 2)dx = 6x2 − 2x + c1 .
Then, integrating y we obtain
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 + c2 = 4
or
c1 + c2 = 3
6 − 2 + c1 = −1
or
c1 = −5.
and
Thus, c1 = −5 and c2 = 8, so y = 2x3 − x2 − 5x + 8.
41. When x = 0 and y =
1
2
, y = −1, so the only plausible solution curve is the one with negative slope at (0, 12 ),
or the black curve.
42. 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.
43. The theorem guarantees a unique (meaning single) solution through any point. Thus, there cannot be two
distinct solutions through any point.
44. When y =
1 4
16 x ,
y = 14 x3 = x( 14 x2 ) = xy 1/2 , and y(2) =
y=
1
16 (16)
= 1. When
0,
x<0
1 4
16 x ,
x≥0
we have
y =
and y(2) =
1
16 (16)
0,
x<0
1 3
4x ,
x≥0
=x
0,
x<0
1 2
4x ,
x≥0
= xy 1/2 ,
= 1. The two different solutions are the same on the interval (0, ∞), which is all that is
required by Theorem 1.1.
45. At t = 0, dP/dt = 0.15P (0) + 20 = 0.15(100) + 20 = 35. Thus, the population is increasing at a rate of 3,500
individuals per year.
12
1.3 Differential Equations as Mathematical Models
If the population is 500 at time t = T then
dP = 0.15P (T ) + 20 = 0.15(500) + 20 = 95.
dt t=T
Thus, at this time, the population is increasing at a rate of 9,500 individuals per year.
EXERCISES 1.3
Differential Equations as Mathematical Models
dP
dP
= kP + r;
= kP − r
dt
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 .
1.
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 .
dP
4.
= 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=
dT /dt
−1
= −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
dT
π = k T − 80 − 30 cos t , t > 0.
dt
12
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
9. The rate at which salt is leaving the tank is
Rout (3 gal/min) ·
x(0) = 1.
A
A
lb/gal =
lb/min.
300
100
Thus dA/dt = A/100. The initial amount is A(0) = 50.
10. The rate at which salt is entering the tank is
Rin = (3 gal/min) · (2 lb/gal) = 6 lb/min.
13
1.3
Differential Equations as Mathematical Models
Since the solution is pumped out at a slower rate, it is accumulating at the rate of (3 − 2)gal/min = 1 gal/min.
After t minutes there are 300 + t gallons of brine in the tank. The rate at which salt is leaving is
A
2A
Rout = (2 gal/min) ·
lb/gal =
lb/min.
300 + t
300 + t
The differential equation is
dA
2A
=6−
.
dt
300 + t
11. The rate at which salt is entering the tank is
Rin = (3 gal/min) · (2 lb/gal) = 6 lb/min.
Since the tank loses liquid at the net rate of
3 gal/min − 3.5 gal/min = −0.5 gal/min,
after t minutes the number of gallons of brine in the tank is 300 − 12 t gallons. Thus the rate at which salt is
leaving is
A
3.5A
7A
Rout =
lb/gal · (3.5 gal/min) =
lb/min =
lb/min.
300 − t/2
300 − t/2
600 − t
The differential equation is
dA
7A
=6−
dt
600 − t
dA
7
+
A = 6.
dt
600 − t
or
12. The rate at which salt is entering the tank is
Rin = (cin lb/gal) · (rin gal/min) = cin rin lb/min.
Now let A(t) denote the number of pounds of salt and N (t) the number of gallons 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 gallons
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 gallons of brine in the tank at time t is N (t) = N0 + (rin − rout )t.
The output rate of salt is then
A
A
Rout =
lb/gal · (rout gal/min) = rout
lb/min.
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
13. The volume of water in the tank at time t is V = Aw h. The differential equation is then
dh
cAh 1 dV
1
−cAh 2gh = −
2gh .
=
=
dt
Aw dt
Aw
Aw
2
2
π
Using Ah = π
=
, Aw = 102 = 100, and g = 32, this becomes
12
36
dh
cπ √
cπ/36 √
64h = −
h.
=−
dt
100
450
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
2
4
the figure in the text we see that r/h = 8/20 so that r = 25 h and V = 13 π 25 h h = 75
πh3 . Differentiating with
4
respect to t we have dV /dt = 25
πh2 dh/dt or
dh
25 dV
=
.
dt
4πh2 dt
14
1.3 Differential Equations as Mathematical Models
2 2
√
From Problem 13 we have dV /dt = −cAh 2gh where c = 0.6, Ah = π 12
, and g = 32. Thus dV /dt =
√
−2π h/15 and
√ 2π h
dh
25
5
−
=
= − 3/2 .
dt
4πh2
15
6h
15. Since i = dq/dt and L d2 q/dt2 + R dq/dt = E(t), we obtain L di/dt + Ri = E(t).
dq
1
16. By Kirchhoff’s second law we obtain R + q = E(t).
dt
C
dv
17. From Newton’s second law we obtain m
= −kv 2 + mg.
dt
18. Since the barrel in Figure 1.35(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
= (62.4) × (volume of water displaced)
= (62.4)π(s/2)2 y = 15.6πs2 y.
It then follows from Newton’s second law that
w d2 y
= −15.6πs2 y
g dt2
d2 y 15.6πs2 g
+
y = 0,
dt2
w
or
where g = 32 and w is the weight of the barrel in pounds.
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
d2 x
dx
= −kx − β
2
dt
dt
or
m
d2 x
dx
+ kx = 0.
+β
2
dt
dt
21. Let x(t) denote the height of the top of the chain at time t with the positive direction upward. The weight of
the portion of chain off the ground is W = (x ft) · (1 lb/ft) = x. The mass of the chain is m = W/g = x/32.
The net force is F = 5 − W = 5 − x. By Newton’s second law,
d x
v =5−x
dt 32
or
x
dv
dx
+v
= 160 − 32x.
dt
dt
Thus, the differential equation is
x
d2 x
dx
+
dt2
dt
2
+ 32x = 160.
22. The force is the weight of the chain, 2L, so by Newton’s second law,
of chain off the ground is m = 2(L − x)/g, we have
d 2(L − x) v = 2L
or
dt
g
(L − x)
15
d
[mv] = 2L. Since the mass of the portion
dt
dv
dx
+v −
= Lg.
dt
dt
1.3
Differential Equations as Mathematical Models
Thus, the differential equation is
(L − x)
d2 x
dx
−
2
dt
dt
2
= Lg.
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
d2 r
k
gR2
=
−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
Mr m
r3 M m/R3
mM
= −k
= −k 3 r.
2
r
r2
R
Now from F = ma = d2 r/dt2 we have
m
d2 r
mM
= −k 3 r
2
dt
R
d2 r
kM
= − 3 r.
2
dt
R
or
dA
= k(M − A).
dt
dA
26. The differential equation is
= k1 (M − A) − k2 A.
dt
27. The differential equation is x (t) = r − kx(t) where k > 0.
25. The differential equation is
−y
28. By the Pythagorean Theorem the slope of the tangent line is y = .
s2 − y 2
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 formula, we get
−2x ± 4x2 + 4y 2
−x ± x2 + y 2
y =
=
.
2y
y
Since dy/dx > 0, the differential equation is
dy
−x + x2 + y 2
=
dx
y
or
y
(x,y) θ
θα
θ
x
α
y
φ
x
dy 2
− x + y 2 + x = 0.
dx
30. The differential equation is dP/dt = kP , so from Problem 37 in Exercises 1.1, P = ekt , and 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 = 6 − A/100. If A(t) attains a maximum, then dA/dt = 0 at this time
and A = 600. If A(t) continues to increase without reaching a maximum, then A (t) > 0 for t > 0 and A cannot
exceed 600. In this case, if A (t) approaches 0 as t increases to infinity, we see that A(t) approaches 600 as t
increases to infinity.
33. This differential equation could describe a population that undergoes periodic fluctuations.
16
1.3 Differential Equations as Mathematical Models
34. (a) As shown in Figure 1.43(b) 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 =
F sin θ
mω 2 x
ω2 x
=
=
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 + 3 + · · · .
2
dt
(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
dm
dr
dV
d 4 3
=ρ
=ρ
πr = ρ 4πr2
dt
dt
dt 3
dt
= ρS
dr
.
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
(b) From Newton’s second law, [mv] = mg, where v is the velocity of the raindrop. Then
dt
m
dv
dm
+v
= mg
dt
dt
or
ρ
4 3 dv
4
πr
+ v(k4πr2 ) = ρ πr3 g.
3
dt
3
Dividing by 4ρπr3 /3 we get
dv 3k
+
v=g
dt
ρr
dv
3k/ρ
v = g, k < 0.
+
dt
kt/ρ + r0
or
37. We assume that the plow clears snow at a constant rate of k cubic miles 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
wx
dy
= k,
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 snowing and let s be the constant rate
in miles per hour at which x increases. Then for t > −t0 , x = s(t + t0 ). The differential equation then becomes
dy
k
1
.
=
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
k
t
y=
ln 1 +
.
ws
t0
y=
17
1.3
Differential Equations as Mathematical Models
Finally, from the fact that when t = 1, y = 2 and when t = 2, y = 3, we obtain
2 3
2
1
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.
dP
dA
38. (1):
= kP is linear
(2):
= kA is linear
dt
dt
dx
dT
(5):
(3):
= k(T − Tm ) is linear
= kx(n + 1 − x) is nonlinear
dt
dt
dX
dA
A
(6):
= k(α − X)(β − X) is nonlinear
(8):
=6−
is linear
dt
dt
100
dh
d2 q
dq
Ah 1
(10):
2gh is nonlinear
(11): L 2 + R + q = E(t) is linear
=−
dt
Aw
dt
dt
C
d2 s
dv
(12):
= −g is linear
(14): m
= mg − kv is linear
2
dt
dt
d2 s
ds
d2 x 64
(15): m 2 + k
− x = 0 is linear
= mg is linear
(16):
dt
dt
dt2
L
(17): linearity or nonlinearity is determined by the manner in which W and T1 involve x.
39. 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.
40. Setting A (t) = −0.002 and solving A (t) = −0.0004332A(t) for A(t), we obtain
A(t) =
A (t)
−0.002
=
≈ 4.6 grams.
−0.0004332
−0.0004332
CHAPTER 1 REVIEW EXERCISES
d
dy
c1 ekx = c1 kekx ;
= ky
dx
dx
d
dy
dy
2.
(5 + c1 e−2x ) = −2c1 e−2x = −2(5 + c1 e−2x − 5);
= −2(y − 5) or
= −2y + 10
dx
dx
dx
d
3.
(c1 cos kx + c2 sin kx) = −kc1 sin kx + kc2 cos kx;
dx
d2
(c1 cos kx + c2 sin kx) = −k 2 c1 cos kx − k 2 c2 sin kx = −k 2 (c1 cos kx + c2 sin kx);
dx2
d2 y
d2 y
2
=
−k
y
or
+ k2 y = 0
dx2
dx2
d
(c1 cosh kx + c2 sinh kx) = kc1 sinh kx + kc2 cosh kx;
4.
dx
d2
(c1 cosh kx + c2 sinh kx) = k 2 c1 cosh kx + k 2 c2 sinh kx = k 2 (c1 cosh kx + c2 sinh kx);
dx2
1.
18
CHAPTER 1 REVIEW EXERCISES
d2 y
= k2 y
dx2
d2 y
− k2 y = 0
dx2
or
y = c1 ex + c2 xex + c2 ex ;
5. y = c1 ex + c2 xex ;
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
y − 2y + 2y = 0
10. a,c
9. b
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.
(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 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
or
x20 − x0 − 2 = (x0 − 2)(x0 + 1) = 0.
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) and y = −2/x + x is a solution of the initial-value
problem xy + y = 2x, y(2) = 1, on the interval (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
3
y
3
2
2
1
1
-3 -2 -1
-1
1
2
3 x
-3 -2 -1
-1
-2
-2
-3
-3
y = x2 + c1
1
2
y = −x2 + c2
19
3 x
CHAPTER 1 REVIEW EXERCISES
(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 =
√
22. The slope of the tangent line is y (−1,4) = 6 4 + 5(−1)3 = 7.
−x2 , x ≤ 0
x2 ,
x > 0.
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
y = x cos x + sin x + cos x
− sin x
cos x
− (sin x) ln(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)
sin2 x
− (cos x) ln(cos x)
cos x
1 − cos2 x
= −x sin x + cos 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 + 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)
x2 y + xy + y = x2 −
+x
+ sin(ln x) = 0.
2
x
x
An interval of definition for the solution is (0, ∞).
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
y = cos(ln x)
−
+ ln(cos(ln x)) −
+ ln x
+
cos(ln x)
x
x
x
x
=−
ln(cos(ln x)) sin(ln x) (ln x) cos(ln x)
+
x
x
20
CHAPTER 1 REVIEW EXERCISES
and
1
sin(ln x)
1
cos(ln x)
y = −x ln(cos(ln x))
+ sin(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
= 2 − ln(cos(ln x)) cos(ln x) +
+ ln(cos(ln x)) sin(ln x)
x
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)
2
=
sin (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.)
27. From the graph we see that estimates for y0 and y1 are y0 = −3 and y1 = 0.
28. The differential equation is
dh
cA0 2gh .
=−
dt
Aw
Using A0 = π(1/24)2 = π/576, Aw = π(2)2 = 4π, and g = 32, this becomes
dh
c √
cπ/576 √
64h =
h.
=−
dt
4π
288
21
2
First-Order Differential Equations
EXERCISES 2.1
Solution Curves Without the Solution
y
3
1.
2.
y
10
2
5
1
-3
-2
-1
1
2
x
0
x
3
-1
-5
-2
-10
-3
3.
y
-5
10
y
4.
4
5
0
4
2
2
x
0
x
0
-2
-2
-4
-4
5.
-2
0
2
4
-4
y
6.
4
2
0
2
4
y
4
2
x
0
x
0
-2
-4
-2
-2
-2
0
2
4
-4
22
-2
0
2
4
2.1
7.
y
8.
4
Solution Curves Without the Solution
y
4
2
2
x
0
x
0
-2
-2
-4
-4
9.
-2
0
2
4
-4
y
10.
4
2
x
2
4
y
4
x
0
-2
-2
-4
-2
0
2
4
-4
y
12.
4
2
-2
0
2
4
y
4
2
x
0
x
0
-2
-2
-4
13.
0
2
0
11.
-2
-2
0
2
4
-4
y
-2
2
4
y
14.
3
0
4
2
2
1
x
0
x
0
-1
-2
-2
-4
-3
-3
-2
-1
0
1
2
3
-4
23
-2
0
2
4
2.1
Solution Curves Without the Solution
15. (a) The isoclines have the form y = −x + c, which are straight
y
3
lines with slope −1.
2
1
-3 -2 -1
1
3 x
2
-1
-2
-3
(b) The isoclines have the form x2 + y 2 = c, which are circles
centered at the origin.
y
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 ∞.
y = x2 − 2y is positive and the portions of solution
curves “outside” the nullcline parabola are increasing. When y > 12 x2 ,
17. When y <
1 2
2x ,
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 differential equations in Problems 1, 3, and 4 superimposed on the corresponding direction
field.
24
2.1
y
Solution Curves Without the Solution
y
y
4
3
4
2
2
2
1
x
0
x
0
x
0
-1
-2
-2
-2
-4
-3
-3 -2 -1 0 1 2
Problem 1
-4
-4
3
4
-2
0
2
Problem 3
-4 -2 0 2
Problem 4
4
(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 located at y = −1, y = 0, and y = 1. The phase portrait is shown at the right.
1
y
(a)
y
(b)
5
4
0
1
3
-1
2
1
-2
1
2
y
(c)
-1
1
(d)
-1
1
2
x
y
1
-2
2
x
x
2
x
-1
-2
-3
-4
-1
-5
20. Writing the differential equation in the form dy/dx = y 2 (1 − y)(1 + y) we see that critical points
are located at y = −1, y = 0, and y = 1. The phase portrait is shown at the right.
y
(a)
1
y
(b)
5
4
0
1
3
2
-1
1
-2
1
y
2
-1
1
2
x
y
(c)
(d)
-2
-2
-1
1
2
x
x
-1
-1
-2
-3
-4
-1
-5
25
x
2.1
Solution Curves Without the Solution
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
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
26
2.1
Solution Curves Without the Solution
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
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 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).
ln 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 (−∞, 0) and (c, ∞). Since f (y) < 0 for 0 < y < c, the graph of
the solution is decreasing on (0, c).
27
2.1
Solution Curves Without the Solution
y
c
c
x
0
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 (−∞, −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 (−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
(π/2)y − sin y = 0 has only three solutions. By inspection
1
we see that the critical points are −π/2, 0, and π/2.
Π
From the graph at the right we see that
Π
2
y
Π
-1
2
y − sin y
π
2
y − sin y
π
Π
2
<0
>0
for y < −π/2
for y > π/2
> 0 for
< 0 for
− π/2 < y < 0
0 < y < π/2.
Π
2
0
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 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
28
2.1
Solution Curves Without the Solution
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
y
5
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
-5
have the same y-coordinates because between critical points they are horizontal
translates of each other.
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.
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.
39. (a) Writing the differential equation in the form
dv
k mg
=
−v
dt
m 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, limt→∞ v = mg/k.
29
b
a
0
mg
k
2.1
Solution Curves Without the Solution
(b) Writing the differential equation in the form
dv
k
mg
mg
k mg
2
=
−v =
−v
+v
dt
m k
m
k
k
we see that the only physically meaningful critical point is mg/k.
From the phase portrait we see that mg/k is an asymptotically stable
critical point. Thus, limt→∞ v = mg/k.
mg
k
40. (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
Β
not known.
Α
(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 → ∞.
Α
(c) When k = 1 and α = β the differential equation is dX/dt = (α − X)2 . For X(t) = α − 1/(t + c) we have
dX/dt = 1/(t + c)2 and
2
1
1
dX
2
(α − X) = α − α −
=
=
.
2
t+c
(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
30
2.2
X
X
2α
α
α
−2/α
Separable Variables
α/2
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→∞
EXERCISES 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.
3. From dy = −e−3x dx we obtain y = 13 e−3x + c.
1
1
1
= x + c or y = 1 −
.
dy = dx we obtain −
(y − 1)2
y−1
x+c
1
4
5. From dy = dx we obtain ln |y| = 4 ln |x| + c or y = c1 x4 .
y
x
1
1
1
6. From 2 dy = −2x dx we obtain − = −x2 + c or y = 2
.
y
y
x + c1
4. From
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
2
1
y
x3
1
9. From y + 2 +
dy = x2 ln x dx we obtain
+ 2y + ln |y| =
ln |x| − x3 + c.
y
2
3
9
1
1
1
2
10. From
=
+ c.
dy =
dx we obtain
(2y + 3)2
(4x + 5)2
2y + 3
4x + 5
1
1
11. From
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 .
12. From 2y dy = −
sin 3x
dx or 2y dy = − tan 3x sec2 3x dx we obtain y 2 = − 16 sec2 3x + c.
cos3 3x
31
2.2
Separable Variables
13. From
14. From
ey
2
(ey + 1)
y
1/2
(1 + y 2 )
−ex
−1
−2
y
= 12 (ex + 1)
3 dx we obtain − (e + 1)
(ex + 1)
x
1/2
dy =
dx we obtain 1 + y 2
= 1 + x2
1/2
(1 + x2 )
dy =
+ c.
1/2
+ c.
1
dS = k dr we obtain S = cekr .
S
1
From
dQ = k dt we obtain ln |Q − 70| = kt + c or Q − 70 = c1 ekt .
Q − 70
P
1
1
1
= t + c or
From
dP
=
+
dP = dt we obtain ln |P | − ln |1 − P | = t + c so that ln
P − P2
P
1−P
1−P
P
c1 et
.
= c1 et . Solving for P we have P =
1−P
1 + c1 et
1
t+2
t+2
From
dN = tet+2 − 1 dt we obtain ln |N | = tet+2 − et+2 − t + c or N = c1 ete −e −t .
N
y−2
x−1
5
5
From
dy =
dx or 1 −
dy = 1 −
dx we obtain y − 5 ln |y + 3| = x − 5 ln |x + 4| + c
y+3
x+4
y+3
x+4
5
x+4
or
= c1 ex−y .
y+3
y+1
x+2
2
5
From
dy =
dx or 1 +
dy = 1 +
dx we obtain y + 2 ln |y − 1| = x + 5 ln |x − 3| + c
y−1
x−3
y−1
x−3
(y − 1)2
or
= c1 ex−y .
(x − 3)5
2
1
x
−1
1 2
From x dx = + c1 .
dy we obtain 2 x = sin y + c or y = sin
2
1 − y2
15. From
16.
17.
18.
19.
20.
21.
22. From
1
1
1
ex
1
.
dx we obtain − = tan−1 ex + c or y = −
dy
=
dx
=
y2
ex + e−x
(ex )2 + 1
y
tan−1 ex + c
1
dx = 4 dt we obtain tan−1 x = 4t + c. Using x(π/4) = 1 we find c = −3π/4. The solution of the
+1
3π
3π
−1
initial-value problem is tan x = 4t −
or x = tan 4t −
.
4
4
1
1
1
1
1
1
1
1
24. From 2
dy = 2
dx or
−
dy =
−
dx we obtain
y −1
x −1
2 y−1 y+1
2 x−1 x+1
y−1
c(x − 1)
ln |y − 1| − ln |y + 1| = ln |x − 1| − ln |x + 1| + ln c or
=
. Using y(2) = 2 we find
y+1
x+1
23. From
x2
c = 1. A solution of the initial-value problem is
y−1
x−1
=
or y = x.
y+1
x+1
1
1
1
1−x
1
25. From dy =
dx =
−
dx we obtain ln |y| = − − ln |x| = c or xy = c1 e−1/x . Using y(−1) = −1
2
2
y
x
x
x
x
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 find c1 = −4.
1 − 2y
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
√
dx
dy
−
=0
2
1−x
1 − y2
and
32
sin−1 x − sin−1 y = c.
2.2
sin
28. From
−1
x − sin
√
3/2 we obtain c = −π/3. Thus, an implicit solution of the initial-value problem is
y = π/3. Solving for y and using an addition formula from trigonometry, we get
√ √
3 1 − x2
π
π
π x
y = sin sin−1 x +
= x cos + 1 − x2 sin = +
.
3
3
3
2
2
Setting x = 0 and y =
−1
Separable Variables
−x
1
dy =
2 dx we obtain
2
1 + (2y)
1 + (x2 )
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
−1
−1 2
tan 2y + tan x = π/4 . Solving for y and using a trigonometric identity we get
π
2y = tan
− tan−1 x2
4
π
1
y = tan
− tan−1 x2
2
4
π
1 tan 4 − tan(tan−1 x2 )
=
2 1 + tan π4 tan(tan−1 x2 )
=
1 1 − x2
.
2 1 + x2
29. (a) The equilibrium solutions y(x) = 2 and y(x) = −2 satisfy the initial conditions y(0) = 2 and y(0) = −2,
respectively. Setting x = 14 and y = 1 in y = 2(1 + ce4x )/(1 − ce4x ) we obtain
1=2
1 + ce
,
1 − ce
1 − ce = 2 + 2ce,
−1 = 3ce,
and c = −
1
.
3e
The solution of the corresponding initial-value problem is
y=2
1 − 13 e4x−1
3 − e4x−1
1 4x−1 = 2 3 + e4x−1 .
1 + 3e
(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
y−2
c
= 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
−3e + e4x
3 − e4x−1
y=2
=
2
.
−3e − e4x
3 + e4x−1
30. Separating variables, we have
dy
dx
=
y2 − y
x
dy
= ln |x| + c.
y(y − 1)
or
33
2.2
Separable Variables
Using partial fractions, we obtain
1
1
−
y−1 y
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.
(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
1
2
= 1/(1 −
1
2
c1 ), so c1 = −2 and y = 1/(1 + 2x).
we have 14 = 1/(1 − 2c1 ), so c1 = − 32 and y = 1/(1 + 32 x) = 2/(2 + 3x).
31. Singular solutions of dy/dx = x 1 − y 2 are y = −1 and y = 1. A singular solution of
(d) Setting x = 2 and y =
1
4
(ex + e−x )dy/dx = y 2 is y = 0.
32. Differentiating ln(x2 + 10) + csc y = c we get
2x
dy
− csc y cot y
= 0,
x2 + 10
dx
or
2x
1
cos y dy
−
·
= 0,
x2 + 10 sin y sin y dx
2x sin2 y dx − (x2 + 10) cos y dy = 0.
Writing the differential equation in the form
dy
2x sin2 y
= 2
dx
(x + 10) cos y
we see that singular solutions occur when sin2 y = 0, or y = kπ, where k is an integer.
33. The singular solution y = 1 satisfies the initial-value problem.
y
1.01
1
-0.004-0.002
0.98
0.97
34
0.002 0.004
x
2.2
34. Separating variables we obtain
−
dy
= dx. Then
(y − 1)2
Separable Variables
y
1.02
1
x+c−1
= x + c and y =
.
y−1
x+c
1.01
Setting x = 0 and y = 1.01 we obtain c = −100. The solution is
y=
x − 101
.
x − 100
-0.004-0.002
0.002 0.004
x
0.99
0.98
dy
35. Separating variables we obtain
= dx. Then
(y − 1)2 + 0.01
y
1.0004
1
x+c
tan
.
10
10
Setting x = 0 and y = 1 we obtain c = 0. The solution is
10 tan−1 10(y − 1) = x + c and y = 1 +
y =1+
1.0002
1
x
tan
.
10
10
-0.004-0.002
0.002 0.004
x
0.9998
0.9996
dy
= dx. Then, from (11) in
(y − 1)2 − 0.01
1
this section of the manual with u = y − 1 and a = 10
, we get
36. Separating variables we obtain
5 ln
y
1.0004
10y − 11
= x + c.
10y − 9
1.0002
Setting x = 0 and y = 1 we obtain c = 5 ln 1 = 0. The solution is
5 ln
-0.004-0.002
10y − 11
= x.
10y − 9
0.002 0.004
x
0.9998
Solving for y we obtain
0.9996
y=
11 + 9ex/5
.
10 + 10ex/5
Alternatively, we can use the fact that
dy
y−1
1
=−
tanh−1
= −10 tanh−1 10(y − 1).
(y − 1)2 − 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).
37. Separating variables, we have
dy
dy
=
=
3
y−y
y(1 − y)(1 + y)
1
1/2
1/2
+
−
y 1−y 1+y
Integrating, we get
ln |y| −
1
1
ln |1 − y| − ln |1 + y| = x + c.
2
2
35
dy = dx.
2.2
Separable Variables
When y > 1, this becomes
1
1
y
ln(y − 1) − ln(y + 1) = ln = x + c.
2
2
y2 − 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).
ln y −
When 0 < y < 1 we have
1
1
y
= x + c.
ln(1 − y) − ln(1 + y) = ln 2
2
1 − y2
√
√
we find c = ln(1/ 3 ). Solving for y we get y2 (x) = ex / e2x + 3 , where −∞ < x < ∞.
ln y −
1
2
Letting x = 0 and y =
When −1 < y < 0 we have
1
1
−y
= x + c.
ln(1 − y) − ln(1 + y) = ln 2
2
1 − y2
√
√
we find c = ln(1/ 3 ). Solving for y we get y3 (x) = −ex / e2x + 3 , where
ln(−y) −
Letting x = 0 and y = − 12
−∞ < x < ∞.
When y < −1 we have
1
1
−y
= x + c.
ln(1 − y) − ln(−1 − y) = ln 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).
ln(−y) −
y
y
y
y
4
4
4
4
2
2
2
2
1
2
3
4
5x
-4 -2
2
4
x
-4 -2
2
4
x
1
-2
-2
-2
-2
-4
-4
-4
-4
38. (a) The second derivative of y is
d y
dy/dx
1/(y − 3)
1
=−
=−
=−
.
2
2
2
dx
(y − 1)
(y − 3)
(y − 3)3
5x
4
6
2
The solution curve is concave down when d y/dx < 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.
3
y
8
2
2
2
4
3
2
-4
-2
2
x
4
-2
(b) Separating variables and integrating we obtain
y
8
(y − 3) dy = dx
1 2
y − 3y = x + c
2
y 2 − 6y + 9 = 2x + c1
6
4
2
(y − 3)2 = 2x + c1
√
y = 3 ± 2x + c1 .
-1
1
-2
36
2
3
4
5 x
2.2
Separable Variables
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 .
√
When y2 (0) = 2 we have c1 = 1 and y2 (x) = 3 − 2x + 1 .
√
When y3 (1) = 2 we have c1 = −1 and y3 (x) = 3 − 2x − 1 .
√
When y4 (−1) = 4 we have c1 = 3 and y4 (x) = 3 + 2x + 3 .
39. (a) Separating variables we have 2y dy = (2x + 1)dx. Integrating gives y 2 = x2 + x + c. When y(−2) = −1 we
√
find c = −1, so y 2 = x2 + x − 1 and y = − x2 + x − 1 . The negative square root is chosen because of the
initial condition.
y
2
(b) From the figure, the largest interval of definition appears to be
approximately (−∞, −1.65).
1
-5 -4 -3 -2 -1
-1
1
2 x
-2
-3
-4
-5
√
(c) Solving x + x − 1 = 0 we get x =
±
5 , so the largest interval of definition is (−∞, − 12 − 12 5 ).
√
The right-hand endpoint of the interval is excluded because y = − x2 + x − 1 is not differentiable at this
point.
2
− 12
1
2
√
40. (a) From Problem 7 the general solution is 3e−2y + 2e3x = c. When y(0) = 0 we find c = 5, so 3e−2y + 2e3x = 5.
Solving for y we get y = − 12 ln 13 (5 − 2e3x ).
y
(b) The interval of definition appears to be approximately (−∞, 0.3).
2
1
x
-2 -1.5 -1 -0.5
-1
-2
(c) Solving 13 (5 − 2e3x ) = 0 we get x = 13 ln( 52 ), so the exact interval of definition is (−∞, 13 ln 52 ).
√
41. (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.
42. (a) Separating variables and integrating we obtain x2 − y 2 = c. For c = 0 the graph is a hyperbola centered at
the origin. All four initial conditions imply c = 0 and y = ±x. Since the differential equation is not defined
for y = 0, solutions are y = ±x, x < 0 and y = ±x, x > 0. The solution for y(a) = a is y = x, x > 0; for
y(a) = −a is y = −x; for y(−a) = a is y = −x, x < 0; and for y(−a) = −a is y = x, x < 0.
(b) Since x/y is not defined when y = 0, the initial-value problem has no solution.
√
(c) Setting x = 1 and y = 2 in x2 − y 2 = c we get c = −3, so y 2 = x2 + 3 and y(x) = x2 + 3 , where
the positive square root is chosen because of the initial condition. The domain is all real numbers since
x2 + 3 > 0 for all x.
37
2.2
Separable Variables
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.
43. Separating variables we have dy/
y
3.5
3
2.5
2
1.5
1
0.5
-6
-4
-2
2
4
6
8
x
√
√
√
44. Separating variables we have dy/( y + y) = dx/( x + x). To integrate dx/( x + x) we substitute u2 = x
and get
√
2u
2
du =
du = 2 ln |1 + u| + c = 2 ln(1 + x ) + c.
u + u2
1+u
Integrating the separated differential equation we have
√
√
√
√
2 ln(1 + y ) = 2 ln(1 + x ) + c or ln(1 + y ) = ln(1 + x ) + ln c1 .
√
Solving for y we get y = [c1 (1 + x ) − 1]2 .
45. We are looking for a function y(x) such that
y2 +
dy
dx
2
= 1.
Using the positive square root gives
dy
dy
= dx =⇒ sin−1 y = x + c.
= 1 − y 2 =⇒ dx
1 − y2
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.
y
46. (a)
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
dy
y2
−1
=√
dx
x2 − 1
and
cosh−1 y = cosh−1 x + c.
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.
47. 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
38
2.2
Separable Variables
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 2.22 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.
48. (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
0
x
-2
-4
-4
(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.
-2
0
2
4
y
4
2
x
0
-2
-4
-4
-2
0
2
4
49. (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:
y=
−2 ±
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.
39
2.2
Separable Variables
1 f x
fx
-4
4
-2
x
2
-2
2
-4
-4
x
-2
-6
-2
-8
-4
-10
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
x
2
equality has been removed since the derivative dy/dx does not exist at the points where
-2
f (x) = 0. The graph of the solution y = φ(x) is given on the right.
-4
-6
-8
-10
50. (a) Separating variables and integrating, we have
y
(−2y + y 2 )dy = (x − x2 )dx
4
and
1
1
1
−y 2 + y 3 = x2 − x3 + c.
3
2
3
Using a CAS we show some contours of
2
-2
f (x, y) = 2y 3 − 6y 2 + 2x3 − 3x2 .
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.
-4
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
(b) The value of c corresponding to y(0) =
-6
-4
3
2
-2
0
2
4
6
y
4
2
interval of definition we find dy/dx for
2y 3 − 6y 2 + 2x3 − 3x2 = −
x
0
x
0
27
.
4
-2
Using implicit differentiation we get y = (x − x2 )/(y 2 − 2y),
which is infinite when y = 0 and y = 2. Letting y = 0 in
-4
-2
0
2
4
6
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).
40
2.3
(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
y
4
2
x
0
-2
2y 3 − 6y 2 + 2x3 − 3x2 = −40.
Linear Equations
-4
Using implicit differentiation we get y = (x − x )/(y − 2y),
which is infinite when y = 0 and y = 2. Letting y = 0 in
2y 3 − 6y 2 + 2x3 − 3x2 = −40 and using a CAS to solve for x
2
2
-6
-8
-4
-2
0
2
4
6
8
10
we get x = −2.29551. The largest interval of definition is approximately (−2.29551, ∞).
EXERCISES 2.3
Linear Equations
d −5x 1. For y − 5y = 0 an integrating factor is e− 5 dx = e−5x so that
y = 0 and y = ce5x for −∞ < x < ∞.
e
dx
d 2x 2. For y + 2y = 0 an integrating factor is e 2 dx = e2x so that
e y = 0 and y = ce−2x for −∞ < x < ∞.
dx
The transient term is ce−2x .
d x
3. For y +y = e3x an integrating factor is e dx = ex so that
[e y] = e4x and y = 14 e3x +ce−x for −∞ < x < ∞.
dx
The transient term is ce−x .
d 4x 4. For y + 4y = 43 an integrating factor is e 4 dx = e4x so that
e y = 43 e4x and y = 13 + ce−4x for
dx
−∞ < x < ∞. The transient term is ce−4x .
2
3
d x3 3
3
5. For y + 3x2 y = x2 an integrating factor is e 3x dx = ex so that
e y = x2 ex and y = 13 + ce−x for
dx
3
−∞ < x < ∞. The transient term is ce−x .
2
d x2 2
2
6. For y + 2xy = x3 an integrating factor is e 2x dx = ex so that
e y = x3 ex and y = 12 x2 − 12 + ce−x
dx
2
for −∞ < x < ∞. The transient term is ce−x .
1
d
1
1
1
c
7. For y + y = 2 an integrating factor is e (1/x)dx = x so that
[xy] = and y = ln x + for 0 < x < ∞.
x
x
dx
x
x
x
d
8. For y − 2y = x2 + 5 an integrating factor is e− 2 dx = e−2x so that
e−2x y = x2 e−2x + 5e−2x and
dx
2x
y = − 12 x2 − 12 x − 11
for −∞ < x < ∞.
4 + ce
d 1
1
1
− (1/x)dx
9. For y − y = x sin x an integrating factor is e
y = sin x and y = cx − x cos x for
= so that
x
x
dx x
0 < x < ∞.
3
2
d 2 10. For y + y = an integrating factor is e (2/x)dx = x2 so that
x y = 3x and y = 32 +cx−2 for 0 < x < ∞.
x
x
dx
4
d 4 11. For y + y = x2 −1 an integrating factor is e (4/x)dx = x4 so that
x y = x6 −x4 and y = 17 x3 − 15 x+cx−4
x
dx
for 0 < x < ∞.
41
2.3
Linear Equations
x
d y = x an integrating factor is e− [x/(1+x)]dx = (x+1)e−x so that
(x + 1)e−x y = x(x+1)e−x
(1 + x)
dx
2x + 3
cex
and y = −x −
+
for −1 < x < ∞.
x+1
x+1
ex
2
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 and
x
x
dx
ce−x
ce−x
1 ex
+
for
0
<
x
<
∞.
The
transient
term
is
.
y=
2 x2
x2
x2
1
1
d
14. For y + 1 +
y = e−x sin 2x an integrating factor is e [1+(1/x)]dx = xex so that
[xex y] = sin 2x and
x
x
dx
1
ce−x
y = − e−x cos 2x +
for 0 < x < ∞. The entire solution is transient.
2x
x
dx 4
d −4 −4
15. For
− x = 4y 5 an integrating factor is e− (4/y)dy = eln y = y −4 so that
y x = 4y and x = 2y 6 +cy 4
dy y
dy
12. For y −
for 0 < y < ∞.
dx 2
d 2 2
2
c
+ x = ey an integrating factor is e (2/y)dy = y 2 so that
y x = y 2 ey and x = ey − ey + 2 ey + 2
dy y
dy
y
y
y
c
for 0 < y < ∞. The transient term is 2 .
y
d
17. For y + (tan x)y = sec x an integrating factor is e tan x dx = sec x so that
[(sec x)y] = sec2 x and
dx
y = sin x + c cos x for −π/2 < x < π/2.
d
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
[(sin x) y] = sec2 x
dx
and y = sec x + c csc x for 0 < x < π/2.
x+2
d
2xe−x
19. For y +
y=
an integrating factor is e [(x+2)/(x+1)]dx = (x + 1)ex , so
[(x + 1)ex y] = 2x and
x+1
x+1
dx
16. For
x2 −x
c
e +
e−x for −1 < x < ∞. The entire solution is transient.
x+1
x+1
4
d 5
[4/(x+2)]dx
For y +
an
integrating
factor
is
e
= (x + 2)4 so that
y=
(x + 2)4 y = 5(x + 2)2
2
x+2
(x + 2)
dx
5
−1
−4
and y = (x + 2) + c(x + 2) for −2 < x < ∞. The entire solution is transient.
3
dr
For
+ r sec θ = cos θ an integrating factor is e sec θ dθ = eln | sec x+tan x| = sec θ + tan θ so that
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 .
dθ
2
dP
d t2 −t 2
For
P = (4t − 2)et −t and
+ (2t − 1)P = 4t − 2 an integrating factor is e (2t−1) dt = et −t so that
e
dt
dt
2
2
P = 2 + cet−t for −∞ < t < ∞. The transient term is cet−t .
1
d 3x e−3x
For y + 3 +
y=
an integrating factor is e [3+(1/x)]dx = xe3x so that
xe y = 1 and
x
x
dx
ce−3x
y = e−3x +
for 0 < x < ∞. The transient term is ce−3x /x.
x
2
x−1
x+1
d x−1
[2/(x2 −1)]dx
=
For y + 2
y =
an integrating factor is e
so that
y = 1 and
x −1
x−1
x+1
dx x + 1
(x − 1)y = x(x + 1) + c(x + 1) for −1 < x < 1.
y=
20.
21.
22.
23.
24.
42
2.3
Linear Equations
1
d
1
c
1
y = ex an integrating factor is e (1/x)dx = x so that
[xy] = ex and y = ex + for 0 < x < ∞.
x
x
dx
x
x
1
2−e
If y(1) = 2 then c = 2 − e and y = ex +
.
x
x
dx 1
1
d 1
− (1/y)dy
For
= so that
− x = 2y an integrating factor is e
x = 2 and x = 2y 2 +cy for 0 < y < ∞.
dy y
y
dy y
49
If y(1) = 5 then c = −49/5 and x = 2y 2 − y.
5
d Rt/L E Rt/L
di R
E
i = e
and
For
+ i=
an integrating factor is e (R/L) dt = eRt/L so that
e
dt
L
L
dt
L
E
E
E
i=
+ ce−Rt/L for −∞ < t < ∞. If i(0) = i0 then c = i0 − E/R and i =
+ i0 −
e−Rt/L .
R
R
R
dT
d −kt
For
−kT = −Tm k an integrating factor is e (−k)dt = e−kt so that
[e T ] = −Tm ke−kt and T = Tm +cekt
dt
dt
for −∞ < t < ∞. If T (0) = T0 then c = T0 − Tm and T = Tm + (T0 − Tm )ekt .
1
d
ln x
For y +
y =
an integrating factor is e [1/(x+1)]dx = x + 1 so that
[(x + 1)y] = ln x and
x+1
x+1
dx
x
x
c
x
x
21
y=
ln x −
+
for 0 < x < ∞. If y(1) = 10 then c = 21 and y =
ln x −
+
.
x+1
x+1 x+1
x+1
x+1 x+1
d
For y + (tan x)y = cos2 x an integrating factor is e tan x dx = eln | sec x| = sec x so that
[(sec x) y] = cos x
dx
and y = sin x cos x + c cos x for −π/2 < x < π/2. If y(0) = −1 then c = −1 and y = sin x cos x − cos x.
25. For y +
26.
27.
28.
29.
30.
31. For y + 2y = f (x) an integrating factor is e2x so that
1 2x
e + c1 , 0 ≤ x ≤ 3
ye2x = 2
c2 ,
x > 3.
1
y
If y(0) = 0 then c1 = −1/2 and for continuity we must have c2 = 12 e6 − 12
so that
1
y=
−2x
),
2 (1 − e
1 6
2 (e
x
5
0≤x≤3
− 1)e−2x , x > 3.
32. For y + y = f (x) an integrating factor is ex so that
x
0≤x≤1
e + c1 ,
x
ye =
x
−e + c2 , x > 1.
1
If y(0) = 1 then c1 = 0 and for continuity we must have c2 = 2e so that
1,
0≤x≤1
y=
1−x
2e
− 1, x > 1.
33. For y + 2xy = f (x) an integrating factor is ex so that
1 x2
e + c1 , 0 ≤ x ≤ 1
x2
ye = 2
x > 1.
c2 ,
y
5
x
3
x
-1
2
If y(0) = 2 then c1 = 3/2 and for continuity we must have c2 = 12 e +
2
y
3
2
so that
1
y=
2
+ 32 e−x ,
2
1
2e
+
3
2
0≤x≤1
e−x ,
2
x > 1.
43
2.3
Linear Equations
34. For
y +
 x

 1 + x2 , 0 ≤ x ≤ 1
1
2x
y=

1 + x2
 −x , x > 1,
1 + x2
y
5
x
-1
an integrating factor is 1 + x2 so that
1 2
0≤x≤1
2 x + c1 ,
2
1+x y =
− 12 x2 + c2 , x > 1.
If y(0) = 0 then c1 = 0 and for continuity we must have c2 = 1 so that
1
1


 2 − 2 (1 + x2 ) , 0 ≤ x ≤ 1
y=

3
1


− , x > 1.
2 (1 + x2 ) 2
35. We first solve the initial-value
problem y + 2y = 4x, y(0) = 3 on the interval [0, 1].
2 dx
The integrating factor is e
= e2x , so
d 2x
[e y] = 4xe2x
dx
e2x y =
y
20
15
4xe2x dx = 2xe2x − e2x + c1
10
y = 2x − 1 + c1 e−2x .
5
−2x
Using the initial condition, we find y(0) = −1+c1 = 3, so c1 = 4 and y = 2x−1+4e
,
−2
−2
0 ≤ x ≤ 1. Now, since y(1) = 2−1+4e = 1+4e , 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
3
x
= e−2 ln x = x−2 , so
4
d −2
[x y] = 4xx−2 =
dx
x
4
x−2 y =
dx = 4 ln x + c2
x
y = 4x2 ln x + c2 x2 .
(We use ln x instead of ln |x| because x > 1.) Using the initial condition we find y(1) = c2 = 1 + 4e−2 , so
y = 4x2 ln x + (1 + 4e−2 )x2 , x > 1. Thus, the solution of the original initial-value problem is
y=
2x − 1 + 4e−2x ,
0≤x≤1
4x2 ln x + (1 + 4e−2 )x2 , x > 1.
See Problem 42 in this section.
36. For y + ex y = 1 an integrating factor is ee . Thus
x
x
x
d ex e y = ee and ee y =
dx
x x
t
x
From y(0) = 1 we get c = e, so y = e−e 0 ee dt + e1−e .
x
t
ee dt + c.
0
When y + ex y = 0 we can separate variables and integrate:
dy
= −ex dx
y
and
44
ln |y| = −ex + c.
2.3
Linear Equations
Thus y = c1 e−e . From y(0) = 1 we get c1 = e, so y = e1−e .
x
x
When y + ex y = ex we can see by inspection that y = 1 is a solution.
37. An integrating factor for y − 2xy = 1 is e−x . Thus
2
2
d −x2
y] = e−x
[e
dx
x
e−x y =
2
e−t dt =
2
0
√
y=
√
π
erf(x) + c
2
2
π x2
e erf(x) + cex .
2
√
√
From y(1) = ( π/2)e erf(1) + ce = 1 we get c = e−1 − 2π erf(1). The solution of the initial-value problem is
√
√
2
π x2
π
−1
y=
e erf(x) + e −
erf(1) ex
2
2
√
π x2
x2 −1
=e
+
e (erf(x) − erf(1)).
2
38. We want 4 to be a critical point, so we use y = 4 − y.
39. (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 1.1 are not satisfied and the initial-value problem does
not have a unique solution.
(b) The differential equation has no solution satisfying y(0) = y0 , y0 > 0.
(c) In this case, since x0 > 0, Theorem 1.1 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 .
40. On the interval (−3, 3) the integrating factor is
2
e x dx/(x −9) = e−
and so
x dx/(9−x2 )
d 9 − x2 y = 0
dx
1
2
= e 2 ln(9−x
and y = √
)
=
9 − x2
c
.
9 − x2
41. 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.
42. 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.
43. (a) Differentiating yc = c/x3 we get
3c
3 c
3
=−
= − yc
4
3
x
x x
x
so a differential equation with general solution yc = c/x3 is xy + 3y = 0. Now
yc = −
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, ∞).
45
2.3
Linear 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
y
3
interval of definition is (0, ∞). The initial-value 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
5
of y = x3 − 2/x3 .
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 .
44. Since e P (x)dx+c = ec e P (x)dx = c1 e P (x)dx , we would have
c1 e P (x)dx y = c2 + c1 e P (x)dx f (x) dx and e P (x)dx y = c3 + e P (x)dx f (x) dx,
which is the same as (6) in the text.
45. We see by inspection that y = 0 is a solution.
46. 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
x0 λ1 (λ2 −λ1 )t
eλ2 t y =
e
+ c2
λ2 − λ 1
x0 λ1 −λ1 t
y=
e
+ c2 e−λ2 t .
λ2 − λ 1
From y(0) = y0 we obtain c2 = (y0 λ2 − y0 λ1 − x0 λ1 )/(λ2 − λ1 ). The solution is
y=
47. Writing the differential equation as
x0 λ1 −λ1 t y0 λ2 − y0 λ1 − x0 λ1 −λ2 t
e
+
e
.
λ2 − λ 1
λ2 − λ1
dE
1
+
E = 0 we see that an integrating factor is et/RC . Then
dt
RC
d t/RC
E] = 0
[e
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 .
46
2.3
Linear Equations
48. (a) An integrating factor for y − 2xy = −1 is e−x . Thus
2
2
d −x2
y] = −e−x
[e
dx
x
√
π
erf(x) + c.
2
0
√
√
From y(0) = π/2, and noting that erf(0) = 0, we get c = π/2. Thus
√
√ √
√
2
π
π
π x2
π x2
y = ex −
erf(x) +
=
e (1 − erf(x)) =
e erfc(x).
2
2
2
2
−x2
e
y=−
−t2
e
dt = −
y
(b) Using a CAS we find y(2) ≈ 0.226339.
5
5
x
49. (a) An integrating factor for
y +
is x2 . Thus
2
10 sin x
y=
x
x3
d 2
sin x
[x y] = 10
dx
x
x
sin t
2
x y = 10
dt + c
t
0
y = 10x−2 Si(x) + cx−2 .
From y(1) = 0 we get c = −10Si(1). Thus
y = 10x−2 Si(x) − 10x−2 Si(1) = 10x−2 (Si(x) − Si(1)).
(b)
y
2
1
1
2
3
4
5
x
-1
-2
-3
-4
-5
(c) 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).
x
−
sin t2 dt
50. (a) The integrating factor for y − (sin x2 )y = 0 is e 0
. Then
x
d − sin t2 dt
y] = 0
[e 0
dx
x
−
sin t2 dt
e 0
y = c1
x
sin t2 dt
y = c1 e 0
.
47
2.3
Linear Equations
Letting t =
π/2 u we have dt =
π/2 du and
√
x
2/π x
π π
π
2
2
2
u du =
S
x
sin t dt =
sin
2 0
2
2
π
0
√
√
√
√
so y = c1 e π/2 S( 2/π x) . Using S(0) = 0 and y(0) = c1 = 5 we have y = 5e π/2 S( 2/π x) .
(b)
y
10
5
-10
-5
x
5
10
(c) From the graph we see that as x → ∞, y(x) oscillates with decreasing amplitudes approaching 9.35672.
√
Since limx→∞ 5S(x) = 12 , we have limx→∞ y(x) = 5e π/8 ≈ 9.357, and since limx→−∞ S(x) = − 12 , we
√
have limx→−∞ y(x) = 5e− π/8 ≈ 2.672.
(d) 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).
EXERCISES 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.
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(y), h (x) = 1 + ln x, and h(y) = x ln x. A solution is −y + y ln x + x ln x = c.
48
2.4 Exact Equations
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 we obtain
3
f = x − 3 ln |x| + xy + h(y), h (y) = 1 − , and h(y) = y − 3 ln |y|. A solution is x + y + xy − 3 ln |xy| = c.
y
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
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.
we
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 .
From fx = 2y sin x cos x − y + 2y 2 exy we obtain f = y sin2 x − xy + 2exy + h(y), h (y) = 0, and h(y) = 0. A
2
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.
2
20. Let M = 1/t + 1/t2 − y/ t2 + y 2 and N = yey + t/ t2 + y 2 so that My = y 2 − t2 / t2 + y 2 = Nt . From
1
t
2
2
2
ft = 1/t + 1/t − y/ t + y we obtain f = ln |t| − − arctan
+ h(y), h (y) = yey , and h(y) = yey − ey .
t
y
A solution is
1
t
ln |t| − − arctan
+ yey − ey = c.
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 − y. 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.
49
2.4
Exact Equations
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 obtain f =
t2
+ h(y),
4y 4
3
3
t2
3
,
and
h(y)
=
−
.
The
solution
is
−
= c. If y(1) = 1 then c = −5/4 and a solution of the
y3
2y 2
4y 4 2y 2
2
t
3
5
initial-value problem is
− 2 =− .
4
4y
2y
4
h (y) =
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 fx = y 2 + y sin x we
−1
obtain f = xy 2 −y cos x+h(y), h (y) =
, and h(y) = − tan−1 y. The solution is xy 2 −y cos x−tan−1 y = c.
1 + y2
π
If y(0) = 1 then c = −1 − π/4 and a solution of the initial-value problem is xy 2 − y cos x − tan−1 y = −1 − .
4
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 .
2y 2
From fx = x2 + 2xy + y 2 − 2y 2 /(x + y)2 we obtain f = x +
+ h(y), h (y) = −1, and h(y) = −y. A
x+y
solution of the differential equation is 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 (y) = 0, and h(y) = 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 2dy/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 . From fx = 10e3x − 6ye3x + 1 we obtain f =
10 3x
10 3x
3x
3x
3 e − 2ye + x + h(y), h (y) = 0, and h(y) = 0. A solution of the differential equation is 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.
50
2.4 Exact Equations
2
37. We note that (My − Nx )/N = 2x/(4 + x2 ), so an integrating factor is 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.
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
−
y2
2
2
+ ln |1 + x| = c.
+
+
2(1 + x)2
(1 + x)2
(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
dx
dx
and
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.
y
4
(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 .
2
-4
Observe in the figure that y1 (0) = −2 and y2 (1) = 1.
-2
y2
2
-2
4
x
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
y
42. (a) Since fy = N (x, y) = xexy +2xy+1/x we obtain f = exy +xy 2 + +h(x) so that fx = yexy +y 2 − 2 +h (x).
x
x
y
Let M (x, y) = yexy + y 2 − 2 .
x
1
−1
(b) Since fx = M (x, y) = y 1/2 x−1/2 + x x2 + y
we obtain f = 2y 1/2 x1/2 + ln x2 + y + g(y) so that
2
1 2
1 2
−1
−1
−1/2 1/2
−1/2 1/2
x +
+ g (x). Let N (x, y) = y
x +
.
x +y
x +y
fy = y
2
2
43. First note that
x
y
d
x2 + y 2 = dx + dy.
2
2
2
x +y
x + y2
Then x dx + y dy = x2 + y 2 dx becomes
x
y
dx + dy = d
x2 + y 2 = dx.
x2 + y 2
x2 + y 2
51
2.4
Exact Equations
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.
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 we have (v 2 − 32x)dx + xv dv = 0. This is not an exact form, but µ(x) = x is an
integrating factor. Multiplying by x we get (xv 2 − 32x2 )dx + x2 v dv = 0. This form is the total differential
1 2 2
32 3
3
of u = 12 x2 v 2 − 32
3 x , so an implicit solution is 2 x v − 3 x = c. Letting x = 3 and v = 0 we find c = −288.
Solving for v we get
v=8
x
9
− 2.
3 x
(b) The chain leaves the platform when x = 8, so the velocity at this time is
8
9
v(8) = 8
−
≈ 12.7 ft/s.
3 64
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
My =
2x3 − 8xy 2
= Nx ,
(x2 + y 2 )3
so the differential equation is exact. Then we have
∂f
2xy
= M (x, y) = 2
= 2xy(x2 + y 2 )−2
∂x
(x + y 2 )2
y
f (x, y) = −y(x2 + y 2 )−1 + g(y) = − 2
+ g(y)
x + y2
∂f
y 2 − x2
y 2 − x2
+ g (y) = N (x, y) = 1 + 2
.
= 2
2
2
∂y
(x + y )
(x + y 2 )2
y
Thus, g (y) = 1 and g(y) = y. The solution is y − 2
= c. When c = 0 the solution is x2 + y 2 = 1.
x + y2
(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.
52
2.5
-3
-2
Solutions by Substitutions
y
3
x
3
y
3
2
2
2
1
1
1
-1
1
2
3
x
-1.5
-10.5 0.511.5y
-3
-2
-1
1
-1
-1
-1
-2
-2
-2
-3
-3
-3
EXERCISES 2.5
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. Letting y = ux we have
(x + ux) dx + x(u dx + x du) = 0
(1 + 2u) dx + x du = 0
dx
du
+
=0
x
1 + 2u
1
ln |x| + ln |1 + 2u| = c
2
y
x2 1 + 2
= c1
x
x2 + 2xy = c1 .
53
2
3
2.5
Solutions by Substitutions
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
v dv
dy
+
=0
2
(v − 1)
y
1
ln |v − 1| −
+ ln |y| = c
v−1
x
1
ln
−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
dv
dy
−
=0
v+2
y
ln |v + 2| − ln |y| = c
x
+ 2 − ln |y| = c
y
ln
x + 2y = c1 y 2 .
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
dx
du
+
=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).
54
2.5
Solutions by Substitutions
7. Letting y = ux we have
(ux − x) dx − (ux + x)(u dx + x du) = 0
u2 + 1 dx + x(u + 1) du = 0
dx
u+1
+ 2
du = 0
x
u +1
1
ln u2 + 1 + tan−1 u = c
2
2
y
y
ln x2
+
1
+ 2 tan−1 = c1
x2
x
ln |x| +
ln x2 + y 2 + 2 tan−1
y
= c1 .
x
8. Letting y = ux we have
(x + 3ux) dx − (3x + ux)(u dx + x du) = 0
u2 − 1 dx + x(u + 3) du = 0
dx
u+3
+
du = 0
x
(u − 1)(u + 1)
ln |x| + 2 ln |u − 1| − ln |u + 1| = c
x(u − 1)2
= c1
u+1
y
y
2
x
− 1 = c1
+1
x
x
(y − x)2 = c1 (y + x).
9. Letting y = ux we have
−ux dx + (x +
√
u x)(u dx + x du) = 0
√
(x + x u ) du + xu3/2 dx = 0
dx
1
−3/2
u
du +
=0
+
u
x
2
2
−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
ux +
x2 − (ux)2 dx − x(udx + xdu) du = 0
x2 − u2 x2 dx − x2 du = 0
x 1 − u2 dx − x2 du = 0,
(x > 0)
dx
du
−√
=0
x
1 − u2
ln x − sin−1 u = c
sin−1 u = ln x + c1
55
2.5
Solutions by Substitutions
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.
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
dx
u du
=0
−
x
1 + u2
1
ln |x| − 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.
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
dv
dy
+
=0
v ln v
y
ln |ln |v|| + ln |y| = c
y ln
56
x
= c1 .
y
2.5
Using y(1) = e we find c1 = −e. The solution of the initial-value problem is y ln
15. From y +
Solutions by Substitutions
x
= −e.
y
1
dw
1
3
3
y = y −2 and w = y 3 we obtain
+ w = . An integrating factor is x3 so that x3 w = x3 + c
x
x
dx
x
x
or y 3 = 1 + cx−3 .
16. From y − y = ex y 2 and w = y −1 we obtain
or y −1 = − 12 ex + ce−x .
dw
+ w = −ex . An integrating factor is ex so that ex w = − 12 e2x + c
dx
17. From y + y = xy 4 and w = y −3 we obtain
xe−3x + 13 e−3x + c or y −3 = x +
1
3
+ ce3x .
dw
− 3w = −3x. An integrating factor is e−3x so that e−3x w =
dx
1
dw
1
18. From y − 1 +
y = y 2 and w = y −1 we obtain
+ 1+
w = −1. An integrating factor is xex so that
x
dx
x
1
c
xex w = −xex + ex + c or y −1 = −1 + + e−x .
x x
1
dw 1
1
1
19. From y − y = − 2 y 2 and w = y −1 we obtain
+ w = 2 . An integrating factor is t so that tw = ln t + c
t
t
dt
t
t
1
c
t
or y −1 = ln t + . Writing this in the form
= ln t + c, we see that the solution can also be expressed in the
t
t
y
form et/y = c1 t.
2
dw
−2t
2t
2t
w=
. An integrating factor is
y=
y 4 and w = y −3 we obtain
−
3 (1 + t2 )
3 (1 + t2 )
dt
1 + t2
1 + t2
1
w
1
so that
=
+ c or y −3 = 1 + c 1 + t2 .
1 + t2
1 + t2
1 + t2
20. From y +
21. From y −
2
dw
3
6
9
y = 2 y 4 and w = y −3 we obtain
+ w = − 2 . An integrating factor is x6 so that
x
x
dx
x
x
x6 w = − 95 x5 + c or y −3 = − 95 x−1 + cx−6 . If y(1) =
1
2
then c =
49
5
and y −3 = − 95 x−1 +
49 −6
.
5 x
dw
3
3
+ w = . An integrating factor is e3x/2 so that e3x/2 w =
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
e3x/2 + c or y 3/2
du
1
du = dx. Thus tan−1 u = x + c or
− 1 = u2 or
dx
1 + u2
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
and (x + y)2 = 2x + c1 .
du
1−u
−1 =
or u du = dx. Thus 12 u2 = x + c or u2 = 2x + c1 ,
dx
u
25. Let u = x + y so that du/dx = 1 + dy/dx. Then
du
− 1 = tan2 u or cos2 u du = dx. Thus 12 u +
dx
1
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 .
26. Let u = x + y so that du/dx = 1 + dy/dx. Then
(1 − sin u)/(1 − sin u) we have
du
1
− 1 = sin u or
du = dx. Multiplying by
dx
1 + sin u
1 − sin u
du = dx or (sec2 u − sec u tan u)du = dx. Thus tan u − sec u = x + c or
cos2 u
tan(x + y) − sec(x + y) = x + c.
57
2.5
Solutions by Substitutions
√
√
du
1
27. Let u = y − 2x + 3 so that du/dx = dy/dx − 2. Then
+ 2 = 2 + u or √ du = dx. Thus 2 u = x + c and
dx
u
√
2 y − 2x + 3 = x + c.
du
28. Let u = y − x + 5 so that du/dx = dy/dx − 1. Then
+ 1 = 1 + eu or e−u du = dx. Thus −e−u = x + c and
dx
−ey−x+5 = x + c.
du
1
29. Let u = x + y so that du/dx = 1 + dy/dx. Then
− 1 = cos u and
du = dx. Now
dx
1 + cos u
1
1 − cos u
1 − cos u
= csc2 u − csc u cot u
=
=
1 + cos u
1 − cos2 u
sin2 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 + 2 − 1.
30. Let u = 3x + 2y so that du/dx = 3 + 2 dy/dx. Then
du
2u
5u + 6
u+2
=3+
=
and
du = dx. Now by
dx
u+2
u+2
5u + 6
long division
u+2
1
4
= +
5u + 6
5 25u + 30
so we have
and
1
5u
+
4
25
1
4
+
5 25u + 30
du = dx
ln |25u + 30| = x + c. Thus
1
4
(3x + 2y) +
ln |75x + 50y + 30| = x + c.
5
25
Setting x = −1 and y = −1 we obtain c =
or
4
25
ln 95. The solution is
1
4
4
(3x + 2y) +
ln |75x + 50y + 30| = x +
ln 95
5
25
25
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
dy
xα M (1, y/x)
M (1, y/x)
= 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
xy
dy
= 5x2 − 2y 2
dx
and divide by xy, so that
dy
x
y
=5 −2 .
dx
y
x
58
2.5
We then identify
F
y
x
=5
y −1
x
−2
y
x
Solutions by Substitutions
.
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 + c
or c = − 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
π
x
π
− ≤ ln ≤
2
5
2
x
−π/2
e
≤ ≤ eπ/2
5
−π/2
5e
≤ x ≤ 5eπ/2 .
y
20
15
10
5
5
10
15
20
x
The interval of definition of the solution is approximately
[1.04, 24.05].
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
dy
dy1
du
=
+
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 .
dw
1
4
(b) Identify P (x) = −4/x , Q(x) = −1/x, and R(x) = 1. Then
+ − +
w = −1. An integrating
dx
x x
−1
2
factor is x3 so that x3 w = − 14 x4 + c or u = − 14 x + cx−3
. Thus, y = + u.
x
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.
2
An integrating factor is e−
dx/x
= 1/x, so that (using integration by parts)
d 1 ln x
u
1 ln x
u = 2 and
=− −
+ c.
dx x
x
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
59
ecx−1
.
x
2.5
Solutions by Substitutions
and let u = v 2 or v = u1/2 . Then
du
dv
1
= u−1/2
,
dx
2
dx
and substituting into the differential equation, we have
1 −1/2 du 1 1/2
= 32u−1/2
u
+ u
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
x +c
3
or
v2 =
64
c
x+ 2 .
3
x
38. Write the differential equation as dP/dt − aP = −bP 2 and let u = P −1 or P = u−1 . Then
dp
du
= −u−2
,
dt
dt
and substituting into the differential equation, we have
−u−2
du
− au−1 = −bu−2
dt
or
The latter differential equation is linear with integrating factor e
du
+ au = b.
dt
a dt
= eat , so
d at
[e u] = beat
dt
and
b at
e +c
a
b
eat P −1 = eat + c
a
b
P −1 = + ce−at
a
1
a
P =
=
.
−at
b/a + ce
b + c1 e−at
eat u =
EXERCISES 2.6
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,
60
2.6
and
A Numerical Method
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
ln |y| = x + c.
and
Thus y = c1 ex and, using y(0) = 1, we find c = 1, so y = ex is the solution of the initial-value problem.
h=0.1
xn
0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
h=0.05
yn
1.0000
1.1000
1.2100
1.3310
1.4641
1.6105
1.7716
1.9487
2.1436
2.3579
2.5937
Actual
Value
1.0000
1.1052
1.2214
1.3499
1.4918
1.6487
1.8221
2.0138
2.2255
2.4596
2.7183
% Rel .
Abs .
Error
Error
0.0000 0.00
0.0052 0.47
0.0114 0.93
0.0189 1.40
0.0277 1.86
0.0382 2.32
0.0506 2.77
0.0650 3.23
0.0820 3.68
0.1017 4.13
0.1245 4.58
xn
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
yn
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
61
Actual
Value
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
% Rel .
Abs .
Error
Error
0.0000 0.00
0.0013 0.12
0.0027 0.24
0.0042 0.36
0.0059 0.48
0.0077 0.60
0.0098 0.72
0.0120 0.84
0.0144 0.96
0.0170 1.08
0.0198 1.20
0.0229 1.32
0.0263 1.44
0.0299 1.56
0.0338 1.68
0.0381 1.80
0.0427 1.92
0.0476 2.04
0.0530 2.15
0.0588 2.27
0.0650 2.39
2.6
A Numerical Method
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
2
2
h=0.1
xn
1.00
1.10
1.20
1.30
1.40
1.50
5.
7.
yn
1.0000
1.2000
1.4640
1.8154
2.2874
2.9278
Actual
Value
1.0000
1.2337
1.5527
1.9937
2.6117
3.4903
Abs .
Error
0.0000
0.0337
0.0887
0.1784
0.3243
0.5625
xn
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
xn
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
xn
yn
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
1.0000
1.1000
1.2155
1.3492
1.5044
1.6849
1.8955
2.1419
2.4311
2.7714
3.1733
yn
0.5000
0.5125
0.5232
0.5322
0.5395
0.5452
0.5496
0.5527
0.5547
0.5559
0.5565
% Rel .
Abs .
Error
Error
0.0000 0.00
0.0079 0.72
0.0182 1.47
0.0314 2.27
0.0483 3.11
0.0702 4.00
0.0982 4.93
0.1343 5.90
0.1806 6.92
0.2403 7.98
0.3171 9.08
h=0.05
yn
1.0000
1.1000
1.2220
1.3753
1.5735
1.8371
h=0.1
xn
0.00
0.10
0.20
0.30
0.40
0.50
62
Actual
Value
1.0000
1.1079
1.2337
1.3806
1.5527
1.7551
1.9937
2.2762
2.6117
3.0117
3.4903
h=0.1
xn
0.00
0.10
0.20
0.30
0.40
0.50
8.
h=0.05
yn
0.5000
0.5250
0.5431
0.5548
0.5613
0.5639
% Rel .
Error
0.00
2.73
5.71
8.95
12.42
16.12
6.
h=0.05
yn
0.0000
0.1000
0.1905
0.2731
0.3492
0.4198
h=0.1
xn
0.00
0.10
0.20
0.30
0.40
0.50
is the solution of the initial-value problem.
h=0.05
h=0.1
xn
0.00
0.10
0.20
0.30
0.40
0.50
−1
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
h=0.05
yn
1.0000
1.1000
1.2159
1.3505
1.5072
1.6902
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
1.1039
1.1619
1.2245
1.2921
1.3651
1.4440
1.5293
1.6217
1.7219
2.6
9.
h=0.1
10.
h=0.05
xn
1.00
1.10
1.20
1.30
1.40
1.50
yn
1.0000
1.0000
1.0191
1.0588
1.1231
1.2194
xn
1.00
1.05
1.10
1.15
1.20
1.25
1.30
1.35
1.40
1.45
1.50
h=0.1
yn
1.0000
1.0000
1.0049
1.0147
1.0298
1.0506
1.0775
1.1115
1.1538
1.2057
1.2696
xn
0.00
0.10
0.20
0.30
0.40
0.50
A Numerical Method
h=0.05
yn
0.5000
0.5250
0.5499
0.5747
0.5991
0.6231
xn
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
yn
0.5000
0.5125
0.5250
0.5375
0.5499
0.5623
0.5746
0.5868
0.5989
0.6109
0.6228
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
Euler
2
4
6
8
y
7
6
5
4
3
2
1
RK4
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
h=0.1
y
h=0.05
y
6
y
6
RK4
5
4
6
RK4
5
Euler
3
Euler
4
4
3
3
2
2
2
1
1
1
1
2
3
4
5
x
1
2
3
4
RK4
5
5
x
Euler
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 above in Problem 11.
63
2.6
A Numerical Method
h=0.25
h=0.1
y
10
y
10
RK4
8
RK4
8
6
6
4
4
Euler
2
Euler
2
0.2
0.4
0.6
0.8
1
x
0.2
0.4
0.6
0.8
1
x
EXERCISES 2.7
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
P (10) = kP (10) =
1
(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 P (10) = 500e10k = 575. Solving for k, we get
1
1
k = 10
ln 575
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
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
−3/7
2000
2000
−7/3
=
P
and
P
=
≈ 201.
0
0
40010/3
40010/3
64
2.7
Linear Models
5. Let A = A(t) be the amount of lead present at time t. From dA/dt = kA and A(0) = 1 we obtain A = ekt .
Using A(3.3) = 1/2 we find k =
1
3.3
ln(1/2). When 90% of the lead has decayed, 0.1 grams will remain. Setting
A(t) = 0.1 we have et(1/3.3) ln(1/2) = 0.1, so
t
1
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 =
T = −(ln 2)/k.
1
2 A0
and solving for t we obtain the half-life
(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(3) = 0.25I0 , then I = I0 ekt ,
k = 13 ln 0.25, and I(15) = 0.00098I0 .
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
11. Assume that A = A0 ekt and k = −0.00012378. If A(t) = 0.145A0 then t ≈15,600 years.
12. From Example 3 in the text, the amount of carbon present at time t is A(t) = A0 e−0.00012378t . Letting t = 660
and solving for A0 we have A(660) = A0 e−0.0001237(660) = 0.921553A0 . Thus, approximately 92% of the original
amount of C-14 remained in the cloth as of 1988.
13. Assume that dT /dt = k(T − 10) so that T = 10 + cekt . If T (0) = 70◦ and T (1/2) = 50◦ then c = 60 and
k = 2 ln(2/3) so that T (1) = 36.67◦ . If T (t) = 15◦ then t = 3.06 minutes.
14. Assume that dT /dt = k(T − 5) so that T = 5 + cekt . If T (1) = 55◦ and T (5) = 30◦ then k = − 14 ln 2 and
c = 59.4611 so that T (0) = 64.4611◦ .
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
65
2.7
Linear Models
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
9
ek2 =
19
9
k2 = ln
19
and T2 (t) = 100 − 19et ln(9/19) . Setting T2 (t) = 99.9 we have
100 − 19et ln(9/19) = 99.9
0.1
et ln(9/19) =
19
ln(0.1/19)
t=
≈ 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) = 70 we find
c = 70 − Tm , so T (t) = Tm + (70 − Tm )ekt . Using the given observations, we obtain
T
1
= Tm + (70 − Tm )ek/2 = 110
2
T (1) = Tm + (70 − Tm )ek = 145.
Then, from the first equation, ek/2 = (110 − Tm )/(70 − Tm ) and
k
k/2 2
e = (e
) =
110 − Tm
70 − Tm
2
=
145 − Tm
70 − Tm
(110 − Tm )2
= 145 − Tm
70 − Tm
2
2
12100 − 220Tm + Tm
= 10150 − 250Tm + Tm
Tm = 390.
The temperature in the oven is 390◦ .
18. (a) The initial temperature of the bath is Tm (0) = 60◦ , so in the short term the temperature of the chemical,
which starts at 80◦ , should decrease or cool. Over time, the temperature of the bath will increase toward
100◦ 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 100◦ .
(b) Adapting the model for Newton’s law of cooling, we have
dT
= −0.1(T − 100 + 40e−0.1t ),
dt
T
100
T (0) = 80.
90
Writing the differential equation in the form
80
dT
+ 0.1T = 10 − 4e−0.1t
dt
we see that it is linear with integrating factor e
70
0.1 dt
66
= e0.1t .
10
20
30
40
50
t
2.7
Linear Models
Thus
d 0.1t
[e T ] = 10e0.1t − 4
dt
e0.1t T = 100e0.1t − 4t + c
and
T (t) = 100 − 4te−0.1t + ce−0.1t .
Now T (0) = 80 so 100 + c = 80, c = −20 and
T (t) = 100 − 4te−0.1t − 20e−0.1t = 100 − (4t + 20)e−0.1t .
The thinner curve verifies the prediction of cooling followed by warming toward 100◦ . The wider curve shows
the temperature Tm of the liquid bath.
19. From dA/dt = 4 − A/50 we obtain A = 200 + ce−t/50 .
−t/50
A = 200 − 170e
If A(0) = 30 then c = −170 and
.
20. From dA/dt = 0 − A/50 we obtain A = ce−t/50 . If A(0) = 30 then c = 30 and A = 30e−t/50 .
21. From dA/dt = 10 − A/100 we obtain A = 1000 + ce−t/100 . If A(0) = 0 then c = −1000 and A(t) =
1000 − 1000e−t/100 .
22. From Problem 21 the number of pounds of salt in the tank at time t is A(t) = 1000 − 1000e−t/100 . The
concentration at time t is c(t) = A(t)/500 = 2 − 2e−t/100 . Therefore c(5) = 2 − 2e−1/20 = 0.0975 lb/gal and
limt→∞ c(t) = 2. Solving c(t) = 1 = 2 − 2e−t/100 for t we obtain t = 100 ln 2 ≈ 69.3 min.
23. From
dA
10A
2A
= 10 −
= 10 −
dt
500 − (10 − 5)t
100 − t
1
we obtain A = 1000 − 10t + c(100 − t)2 . If A(0) = 0 then c = − 10
. The tank is empty in 100 minutes.
24. With cin (t) = 2 + sin(t/4) lb/gal, the initial-value problem is
dA
1
t
+
A = 6 + 3 sin , A(0) = 50.
dt
100
4
The differential equation is linear with integrating factor e dt/100 = et/100 , so
d t/100
t
A(t)] = 6 + 3 sin
[e
et/100
dt
4
150 t/100
t
t
3750 t/100
et/100 A(t) = 600et/100 +
sin −
cos + c,
e
e
313
4
313
4
and
150
t
3750
t
A(t) = 600 +
sin −
cos + ce−t/100 .
313
4
313
4
Letting t = 0 and A = 50 we have 600 − 3750/313 + c = 50 and c = −168400/313. Then
A(t) = 600 +
150
t
3750
t
168400 −t/100
.
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 2.38(a) in the text.
67
2.7
Linear Models
A t
600
A t
600
500
500
400
400
300
300
200
200
100
100
50
100
25. From
150
200
250
300
t
100
200
300
400
500
600
t
dA
4A
2A
=3−
=3−
dt
100 + (6 − 4)t
50 + t
we obtain A = 50 + t + c(50 + t)−2 . If A(0) = 10 then c = −100,000 and A(30) = 64.38 pounds.
26. (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
dA
A
= (3 gal/min)(2 lb/gal) −
lb/gal (3 gal/min)
dt
400
3A
=6−
,
A(100) = 490.625.
400
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)
A
800
600
400
200
200
400
600
t
27. Assume L di/dt + Ri = E(t), L = 0.1, R = 50, and E(t) = 50 so that i =
and limt→∞ i(t) = 3/5.
68
3
5
+ ce−500t . If i(0) = 0 then c = −3/5
2.7
Linear Models
28. 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
sin ωt −
E0 Lω
2
L ω 2 + R2
cos ωt + ce−Rt/L .
E0 Lω
.
L2 ω 2 + R2
29. 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 .
Since i(0) = i0 we obtain c = i0 +
1
30. Assume R dq/dt + (1/C)q = E(t), R = 1000, C = 5 × 10−6 , and E(t) = 200. Then q = 1000
+ ce−200t and
1
i = −200ce−200t . If i(0) = 0.4 then c = − 500
, q(0.005) = 0.003 coulombs, and i(0.005) = 0.1472 amps. We have
q→
1
1000
as t → ∞.
31. 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
60 − 60e−t/10 ,
0 ≤ t ≤ 20
i(t) =
2
−t/10
60 e − 1 e
, t > 20.
32. Separating variables, we obtain
dq
dt
=
E0 − q/C
k1 + k2 t
q
1
−C ln E0 −
ln |k1 + k2 t| + c1
=
C
k2
(E0 − q/C)−C
= c2 .
(k1 + k2 t)1/k2
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
1/k
(k1 + k2 t) 2
k1 2
−1/k2
k1
q −C q0 −C
E0 −
= E0 −
C
C
k + k2 t
1/Ck2
q
q0
k1
E0 −
= E0 −
C
C
k + k2 t
1/Ck2
k1
q = E0 C + (q0 − E0 C)
.
k + k2 t
33. (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 mg −kt/m
v(t) =
.
+ v0 −
e
k
k
(b) As t → ∞ the limiting velocity is mg/k.
(c) From ds/dt = v and s(0) = 0 we obtain
s(t) =
mg
m
mg −kt/m m mg t−
v0 −
e
v0 −
.
+
k
k
k
k
k
34. (a) Integrating d2 s/dt2 = −g we get v(t) = ds/dt = −gt + c. From v(0) = 300 we find c = 300, and we are
given g = 32, so the velocity is v(t) = −32t + 300.
69
2.7
Linear Models
(b) Integrating again and using s(0) = 0 we get s(t) = −16t2 + 300t. The maximum height is attained when
v = 0, that is, at ta = 9.375. The maximum height will be s(9.375) = 1406.25 ft.
35. 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) = 300 we get
mg mg −kt/m
v(t) = −
+ 300 +
e
.
k
k
Integrating and using s(0) = 0 we find
mg
m
mg s(t) = −
t+
300 +
(1 − e−kt/m ).
k
k
k
Setting k = 0.0025, m = 16/32 = 0.5, and g = 32 we have
s(t) = 1,340,000 − 6,400t − 1,340,000e−0.005t
and
v(t) = −6,400 + 6,700e−0.005t .
The maximum height is attained when v = 0, that is, at ta = 9.162. The maximum height will be s(9.162) =
1363.79 ft, which is less than the maximum height in Problem 34.
36. 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 = 0.5, m = (125 + 35)/32 = 5, and g = 32, we have v(t) = 320(1 − e−0.1t ). Integrating,
we find s(t) = 320t + 3200e−0.1t + c1 . Solving s(0) = 0 for c1 we find c1 = −3200, therefore s(t) =
320t + 3200e−0.1t − 3200. At t = 15, when the parachute opens, v(15) = 248.598 and s(15) = 2314.02.
At this time the value of k changes to k = 10 and the new initial velocity is v0 = 248.598. 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 = 5, g = 32, and k = 10, this gives vp (t) = 16 + c2 e−2t . From v(0) = 248.598 we find c2 = 232.598,
so vp (t) = 16 + 232.598e−2t . Integrating, we get sp (t) = 16t − 116.299e−2t + c3 . Solving sp (0) = 0 for c3 ,
we find c3 = 116.299, so sp (t) = 16t − 116.299e−2t + 116.299. Twenty seconds after leaving the plane is five
seconds after the parachute opens. The skydiver’s velocity at this time is vp (5) = 16.0106 ft/s and she has
fallen a total of s(15) + sp (5) = 2314.02 + 196.294 = 2510.31 ft. Her terminal velocity is limt→∞ vp (t) = 16, so
she has very nearly reached her terminal velocity five seconds after the parachute opens. When the parachute
opens, the distance to the ground is 15,000 − s(15) = 15,000 − 2,314 = 12,686 ft. Solving sp (t) = 12,686 we
get t = 785.6 s = 13.1 min. Thus, it will take her approximately 13.1 minutes to reach the ground after her
parachute has opened and a total of (785.6 + 15)/60 = 13.34 minutes after she exits the plane.
37. (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
g
gr04
gρ k gρr04
v(t) = (r0 + bt) −
=
+
.
r
t
−
0
4b
4b(r0 + bt)3
4k
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 .
70
2.7
Linear Models
(c) If r = 0.007 ft when t = 10 s, then solving r(10) = 0.007 for k/ρ, we obtain k/ρ = −0.0003 and r(t) =
0.01 − 0.0003t. Solving r(t) = 0 we get t = 33.3, so the raindrop will have evaporated completely at
33.3 seconds.
38. Separating variables, we obtain dP/P = k cos t dt, so
ln |P | = k sin t + c
and
P = c1 ek sin t .
If P (0) = P0 , then c1 = P0 and P = P0 ek sin t .
39. (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 → ∞.
40. (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 limt→∞ A(t) = k1 M/(k1 + k2 ).
A
Since k2 > 0, the material will never be completely memorized and the larger k2 is, the
less the amount of material will be memorized over time.
M k1
k1 k2
(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
d (k1 +k2 )t A = k1 M e(k1 +k2 )t
e
dt
k1 M (k1 +k2 )t
e(k1 +k2 )t A =
e
+c
k1 + k2
A=
Using A(0) = 0 we find c = −
A→
k1 M
+ ce−(k1 +k2 )t .
k1 + k2
k1 M
k1 M 1 − e−(k1 +k2 )t . As t → ∞,
and A =
k1 + k2
k1 + k2
k1 M
.
k1 + k2
x
41. (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 limt→∞ x(t) = r/k.
r
k
71
2.7
Linear Models
(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
rêk
t
42. The bar removed from the oven has an initial temperature of 300◦ F and, after being removed from the oven,
approaches a temperature of 70◦ F. The bar taken from the room and placed in the oven has an initial temperature
of 70◦ F and approaches a temperature of 300◦ F in the oven. Since the two temperature functions are continuous
they must intersect at some time, t∗ .
43. (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,
0 ≤ t < 4, 6 ≤ t < 10, 12 ≤ t < 16



(4−t)/RC

12e
,
4
≤t<6


(10−t)/RC
, 10 ≤ t < 12
E(t) = 12e


(16−t)/RC

12e
, 16 ≤ t < 18



(22−t)/RC
, 22 ≤ t < 24.
12e
(b)
E
10
5
4
6
10 12
16 18
22 24
t
44. (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.
(ii) The model now becomes
dv
m
= 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.
(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
72
2.7
Linear Models
(b) (i) With m = 3 slugs, the differential equation is
3
dv
1
= (96) ·
dt
2
or
dv
= 16.
dt
Integrating the last equation gives v(t) = 16t + c1 . Since v(0) = 0, we have c1 = 0 and so v(t) = 16t.
(ii) With m = 3 slugs, the differential equation is
√
√
dv
3
3
1
3
= (96) · −
· (96) ·
dt
2
4
2
dv
= 4.
dt
or
In this case v(t) = 4t.
(iii) When the retarding force due to air resistance is taken into account, the differential equation for
velocity v becomes
3
√
√
dv
3
3 1
1
= (96) · −
· (96) ·
− v
dt
2
4
2
4
or
3
dv
1
= 12 − v.
dt
4
The last differential equation is linear and has solution v(t) = 48 + c1 e−t/12 . Since v(0) = 0, we find
c1 = −48, so v(t) = 48 − 48e−t/12 .
45. (a) (i) If s(t) is distance measured down the plane from the highest point, then ds/dt = v. Integrating
ds/dt = 16t gives s(t) = 8t2 + c2 . Using s(0) = 0 then gives c2 = 0. Now the length L of the plane is
L = 50/ sin 30◦ = 100 ft. The time it takes the box to slide completely down the plane is the solution of
s(t) = 100 or t2 = 25/2, so t ≈ 3.54 s.
(ii) Integrating ds/dt = 4t gives s(t) = 2t2 + c2 . Using s(0) = 0 gives c2 = 0, so s(t) = 2t2 and the solution
of s(t) = 100 is now t ≈ 7.07 s.
(iii) Integrating ds/dt = 48 − 48e−t/12 and using s(0) = 0 to determine the constant of integration, we
obtain s(t) = 48t + 576e−t/12 − 576. With the aid of a CAS we find that the solution of s(t) = 100, or
100 = 48t + 576e−t/12 − 576
or
0 = 48t + 576e−t/12 − 676,
is now t ≈ 7.84 s.
(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 = 32 cos 23◦ (tan 23◦ − 3/4) = −0.251493. Integration and the use of the initial condition gives
v(t) = −0.251493t + 1. When the box stops, v(t) = 0 or 0 = −0.251493t + 1 or t = 3.976254 s. From
s(t) = −0.125747t2 + t we find s(3.976254) = 1.988119 ft.
(d) With v0 > 0, v(t) = −0.251493t + v0 and s(t) = −0.125747t2 + v0 t. Because two real positive solutions
of the equation s(t) = 100, or 0 = −0.125747t2 + v0 t − 100, would be physically meaningless, we use
the quadratic formula and require that b2 − 4ac = 0 or v02 − 50.2987 = 0. From this last equality we
find v0 ≈ 7.092164 ft/s. For the time it takes the box to traverse the entire inclined plane, we must
have 0 = −0.125747t2 + 7.092164t − 100. Mathematica gives complex roots for the last equation: t =
28.2001 ± 0.0124458i. But, for
0 = −0.125747t2 + 7.092164691t − 100,
73
2.7
Linear Models
the roots are t = 28.1999 s and t = 28.2004 s. So if v0 > 7.092164, we are guaranteed that the box will slide
completely down the plane.
46. (a) We saw in part (b) of Problem 34 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 35 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.
EXERCISES 2.8
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 limt→∞ N (t) = 2000.
A graph of the solution is shown in part (b).
N
2000
0
(b) Separating variables and integrating we have
1
dN
1
=
−
dN = dt
N (1 − 0.0005N )
N
N − 2000
N
2000
1500
1000
and
500
ln N − ln(N − 2000) = t + c.
5
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 limt→∞ N = a/b = 50,000 and N (1) = 1000 we have a = 0.7033, b = 0.00014, and N =
50,000/(1 + 99e−0.7033t ) .
74
2.8
3. From dP/dt = P 10−1 − 10−7 P
Nonlinear Models
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
3.929a
a/b
=
3.929b + (a − 3.929b)e−at
1 + (a/3.929b − 1)e−at
c
=
,
1 + (c/3.929 − 1)e−at
P (t) =
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,
P (t) =
(b)
Year
1790
1800
1810
1820
1830
1840
1850
1860
1870
1880
1890
1900
1910
1920
1930
1940
1950
Census
Population
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
Predicted
Population
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
Error
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
197.274
.
1 + 49.21e−0.0313t
%
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
The model predicts a population of 159.0 million for 1960 and 167.8 million for 1970. The census populations
for these years were 179.3 and 203.3, respectively. The percentage errors are 12.8 and 21.2, respectively.
75
2.8
Nonlinear Models
5. (a) The differential equation is dP/dt = P (5 − P ) − 4. Solving P (5 − P ) − 4 = 0 for P we
P
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
(b) The differential equation is
P
dP
= P (5 − P ) − 4 = −(P 2 − 5P + 4) = −(P − 4)(P − 1).
dt
Separating variables and integrating, we obtain
4
1
3
dP
= −dt
(P − 4)(P − 1)
1/3
1/3
−
dP = −dt
P −4 P −1
t
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
(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
6. Solving P (5 − P ) −
if P0 <
5
2
25
4
= 0 for P we obtain the equilibrium solution P =
5
2
. For P =
5
2
, dP/dt < 0. Thus,
, 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 <
extinction is t = 4P0 /5(5 − 2P0 ).
5
2
we solve P (t) = 0 for t. We see that the time of
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
√
√
5
3
3
2P0 − 5
√
P (t) = +
tan tan−1
t .
−
2
2
2
3
76
2.8
Nonlinear Models
Solving P (t) = 0 for t we see that the time of extinction is
√
√
√ 2 √
t=
3 tan−1 (5/ 3 ) + 3 tan−1 (2P0 − 5)/ 3 .
3
P
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.
e
t
(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.
P
1êe
t
−bt
(c) From dP/dt = P (a − b ln P ) we obtain −(1/b) ln |a − b ln P | = t + c1 so that P = ea/b e−ce
. If P (0) = P0
then c = (a/b) − ln P0 .
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 grams. 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 grams 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 = −8Ah h /Aw , h(0) = H.
h
10
8
6
4
2
Separating variables and integrating we have
√
dh
8A
8A
√ = − h dt and 2 h = − h t + c.
A
Aw
h
w
√
Using h(0) = H we find c = 2 H , so the solution of the
500
√
initial-value problem is h(t) = (Aw H − 4Ah t)/Aw , where
√
Aw H − 4Ah t ≥ 0. Thus,
√
h(t) = (Aw H − 4Ah t)2 /A2w for 0 ≤ t ≤ Aw H/4Ah .
1000
1500
t
(b) Identifying H = 10, Aw = 4π, and Ah = π/576 we have h(t) = t2 /331,776 − ( 5/2 /144)t + 10. Solving
√
h(t) = 0 we see that the tank empties in 576 10 seconds or 30.36 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 − 4cAh t)2 /A2w . Solving h(t) = 0 with c = 0.6 and the values from Problem 11 we see that
the tank empties in 3035.79 seconds or 50.6 minutes.
77
2.8
Nonlinear Models
13. (a) Separating variables and integrating gives
6h3/2 dh = −5t
and
12 5/2
h
= −5t + c.
5
√
√
2/5
Using h(0) = 20 we find c = 1920 5 , so the solution of the initial-value problem is h(t) = 800 5− 25
.
12 t
√
Solving h(t) = 0 we see that the tank empties in 384 5 seconds or 14.31 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
π(2/12)2 √
2
Ah 2gh = −0.6
64h = − 3/2 .
= −c
dt
Aw
πh2 /3
5h
Separating variables and integrating gives
5h3/2 dh = −2 dt
and
2h5/2 = −2t + c.
Using h(0) = 9 we find c = 486, so the solution of the initial-value problem is h(t) = (243 − t)2/5 . Solving
h(t) = 0 we see that the tank empties in 24.3 seconds or 4.05 minutes.
14. When the height of the water is h,
the radius of the top of the water is
2
5 (20
− h) and
Aw = 4π(20 − h) /25. The differential equation is
2
√
√
dh
h
π(2/12)2
5
Ah 2gh = −0.6
64h = −
.
= −c
dt
Aw
4π(20 − h)2 /25
6 (20 − h)2
Separating variables and integrating we have
√
(20 − h)2
80
2
5
5
√
dh = − dt and 800 h − h3/2 + h5/2 = − t + c.
6
3
5
6
h
√
Using h(0) = 20 we find c = 2560 5/3, so an implicit solution of the initial-value problem is
√
√
80
2
5
2560 5
800 h − h3/2 + h5/2 = − t +
.
3
5
6
3
√
To find the time it takes the tank to empty we set h = 0 and solve for t. The tank empties in 1024 5 seconds
or 38.16 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
mg − kv 2
1
dv
√
g 1 − ( k v/√mg )2
√
mg
k/mg dv
√
√
√
k g 1 − ( k v/ mg )2
√
m
kv
tanh−1 √
kg
mg
√
kv
−1
tanh √
mg
Thus the velocity at time t is
= dt
= dt
= dt
=t+c
=
kg
t + c1 .
m
mg
kg
v(t) =
tanh
t + c1 .
k
m
√
√
Setting t = 0 and v = v0 we find c1 = tanh−1 ( k v0 / mg ).
78
2.8
(b) Since tanh t → 1 as t → ∞, we have v →
Nonlinear Models
mg/k as t → ∞.
(c) Integrating the expression for v(t) in part (a) we obtain an integral of the form du/u:
m
mg
kg
kg
s(t) =
+ c2 .
tanh
t + c1 dt =
ln cosh
t + c1
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
√
tan−1 √
=− t+c
mg
m
mgk
√ kv
gk
−1
tan
=−
t + c1
√
mg
m
mg
gk
tan c1 −
t .
v(t) =
k
m
Setting v(0) = 300, m =
16
32
=
1
2
, g = 32, and k = 0.0003, we find v(t) = 230.94 tan(c1 − 0.138564t) and
c1 = 0.914743. Integrating
v(t) = 230.94 tan(0.914743 − 0.138564t)
we get
s(t) = 1666.67 ln | cos(0.914743 − 0.138564t)| + c2 .
Using s(0) = 0 we find c2 = 823.843. Solving v(t) = 0 we see that the maximum height is attained when
t = 6.60159. The maximum height is s(6.60159) = 823.843 ft.
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
√
m
kv
1
−1
√ √
tanh √
= t + c.
mg − ρV
k mg − ρV
Thus
v(t) =
mg − ρV
tanh
k
√
(c) Since tanh t → 1 as t → ∞, the terminal velocity is
kmg − kρV
t + c1 .
m
(mg − ρV )/k .
79
2.8
Nonlinear Models
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
18. (a) Writing the equation in the form (x −
u du
dx
√
=
x
1 + u2 − 1 + u2
u du
dx
√
√
=
.
x
1 + u2 (1 − 1 + u2 )
√
√
Letting w = 1 − 1 + u2 we have dw = −u du/ 1 + u2 so that
− ln 1 − 1 + u2 = ln |x| + c
−
1
√
= c1 x
1 + u2
c2
1 − 1 + u2 = −
x
c2
y2
1+
= 1+ 2
x
x
1−
1+
(−c2 = 1/c1 )
2c2
y2
c2
+ 22 = 1 + 2 .
x
x
x
Solving for y 2 we have
c2 2
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.
y 2 = 2c2 x + c22 = 4
c 2
x+
(b) Let u = x2 + y 2 so that
du
dy
= 2x + 2y
.
dx
dx
Then
dy
1 du
=
−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
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.
80
2.8
Nonlinear Models
(b) Separating variables and using a CAS to integrate we get
dW
√
= dx
W 4 − 2W
and
− tanh−1
1√
2
4 − 2W
= 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
(b) Letting k = 1/100, separating variables and integrating (with the help
of a CAS), we get
100h(h − 20)
dh = dt
(h − 10)2
and
100(h2 − 10h + 100)
= t + c.
10 − h
h
10
8
6
Using h(0) = 0 we find c = 1000, and solving for h we get h(t) =
√
0.005 t2 + 4000t−t , where the positive square root is chosen because
4
h ≥ 0.
2
t
2000 4000 6000 8000 10000
(c) The volume of the tank is V = 23 π(10)3 feet, so at a rate of π cubic feet per minute, the tank will fill in
2
3
3 (10) ≈ 666.67 minutes ≈ 11.11 hours.
(d) At 666.67 minutes, the depth of the water is h(666.67) = 5.486 feet. 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
lim h(t) = 0.005 lim
t2 + 4000t − t = 0.005 lim √
t→∞
t→∞
t→∞
t2 + 4000t + t
4000t
4000
= 0.005 lim = 0.005(2000) = 10.
= 0.005
t→∞ t 1 + 4000/t + t
1+1
81
2.8
Nonlinear Models
21. (a)
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
(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)
0.136883
.
0.000660944 + 0.0341781e−0.0348391t
P
175
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.
82
2.8
Nonlinear Models
22. (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.
(b) Since f (P ) > 0 for 0 < P < 1.09216, the solution P (t) of
dP/dt = P (1 − P ) + 0.3e−P ,
f
P (0) = P0 ,
2
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.
1
0.5 1 1.5 2 2.5 3 p
-1
-2
(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
p
2
1.5
see that the percentage increase is 9.216%.
1
0.5
2
4
6
8
10 t
23. To find td we solve
m
dv
= mg − kv 2 ,
dt
v(0) = 0
using separation of variables. This gives
v(t) =
Integrating and using s(0) = 0 gives
mg
tanh
k
kg
t.
m
m
kg
s(t) =
ln cosh
t .
k
m
To find the time of descent we solve s(t) = 823.84 and find td = 7.77882. The impact velocity is v(td ) = 182.998,
which is positive because the positive direction is downward.
24. (a) Solving vt = mg/k for k we obtain k = mg/vt2 . The differential equation then becomes
dv
mg
1
dv
m
= mg − 2 v 2 or
= g 1 − 2 v2 .
dt
vt
dt
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
We find the distance by integrating:
s(t) =
vt tanh
gt
v2
gt
dt = t ln cosh
+ c2 .
vt
g
vt
83
2.8
Nonlinear Models
The initial condition s(0) = 0 implies c2 = 0, so
s(t) =
vt2
gt
.
ln cosh
g
vt
In 25 seconds she has fallen 20,000 − 14,800 = 5,200 feet. Using a CAS to solve
32(25)
2
5200 = (vt /32) ln cosh
vt
for vt gives vt ≈ 271.711 ft/s. Then
s(t) =
vt2
gt
= 2307.08 ln(cosh 0.117772t).
ln cosh
g
vt
(b) At t = 15, s(15) = 2,542.94 ft and v(15) = s (15) = 256.287 ft/sec.
25. While the object is in the air its velocity is modeled by the linear differential equation m dv/dt = mg −kv. Using
m = 160, k = 14 , and g = 32, the differential equation becomes dv/dt + (1/640)v = 32. The integrating factor
is e dt/640 = et/640 and the solution of the differential equation is et/640 v = 32et/640 dt = 20,480et/640 + c.
Using v(0) = 0 we see that c = −20,480 and v(t) = 20,480 − 20,480e−t/640 . Integrating we get s(t) = 20,480t +
13,107,200e−t/640 + c. Since s(0) = 0, c = −13,107,200 and s(t) = −13,107,200 + 20,480t + 13,107,200e−t/640 .
To find when the object hits the liquid we solve s(t) = 500 − 75 = 425, obtaining ta = 5.16018. The velocity
at the time of impact with the liquid is va = v(ta ) = 164.482. When the object is in the liquid its velocity is
modeled by the nonlinear differential equation m dv/dt = mg − kv 2 . Using m = 160, g = 32, and k = 0.1 this
becomes dv/dt = (51,200 − v 2 )/1600. Separating variables and integrating we have
√
√
dv
2
v − 160 2
dt
1
√ =
and
ln
t + c.
=
51,200 − v 2
1600
640
1600
v + 160 2
√
Solving v(0) = va = 164.482 we obtain c = −0.00407537. Then, for v < 160 2 = 226.274,
√
√
√
√
v − 160 2
v − 160 2
2t/5−1.8443
√ =e
√ = e 2t/5−1.8443 .
or −
v + 160 2
v + 160 2
Solving for v we get
√
v(t) =
13964.6 − 2208.29e
2t/5
61.7153 + 9.75937e
2t/5
√
Integrating we find
√
s(t) = 226.275t − 1600 ln(6.3237 + e
.
2t/5
) + c.
Solving s(0) = 0 we see that c = 3185.78, so
√
s(t) = 3185.78 + 226.275t − 1600 ln(6.3237 + e
2t/5
).
To find when the object hits the bottom of the tank we solve s(t) = 75, obtaining tb = 0.466273. The time
from when the object is dropped from the helicopter to when it hits the bottom of the tank is ta + tb =
5.62708 seconds.
84
2.9
Modeling with Systems of First-Order DEs
EXERCISES 2.9
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 e y = λ1 x0 e(λ2 −λ1 )t + c2
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
y=
λ1 x0
e−λ1 t − e−λ2 t .
λ 2 − λ1
Substituting this result into the third differential equation we have
dz
λ1 λ2 x0 −λ1 t
− e−λ2 t .
e
=
dt
λ2 − λ1
Integrating we find
z=−
Using z(0) = 0 we find c3 = x0 . Thus
λ2 x0 −λ1 t
λ1 x0 −λ2 t
e
+
e
+ c3 .
λ2 − λ1
λ2 − λ1
z = x0 1 −
λ2
λ1
e−λ1 t +
e−λ2 t .
λ2 − λ1
λ 2 − λ1
2. We see from the graph that the half-life of A is approximately 4.7
days. To determine the half-life 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)
z(t)
25 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
85
2.9
Modeling with Systems of First-Order DEs
z(t) denote the amounts of substances W , X, Y , and Z, respectively. A model for the radioactive series is
dw
dt
dx
dt
dy
dt
dz
dt
= −λ1 w
= λ1 w − λ2 x
= λ2 x − λ3 y
= λ3 y.
5. The system is
1
1
2
1
x2 − x1 · 4 = − x1 + x2 + 6
50
50
25
50
1
1
1
2
2
x2 =
x1 · 4 − x2 − x2 · 3 =
x1 − x2 .
50
50
50
25
25
x1 = 2 · 3 +
6. Let x1 , x2 , and x3 be the amounts of salt in tanks A, B, and C, respectively, so that
1
x2 · 2 −
100
1
x2 =
x1 · 6 +
100
1
x2 · 5 −
x3 =
100
x1 =
1
1
3
x1 · 6 =
x2 − x1
100
50
50
1
1
1
3
7
1
x3 −
x2 · 2 −
x2 · 5 =
x1 −
x2 +
x3
100
100
100
50
100
100
1
1
1
1
x3 −
x3 · 4 =
x2 − x3 .
100
100
20
20
7. (a) A model is
dx1
x2
x1
=3·
−2·
,
dt
100 − t
100 + t
dx2
x1
x2
=2·
−3·
,
dt
100 + t
100 − t
x1 (0) = 100
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
dx2
2(150 − x2 )
3x2
300
2x2
3x2
=
−
=
−
−
dt
100 + t
100 − t
100 + t 100 + t 100 − t
or
dx2
+
dt
2
3
+
100 + t 100 − t
x2 =
300
,
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
so
d
[(100 + t)2 (100 − t)−3 x2 ] = 300(100 + t)(100 − t)−3 .
dt
Using integration by parts, we obtain
1
1
(100 + t)2 (100 − t)−3 x2 = 300 (100 + t)(100 − t)−2 − (100 − t)−1 + c .
2
2
Thus
300
1
1
c(100 − t)3 − (100 − t)2 + (100 + t)(100 − t)
(100 + t)2
2
2
300
=
[c(100 − t)3 + t(100 − t)].
(100 + t)2
x2 =
86
2.9
Modeling with Systems of First-Order DEs
Using x2 (0) = 50 we find c = 5/3000. At t = 30, x2 = (300/1302 )(703 c + 30 · 70) ≈ 47.4 lbs.
8. A model is
dx1
1
= (4 gal/min)(0 lb/gal) − (4 gal/min)
x1 lb/gal
dt
200
dx2
1
1
= (4 gal/min)
x1 lb/gal − (4 gal/min)
x2 lb/gal
dt
200
150
dx3
1
1
= (4 gal/min)
x2 lb/gal − (4 gal/min)
x3 lb/gal
dt
150
100
or
dx1
1
= − x1
dt
50
dx2
1
2
=
x1 − x2
dt
50
75
dx3
1
2
=
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.
9. 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
t
50
10. (a) The population y(t) approaches 10,000, while the population x(t)
100
x,y
10
approaches extinction.
y
5
x
(b) The population x(t) approaches 5,000, while the population y(t)
approaches extinction.
10
20
10
20
10
20
t
x,y
10
x
5
y
(c) The population y(t) approaches 10,000, while the population x(t)
approaches extinction.
t
x,y
10
y
5
x
87
t
2.9
Modeling with Systems of First-Order DEs
(d) The population x(t) approaches 5,000, while the population y(t)
x,y
10
approaches extinction.
x
5
y
10
11. (a)
x,y
10
y
5
x
(b)
20
(c)
40
y
x
5
20
40
y
5
x
20
(d)
t
x,y
10
t
x,y
10
20
40
t
x,y
10
y
5
x
t
20
40
t
In each case the population x(t) approaches 6,000, while the population y(t) approaches 8,000.
12. 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 i3 = q = CR1 i2 − CR2 i3 so that the system is
Li2 + Li3 + R1 i2 = E(t)
1
−R1 i2 + R2 i3 + i3 = 0.
C
13. By Kirchhoff’s first law we have i1 = i2 + i3 . Applying Kirchhoff’s second law to each loop we obtain
E(t) = i1 R1 + L1
di2
+ i2 R2
dt
E(t) = i1 R1 + L2
di3
+ i3 R3 .
dt
and
Combining the three equations, we obtain the system
di2
+ (R1 + R2 )i2 + R1 i3 = E
dt
di3
L2
+ R1 i2 + (R1 + R3 )i3 = E.
dt
L1
14. 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.
15. 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
88
2.9
Modeling with Systems of First-Order DEs
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
di
dr ds
=− −
= −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
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.
16. (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 consider 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.
10
10
5
5
5
10
5
i
s
i
i
s,i
s,i
s,i
s,i
10
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.
89
2.9 Modeling
with Systems
of First-Order DEs
CHAPTER
2 REVIEW
EXERCISES
CHAPTER 2 REVIEW EXERCISES
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
dy
4
= 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.
dy
3.
= (y − 1)2 (y − 3)2
dx
4.
dy
= y(y − 2)2 (y − 4)
dx
5. 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.
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.
6. 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, limt→∞ P (t) = 1.3214.
y
7.
x
8. (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
90
CHAPTER 2 REVIEW EXERCISES
(i) Bernoulli
(j) homogeneous, exact, Bernoulli
(k) linear in x and y, exact, separable, homoge-
(l) exact, linear in y
neous
(m) homogeneous
(n) separable
9. Separating variables and using the identity cos2 x = 12 (1 + cos 2x), we have
cos2 x dx =
y2
y
dy,
+1
1
1
1
x + sin 2x = ln y 2 + 1 + c,
2
4
2
and
2x + sin 2x = 2 ln y 2 + 1 + c.
10. 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.
11. 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
dw
6
9x2
+
w=−
.
dx
6x + 1
6x + 1
An integrating factor is 6x + 1, so
d
[(6x + 1)w] = −9x2 ,
dx
3x3
c
w=−
+
,
6x + 1 6x + 1
and
(6x + 1)y 3 = −3x3 + c.
(Note: The differential equation is also exact.)
12. 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
91
CHAPTER 2 REVIEW EXERCISES
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
13. 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
1
1
tQ = − t5 + t5 ln t + c
25
5
and
Q=−
1 4 1 4
c
t + t ln t + .
25
5
t
14. Letting u = 2x + y + 1 we have
du
=2+
dx
and so the given differential equation is transformed into
du
u
− 2 = 1 or
dx
dy
,
dx
du
2u + 1
=
.
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| = 2x + c1 .
Resubstituting for u gives the solution
4x + 2y + 2 − ln |4x + 2y + 3| = 2x + c1
or
2x + 2y + 2 − ln |4x + 2y + 3| = c1 .
15. 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
3
y=
1 2
x +4
4
y=
1
+ c x2 + 4
4
and
92
4
+c
−4
.
CHAPTER 2 REVIEW EXERCISES
16. 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.
17. 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 < x < 2π is chosen
to include x = 7π/6.
18. Separating variables and integrating we have
dy
= −2(t + 1) dt
y2
1
− = −(t + 1)2 + c
y
1
y=
,
(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.
√
19. (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
√
√
2 y = x + 2 y0 − x0
and y =
1
√
(x + 2 y0 − x0 )2 .
4
√
√
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 , ∞).
Since
20. (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|).
93
CHAPTER 2 REVIEW EXERCISES
(b) Solving for y in part (a) and being sure that the initial condition 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
y
2
1
-2
0 < x < e.
-1
1
2
x
-1
-2
21. The graph of y1 (x) is the portion of the closed black 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 black curve lying in the third
quadrant. Its interval of definition is (−∞, 0).
22. 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.
23. From
dP
= 0.018P and P (0) = 4 billion we obtain P = 4e0.018t so that P (45) = 8.99 billion.
dt
24. Let A = A(t) be the volume of CO2 at time t. From dA/dt = 1.2 − A/4 and A(0) = 16 ft3 we obtain
A = 4.8 + 11.2e−t/4 . Since A(10) = 5.7 ft3 , the concentration is 0.017%. As t → ∞ we have A → 4.8 ft3 or
0.06%.
25. Separating variables, we have
s2 − y 2
dy = −dx.
y
Substituting y = s sin θ, this becomes
s2 − s2 sin2 θ
(s cos θ)dθ = −dx
s sin θ
cos2
s
dθ = − dx
sin θ
s
s
1 − sin2 θ
dθ = −x + c
sin θ
(csc θ − sin θ)dθ = −x + c
s ln | csc θ − cot θ| + s cos θ = −x + c
s
s2 − y 2
s2 − y 2
s ln −
+s
= −x + c.
y
y
s
Letting s = 10, this is
10
10 ln
−
y
100 − y 2
+ 100 − y 2 = −x + c.
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
26. From V dC/dt = kA(Cs − C) and C(0) = C0 we obtain C = Cs + (C0 − Cs )e−kAt/V .
94
CHAPTER 2 REVIEW EXERCISES
27. (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→∞
Moreover,
BT1 + T2
.
1+B
BT1 + T2
BT1 + T2
lim Tm (t) = lim [T2 + B(T1 − T )] = T2 + B T1 −
=
.
t→∞
t→∞
1+B
1+B
(b) The differential equation is
or
dT
= k(T − Tm ) = k(T − T2 − BT1 + BT )
dt
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
BT1 + T2 −k(1+B)t
e
e−k(1+B)t T =
+c
1+B
BT1 + T2
T (t) =
+ cek(1+B)t .
1+B
Since k is negative, limt→∞ T (t) = (BT1 + T2 )/(1 + B).
(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,
B
1
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.
28. (a) By separation of variables and partial fractions,
T − Tm
ln
− 2 tan−1
T + Tm
T
Tm
3
= 4Tm
kt + c.
Then rewrite the right-hand side of the differential equation as
dT
4
4
) = [(Tm + (T − Tm ))4 − Tm
]
= k(T 4 − Tm
dt
4
T − Tm
4
= kTm
1+
−1
Tm
2 T
−
T
T
−
T
m
m
4
1+4
= kTm
+6
· · · − 1 ← binomial expansion
Tm
Tm
95
CHAPTER 2 REVIEW EXERCISES
(b) When T − Tm is small compared to Tm , every term in the expansion after the first two can be ignored,
giving
dT
3
.
≈ k1 (T − Tm ), where k1 = 4kTm
dt
29. We first solve (1 − t/10)di/dt + 0.2i = 4. Separating variables we obtain di/(40 − 2i) =
dt/(10 − t). Then
√
1
− ln |40 − 2i| = − ln |10 − t| + c or
40 − 2i = c1 (10 − t).
2
√
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
4t − 15 t2 , 0 ≤ t < 10
i(t) =
20,
t ≥ 10.
20
10
10
20
The graph of i(t) is given in the figure.
√ √
30. From y 1 + (y )2 = k we obtain dx = ( y/ k − y )dy. If y = k sin2 θ then
1 1
k
dy = 2k sin θ cos θ dθ, dx = 2k
− cos 2θ dθ, and x = kθ − sin 2θ + c.
2 2
2
If x = 0 when θ = 0 then c = 0.
1
1 2
31. Letting c = 0.6, Ah = π( 32
· 12
) , Aw = π · 12 = π, and g = 32, the differential equation becomes
√
√
dh/dt = −0.00003255 h . Separating variables and integrating, we get 2 h = −0.00003255t + c, so h =
√
√
(c1 − 0.00001628t)2 . Setting h(0) = 2, we find c = 2 , so h(t) = ( 2 − 0.00001628t)2 , where h is measured in
feet and t in seconds.
32. One hour is 3,600 seconds, so the hour mark should be placed at
√
h(3600) = [ 2 − 0.00001628(3600)]2 ≈ 1.838 ft ≈ 22.0525 in.
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.
33. In this case Aw = πh2 /4 and the differential equation is
dh
1
=−
h−3/2 .
dt
7680
Separating variables and integrating, we have
1
dt
7680
1
=−
t + c1 .
7680
h3/2 dh = −
2 5/2
h
5
96
time
seconds 0
1
2
3
4
5
6
7
8
9
10
11
12
height
inches 24.0000
22.0520
20.1864
18.4033
16.7026
15.0844
13.5485
12.0952
10.7242
9.4357
8.2297
7.1060
6.0648
CHAPTER 2 REVIEW EXERCISES
√
Setting h(0) = 2 we find c1 = 8 2/5, so that
√
1
2 5/2
8 2
=−
h
t+
,
5
7680
5
√
1
h5/2 = 4 2 −
t,
3072
2/5
√
1
h= 4 2−
.
t
3072
and
In this case h(4 hr) = h(14,400 s) = 11.8515 inches 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 ≈ 17,378 s ≈ 4.83 hr. Thus, this particular conical
water clock can only measure time intervals of less than 4.83 hours.
34. 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
√
√
dh
cπrh2 2g √
8crh2 h
=−
h=−
.
dt
π[f (h)]2
[f (h)]2
h
2
1
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
−1
√
8crh2 h
,
[f (h)] =
a
2
and
r = f (h) = 2rh
1
r
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.
35. From dx/dt = k1 x(α − x) we obtain
1/α
1/α
+
x
α−x
dx = k1 dt
so that x = αc1 eαk1 t /(1 + c1 eαk1 t ). From dy/dt = k2 xy we obtain
ln |y| =
k2
ln 1 + c1 eαk1 t + c or y = c2 1 + c1 eαk1 t
k1
k2 /k1
.
36. In tank A the salt input is
gal
lb
lb
gal
x2 lb
1
7
2
+ 1
= 14 +
x2
.
min
gal
min
100 gal
100
min
The salt output is
3
gal
min
In tank B the salt input is
x1 lb
100 gal
5
The salt output is
1
gal
min
gal
min
x2 lb
100 gal
lb
gal
x1 lb
2
+ 5
=
x1
.
min
100 gal
25
min
x1 lb
100 gal
=
lb
1
x1
.
20
min
lb
gal
x2 lb
1
+ 4
=
x2
.
min
100 gal
20
min
97
CHAPTER 2 REVIEW EXERCISES
The system of differential equations is then
dx1
1
2
= 14 +
x2 − x1
dt
100
25
dx2
1
1
=
x1 − x2 .
dt
20
20
37. From y = −x − 1 + c1 ex we obtain y = y + x so that the differential equation of the orthogonal family is
dy
1
=−
dx
y+x
dx
+ x = −y.
dy
This is a linear differential equation and has integrating factor e dy = ey , so
or
d y
[e x] = −yey
dy
ey x = −yey + ey + c2
x = −y + 1 + c2 e−y .
38. Differentiating the family of curves, we have
y
5
1
1
y = −
=− 2.
2
(x + c1 )
y
The differential equation for the family of orthogonal trajectories is then
y = y 2 . Separating variables and integrating we get
-5
dy
= dx
y2
1
− = x + c1
y
1
y=−
.
x + c1
5 x
-5
98
3
Higher-Order
Differential Equations
EXERCISES 3.1
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 =
and c2 = − 12 . The solution is y = 12 ex − 12 e−x .
1
2
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 3.1 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
x(t) = x0 cos ωt +
sin 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 =
Thus
ωx0 cos ωt0 − x1 sin ωt0
ω
and c2 =
x1 cos ωt0 + ωx0 sin ωt0
.
ω
ωx0 cos ωt0 − x1 sin ωt0
x1 cos ωt0 + ωx0 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
= x0 cos ω(t − t0 ) +
sin ω(t − t0 ).
ω
x(t) =
9. Since a2 (x) = x − 2 and x0 = 0 the problem has a unique solution for −∞ < x < 2.
99
3.1
Preliminary Theory: Linear Equations
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=
1
ex − e−x
2
ex − e−x
e
sinh x = 1
=
= 2
(ex − e−x ).
−1
sinh 1
e −e
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.
(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.
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.
100
3.1
Preliminary Theory: Linear Equations
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.
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 on this interval 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 on this interval 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 on this interval 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 on this interval is
y = c1 + c2 x + c3 cos x + c4 sin x.
101
3.1
Preliminary Theory: Linear Equations
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.
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 homogeneous
1 2
equation, and yp = 15
x − 16 x is a particular solution of the nonhomogeneous 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 .
(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
1
ex + e−x
(y1 + y2 ) =
= cosh x and
2
2
102
ex − e−x
1
(y1 − y2 ) =
= sinh x.
2
2
3.2
Reduction of Order
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
3
x
W (y1 , y2 ) = 2
3x
If x < 0 then y2 = −x3 and
3
x
W (y1 , y2 ) = 2
3x
x3 = 0.
3x2 −x3 = 0.
−3x2 This does not violate Theorem 3.3 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. They are linearly independent
since W x3 , x2 = x4 = 0 for −∞ < x < ∞.
(d) The function y = x3 satisfies y(0) = 0 and 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.
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.
EXERCISES 3.2
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 .
103
3.2
Reduction of Order
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 +
2 u = 0.
x
2
If w = u we obtain the linear first-order equation w + w = 0 which has the integrating factor
x
e2 dx/x = x2 . Now
d 2
[x w] = 0 gives x2 w = c.
dx
1
Therefore w = u = c/x2 and u = c1 /x. A second solution is y2 = xe−x = 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.
If w = u we obtain the linear first-order equation w − 8(tan 4x)w = 0 which has the integrating factor
e−8 tan 4x dx = cos2 4x. Now
d
[(cos2 4x)w] = 0
dx
gives
(cos2 4x)w = c.
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
104
3.2
Reduction of Order
and
y − 25y = e5x (u + 10u ) = 0
or u + 10u = 0.
If w
= u we obtain the linear first-order equation w + 10w = 0 which has the integrating factor
10 dx
e
= e10x . Now
d 10x
[e w] = 0 gives e10x w = c.
dx
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
y =
2 2x/3
e
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
y =
1 x/3
e u + ex/3 u ,
3
2
1
y = ex/3 u + ex/3 u + ex/3 u
3
9
and
5
or u + u = 0.
6
5
If w = u we obtain the linear first-order equation w + 6 w = 0 which has the integrating factor
e(5/6) dx = e5x/6 . Now
d 5x/6
w] = 0 gives e5x/6 w = c.
[e
dx
6y + y − y = ex/3 (6u + 5u ) = 0
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
4
y2 = x
e−
(−7/x) dx
4
x8
dx = x
1
dx = x4 ln |x|.
x
A second solution is y2 = x4 ln |x|.
10. Identifying P (x) = 2/x we have
2
y2 = x
e−
(2/x) dx
2
x4
dx = x
1
x−6 dx = − x−3 .
5
A second solution is y2 = x−3 .
11. Identifying P (x) = 1/x we have
y2 = ln x
e− dx/x
dx
1
dx = ln x
= ln x −
2
2
(ln x)
x(ln x)
ln x
= −1.
A second solution is y2 = 1.
12. Identifying P (x) = 0 we have
y2 = x1/2 ln x
e− 0 dx
1
1/2
dx
=
x
ln
x
−
2
x(ln x)
ln x
A second solution is y2 = x1/2 .
105
= −x1/2 .
3.2
Reduction of Order
13. Identifying P (x) = −1/x we have
e− −dx/x
x
dx = x sin(ln x)
dx
x2 sin2 (ln x)
x2 sin2 (ln x)
y2 = x sin(ln x)
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
2
y2 = x cos(ln x)
= x2 cos(ln x)
e− −3 dx/x
x3
2
cos(ln
x)
dx
=
x
dx
4
2
4
x cos (ln x)
x 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
− 2(1+x)dx/(1−2x−x2 )
ln(1−2x−x2 )
e
e
y2 = (x + 1)
dx
=
(x
+
1)
dx
2
(x + 1)
(x + 1)2
1 − 2x − x2
2
= (x + 1)
dx
=
(x
+
1)
− 1 dx
2
(x + 1)
(x + 1)2
= (x + 1) −
2
− x = −2 − x2 − x.
x+1
A second solution is y2 = x2 + x + 2.
16. Identifying P (x) = −2x/ 1 − x2 we have
2
2
y2 = e− −2x dx/(1−x ) dx = e− ln(1−x ) dx =
1
1 1 + x .
dx = ln 1 − x2
2
1 − x
A second solution is y2 = ln |(1 + x)/(1 − x)|.
17. Define y = u(x)e−2x so
y = −2ue−2x + u e−2x ,
y = u e−2x − 4u e−2x + 4ue−2x
and
y − 4y = e−2x u − 4e−2x u = 0
or u − 4u = 0.
If w = u we obtain the linear first-order equation w − 4w = 0 which has the integrating factor
e−4 dx = e−4x . Now
d −4x
w] = 0 gives e−4x w = c.
[e
dx
Therefore w = u = ce4x and u = c1 e4x . A second solution is y2 = e−2x e4x = e2x . We see by observation that a
particular solution is yp = −1/2. The general solution is
1
y = c1 e−2x + c2 e2x − .
2
18. Define y = u(x) · 1 so
y = u ,
y = u
and y + y = u + u = 1.
106
3.2
Reduction of Order
If w = u we obtain the linear first-order equation w + w = 1 which has the integrating factor e
d x
[e w] = ex
dx
gives
dx
= ex . Now
ex w = ex + c.
Therefore w = u = 1 + ce−x and u = x + c1 e−x + c2 . The general solution is
y = u = x + c1 e−x + c2 .
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−
Now
d −x
[e w] = 5ex gives e−x w = 5ex + c1 .
dx
dx
= e−x .
Therefore w = u = 5e2x + c1 ex and u = 52 e2x + c1 ex + c2 . The general solution is
y = uex =
5 3x
e + c1 e2x + c2 ex .
2
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−
e−2x . Now
d −2x
1
1
w] = xe−3x gives e−2x w = − xe−3x − e−3x + c1 .
[e
dx
3
9
Therefore w = u = − 13 xe−x − 19 e−x + c1 e2x and u =
y = uex =
1
3
2dx
=
xe−x + 49 e−x + c2 e2x + c3 . The general solution is
4
1
x + + c2 e3x + c3 ex .
3
9
21. (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
107
3.2
Reduction of Order
and let y1 = em1 x be a solution. Then a second solution is given by
−bx/a
e
y2 = em1 x
dx
e2m1 x
= em1 x 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 = −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
m1 x
y2 = e
dx = xem1 x .
(c) The functions
1 ix
(e − e−ix )
cos x =
2i
1
sinh x = (ex − e−x )
cosh x =
2
are all expressible in terms of exponential functions.
sin x =
1 ix
(e + e−ix )
2
1 x
(e + e−x )
2
22. 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:
dw 1
= 1−
w
dx
x
dw 1
= 1−
dx
w
x
ln w = x − ln x + c
w = c1
Then u = c1 ex /x and u = c1
second solution is
ex
.
x
ex dx/x. To integrate ex /x we use the series representation for ex . Thus, a
ex
dx
x
1
1
1
= c1 x
1 + x + x2 + x3 + · · · dx
x
2!
3!
1
1
1
= c1 x
+ 1 + x + x2 + · · · dx
x
2!
3!
1 2
1 3
= c1 x ln x + x +
x +
x + ···
2(2!)
3(3!)
1 3
1 4
= c1 x ln x + x2 +
x +
x + ··· .
2(2!)
3(3!)
y2 = xu(x) = c1 x
An interval of definition is probably (0, ∞) because of the ln x term.
108
3.3
Homogeneous Linear Equations with Constant Coefficients
23. (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.
(b) By (5) in the text a second solution is
x+10
− P (x) dx
dx
x
e
e
e (1+10/x)dx
x
x
y2 = y1
dx = e
dx = e
dx
y12
e2x
e2x
x+ln x10
e
x
= ex
x10 e−x dx
dx
=
e
e2x
= 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 .
10
1
1 n
(c) By Corollary (A) of Theorem 3.2, −
y2 =
x is a solution.
10!
n!
n=0
EXERCISES 3.3
Homogeneous Linear Equations with
Constant Coefficients
1. From 4m2 + m = 0 we obtain m = 0 and m = −1/4 so that y = c1 + c2 e−x/4 .
2. From m2 − 36 = 0 we obtain m = 6 and m = −6 so that y = c1 e6x + c2 e−6x .
3. From m2 − m − 6 = 0 we obtain m = 3 and m = −2 so that y = c1 e3x + c2 e−2x .
4. From m2 − 3m + 2 = 0 we obtain m = 1 and m = 2 so that y = c1 ex + c2 e2x .
5. From m2 + 8m + 16 = 0 we obtain m = −4 and m = −4 so that y = c1 e−4x + c2 xe−4x .
6. From m2 − 10m + 25 = 0 we obtain m = 5 and m = 5 so that y = c1 e5x + c2 xe5x .
7. From 12m2 − 5m − 2 = 0 we obtain m = −1/4 and m = 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− 5 )x .
9. From m2 + 9 = 0 we obtain m = 3i and m = −3i so that y = c1 cos 3x + c2 sin 3x.
√
√
√
√
10. From 3m2 + 1 = 0 we obtain m = i/ 3 and m = −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)].
13. From 3m2 + 2m + 1 = 0 we obtain m = −1/3 ±
√
2 i/3 so that
√
√
y = e−x/3 [c1 cos( 2x/3) + c2 sin( 2x/3)].
109
3.3
Homogeneous Linear Equations with Constant Coefficients
14. From 2m2 − 3m + 4 = 0 we obtain m = 3/4 ±
√
23 i/4 so that
√
√
y = e3x/4 [c1 cos( 23x/4) + c2 sin( 23x/4)].
15. From m3 − 4m2 − 5m = 0 we obtain m = 0, m = 5, and m = −1 so that
y = c1 + c2 e5x + c3 e−x .
√
16. From m3 − 1 = 0 we obtain m = 1 and m = −1/2 ± 3 i/2 so that
√
√
y = c1 ex + e−x/2 [c2 cos( 3x/2) + c3 sin( 3x/2)].
17. From m3 − 5m2 + 3m + 9 = 0 we obtain m = −1, m = 3, and m = 3 so that
y = c1 e−x + c2 e3x + c3 xe3x .
18. From m3 + 3m2 − 4m − 12 = 0 we obtain m = −2, m = 2, and m = −3 so that
y = c1 e−2x + c2 e2x + c3 e−3x .
19. From m3 + m2 − 2 = 0 we obtain m = 1 and m = −1 ± i so that
u = c1 et + e−t (c2 cos t + c3 sin t).
√
20. From m3 − m2 − 4 = 0 we obtain m = 2 and m = −1/2 ± 7 i/2 so that
√
√
x = c1 e2t + e−t/2 [c2 cos( 7t/2) + c3 sin( 7t/2)].
21. From m3 + 3m2 + 3m + 1 = 0 we obtain m = −1, m = −1, and m = −1 so that
y = c1 e−x + c2 xe−x + c3 x2 e−x .
22. From m3 − 6m2 + 12m − 8 = 0 we obtain m = 2, m = 2, and m = 2 so that
y = c1 e2x + c2 xe2x + c3 x2 e2x .
√
23. From m4 + m3 + m2 = 0 we obtain m = 0, m = 0, and m = −1/2 ± 3 i/2 so that
√
√
y = c1 + c2 x + e−x/2 [c3 cos( 3x/2) + c4 sin( 3x/2)].
24. From m4 − 2m2 + 1 = 0 we obtain m = 1, m = 1, m = −1, and m = −1 so that
y = c1 ex + c2 xex + c3 e−x + c4 xe−x .
√
√
25. From 16m4 + 24m2 + 9 = 0 we obtain m = ± 3 i/2 and m = ± 3 i/2 so that
√
√
√
√
y = c1 cos( 3x/2) + c2 sin( 3x/2) + c3 x cos( 3x/2) + c4 x sin( 3x/2).
√
26. From m4 − 7m2 − 18 = 0 we obtain m = 3, m = −3, and m = ± 2 i so that
√
√
y = c1 e3x + c2 e−3x + c3 cos 2x + c4 sin 2x.
27. From m5 + 5m4 − 2m3 − 10m2 + m + 5 = 0 we obtain m = −1, m = −1, m = 1, and m = 1, and m = −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 m = 0, m = 0, m = −1/2, and m = 2 ± 2i so that
x = c1 + c2 s + c3 e−s/2 + e2s (c4 cos 2s + c5 sin 2s).
110
3.3
Homogeneous Linear Equations with Constant Coefficients
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
√
3
1
−
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 m = −1 and m = 5, so that y = c1 e−x + c2 e5x . 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−x + 13 e5x−5 .
32. From 4m2 − 4m − 3 = 0 we obtain m = −1/2 and m = 3/2 so that y = c1 e−x/2 + c2 e3x/2 . If y(0) = 1 and
3x/2
y (0) = 5 then c1 + c2 = 1, − 12 c1 + 32 c2 = 5, so c1 = −7/4, c2 = 11/4, and y = − 74 e−x/2 + 11
.
4 e
√
√
√
2
−x/2
33. From m + m + 2 = 0 we obtain m = −1/2 ± 7 i/2 so that y = e
[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 m = 1 and m = 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 .
35. From m3 + 12m2 + 36m = 0 we obtain m = 0, m = −6, and m = −6 so that y = c1 + c2 e−6x + c3 xe−6x . If
y(0) = 0, y (0) = 1, and y (0) = −7 then
−6c2 + c3 = 1,
c1 + c2 = 0,
so c1 = 5/36, c2 = −5/36, c3 = 1/6, and y =
5
36
−
5 −6x
36 e
36c2 − 12c3 = −7,
+ 16 xe−6x .
36. From m3 + 2m2 − 5m − 6 = 0 we obtain m = −1, m = 2, and m = −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 m = 5 and m = 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 solution is y =
√
√
√
√
√
√
c1 e 3x + c2 e− 3x . By (11) the general solution is y = c1 cosh 3x + c2 sinh 3x. For y = c1 e 3x + c2 e− 3x the
111
3.3
Homogeneous Linear Equations with Constant Coefficients
√
√
√
initial conditions imply c1 + c2 = 1, 3c1 − 3c2 = 5. Solving for c1 and c2 we find c1 = 12 (1 + 5 3 ) and c2 =
√
√
√
√
√
√
√
1
y = 12 (1 + 5 3 )e 3x + 12 (1 − 5 3 )e− 3x . For y = c1 cosh 3x + c2 sinh 3x the initial conditions
2 (1 − 5 3 ) so√
√
√
√
√
imply c1 = 1, 3c2 = 5. Solving for c1 and c2 we find c1 = 1 and c2 = 53 3 so y = cosh 3x + 53 3 sinh 3x.
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.
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. Since (m − 4)(m + 5)2 = m3 + 6m2 − 15m − 100 the differential equation is y + 6y − 15y − 100y = 0. The
differential equation is not unique since any constant multiple of the left-hand side of the differential equation
would lead to the auxiliary roots.
50. A third root must be m3 = 3 − i and the auxiliary equation is
1
11
1
m+
[m − (3 + i)][m − (3 − i)] = m +
(m2 − 6x + 10) = m3 − m2 + 7m + 5.
2
2
2
The differential equation is
y −
11 y + 7y + 5y = 0.
2
51. 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 .
52. 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
√
√ √
√ √
√
2
2
2
2
y(x) = e 2 x/2 c1 cos
x + c2 sin
x + e− 2 x/2 c3 cos
x + c4 sin
x .
2
2
2
2
112
3.3
Homogeneous Linear Equations with Constant Coefficients
53. 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
1
π
π = ex − e−x − 2 (cos x) cos
− (sin x) sin
2
2
6
6
√
1
3
1
1
= ex − e−x − 2
cos x − sin x
2
2
2
2
=
1 x 1 −x √
e − e − 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.
54. The auxiliary equation is m2 + α = 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 eα x + c2 e−α x , or
alternatively, y = c1 cosh α x + c2 sinh α x. Again, y(0) = 0 implies c1 = 0 so y = c2 sinh α x. The second
boundary condition implies 0 = y(π/2) = c2 sinh α π/2 or c2 = 0. In this case also, the problem possesses only
the trivial solution.
When λ = α2 > 0 the general solution of the differential equation is y = c1 cos α x + c2 sin α x.
In this case also, y(0) = 0 yields c1 = 0, so that y = c2 sin α x. The second boundary condition implies
0 = c2 sin α π/2. When α π/2 is an integer multiple of π, that is, when α = 2k for k a nonzero integer, the
Case III
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.
55. Applying integration by parts twice we have
1
eax f (x) dx = eax f (x) −
a
1
= eax f (x) −
a
1
= eax f (x) −
a
Collecting the integrals we get
1
ax
e
f (x) − 2 f (x)
a
1
eax f (x) dx
a
1 1 ax 1
e f (x) −
eax f (x) dx
a a
a
1 ax 1
eax f (x) dx.
e f (x) + 2
a2
a
dx =
In order for the technique to work we need to have
1
ax
e
f (x) − 2 f (x)
a
1 ax
1
e f (x) − 2 eax f (x).
a
a
dx = k
eax f (x) dx
or
1 f (x) = kf (x),
a2
where k =
0. This is the second-order differential equation
f (x) −
f (x) + a2 (k − 1)f (x) = 0.
113
3.3
Homogeneous Linear Equations with Constant Coefficients
If k < 1, k = 0, the solution of the differential equation is a pair of exponential functions, in which case
the original integrand is an exponential function and does not require integration by parts for its evaluation.
Similarly, if k = 1, f (x) = 0 and f (x) has the form f (x) = ax+b. In this case a single application of integration
by parts will suffice. Finally, if k > 1, the solution of the differential equation is
√
√
f (x) = c1 cos a k − 1 x + c2 sin a k − 1 x,
and we see that the technique will work for linear combinations of cos αx and sin αx.
√
56. (a) The auxiliary equation is m2 − 64/L = 0 which has roots ±8/ L . Thus, the general solution of the
√
√
differential equation is x = c1 cosh(8t/ L ) + c2 sinh(8t/ L ).
√
(b) Setting x(0) = x0 and x (0) = 0 we have c1 = x0 , 8c2 / L = 0. Solving for c1 and c2 we get c1 = x0 and
√
c2 = 0, so x(t) = x0 cosh(8t/ L ).
√
(c) When L = 20 and x0 = 1, x(t) = cosh(4t/ 5 ). The chain will last touch the peg when x(t) = 10.
√
Solving x(t) = 10 for t we get t1 = 14 5 cosh−1 10 ≈ 1.67326. The velocity of the chain at this instant is
x (t1 ) = 12 11/5 ≈ 17.7989 ft/s.
57. 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 .
58. 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).
59. 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).
√
√
60. 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
y=e
c1 cos
c3 cos
x + c2 sin
x +e
x + c4 sin
x .
2
2
2
2
61. From 2m4 + 3m3 − 16m2 + 15m − 4 = 0 we obtain m = −4, m = 12 , m = 1, and m = 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
so c1 = − 75
, c2 = − 116
3 , c3 =
918
25
, c4 = − 58
5 , and
y=−
4 −4x 116 x/2 918 x 58 x
−
+
e
e
e − xe .
75
3
25
5
114
3.4
Undetermined Coefficients
62. From m4 −3m3 +3m2 −m = 0 we obtain m = 0, m = 1, m = 1, and m = 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
EXERCISES 3.4
Undetermined Coefficients
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 = 65 ,
yp = 65 x + 65 , and
6
6
y = c1 e5x + c2 xe5x + x + .
5
5
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.
1
,
Substituting into the differential equation we obtain −6A = 2 and A − 6B = 0. Then A = − 13 , B = − 18
1
1
yp = − 3 x − 18 , and
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.
115
3.4
Undetermined Coefficients
x
+ −2x − 12
13 e and
11
12 x
4x
2
y = e (c1 cos 2x + c2 sin 2x) + 5x + 4x +
+ −2x −
e .
10
13
√
√
√
√
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
y = c1 cos 3 x + c2 sin 3 x + −4x2 + 4x −
e .
3
Then A = 5, B = 4, C =
11
10
2
, D = −2, E = − 12
13 , yp = 5x + 4x +
11
10
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 − 8B = 1 and 8A − 19B = 0.
19
8
19
8
Then A = − 425
, B = − 425
, yp = − 425
cos 2x − 425
sin 2x, and
y = c1 e3x/2 + c2 e−x/2 −
19
8
cos 2x −
sin 2x.
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.
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 =
1
2
, 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 +
1
4
= 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.
2
Then yc = c1 e2x + c2 e−2x and we assume that
2
yp = (Ax + Bx + C) cos 2x + (Dx + 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.
116
3.4
Undetermined Coefficients
, so yp = − 18 x cos 2x + − 18 x2 +
1
13
1
2x
−2x
y = c1 e + c2 e
− x cos 2x + − x2 +
sin 2x.
8
8
32
Then A = 0, B = − 18 , C = 0, D = − 18 , E = 0, F =
13
32
13
32
sin 2x, and
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, 12A − 15B = −4, 6B − 10C = −1, and
1
53
697
1 4
14 3
53 2
697
2C − 5D = 6. Then A = − 10
, B = 14
75 , C = 250 , D = − 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 =
7
5
, 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
9
−4C − 3D = 0. Then A = − 12 , B = 0, C = − 25
,D=
y = c1 e−x + c2 xe−x −
12
25
, yp = − 12 cos x −
9
25
cos 2x +
12
25
sin 2x, and
1
9
12
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 −24A = 16, 2B + 10C = −2,
1 2
1
19
19
and 20B = −1. Then A = − 23 , B = − 20
, C = − 100
, yp = − 23 − 20
x + 100
x e4x , and
2
19
1 2
y = c1 e−6x + c2 e4x − −
x +
x e4x .
3
20
100
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 obtain −12A = 3, 6B − C = −1,
6
1
6
1
and B + 6C = 0. Then A = − 14 , B = − 37
, C = 37
, yp = − 14 x2 − 37
cos x + 37
sin x, and
1
6
1
y = c1 + c2 x + c3 e6x − x2 −
cos x +
sin x.
4
37
37
117
3.4
Undetermined Coefficients
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 differential equation we obtain 24A = 6 and 6A + 8B = 0.
3
3 2
Then A = 14 , B = − 16
, yp = 14 x3 − 16
x e2x , and
1 3
3
y = c1 e2x + c2 xe2x + c3 e−2x +
x − x2 e2x .
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 = c1 + c2 x + c3 e + c4 e
2
− x3 −
3
1 2 5
x + x e−x .
2
2
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
y=
√
2 sin 2x −
1
.
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 +
− 7x2 − 19x − 37.
e
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.
118
3.4
Undetermined Coefficients
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
y = 2e−2x + 9xe−2x +
x + x e
.
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.
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 +
F0
cos γt.
ω2 − γ 2
From the initial conditions we obtain c1 = −F0 /(ω 2 − γ 2 ) and c2 = 0, so
x=−
F0
F0
cos ωt + 2
cos γt.
ω2 − γ 2
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 + c3 sin 3 x) + x − + xe−2x .
4
8 3
√
23
59
17
From the initial conditions we obtain c1 = − 12 , c2 = − 24 , and c3 = 72 3 , so
√
√
23
17 √
1
5 2
59
y = − e−2x + ex − cos 3 x +
3 sin 3 x + x − + xe−2x .
12
24
72
4
8 3
y = c1 e−2x + ex (c2 cos
√
119
3.4
Undetermined Coefficients
37. We have yc = c1 cos x + c2 sin x and we assume yp = A2 + 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 − e c1 = π.
π
Solving this system we find c1 = 0 and c2 is any real number. The solution of the boundary-value 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
√
c1 + 3c2 + 2 = 0
√
√
c1 cos 3 + c2 sin 3 + 2 = 0.
Solving gives
Thus,
√
√
2(− 3 + sin 3 )
√
√
c1 = √
3 cos 3 − sin 3
and
c2 = √
√
2(1 − cos 3 )
√
√ .
3 cos 3 − sin 3
√
√
√
√
√
2(− 3 + sin 3 ) cos 3x 2(1 − cos 3 ) sin 3x
√
√
√
√
√ + 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 =
5
6
. Thus y = cos 2x +
5
6
sin 2x +
1
3
sin x on [0, π/2]. Now continuity
of y at x = π/2 implies
5
1
π
sin π + sin = c3 cos π + c4 sin π
6
3
2
2
= −c3 . Hence c3 = 3 . Continuity of y at x = π/2 implies
cos π +
or −1 +
1
3
−2 sin π +
5
1
π
cos π + cos = −2c3 sin π + 2c4 cos π
3
3
2
120
3.4
or − 53 = −2c4 . Then c4 =
5
6
Undetermined Coefficients
and the solution of the initial-value problem is
cos 2x + 56 sin 2x + 13 sin x, 0 ≤ x ≤ π/2
y(x) =
2
5
x > π/2.
3 cos 2x + 6 sin 2x,
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,
Solving this system, we find c1 = −2 and c2 =
continuity of y at x = π implies
eπ (−2 cos 3π +
2
3
c1 + 3c2 = 0.
. Thus y = ex (−2 cos 3x +
2
3
sin 3x) + 2 on [0, π]. Now,
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 ). The solution of the initial-value
problem is
x
e (−2 cos 3x + 23 sin 3x) + 2,
0≤x≤π
y(x) =
2
π x−π
(1 + e )e
(−2 cos 3x + 3 sin 3x), x > π.
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).
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.
121
3.4
Undetermined Coefficients
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 =
yp =
1
4
. Thus, a particular solution is
1
1
x cos x + sin3 x.
4
4
45. (a) f (x) = ex sin x. We see that yp → ∞ as x → ∞ and yp → 0 as x → −∞.
(b) f (x) = e−x . We see that yp → ∞ as x → ∞ and yp → ∞ as x → −∞.
(c) f (x) = sin 2x. We see that yp is sinusoidal.
(d) f (x) = 1. We see that yp is constant and simply translates yc vertically.
46. 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,
−8B + 6D = −1,
2B + 4F = 0,
−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
1
3
1 3 1 2 1
yp = − x3 + x2 + x e2x cos 2x +
x + x − x e2x sin 2x.
6
4
8
6
4
8
47. 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 =
122
1
8
,A=
1
8
, and yp = 18 x2 cos x + 18 x3 sin x.
3.5
Variation of Parameters
EXERCISES 3.5
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
u2 =
= 1.
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.
Then
u1 =
1
1
1
1
sin 2x − x = sin x cos x − x
4
2
2
2
1
u2 = − cos2 x.
2
123
3.5
Variation of Parameters
and
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
y = c1 cos x + c2 sin x +
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 −
1
3
sin3 x, and
1
1
cos4 x + sin2 x − sin4 x
3
3
1 2
= c1 cos x + c2 sin x +
cos x + sin2 x cos2 x − sin2 x + sin2 x
3
1
2
= c1 cos x + c2 sin x + cos2 x + sin2 x
3
3
1 1
= c1 cos x + c2 sin x + + sin2 x.
3 3
y = c1 cos x + c2 sin x +
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|
124
3.5
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
u1 =
1 −2x 1
e
+
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
8. 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) = sinh 2x we obtain
1
1
u1 = − e−3x + ex
4
4
u2 =
Then
u1 =
1 −x 1 3x
e − e .
4
4
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
− e2x
e
12
4
4
12
1
= c1 ex + c2 e−x +
e2x − e−2x
6
y = c1 ex + c2 e−x +
= c1 ex + c2 e−x +
1
sinh 2x.
3
9. The auxiliary equation is m2 − 4 = 0, so yc = c1 e2x + c2 e−2x and
2x
e
e−2x = −4.
W = 2x
2e
−2e−2x 125
Variation of Parameters
3.5
Variation of Parameters
Identifying f (x) = e2x /x we obtain u1 = 1/4x and u2 = −e4x /4x. Then
1
ln |x|,
4
1 x e4t
u2 = −
dt
4 x0 t
u1 =
and
2x
y = c1 e
−2x
+ c2 e
1
+
4
2x
e
−2x
x
ln |x| − e
x0
e4t
dt ,
t
x0 > 0.
10. The auxiliary equation is m2 − 9 = 0, so yc = c1 e3x + c2 e−3x and
3x
e
e−3x W = 3x
= −6.
3e
−3e−3x Identifying f (x) = 9x/e3x we obtain u1 = 32 xe−6x and u2 = − 32 x. Then
1 −6x 1 −6x
e
− xe
,
24
4
3
u2 = − x2
4
u1 = −
and
y = c1 e3x + c2 e−3x −
1 −3x 1 −3x 3 2 −3x
− xe
− x e
e
24
4
4
1
= c1 e3x + c3 e−3x − xe−3x (1 − 3x).
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 .
x
1+e
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 = e2x .
W = x
e
xex + ex Identifying f (x) = ex / 1 + x2 we obtain
u1 = −
u2 =
x
xex ex
=−
2x
2
e (1 + x )
1 + x2
e2x
1
ex ex
=
.
(1 + x2 )
1 + x2
126
3.5
Variation of Parameters
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 = −e−3x .
W = −x
−e
−2e−2x Identifying f (x) = sin ex we obtain
u1 =
e−2x sin ex
= ex sin ex
e−3x
e−x sin ex
= −e2x sin ex .
−e−3x
Then u1 = − cos ex , u2 = ex cos x − sin ex , and
u2 =
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 = e2t .
W = t
e 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
y = c1 et + c2 tet + −
tan−1 t +
2
2
1 et + t tan−1 t − ln 1 + t2 tet
2
1
= c1 et + c3 tet + et t2 − 1 tan−1 t − ln 1 + t2 .
2
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
= e−2t .
W = −t
−t
−t −e
−te + e
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
127
3.5
Variation of Parameters
Then
1
1
u1 = − t2 ln t + t2
2
4
u2 = t ln t − t
and
1
y = c1 e−t + c2 te−t − t2 e−t ln t +
2
1
= c1 e−t + c2 te−t + t2 e−t ln t −
2
1 2 −t
t e + t2 e−t ln t − t2 e−t
4
3 2 −t
t e .
4
16. The auxiliary equation is 2m2 + 2m + 1 = 0, so yc = e−x/2 [c1 cos(x/2) + c2 sin(x/2)] and
x
e−x/2 cos
2
W =
1 −x/2
x 1
x
− e
cos − e−x/2 sin
2
2 2
2
1
= e−x .
x 1 x/2
x 2
1 −x/2
cos − e sin e
2
2 2
2
e−x/2 sin
x
2
√
Identifying f (x) = 2 x we obtain
√
√
e−x/2 sin(x/2)2 x
x
= −4ex/2 x sin
−x/2
2
e
√
√
e−x/2 cos(x/2)2 x
x
u2 = −
= 4ex/2 x cos .
−x/2
2
e
u1 = −
Then
√
t
et/2 t sin dt
2
x0
x
√
t
u2 = 4
et/2 t cos dt
2
x0
x
u1 = −4
and
−x/2
y=e
x
x
x
c1 cos + c2 sin
− 4e−x/2 cos
2
2
2
x
t/2
e
x0
√
t
x
t sin dt + 4e−x/2 sin
2
2
x0
17. The auxiliary equation is 3m2 − 6m + 6 = 0, so yc = ex (c1 cos x + c2 sin x) and
ex cos x
ex sin x
= e2x .
W = x
x
x
x
e cos x − e sin x e cos x + e 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
= .
e2x
3
ln(cos x), u2 = 13 x, and
y = c1 ex cos x + c2 ex cos 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
xex/2
= ex .
W = 1 x/2 1 x/2
x/2 e
xe
+
e
2
2
128
x
√
t
et/2 t cos dt.
2
3.5
Variation of Parameters
√
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
To find u1 and u2 we use the substitution v = 1 − x2 and the trig substitution x = sin θ, respectively:
3/2
1 u1 =
1 − x2
12
x
1
u2 =
1 − x2 + sin−1 x.
8
8
Thus
3/2 1 2 x/2 1
1
y = c1 ex/2 + c2 xex/2 + ex/2 1 − x2
+ x e
1 − x2 + xex/2 sin−1 x.
12
8
8
u1 = −
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 x/2
= −1.
− 12 e−x/2 2e
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
y = c1 ex/2 + c2 e−x/2 + x2 ex/2 −
8
1
= c3 ex/2 + c2 e−x/2 + x2 ex/2 −
8
and
y =
1 x/2 1 x/2
+ e
xe
4
4
1 x/2
xe
4
1 x/2 1 −x/2
1
1
1
c3 e
− c2 e
+ x2 ex/2 + xex/2 − ex/2 .
2
2
16
8
4
The initial conditions imply
c3 + c2
=1
1
1
1
c3 − c2 − = 0.
2
2
4
Thus c3 = 3/4 and c2 = 1/4, and
y=
3 x/2 1 −x/2 1 2 x/2 1 x/2
+ e
+ x e
− xe .
e
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
e
e−x 3
= − e−x/2 .
W = 1 x/2
−x 2
−e
2e
Identifying f (x) = (x + 1)/2 we obtain
1 −x/2
(x + 1)
e
3
1
u2 = − ex (x + 1).
3
u1 =
Then
u1 = −e−x/2
1
u2 = − xex .
3
129
2
x−2
3
3.5
Variation of Parameters
Thus
y = c1 ex/2 + c2 e−x − x − 2
and
y =
1 x/2
− c2 e−x − 1.
c1 e
2
The initial conditions imply
c1 − c2 − 2 = 1
1
c1 − c2 − 1 = 0.
2
Thus c1 = 8/3 and c2 = 1/3, and
y=
8 x/2 1 −x
+ e − x − 2.
e
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 = −6e−2x .
W = 2x
2e
−4e−4x Identifying f (x) = 2e−2x − e−x we obtain
1 −4x 1 −3x
− e
e
3
6
1
1
u2 = e3x − e2x .
6
3
u1 =
Then
1 −4x
1
e
+ e−3x
12
18
1 3x 1 2x
e − e .
u2 =
18
6
u1 = −
Thus
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
y = c1 e2x + c2 e−4x −
and
1
1
y = 2c1 e2x − 4c2 e−4x + e−2x − e−x .
2
9
The initial conditions imply
5
=1
36
7
2c1 − 4c2 +
= 0.
18
c1 + c2 −
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
22. The auxiliary equation is m2 − 4m + 4 = (m − 2)2 = 0, so yc = c1 e2x + c2 xe2x and
2x
e
xe2x
= e4x .
W = 2x
2x
2x 2e
2xe + e
130
3.5
Variation of Parameters
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
=1
2c1 + c2 = 0.
Thus c1 = 1 and c2 = −2, and
y = e2x − 2xe2x + e2x x4 − x3 = e2x x4 − x3 − 2x + 1 .
23. Write the equation in the form
y +
1 1
y + 1− 2
x
4x
y = x−1/2
and identify f (x) = x−1/2 . From y1 = x−1/2 cos x and y2 = x−1/2 sin x we compute
x−1/2 cos x
W (y1 , y2 ) = −x−1/2 sin x − 1 x−3/2 cos x
2
1
x−1/2 sin x
= .
1 −3/2
−1/2
x
cos x − 2 x
sin x x
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 .
24. Write the equation in the form
1 1
sec(ln x)
y + 2y =
x
x
x2
2
and identify f (x) = sec(ln x)/x . From y1 = cos(ln x) and y2 = sin(ln x) we compute
cos(ln x) sin(ln x) 1
= .
W = x
sin(ln
x)
cos(ln
x)
−
x
x
y +
131
3.5
Variation of Parameters
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).
25. 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
u1 = W 1 = 0
tan x − cos x
1
u2 = W2 = 0
0
0
0
tan x
sin x cos x = tan x
− sin x sin x cos x = − sin x
− sin x 1
cos x
0 cos2 x − 1
= cos x − sec x.
0 = − sin x tan 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.
26. 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 132
3.5
Variation of Parameters
Identifying f (x) = sec 2x we obtain
0
cos 2x
sin 2x 1
1
1
u1 = W1 = 0
−2 sin 2x
2 cos 2x = sec 2x
4
8
8
sec 2x −4 cos 2x −4 sin 2x 1
0
sin 2x 1
1
1
0
2 cos 2x = −
u2 = W2 = 0
8
8
4
0 sec 2x −4 sin 2x 1
cos 2x
0 1
1
1
u3 = W3 = 0 −2 sin 2x
0 = − tan 2x.
8
8
4
0 −4 cos 2x sec 2x Then
1
ln | sec 2x + tan 2x|
8
1
u2 = − x
4
1
u3 = ln | cos 2x|
8
u1 =
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.
27. 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,
9
2
9
and B = 17
. Thus, yp1 = 17
cos x+ 17
sin x. Next, we consider the differential 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 so A =
2
17
Identifying f (x) = 13 ex tan x we obtain
1
1
u1 = − sin 3x tan 3x = −
9
9
so
u1 = −
Next
u2 =
Thus
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
1 x
1 x
e cos 3x(ln | sec 3x + tan 3x| − sin 3x) −
e sin 3x cos 3x
27
27
1
= − ex (cos 3x) ln | sec 3x + tan 3x|
27
yp2 = −
133
3.5
Variation of Parameters
and the general solution of the original differential equation is
y = ex (c1 cos 3x + c2 sin 3x) + yp1 (x) + yp2 (x).
28. 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,
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 = e2x .
W = x
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
.
29. The interval of definition for Problem 1 is (−π/2, π/2), for Problem 7 is (−∞, ∞), for Problem 9 is (0, ∞), and
for Problem 18 is (−1, 1). In Problem 24 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.
30. 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
dw
5
= − dx
w
x
and
ln |w| = −5 ln x + c.
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
x
1/x2 4
W =
=− .
3
x
2x −2/x
1 −4
Identifying f (x) = 1/x4 we obtain u1 = 14 x−5 and u2 = − 14 x−1 . Then u1 = − 16
x and u2 = − 14 ln x, so
yp = −
1 −4 2 1
1
1
x x − (ln x)x−2 = − x−2 − x−2 ln x.
16
4
16
4
134
3.6
Cauchy-Euler Equation
The general solution is
y = c1 x2 +
c2
1
1
−
− 2 ln x.
x2
16x2
4x
EXERCISES 3.6
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−
6
√
+ c2 x2+
√
6
.
√
8. The auxiliary equation is m2 + 2m − 4 = 0 so that y = c1 x−1+ 5 + c2 x−1− 5 .
9. The auxiliary equation is 25m2 + 1 = 0 so that y = c1 cos 15 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
3
−1/2
y=x
c1 cos
ln x + c2 sin
ln x
.
6
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 .
135
3.6
Cauchy-Euler Equation
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
Identifying f (x) = x3 we obtain u1 = − 15 x4 and u2 = 1/5x. Then u1 = − 25
x , u2 =
y = c1 + c2 x5 −
1
5
ln x, and
1 5 1 5
1
x + x ln x = c1 + c3 x5 + x5 ln x.
25
5
5
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
x−1/2 1 −5/2
−1
−1/2
= x
W (x , x
)=
.
−x−2 − 12 x−3/2 2
Identifying f (x) =
1
2
−
1
2x
we obtain u1 = x − x2 and u2 = x3/2 − x1/2 . Then u1 = 12 x2 − 13 x3 ,
u2 = 25 x5/2 − 23 x3/2 , and
1
2
2
1
1
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 = x.
W (x, x ln 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.
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,
136
and
W2 = ln x/x3 .
3.6
Cauchy-Euler Equation
Then u1 = W1 /W = −(ln x)/2, u2 = W2 /W = (ln x)/2x2 , and integration by parts gives
1
1
x − x ln x
2
2
1 −1
1
u2 = − x ln x − x−1 ,
2
2
u1 =
so
yp = u 1 y 1 + u 2 y 2 =
1
1
1
1
−1
x − x ln x x + − x−1 ln x − x−1 x = − ln x
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 ,
Then u1 = W1 /W = −1/2(x + 1),
gives
W1 = −1/x(x + 1),
W2 = 1/x3 (x + 1).
and
u2 = W2 /W = 1/2x2 (x + 1), and integration (by partial fractions for u2 )
1
u1 = − ln(x + 1)
2
1
1 −1 1
u2 = − x − 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 1
1
ln(x + 1)
1 1
1
ln(x + 1)
= − − x ln x + x ln(x + 1) −
= − + x ln 1 +
−
2 2
2
2x
2 2
x
2x
and
y = yc + yp = c1 x−1 + c2 x −
1 1
1
+ x ln 1 +
2 2
x
−
ln(x + 1)
,
2x
x > 0.
y
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
5x
c1 + c2 = 0
−2c2 = 4.
Thus, c1 = 2, c2 = −2, and y = 2 − 2x−2 . The graph is given to the right.
-10
-20
137
3.6
Cauchy-Euler Equation
26. The auxiliary equation is m2 − 6m + 8 = (m − 2)(m − 4) = 0, so that
2
4
y = c1 x + c2 x
y
3
and y = 2c1 x + 4c2 x .
30
The initial conditions imply
20
4c1 + 16c2 = 32
10
4c1 + 32c2 = 0.
-1
Thus, c1 = 16, c2 = −2, and y = 16x2 − 2x4 . The graph is given to the right.
4 x
-4
-20
-30
27. The auxiliary equation is m2 + 1 = 0, so that
y
3
y = c1 cos(ln x) + c2 sin(ln x)
and
1
1
sin(ln x) + c2 cos(ln x).
x
x
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.
y = −c1
100 x
50
-3
28. The auxiliary equation is m2 − 4m + 4 = (m − 2)2 = 0, so that
y
y = c1 x2 + c2 x2 ln x and y = 2c1 x + c2 (x + 2x ln x).
5
The initial conditions imply c1 = 5 and c2 + 10 = 3. Thus y = 5x2 − 7x2 ln x.
The graph is given to the right.
x
-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) = 0 1/x x
y
15
Identifying f (x) = 1 we obtain u1 = −x ln x and u2 = x. Then
10
u1 = 14 x2 − 12 x2 ln x, u2 = 12 x2 , and
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 =
y = 34 − ln x + 14 x2 . The graph is given to the right.
3
4
5
, c2 = −1, and
5
138
3.6
30. The auxiliary equation is m2 − 6m + 8 = (m − 2)(m − 4) = 0, so that
yc = c1 x2 + c2 x4 and
2
x
x4 = 2x5 .
W = 2x 4x3 Cauchy-Euler Equation
y
0.05
Identifying f (x) = 8x4 we obtain u1 = −4x3 and u2 = 4x. Then
-1
u1 = −x4 , u2 = 2x2 , and y = c1 x2 + c2 x4 + x6 . The initial conditions
imply
1
1
1
c1 + c2 = −
4
16
64
1
3
c1 + c2 = − .
2
16
1
1 2
Thus c1 = 16
, c2 = − 12 , and y = 16
x − 12 x4 + x6 . The graph is given above.
1
x
31. Substituting x = et into the differential equation we obtain
d2 y
dy
+8
− 20y = 0.
2
dt
dt
The auxiliary equation is m2 + 8m − 20 = (m + 10)(m − 2) = 0 so that
y = c1 e−10t + c2 e2t = c1 x−10 + c2 x2 .
32. Substituting x = et into the differential equation we obtain
d2 y
dy
− 10
+ 25y = 0.
2
dt
dt
The auxiliary equation is m2 − 10m + 25 = (m − 5)2 = 0 so that
y = c1 e5t + c2 te5t = c1 x5 + c2 x5 ln x.
33. Substituting x = et into the differential equation we obtain
d2 y
dy
+9
+ 8y = e2t .
2
dt
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
1
e = c1 x−1 + c2 x−8 + x2 .
30
30
34. Substituting x = et into the differential equation we obtain
d2 y
dy
−5
+ 6y = 2t.
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
1
5
5
y = c1 e2t + c2 e3t + t +
= c1 x2 + c2 x3 + ln x + .
3
18
3
18
35. Substituting x = et into the differential equation we obtain
d2 y
dy
−4
+ 13y = 4 + 3et .
2
dt
dt
139
3.6
Cauchy-Euler Equation
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
4
3
+ et
13 10
4
3
= x2 [c1 cos(3 ln x) + c2 sin(3 ln x)] +
+ x.
13 10
y = e2t (c1 cos 3t + c2 sin 3t) +
36. From
it follows that
d2 y
1
= 2
dx2
x
d3 y
1 d
= 2
dx3
x dx
d2 y dy
−
dt2
dt
d2 y dy
−
dt2
dt
−
2
x3
d2 y dy
−
dt2
dt
1 d dy
d2 y
2 d2 y
2 dy
− 2
− 3 2 + 3
2
dt
x dx dt
x dt
x dt
d3 y 1
2 dy
1 d2 y 1
2 d2 y
+ 3
−
−
3
2
2
3
2
dt
x
x dt
x
x dt
x dt
3
d y
d2 y
dy
−
3
+2
.
dt3
dt2
dt
=
1 d
x2 dx
=
1
x2
=
1
x3
Substituting into the differential equation we obtain
2
d3 y
d2 y
dy
d y dy
dy
−
3
+
2
−
−
3
+6
− 6y = 3 + 3t
dt3
dt2
dt
dt2
dt
dt
or
d3 y
d2 y
dy
−
6
+ 11
− 6y = 3 + 3t.
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
17 1
17 1
y = c1 et + c2 e2t + c3 e3t −
− t = c1 x + c2 x2 + c3 x3 −
− ln x.
12 2
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
dy dx
dy
=
=−
dt
dx dt
dx
and
d2 y
d
dy
d
d2 y dx
d2 y
d dy
dy dx
=
−
= − (y ) = −
=− 2
= 2.
=
2
dt
dt dt
dt
dx
dt
dx dt
dx dt
dx
37. The differential equation and initial conditions become
2
2 d y
4t
+ y = 0; y(t) = 2,
2
dt
t=1
y (t) The auxiliary equation is 4m2 − 4m + 1 = (2m − 1)2 = 0, so that
y = c1 t1/2 + c2 t1/2 ln t
and y =
= −4.
t=1
1 −1/2
1
+ c2 t−1/2 + t−1/2 ln t .
c1 t
2
2
140
3.6
Cauchy-Euler Equation
The initial conditions imply c1 = 2 and 1 + c2 = −4. Thus
y = 2t1/2 − 5t1/2 ln t = 2(−x)1/2 − 5(−x)1/2 ln(−x),
38. The differential equation and initial conditions become
d2 y
dy
t
− 4t
+ 6y = 0;
dt2
dt
2
y(t) = 8,
t=2
y (t) x < 0.
= 0.
t=2
The auxiliary equation is m2 − 5m + 6 = (m − 2)(m − 3) = 0, so that
y = c1 t2 + c2 t3
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 = 6x2 + 2x3 ,
x < 0.
39. Letting u = x + 2 we obtain dy/dx = dy/du and, using the Chain Rule,
d2 y
d dy
d2 y
d2 y du
d2 y
=
(1)
=
.
=
=
dx2
dx du
du2 dx
du2
du2
Substituting into the differential equation we obtain
u2
d2 y
dy
+u
+ y = 0.
du2
du
The auxiliary equation is m2 + 1 = 0 so that
y = c1 cos(ln u) + c2 sin(ln u) = c1 cos[ ln(x + 2)] + c2 sin[ ln(x + 2)].
40. 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.
41. 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.
√
42. The function y(x) = − x cos(ln x) is defined for x > 0 and has x-intercepts where ln x = π/2 + kπ for k an
integer or where x = eπ/2+kπ . Solving π/2 + kπ = 0.5 we get k ≈ −0.34, so eπ/2+kπ < 0.5 for all negative
integers and the graph has infinitely many x-intercepts in the interval (0, 0.5).
141
3.6
Cauchy-Euler Equation
43. 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 .
44. 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)].
45. The auxiliary equation is m(m − 1)(m − 2)(m − 3) + 6m(m − 1)(m − 2) + 3m(m − 1) − 3m + 4 = 0, so that
√
√
m1 = m2 = 2 and m3 = m4 = − 2 . The general solution of the differential equation is
√
y = c1 x
2
√
+ c2 x
2
√
ln x + c3 x−
2
√
+ c4 x−
2
ln x.
46. 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)].
47. The auxiliary equation
m(m − 1)(m − 2) − m(m − 1) − 2m + 6 = m3 − 4m2 + m + 6 = 0
has roots m1 = −1, m2 = 2, and m3 = 3, so yc = c1 x−1 + c2 x2 + c3 x3 . With y1 = x−1 , y2 = x2 , y3 = x3 , and
the identification f (x) = 1/x, we get from (10) of Section 4.6 in the text
W1 = x3 ,
W2 = −4,
W3 = 3/x,
and
W = 12x.
Then u1 = W1 /W = x2 /12, u2 = W2 /W = −1/3x, u3 = 1/4x2 , and integration gives
u1 =
x3
,
36
so
yp = u 1 y 1 + u 2 y 2 + u 3 y3 =
and
1
u2 = − ln x,
3
and
u3 = −
1
1
x3 −1
x + x2 − ln x + x3 −
36
3
4x
2
1
y = yc + yp = c1 x−1 + c2 x2 + c3 x3 − x2 − x2 ln x,
9
3
142
1
,
4x
2
1
= − x2 − x2 ln x,
9
3
x > 0.
3.7 Nonlinear Equations
EXERCISES 3.7
Nonlinear 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
y1 y1 = 1 · 0 = 0 =
Also, y2 = 2x and y2 = 2, so
1 2
1
(0) = (y1 )2 .
2
2
1
1
(2x)2 = (y2 )2 .
2
2
2
= (c1 · 1 + c2 x )(c1 · 0 + 2c2 ) = 2c2 (c1 + c2 x2 ) and
y2 y2 = x2 (2) = 2x2 =
However, if y = c1 y1 + c2 y2 , we have yy 1
2 [c1
· 0 + c2 (2x)] =
2
2c22 x2 .
Thus yy =
1 2
2 (y ) .
1 2
2 (y )
=
3. Let u = y so that u = y . The equation becomes u = −u − 1 which is separable. Thus
du
= −dx =⇒ tan−1 u = −x + c1 =⇒ y = tan(c1 − x) =⇒ y = ln | cos(c1 − x)| + c2 .
u2 + 1
4. Let u = y so that u = y . The equation becomes u = 1 + u2 . Separating variables we obtain
du
= dx =⇒ tan−1 u = x + c1 =⇒ u = tan(x + c1 ) =⇒ y = − ln | cos(x + c1 )| + c2 .
1 + u2
5. Let u = y so that u = y . The equation becomes x2 u + u2 = 0. Separating variables we obtain
1
x
1
du
1
1
dx
1
c1 x + 1
=
=
+
c
=⇒
u
=
−
−1
=
−
=⇒
−
=
1
2
2
u
x
u
x
x
c1 x + 1/c1
c1 c1 x + 1
1
1
=⇒ y = 2 ln |c1 x + 1| − x + c2 .
c1
c1
6. Let u = y so that y = u du/dy. The equation becomes (y + 1)u du/dy = u2 . Separating variables we obtain
du
dy
=
=⇒ ln |u| = ln |y + 1| + ln c1 =⇒ u = c1 (y + 1)
u
y+1
dy
dy
=⇒
= c1 (y + 1) =⇒
= c1 dx
dx
y+1
=⇒ ln |y + 1| = c1 x + c2 =⇒ y + 1 = c3 ec1 x .
7. Let u = y so that y = u du/dy. The equation becomes u du/dy + 2yu3 = 0. Separating variables we obtain
du
1
1
1
+ 2y dy = 0 =⇒ − + y 2 = c =⇒ u = 2
=⇒ y = 2
u2
u
y + c1
y + c1
1
=⇒ y 2 + c1 dy = dx =⇒ y 3 + c1 y = x + c2 .
3
143
3.7
Nonlinear Equations
8. Let u = y so that y = u du/dy. The equation becomes y 2 u du/dy = u. Separating variables we obtain
du =
dy
y
1
c1 y − 1
=⇒
dy = dx
=⇒ u = − + c1 =⇒ y =
y2
y
y
c1 y − 1
1
1
1
1
=⇒
1+
y + 2 ln |y − 1| = x + c2 .
dy = dx (for c1 = 0) =⇒
c1
c1 y − 1
c1
c1
If c1 = 0, then y dy = −dx and another solution is 12 y 2 = −x + c2 .
y
9. (a)
10
−π/2
x
3π/2
−10
(b) Let u = y so that y = u du/dy. The equation becomes u du/dy + yu = 0. Separating variables we obtain
1
1
du = −y dy =⇒ u = − y 2 + c1 =⇒ y = − y 2 + c1 .
2
2
When x = 0, y = 1 and y = −1 so −1 = −1/2 + c1 and c1 = −1/2. Then
dy
1
1
1
dy
1
= − y2 −
=⇒ 2
= − dx =⇒ tan−1 y = − x + c2
dx
2
2
y +1
2
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
y = tan
− x .
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.
10. 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 =⇒ sin−1 u = x + c1 =⇒ u = sin(x + c1 )
1 − u2
−2π
=⇒ y = sin(x + c1 ).
√
√
When x = π/2, y = 3/2, so 3/2 = sin(π/2 + c1 ) and c1 = −π/6. Thus
π
π
y = sin x −
=⇒ y = − cos x −
+ c2 .
6
6
144
y
2
x
2π
3.7 Nonlinear Equations
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:
du
√
= −dx =⇒ cos−1 u = x + c1
1 − u2
=⇒ u = cos(x + c1 ) =⇒ y = cos(x + c1 ).
−2π
√
√
When x = π/2, y = 3/2, so 3/2 = cos(π/2 + c1 ) and c1 = −π/3. Thus
π
π
y = cos x −
=⇒ y = sin x −
+ c2 .
3
3
y
1
x
2π
−1
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).
11. 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
c1
[x w] = −2x =⇒ x2 w = −x2 + c1 =⇒ w = −1 + 2
dx
x
c1
x
=⇒ u−2 = −1 + 2 =⇒ u = √
x
c1 − x2
dy
x
=⇒
=⇒
y
=
−
c1 − x2 + c2
=√
dx
c1 − x2
=⇒ c1 − x2 = (c2 − y)2 =⇒ x2 + (c2 − y)2 = c1 .
12. 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
1
1
1
c1 − x2
2x
=⇒ y = − ln c1 − x2 + c2 .
[xw] = −x =⇒ w = − x + c =⇒
=
=⇒ u =
2
dx
2
x
u
2x
c1 − x
In Problems 13-16 the thinner curve is obtained using a numerical solver, while the thicker curve is the graph of the
Taylor polynomial.
13. We look for a solution of the form
1
1
1
1
y(x) = y(0) + y (0)x + y (0)x2 + y (0)x3 + y (4) (0)x4 + y (5) (0)x5 .
2!
3!
4!
5!
y
40
From y (x) = x + y 2 we compute
y (x) = 1 + 2yy 30
y (4) (x) = 2yy + 2(y )2
y (5) (x) = 2yy + 6y y .
20
Using y(0) = 1 and y (0) = 1 we find
y (0) = 1,
y (0) = 3,
y (4) (0) = 4,
y (5) (0) = 12.
10
An approximate solution is
1
1
1
1
y(x) = 1 + x + x2 + x3 + x4 + x5 .
2
2
6
10
145
0.5 1 1.5 2 2.5 3 x
3.7
Nonlinear Equations
14. We look for a solution of the form
1
1
1
1
y(x) = y(0) + y (0)x + y (0)x2 + y (0)x3 + y (4) (0)x4 + y (5) (0)x5 .
2!
3!
4!
5!
y
10
From y (x) = 1 − y 2 we compute
y (x) = −2yy y
(4)
5
2
(x) = −2yy − 2(y )
y (5) (x) = −2yy − 6y y .
0.5 1 1.5 2 2.5 3
x
Using y(0) = 2 and y (0) = 3 we find
y (0) = −3,
y (0) = −12,
y (4) (0) = −6,
y (5) (0) = 102.
-5
An approximate solution is
3
1
17
y(x) = 2 + 3x − x2 − 2x3 − x4 + x5 .
2
4
20
15. We look for a solution of the form
1
1
1
1
y(x) = y(0) + y (0)x + y (0)x2 + y (0)x3 + y (4) (0)x4 + y (5) (0)x5 .
2!
3!
4!
5!
-10
y
40
From y (x) = x2 + y 2 − 2y we compute
30
y (x) = 2x + 2yy − 2y y (4) (x) = 2 + 2(y )2 + 2yy − 2y y (5) (x) = 6y y + 2yy − 2y (4) .
20
Using y(0) = 1 and y (0) = 1 we find
y (0) = −1,
y (0) = 4,
y (4) (0) = −6,
y (5) (0) = 14.
10
An approximate solution is
1
2
1
7
y(x) = 1 + x − x2 + x3 − x4 + x5 .
2
3
4
60
0.5 1 1.5 2 2.5 3 3.5x
16. We look for a solution of the form
1 1
1
y (0)x2 + y (0)x3 + y (4) (0)x4
2!
3!
4!
1 (5)
1 (6)
5
6
+ y (0)x + y (0)x .
5!
6!
y(x) = y(0) + y (0)x +
From y (x) = ey we compute
y
10
8
6
y (x) = ey y 4
y (4) (x) = ey (y )2 + ey y 2
y (5) (x) = ey (y )3 + 3ey y y + ey y y (6) (x) = ey (y )4 + 6ey (y )2 y + 3ey (y )2 + 4ey y y + ey y (4) .
1
-2
146
2
3
4
5x
3.7 Nonlinear Equations
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.
An approximate solution is
1
1
1
1
1
y(x) = −x + x2 − x3 + x4 + x5 + x6 .
2
6
12
24
45
17. 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
sec2 θ
sec2 θ
=
dx
=⇒
dθ
=
x
=⇒
dθ = x
3/2
3/2
sec3 θ
(1 + u2 )
1 + tan2 θ
u
=⇒
cos θ dθ = x =⇒ sin θ = x =⇒ √
=x
1 + u2
y
x2
=⇒ = x =⇒ (y )2 = x2 1 + (y )2 =
1 − x2
1 + (y )2
x
=⇒ y = √
(for x > 0) =⇒ y = − 1 − x2 .
1 − x2
18. 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.
19. Letting u = y , separating variables, and integrating we have
du du
= dx, and
= 1 + u2 , √
dx
1 + u2
sinh−1 u = x + c1 .
Then
u = y = sinh(x + c1 ),
y = cosh(x + c1 ) + c2 ,
and y = sinh(x + c1 ) + c2 x + c3 .
20. If the constant −c21 is used instead of c21 , then, using partial fractions,
x + c1
1
1
dx
1
1
dx =
y=−
=−
−
ln x2 − c21
2c1
x − c1
x + c1
2c1 x − c1
+c2 .
Alternatively, the inverse hyperbolic tangent can be used.
21. 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
k2
1 2
1 2
k2
k2
u
+
c
=⇒
v
+ c.
dx
=⇒
=
=
x2
2
x
2
x
When t = 0, x = x0 and v = 0 so 0 = (k 2 /x0 ) + c and c = −k 2 /x0 . Then
√
x0 − x
1 2
1
1
dx
.
v = k2
−
= −k 2
and
2
x x0
dt
xx0
u du = −
147
3.7
Nonlinear Equations
Separating variables we have
√
xx0
x
1 x0
−
dx = k 2 dt =⇒ t = −
dx.
x0 − x
k
2
x0 − x
Using Mathematica to integrate we obtain
1 x0
x
x0
−1 (x0 − 2x)
t=−
− x(x0 − x) −
tan
k
2
2
2x
x0 − x
1 x0 x0 − 2x
x0
=
x(x0 − x) +
tan−1 .
k
2
2
2 x(x0 − x)
22.
x
x
2
x
2
10
-2
20
2
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
1
-1
x
1
x1 = 0
10
1
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.
EXERCISES 3.8
Linear Models: Initial-Value Problems
1. From 18 x + 16x = 0 we obtain
√
√
x = c1 cos 8 2 t + c2 sin 8 2 t
√
√
so that the period of motion is 2π/8 2 = 2 π/8 seconds.
2. From 20x + kx = 0 we obtain
1
x = c1 cos
2
k
1
t + c2 sin
5
2
k
t
5
so that the frequency 2/π = 14 k/5 π 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 34 x + 72x = 0, x(0) = −1/4, and x (0) = 0 we obtain x = − 14 cos 4 6 t.
148
3.8
4. From 34 x + 72x = 0, x(0) = 0, and x (0) = 2 we obtain x =
√
6
12
5. From 58 x + 40x = 0, x(0) = 1/2, and x (0) = 0 we obtain x =
Linear Models: Initial-Value Problems
√
sin 4 6 t.
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 ft/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
(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 obtain
√
1
5
x = − cos 4t − sin 4t =
cos(4t − 3.605).
2
2
√
The period is π/2 seconds and the amplitude is 5/2 feet. In 4π seconds it will make 8 complete cycles.
9. From 14 x + x = 0, x(0) = 1/2, and x (0) = 3/2 we obtain
√
1
13
3
x = cos 2t + sin 2t =
sin(2t + 0.588).
2
4
4
10. From 1.6x + 40x = 0, x(0) = −1/3, and x (0) = 5/4 we obtain
1
1
5
x = − cos 5t + sin 5t =
sin(5t − 0.927).
3
4
12
If x = 5/24 then t = 15 π6 + 0.927 + 2nπ and t = 15 5π
6 + 0.927 + 2nπ for n = 0, 1, 2, . . . .
11. From 2x + 200x = 0, x(0) = −2/3, and x (0) = 5 we obtain
(a) x = − 23 cos 10t +
1
2
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 =
(j) If x = 5/12 then
(k) If x = 5/12 and
1
10 (0.927 + nπ) for n = 0, 1, 2, . . .. The velocity at these times is x = ±8.33 ft/s.
1
1
t = 10 (π/6 + 0.927 + 2nπ) and t = 10 (5π/6 + 0.927 + 2nπ) for n = 0, 1, 2, . . . .
1
(5π/6 + 0.927 + 2nπ) for n = 0, 1, 2, . . . .
x < 0 then t = 10
√
12. From x + 9x = 0, x(0) = −1, and x (0) = − 3 we obtain
√
3
4π
2
x = − cos 3t −
sin 3t = √ sin 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, . . . .
149
3.8
Linear Models: Initial-Value Problems
13. From k1 = 40 and k2 = 120 we compute the effective spring constant k = 4(40)(120)/160 = 120. Now,
m = 20/32 so k/m = 120(32)/20 = 192 and x + 192x = 0. Using x(0) = 0 and x (0) = 2 we obtain
√
√
x(t) = 123 sin 8 3 t.
14. Let m be the mass and k1 and k2 the spring constants. Then k = 4k1 k2 /(k1 + k2 ) is the effective spring constant
of the system. Since the initial mass stretches one spring 13 foot and another spring 12 foot, using F = ks, we
= 12 k2 or 2k1 = 3k2 . The given period of the combined system is 2π/ω = π/15, so ω = 30. Since a
mass weighing 8 pounds is 14 slug, we have from w2 = k/m
have
1
3 k1
302 =
k
= 4k
1/4
or k = 225.
We now have the system of equations
4k1 k2
= 225
k1 + k2
2k1 = 3k2 .
Solving the second equation for k1 and substituting in the first equation, we obtain
4(3k2 /2)k2
12k22
12k2
=
=
= 225.
3k2 /2 + k2
5k2
5
Thus, k2 = 375/4 and k1 = 1125/8. Finally, the weight of the first mass is
32m =
k1
1125/8
375
=
=
≈ 46.88 lb.
3
3
8
15. 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.
16. 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.
17. (a) above
(b) heading upward
18. (a) below
(b) from rest
19. (a) below
(b) heading upward
20. (a) above
(b) heading downward
+ 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 feet.
√
√
√ √
22. 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 feet.
21. From
1 8x
23. (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 + x + 16x = 0, x(0) = 1, and x (0) = −12 then x = − 23 e−2t + 53 e−8t .
24. (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 =
0.384 second and the extreme displacement is x = −0.232 meter.
150
1
6
ln 10 ≈
3.8
Linear Models: Initial-Value Problems
25. (a) From 0.1x + 0.4x + 2x = 0, x(0) = −1, and x (0) = 0 we obtain x = e−2t − cos 4t −
√
5 −2t
(b) x =
e
sin(4t + 4.25)
2
1
2
sin 4t .
(c) If x = 0 then 4t + 4.25 = 2π, 3π, 4π, . . . so that the first time heading upward is t = 1.294 seconds.
26. (a) From 14 x + x + 5x = 0, x(0) = 1/2, and x (0) = 1 we obtain x = e−2t 12 cos 4t + 12 sin 4t .
1
π
(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
.5
.5
2
t
-.5
-1
27. From
5 16 x
+ βx + 5x = 0 we find that the roots of the auxiliary equation are m = − 85 β ±
4
5
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.
√
28. 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
2 2
2 2
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.
x = e−2βt/3 c1 cosh
29. If 12 x + 12 x + 6x = 10 cos 3t, x(0) = −2, and x (0) = 0 then
√
√
47
47
xc = e−t/2 c1 cos
t + c2 sin
t
2
2
and xp =
10
3 (cos 3t
+ sin 3t) so that the equation of motion is
√
√
4
47
47
64
10
−t/2
x=e
− cos
t − √ sin
t + (cos 3t + sin 3t).
3
2
2
3
3 47
30. (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.
151
3.8
Linear Models: Initial-Value Problems
x
(b)
(c)
steady-state
3
2
-3
4
6
x
3
x=xc+xp
t
2
4
6
t
-3
transient
31. 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
32. 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 xp =
24 −t
7 −t
− 625
e cos 4t − 625
e sin 4t so that
x=
1 −t
1 −4t
e (24 + 100t) −
e (24 cos 4t + 7 sin 4t).
625
625
As t → ∞ the displacement x → 0.
33. 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 =
1 −2t
2e
cos 4t − 2e−2t sin 4t so that
1
9
1
x = − cos 4t + sin 4t + e−2t cos 4t − 2e−2t sin 4t.
2
4
2
√
85
4
34. Since x =
sin(4t − 0.219) −
√
17 −2t
2 e
sin(4t − 2.897), the amplitude approaches
√
85/4 as t → ∞.
35. (a) By Hooke’s law the external force is F (t) = kh(t) so that mx + βx + kx = kh(t).
(b) From
xp =
1 2 x + 2x + 4x
56
32
13 cos t + 13 sin t
= 20 cos t, x(0) = 0, and x (0) = 0 we obtain xc = e−2t (c1 cos 2t + c2 sin 2t) and
so that
−2t
x=e
−
72
56
32
56
cos 2t −
sin 2t +
cos t +
sin t.
13
13
13
13
36. (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 trig identity
x=
(b) If x =
1
3
(c) If x =
2
1
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, . . . .
− 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 many other values of t for which x = 0.
√
√
(d) x(π/6 + nπ/2) = 3/2 cm and x(π/3 + nπ/2) = − 3/2 cm.
(e)
8
3
cos 4t −
8
3
cos 8t =
8
3 (1
x
1
1
2
3
t
-1
152
3.8
Linear Models: Initial-Value Problems
37. 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 =
3
5
4 t sin 2t + 4 t cos 2t, and
1
3
5
x = − cos 2t − sin 2t + t sin 2t + t cos 2t.
8
4
4
38. 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
x = 2 cos 3t +
5
5
sin 3t − t cos 3t.
18
6
39. (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=−
(b) lim
γ→ω
ω2
F0
F0
cos ωt + 2
cos γt.
2
−γ
ω − γ2
F0
−F0 t sin γt
F0
(cos γt − cos ωt) = lim
=
t sin ωt.
γ→ω
ω2 − γ 2
−2γ
2ω
40. 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.
41. (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 =
1
2 [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.
42. 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
π
−1
9π
43. (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)
= Ae−λt sin[ ω 2 − λ2 t + φ],
where A = c21 + c22 , sin φ = c1 /A, and cos φ = c2 /A. Now
xp (t) =
(ω 2
F0 (ω 2 − γ 2 )
F0 (−2λγ)
sin γt + 2
cos γt = A sin(γt + θ),
− γ 2 )2 + 4λ2 γ 2
(ω − γ 2 )2 + 4λ2 γ 2
where
sin θ =
(ω 2
F0 (−2λγ)
− γ 2 )2 + 4λ2 γ 2
F0
ω 2 − γ 2 + 4λ2 γ 2
−2λγ
=
2
(ω − γ 2 )2 + 4λ2 γ 2
and
153
3.8
Linear Models: Initial-Value Problems
cos θ =
(ω 2
F0 (ω 2 − γ 2 )
− γ 2 )2 + 4λ2 γ 2
=
F0
(ω 2 − γ 2 )2 + 4λ2 γ 2
ω2 − γ 2
(ω 2
− γ 2 )2 + 4λ2 γ 2
.
√
(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
g
ω 2 − 2λ2 = F0 /2λ ω 2 − λ2 .
√
(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.
g
β
2.00
1.00
0.75
0.50
0.25
γ1
1.41
1.87
1.93
1.97
1.99
g
0.58
1.03
1.36
2.02
4.01
β=1/4
4
β=0
3
2
1
1
44. (a) For n = 2, sin2 γt =
γ1 = ω/2.
1
2 (1
2
β=1/2
β=0
β=3/4
β=0
β=1
β=1
β=2
β=2
γ
3
4
− cos 2γt). The system is in pure resonance when 2γ1 /2π = ω/2π, or when
(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 =
1
[sin 3γt − sin γt]
2
and
sin3 γt =
1
3
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 γ3 = ω/3, the system will be in pure resonance.
(c)
γ1=1/2
x
10
γ1=1
x
10
n=2
x
n=3
10
5
5
10
20
30
t
γ2=1/3
n=3
5
10
20
30
t
20
-5
-5
-5
-10
-10
-10
154
40
t
3.8
Linear Models: Initial-Value Problems
+ 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
5
−20t
5 cos 40t + sin 40t = 25 + 25/4 e−20t sin(40t + 1.1071)
q(t) = e
2
45. Solving
1 20 q
and q(0.01) ≈ 4.5676 coulombs.
0.0509 second.
The charge is zero for the first time when 40t + 1.1071 = π or t ≈
46. 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.
47. Solving
5 3q
+ 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.
48. 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.
√ √
49. 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) =
150
100
cos t +
sin t
13
13
and the steady-state current is
ip (t) = −
50. From
and Z =
√
E0
ip (t) =
Z
100
150
sin t +
cos t.
13
13
R
X
sin γt −
cos γt
Z
Z
X 2 + R2 we see that the amplitude of ip (t) is
E02 R2
E02 X 2
E0 2
E0
A=
+
=
R + X2 =
.
4
4
2
Z
Z
Z
Z
51. 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
155
3.8
Linear Models: Initial-Value Problems
E0
100
=
≈ 4.1603.
Z
Z
From Problem 50, 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).
52. 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) = −
1
3
4
1
sin 60t −
cos 60t +
sin 40t +
cos 40t
26
52
17
17
and the steady-state current is
ip (t) = −
30
45
160
40
cos 60t +
sin 60t +
cos 40t −
sin 40t.
13
13
17
17
+ 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
53. Solving
1 2q
1
3
q(t) = − e−10t (cos 10t + sin 10t) + .
2
2
As t → ∞, q(t) → 3/2.
54. In Problem 50 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.
√
55. By Problem 50 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 .
56. 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.
Equating coefficients we obtain A = 100/(100 − γ 2 ) and B = 0. Thus, qp (t) =
conditions q(0) = q (0) = 0 imply c1 = 0 and c2 = −10γ/(100 − γ 2 ). The charge is
q(t) =
10
(10 sin γt − γ sin 10t)
100 − γ 2
156
100
sin γt. The initial
100 − γ 2
3.9
and the current is
i(t) =
Linear Models: Boundary-Value Problems
100γ
(cos γt − cos 10t).
100 − γ 2
57. In an LC-series circuit there is no resistor, so the differential equation is
d2 q
1
+ q = E(t).
dt2
C
√
√
Then q(t) = c1 cos t/ LC + c2 sin t/ LC + qp (t) where qp (t) = A sin γt + B cos γt. Substituting qp (t) into
L
the differential equation we find
1
1
2
− Lγ 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 √
1
1
E0 C
cos γt.
t + c2 sin √
t+
1 − LCγ 2
LC
LC
√
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
i(t) = − √
sin γt
sin √
t+ √
cos √
t−
1 − LCγ 2
LC
LC
LC
LC
1
1
E0 C
1
E0 Cγ
= i0 cos √
sin √
sin γt.
t− √
q0 −
t−
2
1 − LCγ
1 − LCγ 2
LC
LC
LC
√
58. 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 +
=
E0
(kt cos kt + sin kt)
2kL
E0
E0
− q0 k sin kt + i0 cos kt +
t cos kt.
2kL
2L
EXERCISES 3.9
Linear Models: Boundary-Value Problems
1. (a) The general solution is
y(x) = c1 + c2 x + c3 x2 + c4 x3 +
157
w0 4
x .
24EI
3.9
Linear Models: Boundary-Value Problems
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
2c3 + 6c4 L +
L =0
2EI
w0
6c4 +
L = 0.
EI
Solving, we obtain c3 = w0 L2 /4EI and c4 = −w0 L/6EI. The deflection is
w0
y(x) =
(6L2 x2 − 4Lx3 + x4 ).
24EI
x
(b)
0.2 0.4 0.6 0.8 1
1
2
3 y
2. (a) The general solution is
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
c2 L + c4 L3 +
L =0
24EI
w0 2
6c4 L +
L = 0.
2EI
y(x) = c1 + c2 x + c3 x2 + c4 x3 +
Solving, we obtain c2 = w0 L3 /24EI and c4 = −w0 L/12EI. The deflection is
w0
y(x) =
(L3 x − 2Lx3 + x4 ).
24EI
(b)
0.2 0.4 0.6 0.8 1
x
1 y
3. (a) The general solution is
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
c3 L2 + c4 L3 +
L =0
24EI
w0 2
2c3 + 6c4 L +
L = 0.
2EI
y(x) = c1 + c2 x + c3 x2 + c4 x3 +
158
3.9
Linear Models: Boundary-Value Problems
Solving, we obtain c3 = w0 L2 /16EI and c4 = −5w0 L/48EI. The deflection is
w0
y(x) =
(3L2 x2 − 5Lx3 + 2x4 ).
48EI
(b)
0.2 0.4 0.6 0.8
1
1
x
y
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
c3 L2 + c4 L3 +
L =0
EIπ 3
2c3 + 6c4 L = 0.
y(x) = c1 + c2 x + c3 x2 + c4 x3 +
Solving, we obtain c3 = 3w0 L2 /2EIπ 3 and c4 = −w0 L/2EIπ 3 . The deflection is
w0 L
2L3
π
2
2
3
y(x) =
−2L
x
+
3Lx
−
x
+
sin x .
3
2EIπ
π
L
(b)
0.2 0.4 0.6 0.8
1
x
1 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
c2 L + c4 L3 +
L5 = 0
120EI
w0 3
6c4 L +
L = 0.
6EI
Solving, we obtain c2 = 7w0 L4 /360EI and c4 = −w0 L2 /36EI. The deflection is
w0
y(x) =
(7L4 x − 10L2 x3 + 3x5 ).
360EI
y(x) = c1 + c2 x + c3 x2 + c4 x3 +
159
3.9
Linear Models: Boundary-Value Problems
(b)
1 x
0.2 0.4 0.6 0.8
1
y
(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
P
P
w0 2 w0 EI
y = c1 cosh
.
x + c2 sinh
x+
x +
EI
EI
2P
P2
Setting y(0) = 0 we obtain c1 = −w0 EI/P 2 , so that
w0 EI
w0 2 w0 EI
P
P
y=−
x + c2 sinh
x+
x +
cosh
.
2
P
EI
EI
2P
P2
Setting y (L) = 0 we find
c2 =
P w0 EI
sinh
EI P 2
P
w0 L
L−
EI
P
P
cosh
EI
P
L.
EI
8. The general solution of the differential equation is
w0 2 w0 EI
P
P
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 EI
P
P
w0 2 w0 EI
y=−
cos
.
x + c2 sin
x+
x +
P2
EI
EI
2P
P2
Setting y (L) = 0 we find
c2 =
−
P w0 EI
sin
EI P 2
w0 L
P
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, . . . .
160
3.9
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
(2n − 1)π
2
(2n − 1)2 π 2
, n = 1, 2, 3, . . . .
4L2
(2n − 1)π
The eigenvalues (2n − 1)2 π 2 /4L2 correspond to the eigenfunctions cos
x for n = 1, 2, 3, . . . .
2L
12. For λ ≤ 0 the only solution of the boundary-value problem is y = 0. For λ = α2 > 0 we have
αL =
or λ = α2 =
y = c1 cos αx + c2 sin αx.
Since y(0) = 0 implies c1 = 0, y = c2 sin x dx. Now
π
π
y
= c2 α cos α = 0
2
2
gives
π
(2n − 1)π
α =
or λ = α2 = (2n − 1)2 , n = 1, 2, 3, . . . .
2
2
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, . . . .
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.
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 161
(1)
3.9
Linear Models: Boundary-Value Problems
Then
n2
; n = 1, 2, 3, . . . .
4
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
2απ = nπ
or
λ = α2 =
(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
1
−2 ± 4 − 4(λ + 1) = −1 ± α.
2
m=
For λ = −α2 < 0 we have
y = e−x (c1 cosh αx + c2 sinh αx) .
The boundary conditions imply
y(0) = c1 = 0
y(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
gives
n2 π 2
, n = 1, 2, 3, . . . .
25
n2 π 2
nπ
The eigenvalues λn =
correspond to the eigenfunctions yn = e−x sin
x for n = 1, 2, 3, . . . .
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 .
5α = nπ
or λ = α2 =
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 λ = n2 π 2 − 1, n = 1, 2, 3, . . . .
The eigenvalues n2 π 2 − 1 correspond to the eigenfunctions cos nπx for n = 0, 1, 2, . . . .
162
3.9
Linear Models: Boundary-Value Problems
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,
λ = α 2 = n2 ,
n = 1, 2, 3, . . .
and y = 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.
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 λ = α2 =
(2n − 1)2 π 2 /4 for n = 1, 2, 3, . . . . The corresponding eigenfunctions are
2n − 1
yn = sin
π ln x .
2
19. 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.
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 or
2α2 sinh α sin α = 0.
−α2 sin α α2 sinh 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
y = sin nπx.
20. 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.
163
3.9
Linear Models: Boundary-Value Problems
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
c2 cos απ + c4 cosh απ = 0
−c2 λ cos απ + c4 λ2 cosh απ = 0.
2
The determinant of the coefficients is 2α2 cosh α cos α = 0. But since α > 0, the only way that this is satisfied
is to have cos απ = 0 or α = (2n − 1)/2, n = 1, 2, 3, . . . . The eigenvalues and corresponding eigenfunctions are
4
2n − 1
2n − 1
λn = α 4 =
, n = 1, 2, 3, . . .
and
y = cos
x.
2
2
y
21. 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 .
L
x
22. (a) The general solution of the differential equation is
P
P
y = c1 cos
x + c2 sin
x + δ.
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.
(b) If δ = 0, the boundary conditions give, in turn, c1 = −δ and c2 = 0. Then
P
y = δ 1 − cos
x .
EI
In order to satisfy the boundary condition y(L) = δ we must have
P
P
δ = δ 1 − cos
L
or cos
L = 0.
EI
EI
This gives
P/EI L = nπ/2 for n = 1, 2, 3, . . . . The smallest value of Pn , the Euler load, is then
π
P1
1 π 2 EI
L=
or P1 =
.
EI
2
4
L2
23. 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 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.
164
3.9
Linear Models: Boundary-Value Problems
24. (a) The boundary-value problem is
d4 y
d2 y
+ λ 2 = 0,
4
dx
dx
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 or
tan β = β,
α cos αL 1 = 0
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 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
yn (x) = c2 sin αn x + c3 x = c2 sin
x −
cos βn x .
L
L
(b) With L = 1 and c2 appropriately chosen, the general shape of the first buckling mode,
y1 (x) = c2 sin
4.4934
x −
L
4.4934
cos(4.4934) x ,
L
is shown below.
y1
0.2
25. The general solution is
0.4
0.6
0.8
1
x
ρ
ρ
y = c1 cos
ωx + c2 sin
ωx.
T
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
y(x) = c2 sin
nπ
x,
L
n = 1, 2, 3, . . . ,
where c2 = 0.
26. (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
165
3.9
Linear Models: Boundary-Value Problems
are
1 1
1 1
m1 = − −
4ρω 2 − 1 i, m2 = − +
4ρω 2 − 1 i
2 2
2 2
when ρω 2 > 0.25. Thus
y = c1 x−1/2 cos(λ ln x) + c2 x−1/2 sin(λ ln x)
where λ = 12 4ρω 2 − 1. Applying y(1) = 0 gives c1 = 0 and consequently
y = c2 x−1/2 sin(λ ln x).
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
1
λn =
4ρωn2 − 1 = nπ.
2
Solving for ωn gives
1 2 2
ωn =
(4n π + 1)/ρ .
2
The corresponding solutions are
yn (x) = c2 x−1/2 sin(nπ ln x).
(b)
y
y
1
y
1
n=1
1
n=3
n=2
ex
1
ex
1
-1
-1
1
ex
-1
27. 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
c1 =
ab and c2 =
.
b−a
b−a
Thus
u(r) =
u0 − u1
b−a
ab u1 b − u0 a
+
.
r
b−a
28. 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)
u1 ln a − u0 ln b u0 − u1
u0 ln(r/b) − u1 ln(r/a)
+
ln r =
.
ln(a/b)
ln(a/b)
ln(a/b)
29. The solution of the initial-value problem
x + ω 2 x = 0,
x(0) = 0, x (0) = v0 , ω 2 = 10/m
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
166
3.9
Linear Models: Boundary-Value Problems
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
to t = 1; and so on.
2
3
, the mass will pass through x = 0 twice (t =
1
3
and t = 23 ) prior
30. The initial-value problem is
x +
2 k
x + x = 0,
m
m
x(0) = 0, x (0) = v0 .
√
With k = 10, the auxiliary equation has roots γ = −1/m ± 1 − 10m/m. Consider the three cases:
. 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.
(i) m =
1
10
(ii) 1 − 10m > 0 or 0 < m <
1
10
γ1 = −
. The roots are
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
γ1 c1 + γ2 c2 = 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
and the solution of the differential equation is
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
mv0
1 √
−t/m
x(t) = √
sin
10m − 1 t ,
e
m
10m − 1
x(t) = c1 e−t/m cos
The condition x(1) = 0 implies
√
mv0
1√
e−1/m sin
10m − 1 = 0
m
10m − 1
1√
10m − 1 = 0
sin
m
1√
10m − 1 = nπ
m
10m − 1
= n2 π 2 , n = 1, 2, 3, . . .
m2
(n2 π 2 )m2 − 10m + 1 = 0
167
3.9
Linear Models: Boundary-Value Problems
√
√
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
m1 =
and
m2 =
.
π2
π2
31. (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 will be no solutions.
(d) The boundary-value problem will have the trivial solution when y0 = y1 = 0. This solution will not be
unique.
32. (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.
33. (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.
168
3.9
Linear Models: Boundary-Value Problems
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, 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.
y
(c)
y
3
-p
3
px
-p
-3
y = 2 sin 3x
2
px
-3
√
y = sin 4x − 2 cos 3x
√
α x + c2 sin α x. Setting y(0) = 0 we find c1 = 0, so that
34. For λ = α > 0 the general solution is y = c1 cos
√
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 .
35. We see from the graph that tan x = −x has infinitely many roots. Since
λn = αn2 , there are no new eigenvalues when αn < 0. For λ = 0, the
differential equation y = 0 has general solution y = c1 x + c2 . The boundary
tan x
5
2.5
conditions imply c1 = c2 = 0, so y = 0.
2
4
6
8 10 12 x
-2.5
-5
-7.5
-10
36. 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).
169
3.9
Linear Models: Boundary-Value Problems
37. 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
y
1
point with approximate x-coordinate α1 = 1.915. Thus, there is a
single negative eigenvalue λ1 = −α12 ≈ −3.667 and the corresponding
eigenfuntion is y1 = sinh 1.915x.
1
x
2
For λ = 0 the only solution of the boundary-value 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 α cos 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.
38. For λ = α4 , α > 0, the solution of the differential equation is
y = c1 cos αx + c2 sin αx + c3 cosh αx + c4 sinh αx.
y
1
The boundary conditions y(0) = 0, y (0) = 0, y(1) = 0, y (1) = 0 give,
in turn,
c1 + c3 = 0
2
4
6
8
10
12
x
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 =0
cos α − cosh α 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. 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. Using the third equation in the
system to eliminate c2 , we find that the eigenfunctions are
yn = (− sin αn + sinh αn )(cos αn x − cosh αn x) + (cos αn − cosh αn )(sin αn x − sinh αn x).
170
3.10
Nonlinear Models
EXERCISES 3.10
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.
x
2
1
4
2
6
8
-1
-2
2. The solutions are not periodic.
x
10
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
x
10
8
solution.
6
4
2
4
2
6
8
10
-2
4. Both solutions have periods of approximately 6.3.
x
3
2
1
-1
-2
-3
171
2
4
6
8
10
t
t
3.10 Nonlinear Models
5. From the graph we see that |x1 | ≈ 1.2.
x
4
3
2
1
x1=1.2
x1=1.1
t
−1
5
6. From the graphs we see that the interval is approximately
(−0.8, 1.1).
10
x
3
2
1
5
10
t
−1
7. Since
xe0.01x = x[1 + 0.01x +
for small values of x, a linearization is
x
8.
1
(0.01x)2 + · · · ] ≈ x
2!
d2 x
+ x = 0.
dt2
3
t
5
10
15
−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.
172
3.10
9. This is a damped hard spring, so x will approach 0 as t
Nonlinear Models
x
approaches ∞.
2
4
2
6
8
t
-2
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.
x
5
4
3
2
1
2
4
t
-1
-2
11.
x
15
x
k1 = 0.01
10
2
5
1
10
20
t
30
10
-5
-1
-10
-2
-15
-3
x
k1 = 1
3
x
k1 = 20
3
3
2
2
1
1
5
10
t
-1
-2
-2
-3
-3
t
k1 = 100
1
-1
20
2
3
t
When k1 is very small the effect of the nonlinearity is greatly diminished, and the system is close to pure
resonance.
173
3.10 Nonlinear Models
12. (a)
x
x
40
40
20
20
20
40
60
80
t
100
20
-20
40
60
80
100
t
-20
k 0.000465
-40
k 0.000466
-40
The system appears to be oscillatory for −0.000465 ≤ k1 < 0 and nonoscillatory for k1 ≤ −0.000466.
(b)
x
x
3
3
2
2
1
1
20
40
60
80
100 120 140
t
20
-1
-2
40
60
80
100 120 140
t
-1
k 0.3493
k 0.3494
-2
-3
-3
The system appears to be oscillatory for −0.3493 ≤ k1 < 0 and nonoscillatory for k1 ≤ −0.3494.
13. For λ2 − ω 2 > 0 we choose λ = 2 and ω = 1 with
x(0) = 1 and x (0) = 2. For λ2 − ω 2 < 0 we choose
λ = 1/3 and ω = 1 with x(0) = −2 and x (0) = 4. In
θ
3
λ=1/3, ω=1
both cases the motion corresponds to the overdamped
λ=2, ω=1
and underdamped cases for spring/mass systems.
t
5
10
15
−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/dt, 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
R2
− 2gR + v02 .
y
√
(b) As y → ∞ we assume that v → 0+ . Then v02 = 2gR and v0 = 2gR .
v 2 = 2g
(c) Using g = 32 ft/s and R = 4000(5280) ft we find
v0 = 2(32)(4000)(5280) ≈ 36765.2 ft/s ≈ 25067 mi/hr.
(d) v0 =
2(0.165)(32)(1080) ≈ 7760 ft/s ≈ 5291 mi/hr
174
3.10
Nonlinear Models
15. (a) Intuitively, one might expect that only half of a 10-pound chain could be lifted by a 5-pound vertical force.
√
(b) Since x = 0 when t = 0, and v = dx/dt = 160 − 64x/3 , we have v(0) = 160 ≈ 12.65 ft/s.
(c) Since x should always be positive, we solve x(t) = 0, getting t = 0 and t = 32 5/2 ≈ 2.3717. Since the
graph of x(t) is a parabola, the maximum value occurs at tm = 34 5/2 . (This can also be obtained by
solving x (t) = 0.) At this time the height of the chain is x(tm ) ≈ 7.5 ft. This is higher than predicted
because of the momentum generated by the force. When the chain is 5 feet high it still has a positive
velocity of about 7.3 ft/s, which keeps it going higher for a while.
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
v
1
dv =
dx
v 2 + Lg
L−x
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
and
Again, separating variables and integrating we obtain
L−x
dx = dt
√
and
2Lx − x2
√
= t + c1 .
Lg
Lg(2Lx − x2 )
√
√
Since x(0) = 0, we have c1 = 0 and 2Lx − x2 / Lg = t. Solving for x we get
√
Lgt
dx
2
2
x(t) = L − L − 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.
17. (a) The weight of x feet of the chain is 2x, so the corresponding mass is m = 2x/32 = x/16. The only force
acting on the chain is the weight of the portion of the chain hanging over the edge of the platform. Thus,
by Newton’s second law,
d
d x
1
dv
dx
1
dv
(mv) =
v =
x
+v
=
x
+ v 2 = 2x
dt
dt 16
16
dt
dt
16
dt
and x dv/dt + v 2 = 32x.
xv dv/dx + v 2 = 32x.
Now, by the Chain Rule, dv/dt = (dv/dx)(dx/dt) = v dv/dx, so
(b) We separate variables and write the differential equation as (v 2 − 32x) dx + xv dv = 0. This is not an exact
form, but µ(x) = x is an integrating factor. Multiplying by x we get (xv 2 − 32x2 ) dx + x2 v dv = 0. This
1 2 2
32 3
3
form is the total differential of u = 12 x2 v 2 − 32
3 x , so an implicit solution is 2 x v − 3 x = c. Letting x = 3
and v = 0 we find c = −288. Solving for v we get
√
dx
8 x3 − 27
√
, 3 ≤ x ≤ 8.
=v=
dt
3x
(c) Separating variables and integrating we obtain
√
x
8
dx = √ dt
3
x3 − 27
and
3
175
x
√
s
8
ds = √ t + c.
3
s3 − 27
3.10 Nonlinear Models
Since x = 3 when t = 0, we see that c = 0 and
√ x
3
s
√
t=
ds.
3
8 3
s − 27
We want to find t when x = 7. Using a CAS we find t(7) = 0.576 seconds.
18. (a) There are two forces acting on the chain as it falls from the platform. One is the force due to gravity on
the portion of the chain hanging over the edge of the platform. This is F1 = 2x. The second is due to the
motion of the portion of the chain stretched out on the platform. By Newton’s second law this is
F2 =
=
From
d
d (8 − x)2
d 8−x
[mv] =
v =
v
dt
dt
32
dt 16
8 − x dv
1 dx
1
dv
− v
=
(8 − x)
− v2 .
16 dt
16 dt
16
dt
d
[mv] = F1 − F2 we have
dt
d 2x
1
dv
v = 2x −
(8 − x)
− v2
dt 32
16
dt
x dv
1 dx
1
dv
+
v
= 2x −
(8 − x)
− v2
16 dt
16 dt
16
dt
x
dv
dv
+ v 2 = 32x − (8 − x)
+ v2
dt
dt
x
dv
dv
dv
= 32x − 8
+x
dt
dt
dt
dv
= 32x.
dt
By the Chain Rule, dv/dt = (dv/dx)(dx/dt) = v dv/dx, so
8
8
dv
dv
dv
= 8v
= 32x and v
= 4x.
dt
dx
dx
(b) Integrating v dv = 4x dx we get 12 v 2 = 2x2 + c. Since v = 0 when x = 3, we have c = −18. Then
√
v 2 = 4x2 − 36 and v = 4x2 − 36 . Using v = dx/dt, separating variables, and integrating we obtain
dx
x
= 2 dt and cosh−1 = 2t + c1 .
2
3
x −9
Solving for x we get x(t) = 3 cosh(2t + c1 ). Since x = 3 when t = 0, we have cosh c1 = 1 and c1 = 0. Thus,
√
x(t) = 3 cosh 2t. Differentiating, we find v(t) = dx/dt = 6 sinh 2t.
(c) To find the time when the back end of the chain leaves the platform we solve x(t) = 3 cosh 2t = 8. This
gives t1 = 12 cosh−1 83 ≈ 0.8184 seconds. The velocity at this instant is
√
8
v(t1 ) = 6 sinh cosh−1
= 2 55 ≈ 14.83 ft/s.
3
(d) Replacing 8 with L and 32 with g in part (a) we have L dv/dt = gx. Then
dv
dv
dv
g
= Lv
= gx and v
= x.
dt
dx
dx
L
Integrating we get 12 v 2 = (g/2L)x2 + c. Setting x = x0 and v = 0, we find c = −(g/2L)x20 . Solving for v
we find
g 2 g 2
v(x) =
x − x0 .
L
L
L
176
3.10
Nonlinear Models
Then the velocity at which the end of the chain leaves the edge of the platform is
g 2
v(L) =
(L − x20 ) .
L
19. Let (x, y) be the coordinates of S2 on the curve C. The slope at (x, y) is then
dy/dx = (v1 t − y)/(0 − x) = (y − v1 t)/x or xy − y = −v1 t.
Differentiating with respect to x and using r = v1 /v2 gives
dt
dx
dt ds
xy = −v1
ds dx
1
xy = −v1 (− 1 + (y )2 )
v2
xy = r 1 + (y )2 .
xy + y − y = −v1
Letting u = y and separating variables, we obtain
du
= r 1 + u2
dx
du
r
√
= dx
2
x
1+u
x
sinh−1 u = r ln x + ln c = ln(cxr )
u = sinh(ln cxr )
dy
1
1
=
cxr − r
dx
2
cx
.
At t = 0, dy/dx = 0 and x = a, so 0 = car − 1/car . Thus c = 1/ar and
dy
1 x r a r 1 x r x −r
.
=
−
−
=
dx
2 a
x
2
a
a
If r > 1 or r < 1, integrating gives
y=
a
1 x 1−r
1 x 1+r
+ c1 .
−
2 1+r a
1−r a
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
a
ar
1 x 1+r
1 x 1−r
y=
+
−
.
2 1+r a
1−r a
1 − r2
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 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
y=
1 x2
1
1 a 1
1 1 2
1 a
− ln x −
− ln a =
(x − a2 ) + ln
.
2 2a a
2 2 a
2 2a
a x
Since y → ∞ as x → 0+ , S2 will never catch up with S1 .
177
3.10 Nonlinear Models
20. (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
= r2
2
dθ
dt
+
dr
dt
2
.
Using ds/dt = 30 and dr/dt = 15 then gives
900 = 225t
2
675 = 225t2
dθ
dt
2
dθ
dt
2
+ 225
√
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
θ(t) =
Thus r = 3eθ/
√
3
√
r
1
3 ln
− ln
15
5
=
√
3 ln
r
.
3
, whose graph is a logarithmic spiral.
(c) The time for S1 to go from (9, 0) to (3, 0) =
1
5
hour. Now S1 must intercept the path of S2 for some angle
β, where 0 < β < 2π. At the time of interception t2 we have 15t2 = 3eβ/
is then
t=
√
3
√
or t = 15 eβ/
3
. The total time
√
1 1 β/√3
1
< (1 + e2π/ 3 ).
+ e
5 5
5
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
1.25
approximation is accurate to one decimal place for θ1 = 0.67
1
and to two decimal places for θ1 = 0.31.
1.5
0.75
0.5
0.25
0.25 0.5 0.75
178
1
1.25 1.5
x
3.10
(b)
Θ
Nonlinear Models
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
23. (a) Write the differential equation as
2
3
4
5
6
θ
2
d2 θ
+ ω 2 sin θ = 0,
dt2
moon
earth
2
where ω = g/l. To test for differences between the earth
and the moon we take l = 3, θ(0) = 1, and θ (0) = 2.
t
5
Using g = 32 on the earth and g = 5.5 on the moon we
-2
obtain the graphs shown in the figure. Comparing the
apparent periods of the graphs, we see that the pendulum oscillates faster on the earth than on the moon.
(b) The amplitude is greater on the moon than on the earth.
(c) The linear model is
θ
d2 θ
+ ω 2 θ = 0,
dt2
2
moon
earth
where ω 2 = g/l. When g = 32, l = 3, θ(0) = 1, and
θ (0) = 2, the solution is
θ(t) = cos 3.266t + 0.612 sin 3.266t.
t
5
-2
When g = 5.5 the solution is
θ(t) = cos 1.354t + 1.477 sin 1.354t.
As in the nonlinear case, the pendulum oscillates faster on the earth than on the moon and still has greater
amplitude on the moon.
24. (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
θ(t) = (π/12) cos t − (1/3) sin t.
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
√ √
θ (0) = − sin(π/12) = − 2 ( 3 − 1)/4
and
√ √
θ (0) = − cos θ(0) · θ (0) = − cos(π/12)(−1/3) = 2 ( 3 + 1)/12.
179
3.10 Nonlinear Models
Thus
θ(t) =
π
1
− t−
12 3
√
√
√ √
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
√ √
π
2 ( 3 − 1) 2
1
− t−
t =0
12 3
8
and using the positive root we obtain t1 ≈ 0.63088.
(e) Setting
√ √
√ √
π
2 ( 3 − 1) 2
2 ( 3 + 1) 3
1
− t−
t +
t =0
12 3
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
(g) 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
25. From the table below we see that the pendulum first passes the vertical position between 1.7 and 1.8 seconds.
To refine our estimate of t1 we estimate the solution of the differential equation on [1.7, 1.8] using a step size of
h = 0.01. From the resulting table we see that t1 is between 1.76 and 1.77 seconds. Repeating the process with
h = 0.001 we conclude that t1 ≈ 1.767. Then the period of the pendulum is approximately 4t1 = 7.068. The
error when using t1 = 2π is 7.068 − 6.283 = 0.785 and the percentage relative error is (0.785/7.068)100 = 11.1.
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
h=0.01
θn
0.78540
0.78523
0.78407
0.78092
0.77482
0.76482
0.75004
0.72962
0.70275
0.66872
0.62687
0.57660
0.51744
0.44895
0.37085
0.28289
0.18497
0.07706
-0.04076
-0.16831
-0.30531
tn
1.70
1.71
1.72
1.73
1.74
1.75
1.76
1.77
1.78
1.79
1.80
θn
0.07706
0.06572
0.05428
0.04275
0.03111
0.01938
0.00755
-0.00438
-0.01641
-0.02854
-0.04076
h=0.001
1.763
1.764
1.765
1.766
1.767
1.768
1.769
1.770
0.00398
0.00279
0.00160
0.00040
-0.00079
-0.00199
-0.00318
-0.00438
180
3.11 Solving Systems of Linear Equations
EXERCISES 3.11
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 −
1
4
, Dy = 14 D2 x, and (D2 + 1)x = 2. The solution is
x = c1 cos t + c2 sin t + 2
1
1
1
y = c2 cos t − c1 sin t − .
4
4
4
5. From (D2 + 5)x − 2y = 0 and −2x + (D2 + 2)y = 0 we obtain y = 12 (D2 + 5)x, D2 y =
(D2 + 1)(D2 + 6)x = 0. The solution is
√
√
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
1
4
2 (D
+ 5D2 )x, and
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
y = c1 cos 5 t + c2 sin 5 t −
5
√
√
√
√
2
5
5
2
3
x = − c1 −
c2 cos 5 t +
c1 − c2 sin 5 t + .
3
3
3
3
5
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
181
3.11 Solving Systems of Linear Equations
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
4
y = c1 + c2 sin t + c3 cos t − e3t
15
17
x = c4 + c5 sin t + c6 cos t + e3t .
15
3t
Substituting into (D + 1)x + (D − 1)y = 4e 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
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 = −c1 , c5 = −c2 , c6 = −3c3 , and
1
y = −c1 − c2 e−t − 3c3 e−3t + t − t2 .
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
√
t
−t/2
x = c1 e + e
√ 3
3
c2 cos
t + c3 sin
t
2
2
y=
√
√
√
√
3
3
3
3
3
3
−t/2
−t/2
− c2 −
c3 e
t+
c2 − c3 e
t.
cos
sin
2
2
2
2
2
2
182
3.11 Solving Systems of Linear Equations
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
and Dy = −3c1 e4t + 5et 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 .
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
1
2
y = c1 + c2 e−t cos t + c3 e−t sin t + cos t + sin t
5
5
1
2
x = (c2 + c3 )e−t sin t + (c2 − c3 )e−t cos t + sin t − cos t.
5
5
183
3.11 Solving Systems of Linear Equations
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,
√
√ 3
3
x = c1 et + e−t/2 c2 sin
t + c3 cos
t ,
2
2
√
√
√
√
3
3
3
3
1
1
−t/2
− c2 −
sin
c3 e
t+
c2 − c3 e−t/2 cos
t,
2
2
2
2
2
2
t
y = c1 e +
and
√
√
√
√
3
3
3
3
1
1
−t/2
− c2 +
sin
c3 e
t+ −
c2 − c3 e−t/2 cos
t.
2
2
2
2
2
2
t
z = c1 e +
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 2t
c1 e − c2 cos t − c3 sin t
2
z = −2c1 e2t − c3 cos t + c2 sin t.
y=
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 −t 1 −2t 1 3t
z = − c1 e − c2 e
+ c3 e .
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 .
184
3.11 Solving Systems of Linear 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
√
√ y = et c1 cos 2 t + c2 sin 2 t + 1
and
√
√
√ 2
√
1 t c1 − 2 c2 cos 2 t +
2 c1 + c2 sin 2 t + .
e
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
x = e − cos 2 t −
sin 2 t +
3
6
3
√
√
√
2
y = et − cos 2 t +
sin 2 t + 1.
2
x=
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 .
24. From Newton’s second law in the x-direction we have
d2 x
dx
1 dx
m 2 = −k cos θ = −k
= −|c| .
dt
v dt
dt
In the y-direction we have
d2 y
1 dy
dy
= −mg − k sin θ = −mg − k
= −mg − |c| .
2
dt
v dt
dt
From mD2 x + |c|Dx = 0 we have D(mD + |c|)x = 0 so that (mD
+
|c|)x
=
c
or
(D + |c|/m)x = c2 . This is a
1
m
linear first-order differential equation. An integrating factor is e
|c|dt/m
d |c|t/m
x] = c2 e|c|t/m
[e
dt
185
= e|c|t/m so that
3.11 Solving Systems of Linear Equations
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 .
|c|dt/m
This is a linear first-order differential equation with integrating factor e
= e|c|t/m . Thus
d |c|t/m
y] = (−gt + c2 )e|c|t/m
[e
dt
mg |c|t/m m2 g |c|t/m
e|c|t/m y = −
+ 2 e
+ c3 e|c|t/m + c4
te
|c|
c
and
y(t) = −
mg
m2 g
t + 2 + c3 + c4 e−|c|t/m .
|c|
c
25. 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.
26. (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
dx2
15 −t/50
2
=
e
− x2
dt
50
75
2
dx2
3 −t/50
+ x2 =
e
.
dt
75
10
This differential equation is linear and has the integrating factor e 2 dt/75 = e2t/75 . Then
or
d 2t/75
3 −t/50+2t/75
3 t/150
x2 ] =
=
[e
e
e
dt
10
10
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
or
dx3
1
90 −t/50 70 −2t/75
− e
− x3
=
e
dt
75
75
25
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
14
6
14
x3 ] = e−t/50+t/25 − e−2t/75+t/25 = et/50 − et/75 ,
[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 .
186
3.11 Solving Systems of Linear Equations
(b)
pounds 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 pounds of salt
after 224.4 minutes.
27. (a) Write the system as
(D2 + 3)x1 −
x2 = 0
−2x1 + (D2 + 2)x2 = 0.
Then
(D2 + 2)(D2 + 3)x1 − 2x1 = (D2 + 1)(D2 + 4)x1 = 0,
and
x1 (t) = c1 cos t + c2 sin t + c3 cos 2t + c4 sin 2t.
Since x2 = (D2 + 3)x1 , we have
x2 (t) = 2c1 cos t + 2c2 sin t − c3 cos 2t − c4 sin 2t.
The initial conditions imply
x1 (0) = c1 + c3 = 2
x1 (0) = c1 + 2c4 = 1
x2 (0) = 2c1 − c3 = −1
x2 (0) = 2c2 − 2c4 = 1,
so c1 =
1
3
, c2 =
2
3
, c3 =
5
3
, and c4 =
1
6
. Thus
1
cos t +
3
2
x2 (t) = cos t +
3
x1 (t) =
(b)
x1
3
2
1
-1
-2
-3
2
sin t +
3
4
sin t −
3
5
cos 2t +
3
5
cos 2t −
3
1
sin 2t
6
1
sin 2t.
6
x2
3
2
1
5
10
15
20
t
-1
-2
-3
5
10
15
20
t
In this problem the motion appears to be periodic with period 2π. In Figure 3.59 of Example 4 in the text
the motion does not appear to be periodic.
187
3.11 Solving Systems of Linear Equations
(c)
x2
3
2
1
-2
-1
1
2 x1
-1
-2
CHAPTER 3 REVIEW EXERCISES
1. y = 0
2. Since yc = c1 ex + c2 e−x , a particular solution for y − y = 1 + ex is yp = A + Bxex .
3. 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. True
5. 8 ft, since k = 4
6. 2π/5, since 14 x + 6.25x = 0
7. 5/4 m, since x = − cos 4t + 34 sin 4t
√
√
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. The set is linearly independent over (−∞, ∞) and linearly dependent over (0, ∞).
10. (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.
188
CHAPTER 3 REVIEW EXERCISES
11. (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.
12. (a) 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.
(b) 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.
√
13. From m2 − 2m − 2 = 0 we obtain m = 1 ± 3 so that
y = c1 e(1+
√
3 )x
√
+ c2 e(1−
3 )x
.
√
14. From 2m2 + 2m + 3 = 0 we obtain m = −1/2 ± ( 5/2)i so that
√
√ 5
5
−x/2
c1 cos
y=e
x + c2 sin
x .
2
2
15. From m3 + 10m2 + 25m = 0 we obtain m = 0, m = −5, and m = −5 so that
y = c1 + c2 e−5x + c3 xe−5x .
16. 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 .
√
17. From 3m3 + 10m2 + 15m + 4 = 0 we obtain m = −1/3 and m = −3/2 ± ( 7/2)i so that
√
√ 7
7
−x/3
−3x/2
y = c1 e
+e
c2 cos
x + c3 sin
x .
2
2
√
18. 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.
19. Applying D4 to the differential equation we obtain D4 (D2 − 3D + 5) = 0. Then
√
√
11
11
3x/2
y=e
c1 cos
x + c2 sin
x + c3 + c4 x + c5 x2 + c6 x3
2
2
yc
2
3
and yp = A + Bx + Cx + Dx . Substituting yp into the differential equation yields
(5A − 3B + 2C) + (5B − 6C + 6D)x + (5C − 9D)x2 + 5Dx3 = −2x + 4x3 .
Equating coefficients gives A = −222/625, B = 46/125, C = 36/25, and D = 4/5. The general solution is
√
√
222
46
36
11
11
4
x + c2 sin
x −
+
x + x2 + x3 .
y = e3x/2 c1 cos
2
2
625 125
25
5
189
CHAPTER 3 REVIEW EXERCISES
20. 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
2 x
3 x
4 x
and yp = Ax e + Bx e + Cx e . 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
21. 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
22. 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 .
23. 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
ex sin x
ex cos x
= −e2x .
W = x
e cos x + ex sin x −ex sin x + ex 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|.
24. 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 190
CHAPTER 3 REVIEW EXERCISES
Identifying f (x) = 2ex /(ex + e−x ) we obtain
u1 =
1
ex
=
ex + e−x
1 + e2x
u2 = −
e2x
e3x
ex
x
=
−
=
−e
+
.
ex + e−x
1 + e2x
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 .
25. The auxiliary equation is 6m2 − m − 1 = 0 so that
y = c1 x1/2 + c2 x−1/3 .
26. 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 .
27. 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.
28. 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
29. (a) 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.
(b) 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 .
30. (a) 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).
(b) 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.
31. (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.
191
CHAPTER 3 REVIEW EXERCISES
(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.
32. 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 ).
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 .
33. 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.
34. 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 .
35. 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 −
1
2
sin x and
y = c1 ex + c2 e−x − x −
1
sin x.
2
Setting y(0) = 2 and y (0) = 3 we obtain
Solving this system we find c1 =
13
4
c1 + c2 = 2
3
c1 − c2 − = 3.
2
5
and c2 = − 4 . The solution of the initial-value problem is
y=
13 x 5 −x
1
e − e − x − sin x.
4
4
2
36. 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 cos2 x
2
u2 = tan x.
Thus
y = c1 cos x + c2 sin x −
1
cos x sec2 x + sin x tan x
2
192
CHAPTER 3 REVIEW EXERCISES
= 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
37. Let u = y so that u = y . The equation becomes u du/dx = 4x. Separating variables we obtain
1
u du = 4x dx =⇒ u2 = 2x2 + c1 =⇒ u2 = 4x2 + c2 .
2
When x = 1, y = u = 2, so 4 = 4 + c2 and c2 = 0. Then
dy
dy
u2 = 4x2 =⇒
= 2x or
= −2x
dx
dx
=⇒ y = x2 + c3 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.
38. 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 =⇒ u2 = y 3 + c1 .
When x = 0, y = 1 and y = u = 1 so 1 = 1 + c1 and c1 = 0. Then
2
dy
dy
2
3
u = y =⇒
= y 3/2 =⇒ y −3/2 dy = dx
= y 3 =⇒
dx
dx
4
=⇒ −2y −1/2 = x + c2 =⇒ y =
.
(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 .
39. (a) The auxiliary equation is 12m4 + 64m3 + 59m2 − 23m − 12 = 0 and has roots −4, − 32 , − 13 , and
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
3
1
1
−4c1 − c2 − c3 + c4
2
3
2
9
1
1
16c1 + c2 + c3 + c4
4
9
4
27
1
1
−64c1 − c2 − c3 + c4
8
27
8
193
= −1
=2
=5
= 0.
1
2.
The
CHAPTER 3 REVIEW EXERCISES
73
3726
257
Using a CAS we find c1 = − 495
, c2 = 109
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
y=−
e
e
e
e .
+
−
+
495
35
385
45
40. Consider xy + y = 0 and look for a solution of the form y = xm . Substituting
into the differential equation we have
xy + y = m(m − 1)xm−1 + mxm−1 = m2 xm−1 .
Identifying f (x) = −x
1
2
3
4
5 x
-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. The Wronskian is
1 ln x 1
= .
W =
0 1/x x
−1/2
y
-2
-3
-4
-5
we obtain
u1 =
√
x−1/2 ln x √
−x−1/2
= x ln x and u2 =
= − x,
1/x
1/x
so that
u1 = x3/2
2
3
ln x −
4
9
2
and u2 = − x3/2 .
3
Then
2
4
4 2 3/2
ln x −
− x ln x = − x3/2
3
9
3
9
and the general solution of the differential equation is
yp = x3/2
4
y = c1 + c2 ln x − x3/2 .
9
The initial conditions are y(1) = 0 and y (1) = 0. These imply that c1 =
initial-value problem is
4 2
4
y = + ln x − x3/2 .
9 3
9
The graph is shown above.
4
9
and c2 =
2
3
. The solution of the
41. 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 = 52 so that
3
5
x = −c2 et − c1 e2t + .
2
2
42. From (D − 2)x − y = t − 2 and −3x + (D − 4)y = −4t we obtain (D − 1)(D − 5)x = 9 − 8t. Then
and
8
3
x = c1 et + c2 e5t − t −
5
25
y = (D − 2)x − t + 2 = −c1 et + 3c2 e5t +
16 11
+ t.
25 25
43. 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
194
CHAPTER 3 REVIEW EXERCISES
y = (D − 2)x + et = −c1 et + 3c2 e5t − tet + 2et .
44. 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 −
and
2
7
cos 2t + sin 2t
3
3
1
1
x = − (D + 3)y + cos 2t
5
5
1
1
1
5
3
3
=
c1 − c2 sin t + − c2 − c1 cos t − sin 2t − cos 2t.
5
5
5
5
3
3
45. 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 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 pounds.
46. (a) Solving 38 x + 6x = 0 subject to x(0) = 1 and x (0) = −4 we obtain
√
x = cos 4t − sin 4t = 2 sin (4t + 3π/4) .
(b) The amplitude is
√
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.
47. From mx + 4x + 2x = 0 we see that nonoscillatory motion results if 16 − 8m ≥ 0 or 0 < m ≤ 2.
48. From x + βx + 64x = 0 we see that oscillatory motion results if β 2 − 256 < 0 or 0 ≤ β < 16.
49. From q + 104 q = 100 sin 50t, q(0) = 0, and q (0) = 0 we obtain qc = c1 cos 100t + c2 sin 100t, qp =
and
1
(a) q = − 150
sin 100t +
1
75
(b) i = − 23 cos 100t +
cos 50t, and
2
3
sin 50t,
(c) q = 0 when sin 50t(1 − cos 50t) = 0 or t = nπ/50 for n = 0, 1, 2, . . . .
50. By Kirchhoff’s second law,
d2 q
dq
1
+R
+ q = E(t).
dt2
dt
C
Using q (t) = i(t) we can write the differential equation in the form
L
L
di
1
+ Ri + q = E(t).
dt
C
Then differentiating we obtain
L
d2 i
di
1
+R
+ i = E (t).
dt2
dt C
51. 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,
195
1
75
sin 50t,
CHAPTER 3 REVIEW EXERCISES
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.
52. (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) =
2ω 2 r0 + 2ωv0 − g ωt 2ω 2 r0 − 2ωv0 + g −ωt
g
e +
e
+
sin ωt.
4ω 2
4ω 2
2ω 2
(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.
(d)
r
17
20
16.1
16
10
2
4
6
8
10
12
14
-10
-20
0
10
15
196
t
CHAPTER 3 REVIEW EXERCISES
(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
t
-20
1.55007
-20
2.35494
-20
3.43088
20
6.11627
20
4.22339
When v0 = 16 the bead never leaves the rod.
53. 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,
m
d2 x
= −kx
dt2
or
d2 x
k
+ x = 0.
dt2
m
54. 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
and for motion to the left (x < 0),
d2 x
= −kx + µmg.
dt2
Traditionally, these two equations are written as one expression
m
m
d2 x
+ fx sgn(x ) + kx = 0,
dt2
where fk = µmg and
sgn(x ) =
1, x > 0
−1, x < 0.
197
4
The Laplace Transform
EXERCISES 4.1
Definition of the Laplace Transform
1
∞
1 −st 1 −st {f (t)} =
−e dt +
e dt = e
−se
s
0
1
0
1
2 −s 1
1 −s
1 −s 1
= e − , s>0
= e − − 0− e
s
s
s
s
s
2
2
4
4
{f (t)} =
4e−st dt = − e−st = − (e−2s − 1), s > 0
s
s
0
0
∞
1
1
∞
1 −st
1 −st 1 −st −st
−st
{f (t)} =
te dt +
e dt = − te
− 2e
−se
s
s
0
1
0
1
1
1
1 −s
1 −s
1
−s
−s
− 0 − 2 − (0 − e ) = 2 (1 − e ), s > 0
= − e − 2e
s
s
s
s
s
1
1
2
2
1
{f (t)} =
(2t + 1)e−st dt = − te−st − 2 e−st − e−st s
s
s
0
0
2
1
2
2
1
1
2
= (1 − 3e−s ) + 2 (1 − e−s ),
= − e−s − 2 e−s − e−s − 0 − 2 −
s
s
s
s
s
s
s
π
π
s
1
{f (t)} =
e−st sin t − 2
e−st cos t (sin t)e−st dt = − 2
s +1
s +1
0
0
1
1
1
= 0+ 2
e−πs − 0 − 2
= 2
(e−πs + 1), s > 0
s +1
s +1
s +1
∞
∞
s
1
−st
−st
−st
{f (t)} =
e
e
(cos t)e dt = − 2
cos t + 2
sin t s
+
1
s
+
1
π/2
π/2
1
1
e−πs/2 = − 2
e−πs/2 , s > 0
=0− 0+ 2
s +1
s +1
0, 0 < t < 1
f (t) =
t, t > 1
∞
∞
1 −st
1 −st 1 −s
1 −s
−st
{f (t)} =
te
dt = − te
− 2e
= s e + s2 e , s > 0
s
s
1
1
0,
0<t<1
f (t) =
2t − 2, t > 1
∞
∞
1
1
2
{f (t)} = 2
(t − 1)e−st dt = 2 − (t − 1)e−st − 2 e−st = 2 e−s , s > 0
s
s
s
1.
2.
3.
4.
5.
6.
7.
8.
1
−st
∞
−st
1
1
198
s>0
4.1
Definition of the Laplace Transform
1 − t, 0 < t < 1
so
0,
t>1
1
1
∞
1
1
1 −st −st
−st
−st
−st
{f (t)} =
(1 − t)e
dt +
0e
dt =
(1 − t)e
dt = − (1 − t)e
+ 2e
s
s
0
1
0
0
1 −s 1
1
= 2e + − 2, s>0
s
s s
9. The function is f (t) =

 0, 0 < t < a
10. f (t) = c, a < t < b ;

0, t > b
11.
∞
{f (t)} =
{f (t)} =
e
12.
dt = e
(1−s)t
e
−2t−5 −st
e
e
−5
∞
{f (t)} =
4t −st
te e
dt =
0
e
∞
te
t2 e−2t e−st dt =
∞
e−t (sin t)e−st dt =
0
∞
−(s + 1) −(s+1)t
1
e
e−(s+1)t cos t
sin t −
(s + 1)2 + 1
(s + 1)2 + 1
1
1
= 2
, s > −1
=
(s + 1)2 + 1
s + 2s + 2
∞
{f (t)} =
et (cos t)e−st dt =
0
∞
1−s
1
e(1−s)t cos t +
e(1−s)t sin t
(1 − s)2 + 1
(1 − s)2 + 1
1−s
s−1
=−
= 2
, s>1
(1 − s)2 + 1
s − 2s + 2
17.
∞
{f (t)} =
0
∞
2
=
,
(s
+
2)3
0
∞
0
(cos t)e(1−s)t dt
0
=
∞
(sin t)e−(s+1)t dt
0
=
16.
s > −2
t2 e−(s+2)t dt
1 2 −(s+2)t
2
2
−
te−(s+2)t −
e−(s+2)t
t e
s+2
(s + 2)2
(s + 2)3
∞
{f (t)} =
1
1
e(4−s)t
te(4−s)t −
4−s
(4 − s)2
0
−
15.
dt =
∞
e−5 −(s+2)t e−5
dt = −
=
e
,
s+2
s+2
0
∞
0
t(cos t)e−st dt
0
=
=
st
s2 − 1
− 2
−
s + 1 (s2 + 1)2
s2 − 1
2
(s2 + 1)
s>1
s>4
0
=
(4−s)t
0
1
=
,
(4 − s)2
{f (t)} =
−(s+2)t
∞
b
c −st c −sa
dt = − e
− e−sb ), s > 0
= s (e
s
a
∞
e7
e7 (1−s)t e7
e
=
,
dt =
=
0
−
1−s
1−s
s−1
0
0
14.
∞
dt = e
0
13.
ce
0
∞
{f (t)} =
∞
7
0
−st
a
t+7 −st
e
b
,
−st
(cos t)e
+
2s
t
+
s2 + 1 (s2 + 1)2
s>0
199
∞
−st
(sin t)e
0
s > −2
4.1
Definition of the Laplace Transform
18.
{f (t)} =
∞
t(sin t)e−st dt
0
t
2s
− 2
−
2
s + 1 (s + 1)2
=
=
2s
,
2
(s2 + 1)
{2t4 } = 2
21.
{4t − 10} =
23.
{t2 + 6t − 3} =
25.
{t3 + 3t2 + 3t + 1} =
27.
{1 + e4t } =
29.
{1 + 2e2t + e4t } =
31.
{4t2 − 5 sin 3t} = 4
33.
1
{sinh kt} =
2
34.
{cosh kt} =
35.
36.
37.
38.
(cos t)e
(sin t)e
0
5!
s6
{t5 } =
22.
{7t + 3} =
24.
{−4t2 + 16t + 9} = −4
26.
{8t3 − 12t2 + 6t − 1} = 8
28.
{t2 − e−9t + 5} =
1
2
1
+
+
s s−2 s−4
30.
{e2t − 2 + e−2t } =
1
2
1
− +
s−2 s s+2
2
3
−5 2
s3
s +9
32.
{cos 5t + sin 2t} =
s
2
+
s2 + 25 s2 + 4
2
6
3
+ 2−
s3
s
s
3!
2
3
1
+3 3 + 2 +
s4
s
s
s
1
1
+
s s−4
1
2
∞
−st
20.
4
10
−
s2
s
{e
−
s2 − 1
st
+
2
s + 1 (s2 + 1)2
s>0
4!
s5
19.
−st
kt
−kt
−e
1
1
1
k
}=
−
= 2
2 s−k s+k
s − k2
3
7
+
s2
s
2
16 9
+ 2 +
s3
s
s
3!
2
6
1
− 12 3 + 2 −
s4
s
s
s
5
2
1
+
−
s3
s+9 s
s
s2 − k 2
t
−t
1 2t 1
1
1
t
t e −e
{e sinh t} =
e
=
e −
=
−
2
2
2
2(s − 2) 2s
t
−t
e +e
1
1 1 −2t
1
{e−t cosh t} =
=
e−t
=
+ e
+
2
2 2
2s 2(s + 2)
1
2
{sin 2t cos 2t} =
sin 4t = 2
2
s + 16
1 1
1
1 s
{cos2 t} =
+ cos 2t =
+
2 2
2s 2 s2 + 4
{ekt + ekt } =
39. From the addition formula for the sine function, sin(4t + 5) = sin 4t cos 5 + cos 4t sin 5 so
{sin(4t + 5)} = (cos 5)
{sin 4t} + (sin 5)
{cos 4t} = (cos 5)
s2
4
s
4 cos 5 + (sin 5)s
+ (sin 5) 2
=
.
+ 16
s + 16
s2 + 16
40. From the addition formula for the cosine function,
√
π
3
π
π
1
cos t −
= cos t cos + sin t sin =
cos t + sin t
6
6
6
2
2
so
√
π 3
1
cos t −
=
{cos t} +
{sin t}
6
2
2
√
√
1 1
1 3s + 1
3 s
+
=
.
=
2 s2 + 1 2 s2 + 1
2 s2 + 1
200
4.1
41. (a) Using integration by parts for α > 0,
∞
∞
α −t
α −t Γ(α + 1) =
t e dt = −t e + α
0
0
(b) Let u = st so that du = s dt. Then
∞
α
−st α
{t } =
e t dt =
0
42. (a)
−1/2
{t
Γ(1/2)
} = 1/2 =
s
∞
e−u
0
π
s
tα−1 e−t dt = αΓ(α).
0
u α 1
1
du = α+1 Γ(α + 1),
s
s
s
{t1/2 } =
(b)
∞
Definition of the Laplace Transform
√
π
Γ(3/2)
=
3/2
3/2
s
2s
43. Let F (t) = t1/3 . Then F (t) is of exponential order, but f (t) = F (t) =
hence is not of exponential order. Let
2
2
f (t) = 2tet cos et =
{t3/2 } =
(c)
1 −2/3
3t
α > −1.
√
Γ(5/2)
3 π
=
s5/2
4s5/2
is unbounded near t = 0 and
2
d
sin et .
dt
This function is not of exponential order, but we can show that its Laplace transform exists. Using integration
by parts we have
∞
a
a
d
t2
t2
−st
t2
−st
t2 −st
t2
{2te cos e } =
dt = lim e
e
sin e + s
e
sin e dt
sin e
a→∞
dt
0
0
0
∞
2
2
= − sin 1 + s
e−st sin et dt = s {sin et } − sin 1.
0
2
Since sin et is continuous and of exponential order,
2
{sin et } exists, and therefore
2
2
{2tet cos et } exists.
44. The relation will be valid when s is greater than the maximum of c1 and c2 .
2
45. Since et is an increasing function and t2 > ln M + ct for M > 0 we have et > eln M +ct = M ect for t sufficiently
2
large and for any c. Thus, et is not of exponential order.
46. Assuming that (c) of Theorem 4.1 is applicable with a complex exponent, we have
{e(a+ib)t } =
1
1
(s − a) + ib
s − a + ib
.
=
=
s − (a + ib)
(s − a) − ib (s − a) + ib
(s − a)2 + b2
By Euler’s formula, eiθ = cos θ + i sin θ, so
{e(a+ib)t } =
=
=
{eat eibt } =
{eat cos bt} + i
{eat (cos bt + i sin bt)}
{eat sin bt}
s−a
b
+i
.
(s − a)2 + b2
(s − a)2 + b2
Equating real and imaginary parts we get
{eat cos bt} =
s−a
(s − a)2 + b2
and
{eat sin bt} =
b
.
(s − a)2 + b2
47. 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.
201
4.1
Definition of the Laplace Transform
{tn−1 } = (n − 1)!/sn . Then, using the definition of the Laplace transform and integration by
48. Assume that
parts, we have
∞
1
n ∞ −st n−1
e−st tn dt = − e−st tn +
e t
dt
s
s 0
0
0
n
n (n − 1)!
n!
=0+
= n+1 .
{tn−1 } =
s
s
sn
s
{tn } =
EXERCISES 4.2
1.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
=
1
s4
1
2
2
s3
=
The Inverse Transform and
Transforms of Derivatives
1 2
t
2
1
1
3!
= t3
6
s4
6
1
1
48
48 4!
=
= t − 2t4
−
−
·
s2
s5
s2
24 s5
2 1
1
5!
2
2
4 3!
1
1 5
=
4· 2 − · 4 +
− 3
· 6 = 4t − t3 +
t
s s
s
6 s
120 s
3
120
3
1
(s + 1)3
3 2
1 3!
1
1
=
= 1 + 3t + t2 + t3
+
+
+
3
·
·
·
4
2
3
4
s
s
s
2 s
6 s
2
6
1
(s + 2)2
2
1
=
+ 4 · 2 + 2 · 3 = 1 + 4t + 2t2
3
s
s
s
s
1
1
1
= t − 1 + e2t
− +
2
s
s s−2
4
1
1
6
1 1 4!
1
+ 5−
=
4· + · 5 −
= 4 + t4 − e−8t
s s
s+8
s 4 s
s+8
4
1
1
1
1
=
= e−t/4
4s + 1
4
s + 1/4
4
1
1
1
1
=
·
= e2t/5
5s − 2
5 s − 2/5
5
5
7
5
5
=
·
= sin 7t
s2 + 49
7 s2 + 49
7
10s
= 10 cos 4t
s2 + 16
4s
s
1
=
= cos t
4s2 + 1
s2 + 1/4
2
1
1
1/2
1
1
=
·
= sin t
4s2 + 1
2 s2 + 1/4
2
2
2.
1
s3
∞
=
202
4.2
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
s
3
−
2
·
= 2 cos 3t − 2 sin 3t
s2 + 9
s2 + 9
√
√ √
√
s+1
2
2
s
1
=
+√ 2
= cos 2t +
sin 2 t
s2 + 2
s2 + 2
s
+
2
2
2
1
1 1 1
1
1 1
=
·
−
·
= − e−3t
s2 + 3s
3 s 3 s+3
3 3
s+1
1 1 5
1
1 5
=
− · + ·
= − + e4t
2
s − 4s
4 s 4 s−4
4 4
s
3
1
1
3
1
1
=
·
+ ·
= et + e−3t
2
s + 2s − 3
4 s−1 4 s+3
4
4
1
1
1
1
1
1
1
=
·
− ·
= e4t − e−5t
s2 + s − 20
9 s−4 9 s+5
9
9
1
1
0.9s
=
(0.3) ·
+ (0.6) ·
= 0.3e0.1t + 0.6e−0.2t
(s − 0.1)(s + 0.2)
s − 0.1
s + 0.2
√ √
√
√
√
3
s−3
s
√
√
− 3· 2
= cosh 3 t − 3 sinh 3 t
=
2
s −3
s −3
(s − 3 )(s + 3 )
s
1
1
1
1
1
1
1
=
·
−
+ ·
= e2t − e3t + e6t
(s − 2)(s − 3)(s − 6)
2 s−2 s−3 2 s−6
2
2
s2 + 1
1 1
1
1
1
5
1
=
· −
− ·
+ ·
s(s − 1)(s + 1)(s − 2)
2 s s−1 3 s+1 6 s−2
1
5
1
= − et − e−t + e2t
2
3
6
√
1
1
1 1 1 s
1 1
5t
=
=
·
−
=
−
cos
s3 + 5s
s(s2 + 5)
5 s 5 s2 + 5
5 5
s
1
s
1
2
1
1
1
1
1
=
·
+
·
−
·
= cos 2t + sin 2t − e−2t
(s2 + 4)(s + 2)
4 s2 + 4 4 s2 + 4 4 s + 2
4
4
4
2s − 4
2s − 4
4
3
s
3
=
=
−
+
+
+
(s2 + s)(s2 + 1)
s(s + 1)(s2 + 1)
s s + 1 s2 + 1 s2 + 1
2s − 6
s2 + 9
The Inverse Transform and Transforms of Derivatives
=
2·
= −4 + 3e−t + cos t + 3 sin t
√
√ √
√
1
3
3
1
1
1
1
√ · 2
=
− √ · 2
= √ sinh 3 t − √ sin 3 t
4
s −9
6 3 s −3 6 3 s +3
6 3
6 3
1
1
1
1
1
1
1
1
2
=
·
−
·
=
·
−
·
(s2 + 1)(s2 + 4)
3 s2 + 1 3 s2 + 4
3 s2 + 1 6 s2 + 4
1
1
= sin t − sin 2t
3
6
6s + 3
s
1
s
1
2
+
−2· 2
− ·
=
2· 2
(s2 + 1)(s2 + 4)
s + 1 s2 + 1
s + 4 2 s2 + 4
1
= 2 cos t + sin t − 2 cos 2t − sin 2t
2
The Laplace transform of the initial-value problem is
1
s {y} − y(0) −
{y} = .
s
203
4.2
The Inverse Transform and Transforms of Derivatives
Solving for
{y} we obtain
1
1
{y} = − +
.
s s−1
Thus
y = −1 + et .
32. The Laplace transform of the initial-value problem is
2s
Solving for
{y} − 2y(0) +
{y} = 0.
{y} we obtain
6
3
=
.
2s + 1
s + 1/2
{y} =
Thus
y = 3e−t/2 .
33. The Laplace transform of the initial-value problem is
s
Solving for
{y} − y(0) + 6
{y} =
1
.
s−4
{y} we obtain
{y} =
1
2
1
1
19
1
+
=
·
+
·
.
(s − 4)(s + 6) s + 6
10 s − 4 10 s + 6
Thus
y=
1 4t 19 −6t
e + e .
10
10
34. The Laplace transform of the initial-value problem is
s
Solving for
{y} −
{y} =
s2
2s
.
+ 25
{y} we obtain
{y} =
2s
1
1
1
s
5
5
=
·
−
+
· 2
.
2
2
(s − 1)(s + 25)
13 s − 1 13 s + 25 13 s + 25
Thus
y=
1 t
1
5
e −
cos 5t +
sin 5t.
13
13
13
35. The Laplace transform of the initial-value problem is
s2
Solving for
{y} − sy(0) − y (0) + 5 [s
{y} − y(0)] + 4
{y} = 0.
{y} we obtain
{y} =
s+5
4 1
1 1
=
−
.
s2 + 5s + 4
3 s+1 3 s+4
Thus
y=
4 −t 1 −4t
e − e .
3
3
36. The Laplace transform of the initial-value problem is
s2
{y} − sy(0) − y (0) − 4 [s
{y} − y(0)] =
204
3
6
−
.
s−3 s+1
4.2
Solving for
The Inverse Transform and Transforms of Derivatives
{y} we obtain
{y} =
=
6
3
s−5
−
+ 2
2
2
(s − 3)(s − 4s) (s + 1)(s − 4s) s − 4s
5 1
2
3
1
11
1
· −
− ·
+
·
.
2 s s − 3 5 s + 1 10 s − 4
Thus
y=
5
3
11
− 2e3t − e−t + e4t .
2
5
10
37. The Laplace transform of the initial-value problem is
s2
Solving for
{y} − sy(0) +
{y} =
2
.
s2 + 2
{y} we obtain
{y} =
2
10s
10s
2
2
+
= 2
+
−
.
(s2 + 1)(s2 + 2) s2 + 1
s + 1 s2 + 1 s2 + 2
Thus
y = 10 cos t + 2 sin t −
√
√
2 sin 2 t.
38. The Laplace transform of the initial-value problem is
{y} + 9
s2
Solving for
{y} =
1
.
s−1
{y} we obtain
{y} =
1
1
1
1
1
1
s
=
·
−
· 2
−
· 2
.
2
(s − 1)(s + 9)
10 s − 1 10 s + 9 10 s + 9
Thus
y=
1 t
1
1
e −
sin 3t −
cos 3t.
10
30
10
39. The Laplace transform of the initial-value problem is
2 s3
{y} − s2 (0) − sy (0) − y (0) + 3 s2
Solving for
{y} − sy(0) − y (0) − 3[s
{y} − y(0)] − 2
{y} =
1
.
s+1
{y} we obtain
{y} =
2s + 3
1 1
5
1
8
1
1 1
=
+
−
+
.
(s + 1)(s − 1)(2s + 1)(s + 2)
2 s + 1 18 s − 1 9 s + 1/2 9 s + 2
Thus
y=
1 −t
5
8
1
e + et − e−t/2 + e−2t .
2
18
9
9
40. The Laplace transform of the initial-value problem is
s3
{y} − s2 (0) − sy (0) − y (0) + 2 s2
Solving for
{y} − sy(0) − y (0) − [s
{y} − y(0)] − 2
{y} we obtain
{y} =
=
s2 + 12
(s − 1)(s + 1)(s + 2)(s2 + 9)
13 1
13 1
16 1
3
s
1
3
−
+
+
−
.
60 s − 1 20 s + 1 39 s + 2 130 s2 + 9 65 s2 + 9
205
{y} =
3
.
s2 + 9
4.2
The Inverse Transform and Transforms of Derivatives
Thus
y=
13 t 13 −t 16 −2t
1
3
e − e + e
cos 3t −
sin 3t.
+
60
20
39
130
65
41. The Laplace transform of the initial-value problem is
{y} +
s
Solving for
{y} =
s+3
.
s2 + 6s + 13
{y} we obtain
s+3
1
1
1
s+1
= ·
− · 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
{y} =
Thus
y=
1 −t 1 −3t
1
e − e
cos 2t + e−3t sin 2t.
4
4
4
42. The Laplace transform of the initial-value problem is
s2
Solving for
{y} − s · 1 − 3 − 2[s
{y} − 1] + 5
{y} = (s2 − 2s + 5)
{y} − s − 1 = 0.
{y} we obtain
{y} =
s+1
s−1+2
s−1
2
=
=
+
.
s2 − 2s + 5
(s − 1)2 + 22
(s − 1)2 + 22
(s − 1)2 + 22
Thus
y = et cos 2t + et sin 2t.
43. (a) Differentiating f (t) = teat we get f (t) = ateat + eat so
f (0) = 0. Writing the equation as
{teat } +
a
and solving for
{ateat + eat } = s
{eat } = s
{teat }, where we have used
{teat }
{teat } we get
{teat } =
1
s−a
{eat } =
1
.
(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
{−k 2 t sin t + 2k cos kt} = s2
{t sin kt}
where we have used f (0) = 0 and f (0) = 0. Writing the above equation as
−k 2
and solving for
{t sin kt} + 2k
{cos kt} = s2
{t sin kt}
{t sin kt} gives
{t sin kt} =
44. Let f1 (t) = 1 and f2 (t) =
2k
s2 + k 2
1,
t ≥ 0, t = 1
0,
t=1
{cos kt} =
. Then
s
2k
2ks
= 2
.
s2 + k 2 s2 + k 2
(s + k 2 )2
{f1 (t)} =
206
{f2 (t)} = 1/s, but f1 (t) = f2 (t).
4.3 Translation Theorems
45. For y − 4y = 6e3t − 3e−t the transfer function is W (s) = 1/(s2 − 4s). The zero-input response is
s−5
5 1 1
1
5 1
y0 (t) =
=
·
−
·
= − e4t ,
s2 − 4s
4 s 4 s−4
4 4
and the zero-state response is
y1 (t) =
=
=
6
3
−
(s − 3)(s2 − 4s) (s + 1)(s2 − 4s)
27
1
2
5 1 3
1
·
−
+ · − ·
20 s − 4 s − 3 4 s 5 s + 1
27 4t
5 3
e − 2e3t + − e−t .
20
4 5
46. From Theorem 4.4, if f and f are continuous and of exponential order,
Theorem 4.5, lims→∞ {f (t)} = 0 so
lim [sF (s) − f (0)] = 0
and
s→∞
For f (t) = cos kt,
lim sF (s) = lim s
s→∞
s→∞
s2
{f (t)} = sF (s) − f (0). From
lim F (s) = f (0).
s→∞
s
= 1 = f (0).
+ k2
EXERCISES 4.3
Translation Theorems
1.
3.
5.
6.
7.
9.
10.
11.
te10t =
t3 e−2t =
1
(s − 10)2
2.
3!
(s + 2)4
2 t et + e2t
=
e2t (t − 1)2 =
et sin 3t =
4.
te2t + 2te3t + te4t =
t2 e2t − 2te2t + e2t =
3
(s − 1)2 + 9
{(1 − et + 3e−4t ) cos 5t} =
te−6t =
t10 e−7t =
1
(s + 6)2
10!
(s + 7)11
2
1
1
+
+
2
2
(s − 2)
(s − 3)
(s − 4)2
2
1
2
−
+
(s − 2)3
(s − 2)2
s−2
8.
e−2t cos 4t =
{cos 5t − et cos 5t + 3e−4t cos 5t} =
s+2
(s + 2)2 + 16
s
s−1
3(s + 4)
−
+
s2 + 25 (s − 1)2 + 25 (s + 4)2 + 25
t
t
5
9
4
e3t 9 − 4t + 10 sin
+
=
9e3t − 4te3t + 10e3t sin
=
−
2
2
s − 3 (s − 3)2
(s − 3)2 + 1/4
1
2
1
1
=
= t2 e−2t
(s + 2)3
2 (s + 2)3
2
207
4.3
Translation Theorems
12.
1
(s − 1)4
=
1
6
3!
(s − 1)4
=
1 3 t
t e
6
1
1
=
= e3t sin t
− 6s + 10
(s − 3)2 + 12
1
1
1
2
= e−t sin 2t
=
2
2
2
s + 2s + 5
2 (s + 1) + 2
2
s
1
s+2
= e−2t cos t − 2e−2t sin t
−2
=
s2 + 4s + 5
(s + 2)2 + 12
(s + 2)2 + 12
2s + 5
1
1
(s + 3)
5
= 2e−3t cos 5t − e−3t sin 5t
−
=
2
2
2
2
2
2
s + 6s + 34
(s + 3) + 5
5 (s + 3) + 5
5
s
s+1−1
1
1
=
=
= e−t − te−t
−
(s + 1)2
(s + 1)2
s + 1 (s + 1)2
5(s − 2) + 10
5
5s
10
=
=
= 5e2t + 10te2t
+
(s − 2)2
(s − 2)2
s − 2 (s − 2)2
5
2s − 1
5
3
3
1
4
2
=
= 5 − t − 5e−t − 4te−t − t2 e−t
−
−
−
−
s2 (s + 1)3
s s2
s + 1 (s + 1)2
2 (s + 1)3
2
1
(s + 1)2
2
1
1
3!
=
= te−2t − t2 e−2t + t3 e−2t
−
+
(s + 2)4
(s + 2)2
(s + 2)3
6 (s + 2)4
6
13.
s2
14.
15.
16.
17.
18.
19.
20.
21. The Laplace transform of the differential equation is
s
Solving for
{y} − y(0) + 4
{y} =
1
.
s+4
{y} we obtain
{y} =
1
2
+
.
2
(s + 4)
s+4
Thus
y = te−4t + 2e−4t .
22. The Laplace transform of the differential equation is
s
Solving for
{y} −
{y} =
1
1
+
.
s (s − 1)2
{y} we obtain
{y} =
1
1
1
1
1
=− +
.
+
+
3
s(s − 1) (s − 1)
s s − 1 (s − 1)3
Thus
1
y = −1 + et + t2 et .
2
23. The Laplace transform of the differential equation is
s2
Solving for
{y} − sy(0) − y (0) + 2 s
{y} − y(0) +
{y} we obtain
{y} =
s+3
2
1
+
=
.
(s + 1)2
s + 1 (s + 1)2
Thus
y = e−t + 2te−t .
208
{y} = 0.
4.3 Translation Theorems
24. The Laplace transform of the differential equation is
{y} − sy(0) − y (0) − 4 [s
s2
{y} − y(0)] + 4
{y} =
6
.
(s − 2)4
1
1 5 2t
5!
. Thus, y =
t e .
20 (s − 2)6
20
25. The Laplace transform of the differential equation is
Solving for
{y} we obtain
{y} − sy(0) − y (0) − 6 [s
s2
Solving for
{y} =
{y} − y(0)] + 9
{y} =
1
.
s2
{y} we obtain
{y} =
1 + s2
1
10
1
2 1 1 1
2
+
+
=
−
.
2
2
2
s (s − 3)
27 s 9 s
27 s − 3
9 (s − 3)2
Thus
y=
2
1
2
10
+ t − e3t + te3t .
27 9
27
9
26. The Laplace transform of the differential equation is
s2
Solving for
{y} − sy(0) − y (0) − 4 [s
{y} − y(0)] + 4
{y} =
6
.
s4
{y} we obtain
{y} =
s5 − 4s4 + 6
13
1
3 1 9 1
3 2
1 3!
1 1
+
−
=
+
+
+
.
s4 (s − 2)2
4 s 8 s2
4 s3
4 s4
4 s−2
8 (s − 2)2
Thus
y=
3 9
1
1
13
3
+ t + t2 + t3 + e2t − te2t .
4 8
4
4
4
8
27. The Laplace transform of the differential equation is
s2
Solving for
{y} − sy(0) − y (0) − 6 [s
{y} = 0.
{y} we obtain
{y} = −
Thus
3
3
2
=−
.
s2 − 6s + 13
2 (s − 3)2 + 22
3
y = − e3t sin 2t.
2
28. The Laplace transform of the differential equation is
2 s2 {y} − sy(0) + 20 s
Solving for
{y} − y(0)] + 13
{y} − y(0) + 51
{y} = 0.
{y} we obtain
{y} =
2s2
Thus
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
√
√
√
y = 2e−5t cos(t/ 2 ) + 10 2 e−5t sin(t/ 2 ).
29. The Laplace transform of the differential equation is
s2
{y} − sy(0) − y (0) − [s
{y} − y(0)] =
209
s−1
.
(s − 1)2 + 1
4.3
Translation Theorems
Solving for
{y} we obtain
{y} =
s(s2
1
1 1 1
s−1
1
1
=
−
+
.
2
− 2s + 2)
2 s 2 (s − 1) + 1 2 (s − 1)2 + 1
Thus
1 1 t
1
− e cos t + et sin t.
2 2
2
y=
30. The Laplace transform of the differential equation is
{y} − sy(0) − y (0) − 2 [s
s2
Solving for
{y} − y(0)] + 5
{y} =
1
1
.
+
s s2
{y} we obtain
{y} =
=
4s2 + s + 1
−7s/25 + 109/25
7 1 1 1
+
=
+
s2 (s2 − 2s + 5)
25 s 5 s2
s2 − 2s + 5
7 1 1 1
7
51
s−1
2
−
+
.
+
25 s 5 s2
25 (s − 1)2 + 22
25 (s − 1)2 + 22
Thus
y=
7
51
1
7
+ t − et cos 2t + et sin 2t.
25 5
25
25
31. Taking the Laplace transform of both sides of the differential equation and letting c = y(0) we obtain
{y } +
{y} − sy(0) − y (0) + 2s
2
s
s2
{2y } +
{y} = 0
{y} − 2y(0) +
{y} = 0
{y} − cs − 2 + 2s
{y} − 2c +
s2 + 2s + 1
{y} = 0
{y} = cs + 2c + 2
2c + 2
cs
+
{y} =
2
(s + 1)
(s + 1)2
=c
=
Therefore,
y(t) = c
1
s+1
+ (c + 2)
1
(s + 1)2
s+1−1
2c + 2
+
(s + 1)2
(s + 1)2
c
c+2
.
+
s + 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
{y } +
2
s
{8y } +
{20y} = 0
{y} − y (0) + 8s
{y} + 20
{y} = 0
{y} − c + 8s
{y} + 20
{y} = 0
(s + 8s + 20)
{y} = c
s2
2
{y} =
210
c
c
=
.
s2 + 8s + 20
(s + 4)2 + 4
4.3 Translation Theorems
Therefore,
y(t) =
c
(s + 4)2 + 4
=
c −4t
e
sin 2t = c1 e−4t sin 2t.
2
To find c we let y (π) = 0. Then 0 = y (π) = ce−4π and c = 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 m = W/g = 4/32 = 18 slug, and 4 = 2k so that k = 2 lb/ft.
Thus, the differential equation is x + 7x + 16x = 0. The initial conditions are x(0) = −3/2 and x (0) = 0.
The Laplace transform of the differential equation is
s2
Solving for
3
{x} + s + 7s
2
{x} +
21
+ 16
2
{x} = 0.
{x} we obtain
√
√
−3s/2 − 21/2
15/2
3
s + 7/2
7 15
√
√
{x} = 2
=−
−
.
s + 7s + 16
2 (s + 7/2)2 + ( 15/2)2
10 (s + 7/2)2 + ( 15/2)2
Thus
√
√
√
3 −7t/2
15
15
7 15 −7t/2
x=− e
cos
sin
t−
e
t.
2
2
10
2
34. The differential equation is
d2 q
dq
+ 20 + 200q = 150,
2
dt
dt
The Laplace transform of this equation is
s2
Solving for
{q} + 20s
q(0) = q (0) = 0.
{q} + 200
150
.
s
{q} =
{q} we obtain
{q} =
s(s2
150
3 1 3
s + 10
10
3
=
−
−
.
2
2
+ 20s + 200)
4 s 4 (s + 10) + 10
4 (s + 10)2 + 102
Thus
q(t) =
3 3 −10t
3
cos 10t − e−10t sin 10t
− e
4 4
4
and
i(t) = q (t) = 15e−10t sin 10t.
35. The differential equation is
d2 q
dq
E0
+ 2λ + ω 2 q =
,
2
dt
dt
L
The Laplace transform of this equation is
s2
or
Solving for
{q} + 2λs
{q} + ω 2
s2 + 2λs + ω 2
q(0) = q (0) = 0.
{q} =
{q} =
E0 1
L s
E0 1
.
L s
{q} and using partial fractions we obtain
E0 1/ω 2
(1/ω 2 )s + 2λ/ω 2
s + 2λ
E0 1
{q} =
− 2
−
=
.
L
s
s + 2λs + ω 2
Lω 2 s s2 + 2λs + ω 2
211
4.3
Translation Theorems
For λ > ω we write s2 + 2λs + ω 2 = (s + λ)2 − λ2 − ω 2 , so (recalling that ω 2 = 1/LC)
1
s+λ
λ
{q} = E0 C
−
−
.
s (s + λ)2 − (λ2 − ω 2 ) (s + λ)2 − (λ2 − ω 2 )
Thus for λ > ω,
λ
q(t) = E0 C 1 − e−λt cosh λ2 − ω 2 t − √
sinh λ2 − ω 2 t .
λ2 − ω 2
For λ < ω we write s2 + 2λs + ω 2 = (s + λ)2 + ω 2 − λ2 , so
s+λ
λ
1
{q} = E0 C
−
−
.
s (s + λ)2 + (ω 2 − λ2 ) (s + λ)2 + (ω 2 − λ2 )
Thus for λ < ω,
λ
q(t) = E0 C 1 − e−λt cos ω 2 − λ2 t − √
sin ω 2 − λ2 t .
ω 2 − λ2
For λ = ω, s2 + 2λ + ω 2 = (s + λ)2 and
E0
E0 1
E0 1/λ2
1
1/λ2
1/λ
1
λ
{q} =
=
.
=
−
−
−
−
L s(s + λ)2
L
s
s + λ (s + λ)2
Lλ2 s s + λ (s + λ)2
Thus for λ = ω,
q(t) = E0 C 1 − e−λt − λte−λt .
36. The differential equation is
dq
1
+ q = E0 e−kt , q(0) = 0.
dt
C
The Laplace transform of this equation is
R
Rs
Solving for
{q} +
1
C
{q} = E0
1
.
s+k
{q} we obtain
{q} =
E0 C
E0 /R
=
.
(s + k)(RCs + 1)
(s + k)(s + 1/RC)
When 1/RC = k we have by partial fractions
E0 1/(1/RC − k) 1/(1/RC − k)
E0
1
1
1
{q} =
−
=
−
.
R
s+k
s + 1/RC
R 1/RC − k s + k s + 1/RC
Thus
q(t) =
E0 C −kt
− e−t/RC .
e
1 − kRC
When 1/RC = k we have
{q} =
Thus
q(t) =
37.
38.
(t − 1)
e2−t
e−s
(t − 1) = 2
s
(t − 2) =
e−(t−2)
E0
1
.
R (s + k)2
E0 −kt
E0 −t/RC
=
.
te
te
R
R
e−2s
(t − 2) =
s+1
212
4.3 Translation Theorems
39.
t
(t − 2) =
{(t − 2)
(t − 2) + 2
(t − 2)} =
e−2s
2e−2s
+
2
s
s
Alternatively, (16) of this section could be used:
{t
40.
(t − 1) = 3
(3t + 1)
−2s
−2s
(t − 2)} = e
(t − 1)
{t + 2} = e
(t − 1) + 4
1
2
+
2
s
s
.
3e−s
4e−s
(t − 1) = 2 +
s
s
Alternatively, (16) of this section could be used:
{(3t + 1)
41.
(t − π) =
cos 2t
{cos 2(t − π)
(t − 1)} = e−s
(t − π)} =
{3t + 4} = e−s
3
4
+
s2
s
.
se−πs
s2 + 4
Alternatively, (16) of this section could be used:
{cos 2t
42.
sin t
t−
π =
2
(t − π)} = e−πs
π
cos t −
2
{cos 2(t + π)} = e−πs
t−
44.
45.
46.
47.
48.
e−2s
s3
49. (c)
56.
57.
{cos t} = e−πs/2
s
.
s2 + 1
1 2 −2s
1
= (t − 2)2 (t − 2)
· 3e
2 s
2
(1 + e−2s )2
1
2e−2s
e−4s
=
+
+
= e−2t + 2e−2(t−2) (t − 2) + e−2(t−4) (t − 4)
s+2
s+2
s+2
s+2
−πs e
= sin(t − π) (t − π) = − sin t (t − π)
s2 + 1
−πs/2 se
π π
π
=
cos
2
t
−
t
−
=
−
cos
2t
t
−
s2 + 4
2
2
2
e−s
e−s
e−s
=
−
= (t − 1) − e−(t−1) (t − 1)
s(s + 1)
s
s+1
−2s
e−2s
e−2s
e
e−2s
=
−
− 2 +
= − (t − 2) − (t − 2) (t − 2) + et−2 (t − 2)
s2 (s − 1)
s
s
s−1
43.
55.
s
.
s2 + 4
π se−πs/2
= 2
2
s +1
Alternatively, (16) of this section could be used:
π π sin t
t−
sin t +
= e−πs/2
= e−πs/2
2
2
{cos 2t} = e−πs
=
50. (e)
51. (f )
52. (b)
2 4
(t − 3) = − e−3s
s s
1 e−4s
e−5s
1 − (t − 4) + (t − 5) = −
+
s
s
s
2
t
(t − 1) =
(t − 1)2 + 2t − 1
(t − 1) =
2
2
1 −s
=
+
+
e
s3
s2
s
53. (a)
54. (d)
2−4
213
(t − 1)2 + 2(t − 1) − 1
(t − 1)
4.3
Translation Theorems
Alternatively, by (16) of this section,
(t − 1)} = e−s
{t2
58.
59.
60.
61.
62.
sin t
(t − 2) =
t−t
t−
3π
2
3π
− cos t −
2
=
f (t) =
f (t) =
t−
3π
2
=−
2
2
1
+ 2+
s3
s
s
.
se−3πs/2
s2 + 1
1
e−2s
2e−2s
(t − 2) = 2 − 2 −
s
s
s
−2πs
1
e
sin t − sin(t − 2π) (t − 2π) = 2
−
s + 1 s2 + 1
t − (t − 2)
(t − 2π) =
sin t − sin t
{t2 + 2t + 1} = e−s
(t − 2) − 2
(t − a) −
e−as
e−bs
(t − b) =
−
s
s
(t − 1) +
(t − 2) +
e−s
e−2s
e−3s
1 e−s
(t − 3) + · · · =
+
+
+ ··· =
s
s
s
s 1 − e−s
63. The Laplace transform of the differential equation is
{y} − y(0) +
s
Solving for
{y} we obtain
{y} =
5 −s
e .
s
5e−s
1
−s 1
{y} =
= 5e
−
.
s(s + 1)
s s+1
Thus
y=5
(t − 1) − 5e−(t−1)
(t − 1).
64. The Laplace transform of the differential equation is
s
Solving for
{y} − y(0) +
{y} =
1 2 −s
− e .
s s
{y} we obtain
1
2e−s
1
1
1
−s 1
{y} =
−
= −
− 2e
−
.
s(s + 1) s(s + 1)
s s+1
s s+1
Thus
y = 1 − e−t − 2 1 − e−(t−1) (t − 1).
65. The Laplace transform of the differential equation is
s
Solving for
{y} − y(0) + 2
{y} =
1
s+1
− e−s 2 .
2
s
s
{y} we obtain
{y} =
1
1 1
1 1
1 1 1 1
1 1
−s s + 1
−s 1 1
+
−
−
e
=
−
+
−
e
+
.
s2 (s + 2)
s2 (s + 2)
4 s 2 s2
4 s+2
4 s 2 s2
4 s+2
Thus
1 1
1 1
1
1
y = − + t + e−2t −
+ (t − 1) − e−2(t−1)
4 2
4
4 2
4
(t − 1).
66. The Laplace transform of the differential equation is
s2
{y} − sy(0) − y (0) + 4
214
{y} =
1 e−s
−
.
s
s
4.3 Translation Theorems
Solving for
{y} we obtain
{y} =
1−s
1
1 1 1 s
1 2
1 s
−s
−s 1 1
−
e
=
−
−
−
e
−
.
s(s2 + 4)
s(s2 + 4)
4 s 4 s2 + 4 2 s2 + 4
4 s 4 s2 + 4
Thus
y=
1 1
1
1 1
− cos 2t − sin 2t −
− cos 2(t − 1) (t − 1).
4 4
2
4 4
67. The Laplace transform of the differential equation is
s2
Solving for
{y} − sy(0) − y (0) + 4
{y} we obtain
{y} =
Thus
y = cos 2t +
{y} = e−2πs
1
.
s2 + 1
s
1
1 2
−2πs 1
+
e
−
.
s2 + 4
3 s2 + 1 6 s2 + 4
1
1
sin(t − 2π) − sin 2(t − 2π) (t − 2π).
3
6
68. The Laplace transform of the differential equation is
{y} − sy(0) − y (0) − 5 [s
s2
Solving for
{y} − y(0)] + 6
{y} =
e−s
.
s
{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
{y} = e−s
Thus
1 1 2(t−1) 1 3(t−1)
y=
− e
+ e
6 2
3
(t − 1) − e2t + e3t .
69. The Laplace transform of the differential equation is
s2
Solving for
{y} − sy(0) − y (0) +
{y} we obtain
−πs
{y} = e
{y} =
e−2πs
e−πs
−
.
s
s
1
s
s
1
−2πs 1
−
−e
−
+ 2
.
s s2 + 1
s s2 + 1
s +1
Thus
y = [1 − cos(t − π)]
(t − π) − [1 − cos(t − 2π)]
(t − 2π) + sin t.
70. The Laplace transform of the differential equation is
s2
Solving for
{y} − sy(0) − y (0) + 4 s
{y} − y(0) + 3
{y} we obtain
{y} =
1 e−2s
e−4s
e−6s
−
−
+
.
s
s
s
s
1 1 1 1
1 1
1 1
1 1
−2s 1 1
{y} =
−
+
−e
−
+
3 s 2 s+1 6 s+3
3 s 2 s+1 6 s+3
1 1
1 1
1 1 1 1
1 1 1 1
− e−4s
−
+
+ e−6s
−
+
.
3 s 2 s+1 6 s+3
3 s 2 s+1 6 s+3
215
4.3
Translation Theorems
Thus
y=
1 1 −t 1 −3t
1 1 −(t−2) 1 −3(t−2)
(t − 2)
−
+ e
− e + 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
(t − 4) +
(t − 6).
+ e
+ e
3 2
6
3 2
6
71. 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
f (t) =
20t, 0 ≤ t < 5
t≥5
0,
and 20t = 20(t − 5) + 100 we can write
f (t) = 20t − 20t
(t − 5) = 20t − 20(t − 5)
(t − 5) − 100
(t − 5).
The Laplace transform of the differential equation is
{x} + 16
s2
Solving for
{x} =
20 20 −5s 100 −5s
− 2e
−
e .
s2
s
s
{x} we obtain
20
100
20
−
e−5s −
e−5s
s2 (s2 + 16) s2 (s2 + 16)
s(s2 + 16)
5
5 1
25 1 25
4
s
−5s
=
−
−
·
·
1−e
· −
·
e−5s .
4 s2
16 s2 + 16
4 s
4 s2 + 16
{x} =
Thus
5
5
5
5
25 25
x(t) = t −
sin 4t − (t − 5) −
sin 4(t − 5)
(t − 5) −
−
cos 4(t − 5)
(t − 5)
4
16
4
16
4
4
=
5
5
5
t−
sin 4t − t
4
16
4
(t − 5) +
5
sin 4(t − 5)
16
(t − 5) +
25
cos 4(t − 5)
4
(t − 5).
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
f (t) =
sin t, 0 ≤ t < 2π
0,
t ≥ 2π
and sin t = sin(t − 2π) we can write
f (t) = sin t − sin(t − 2π) (t − 2π).
The Laplace transform of the differential equation is
s2
Solving for
{x} + 16
{x} =
1
1
−
e−2πs .
s2 + 1 s2 + 1
{x} we obtain
1
1
−
e−2πs
(s2 + 16) (s2 + 1) (s2 + 16) (s2 + 1)
−1/15
1/15
−1/15
1/15
= 2
+ 2
− 2
+ 2
e−2πs .
s + 16 s + 1
s + 16 s + 1
{x} =
216
4.3 Translation Theorems
Thus
x(t) = −
=
1
1
1
sin 4t +
sin t +
sin 4(t − 2π)
60
15
60
1
− 60
sin 4t +
1
15
(t − 2π) −
1
sin(t − 2π)
15
(t − 2π)
sin t, 0 ≤ t < 2π
t ≥ 2π.
0,
73. The differential equation is
2.5
dq
+ 12.5q = 5
dt
(t − 3).
{q} + 5
2 −3s
e .
s
The Laplace transform of this equation is
s
Solving for
{q} =
{q} we obtain
2
e−3s =
s(s + 5)
{q} =
Thus
q(t) =
2
5
2 1 2
1
· − ·
5 s 5 s+5
2
(t − 3) − e−5(t−3)
5
e−3s .
(t − 3).
74. The differential equation is
dq
+ 10q = 30et − 30et (t − 1.5).
dt
The Laplace transform of this equation is
10
{q} − q0 +
s
Solving for
{q} we obtain
{q} =
Thus
q0 −
q(t) =
q0 −
3
2
3
2
·
{q} =
3
3e1.5 −1.5s
.
−
e
s − 1 s − 1.5
1
3
1
+ ·
− 3e1.5
s+1 2 s−1
−2/5
2/5
+
s+1
s − 1.5
3
6
e−t + et + e1.5 e−(t−1.5) − e1.5(t−1.5)
2
5
e−1.5s .
(t − 1.5).
75. (a) The differential equation is
di
3π
3π
+ 10i = sin t + cos t −
t−
,
dt
2
2
i(0) = 0.
The Laplace transform of this equation is
s
Solving for
{i} + 10
{i} =
1
se−3πs/2
+
.
s2 + 1
s2 + 1
{i} we obtain
1
s
+
e−3πs/2
(s2 + 1)(s + 10) (s2 + 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
{i} =
Thus
i(t) =
1 −10t
− cos t + 10 sin t
e
101
1
3π
3π
3π
+
−10e−10(t−3π/2) + 10 cos t −
+ sin t −
t−
.
101
2
2
2
217
4.3
Translation Theorems
(b)
i
0.2
1
2
5
4
3
t
6
-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.
76. (a) The differential equation is
dq
1
+
q = E0 [
dt
0.01
(t − 1) −
(t − 3)],
q(0) = 0
dq
+ 100q = E0 [
dt
The Laplace transform of this equation is
(t − 1) −
(t − 3)],
q(0) = 0.
50
or
50
{q} + 100
50s
Solving for
{q} = E0
1 −s 1 −3s
.
e − e
s
s
{q} we obtain
E0
1 1
e−s
e−3s
E0 1 1
1
1
−s
−3s
{q} =
.
−
=
−
e −
−
e
50 s(s + 2) s(s + 2)
50 2 s s + 2
2 s s+2
Thus
q(t) =
(b)
E0 (t − 1) − 1 − e−2(t−3)
(t − 3) .
1 − e−2(t−1)
100
q
1
1
2
4
3
5
6
t
The maximum value of q(t) is approximately 1 at t = 3.
77. The differential equation is
d4 y
= w0 [1 − (x − L/2)].
dx4
Taking the Laplace transform of both sides and using y(0) = y (0) = 0 we obtain
w0 1 s4 {y} − sy (0) − y (0) =
1 − e−Ls/2 .
EI s
Letting y (0) = c1 and y (0) = c2 we have
c1
c2
w0 1 −Ls/2
{y} = 3 + 4 +
1
−
e
s
s
EI s5
so that
4 1
L
1
1 w0
L
2
3
4
y(x) = c1 x + c2 x +
x − x−
.
x−
2
6
24 EI
2
2
EI
To find c1 and c2 we compute
y (x) = c1 + c2 x +
2
1 w0
L
x2 − x −
2 EI
2
and
218
x−
L
2
4.3 Translation Theorems
y (x) = c2 +
w0
L
L
x− x−
x−
.
EI
2
2
Then y (L) = y (L) = 0 yields the system
1 w0
c1 + c2 L +
L2 −
2 EI
2
L
3 w0 L2
= c1 + c2 L +
=0
2
8 EI
w0
c2 +
EI
1 w0 L
L
= c2 +
= 0.
2
2 EI
Solving for c1 and c2 we obtain c1 = 18 w0 L2 /EI and c2 = − 12 w0 L/EI. Thus
y(x) =
w0
EI
1
1
1
1 2 2
L x − Lx3 + x4 −
16
12
24
24
x−
L
2
4
x−
L
2
.
78. The differential equation is
d4 y
= w0 [ (x − L/3) − (x − 2L/3)].
dx4
Taking the Laplace transform of both sides and using y(0) = y (0) = 0 we obtain
w0 1 −Ls/3
s4 {y} − sy (0) − y (0) =
− e−2Ls/3 .
e
EI s
EI
Letting y (0) = c1 and y (0) = c2 we have
{y} =
c1
c2
w0 1 −Ls/3
−2Ls/3
e
+
+
−
e
s3
s4
EI s5
so that
y(x) =
1
1
1 w0
c1 x2 + c2 x3 +
2
6
24 EI
x−
L
3
4
x−
L
3
4
2L
− x−
3
x−
2L
3
.
To find c1 and c2 we compute
y (x) = c1 + c2 x +
and
w0
y (x) = c2 +
EI
1 w0
2 EI
x−
L
3
2
x−
L
3
2
2L
− x−
3
x−
2L
3
L
L
2L
2L
x−
x−
− x−
x−
.
3
3
3
3
Then y (L) = y (L) = 0 yields the system
2 2
2L
1 w0
L
1 w0 L2
c1 + c2 L +
= c1 + c2 L +
=0
−
2 EI
3
3
6 EI
c2 +
w0
EI
1 w0 L
2L L
−
= 0.
= c2 +
3
3
3 EI
Solving for c1 and c2 we obtain c1 = 16 w0 L2 /EI and c2 = − 13 w0 L/EI. Thus
y(x) =
w0
EI
1
1
1 2 2
L x − Lx3 +
12
18
24
79. The differential equation is
EI
x−
L
3
4
x−
d4 y
2w0 L
L
=
−
x
+
x
−
dx4
L
2
2
219
L
3
4
2L
− x−
3
x−
L
2
.
x−
2L
3
.
4.3
Translation Theorems
Taking the Laplace transform of both sides and using y(0) = y (0) = 0 we obtain
{y} − sy (0) − y (0) =
s4
2w0 L
1
1
− 2 + 2 e−Ls/2 .
EIL 2s s
s
Letting y (0) = c1 and y (0) = c2 we have
{y} =
so that
c1
c2
2w0
1
1
L
+ 4+
− 6 + 6 e−Ls/2
3
5
s
s
EIL 2s
s
s
5 1
L
1
2w0 L 4
1 5
1
L
x−
c1 x2 + c2 x3 +
x −
x +
x−
2
6
EIL 48
120
120
2
2
5 1
5L 4
L
1
w0
L
= c1 x2 + c2 x3 +
x − x5 + x −
.
x−
2
6
60EIL 2
2
2
y(x) =
To find c1 and c2 we compute
y (x) = c1 + c2 x +
and
y (x) = c2 +
3
w0
L
30Lx2 − 20x3 + 20 x −
60EIL
2
2
w0
L
60Lx − 60x2 + 60 x −
60EIL
2
x−
x−
L
2
L
2
.
Then y (L) = y (L) = 0 yields the system
c1 + c2 L +
w0
5
5w0 L2
30L3 − 20L3 + L3 = c1 + c2 L +
=0
60EIL
2
24EI
w0
w0 L
c2 +
[60L2 − 60L2 + 15L2 ] = c2 +
= 0.
60EIL
4EI
Solving for c1 and c2 we obtain c1 = w0 L2 /24EI and c2 = −w0 L/4EI. Thus
5
w0 L2 2
5L 4
w0 L 3
w0
L
y(x) =
x −
x +
x − x5 + x −
48EI
24EI
60EIL 2
2
x−
80. The differential equation is
d4 y
= w0 [1 − (x − L/2)].
dx4
Taking the Laplace transform of both sides and using y(0) = y (0) = 0 we obtain
w0 1 s4 {y} − sy (0) − y (0) =
1 − e−Ls/2 .
EI s
EI
Letting y (0) = c1 and y (0) = c2 we have
{y} =
so that
y(x) =
c1
c2
w0 1 −Ls/2
1
−
e
+
+
s3
s4
EI s5
4
1
1
1 w0
L
c1 x2 + c2 x3 +
x4 − x −
2
6
24 EI
2
x−
L
2
.
To find c1 and c2 we compute
2
1 w0
L
2
x − x−
y (x) = c1 + c2 x +
2 EI
2
220
L
x−
2
.
L
2
.
4.3 Translation Theorems
Then y(L) = y (L) = 0 yields the system
1
1
1 w0
c1 L2 + c2 L3 +
L4 −
2
6
24 EI
4
1
L
1
5w0
= c1 L2 + c2 L3 +
L4 = 0
2
2
6
128EI
1 w0
c1 + c2 L +
L2 −
2 EI
57
w0 L2 /EI and c2 = − 128
w0 L/EI. Thus
4 L
19
1
1
9 2 2
L
x−
L x −
Lx3 + x4 −
x−
.
256
256
24
24
2
2
9
128
Solving for c1 and c2 we obtain c1 =
y(x) =
w0
EI
2
L
3w0 2
= c1 + c2 L +
L = 0.
2
8EI
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 = 70 +
Hence for t ≥ 0,
Tm =
300 − 70
t = 70 + 57.5t.
4−0
70 + 57.5t, 0 ≤ t < 4
t ≥ 4.
300,
In terms of the unit step function,
Tm = (70 + 57.5t)[1 −
(t − 4)] + 300
(t − 4) = 70 + 57.5t + (230 − 57.5t)
(t − 4).
The initial-value problem is then
dT
= k[T − 70 − 57.5t − (230 − 57.5t)
dt
(b) Let t(s) =
or
(t − 4)],
T (0) = 70.
{T (t)}. Transforming the equation, using 230 − 57.5t = −57.5(t − 4) and Theorem 4.7, gives
70 57.5 57.5 −4s
st(s) − 70 = k t(s) −
− 2 + 2 e
s
s
s
t(s) =
70
70k
57.5k
57.5k
−
− 2
+ 2
e−4s .
s − k s(s − k) s (s − k) s (s − k)
After using partial functions, the inverse transform is then
1
1
1 kt
1 k(t−4)
T (t) = 70 + 57.5
− 57.5
+t− e
+t−4− e
k
k
k
k
(t − 4).
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
1
1 20k
1 16k
300 = 70 + 57.5
− 57.5
.
+ 20 − e
+ 16 − e
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) → 300 as t increases. Using T (20) = 299 instead, we find, with the help of a
CAS, that k ≈ −0.3.
82. In order to apply Theorem 4.7 we need the function to have the form f (t − a)
rewrite the functions given in the forms shown below.
221
(t − a). To accomplish this
4.3
Translation Theorems
(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
{tekti } = 1/(s − ki)2 . Then, using Euler’s formula,
83. (a) From Theorem 4.6 we have
{tekti } =
=
{t cos kt + it sin kt} =
{t cos kt} + i
{t sin kt}
1
(s + ki)2
s2 − k 2
2ks
= 2
= 2
+i 2
.
2
2
2
(s − ki)
(s + k )
(s + k 2 )2
(s + k 2 )2
Equating real and imaginary parts we have
{t cos kt} =
s2 − k 2
(s2 + k 2 )2
{t sin kt} =
and
(s2
2ks
.
+ k 2 )2
(b) The Laplace transform of the differential equation is
s2
{x} we obtain
Solving for
{x} + ω 2
{x} =
s2
s
.
+ ω2
{x} = s/(s2 + ω 2 )2 . Thus x = (1/2ω)t sin ωt.
EXERCISES 4.4
Additional Operational Properties
−10t
d
}=−
ds
1.
{te
3.
{t cos 2t} = −
5.
{t2 sinh t} =
6.
7.
8.
d
ds
1
s + 10
s
s2 + 4
1
=
(s + 10)2
=
s2 − 4
2
(s2 + 4)
2.
d3
{t e } = (−1)
ds3
4.
{t sinh 3t} = −
3 t
3
d
ds
1
s−1
3
s2 − 9
=
=
6
(s − 1)4
6s
1
d2
6s2 + 2
=
3
2
2
ds
s −1
(s2 − 1)
2s s2 − 3
s
d2
d
1 − s2
2
=
{t cos t} = 2
=
3
ds
s2 + 1
ds (s2 + 1)2
(s2 + 1)
2t
6
12(s − 2)
d
=
te sin 6t = −
2
ds (s − 2)2 + 36
[(s − 2)2 + 36]
−3t
d
s+3
(s + 3)2 − 9
te
cos 3t = −
=
2
2
ds (s + 3) + 9
[(s + 3)2 + 9]
9. The Laplace transform of the differential equation is
s
Solving for
{y} +
{y} =
2s
.
(s2 + 1)2
{y} we obtain
{y} =
2s
1 1
1 s
1
1 1
s
−
+
+
=−
+ 2
.
(s + 1)(s2 + 1)2
2 s + 1 2 s2 + 1 2 s2 + 1 (s2 + 1)2
(s + 1)2
222
2
(s2 − 9)
4.4
Thus
1
y(t) = − e−t −
2
1
= − e−t +
2
Additional Operational Properties
1
1
1
1
sin t + cos t + (sin t − t cos t) + t sin t
2
2
2
2
1
1
1
cos t − t cos t + t sin t.
2
2
2
10. The Laplace transform of the differential equation is
{y} −
s
Solving for
{y} =
2(s − 1)
.
((s − 1)2 + 1)2
{y} we obtain
{y} =
2
.
((s − 1)2 + 1)2
Thus
y = et sin t − tet cos t.
11. The Laplace transform of the differential equation is
{y} − sy(0) − y (0) + 9
s2
Letting y(0) = 2 and y (0) = 5 and solving for
{y} =
s
.
s2 + 9
{y} =
{y} we obtain
2s + 5s + 19s − 45
5
s
2s
+ 2
+ 2
= 2
.
2
2
(s + 9)
s + 9 s + 9 (s + 9)2
3
2
Thus
5
1
sin 3t + t sin 3t.
3
6
y = 2 cos 3t +
12. The Laplace transform of the differential equation is
s2
Solving for
{y} − sy(0) − y (0) +
{y} =
1
.
s2 + 1
{y} we obtain
{y} =
s3 − s2 + s
s
1
1
= 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
{y} − sy(0) − y (0) + 16
{y} =
{cos 4t − cos 4t
(t − π)}
or by (16) of Section 4.3 in the text,
(s2 + 16)
s
− e−πs
+ 16
s
=1+ 2
− e−πs
s + 16
{y} = 1 +
s2
Thus
{y} =
and
y=
{cos 4(t + π)}
{cos 4t} = 1 +
s2
s
s
− 2
e−πs .
+ 16 s + 16
1
s
s
− 2
e−πs
+
s2 + 16 (s2 + 16)2
(s + 16)2
1
1
1
sin 4t + t sin 4t − (t − π) sin 4(t − π) (t − π).
4
8
8
223
4.4
Additional Operational Properties
14. The Laplace transform of the differential equation is
s2
{y} − sy(0) − y (0) +
(s2 + 1)
or
Thus
{y} =
1−
t−
π
π + sin t
t−
2
2
π 1 1 −πs/2
sin t +
+ e−πs/2
− e
s s
2
1 1 −πs/2
=s+ − e
+ e−πs/2 {cos t}
s s
1 1
s
= s + − e−πs/2 + 2
e−πs/2 .
s s
s +1
{y} = s +
s
1
1
s
+
−
e−πs/2 + 2
e−πs/2
s2 + 1 s(s2 + 1) s(s2 + 1)
(s + 1)2
1
s
1
s
s
s
+ −
−
−
e−πs/2 + 2
= 2
e−πs/2
s + 1 s s2 + 1
s s2 + 1
(s + 1)2
1
s
1
s
= −
e−πs/2
− 2
e−πs/2 + 2
s
s s +1
(s + 1)2
{y} =
and
15.
π π 1 π
π
y = 1 − 1 − cos t −
t−
+
t−
sin t −
2
2
2
2
2
π 1 π
π
= 1 − (1 − sin t)
t−
−
t−
cos t
t−
.
2
2
2
2
16.
y
1
t−
π
2
y
4
0.5
2
1
2
3
4
5
6
t
1
2
3
4
5
6
t
-2
-0.5
-4
-1
17. From (7) of Section 4.2 in the text along with Theorem 4.8,
dY
d
d
{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
{ty } = −
in Y (s):
4
3
4
or
Y+ Y =− 5 .
3
s
s
s
The solution of the latter equation is Y (s) = 4/s4 + c/s3 , so
s2 Y + 3sY = −
y(t) =
{Y (s)} =
2 3 c 2
t + t .
3
2
18. From Theorem 4.8 in the text
dY
d
d
{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):
3
10
Y+
− 2s Y = − .
s
s
{ty } = −
224
4.4
Additional Operational Properties
Using the integrating factor s3 e−s , the last equation yields
2
c 2
5
+ 3 es .
s3
s
But if Y (s) is the Laplace transform of a piecewise-continuous function of exponential order, we must have, in
view of Theorem 4.5, lims→∞ Y (s) = 0. In order to obtain this condition we require c = 0. Hence
5
5
y(t) =
= t2 .
s3
2
Y (s) =
19.
21.
1 3!
6
1 ∗ t3 =
= 5
4
s s
s
20.
e−t ∗ et cos t =
22.
t
τ
23.
e dτ
=
0
t
24.
cos τ dτ
1
s
{et } =
t
25.
e−τ cos τ dτ
0
t
26.
τ sin τ dτ
0
=
=
1
s
1
s
t−τ
τe
dτ
=
sin τ cos(t − τ ) dτ
28.
30.
31.
32.
33.
34.
1
(s − 2)(s2 + 1)
s
1
= 2
+ 1)
s +1
s(s2
{et } =
=
1
s2 (s − 1)
{sin t}
{cos t} =
0
29.
e2t ∗ sin t =
1
s+1
s+1
e−t cos t =
=
s (s + 1)2 + 1
s (s2 + 2s + 2)
1
d
1
1 −2s
2
{t sin t} =
−
=−
=
2
s
ds s2 + 1
s (s2 + 1)2
(s2 + 1)
t
2
s3 (s − 1)2
{t}
0
t2 ∗ tet =
1
s(s − 1)
{cos t} =
t
27.
1
s
=
0
s−1
(s + 1) [(s − 1)2 + 1]
s
(s2
2
+ 1)
t
t
d
d 1 1
3s2 + 1
t
sin τ dτ = −
sin τ dτ = −
=
2
2
ds
ds s s + 1
s2 (s2 + 1)
0
0
t
t
1
3s + 1
d
d 1
−τ
−τ
= 2
t
τ e dτ = −
τ e dτ = −
2
ds
ds s (s + 1)
s (s + 1)3
0
0
t
1
1/(s − 1)
eτ dτ = et − 1
=
=
s(s − 1)
s
0
t
1
1/s(s − 1)
(eτ − 1)dτ = et − t − 1
=
=
s2 (s − 1)
s
0
t
1
1
1/s2 (s − 1)
(eτ − τ − 1)dτ = et − t2 − t − 1
=
=
3
s (s − 1)
s
2
0
1
Using
= teat , (8) in the text gives
(s − a)2
t
1
1
=
τ eaτ dτ = 2 (ateat − eat + 1).
2
s(s − a)
a
0
35. (a) The result in (4) in the text is
F (s) =
{F (s)G(s)} = f ∗ g, so identify
(s2
2k 3
+ k 2 )2
and
225
G(s) =
4s
.
s2 + k 2
4.4
Additional Operational Properties
Then
f (t) = sin kt − kt cos kt
so
8k 3 s
(s2 + k 2 )3
and
g(t) = 4 cos kt
t
{F (s)G(s)} = f ∗ g = 4
=
f (τ )g(t − τ )dt
0
t
(sin kτ − kτ cos kτ ) cos k(t − τ )dτ.
=4
0
Using a CAS to evaluate the integral we get
8k 3 s
= t sin kt − kt2 cos kt.
(s2 + k 2 )3
(b) Observe from part (a) that
and from Theorem 4.8 that
t(sin kt − kt cos kt) =
8k 3 s
,
+ k 2 )3
(s2
tf (t) = −F (s). We saw in (5) in the text that
{sin kt − kt cos kt} = 2k 3 /(s2 + k 2 )2 ,
so
d
2k 3
8k 3 s
t(sin kt − kt cos kt) = −
= 2
.
2
2
2
ds (s + k )
(s + k 2 )3
36. The Laplace transform of the differential equation is
s2
{y} +
y
1
2s
{y} = 2
.
+
(s + 1) (s2 + 1)2
50
Thus
1
2s
+ 2
(s2 + 1)2
(s + 1)3
and, using Problem 35 with k = 1,
{y} =
y=
5
-50
1
1
(sin t − t cos t) + (t sin t − t2 cos t).
2
4
37. The Laplace transform of the given equation is
{f } +
Solving for
{f } we obtain
{f } =
s2
{t}
{f } =
{t}.
1
. Thus, f (t) = sin t.
+1
38. The Laplace transform of the given equation is
{f } =
Solving for
{f } we obtain
{f } =
{2t} − 4
{sin t}
{f }.
√
2s2 + 2
5
8
2 1
√
+
=
.
s2 (s2 + 5)
5 s2
5 5 s2 + 5
Thus
f (t) =
39. The Laplace transform of the given equation is
{f } =
√
2
8
t + √ sin 5 t.
5
5 5
tet +
226
{t}
{f }.
10
15
t
4.4
Solving for
Additional Operational Properties
{f } we obtain
{f } =
s2
1 1
3
1
2
1
1 1
=
+
.
+
−
(s − 1)3 (s + 1)
8 s − 1 4 (s − 1)2
4 (s − 1)3
8 s+1
Thus
1 t 3 t 1 2 t 1 −t
e + te + t e − e
8
4
4
8
f (t) =
40. The Laplace transform of the given equation is
{f } + 2
Solving for
{cos t}
{f } = 4
e−t +
{sin t}.
{f } we obtain
{f } =
4s2 + s + 5
4
2
7
=
+4
.
−
(s + 1)3
s + 1 (s + 1)2
(s + 1)3
Thus
f (t) = 4e−t − 7te−t + 4t2 e−t .
41. The Laplace transform of the given equation is
{f } +
Solving for
{f } we obtain
{f } =
{1}
{f } =
{1}.
1
. Thus, f (t) = e−t .
s+1
42. The Laplace transform of the given equation is
{f } =
Solving for
{cos t} +
e−t
{f }.
{f } we obtain
{f } =
s
1
+
.
s2 + 1 s2 + 1
Thus
f (t) = cos t + sin t.
43. The Laplace transform of the given equation is
{f } =
=
Solving for
{1} +
1
8
1
+ 2+
s s
3
{t} −
{t3 }
t
8
(t − τ )3 f (τ ) dτ
3 0
1
16
1
{f } = + 2 + 4
{f }.
s s
s
{f } we obtain
{f } =
s2 (s + 1)
1 1
3 1
1 2
1 s
=
+
+
+
.
4
2
s − 16
8 s + 2 8 s − 2 4 s + 4 2 s2 + 4
Thus
f (t) =
1 −2t 3 2t 1
1
+ e + sin 2t + cos 2t.
e
8
8
4
2
44. The Laplace transform of the given equation is
{t} − 2
Solving for
{f } we obtain
{f } =
{f } =
et − e−t
{f }.
s2 − 1
1 1
1 3!
=
−
.
4
2
2s
2 s
12 s4
227
4.4
Additional Operational Properties
Thus
f (t) =
1
1
t − t3 .
2
12
45. The Laplace transform of the given equation is
{y} − y(0) =
s
Solving for
{1} −
{sin t} −
{1}
{y}.
{f } we obtain
{y} =
s2 − s + 1
1
1
2s
= 2
.
−
(s2 + 1)2
s + 1 2 (s2 + 1)2
Thus
1
t sin t.
2
y = sin t −
46. The Laplace transform of the given equation is
{y} − y(0) + 6
s
Solving for
{f } we obtain
47. The differential equation is
0.1
di
1
+ 3i +
dt
0.05
or
di
+ 30i + 200
dt
{y} =
{y} + 9
{1}
{y} =
1
. Thus, y = te−3t .
(s + 3)2
i
30
t
i(τ )dτ = 100
(t − 2)
(t − 1) −
0
i(τ )dτ = 1000
t
(t − 2) ,
(t − 1) −
0
{i} − y(0) + 30
200
{i} +
s
{i} we obtain
{i} =
1000e−s − 1000e−2s
=
s2 + 30s + 200
i(t) = 100 e−10(t−1) − e−20(t−1)
di
1
+i+
dt
0.02
or
di
+ 200i + 10,000
dt
-20
-30
1000 −s
{i} =
(e − e−2s ).
s
100
100
−
(e−s − e−2s ).
s + 10 s + 20
(t − 1) − 100 e−10(t−2) − e−20(t−2)
(t − 2).
i
48. The differential equation is
0.005
0.5 1 1.5 2 2.5 3 t
-10
s
Thus
20
10
where i(0) = 0. The Laplace transform of the differential equation is
Solving for
{1}.
t
i(τ )dτ = 100 t − (t − 1)
(t − 1)
0
t
2
1.5
i(τ )dτ = 20,000 t − (t − 1)
(t − 1) ,
1
0
where i(0) = 0. The Laplace transform of the differential equation is
1 −s
10,000
1
s {i} + 200 {i} +
{i} = 20,000 2 − 2 e
.
s
s
s
228
0.5
0.5
1
1.5
2 t
4.4
Solving for
Additional Operational Properties
{i} we obtain
20,000
2
2
200
−s
{i} =
(1 − e−s ).
(1 − e ) =
−
−
s(s + 100)2
s s + 100 (s + 100)2
Thus
i(t) = 2 − 2e−100t − 200te−100t − 2
49.
50.
{f (t)} =
{f (t)} =
1
1 − e−2as
1
1 − e−2as
a
e−st dt −
0
a
2a
a
e−st dt =
0
(t − 1) + 2e−100(t−1)
(t − 1) + 200(t − 1)e−100(t−1)
(t − 1).
(1 − e−as )2
1 − e−as
e−st dt =
=
−2as
s(1 − e
)
s(1 + e−as )
1
s(1 + e−as )
51. Using integration by parts,
1
1 − e−bs
{f (t)} =
52.
53.
54.
1
{f (t)} =
1 − e−2s
{f (t)} =
{f (t)} =
1
1 − e−πs
1
−st
te
0
π
1
1 − e−2πs
2
0
−st
(2 − t)e
dt +
b
e−st sin t dt =
π
dt =
1
1
− bs
bs e − 1
.
1 − e−s
s2 (1 − e−2s )
eπs/2 + e−πs/2
πs
1
1
·
coth
= 2
s2 + 1 eπs/2 − e−πs/2
s +1
2
e−st sin t dt =
0
a −st
a
te dt =
b
s
1
0
1
1
·
s2 + 1 1 − e−πs
55. The differential equation is L di/dt + Ri = E(t), where i(0) = 0. The Laplace transform of the equation is
Ls
From Problem 49 we have
{i} + R
{E(t)}.
{E(t)} = (1 − e−s )/s(1 + e−s ). Thus
(Ls + R)
and
{i} =
{i} =
1 − e−s
s(1 + e−s )
1
1 − e−s
1 1 − e−s
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
{i} =
Therefore,
2 1 (t − 1)
1 − e−Rt/L −
1 − e−R(t−1)/L
R
R
2
2
+
(t − 2) −
1 − e−R(t−2)/L
1 − e−R(t−3)/L
R
R
∞ 2 !
1 =
1 − e−R(t−n)/L
(t − n).
1 − e−Rt/L +
R
R n=1
i(t) =
229
(t − 3) + · · ·
4.4
Additional Operational Properties
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
56. The differential equation is L di/dt + Ri = E(t), where i(0) = 0. The Laplace transform of the equation is
Ls
From Problem 51 we have
{E(t)} =
{i} + R
{i} =
{E(t)}.
1 1
1 1
1
1
− s
= 2−
.
s s e −1
s
s es − 1
Thus
(Ls + R)
{i} =
1
1 1
−
2
s
s es − 1
and
1
1
1
1
1
−
L s2 (s + R/L) L s(s + R/L) es − 1
−s
1 1
L
1
1 1
1
L 1
=
+
−
−
e + e−2s + e−3s + · · · .
−
2
R s
R s R s + R/L
R s s + R/L
{i} =
Therefore
L
L −Rt/L
−
t− + e
R R
1 −
1 − e−R(t−2)/L
R
1
L
L
=
t − + e−Rt/L −
R
R R
1
i(t) =
R
1 (t − 1)
1 − e−R(t−1)/L
R
1 (t − 2) −
(t − 3) − · · ·
1 − e−R(t−3)/L
R
∞
1 !
1 − e−R(t−n)/L
(t − n).
R 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
57. The differential equation is x + 2x + 10x = 20f (t), where f (t) is the meander function in Problem 49 with
230
4.4
Additional Operational Properties
a = π. Using the initial conditions x(0) = x (0) = 0 and taking the Laplace transform we obtain
(s2 + 2s + 10)
Then
{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 !
=
(−1)n e−nπs .
+
s
s n=1
{x(t)} =
∞
!
20
40
(−1)n e−nπs
+
s(s2 + 2s + 10) s(s2 + 2s + 10) n=1
∞
!
2
2s + 4
4s + 8
n 4
= − 2
(−1)
+
−
e−nπs
s s + 2s + 10 n=1
s s2 + 2s + 10
∞
!
(s + 1) + 1
2 2(s + 1) + 2
n 1
+4
−
e−nπs
= −
(−1)
2+9
s
(s + 1)2 + 9
s
(s
+
1)
n=1
and
∞
!
1
(−1)n 1 − e−(t−nπ) cos 3(t − nπ)
x(t) = 2 1 − e−t cos 3t − e−t sin 3t + 4
3
n=1
1
− e−(t−nπ) sin 3(t − nπ)
(t − nπ).
3
The graph of x(t) on the interval [0, 2π) is shown below.
x
3
π
2π
t
−3
58. 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)
{x(t)} =
=
Then
{x(t)} =
5
5
1
= (1 − e−πs + e−2πs − e−3πs + e−4πs − · · ·)
−πs
s 1+e
s
∞
5!
(−1)n e−nπs .
s n=0
∞
∞
!
!
5
1
1
1
n −nπs
n
e−nπs
(−1)
e
=
5
(−1)
−
−
2
s(s + 1)2 n=0
s
s
+
1
(s
+
1)
n=0
231
4.4
Additional Operational Properties
and
x(t) = 5
∞
!
(−1)n (1 − e−(t−nπ) − (t − nπ)e−(t−nπ) )
(t − nπ).
n=0
The graph of x(t) on the interval [0, 4π) is shown below.
x
5
2π
4π
t
−5
59. f (t) = −
1
t
d
1
[ln(s − 3) − ln(s + 1)] = −
ds
t
1
1
−
s−3 s+1
=−
1 3t
e − e−t
t
60. 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) =
{J0 (t)}, where J0 has a derivative that is continuous
and of exponential order, implies by Problem 46 of Exercises 4.2 that
1 = J0 (0) = lim sY (s) = c lim √
s→∞
s→∞
s2
s
=c
+ k2
so c = 1 and
Y (s) = √
1
s2 + 1
{J0 (t)} = √
or
1
.
s2 + 1
61. (a) Using Theorem 4.8, the Laplace transform of the differential equation is
−
d 2
d
[s Y − sy(0) − y (0)] + sY − y(0) +
[sY − y(0)] + nY
ds
ds
d
d
= − [s2 Y ] + sY +
[sY ] + nY
ds
ds
dY
dY
= −s2
− 2sY + sY + s
+ Y + nY
ds
ds
dY
2
= (s − s )
+ (1 + n − s)Y = 0.
ds
Separating variables, we find
dY
1+n−s
=
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
232
ds
4.4
Additional Operational Properties
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:
1
n = 0 : L0 (t) =
=1
s
1
s−1
1
=
=1−t
n = 1 : L1 (t) =
−
s2
s s2
1
1
2
(s − 1)2
1
n = 2 : L2 (t) =
=
− 2 + 3 = 1 − 2t + t2
3
s
s s
s
2
3
1
3
(s − 1)
3
1
1
3
=
n = 3 : L3 (t) =
+ 3 − 4 = 1 − 3t + t2 − t3
−
s4
s s2
s
s
2
6
4
1
4
(s − 1)
6
4
1
=
−
n = 4 : L4 (t) =
+ 3− 4+ 5
s5
s s2
s
s
s
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,
t n
e d n −t
1
1
{et f (n) (t)} =
{f (n) (t)} s→s−1
=
t
e
n
n! dt
n!
n!
1 n
s
=
{tn e−t } − sn−1 f (0) − sn−2 f (0) − · · · − f (n−1) (0)
n!
s→s−1
1 n
n −t
=
{t e }
s
n!
s→s−1
n!
1 n
(s − 1)n
=
=
= Y,
s
n+1
n!
(s + 1)
sn+1
s→s−1
where Y =
{Ln (t)}. Thus
Ln (t) =
et dn n −t
(t e ),
n! dtn
n = 0, 1, 2, . . . .
62. 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 ) ==
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
63. The solution is
√
1
1
y(t) = et − e−t/2 cos 15 t −
6
6
233
√
3/5 −t/2
sin 15 t.
e
6
4.4
Additional Operational Properties
64. The solution is
q(t) = 1 − cos t + (6 − 6 cos t)
(t − 3π) − (4 + 4 cos t)
(t − π).
q
5
-5
Π
3Π
t
5Π
EXERCISES 4.5
The Dirac Delta Function
1. The Laplace transform of the differential equation yields
{y} =
1 −2s
e
s−3
so that
y = e3(t−2)
(t − 2).
2. The Laplace transform of the differential equation yields
{y} =
2
e−s
+
s+1 s+1
so that
y = 2e−t + e−(t−1)
(t − 1).
3. The Laplace transform of the differential equation yields
1 {y} = 2
1 + e−2πs
s +1
so that
y = sin t + sin t
(t − 2π).
4. The Laplace transform of the differential equation yields
1
4
{y} =
e−2πs
2
4 s + 16
so that
1
1
y = sin 4(t − 2π) (t − 2π) = sin 4t
4
4
(t − 2π).
5. The Laplace transform of the differential equation yields
1 −πs/2
{y} = 2
e
+ e−3πs/2
s +1
so that
π π
3π
3π
y = sin t −
t−
+ sin t −
t−
2
2
2
2
π
π
= − cos t
t−
+ cos t
t−
.
2
2
234
4.5
The Dirac Delta Function
6. The Laplace transform of the differential equation yields
{y} =
s
1
+
(e−2πs + e−4πs )
s2 + 1 s2 + 1
so that
(t − 2π) +
y = cos t + sin t[
(t − 4π)].
7. The Laplace transform of the differential equation yields
1
1 1 1 1
−s
{y} = 2
(1 + e ) =
−
(1 + e−s )
s + 2s
2 s 2 s+2
so that
1 1
1 1 −2(t−1)
y = − e−2t +
− e
2 2
2 2
(t − 1).
8. The Laplace transform of the differential equation yields
{y} =
s+1
1
3 1 1 1
1 1 −2s
3 1
1 1
−2s
+
e
−
−
−
e
=
+
s2 (s − 2) s(s − 2)
4 s − 2 4 s 2 s2
2 s−2 2 s
so that
y=
3 2t 3 1
1
1
e − − t + e2(t−2) −
4
4 2
2
2
(t − 2).
9. The Laplace transform of the differential equation yields
{y} =
1
e−2πs
(s + 2)2 + 1
so that
y = e−2(t−2π) sin t
(t − 2π).
10. The Laplace transform of the differential equation yields
{y} =
1
e−s
(s + 1)2
so that
y = (t − 1)e−(t−1)
(t − 1).
11. The Laplace transform of the differential equation yields
{y} =
=
s2
4+s
e−πs + e−3πs
+ 2
+ 4s + 13
s + 4s + 13
−πs
2
3
3
s+2
1
e
+
+
+ e−3πs
2
2
2
2
2
2
3 (s + 2) + 3
(s + 2) + 3
3 (s + 2) + 3
so that
y=
2 −2t
1
sin 3t + e−2t cos 3t + e−2(t−π) sin 3(t − π)
e
3
3
1
+ e−2(t−3π) sin 3(t − 3π) (t − 3π).
3
(t − π)
12. The Laplace transform of the differential equation yields
{y} =
1
e−2s + e−4s
+
2
(s − 1) (s − 6) (s − 1)(s − 6)
=−
−2s
1
1
1
1
1
1
1 1
1 1
+
+ e−4s
−
+
−
+
e
2
25 s − 1 5 (s − 1)
25 s − 6
5 s−1 5 s−6
235
4.5
The Dirac Delta Function
so that
1
1
1
1
1
y = − et − tet + e6t + − et−2 + e6(t−2)
25
5
25
5
5
1
1
(t − 2) + − et−4 + e6(t−4)
5
5
(t − 4).
13. The Laplace transform of the differential equation yields
{y} =
1 2 1 3! 1 P0 3! −Ls/2
y (0) +
y (0) +
e
3
4
2 s
6 s
6 EI s4
so that
y=
1 1
1 P0
y (0)x2 + y (0)x3 +
2
6
6 EI
x−
L
2
3
x−
L
2
.
Using y (L) = 0 and y (L) = 0 we obtain
3
1 P0 L 2 1 P 0 3 1 P 0
L
y=
x −
x +
x−
4 EI
6 EI
6 EI
2

P0 L 2 1 3
L



 EI 4 x − 6 x , 0 ≤ x < 2
=
2

1
L
L

 P0 L

x−
,
≤ x ≤ L.
4EI 2
12
2
14. From Problem 13 we know that
1
1
1 P0
y = y (0)x2 + y (0)x3 +
2
6
6 EI
L
x−
2
3
L
x−
2
L
x−
2
.
Using y(L) = 0 and y (L) = 0 we obtain
1 P0 L 2
1
y=
x −
16 EI
12

P0 L 2



 EI 16 x −
=


 P0 L x2 −

EI 16
3 L
P0 3 1 P0
L
x−
x +
x−
EI
6 EI
2
2
L
1 3
x ,
0≤x<
12
2
3
1 P0
1 3
L
L
,
x +
x−
≤ x ≤ L.
12
6 EI
2
2
15. 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.
236
4.6
Systems of Linear Differential Equations
EXERCISES 4.6
Systems of Linear Differential Equations
1. Taking the Laplace transform of the system gives
s
s
{x} = −
{x} +
{y} − 1 = 2
{y}
{x}
so that
{x} =
1
1 1
1 1
=
−
(s − 1)(s + 2)
3 s−1 3 s+2
{y} =
1
2
2 1
1 1
+
=
+
.
s s(s − 1)(s + 2)
3 s−1 3 s+2
and
Then
x=
1 t 1 −2t
e − e
3
3
and
y=
2 t 1 −2t
e + e .
3
3
2. Taking the Laplace transform of the system gives
s
{x} − 1 = 2
s
{y} − 1 = 8
1
s−1
1
{x} − 2
s
{y} +
so that
{y} =
and
y=
s3 + 7s2 − s + 1
1 1
8
1
173 1
53
1
=
−
+
−
2
s(s − 1)(s − 16)
16 s 15 s − 1
96 s − 4 160 s + 4
1
8
173 4t
53 −4t
− et +
e −
e .
16 15
96
160
Then
x=
1 1
173 4t
53 −4t
1
1
y + t = t − et +
e +
e .
8
8
8
15
192
320
3. Taking the Laplace transform of the system gives
s
{x} + 1 =
s
{y} − 2 = 5
{x} − 2
{y}
{x} −
{y}
so that
{x} =
−s − 5
s
5 3
=− 2
−
2
s +9
s + 9 3 s2 + 9
and
x = − cos 3t −
5
sin 3t.
3
Then
y=
1
7
1
x − x = 2 cos 3t − sin 3t.
2
2
3
237
4.6
Systems of Linear Differential Equations
4. Taking the Laplace transform of the system gives
{x} + s
{y} =
1
s
{x} + (s − 1)
{y} =
1
s−1
(s + 3)
(s − 1)
so that
{y} =
5s − 1
4
1
1 1 1 1
+
+
=−
3s(s − 1)2
3 s 3 s − 1 3 (s − 1)2
{x} =
1 − 2s
1
1
1 1 1 1
−
−
=
.
3s(s − 1)2
3 s 3 s − 1 3 (s − 1)2
and
Then
x=
1 1 t 1 t
− e − te
3 3
3
1 1
4
y = − + et + tet .
3 3
3
and
5. Taking the Laplace transform of the system gives
(2s − 2)
(s − 3)
{x} + s
{x} + (s − 3)
1
s
2
{y} =
s
{y} =
so that
{x} =
−s − 3
1 1 5 1
2
=−
+
−
s(s − 2)(s − 3)
2 s 2 s−2 s−3
{y} =
3s − 1
1 1 5 1
8 1
=−
−
+
.
s(s − 2)(s − 3)
6 s 2 s−2 3 s−3
and
Then
1 5
x = − + e2t − 2e3t
2 2
1 5
8
y = − − e2t + e3t .
6 2
3
and
6. Taking the Laplace transform of the system gives
(s + 1)
{x} − (s − 1)
{y} = −1
s
{x} + (s + 2)
{y} = 1
so that
s + 1/2
s + 1/2
√
=
s2 + s + 1
(s + 1/2)2 + ( 3/2)2
√
√
−3/2
3/2
√
{x} = 2
.
=− 3
2
s +s+1
(s + 1/2) + ( 3/2)2
{y} =
and
Then
√
−t/2
y=e
3
cos
t
2
and
√
x = − 3 e−t/2 sin
7. Taking the Laplace transform of the system gives
(s2 + 1)
−
{x} −
{x} + (s2 + 1)
{y} = −2
{y} = 1
so that
{x} =
−2s2 − 1
1 1
3 1
=− 2 −
s4 + 2s2
2 s
2 s2 + 2
and
238
√
3
t.
2
4.6
Systems of Linear Differential Equations
√
1
3
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
{x} +
{y} = 1
{x} − (s + 1)
{y} = 1
(s + 1)
4
so that
{x} =
s+2
s+1
2
1
=
+
s2 + 2s + 5
(s + 1)2 + 22
2 (s + 1)2 + 22
and
{y} =
Then
s2
−s + 3
s+1
2
=−
+2
.
2
2
+ 2s + 5
(s + 1) + 2
(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
d2 x
1
= t2 + 2t
2
dt
2
d2 y
1
= t2 − 2t.
dt2
2
Taking the Laplace transform of the system gives
1
1 3!
1 4!
{x} = 8 +
+
5
s 24 s
3 s4
and
{y} =
so that
x=8+
1 4 1 3
t + t
24
3
1 4!
1 3!
−
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)
(s + 2)
{x} + s3
{x} − 2s3
{y} =
s2
6
+1
{y} = 0
so that
{x} =
4
4 1
4 s
8 1
=
−
−
(s − 2)(s2 + 1)
5 s − 2 5 s2 + 1 5 s2 + 1
{y} =
2s + 4
2
1 1
1
2
6 s
8 1
= − 2 −2 3 +
−
+
.
s3 (s − 2)(s2 + 1)
s s
s
5 s − 2 5 s2 + 1 5 s2 + 1
and
Then
x=
and
4 2t 4
8
e − cos t − sin t
5
5
5
1
6
8
y = 1 − 2t − 2t2 + e2t − cos t + sin t.
5
5
5
239
4.6
Systems of Linear Differential Equations
11. Taking the Laplace transform of the system gives
{x} + 3(s + 1)
s2
s2
so that
{x} = −
{x} + 3
{y} = 2
{y} =
1
(s + 1)2
2s + 1
1
1
1 2
1
= + 2+
.
−
3
+ 1)
s s
2 s
s+1
s3 (s
Then
1
x = 1 + t + t2 − e−t
2
and
y=
1
1
1
1 −t 1 te − x = te−t + e−t − .
3
3
3
3
3
12. Taking the Laplace transform of the system gives
{x} + 2
{y} =
2e−s
s
{x} + (s + 1)
{y} =
1 e−s
+
2
s
(s − 4)
−3
so that
{x} =
−1/2
1
+ e−s
(s − 1)(s − 2)
(s − 1)(s − 2)
1
1 1
1 1
1
=
−
+ e−s −
+
2 s−1 2 s−2
s−1 s−2
and
{y} =
e−s
−s/2 + 2
s/4 − 1
+
+ e−s
s
(s − 1)(s − 2)
(s − 1)(s − 2)
1 1
3 1
1
3 1
−s 1
−
+e
−
+
.
=
4 s−1 2 s−2
s 2 s−1 s−2
Then
x=
1 t 1 2t t−1
+ e2(t−1)
e − e + −e
2
2
y=
3 t 1 2t
3
e − e + 1 − et−1 + e2(t−1)
4
2
2
and
13. The system is
(t − 1)
x1 = −3x1 + 2(x2 − x1 )
x2 = −2(x2 − x1 )
x1 (0) = 0
x1 (0) = 1
x2 (0) = 1
x2 (0) = 0.
Taking the Laplace transform of the system gives
{x1 } − 2
{x2 } = 1
{x1 } + (s2 + 2)
{x2 } = s
(s2 + 5)
−2
240
(t − 1).
4.6
so that
and
Systems of Linear Differential Equations
√
2 s
1 1
2 s
4
6
s2 + 2s + 2
{x1 } = 4
=
+
−
+ √ 2
2
2
2
2
s + 7s + 6
5 s +1 5 s +1 5 s +6 5 6 s +6
√
6
s3 + 5s + 2
4 s
2 1
1 s
2
{x2 } = 2
=
+
+
− √ 2
.
2
2
2
2
(s + 1)(s + 6)
5 s +1 5 s +1 5 s +6 5 6 s +6
Then
x1 =
√
√
2
4
1
2
cos t + sin t − cos 6 t + √ sin 6 t
5
5
5
5 6
x2 =
√
√
4
2
2
1
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
{x1 } −
{x2 } = −1
{x1 } − (2 + s2 )
{x2 } = −1
(2 + s2 )
so that
1
+3
and
√
1
x1 = − √ sin 3 t
3
and
{x1 } = −
Then
s2
{x2 } =
s2
1
.
+3
√
1
x2 = √ sin 3 t.
3
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
{i2 } + 500
{i3 } =
10,000
s
{i2 } + (s + 400)
{i3 } =
8,000
s
(s + 500)
400
so that
{i3 } =
Then
i3 =
80 80 −900t
− e
9
9
s2
and
8,000
80 1 80
1
=
−
.
+ 900s
9 s
9 s + 900
i2 = 20 − 0.0025i3 − i3 =
241
100 100 −900t
.
−
e
9
9
4.6
Systems of Linear Differential Equations
(c) i1 = i2 + i3 = 20 − 20e−900t
16. (a) Taking the Laplace transform of the system
i2 + i3 + 10i2 = 120 − 120
−10i2
+
5i3
(t − 2)
+ 5i3 = 0
gives
{i2 } + s
(s + 10)
−10s
so that
and
120 1 − e−2s
s
{i3 } = 0
{i2 } =
48
120(s + 1)
60
12 −2s
=
1
−
e
−
+
1 − e−2s
2
(3s + 11s + 10)s
s + 5/3 s + 2
s
{i3 } =
240
240
240 −2s
=
1
−
e
−
1 − e−2s .
2
3s + 11s + 10
s + 5/3 s + 2
and
Then
{i2 } + 5(s + 1)
{i3 } =
i2 = 12 + 48e−5t/3 − 60e−2t − 12 + 48e−5(t−2)/3 − 60e−2(t−2) (t − 2)
i3 = 240e−5t/3 − 240e−2t − 240e−5(t−2)/3 − 240e−2(t−2) (t − 2).
(b) i1 = i2 + i3 = 12 + 288e−5t/3 − 300e−2t − 12 + 288e−5(t−2)/3 − 300e−2(t−2)
(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)
6
{i2 } + 6
{i2 } + (s + 6)
50
+1
50
{i3 } = 2
s +1
{i3 } =
s2
so that
{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 =
85
20 −2t
375 −15t 145
e
e
cos t +
sin t
+
+
13
1469
113
113
25
11
30 −2t
250 −15t 280
1
810
+
−
sin t − i2 − i2 =
e
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
242
4.6
Systems of Linear Differential Equations
gives
s
−200
120
s
{i2 } = 0
{i1 } + 100
{i2 } =
{i1 } + (s + 200)
so that
{i2 } =
24,000
6
6 1 6
s + 100
100
−
.
=
−
s(s2 + 200s + 20,000)
5 s 5 (s + 100)2 + 1002
5 (s + 100)2 + 1002
Then
i2 =
and
6 6 −100t
6
cos 100t − e−100t sin 100t
− e
5 5
5
i1 = 0.005i2 + i2 =
6 6 −100t
cos 100t.
− e
5 5
19. Taking the Laplace transform of the system
2i1 + 50i2 = 60
0.005i2 + i2 − i1 = 0
gives
{i1 } + 50
2s
−200
{i1 } + (s + 200)
60
s
{i2 } = 0
{i2 } =
so that
{i2 } =
6,000
s(s2 + 200s + 5,000)
√
√
6 1 6
6 2
s + 100
50 2
√
√
=
−
.
−
5 s 5 (s + 100)2 − (50 2 )2
5 (s + 100)2 − (50 2 )2
Then
and
√
√
√
6 6 −100t
6 2 −100t
i2 = − e
e
cosh 50 2 t −
sinh 50 2 t
5 5
5
√
√
√
6 6 −100t
9 2 −100t
i1 = 0.005i2 + i2 = − e
e
cosh 50 2 t −
sinh 50 2 t.
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
E(t) = iR1 +
1
q,
C
so that
i1 =
1
1
q.
E(t) −
R1
R1 C
Then
dq
1
1
= i1 − i3 =
q − i3
E(t) −
dt
R1
R1 C
and
R1
dq
1
+ q + R1 i3 = E(t).
dt
C
243
(1)
4.6
Systems of Linear Differential Equations
Summing the voltage drops across the longer loop gives
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
di3
1
+ R2 i3 − q = 0.
dt
C
(b) Using L = R1 = R2 = C = 1, E(t) = 50e−t (t − 1) = 50e−1 e−(t−1)
the Laplace transform of the system we obtain
{i3 } =
50e−1 −s
e
s+1
(s + 1)
{q} +
(s + 1)
{i3 } −
{q} = 0,
{q} =
50e−1 e−s
(s + 1)2 + 1
so that
(t − 1), q(0) = i3 (0) = 0, and taking
and
q(t) = 50e−1 e−(t−1) sin(t − 1)
(t − 1) = 50e−t sin(t − 1)
(t − 1).
21. (a) Taking the Laplace transform of the system
4θ1 + θ2 + 8θ1 = 0
θ1 + θ2 + 2θ2 = 0
gives
4 s2 + 2
s2
so that
{θ1 } + s2
{θ1 } + s2 + 2
3s2 + 4 s2 + 4
or
{θ2 } =
Then
θ2 =
θ1 =
(b)
θ1
{θ2 } = 0
{θ2 } = −3s3
s
3 s
1
−
.
2
2 s + 4/3 2 s2 + 4
2
1
3
cos √ t − cos 2t
2
2
3
so that
{θ2 } = 3s
θ1 = −θ2 − 2θ2
and
3
2
1
cos √ t + cos 2t.
4
4
3
θ2
2
2
1
1
−1
3π
6π t
−1
−2
−2
244
3π
6π t
4.6
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) The Lissajous curve is plotted for 0 ≤ t ≤ 30.
θ2
2
1
-1-0.5
0.5 1
θ1
-1
-2
(d)
t=0
t=1
t=2
t=3
t=4
t=5
t=6
t=7
t=8
t
θ1
θ2
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=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
(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
245
4.6
Systems of Linear Differential Equations
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 4 REVIEW EXERCISES
1.
1
{f (t)} =
−st
te
0
2.
∞
dt +
(2 − t)e−st dt =
1
4
{f (t)} =
e−st dt =
2
1
2
− 2 e−s
s2
s
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 4.5 in the text.)
6. False; consider f (t) = 1 and g(t) = 1.
7.
8.
9.
10.
11.
12.
e−7t =
te−7t =
14.
15.
16.
1
(s + 7)2
2
s2 + 4
−3t
2
e
sin 2t =
(s + 3)2 + 4
d
2
4s
{t sin 2t} = −
= 2
ds s2 + 4
(s + 4)2
{sin 2t} =
{sin 2t
13.
1
s+7
(t − π)} =
(t − π)} =
s2
2
e−πs
+4
1 5!
1
= t5
6 s6
6
1
1
1
1
=
= et/3
3s − 1
3 s − 1/3
3
1
1
2
1
=
= t2 e5t
(s − 5)3
2
(s − 5)3
2
√
1
1 √
1
1
1
1
1
√
√
√
√
=
−
+
= − √ e− 5 t + √ e 5 t
2
s −5
2 5 s+ 5 2 5 s− 5
2 5
2 5
20
s6
{sin 2(t − π)
=
246
CHAPTER 4 REVIEW EXERCISES
s
=
s2 − 10s + 29
1 −5s
= (t − 5)
e
s2
s + π −s
=
e
s2 + π 2
17.
18.
19.
5
s−5
2
+
2
2
(s − 5) + 2
2 (s − 5)2 + 22
21.
22.
s
π
e−s + 2
e−s
s2 + π 2
s + π2
= cos π(t − 1)
1
1 L
= 2
L2 s2 + n2 π 2
L nπ
−5t exists for s > −5.
e
23.
24.
5
= e5t cos 2t + e5t sin 2t
2
(t − 5)
20.
(t − 1) + sin π(t − 1) (t − 1)
1
nπ/L
nπ
=
sin
t
s2 + (n2 π 2 )/L2
Lnπ
L
d
te8t f (t) = − F (s − 8).
ds
{eat f (t − k) (t − k)} = e−ks {ea(t+k) f (t)} = e−ks eak {eat f (t)} = e−k(s−a) F (s − a)
t
1
F (s − a)
eaτ f (τ ) dτ =
{eat f (t)} =
, whereas
s
s
0
t
t
F (s) F (s − a)
at
e
.
f (τ ) dτ =
f (τ ) dτ =
=
s
s−a
0
0
s→s−a
s→s−a
25. f (t)
(t − t0 )
26. f (t) − f (t)
27. f (t − t0 )
(t − t0 )
(t − t0 )
28. f (t) − f (t)
(t − t0 ) + f (t)
(t − t1 )
29. f (t) = t − [(t − 1) + 1] (t − 1) +
(t − 1) −
(t − 4) = t − (t − 1)
(t − 1) −
(t − 4)
1
1
1
− 2 e−s − e−4s
2
s
s
s
t
1
1
1 −4(s−1)
e f (t) =
−
e−(s−1) −
e
2
2
(s − 1)
(s − 1)
s−1
{f (t)} =
(t − π) − sin t (t − 3π) = − sin(t − π) (t − π) + sin(t − 3π)
1
1
{f (t)} = − 2
e−πs + 2
e−3πs
s +1
s +1
t
1
1
e f (t) = −
e−π(s−1) +
e−3π(s−1)
(s − 1)2 + 1
(s − 1)2 + 1
30. f (t) = sin t
31. f (t) = 2 − 2
(t − 2) + [(t − 2) + 2] (t − 2) = 2 + (t − 2)
2
1
{f (t)} = + 2 e−2s
s s
t
2
1
e−2(s−1)
e f (t) =
+
s − 1 (s − 1)2
32. f (t) = t − t
(t − 2)
(t − 1) + (2 − t) (t − 1) − (2 − t) (t − 2) = t − 2(t − 1)
1
2
1
{f (t)} = 2 − 2 e−s + 2 e−2s
s
s
s
t
1
2
1
e f (t) =
−
e−(s−1) +
e−2(s−1)
(s − 1)2
(s − 1)2
(s − 1)2
247
(t − 3π)
(t − 1) + (t − 2)
(t − 2)
CHAPTER 4 REVIEW EXERCISES
33. Taking the Laplace transform of the differential equation we obtain
{y} =
so that
5
2
1
+
(s − 1)2
2 (s − 1)3
1
y = 5tet + t2 et .
2
34. Taking the Laplace transform of the differential equation we obtain
{y} =
=
(s −
1
− 8s + 20)
1)2 (s2
6
6
5
1
1
1
s−4
2
−
+
+
2
2
2
169 s − 1 13 (s − 1)
169 (s − 4) + 2
338 (s − 4)2 + 22
so that
y=
6 t
1
6 4t
5 4t
e + tet −
e cos 2t +
e sin 2t.
169
13
169
338
35. Taking the Laplace transform of the given differential equation we obtain
s3 + 6s2 + 1
1
2
−
e−2s −
e−2s
+ 1)(s + 5) s2 (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
1
1
1
2 1 1
· − ·
+
·
e−2s
−
5 s 2 s + 1 10 s + 5
{y} =
so that
y=−
s2 (s
6
13
4
1
3
1
+ t + e−t − e−5t −
(t − 2) − (t − 2) (t − 2)
25 5
2
50
25
5
1 −(t−2)
9 −5(t−2)
+ e
(t − 2) −
(t − 2).
e
4
100
36. Taking the Laplace transform of the differential equation we obtain
{y} =
s3 + 2
2 + 2s + s2 −s
− 3
e
− 5)
s (s − 5)
s3 (s
=−
2 1
1 2
127 1
1 2
37
2 1
37 1 12 1
1
−
+
−
+
−
−
−
−
e−s
125 s 25 s2
5 s3
125 s − 5
125 s 25 s2
5 s3
125 s − 5
so that
2
127 5t
37
37 5(t−1)
2
1
12
1
y=−
− t − t2 +
e − −
− (t − 1) − (t − 1)2 +
e
125 25
5
125
125 25
5
125
37. Taking the Laplace transform of the integral equation we obtain
{y} =
so that
1
1 2
1
+
+
s s2
2 s3
1
y(t) = 1 + t + t2 .
2
38. Taking the Laplace transform of the integral equation we obtain
(
{f })2 = 6 ·
6
s4
or
248
{f } = ±6 ·
1
s2
(t − 1).
CHAPTER 4 REVIEW EXERCISES
so that f (t) = ±6t.
39. Taking the Laplace transform of the system gives
s
4
{x} +
{y} =
{x} + s
1
+1
s2
{y} = 2
so that
{x} =
s2 − 2s + 1
1 1 1 1
9 1
=−
+
+
.
s(s − 2)(s + 2)
4 s 8 s−2 8 s+2
Then
1 1
9
x = − + e2t + e−2t
4 8
8
and y = −x + t =
9 −2t 1 2t
− e + t.
e
4
4
40. Taking the Laplace transform of the system gives
s2
{x} + s2
2s
{x} + s2
1
s−2
1
{y} = −
s−2
{y} =
so that
{x} =
2
1
1 1 1 1
−
+
=
s(s − 2)2
2 s 2 s − 2 (s − 2)2
{y} =
−s − 2
1
3 1 1 1
3 1
−
−
=−
+
.
s2 (s − 2)2
4 s 2 s2
4 s − 2 (s − 2)2
and
Then
x=
1 1 2t
− e + te2t
2 2
41. The integral equation is
3
3 1
and y = − − t + e2t − te2t .
4 2
4
t
i(τ ) dτ = 2t2 + 2t.
10i + 2
0
Taking the Laplace transform we obtain
4
s
s+2
9
2
9
2
2
45
9
{i} =
= 2
=− + 2 +
=− + 2 +
.
+
s3
s2 10s + 2
s (5s + 2)
s s
5s + 1
s s
s + 1/5
Thus
i(t) = −9 + 2t + 9e−t/5 .
42. The differential equation is
1 d2 q
dq
+ 10 + 100q = 10 − 10
2 dt2
dt
Taking the Laplace transform we obtain
{q} =
=
s(s2
(t − 5).
20
1 − e−5s
+ 20s + 200)
1 1
1
1
s + 10
10
−
−
1 − e−5s
2
2
2
2
10 s 10 (s + 10) + 10
10 (s + 10) + 10
249
CHAPTER 4 REVIEW EXERCISES
so that
q(t) =
1
1
1
− e−10t cos 10t − e−10t sin 10t
10 10
10
1
1 −10(t−5)
1 −10(t−5)
−
cos 10(t − 5) − e
sin 10(t − 5) (t − 5).
− e
10 10
10
43. Taking the Laplace transform of the given differential equation we obtain
2w0
c1 2!
1
1
c2 3!
L 4!
5!
5!
{y} =
+
· 5−
· 6+
· 6 e−sL/2 +
·
·
EIL 48 s
120 s
120 s
2 s3
6 s4
so that
y=
2w0 L 4
1 5
1
x −
x +
EIL 48
120
120
x−
L
2
5
x−
L
2
+
c1 2 c2 3
x + x
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
w0
L
L2 3 L3 2 1
1
y=
− x5 + x4 −
x +
x +
12EIL
5
2
2
4
5
L
x−
2
5
L
x−
2
.
44. 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
s4
{y} − s3 y(0) − s2 y (0) − sy(0) − y (0) + 4
and
{y} =
{y} =
{δ(t − π/2)}
c1
2s
c2
4
w0
4
·
+
·
+
·
e−sπ/2 .
2 s4 + 4
4 s4 + 4 4EI s4 + 4
From the table of transforms we get
c1
c2
y=
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 −
4EI
2
2
2
2
x−
π
2
x−
π
.
2
Using y(π) = 0 and y (π) = 0 we find
c1 =
Hence
y=
w0 sinh π2
,
EI sinh π
c2 = −
w0 cosh π2
.
EI sinh π
w0 sinh π2
w0 cosh π2
sin x sinh x −
(sin x cosh x − cos x sinh x)
2EI sinh π
4EI sinh π
w0 π
π
π
π +
sin x −
cosh x −
− cos x −
sinh x −
4EI
2
2
2
2
45. (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 .
250
CHAPTER 4 REVIEW EXERCISES
If we add the two equations, we get
Θ1 (s) + Θ2 (s) = (θ0 + ψ0 )
s
s2 + ω 2
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 + ψ0
s
s
s(s2 + ω 2 + K)θ0 + Ksψ0
θ0 − ψ0
=
,
+
2
2
2
2
2
2
2
(s + ω )(s + ω + 2K)
2
s +ω
2
s + ω 2 + 2K
so
θ0 + ψ0
θ0 − ψ0
cos ωt +
cos ω 2 + 2K t.
2
2
Then from θ2 (t) = −θ1 (t) + (θ0 + ψ0 ) cos ωt we get
θ0 + ψ0
θ0 − ψ0
θ2 (t) =
cos ωt −
cos ω 2 + 2K t.
2
2
θ1 (t) =
(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
θ1 (t) = θ0 cos ω 2 + 2K t,
θ2 (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.
251
5
Series Solutions of Linear
Differential Equations
EXERCISES 5.1
Solutions About Ordinary Points
n+1 n+1
2
x
/(n + 1) 2n
1. lim = lim
|x| = 2|x|
n
n
n→∞
n→∞
2 x /n
n+1
The series is absolutely convergent for 2|x| < 1 or |x| < 12 . The radius of convergence is R = 12 . At x = − 12 , the
∞
∞
series n=1 (−1)n /n converges by the alternating series test. At x = 12 , the series n=1 1/n is the harmonic
series which diverges. Thus, the given series converges on [− 12 , 12 ).
100n+1 (x + 7)n+1 /(n + 1)! = lim 100 |x + 7| = 0
2. lim n→∞ n + 1
n→∞
100n (x + 7)n /n!
The radius of convergence is R = ∞. The series is absolutely convergent on (−∞, ∞).
3. By the ratio test,
(x − 5)n+1 /10n+1 = lim 1 |x − 5| = 1 |x − 5|.
lim
n→∞ (x − 5)n /10n n→∞ 10
10
1
The series is absolutely convergent for 10
|x − 5| < 1, |x − 5| < 10, or on (−5, 15). The radius of convergence is
∞
∞
n
R = 10. At x = −5, the series n=1 (−1) (−10)n /10n = n=1 1 diverges by the nth term test. At x = 15, the
∞
∞
series n=1 (−1)n 10n /10n = n=1 (−1)n diverges by the nth term test. Thus, the series converges on (−5, 15).
(n + 1)!(x − 1)n+1 = lim (n + 1)|x − 1| = ∞, x = 1
4. lim n→∞
n→∞
n!(x − 1)n
0, x = 1
The radius of convergence is R = 0 and the series converges only for x = 1.
x3
x5
x7
x2
x4
x6
2x3
2x5
4x7
5. sin x cos x = x −
+
−
+ ···
1−
+
−
+ ··· = x −
+
−
+ ···
6
120 5040
2
24 720
3
15
315
x2
x3
x4
x2
x4
x3
x4
6. e−x cos x = 1 − x +
−
+
− ···
1−
+
− ··· = 1 − x +
−
+ ···
2
6
24
2
24
3
6
1
1
5x4
61x6
x2
=
+
+
+ ···
=1+
2
4
6
cos x
2
4!
6!
1 − x2 + x − x + · · ·
4!
6!
Since cos(π/2) = cos(−π/2) = 0, the series converges on (−π/2, π/2).
1−x
1 3
3
3
8.
= − x + x2 − x3 + · · ·
2+x
2 4
8
16
Since the function is undefined at x = −2, the series converges on (−2, 2).
7.
9. Let k = n + 2 so that n = k − 2 and
∞
n=1
ncn xn+2 =
∞
(k − 2)ck−2 xk .
k=3
252
5.1
Solutions About Ordinary Points
10. Let k = n − 3 so that n = k + 3 and
∞
(2n − 1)cn xn−3 =
n=3
11.
∞
2ncn xn−1 +
n=1
∞
∞
(2k + 5)ck+3 xk .
k=0
∞
6cn xn+1 = 2 · 1 · c1 x0 +
n=0
2ncn xn−1 +
n=2
∞
n=0
k=n−1
= 2c1 +
∞
= 2c1 +
k=n+1
∞
2(k + 1)ck+1 xk +
k=1
∞
6cn xn+1
6ck−1 xk
k=1
[2(k + 1)ck+1 + 6ck−1 ]xk
k=1
12.
∞
∞
n(n − 1)cn xn + 2
n=2
n=2
ncn xn
n=1
= 2 · 2 · 1c2 x + 2 · 3 · 2c3 x + 3 · 1 · c1 x +
0
∞
n(n − 1)cn xn−2 + 3
1
1
∞
n(n − 1)cn x +2
n
n=2
∞
n(n − 1)cn x
n−2
n=4
= 4c2 + (3c1 + 12c3 )x +
= 4c2 + (3c1 + 12c3 )x +
∞
k(k − 1)ck xk + 2
k=2
∞
k=2
∞
∞
n=2
k=n
= 4c2 + (3c1 + 12c3 )x +
+3
∞
k=n−2
(k + 2)(k + 1)ck+2 xk + 3
k=2
∞
ncn xn
k=n
kck xk
k=2
k(k − 1) + 3k ck + 2(k + 2)(k + 1)ck+2 xk
k(k + 2)ck + 2(k + 1)(k + 2)ck+2 xk
k=2
13. y =
∞
(−1)n+1 xn−1 ,
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
=
∞
∞
(−1)n+1 xn−1
n=1
(−1)n+1 (n − 1)xn−1 +
n=2
∞
(−1)n+1 (n − 1)xn−2 +
n=2
= −x0 + x0 +
∞
∞
(−1)n+1 (n − 1)xn−2 +
n=3
k=n−1
=
=
∞
k=1
∞
(−1)k+2 kxk +
∞
14. y =
k=n−2
(−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
∞
(−1)n 2n 2n−1
x
,
22n (n!)2
n=1
y =
(−1)n+1 xn−1
n=1
(−1)n+1 (n − 1)xn−1 +
n=2
∞
∞
(−1)n 2n(2n − 1) 2n−2
x
22n (n!)2
n=1
253
∞
n=2
(−1)n+1 xn−1
k=n−1
5.1
Solutions About Ordinary Points
xy + y + xy =
∞
∞
∞
(−1)n 2n(2n − 1) 2n−1 (−1)n 2n 2n−1 (−1)n 2n+1
x
+
x
+
x
22n (n!)2
22n (n!)2
22n (n!)2
n=1
n=1
n=0
k=n
∞ k=n
k=n+1
(−1) 2k(2k − 1) (−1) 2k
(−1)
x2k−1
+ 2k
+ 2k−2
2k
2
2
2 (k!)
2 (k!)
2
[(k − 1)!]2
k=1
∞ (−1)k (2k)2
(−1)k
x2k−1
=
− 2k−2
22k (k!)2
2
[(k − 1)!]2
k=1
∞
(2k)2 − 22 k 2 2k−1
x
=
(−1)k
=0
22k (k!)2
=
k
k
k−1
k=1
15. 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.
16. 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.
∞
17. Substituting y = n=0 cn xn into the differential equation we have
y − xy =
∞
n(n − 1)cn xn−2 −
n=2
∞
n=0
k=n−2
= 2c2 +
∞
cn xn+1 =
∞
(k + 2)(k + 1)ck+2 xk −
k=0
∞
ck−1 xk
k=1
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 =
1
ck−1 ,
(k + 2)(k + 1)
k = 1, 2, 3, . . . .
Choosing c0 = 1 and c1 = 0 we find
1
6
c4 = c5 = 0
1
c6 =
180
c3 =
and so on. For c0 = 0 and c1 = 1 we obtain
c3 = 0
1
c4 =
12
c5 = c6 = 0
1
c7 =
504
and so on. Thus, two solutions are
1
1 6
y1 = 1 + x3 +
x + ···
6
180
and
254
y2 = x +
1 4
1 7
x +
x + ··· .
12
504
5.1
18. Substituting y =
∞
n=0
2
y +x y =
Solutions About Ordinary Points
cn xn into the differential equation we have
∞
n(n − 1)cn x
n−2
n=2
+
∞
n+2
cn x
n=0
k=n−2
=
∞
k
(k + 2)(k + 1)ck+2 x +
k=0
∞
ck−2 xk
k=2
k=n+2
∞
= 2c2 + 6c3 x +
[(k + 2)(k + 1)ck+2 + ck−2 ]xk = 0.
k=2
Thus
c2 = 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
1
12
c5 = c6 = c7 = 0
1
c8 =
672
c4 = −
and so on. For c0 = 0 and c1 = 1 we obtain
c4 = 0
1
20
c6 = c7 = c8 = 0
1
c9 =
1440
c5 = −
and so on. Thus, two solutions are
y1 = 1 −
19. Substituting y =
∞
n=0
1 4
1 8
x +
x − ···
12
672
and
y2 = x −
1 5
1 9
x +
x − ··· .
20
1440
cn xn into the differential equation we have
y − 2xy + y =
∞
n=2
n(n − 1)cn xn−2 − 2
∞
n=1
k=n−2
=
∞
ncn xn +
k=0
= 2c2 + c0 +
k=n
(k + 2)(k + 1)ck+2 xk − 2
k=n
kck xk +
k=1
∞
cn xn
n=0
∞
∞
∞
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
255
5.1
Solutions About Ordinary Points
and
1
c2 = − c0
2
2k − 1
ck+2 =
ck ,
(k + 2)(k + 1)
k = 1, 2, 3, . . . .
Choosing c0 = 1 and c1 = 0 we find
1
2
c3 = c5 = c7 = · · · = 0
1
c4 = −
8
7
c6 = −
240
c2 = −
and so on. For c0 = 0 and c1 = 1 we obtain
c2 = c4 = c6 = · · · = 0
1
c3 =
6
1
c5 =
24
1
c7 =
112
and so on. Thus, two solutions are
1
1
7 6
y1 = 1 − x2 − x4 −
x − ···
2
8
240
20. Substituting y =
∞
n=0
1
1
1 7
y2 = x + x3 + x5 +
x + ··· .
6
24
112
and
cn xn into the differential equation we have
y − xy + 2y =
∞
n(n − 1)cn xn−2 −
n=2
∞
n=1
k=n−2
=
∞
ncn xn + 2
k=n
k=0
∞
k=n
kck xk + 2
k=1
= 2c2 + 2c0 +
∞
∞
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 =
cn xn
n=0
(k + 2)(k + 1)ck+2 xk −
∞
k−2
ck ,
(k + 2)(k + 1)
256
k = 1, 2, 3, . . . .
5.1
Solutions About Ordinary Points
Choosing c0 = 1 and c1 = 0 we find
c2 = −1
c3 = c5 = c7 = · · · = 0
c4 = 0
c6 = c8 = c10 = · · · = 0.
For c0 = 0 and c1 = 1 we obtain
c2 = c4 = c6 = · · · = 0
1
c3 = −
6
1
c5 = −
120
and so on. Thus, two solutions are
y1 = 1 − x2
21. Substituting y =
∞
n=0
and
1
1 5
y2 = x − x3 −
x − ··· .
6
120
cn xn into the differential equation we have
y + x2 y + xy =
∞
n(n − 1)cn xn−2 +
n=2
∞
n=1
k=n−2
=
∞
ncn xn+1 +
n=0
k=n+1
(k + 2)(k + 1)ck+2 xk +
k=0
∞
∞
cn xn+1
k=n+1
(k − 1)ck−1 xk +
k=2
= 2c2 + (6c3 + c0 )x +
∞
Thus
c2 = 0
6c3 + c0 = 0
(k + 2)(k + 1)ck+2 + kck−1 = 0
and
c2 = 0
1
c3 = − c0
6
k
ck−1 ,
(k + 2)(k + 1)
Choosing c0 = 1 and c1 = 0 we find
1
6
c4 = c5 = 0
1
c6 =
45
c3 = −
257
ck−1 xk
k=1
[(k + 2)(k + 1)ck+2 + kck−1 ]xk = 0.
k=2
ck+2 = −
∞
k = 2, 3, 4, . . . .
5.1
Solutions About Ordinary Points
and so on. For c0 = 0 and c1 = 1 we obtain
c3 = 0
1
6
c5 = c6 = 0
5
c7 =
252
c4 = −
and so on. Thus, two solutions are
1
1
y1 = 1 − x3 + x6 − · · ·
6
45
22. Substituting y =
∞
n=0
1
5 7
x − ··· .
y2 = x − x4 +
6
252
and
cn xn into the differential equation we have
y + 2xy + 2y =
∞
n(n − 1)cn xn−2 + 2
n=2
∞
n=1
k=n−2
=
∞
ncn xn + 2
k=n
(k + 2)(k + 1)ck+2 xk + 2
k=0
k=n
kck xk + 2
k=1
= 2c2 + 2c0 +
∞
Thus
2c2 + 2c0 = 0
(k + 2)(k + 1)ck+2 + 2(k + 1)ck = 0
and
c2 = −c0
2
ck ,
k+2
k = 1, 2, 3, . . . .
Choosing c0 = 1 and c1 = 0 we find
c2 = −1
c3 = c5 = c7 = · · · = 0
c4 =
1
2
c6 = −
∞
ck xk
k=0
[(k + 2)(k + 1)ck+2 + 2(k + 1)ck ]xk = 0.
k=1
ck+2 = −
cn xn
n=0
∞
∞
1
6
and so on. For c0 = 0 and c1 = 1 we obtain
c2 = c4 = c6 = · · · = 0
2
c3 = −
3
4
c5 =
15
8
c7 = −
105
258
5.1
Solutions About Ordinary Points
and so on. Thus, two solutions are
1
1
2
4
8 7
y1 = 1 − x2 + x4 − x6 + · · · and y2 = x − x3 + x5 −
x + ··· .
2
6
3
15
105
∞
23. Substituting y = n=0 cn xn into the differential equation we have
(x − 1)y + y =
∞
n(n − 1)cn xn−1 −
n=2
∞
n(n − 1)cn xn−2 +
n=2
=
∞
k=n−2
(k + 1)kck+1 xk −
k=1
∞
= −2c2 + c1 +
k=n−1
(k + 2)(k + 1)ck+2 xk +
k=0
∞
ncn xn−1
n=1
k=n−1
∞
∞
(k + 1)ck+1 xk
k=0
[(k + 1)kck+1 − (k + 2)(k + 1)ck+2 + (k + 1)ck+1 ]xk = 0.
k=1
Thus
−2c2 + c1 = 0
(k + 1)2 ck+1 − (k + 2)(k + 1)ck+2 = 0
and
1
c1
2
k+1
=
ck+1 ,
k+2
c2 =
ck+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
1
1
c2 = ,
c3 = ,
c4 = ,
2
3
4
and so on. Thus, two solutions are
1
1
1
y1 = 1
and
y2 = x + x2 + x3 + x4 + · · · .
2
3
4
∞
n
24. Substituting y = n=0 cn x into the differential equation we have
(x + 2)y + xy − y =
∞
n=2
n(n − 1)cn xn−1 +
∞
2n(n − 1)cn xn−2 +
n=2
k=n−1
=
∞
(k + 1)kck+1 xk +
k=1
= 4c2 − c0 +
n=1
k=n−2
∞
ncn xn −
∞
∞
k=n
kck xk −
k=1
Thus
4c2 − c0 = 0
(k + 1)kck+1 + 2(k + 2)(k + 1)ck+2 + (k − 1)ck = 0,
1
c0
4
(k + 1)kck+1 + (k − 1)ck
,
=−
2(k + 2)(k + 1)
k = 1, 2, 3, . . .
c2 =
ck+2
∞
ck xk
k=0
(k + 1)kck+1 + 2(k + 2)(k + 1)ck+2 + (k − 1)ck xk = 0.
k=1
and
cn xn
n=0
k=n
2(k + 2)(k + 1)ck+2 xk +
k=0
∞
∞
259
k = 1, 2, 3, . . . .
5.1
Solutions About Ordinary Points
Choosing c0 = 1 and c1 = 0 we find
c1 = 0,
c2 =
1
,
4
c3 = −
1
,
24
c4 = 0,
c5 =
1
480
and so on. For c0 = 0 and c1 = 1 we obtain
c2 = 0
c3 = 0
c4 = c5 = c6 = · · · = 0.
Thus, two solutions are
1
1
1 5
y1 = c0 1 + x2 − x3 +
x + ···
4
24
480
25. Substituting y =
∞
n=0
and
y2 = c1 x.
cn xn into the differential equation we have
y − (x + 1)y − y =
∞
n(n − 1)cn xn−2 −
n=2
n=1
k=n−2
=
∞
∞
ncn xn −
k=n
k=0
ncn xn−1 −
n=1
(k + 2)(k + 1)ck+2 xk −
∞
∞
k=1
= 2c2 − c1 − c0 +
∞
k=n−1
kck xk −
cn xn
n=0
∞
∞
k=n
(k + 1)ck+1 xk −
k=0
∞
ck xk
k=0
[(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, . . . .
Choosing c0 = 1 and c1 = 0 we find
c2 =
1
,
2
c3 =
1
,
6
c4 =
1
,
6
1
,
2
c3 =
1
,
2
c4 =
1
,
4
and so on. For c0 = 0 and c1 = 1 we obtain
c2 =
and so on. Thus, two solutions are
1
1
1
y1 = 1 + x2 + x3 + x4 + · · ·
2
6
6
and
260
1
1
1
y2 = x + x2 + x3 + x4 + · · · .
2
2
4
5.1
26. Substituting y =
∞
n=0
Solutions About Ordinary Points
cn xn into the differential equation we have
x2 + 1 y − 6y =
∞
n(n − 1)cn xn +
n=2
∞
n(n − 1)cn xn−2 − 6
n=2
=
∞
k=n−2
k(k − 1)ck xk +
k=2
∞
cn xn
n=0
k=n
∞
k=n
(k + 2)(k + 1)ck+2 xk − 6
k=0
∞
ck xk
k=0
= 2c2 − 6c0 + (6c3 − 6c1 )x +
∞
k 2 − 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
c3 = c1
ck+2 = −
k−3
ck ,
k+1
k = 2, 3, 4, . . . .
Choosing c0 = 1 and c1 = 0 we find
c2 = 3
c3 = c5 = c7 = · · · = 0
c4 = 1
c6 = −
1
5
and so on. For c0 = 0 and c1 = 1 we obtain
c2 = c4 = c6 = · · · = 0
c3 = 1
c5 = c7 = c9 = · · · = 0.
Thus, two solutions are
27. Substituting y =
∞
n=0
1
y1 = 1 + 3x2 + x4 − x6 + · · ·
5
y2 = x + x3 .
cn xn into the differential equation we have
x2 + 2 y + 3xy − y =
∞
n(n − 1)cn xn + 2
n=2
∞
n=2
k=n
=
and
∞
k=2
k(k − 1)ck xk + 2
∞
n(n − 1)cn xn−2 + 3
∞
n=1
k=n−2
(k + 2)(k + 1)ck+2 xk + 3
k=0
= (4c2 − c0 ) + (12c3 + 2c1 )x +
∞
kck xk −
∞
∞
cn xn
n=0
k=n
k=1
∞
ncn xn −
k=n
ck xk
k=0
2(k + 2)(k + 1)ck+2 + k 2 + 2k − 1 ck xk = 0.
k=2
261
5.1
Solutions About Ordinary Points
Thus
4c2 − c0 = 0
12c3 + 2c1 = 0
2(k + 2)(k + 1)ck+2 + k 2 + 2k − 1 ck = 0
and
1
c0
4
1
c3 = − c1
6
k 2 + 2k − 1
ck+2 = −
ck ,
2(k + 2)(k + 1)
c2 =
k = 2, 3, 4, . . . .
Choosing c0 = 1 and c1 = 0 we find
1
4
c3 = c5 = c7 = · · · = 0
7
c4 = −
96
c2 =
and so on. For c0 = 0 and c1 = 1 we obtain
c2 = c4 = c6 = · · · = 0
1
c3 = −
6
7
c5 =
120
and so on. Thus, two solutions are
1
7
y1 = 1 + x2 − x4 + · · ·
4
96
28. Substituting y =
∞
n=0
1
7 5
y2 = x − x3 +
x − ··· .
6
120
and
cn xn into the differential equation we have
x2 − 1 y + xy − y =
∞
n=2
n(n − 1)cn xn −
∞
n(n − 1)cn xn−2 +
n=2
k=n
=
∞
k=2
∞
n=1
k=n−2
k(k − 1)ck xk −
∞
k=n
(k + 2)(k + 1)ck+2 xk +
k=0
= (−2c2 − c0 ) − 6c3 x +
ncn xn −
∞
k=1
cn xn
n=0
k=n
kck xk −
∞
ck xk
k=0
−(k + 2)(k + 1)ck+2 + k 2 − 1 ck xk = 0.
k=2
Thus
−2c2 − c0 = 0
−6c3 = 0
−(k + 2)(k + 1)ck+2 + (k − 1)(k + 1)ck = 0
262
∞
∞
5.1
and
Solutions About Ordinary Points
1
c2 = − c0
2
c3 = 0
ck+2 =
k−1
ck ,
k+2
k = 2, 3, 4, . . . .
Choosing c0 = 1 and c1 = 0 we find
1
2
c3 = c5 = c7 = · · · = 0
1
c4 = −
8
c2 = −
and so on. For c0 = 0 and c1 = 1 we obtain
c2 = c4 = c6 = · · · = 0
c3 = c5 = c7 = · · · = 0.
Thus, two solutions are
29. Substituting y =
∞
n=0
1
1
y1 = 1 − x2 − x4 − · · ·
2
8
and
y2 = x.
cn xn into the differential equation we have
(x − 1)y − xy + y =
∞
n(n − 1)cn xn−1 −
n=2
∞
n(n − 1)cn xn−2 −
n=2
k=n−1
=
∞
(k + 1)kck+1 xk −
k=1
= −2c2 + c0 +
n=1
k=n−2
∞
(k + 2)(k + 1)ck+2 xk −
k=0
∞
∞
∞
kck xk +
k=1
cn xn
k=n
∞
ck xk
k=0
[−(k + 2)(k + 1)ck+2 + (k + 1)kck+1 − (k − 1)ck ]xk = 0.
Thus
−2c2 + c0 = 0
−(k + 2)(k + 1)ck+2 + (k − 1)kck+1 − (k − 1)ck = 0
1
c0
2
kck+1
(k − 1)ck
=
−
,
k+2
(k + 2)(k + 1)
c2 =
ck+2
∞
n=0
k=n
k=1
and
ncn xn +
k = 1, 2, 3, . . . .
Choosing c0 = 1 and c1 = 0 we find
1
1
,
c3 = ,
c4 = 0,
2
6
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
and
1
y = C1 x + x2 + · · · + C2 .
2
c2 =
263
5.1
Solutions About Ordinary Points
The initial conditions imply C1 = −2 and C2 = 6, so
1
1
y = −2 1 + x2 + x3 + · · · + 6x = 8x − 2ex .
2
6
∞
30. Substituting y =
n=0
cn xn into the differential equation we have
(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 + 1)kck+1 xk +
k=1
∞
(k + 2)(k + 1)ck+2 xk − 2
= 2c2 − 2c1 + c0 +
∞
∞
n=1
k=n−1
k=0
∞
ncn xn−1 +
ncn xn +
∞
n=0
k=n
(k + 1)ck+1 xk +
k=0
k=n
∞
kck xk +
k=1
[(k + 2)(k + 1)ck+2 − (k + 1)ck+1 + (k + 1)ck ]xk = 0.
k=1
Thus
2c2 − 2c1 + c0 = 0
(k + 2)(k + 1)ck+2 − (k + 1)ck+1 + (k + 1)ck = 0
and
1
c2 = c1 − c0
2
1
1
ck+1 −
ck ,
ck+2 =
k+2
k+2
k = 1, 2, 3, . . . .
Choosing c0 = 1 and c1 = 0 we find
1
c2 = − ,
2
1
c3 = − ,
6
c4 =
1
,
12
and so on. For c0 = 0 and c1 = 1 we obtain
c2 = 1,
c3 = 0,
1
c4 = − ,
4
and so on. Thus,
1
1
1
1
y = C1 1 − x2 − x3 + x4 + · · · + C2 x + x2 − x4 + · · ·
2
6
12
4
and
1
1
y = C1 −x − x2 + x3 + · · · + C2 1 + 2x − x3 + · · · .
2
3
The initial conditions imply C1 = 2 and C2 = −1, so
1 2 1 3
1 4
1 4
2
y = 2 1 − x − x + x + ··· − x + x − x + ···
2
6
12
4
1
5
= 2 − x − 2x2 − x3 + x4 + · · · .
3
12
264
cn xn
∞
k=0
ck xk
5.1
31. Substituting y =
∞
n=0
Solutions About Ordinary Points
cn xn into the differential equation we have
y − 2xy + 8y =
∞
n(n − 1)cn xn−2 − 2
n=2
∞
n=1
k=n−2
=
∞
ncn xn + 8
∞
n=0
k=n
(k + 2)(k + 1)ck+2 xk − 2
k=0
∞
k=n
kck xk + 8
k=1
= 2c2 + 8c0 +
∞
cn xn
∞
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
4
c4 =
3
c6 = c8 = c10 = · · · = 0.
For c0 = 0 and c1 = 1 we obtain
c2 = c4 = c6 = · · · = 0
c3 = −1
1
c5 =
10
and so on. Thus,
4
1
y = C1 1 − 4x2 + x4 + C2 x − x3 + x5 + · · ·
3
10
and
y = C1
16
−8x + x3
3
+ C2
1 4
1 − 3x + x + · · · .
2
2
The initial conditions imply C1 = 3 and C2 = 0, so
4
y = 3 1 − 4x2 + x4 = 3 − 12x2 + 4x4 .
3
265
5.1
Solutions About Ordinary Points
32. Substituting y =
∞
n=0
cn xn into the differential equation we have
(x2 + 1)y + 2xy =
∞
n=2
n(n − 1)cn xn +
∞
n(n − 1)cn xn−2 +
n=2
k=n
=
∞
∞
n=1
k=n−2
k(k − 1)ck xk +
k=2
∞
2ncn xn
k=n
(k + 2)(k + 1)ck+2 xk +
k=0
∞
∞
2kck xk
k=1
k(k + 1)ck + (k + 2)(k + 1)ck+2 xk = 0.
= 2c2 + (6c3 + 2c1 )x +
k=2
Thus
2c2 = 0
6c3 + 2c1 = 0
k(k + 1)ck + (k + 2)(k + 1)ck+2 = 0
and
c2 = 0
1
c3 = − c1
3
k
ck+2 = −
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
1
3
c4 = c6 = c8 = · · · = 0
1
c5 = −
5
1
c7 =
7
c3 = −
and so on. Thus
y = C0 + C1
1 3 1 5 1 7
x − x + x − x + ···
3
5
7
and
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
cn xn into the differential equation we have
∞
1
1 5
y + (sin x)y =
n(n − 1)cn xn−2 + x − x3 +
x − · · · c0 + c1 x + c2 x2 + · · ·
6
120
n=2
1
2
3
2
3
= 2c2 + 6c3 x + 12c4 x + 20c5 x + · · · + c0 x + c1 x + c2 − c0 x + · · ·
6
1
= 2c2 + (6c3 + c0 )x + (12c4 + c1 )x2 + 20c5 + c2 − c0 x3 + · · · = 0.
6
33. Substituting y =
n=0
266
5.1
Solutions About Ordinary Points
Thus
2c2 = 0
6c3 + c0 = 0
12c4 + c1 = 0
1
20c5 + c2 − c0 = 0
6
c2 = 0
and
1
c3 = − c0
6
1
c4 = − c1
12
1
1
c0 .
c5 = − c2 +
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
34. Substituting y =
1
1 5
y1 = 1 − x3 +
x + ···
6
120
∞
n=0
and
y2 = x −
1 4
x + ··· .
12
cn xn into the differential equation we have
y + ex y − y =
∞
n(n − 1)cn xn−2
n=2
∞
1
1
+ 1 + x + x2 + x3 + · · · c1 + 2c2 x + 3c3 x2 + 4c4 x3 + · · · −
cn xn
2
6
n=0
2
3
= 2c2 + 6c3 x + 12c4 x + 20c5 x + · · ·
1
2
+ c1 + (2c2 + c1 )x + 3c3 + 2c2 + c1 x + · · · − [c0 + c1 x + c2 x2 + · · ·]
2
1
= (2c2 + c1 − c0 ) + (6c3 + 2c2 )x + 12c4 + 3c3 + c2 + c1 x2 + · · · = 0.
2
Thus
2c2 + c1 − c0 = 0
6c3 + 2c2 = 0
1
12c4 + 3c3 + c2 + c1 = 0
2
267
5.1
Solutions About Ordinary Points
and
1
1
c0 − c1
2
2
1
c3 = − c2
3
1
1
1
c4 = − c3 + c2 − c1 .
4
12
24
c2 =
Choosing c0 = 1 and c1 = 0 we find
c2 =
1
,
2
1
c3 = − ,
6
c4 = 0
and so on. For c0 = 0 and c1 = 1 we obtain
1
c2 = − ,
2
c3 =
1
,
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
35. 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.
36. Substituting y =
∞
n=0
y − xy =
cn xn into the first differential equation leads to
∞
n(n − 1)cn xn−2 −
n=2
n=0
k=n−2
= 2c2 +
∞
∞
cn xn+1 =
∞
(k + 2)(k + 1)ck+2 xk −
k=0
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 =
ck+2 =
1
2
ck−1
,
(k + 2)(k + 1)
k = 1, 2, 3, . . . .
Let c0 and c1 be arbitrary and iterate to find
1
2
1
c3 = c0
6
1
c4 =
c1
12
1
1
c5 =
c2 =
20
40
c2 =
268
∞
k=1
ck−1 xk
5.1
Solutions About Ordinary Points
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 =
∞
n=0
cn xn into the second differential equation leads to
y − 4xy − 4y =
∞
n(n − 1)cn xn−2 −
n=2
∞
n=1
k=n−2
=
∞
4ncn xn −
∞
n=0
k=n
(k + 2)(k + 1)ck+2 xk −
k=0
∞
= 2c2 − 4c0 +
k=n
4kck xk −
k=1
∞
4cn xn
∞
k=0
(k + 2)(k + 1)ck+2 − 4(k + 1)ck xk
k=1
∞
1 k
x
=e =1+
x .
k!
k=1
Thus
2c2 − 4c0 = 1
(k + 2)(k + 1)ck+2 − 4(k + 1)ck =
1
k!
and
1
+ 2c0
2
1
4
=
+
ck ,
(k + 2)! k + 2
c2 =
ck+2
4ck xk
k = 1, 2, 3, . . . .
Let c0 and c1 be arbitrary and iterate to find
1
+ 2c0
2
4
4
1
1
+ c1 =
+ c1
c3 =
3! 3
3! 3
c2 =
c4 =
1
1
13
4
1
+ 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 =
1
1
261 4
4
4 · 13 8
+ c4 =
+
+ c0 =
+ c0
6! 6
6!
6 · 4!
6
6!
3
c7 =
4
4 · 17
64
64
1
1
409
+ c5 =
+
+
c1 =
+
c1
7! 7
7!
7 · 5!
105
7!
105
and so on. The solution is
269
5.1
Solutions About Ordinary Points
1
1
13
17 16
4
2
3
4
y = c0 + c1 x +
+ 2c0 x +
+ c1 x +
+ 2c0 x +
+ c1 x5
2
3! 3
4!
5!
15
64
261 4
409
+ c0 x6 +
+
c1 x7 + · · ·
+
6!
3
7!
105
4 3 16 5
4 6
64 7
2
4
= c0 1 + 2x + 2x + x + · · · + c1 x + x + x +
x + ···
3
3
15
105
1
1
13
17
261 6 409 7
+ x2 + x3 + x4 + x5 +
x +
x + ··· .
2
3!
4!
5!
6!
7!
37. 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.
38. If x > 0 and y > 0, then y = −xy < 0 and the graph of a solution curve is concave down. Thus, whatever
portion of a solution curve lies in the first quadrant is concave down. When x > 0 and y < 0, y = −xy > 0,
so whatever portion of a solution curve lies in the fourth quadrant is concave up.
∞
39. (a) Substituting y = n=0 cn xn into the differential equation we have
y + xy + y =
∞
n(n − 1)cn xn−2 +
n=2
∞
n=1
k=n−2
=
∞
ncn xn +
k=n
k=0
∞
k=n
kck xk +
k=1
= (2c2 + c0 ) +
∞
cn xn
n=0
(k + 2)(k + 1)ck+2 xk +
∞
∞
ck xk
k=0
(k + 2)(k + 1)ck+2 + (k + 1)ck xk = 0.
k=1
Thus
2c2 + c0 = 0
(k + 2)(k + 1)ck+2 + (k + 1)ck = 0
and
1
c2 = − c0
2
1
ck+2 = −
ck ,
k+2
k = 1, 2, 3, . . . .
Choosing c0 = 1 and c1 = 0 we find
1
2
c3 = c5 = c7 = · · · = 0
1
1 1
c4 = − −
= 2
4
2
2 ·2
1 1 1
c6 = −
=− 3
6 22 · 2
2 · 3!
c2 = −
and so on. For c0 = 0 and c1 = 1 we obtain
270
5.1
Solutions About Ordinary Points
c2 = c4 = c6 = · · · = 0
1
2
c3 = − = −
3
3!
1 1
1
4·2
c5 = − −
=
=
5
3
5·3
5!
6·4·2
14 · 2
c7 = −
=−
7 5!
7!
and so on. Thus, two solutions are
y1 =
∞
(−1)k
k=0
2k · k!
x2k
and
y2 =
∞
(−1)k 2k k!
k=0
(2k + 1)!
x2k+1 .
(b) For y1 , S3 = S2 and S5 = S4 , so we plot S2 , S4 , S6 , S8 , and S10 .
y
y
4
2
-4 -2
-2
y
4
N=2
4
2
x
2 4
-4 -2
-2
-4
2
2 4
x
-4 -2
-2
N=4
-4
N=6
2 4
x
y
y
4
4
2
2
-4 -2
-2
-4
2 4
x
N=8
-4 -2
-2
-4
-4
y
y
N=10
2 4
x
2 4
x
For y2 , S3 = S4 and S5 = S6 , so we plot S2 , S4 , S6 , S8 , and S10 .
y
y
4
4
2
-4 -2
-2
2
x
2 4
N=2
4
N=4
-4 -2
-2
-4
2 4
-4 -2
-2
y2
4
4
2
2
2
4
2
2 4
x
N=6
-4 -2
-2
-4
y1
-2
4
2
x
-4
(c)
-4
y
x
-4
-2
-4
2
-2
-2
-4
-4
4
4
N=8
2 4
2
x
-4 -2
-2
N=10
-4
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].
∞ k
∞
∞
−x2 /2
2
k
k 2k
k
(d) From ex =
=
k=0 x /k! we see that e
k=0 (−x /2) /k! =
k=0 (−1) x /2 k! . From (5) of
Section 3.2 we have
271
5.1
Solutions About Ordinary Points
−x2 /2
2
2
e−x /2
e
−x2 /2
−x2 /2
ex /2 dx
dx
=
e
dx
=
e
y12
e−x2
(e−x2 /2 )2
∞ ∞
∞
∞
(−1)k
(−1)k 2k 1
1
2k
2k
2k
=
x
x dx =
x
x dx
2k k!
2k k!
2k k!
2k k!
k=0
k=0
k=0
k=0
∞
∞
(−1)k
1
2k
2k+1
=
x
x
2k k!
(2k + 1)2k k!
y2 = y1
e−
x dx
dx = e−x
k=0
2
/2
k=0
1
1
1
1 3
1
1
= 1 − x2 + 2 x4 − 3
x6 + · · · x +
x +
x5 +
x7 + · · ·
2
3
2
2 ·2
2 · 3!
3·2
5·2 ·2
7 · 2 · 3!
∞
k
k
(−1) 2 k!
2
4·2 5 6·4·2 7
= x − x3 +
x −
x + ··· =
x2k+1 .
3!
5!
7!
(2k + 1)!
k=0
40. (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 = (2c2 + c0 ) + (6c3 + c1 )x + 12c4 + c2 − c0 x2 + 20c5 + c3 − c1 x3
2
2
1
1 4 1
1 5
+ 30c6 + c4 + c0 − c2 x + 42c7 + c5 + c1 − c3 x + · · · .
24
2
24
2
Then
30c6 + c4 +
1
1
c0 − c2 = 0
24
2
and
42c7 + c5 +
1
1
c1 − c3 = 0,
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
y2 (x) = x − x3 + x5 −
x + ··· .
6
30
5040
(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
y = y1 + y2 = 1 + x − x2 − x3 + x4 + x5 − x6 −
x + ··· .
2
6
12
30
80
5040
272
5.2
(c)
y
y
y
4
4
4
2
2
2
-6 -4 -2
2
4
6
x
-6 -4 -2
2
4
6
x
-6 -4 -2
-2
-2
-2
-4
-4
-4
y
y
4
4
2
2
2
2
4
6
x
-6 -4 -2
2
4
6
x
2
4
6
x
y
4
-6 -4 -2
(d)
Solutions About Singular Points
2
4
6
x
-6 -4 -2
-2
-2
-2
-4
-4
-4
y
6
4
2
-6 -4 -2
-2
2
4
6
x
-4
-6
EXERCISES 5.2
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
273
5.2
Solutions About Singular Points
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(x − 1)2
(x − 1)2 y + 5(x − 1)y +
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 .
13. We identify P (x) = 5/3x + 1 and Q(x) = −1/3x2 , so that p(x) = xP (x) =
Then a0 = 53 , b0 = − 13 , and the indicial equation is
5
3
+ x and q(x) = x2 Q(x) = − 13 .
5
2
1
1
1
1
r(r − 1) + r − = r2 + r − = (3r2 + 2r − 1) = (3r − 1)(r + 1) = 0.
3
3
3
3
3
3
and −1. Since these do not differ by an integer we expect to find two series solutions
using the method of Frobenius.
The indicial roots are
1
3
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 = n=0 cn xn+r into the differential equation and collecting terms, we obtain
2xy − y + 2y = 2r2 − 3r c0 xr−1 +
∞
[2(k + r − 1)(k + r)ck − (k + r)ck + 2ck−1 ]xk+r−1 = 0,
k=1
which implies
2r2 − 3r = r(2r − 3) = 0
and
(k + r)(2k + 2r − 3)ck + 2ck−1 = 0.
274
5.2
Solutions About Singular Points
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 ,
c3 =
4
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
2
c1 = − c0 ,
c2 =
c3
c0 ,
5
35
and so on. The general solution on (0, ∞) is
4 3
2
3/2
y = C1 1 + 2x − 2x + x + · · · + C2 x
1−
9
16. Substituting y =
∞
n=0
=−
4
c0 ,
945
2
4 3
2 2
x+ x −
x + ··· .
5
35
945
cn xn+r into the differential equation and collecting terms, we obtain
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,
2r2 + 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 = −
ck−2
,
(2k − 3)k
k = 2, 3, 4, . . . ,
and
1
c2 = − c0 ,
2
and so on. For r = 0 the recurrence relation is
ck = −
c3 = 0,
ck−2
,
k(2k + 3)
c4 =
1
c0 ,
40
k = 2, 3, 4, . . . ,
and
1
1
c0 ,
c0 ,
c3 = 0,
c4 =
14
616
and so on. The general solution on (0, ∞) is
1
1 4
1
1
y = C1 x−3/2 1 − x2 + x4 + · · · + C2 1 − x2 +
x + ··· .
2
40
14
616
c2 = −
275
5.2
Solutions About Singular Points
∞
cn xn+r into the differential equation and collecting terms, we obtain
∞ 1 7
1
2
r−1
4(k + r)(k + r − 1)ck + (k + r)ck + ck−1 xk+r−1
4xy + y + y = 4r − r c0 x
+
2
2
2
17. Substituting y =
n=0
k=1
= 0,
which implies
7
7
4r − r = r 4r −
=0
2
2
2
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)
and
c1 = −2c0 ,
c2 =
k = 1, 2, 3, . . . ,
2
c0 ,
9
c3 = −
4
c0 ,
459
and so on. For r = 7/8 the recurrence relation is
ck = −
and
c1 = −
2ck−1
,
(8k + 7)k
2
c0 ,
15
c2 =
k = 1, 2, 3, . . . ,
2
c0 ,
345
c3 = −
4
c0 ,
32,085
and so on. The general solution on (0, ∞) is
2
2
4 3
4
2 2
y = C1 1 − 2x + x2 −
x + · · · + C2 x7/8 1 − x +
x −
x3 + · · · .
9
459
15
345
32,085
∞
18. Substituting y = n=0 cn xn+r into the differential equation and collecting terms, we obtain
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,
2r2 + 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 = −
and
ck−2
,
k(2k − 1)
1
c2 = − c0 ,
6
k = 2, 3, 4, . . . ,
c3 = 0,
276
c4 =
1
c0 ,
168
5.2
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
1
1
c0 ,
c0 ,
c3 = 0,
c4 =
10
360
and so on. The general solution on (0, ∞) is
1 2
1 4
1 2
1 4
1/2
y = C1 x
1− x +
x + · · · + C2 x 1 − x +
x + ··· .
6
168
10
360
∞
19. Substituting y = n=0 cn xn+r into the differential equation and collecting terms, we obtain
c2 = −
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−1
ck =
, k = 1, 2, 3, . . . ,
3k − 1
and
1
1
1
c2 =
c3 =
c0 ,
c0 ,
c0 ,
2
10
80
and so on. For r = 1/3 the recurrence relation is
ck−1
ck =
, k = 1, 2, 3, . . . ,
3k
and
1
1
1
c1 = c0 ,
c2 =
c3 =
c0 ,
c0 ,
3
18
162
and so on. The general solution on (0, ∞) is
1
1
1
1 3
1
1
y = C1 1 + x + x2 + x3 + · · · + C2 x1/3 1 + x + x2 +
x + ··· .
2
10
80
3
18
162
∞
20. Substituting y = n=0 cn xn+r into the differential equation and collecting terms, we obtain
∞ 2
2
2
2 2
r
(k + r)(k + r − 1)ck + ck − ck−1 xk+r
x y − x−
y = r −r+
c0 x +
9
9
9
c1 =
k=1
= 0,
which implies
2
r −r+ =
9
2
and
2
r−
3
1
r−
3
=0
2
(k + r)(k + r − 1) +
ck − ck−1 = 0.
9
277
5.2
Solutions About Singular Points
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
3
c0 ,
4
c2 =
k = 1, 2, 3, . . . ,
and
c1 =
9
c0 ,
56
c3 =
9
c0 ,
560
and so on. For r = 1/3 the recurrence relation is
ck =
3ck−1
,
3k 2 − k
3
c0 ,
2
c2 =
k = 1, 2, 3, . . . ,
and
c1 =
9
c0 ,
20
c3 =
9
c0 ,
160
and so on. The general solution on (0, ∞) is
3
3
9 3
9 3
9
9
y = C1 x2/3 1 + x + x2 +
x + · · · + C2 x1/3 1 + x + x2 +
x + ··· .
4
56
560
2
20
160
21. Substituting y =
∞
n=0
cn xn+r into the differential equation and collecting terms, we obtain
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
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
c1 =
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
c1 =
4
c0 ,
7
c2 =
4
c0 ,
21
c3 =
32
c0 ,
693
and so on. The general solution on (0, ∞) is
1
32 3
1
4
1
4
y = C1 1 + x − x2 − x3 + · · · + C2 x5/2 1 + x + x2 +
x + ··· .
3
6
6
7
21
693
278
5.2
Solutions About Singular Points
∞
cn xn+r into the differential equation and collecting terms, we obtain
4
4
5
x2 y + xy + x2 −
y = r2 −
c0 xr + r2 + 2r +
c1 xr+1
9
9
9
∞ 4
(k + r)(k + r − 1)ck + (k + r)ck − ck + ck−2 xk+r
+
9
22. Substituting y =
n=0
k=2
= 0,
which implies
r2 −
and
4
2
2
= r+
r−
= 0,
9
3
3
5
r2 + 2r +
c1 = 0,
9
4
(k + r)2 −
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
ck = −
9ck−2
,
3k(3k − 4)
k = 2, 3, 4, . . . ,
and
3
c2 = − c0 ,
4
and so on. For r = 2/3 the recurrence relation is
ck = −
and
c2 = −
c3 = 0,
9ck−2
,
3k(3k + 4)
3
c0 ,
20
c3 = 0,
c4 =
9
c0 ,
128
k = 2, 3, 4, . . . ,
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
y = C1 x
1− x +
1− x +
x + · · · + C2 x
x + ··· .
4
128
20
1,280
∞
23. Substituting y = n=0 cn xn+r into the differential equation and collecting terms, we obtain
9x2 y + 9x2 y + 2y = 9r2 − 9r + 2 c0 xr +
∞
[9(k + r)(k + r − 1)ck + 2ck + 9(k + r − 1)ck−1 ]xk+r = 0,
k=1
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 = −
and
(3k − 2)ck−1
,
k(3k − 1)
1
c1 = − c0 ,
2
c2 =
1
c0 ,
5
279
k = 1, 2, 3, . . . ,
c3 = −
7
c0 ,
120
5.2
Solutions About Singular Points
and so on. For r = 2/3 the recurrence relation is
ck = −
(3k − 1)ck−1
,
k(3k + 1)
k = 1, 2, 3, . . . ,
and
1
5
1
c1 = − c0 ,
c2 =
c3 = − c0 ,
c0 ,
2
28
21
and so on. The general solution on (0, ∞) is
1
1
7 3
1 3
1 2
5 2
1/3
2/3
y = C1 x
1− x+ x −
1 − x + x − x + ··· .
x + · · · + C2 x
2
5
120
2
28
21
∞
24. Substituting y = n=0 cn xn+r into the differential equation and collecting terms, we obtain
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 = 0,
k=1
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 ,
c3 =
4
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
2
4
c1 = − c0 ,
c2 =
c3 = −
c0 ,
c0 ,
5
35
945
and so on. The general solution on (0, ∞) is
2
2
4
4 3
y = C1 x−1 1 + 2x − 2x2 + x3 + · · · + C2 x1/2 1 − x + x2 −
x + ··· .
9
5
35
945
∞
25. Substituting y = n=0 cn xn+r into the differential equation and collecting terms, we obtain
xy + 2y − xy = r + r c0 x
2
r−1
2
r
+ r + 3r + 2 c1 x +
∞
[(k + r)(k + r − 1)ck + 2(k + r)ck − ck−2 ]xk+r−1 = 0,
k=2
which implies
r2 + r = r(r + 1) = 0,
r2 + 3r + 2 c1 = 0,
and
(k + r)(k + r + 1)ck − ck−2 = 0.
280
5.2
Solutions About Singular Points
The indicial roots are r1 = 0 and r2 = −1, so c1 = 0. For r1 = 0 the recurrence relation is
ck−2
ck =
, k = 2, 3, 4, . . . ,
k(k + 1)
and
1
c0
3!
c3 = c5 = c7 = · · · = 0
1
c4 = c0
5!
1
c2n =
c0 .
(2n + 1)!
c2 =
For r2 = −1 the recurrence relation is
ck =
and
ck−2
,
k(k − 1)
k = 2, 3, 4, . . . ,
1
c0
2!
c3 = c5 = c7 = · · · = 0
1
c4 = c0
4!
1
c0 .
c2n =
(2n)!
c2 =
The general solution on (0, ∞) is
y = C1
∞
1
1
x2n + C2 x−1
x2n
(2n
+
1)!
(2n)!
n=0
n=0
∞
∞
∞
1
1
1
2n+1
2n
=
+ C2
C1
x
x
x
(2n + 1)!
(2n)!
n=0
n=0
=
∞
1
[C1 sinh x + C2 cosh x].
x
cn xn+r into the differential equation and collecting terms, we obtain
1
1
3
2 2
2
r
2
x y + xy + x −
y= r −
c0 x + r + 2r +
c1 xr+1
4
4
4
∞ 1
(k + r)(k + r − 1)ck + (k + r)ck − ck + ck−2 xk+r
+
4
26. Substituting y =
n=0
k=2
= 0,
which implies
r2 −
and
1
1
1
= r−
r+
= 0,
4
2
2
3
r2 + 2r +
c1 = 0,
4
1
(k + r)2 −
ck + ck−2 = 0.
4
281
5.2
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 = −
ck−2
,
k(k + 1)
k = 2, 3, 4, . . . ,
and
1
c0
3!
c3 = c5 = c7 = · · · = 0
1
c4 = c0
5!
(−1)n
c2n =
c0 .
(2n + 1)!
c2 = −
For r2 = −1/2 the recurrence relation is
ck = −
ck−2
,
k(k − 1)
k = 2, 3, 4, . . . ,
and
1
c0
2!
c3 = c5 = c7 = · · · = 0
1
c4 = c0
4!
(−1)n
c0 .
c2n =
(2n)!
c2 = −
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
n=0
= C1 x−1/2
∞
∞
(−1)n 2n+1
(−1)n 2n
+ C2 x−1/2
x
x
(2n + 1)!
(2n)!
n=0
n=0
= x−1/2 [C1 sin x + C2 cos x].
27. Substituting y =
∞
n=0
cn xn+r into the differential equation and collecting terms, we obtain
xy − xy + y = r2 − r c0 xr−1 +
∞
[(k + r + 1)(k + r)ck+1 − (k + r)ck + ck ]xk+r = 0
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
ck+1 =
kck
,
(k + 2)(k + 1)
282
k = 0, 1, 2, . . . ,
5.2
Solutions About Singular Points
and one solution is y1 = c0 x. A second solution is
− −1 dx
x
1 2
e
e
1
1 3
y2 = x
1 + x + x + x + · · · dx
dx = x
dx = x
x2
x2
x2
2
3!
1
1
1 2
1
1 2
1 1
1
1 3
=x
+ + + x + x + · · · dx = x − + ln x + x + x + x + · · ·
x2
x 2 3!
4!
x
2
12
72
1
1
1
= x ln x − 1 + x2 + x3 + x4 + · · · .
2
12
72
The general solution on (0, ∞) is
y = C1 x + C2 y2 (x).
∞
28. Substituting y = n=0 cn xn+r into the differential equation and collecting terms, we obtain
3
y + 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
2ck−2
ck =
, k = 2, 3, 4, . . . ,
k(k + 2)
and
1
c2 = c0
4
c3 = c5 = c7 = · · · = 0
1
c4 =
c0
48
1
c0 .
c6 =
1,152
The result is
1 2
1 4
1
y1 = c0 1 + x + x +
x6 + · · · .
4
48
1,152
A second solution is
− (3/x)dx
e
dx
y2 = y 1
dx
=
y
1
2
1 2
1 4
3
y12
x 1 + 4 x + 48
x + ···
dx
1
1 2
7 4
19 6
x
x
x
1
−
=
y
= y1
+
+
+
·
·
·
dx
1
5 4
7
x3
2
48
576
x3 1 + 12 x2 + 48
x + 576
x6 + · · ·
1
1
19 4
1
1
7
19 3
7 2
−
+
·
·
·
dx
=
y
−
−
+
·
·
·
= y1
−
+
x
−
x
ln
x
+
x
x
1
x3
2x 48
576
2x2
2
96
2,304
1
1
7
19 4
= − y1 ln x + y − 2 + x2 −
x + ··· .
2
2x
96
2,304
283
5.2
Solutions About Singular Points
The general solution on (0, ∞) is
y = C1 y1 (x) + C2 y2 (x).
29. Substituting y =
∞
n=0
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 =
One solution is
y1 = c0
A second solution is
ck−1
,
k
k = 1, 2, 3, . . . .
1 2
1 3
1 + x + x + x + · · · = c0 ex .
2
3!
ex /x
1 −x
x
e dx
dx
=
e
2x
2x
e
e
x
1
1
1 3
1 2
1
1 2
x
x
=e
1 − x + x − x + · · · dx = e
− 1 + x − x + · · · dx
x
2
3!
x
2
3!
∞
1 2
1 3
(−1)n+1 n
= ex ln x − x +
x −
x + · · · = ex ln x − ex
x .
2·2
3 · 3!
n · n!
n=1
y2 = y 1
e−
(1/x−1)dx
dx = ex
The general solution on (0, ∞) is
y = C1 ex + C2 ex
30. Substituting y =
∞
n=0
∞
(−1)n+1 n
ln x −
x
n · n!
n=1
.
cn xn+r into the differential equation and collecting terms, we obtain
xy + y + y = r2 c0 xr−1 +
∞
[(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 = −
One solution is
ck−1
,
k2
k = 1, 2, 3, . . . .
∞
1
1 3
1 4
(−1)n n
y1 = c0 1 − x + 2 x2 −
x
+
x
−
·
·
·
=
c
x .
0
2
2
2
(3!)
(4!)
(n!)2
n=0
284
5.2
A second solution is
y2 = y 1
e−
(1/x)dx
y12
dx = y1
Solutions About Singular Points
dx
x 1−x+
1 2
4x
−
1 3
36 x
+ ···
2
dx
35 4
x 1 − 2x + 32 x2 − 59 x3 + 288
x − ···
1
5 2 23 3 677 4
= y1
1 + 2x + x + x +
x + · · · dx
x
2
9
288
1
677 3
5
23
= y1
+ 2 + x + x2 +
x + · · · dx
x
2
9
288
5
23
677 4
= y1 ln x + 2x + x2 + x3 +
x + ···
4
27
1,152
5
23
677 4
= y1 ln x + y1 2x + x2 + x3 +
x + ··· .
4
27
1,152
= y1
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 = (r2 − 7r)c0 xr−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,
k = 1, 2, 3, . . . ,
or
ck = −
k+3
ck−1 ,
k(k + 7)
k = 1, 2, 3, . . . .
Taking c0 = 0 we obtain
1
c1 = − c0
2
5
c0
c2 =
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,
285
k = 1, 2, 3, . . . .
5.2
Solutions About Singular Points
Then
−6c1 − 3c0 = 0
−10c2 − 2c1 = 0
−12c3 − c2 = 0
−12c4 + 0c3 = 0 =⇒ c4 = 0
−10c5 + c4 = 0 =⇒ c5 = 0
−6c6 + 2c5 = 0 =⇒ c6 = 0
0c7 + 3c6 = 0 =⇒ c7 is arbitrary
and
ck = −
Taking c0 = 0 and c7 = 0 we obtain
k−4
ck−1 ,
k(k − 7)
k = 8, 9, 10, . . . .
1
c1 = − c0
2
1
c2 =
c0
10
1
c3 = −
c0
120
c4 = c5 = c6 = · · · = 0.
Taking c0 = 0 and c7 = 0 we obtain
c1 = c2 = c3 = c4 = c5 = c6 = 0
1
c8 = − c7
2
5
c9 =
c7
36
1
c10 = − c7 ,
36
and so on. In this case we obtain the two solutions
1
1
1 3
1
5
1
y1 = 1 − x + x2 −
x
and
y2 = x7 − x8 + x9 − x10 + · · · .
2
10
120
2
36
36
∞
32. Substituting y = n=0 cn xn+r 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 + 3(k + r)ck − 2ck−1 ]xk+r−1
k=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
286
5.2
Solutions About Singular Points
or
k+1
ck−1 ,
k
ck =
k = 1, 2, 3, . . . .
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, . . . ,
−(k − 4)ck + (k − 3)ck−1 = 0,
k = 1, 2, 3, . . . .
or
Then
3c1 − 2c0 = 0
2c2 − c1 = 0
c3 + 0c2 = 0 ⇒ c3 = 0
0c4 + c3 = 0 ⇒ c4 is arbitrary
and
ck =
(k − 3)ck−1
,
k−4
k = 5, 6, 7, . . . .
Taking c0 = 0 and c4 = 0 we obtain
2
c0
3
1
c2 = c0
3
c3 = c4 = c5 = · · · = 0.
c1 =
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
2
1
y1 = 1 + x + x2
3
3
y2 = x4 + 2x5 + 3x6 + 4x7 + · · · .
and
33. (a) From t = 1/x we have dt/dx = −1/x2 = −t2 . Then
dy
dy
dy dt
=
= −t2
dx
dt dx
dt
and
d2 y
d
=
2
dx
dx
dy
dx
=
d
dx
−t2
Now
x4
d2 y
1
+ λy = 4
dx2
t
dy
dt
t4
= −t2
dy
d2 y dt
−
2
dt dx
dt
d2 y
dy
+ 2t3
dt2
dt
287
+ λy =
2t
dt
dx
= t4
d2 y
dy
+ 2t3
.
2
dt
dt
d2 y 2 dy
+
+ λy = 0
dt2
t dt
5.2
Solutions About Singular Points
becomes
t
(b) Substituting y =
t
∞
n=0 cn t
n+r
d2 y
dy
+2
+ λty = 0.
2
dt
dt
into the differential equation and collecting terms, we obtain
d2 y
dy
+2
+ λty = (r2 + r)c0 tr−1 + (r2 + 3r + 2)c1 tr
2
dt
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,
r2 + 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
ck = −
λck−2
,
k(k + 1)
k = 2, 3, 4, . . . ,
and
λ
c0
3!
c3 = c5 = c7 = · · · = 0
c2 = −
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
c0
2!
c3 = c5 = c7 = · · · = 0
c2 = −
c4 =
λ2
c0
4!
..
.
c2n = (−1)n
λn
c0 .
(2n)!
288
5.2
Solutions About Singular Points
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)n √
(−1)n √
1
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
y(x) = C1 x sin
+ C2 x cos
.
x
x
34. (a) From the boundary conditions y(a) = 0, y(b) = 0 we find
√
√
λ
λ
C1 sin
+ C2 cos
=0
a
a
√
√
λ
λ
C1 sin
+ C2 cos
= 0.
b
b
Since this is a homogeneous system of linear equations, it will have nontrivial solutions for C1 and C2 if
√
λ
sin
a
√
λ
sin
b
√ λ
√
√
√
√
cos
λ
λ
λ
λ
a cos
−
cos
sin
=
sin
√ a
b
a
b
λ
cos
b
√
√ √ b−a
λ
λ
= sin
λ
−
= sin
= 0.
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 b 4
=
.
2
L
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
λ
sin( λ/a)
√
y = C1 x sin
cos
−
x
x
cos( λ/a)
√
√
√
√ λ
λ
λ
λ
= C3 x sin
cos
− cos
sin
x
a
x
a
√
1
1
= C3 x sin λ
−
,
x a
and
yn (x) = C3 x sin
nπab
L
1
1
−
x a
= C3 x sin
nπab a
nπab a
− 1 = C4 x sin
1−
.
La x
L
x
289
5.2
Solutions About Singular Points
(b) When n = 1, b = 11, and a = 1, we have,
for C4 = 1,
1
y1 (x) = x sin 1.1π 1 −
x
y
2
.
1
1
3
5
7
9
11
x
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 =
∞
n+r
n=0 cn x
into the first differential equation and collecting terms, we obtain
x3 y + y = c0 xr +
∞
[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 =
∞
n+r
n=0 cn x
into the second differential equation and collecting terms, we obtain
x2 y + (3x − 1)y + y = −rc0 +
∞
[(k + r + 1)2 ck − (k + r + 1)ck+1 ]xk+r = 0,
k=0
which implies
−rc0 = 0
(k + r + 1) ck − (k + r + 1)ck+1 = 0.
2
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 equation is then
∞
y = n=0 (n + 1)!xn , which converges only at x = 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 (12) in the text. Then the indicial equation is
r(r − 1) +
b
c
r+ =0
a
a
or
ar2 + (b − a)r + c = 0,
which is also the auxiliary equation of ax2 y + bxy + cy = 0.
290
5.3
Special Functions
EXERCISES 5.3
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 = 5 and ν = 23 . Then the general solution is y = c1 J2/3 (5x) + c2 J−2/3 (5x).
√
√
√
10. We identify α = 2 and ν = 8. Then the general solution is y = c1 J8 ( 2x) + c2 Y8 ( 2x).
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
1
x2 y + 2xy + α2 x2 y = x3/2 v (x) + x1/2 v (x) + α2 x3/2 − x−1/2 v(x) = 0.
4
Multiplying by x1/2 we obtain
1
x v (x) + xv (x) + α x −
4
2 2 2
v(x) = 0,
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).
√
12. If y = x v(x) then
1
y = x1/2 v (x) + x−1/2 v(x)
2
1
1/2 y = x v (x) + x−1/2 v (x) − x−3/2 v(x)
4
and
1
1
1
x2 y + α2 x2 − ν 2 +
y = x5/2 v (x) + x3/2 v (x) − x1/2 v(x) + α2 x2 − ν 2 +
x1/2 v(x)
4
4
4
= 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).
291
5.3
Special Functions
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 =
c = 1, b = 1, and p = 32 , so, by (19) in the text, the general solution is
1
2
,
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)].
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
y = x−1 c1 J1/2
x + c2 Y1/2
x
2
2
or
−1
y=x
1 2
1 2
x + C2 J−1/2
x
C1 J1/2
.
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 =
1
9
, and p =
2
3
, so, by (19) in the text, the general solution is
y = c1 J2/3
or
y = C1 J2/3
1 3
x
9
1 3
x
9
+ c2 Y2/3
1 3
x
9
+ C2 J−2/3
292
1 3
x .
9
5.3
Special Functions
21. Using the fact that i2 = −1, along with the definition of Jν (x) in (7) in the text, we have
Iν (x) = i−ν Jν (ix) = i−ν
∞
(−1)n
n!Γ(1 + ν + n)
n=0
=
x 2n+ν
(−1)n
i2n+ν−ν
n!Γ(1 + ν + n)
2
n=0
=
x 2n+ν
(−1)n
(i2 )n
n!Γ(1 + ν + n)
2
n=0
=
x 2n+ν
(−1)2n
n!Γ(1 + ν + n) 2
n=0
=
x 2n+ν
1
,
n!Γ(1 + ν + n) 2
n=0
ix
2
2n+ν
∞
∞
∞
∞
which is a real function.
22. (a) The differential equation has the form of (18) in the text with
1
2
2c − 2 = 2 =⇒ c = 2
1
1
b2 c2 = −β 2 c2 = −1 =⇒ β =
and b = i
2
2
1
2
2 2
a − p c = 0 =⇒ p = .
4
1 − 2a = 0 =⇒ a =
Then, by (19) in the text,
1 2
1 2
y = x1/2 c1 J1/4
.
ix + c2 J−1/4
ix
2
2
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
(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 =⇒ a = 0
2c − 2 = 2 =⇒ c = 2
1√
b2 c2 = −β 2 c2 = −7 =⇒ β =
7
2
a2 − p2 c2 = 0 =⇒ p = 0.
Then, by (19) in the text,
y = c1 J0
1√
7 ix2
2
+ c2 Y0
and b =
1√
7 ix2 .
2
In terms of real functions the general solution can be written
1√ 2
1√ 2
y = C1 I0
7x + C2 K0
7x .
2
2
293
1√
7i
2
5.3
Special Functions
23. The differential equation has the form of (18) in the text with
1
2
2c − 2 = 0 =⇒ c = 1
1 − 2a = 0 =⇒ a =
b2 c2 = 1 =⇒ b = 1
1
a2 − p2 c2 = 0 =⇒ p = .
2
Then, by (19) 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 (18) in the text
with
3
1 − 2a = 4 =⇒ a = −
2
2c − 2 = 0 =⇒ c = 1
b2 c2 = 1 =⇒ b = 1
1
a2 − p2 c2 = 2 =⇒ p = .
2
Then, by (19), (23), and (24) in the text,
−3/2
y=x
−3/2
[c1 J1/2 (x) + c2 J−1/2 (x)] = x
c1
2
sin x + c2
πx
2
1
1
cos x = C1 2 sin x + C2 2 cos x.
πx
x
x
1 2
25. Write the differential equation in the form y + (2/x)y + ( 16
x − 3/4x2 )y = 0. This is the form of (18) in the
text with
1
1 − 2a = 2 =⇒ a = −
2
2c − 2 = 2 =⇒ c = 2
1
1
b2 c2 =
=⇒ b =
16
8
3
1
2
2 2
a − p c = − =⇒ p = .
4
2
Then, by (19) in the text,
1 2
1 2
−1/2
y=x
x + c2 J−1/2
x
c1 J1/2
8
8
16
16
1 2
1 2
−1/2
=x
c1
x + c2
x
sin
cos
πx2
8
πx2
8
1 2
1 2
−3/2
−3/2
= C1 x
x + C2 x
x .
sin
cos
8
8
26. Write the differential equation in the form y − (1/x)y + (4 + 3/4x2 )y = 0. This is the form of (18) in the text
with
1 − 2a = −1 =⇒ a = 1
2c − 2 = 0 =⇒ c = 1
b2 c2 = 4 =⇒ b = 2
3
1
a2 − p2 c2 =
=⇒ p = .
4
2
294
5.3
Special Functions
Then, by (19) in the text,
y = x[c1 J1/2 (2x) + c2 J−1/2 (2x)]
2
2
= 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) = −
=−
=
∞
x 2n+ν
(−1)n ν
(−1)n x 2n+ν−1
+x
n!Γ(1 + ν + n) 2
n!Γ(ν + n) 2
n=0
n=0
∞
x 2n+ν (−1)n (ν + n)
x x 2n+ν−1
(−1)n ν
+
·2
n!Γ(1 + ν + n) 2
n!Γ(1 + ν + n)
2
2
n=0
n=0
∞
∞
∞
(−1)n (2n + ν) x 2n+ν
= xJν (x).
n!Γ(1
+
ν
+
n)
2
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)
29. Letting ν = 1 in (21) in the text we have
d
xJ0 (x) =
[xJ1 (x)]
dx
J0 (x)
30. From (20) we obtain
√
31. Since Γ( 12 ) = π and
we obtain
or
so
x
2νJν (x) = xJν+1 (x) + xJν−1 (x).
r=x
rJ0 (r) dr = rJ1 (r) = xJ1 (x).
r=0
0
= −J1 (x), and from (21) we obtain
J0 (x)
1
(2n − 1)! √
Γ 1− +n =
π
2
(n − 1)!22n−1
= J−1 (x). Thus J0 (x) = J−1 (x) = −J1 (x).
n = 1, 2, 3, . . . ,
∞
x 2n−1/2
(−1)n
1 x −1/2 (−1)n (n − 1)!22n−1 x2n−1/2
√
=
+
n!(2n − 1)!22n−1/2 π
n!Γ(1 − 12 + n) 2
Γ( 12 ) 2
n=0
n=1
∞
∞
1
2 (−1)n 21/2 x−1/2 2n
2
2 (−1)n 2n
2
√ x =
=√
+
+
x =
cos x.
πx
πx n=1 (2n)!
πx
π x n=1 2n(2n − 1)! π
J−1/2 (x) =
∞
32. (a) By Problem 28, with ν = 1/2, we obtain J1/2 (x) = xJ3/2 (x) + xJ−1/2 (x) so that
2 sin x
J3/2 (x) =
− cos x ;
πx
x
with ν = −1/2 we obtain −J−1/2 (x) = xJ1/2 (x) + xJ−3/2 (x) so that
2 cos x
J−3/2 (x) = −
+ sin x ;
πx
x
and with ν = 3/2 we obtain 3J3/2 (x) = xJ5/2 (x) + xJ1/2 (x) so that
2 3 sin x 3 cos x
J5/2 (x) =
−
−
sin
x
.
πx
x2
x
295
5.3
Special Functions
(b)
y
1
0.5
y
1
0.5
ν = 1/2
5
10
15
ν = −1/2
20 x
-0.5
-1
5
5
20 x
10
15
20 x
5
10
15
20 x
-0.5
-1
2
s=
α
k −αt/2
,
e
m
dx
dx
k −αt/2
dx ds
dx 2 k α −αt/2
=
=
=
−
e
−
e
dt
ds dt
dt α m
2
ds
m
d dx
k −αt/2
dx
dx α k −αt/2
+
=
e
−
e
dt
ds 2 m
dt ds
m
dx α k −αt/2
d2 x ds
k −αt/2
=
+ 2
e
−
e
ds 2 m
ds dt
m
dx α k −αt/2
d2 x k −αt
=
+ 2
.
e
e
ds 2 m
ds
m
d2 x
d
=
2
dt
dt
Then
d2 x
d2 x mα
m 2 + ke−αt x = ke−αt 2 +
dt
ds
2
Multiplying by 22 /α2 m we have
22 k −αt d2 x
2
e
+
2
2
α m
ds
α
or, since s = (2/α) k/m e−αt/2 ,
s2
34. Differentiating y = x1/2 w
k −αt/2 dx
e
+ ke−αt x = 0.
m
ds
k −αt/2 dx
22 k −αt
e
+ 2
e x=0
m
ds
α m
d2 x
dx
+s
+ s2 x = 0.
ds2
ds
2
3/2
3 αx
with respect to 23 αx3/2 we obtain
2 3/2
1
2 3/2
αx1/2 + x−1/2 w
y = x1/2 w
αx
αx
3
2
3
2 3/2
2 3/2
1/2
y = αxw
αx + αw
αx
αx
3
3
1
2 3/2
1 −3/2
2 3/2
+ αw
− x
.
w
αx
αx
2
3
4
3
Then, after combining terms and simplifying, we have
3
1
y + α2 xy = α αx3/2 w + w + αx3/2 −
w = 0.
2
4αx3/2
296
ν = 3/2
5
ν = 5/2
33. Letting
and
15
-0.5
-1
y
1
0.5
ν = −3/2
-0.5
-1
and
10
-0.5
-1
y
1
0.5
we have
y
1
0.5
10
15
20 x
5.3
Letting t = 23 αx3/2 or αx3/2 = 32 t this differential equation becomes
3 α 2 1
t w (t) + tw (t) + t2 −
w(t) = 0,
2 t
9
Special Functions
t > 0.
35. (a) By 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
= c1 x1/2 J1/3
+ c2 x1/2 J−1/3
.
y = x1/2 w
αx
αx
αx
3
3
3
(b) Airy’s equation, y + α2 xy = 0, has the form of (18) in the text with
1
2
3
2c − 2 = 1 =⇒ c =
2
2
2 2
2
b c = α =⇒ b = α
3
1
2
2 2
a − p c = 0 =⇒ p = .
3
1 − 2a = 0 =⇒ a =
Then, by (19) in the text,
1/2
y=x
2 3/2
2 3/2
c1 J1/3
+ c2 J−1/3
.
αx
αx
3
3
36. The general solution of the differential equation is
y(x) = c1 J0 (αx) + c2 Y0 (αx).
In order to satisfy the conditions that limx→0+ y(x) and limx→0+ y (x) are finite we are forced to define c2 = 0.
Thus, y(x) = c1 J0 (αx). The second boundary condition, y(2) = 0, implies c1 = 0 or J0 (2α) = 0. In order to
have a nontrivial solution we require that J0 (2α) = 0. From Table 5.1, 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).
37. (a) The differential equation y + (λ/x)y = 0 has the form of (18) in the text with
1
2
1
2c − 2 = −1 =⇒ c =
2
√
2 2
b c = λ =⇒ b = 2 λ
1 − 2a = 0 =⇒ a =
a2 − p2 c2 = 0 =⇒ p = 1.
Then, by (19) 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,
297
5.3
Special Functions
√
or, letting t = 2 x ,
Now, if y =
√
√
xJ1 (2 x ), we have
y =
and
√
√
√
√
4xJ1 (2 x ) + 2 xJ1 (2 x ) + (4x − 1)J1 (2 x ) = 0.
√
√
√
√
√
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
y = x−1/2 J1 (2 x ) +
J1 (2 x ) − x−3/2 J1 (2 x ).
2x
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.
38. We see from the graphs below that the graphs of the modified Bessel functions are not oscillatory, while those
of the Bessel functions, shown in Figures 5.3 and 5.4 in the text, are oscillatory.
I0
I1
I2
20
20
20
15
15
15
10
10
10
5
5
5
1
2
3
4
5 x
1
2
3
5 x
4
K0
5
K1
5
K2
5
4
4
4
3
3
3
2
2
2
1
1
1
1
2
3
4
5 x
1
2
3
4
5 x
39. (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.
298
1
2
3
4
5 x
1
2
3
4
5 x
5.3
Special Functions
Using Cramer’s rule we obtain
c1 =
Y0 (10) − Y0 (10)
J0 (10)Y0 (10) − J0 (10)Y0 (10)
c2 =
J0 (10) − J0 (10)
.
J0 (10)Y0 (10) − J0 (10)Y0 (10)
and
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
50
40. (a) Identifying α =
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(t) = c1 x J1/3
+ c2 x J−1/3
.
x
x
3
3
Solving the system x(0.1) = 1, x (0.1) = − 12 we find c1 = −0.809264 and c2 = 0.782397.
x
(b)
1
t
−1
50
150
100
200
41. (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
y = c1 t J1/3
αt
αt
+ c2 t J−1/3
= c1 θ1 (t) + c2 θ2 (t).
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
.
θ(t) = c2 t J−1/3
αt
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 .
299
5.3
Special Functions
(c) Using Mathematica, the first positive root of J−1/3 (x) is x1 ≈ 1.86635. Thus 23 αL3/2 = 1.86635 implies
1/3
2/3 9EI
3(1.86635)
=
L=
(1.86635)2
2α
4δg
1/3
9(2.6 × 107 )π(0.05)4 /4
2
=
(1.86635)
≈ 76.9 in.
4(0.28)π(0.05)2
42. (a) Writing the differential equation in the form xy + (P L/M )y = 0, we identify λ = P L/M .
Problem 37 the solution of this differential equation is
√
√
y = c1 x J1 2 P Lx/M + c2 x Y1 2 P Lx/M .
From
Now J1 (0) = 0, so y(0) = 0 implies c2 = 0 and
√
y = c1 x J1 2 P Lx/M .
√
(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
√
y1 (x) = c1 x J1 2
x .
= c1 x J1
L
L
(c) For c1 = 1 and L = 1 the graph of y1 =
is shown.
√
√
x J1 (3.8317 x )
y
0.3
0.2
0.1
x
0.2
0.4
0.6
0.8
1
43. (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
+ 2v
+ gθ = 0.
dt2
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/dx) dx
d2 θ
d(dθ/dt)
=
=
=
.
dt2
dt
dx
dt
dx2
Thus, the differential equation becomes
x
d2 θ
dθ
g
+2
+ θ=0
2
dx
dx v
or
300
2 dθ
d2 θ
g
+
+
θ = 0.
2
dx
x dx vx
5.3
Special Functions
(c) The differential equation in part (b) has the form of (18) in the text with
1 − 2a = 2 =⇒ a = −
1
2
1
2
g
g
b2 c2 =
=⇒ b = 2
v
v
2c − 2 = −1 =⇒ c =
a2 − p2 c2 = 0 =⇒ p = 1.
Then, by (19) in the text,
g 1/2
g 1/2
θ(x) = x
c1 J1 2
+ c2 Y1 2
x
x
v
v
2
2
v
θ(t) =
c1 J1
g(l0 + vt) + c2 Y1
g(l0 + vt)
.
l0 + vt
v
v
−1/2
or
(d) To simplify calculations, let
2
g 1/2
u=
g(l0 + vt) = 2
x ,
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,
dθ
d −1
d −1
= 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
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
The solution is then
θ=
and
C2 =
π 2
u θ0 J2 (u0 ).
2 0
π 2
J1 (u)
Y1 (u)
u0 θ0 −Y2 (u0 )
+ J2 (u0 )
.
2
u
u
301
(2)
5.3
Special Functions
√
Returning to u = (2/v) g(l0 + vt) and u0 = (2/v) gl0 , we have


2
2
√
J1
Y
g(l
+
vt)
g(l
+
vt)
0
1
0

π gl0 θ0 
2
v
v
−Y2 2 gl0
.
√
√
θ(t) =
gl
+
J
2
0


v
v
v
l0 + vt
l0 + vt
1
radian, and v = 60
ft/s, the above function is
√
√
J1 (480 2(1 + t/60))
Y1 (480 2(1 + t/60))
θ(t) = −1.69045
− 2.79381
.
1 + t/60
1 + t/60
(e) When l0 = 1 ft, θ0 =
1
10
The plots of θ(t) on [0, 10], [0, 30], and [0, 60] are
Θ t
0.1
Θ t
0.1
Θ t
0.1
0.05
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 l increases. The
20
30
40
50
60
t
Θ t
0.1
graph of θ(t) on [0, 300] is shown.
0.05
50
100
150
200
-0.05
-0.1
44. (a) From (26) in the text, we have
6·7 2 4·6·7·9 4
2 · 4 · 6 · 7 · 9 · 11 6
P6 (x) = c0 1 −
x +
x =
x ,
2!
4!
6!
where
c0 = (−1)3
Thus,
5
P6 (x) = −
16
1·3·5
5
=− .
2·4·6
16
231 6
1 − 21x + 63x −
x
5
2
4
=
1
(231x6 − 315x4 + 105x2 − 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
P7 (x) = c1 x −
x +
x −
x
3!
5!
7!
where
c1 = (−1)3
Thus
35
P7 (x) = −
16
1·3·5·7
35
=− .
2·4·6
16
99
429 7
x − 9x + x5 −
x
5
35
3
=
1
(429x7 − 693x5 + 315x3 − 35x).
16
(b) P6 (x) satisfies 1 − x2 y − 2xy + 42y = 0 and P7 (x) satisfies 1 − x2 y − 2xy + 56y = 0.
302
250
300
t
5.3
Special Functions
45. The recurrence relation can be written
Pk+1 (x) =
k = 1: P2 (x) =
k = 2: P3 (x) =
k = 3: P4 (x) =
k = 4: P5 (x) =
k = 5: P6 (x) =
k = 6: P7 (x) =
=
2k + 1
k
xPk (x) −
Pk−1 (x),
k+1
k+1
k = 2, 3, 4, . . . .
3 2 1
x −
2
2
5
3
3 2 1
2
5
x
x −
− x = x3 − x
3
2
2
3
2
2
7
5 3 3
3 3 2 1
35 4 30 2 3
x
x − x −
x −
=
x − x +
4
2
2
4 2
2
8
8
8
35 4 30 2 3
4 5 3 3
63 5 35 3 15
9
x
x − x +
−
x − x =
x − x + x
5
8
8
8
5 2
2
8
4
8
11
5
63 5 35 3 15
5 35 4 30 2 3
231 6 315 4 105 2
x
x − x + x −
x − x +
=
x −
x +
x −
6
8
4
8
6 8
8
8
16
16
16
16
13
5
231 6 315 4 105 2
6 63 5 35 3 15
x
x −
x +
x −
−
x − x + x
7
16
16
16
16
7 8
4
8
429 7 693 5 315 3 35
x −
x +
x − x
16
16
16
16
46. If x = cos θ then
dy
dy
= − sin θ
,
dθ
dx
d2 y
d2 y
dy
= sin2 θ 2 − cos θ
,
2
dθ
dx
dx
and
d2 y
dy
sin θ 2 + cos θ
+ n(n + 1)(sin θ)y = sin θ
dθ
dθ
d2 y
dy
− 2 cos θ
1 − cos θ
+ n(n + 1)y = 0.
dx2
dx
2
That is,
1 − x2
dy
d2 y
− 2x
+ n(n + 1)y = 0.
2
dx
dx
47. 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).
48. Using a CAS we find
1 d
P1 (x) =
(x2 − 1)1 = x
2 dx
1 d2
P2 (x) = 2
(x2 − 1)2 =
2 2! dx2
1 d3
P3 (x) = 3
(x2 − 1)3 =
2 3! dx3
1 d4
P4 (x) = 4
(x2 − 1)4 =
2 4! dx4
1 d5
P5 (x) = 5
(x2 − 1)5 =
2 5! dx5
1 d6
P6 (x) = 6
(x2 − 1)6 =
2 6! dx6
1 d7
P7 (x) = 7
(x2 − 1)7 =
2 7! dx7
1
(3x2 − 1)
2
1
(5x3 − 3x)
2
1
(35x4 − 30x2 + 3)
8
1
(63x5 − 70x3 + 15x)
8
1
(231x6 − 315x4 + 105x2 − 5)
16
1
(429x7 − 693x5 + 315x3 − 35x)
16
303
5.3
Special Functions
P2
1
P1
1
49.
0.5
-1 -0.5
P3
1
0.5
0.5
1 x
-1 -0.5
P4
1
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
-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
0.5
0.5
0.5
1 x
-1 -0.5
-0.5
-0.5
-0.5
-1
-1
-1
0.5
1 x
50. 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.
CHAPTER 5 REVIEW EXERCISES
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 guarantees the existence of two
∞
power series solutions y = n=1 cn xn of the differential equation that 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 ≤
1
2
.
304
CHAPTER 5 REVIEW EXERCISES
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
is to choose, in turn, c0 = 0, c1 = 0 and c0 = 0, c1 = 0. Assuming that c0 =
0, c1 = 0, we have
1
c2 = − c0
2
2
1
c3 = − c2 = c0
3
3
1
1
1
c4 = − c3 − c2 = − c0
2
12
8
2
1
1
1
c5 = − c4 + c2 − c3 =
c0 ;
5
30
20
60
whereas the assumption that c0 = 0, c1 = 0 implies
c2 = −c1
2
c3 = − c2 −
3
1
c4 = − c3 +
2
2
c5 = − c4 +
5
1
1
c1 = c1
6
2
1
1
5
c1 − c2 = − c1
36
12
36
1
1
1
c2 − c3 = −
c1 .
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
y2 (x) = c1 x − x2 + x3 − x4 −
x + ··· .
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
+
− ···
x−
sin x
2x5
x3
6
120
f (x) =
=
+
+ ··· .
=x+
2
4
cos x
3
15
x
x
1−
+
− ···
2
24
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.
∞
9. Substituting y = n=0 cn xn+r into the differential equation we obtain
2xy + y + y = 2r2 − r c0 xr−1 +
∞
[2(k + r)(k + r − 1)ck + (k + r)ck + ck−1 ]xk+r−1 = 0
k=1
305
CHAPTER 5 REVIEW EXERCISES
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−1
ck = −
, k = 1, 2, 3, . . . ,
k(2k − 1)
so
1
1
c1 = −c0 ,
c2 = c0 ,
c3 = − c0 .
6
90
For r = 1/2 the recurrence relation is
ck−1
ck = −
, k = 1, 2, 3, . . . ,
k(2k + 1)
so
1
1
1
c2 =
c3 = −
c1 = − c0 ,
c0 ,
c0 .
3
30
630
Two linearly independent solutions are
1
1
y1 = 1 − x + x2 − x3 + · · ·
6
90
and
1
1 3
1
y2 = x1/2 1 − x + x2 −
x + ··· .
3
30
630
∞
10. Substituting y = n=0 cn xn into the differential equation we have
y − xy − y =
∞
n(n − 1)cn xn−2 −
n=2
∞
n=1
k=n−2
=
∞
ncn xn −
∞
n=0
k=n
(k + 2)(k + 1)ck+2 xk −
k=0
∞
k=n
kck xk −
k=1
= 2c2 − c0 +
∞
2c2 − c0 = 0
(k + 2)(k + 1)ck+2 − (k + 1)ck = 0
and
1
c0
2
1
=
ck ,
k+2
c2 =
ck+2
∞
ck xk
k=0
[(k + 2)(k + 1)ck+2 − (k + 1)ck ]xk = 0.
k=1
Thus
cn xn
k = 1, 2, 3, . . . .
Choosing c0 = 1 and c1 = 0 we find
1
2
c3 = c5 = c7 = · · · = 0
1
c4 =
8
1
c6 =
48
c2 =
306
CHAPTER 5 REVIEW EXERCISES
and so on. For c0 = 0 and c1 = 1 we obtain
c2 = c4 = c6 = · · · = 0
1
c3 =
3
1
c5 =
15
1
c7 =
105
and so on. Thus, two solutions are
1
1
1
y1 = 1 + x2 + x4 + x6 + · · ·
2
8
48
and
11. Substituting y =
∞
n=0
1
1
1 7
y2 = x + x3 + x5 +
x + ···.
3
15
105
cn xn into the differential equation we obtain
(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
,
2
c3 =
1
,
2
c4 =
5
8
c2 = 0,
c3 =
1
,
2
c4 =
1
4
c2 =
and so on. For c0 = 0 and c1 = 1 we obtain
and so on. Thus, two solutions are
3
1
5
y1 = 1 + x2 + x3 + x4 + · · ·
2
2
8
and
12. Substituting y =
∞
n=0
1
1
y2 = x + x3 + x4 + · · · .
2
4
cn xn into the differential equation we obtain
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
1
6
c4 = c7 = c10 = · · · = 0
c3 = −
c5 = c8 = c11 = · · · = 0
1
c6 = −
90
307
CHAPTER 5 REVIEW EXERCISES
and so on. For c0 = 0 and c1 = 1 we obtain
c3 = 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 =
∞
n+r
n=0 cn x
and y2 = x.
into the differential equation, we obtain
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,
Then
k = 1, 2, 3, . . . .
c1 − c0 = 0
2c2 − c1 = 0
0c3 − 0c2 = 0 =⇒ c3 is arbitrary
and
ck =
1
ck−1 ,
k
k = 4, 5, 6, . . . .
Taking c0 = 0 and c3 = 0 we obtain
c1 = c0
1
c2 = c0
2
c3 = c4 = c5 = · · · = 0.
Taking c0 = 0 and c3 = 0 we obtain
c0 = c1 = c2 = 0
1
6
c4 = c3 = c3
4
4!
1
6
c3 = c3
c5 =
5·4
5!
1
6
c6 =
c3 = c3 ,
6·5·4
6!
and so on. In this case we obtain the two solutions
and
1
y1 = 1 + x + x2
2
y2 = x3 +
6 4
6
6
1
x + x5 + x6 + · · · = 6ex − 6 1 + x + x2 .
4!
5!
6!
2
308
CHAPTER 5 REVIEW EXERCISES
14. Substituting y =
∞
n
n=0 cn x
into the differential equation we have
∞
1 2
1 4
1 6
2
3
4
x + · · · (2c2 + 6c3 x + 12c4 x + 20c5 x + 30c6 x + · · ·) +
(cos x)y + y = 1 − x + x −
cn xn
2
24
720
n=0
1
= 2c2 + 6c3 x + (12c4 − c2 )x2 + (20c5 − 3c3 )x3 + 30c6 − 6c4 + c2 x4 + · · ·
12
+ [c0 + c1 x + c2 x2 + c3 x3 + c4 x4 + · · · ]
1
= (c0 + 2c2 ) + (c1 + 6c3 )x + 12c4 x2 + (20c5 − 2c3 )x3 + 30c6 − 5c4 + c2 x4 + · · ·
12
= 0.
Thus
c0 + 2c2 = 0
c1 + 6c3 = 0
12c4 = 0
20c5 − 2c3 = 0
1
30c6 − 5c4 + c2 = 0
12
and
1
c2 = − c0
2
1
c3 = − c1
6
c4 = 0
1
c5 =
c3
10
1
1
c6 = c4 −
c2 .
6
360
Choosing c0 = 1 and c1 = 0 we find
1
1
c2 = − , c3 = 0, c4 = 0, c5 = 0, c6 =
2
720
and so on. For c0 = 0 and c1 = 1 we find
1
1
c2 = 0, c3 = − , c4 = 0, c5 = − , c6 = 0
6
60
and so on. Thus, two solutions are
1
1 6
y1 = 1 − x2 +
x + ···
2
720
15. Substituting y =
∞
n
n=0 cn x
1
1
y2 = x − x3 − x5 + · · · .
6
60
and
into the differential equation we have
y + xy + 2y =
∞
n=2
n(n − 1)cn xn−2 +
n=1
k=n−2
=
∞
∞
ncn xn + 2
= 2c2 + 2c0 +
k=n
k=0
∞
k=1
∞
cn xn
n=0
(k + 2)(k + 1)ck+2 xk +
∞
k=n
kck xk + 2
∞
ck xk
k=0
[(k + 2)(k + 1)ck+2 + (k + 2)ck ]xk = 0.
k=1
309
CHAPTER 5 REVIEW EXERCISES
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 = c7 = · · · = 0
1
c4 =
3
1
c6 = −
15
and so on. For c0 = 0 and c1 = 1 we obtain
c2 = c4 = c6 = · · · = 0
1
c3 = −
2
1
c5 =
8
1
c7 = −
48
and so on. Thus, the general solution is
1
1
1
1
1
y = C0 1 − x2 + x4 − x6 + · · · + C1 x − x3 + x5 − x7 + · · ·
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
∞
16. Substituting y = n=0 cn xn into the differential equation we have
(x + 2)y + 3y =
∞
n=2
n(n − 1)cn xn−1 + 2
∞
n(n − 1)cn xn−2 + 3
n=2
k=n−1
=
∞
k=n−2
(k + 1)kck+1 xk + 2
k=1
= 4c2 + 3c0 +
∞
cn xn
n=0
k=n
(k + 2)(k + 1)ck+2 xk + 3
k=0
∞
∞
∞
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
310
CHAPTER 5 REVIEW EXERCISES
and
3
c2 = − 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
1
8
1
c4 =
16
c3 =
c5 = −
9
320
and so on. For c0 = 0 and c1 = 1 we obtain
c2 = 0
c3 = −
1
4
1
16
c5 = 0
c4 =
and so on. Thus, the general solution is
3
1
1
1
9 5
1
y = C0 1 − x2 + x3 + x4 −
x + · · · + C1 x − x3 + x4 + · · ·
4
8
16
320
4
16
and
3
3 2 1 3
3 2 1 3
9 4
y = C0 − x + x + x − x + · · · + C1 1 − x + x + · · · .
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.
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
+ (t + 1)
+ y = 0,
y(0) = −6, y (0) = 3.
dt2
dt
∞
Substituting y = n=0 cn tn into the differential equation, we have
∞
∞
∞
∞
d2 y
dy
n−2
n
n−1
+
y
=
+
(t
+
1)
n(n
−
1)c
t
+
nc
t
+
nc
t
+
cn tn
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
= 2c2 + c1 + c0 +
∞
k=1
∞
k=n−1
kck tk +
∞
k=0
k=n
(k + 1)ck+1 tk +
∞
ck tk
k=0
[(k + 2)(k + 1)ck+2 + (k + 1)ck+1 + (k + 1)ck ]tk = 0.
k=1
311
CHAPTER 5 REVIEW EXERCISES
2c2 + c1 + c0 = 0
Thus
(k + 2)(k + 1)ck+2 + (k + 1)ck+1 + (k + 1)ck = 0
and
c1 + c0
2
ck+1 + ck
=−
,
k+2
c2 = −
ck+2
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
1
c3 = − ,
6
c4 =
1
,
6
and so on. Thus, the general solution is
1 2 1 3
1 2 1 3 1 4
1 4
y = c0 1 − t + t + t + · · · + c1 t − t − t + t + · · · .
2
6
12
2
6
6
The initial conditions then imply c0 = −6 and c1 = 3. Thus the solution of the initial-value problem is
1
1
1
2
3
4
y = −6 1 − (x − 1) + (x − 1) + (x − 1) + · · ·
2
6
12
1
1
1
2
3
4
+ 3 (x − 1) − (x − 1) − (x − 1) + (x − 1) + · · · .
2
6
6
19. Writing the differential equation in the form
y +
1 − cos x y + xy = 0,
x
and noting that
1 − cos x
x x3
x5
= −
+
− ···
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
y +
x
y=0
ex − 1 − x
and noting that
x
2 2
x
x2
=
−
+
+
− ···
ex − 1 − x
x 3 18 270
we see that x = 0 is a singular point of the differential equation. Since
2x2
x3
x4
x
2
x
=
2x
−
+
+
− ··· ,
ex − 1 − x
3
18 270
we conclude that x = 0 is a regular singular point.
312
CHAPTER 5 REVIEW EXERCISES
21. Substituting y =
∞
n
n=0 cn x
y + x2 y + 2xy =
into the differential equation we have
∞
n=2
n(n − 1)cn xn−2 +
∞
n=1
k=n−2
=
∞
ncn xn+1 + 2
∞
n=0
k=n+1
(k + 2)(k + 1)ck+2 xk +
k=0
∞
cn xn+1
k=n+1
(k − 1)ck−1 xk + 2
k=2
= 2c2 + (6c3 + 2c0 )x +
∞
∞
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
(k + 2)(k + 1)ck+2 + (k + 1)ck−1 = 0,
k = 4, 5, 6, . . . ,
and
c2 =
5
2
1
1
c3 = − c0 −
3
3
1
c4 = − c1
4
1 1
1 1 5
c5 = − c2 = −
=0
2 5
2 5 2
1
ck+2 = −
ck−1 .
k+2
Using the recurrence relation, we find
1
1
1
1
c6 = − c3 =
(c0 + 1) = 2
c0 + 2
6
3·6
3 · 2!
3 · 2!
1
1
c7 = − c4 =
c1
7
4·7
c8 = c11 = c14 = · · · = 0
1
1
1
c9 = − c6 = − 3
c0 − 3
9
3 · 3!
3 · 3!
1
1
c10 = − c7 = −
c1
10
4 · 7 · 10
1
1
1
c0 + 4
c12 = − c9 = 4
12
3 · 4!
3 · 4!
1
1
c13 = − c0 =
c1
13
4 · 7 · 10 · 13
313
CHAPTER 5 REVIEW EXERCISES
and so on. Thus
22. (a) From y = −
1 3
1
1
1
6
9
12
y = c0 1 − x + 2
x − 3
x + 4
x − ···
3
3 · 2!
3 · 3!
3 · 4!
1
1 7
1
1
+ c1 x − x4 +
x −
x10 +
x13 − · · ·
4
4·7
4 · 7 · 10
4 · 7 · 10 · 13
5
1
1
1
1
+ x2 − x3 + 2
x6 − 3
x9 + 4
x12 − · · · .
2
3
3 · 2!
3 · 3!
3 · 4!
1 du
we obtain
u dx
dy
1
1 d2 u
+ 2
=−
dx
u dx2
u
du
dx
2
.
Then dy/dx = x2 + y 2 becomes
−
1 d2 u
1
+ 2
2
u dx
u
du
dx
2
= x2 +
1
u2
du
dx
2
,
d2 u
+ x2 u = 0.
dx2
so
(b) The differential equation u + x2 u = 0 has the form (18) in the text with
1
2
2c − 2 = 2 =⇒ c = 2
1
b2 c2 = 1 =⇒ b =
2
1
a2 − p2 c2 = 0 =⇒ p = .
4
1 − 2a = 0 =⇒ a =
Then, by (19) in the text,
1/2
u=x
(c) We have
1 2
1 2
c1 J1/4
.
x + c2 J−1/4
x
2
2
1 du
dw dt
1
d 1/2
1
1
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
dw
dw
1
1
1
dw
= − 1/2
x3/2
+ 1/2 w = −
2x2
+w =−
4t
+w .
dt
2xw
dt
2xw
dt
x w
2x
y=−
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) + c1 J1/4 (t) + c2 J−1/4 (t)
4t
4t
1 2
1 2
2
2
= 4c1 tJ−3/4 (t) − 4c2 tJ3/4 (t) = 2c1 x J−3/4
x − 2c2 x J3/4
x ,
2
2
so
y=−
2c1 x2 J−3/4 ( 12 x2 ) − 2c2 x2 J3/4 ( 12 x2 )
−c1 J−3/4 ( 12 x2 ) + c2 J3/4 ( 12 x2 )
=
x
.
2x[c1 J1/4 ( 12 x2 ) + c2 J−1/4 ( 12 x2 )]
c1 J1/4 ( 12 x2 ) + c2 J−1/4 ( 12 x2 )
314
CHAPTER 5 REVIEW EXERCISES
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) Equations (10) and (24) of Section 5.3 in the text imply
cos π2 J1/2 (x) − J−1/2 (x)
2
Y1/2 (x) =
cos x.
= −J−1/2 (x) = −
π
sin 2
πx
(b) From (15) of Section 5.3 in the text
I1/2 (x) = i−1/2 J1/2 (ix)
I−1/2 (x) = i1/2 J−1/2 (ix)
and
so
I1/2 (x) =
and
I−1/2 (x) =
∞
2 1
x2n+1 =
πx n=0 (2n + 1)!
∞
2 1
x2n =
πx n=0 (2n)!
2
sinh x
πx
2
cosh x.
πx
(c) Equation (16) of Section 5.3 in the text and part (b) imply
π I−1/2 (x) − I1/2 (x)
π
K1/2 (x) =
=
2
sin π2
2
=
2
cosh x −
πx
2
sinh x
πx
π ex + e−x
π −x
ex − e−x
−
=
e .
2x
2
2
2x
24. (a) Using formula (5) of Section 3.2 in the text, we find that a second solution of (1 − x2 )y − 2xy = 0 is
y2 (x) = 1 ·
=
e
2x dx/(1−x2 )
e− ln(1−x ) dx
2
dx =
12
dx
1
= ln
2
1−x
2
1+x
1−x
,
where partial fractions was used to obtain the last integral.
(b) Using formula (5) of Section 3.2 in the text, we find that a second solution of
(1 − x2 )y − 2xy + 2y = 0 is
e
2x dx/(1−x2 )
e− ln(1−x )
dx
x2
x2
dx
1
1+x
1
x
1+x
=x
dx
=
x
ln
−
=
ln
− 1,
x2 (1 − x2 )
2
1−x
x
2
1−x
y2 (x) = x ·
dx = x
where partial fractions was used to obtain the last integral.
315
2
CHAPTER 5 REVIEW EXERCISES
(c)
y2
2
y2
2
1
1
1x
-1
y2 (x) =
1x
-1
-1
-1
-2
-2
1
ln
2
1+x
1−x
y2 =
x
ln
2
1+x
1−x
−1
25. (a) By the binomial theorem we have
−1/2
1 + t2 − 2xt
1 2
(−1/2)(−3/2) 2
(−1/2)(−3/2)(−5/2) 2
t − 2xt +
(t − 2xt)2 +
(t − 2xt)3 + · · ·
2
2!
3!
1
3
5
= 1 − (t2 − 2xt) + (t2 − 2xt)2 − (t2 − 2xt)3 + · · ·
2
8
16
∞
1
1
= 1 + xt + (3x2 − 1)t2 + (5x3 − 3x)t3 + · · · =
Pn (x)tn .
2
2
n=0
=1−
(b) Letting x = 1 in (1 − 2xt + t2 )−1/2 , we have
(1 − 2t + t2 )−1/2 = (1 − t)−1 =
From part (a) we have
∞
1
= 1 + t + t2 + t 3 + . . .
1−t
Pn (1)tn = (1 − 2t + t2 )−1/2 =
n=0
∞
(|t| < 1) =
∞
tn .
n=0
tn .
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
= 1 − t + t2 − 3t3 + . . .
1+t
∞
∞
=
(−1)n tn =
Pn (−1)tn ,
(1 + 2t + t2 )−1/2 = (1 + t)−1 =
n=0
n=0
n
so that Pn (−1) = (−1) .
316
(|t| < 1)
6
Numerical Solutions of Ordinary
Differential Equations
EXERCISES 6.1
Euler Methods and Error Analysis
1.
h=0.1
xn
1.00
1.10
1.20
1.30
1.40
1.50
3.
yn
5.0000
3.9900
3.2546
2.7236
2.3451
2.0801
h=0.1
xn
0.00
0.10
0.20
0.30
0.40
0.50
2.
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
5.0000
4.4475
3.9763
3.5751
3.2342
2.9452
2.7009
2.4952
2.3226
2.1786
2.0592
xn
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
xn
0.00
0.10
0.20
0.30
0.40
0.50
4.
h=0.05
yn
0.0000
0.1005
0.2030
0.3098
0.4234
0.5470
h=0.1
yn
0.0000
0.0501
0.1004
0.1512
0.2028
0.2554
0.3095
0.3652
0.4230
0.4832
0.5465
yn
2.0000
1.6600
1.4172
1.2541
1.1564
1.1122
h=0.1
xn
0.00
0.10
0.20
0.30
0.40
0.50
317
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
h=0.05
yn
1.0000
1.1110
1.2515
1.4361
1.6880
2.0488
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
1.0526
1.1113
1.1775
1.2526
1.3388
1.4387
1.5556
1.6939
1.8598
2.0619
6.1
5.
Euler Methods and Error Analysis
h=0.1
xn
0.00
0.10
0.20
0.30
0.40
0.50
7.
9.
yn
0.0000
0.0952
0.1822
0.2622
0.3363
0.4053
h=0.1
xn
0.00
0.10
0.20
0.30
0.40
0.50
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
0.0488
0.0953
0.1397
0.1823
0.2231
0.2623
0.3001
0.3364
0.3715
0.4054
yn
0.5000
0.5215
0.5362
0.5449
0.5490
0.5503
xn
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
yn
0.5000
0.5116
0.5214
0.5294
0.5359
0.5408
0.5444
0.5469
0.5484
0.5492
0.5495
xn
1.00
1.05
1.10
1.15
1.20
1.25
1.30
1.35
1.40
1.45
1.50
yn
1.0000
1.0024
1.0100
1.0228
1.0414
1.0663
1.0984
1.1389
1.1895
1.2526
1.3315
yn
0.0000
0.0050
0.0200
0.0451
0.0805
0.1266
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
h=0.05
yn
1.0000
1.1079
1.2337
1.3806
1.5529
1.7557
h=0.1
xn
0.00
0.10
0.20
0.30
0.40
0.50
318
h=0.05
h=0.1
xn
0.00
0.10
0.20
0.30
0.40
0.50
10.
h=0.05
yn
1.0000
1.0095
1.0404
1.0967
1.1866
1.3260
h=0.1
xn
0.00
0.10
0.20
0.30
0.40
0.50
8.
h=0.05
h=0.1
xn
1.00
1.10
1.20
1.30
1.40
1.50
6.
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
1.0519
1.1079
1.1684
1.2337
1.3043
1.3807
1.4634
1.5530
1.6503
1.7560
h=0.05
yn
0.5000
0.5250
0.5498
0.5744
0.5986
0.6224
xn
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
yn
0.5000
0.5125
0.5250
0.5374
0.5498
0.5622
0.5744
0.5866
0.5987
0.6106
0.6224
6.1
Euler Methods and Error Analysis
11. To obtain the analytic solution use the substitution u = x + y − 1. The resulting differential equation in u(x)
will be separable.
h=0.1
h=0.05
xn
yn
0.00
0.10
0.20
0.30
0.40
0.50
12. (a)
2.0000
2.1220
2.3049
2.5858
3.0378
3.8254
Actual
Value
2.0000
2.1230
2.3085
2.5958
3.0650
3.9082
xn
yn
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
Actual
Value
2.0000
2.1230
2.3085
2.5958
3.0650
3.9082
2.5958
2.7997
3.0650
3.4189
3.9082
y
(b)
20
15
10
5
1.1
1.2 1.3 1.4
xn
1.00
1.10
1.20
1.30
1.40
Euler
1.0000
1.2000
1.4938
1.9711
2.9060
Imp . Euler
1.0000
1.2469
1.6430
2.4042
4.5085
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)
(0.1)2
h2
= 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 0.214 we
see that this is the case.
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)
(0.1)3
h3
= 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.
319
6.1
Euler Methods and Error Analysis
(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.
(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
y (c)
h2
≤ 19(0.1)2 (1) = 0.19.
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
y (c) h ≤ 19(0.1)3 (1) = 0.019.
6
320
6.2
Runge-Kutta Methods
(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.
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
1
h3
=
.
6
(c + 1)3 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
y (c)
h3
(0.1)3
≤ (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.
∗
21. Because yn+1
depends on yn and is used to determine yn+1 , all of the yn∗ cannot be computed at one time
independently of the corresponding yn values. For example, the computation of y4∗ involves the value of y3 .
EXERCISES 6.2
Runge-Kutta Methods
1.
xn
0.00
0.10
0.20
0.30
0.40
0.50
yn
2.0000
2.1230
2.3085
2.5958
3.0649
3.9078
Actual
Value
2.0000
2.1230
2.3085
2.5958
3.0650
3.9082
321
6.2
Runge-Kutta Methods
2. In this problem we use h = 0.1. Substituting w2 =
in (4) in the text, we obtain
w1 = 1 − w2 =
1
,
4
α=
2
1
= ,
2w2
3
3
4
into the equations
xn
and β =
0.00
0.10
0.20
0.30
0.40
0.50
2
1
= .
2w2
3
The resulting second-order Runge-Kutta method is
yn+1 = yn + h
1
3
k1 + k2
4
4
= yn +
Second Order Improved
Runge Kutta
Euler
2.0000
2.0000
2.1213
2.1220
2.3030
2.3049
2.5814
2.5858
3.0277
3.0378
3.8002
3.8254
h
(k1 + 3k2 )
4
where
k1 = f (xn , yn )
2
2
k2 = f 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.
xn
1.00
1.10
1.20
1.30
1.40
1.50
5.
xn
0.00
0.10
0.20
0.30
0.40
0.50
7.
xn
0.00
0.10
0.20
0.30
0.40
0.50
9.
xn
0.00
0.10
0.20
0.30
0.40
0.50
yn
4.
5.0000
3.9724
3.2284
2.6945
2.3163
2.0533
yn
0.00
0.10
0.20
0.30
0.40
0.50
6.
0.0000
0.1003
0.2027
0.3093
0.4228
0.5463
yn
xn
0.00
0.10
0.20
0.30
0.40
0.50
8.
0.0000
0.0953
0.1823
0.2624
0.3365
0.4055
yn
xn
xn
0.00
0.10
0.20
0.30
0.40
0.50
10.
0.5000
0.5213
0.5358
0.5443
0.5482
0.5493
xn
0.00
0.10
0.20
0.30
0.40
0.50
322
yn
2.0000
1.6562
1.4110
1.2465
1.1480
1.1037
yn
1.0000
1.1115
1.2530
1.4397
1.6961
2.0670
yn
0.0000
0.0050
0.0200
0.0451
0.0805
0.1266
yn
1.0000
1.1079
1.2337
1.3807
1.5531
1.7561
6.2
xn
11.
yn
1.00
1.10
1.20
1.30
1.40
1.50
12.
1.0000
1.0101
1.0417
1.0989
1.1905
1.3333
13. (a) Write the equation in the form
dv
= 32 − 0.125v 2 = f (t, v).
dt
(b)
v
40
xn
Runge-Kutta Methods
yn
0.00
0.10
0.20
0.30
0.40
0.50
0.5000
0.5250
0.5498
0.5744
0.5987
0.6225
tn
0.0
1.0
2.0
3.0
4.0
5.0
vn
0.0000
25.2570
32.9390
34.9770
35.5500
35.7130
30
20
10
1
2
3
4
5
6
t
(c) Separating variables and using partial fractions we have
1
1
1
√
√
√
√
+√
dv = dt
2 32
32 − 0.125 v
32 + 0.125 v
and
√
1
√
√
√
√
√
ln | 32 + 0.125 v| − ln | 32 − 0.125 v| = t + c.
2 32 0.125
Since v(0) = 0 we find c = 0. Solving for v we obtain
√
√
16 5 (e 3.2 t − 1)
√
v(t) =
e 3.2 t + 1
and v(5) ≈ 35.7678. Alternatively, the solution can be expressed as
mg
kg
v(t) =
tanh
t.
k
m
14. (a)
t days A observed A approximated A(t)
1
2.78
1.93
2
13.53
12.50
3
36.30
36.46
4
47.50
47.23
(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.
5
49.40
49.00
50
40
30
20
10
0
323
1
2
3
4
5
t
6.2
Runge-Kutta Methods
(c) Let α = 2.128 and β = 0.0432. Separating variables we obtain
dA
= dt
A(α − βA)
1 1
β
+
dA = dt
α A α − βA
1
[ln A − ln(α − βA)] = t + c
α
A
ln
= 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) =
αeα(t+c)
α
α
=
=
.
β + e−αc e−αt
1 + βeα(t+c)
β + e−α(t+c)
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) ≈
t days A observed A actual 15. (a)
1
2.78
1.93
xn
1.00
1.05
1.10
1.15
1.20
1.25
1.30
1.35
h0.05
1.0000
1.1112
1.2511
1.4348
1.6934
2.1047
2.9560
7.8981
1.40
1.0608 1015
2
13.53
12.50
2.128
.
0.0432 + 8.8235e−2.128t
3
36.30
36.46
4
47.50
47.23
5
49.40
49.00
(b)
h0.1
1.0000
y
20
1.2511
15
1.6934
10
5
2.9425
1.1
903.0282
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)
(0.1)5
h5
= 32e2c
= 0.000002667e2c .
120
120
324
1.2
1.3
1.4
x
6.2
Runge-Kutta Methods
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 one-sixteenth 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.
(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 one-sixteenth 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.
y(1.5) ≈ 2.053222989.
With h = 0.05 we obtain
19. (a) Using y (5) = 24/(x + 1)5 we see that the local truncation error is
y (5) (c)
h5
h5
1
=
.
5
120
(c + 1) 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)
(0.1)5
h5
≤ (1)
= 0.000002.
5
5
(c) Using the RK4 method with h = 0.1 we obtain y(0.5) ≈ 0.405465168.
y(0.5) ≈ 0.405465111.
With h = 0.05 we obtain
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 (xn+1 , yn+1
). The second-order Runge-Kutta
methods require 2 function evaluations per step, 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
325
6.2
Runge-Kutta Methods
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.
xn
0.000
0.025
0.050
0.075
0.100
0.125
0.150
0.175
0.200
Imp . Euler
Euler
h0.025
h0.05
2.0000
2.0000
2.0250
2.0526
2.0553
2.0830
2.1165
2.1228
2.1535
2.1943
2.2056
2.2395
2.2895
2.3075
RK4
h0.1
2.0000
2.1230
2.3085
Actual
2.0000
2.0263
2.0554
2.0875
2.1230
2.1624
2.2061
2.2546
2.3085
21. (a) For y + y = 10 sin 3x an integrating factor is ex so that
y
5
d x
[e y] = 10ex sin 3x =⇒ ex y = ex sin 3x − 3ex cos 3x + c
dx
=⇒ y = sin 3x − 3 cos 3x + ce−x .
When x = 0, y = 0, so 0 = −3 + c and c = 3. The solution is
2
x
y = sin 3x − 3 cos 3x + 3e−x .
Using Newton’s method we find that x = 1.53235 is the only positive root in [0, 2].
−5
(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
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
yn
0.0000
0.1440
0.5448
1.1409
1.8559
2.6049
3.3019
3.8675
4.2356
4.3593
4.2147
xn
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
yn
y
4.2147
3.8033
3.1513
2.3076
1.3390
0.3243
0.6530
1.5117
2.1809
2.6061
2.7539
5
2
−5
326
x
6.3
Multistep Methods
EXERCISES 6.3
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−
so that
d −x
[e y] = (x − 1)e−x
dx
and
y = ex (−xe−x + c) = −x + cex .
dx
= e−x ,
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
0.0
0.2
0.4
0.6
0.8
yn
1.00000000
1.02140000
1.09181796
1.22210646
1.42552788
Actual
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;
(* define the differential equation *)
(* set the step size *)
a = 0; y0 = 1; b = 1;
f[x, y]
(* set the initial condition and the interval *)
(* display the DE *)
Clear[k1, k2, k3, k4, x, y, u, v]
x = u[0] = a;
y = v[0] = y0;
n = 0;
While[x < a + 3h,
n = n + 1;
(* use RK4 to compute the first 3 values after y(0) *)
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];
327
6.3
Multistep Methods
While[x ≤ b,
(* use Adams-Bashforth-Moulton *)
p3 = f[u[n - 3], v[n - 3]];
p2 = f[u[n - 2], v[n - 2]];
p1 = f[u[n - 1], v[n - 1]];
p0 = f[u[n], v[n]];
pred = y + (h/24)(55p0 - 59p1 + 37p2 - 9p3);
(* predictor *)
x = x + h;
p4 = f[x, pred];
y = y + (h/24)(9p4 + 19p0 - 5p1 + p2);
n = n + 1;
u[n] = x;
v[n] = y]
(* corrector *)
(*display the table *)
TableForm[Prepend[Table[{u[n], v[n]}, {n, 0, (b-a)/h}], {"x(n)", "y(n)"}]];
3. The first predictor is y4∗ = 0.73318477.
xn
0.0
0.2
0.4
0.6
0.8
yn
1.00000000
0.73280000
0.64608032
0.65851653
0.72319464
init. cond.
RK4
RK4
RK4
ABM
4. The first predictor is y4∗ = 1.21092217.
xn
0.0
0.2
0.4
0.6
0.8
yn
2.00000000
1.41120000
1.14830848
1.10390600
1.20486982
init. cond.
RK4
RK4
RK4
ABM
5. The first predictor for h = 0.2 is y4∗ = 1.02343488.
xn
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
h0.2
0.00000000 init. cond.
0.20270741 RK4
0.42278899 RK4
0.68413340 RK4
1.02969040 ABM
1.55685960 ABM
h0.1
0.00000000
0.10033459
0.20270988
0.30933604
0.42279808
0.54631491
0.68416105
0.84233188
1.02971420
1.26028800
1.55762558
init. cond.
RK4
RK4
RK4
ABM
ABM
ABM
ABM
ABM
ABM
ABM
328
6.3
6. The first predictor for h = 0.2 is y4∗ = 3.34828434.
xn
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
h0.2
1.00000000 init. cond.
1.44139950 RK4
1.97190167 RK4
2.60280694 RK4
3.34860927 ABM
4.22797875 ABM
h0.1
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
7. The first predictor for h = 0.2 is y4∗ = 0.13618654.
xn
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
h0.2
0.00000000 init. cond.
0.00262739 RK4
0.02005764 RK4
0.06296284 RK4
0.13598600 ABM
0.23854783 ABM
h0.1
0.00000000
0.00033209
0.00262486
0.00868768
0.02004821
0.03787884
0.06294717
0.09563116
0.13596515
0.18370712
0.23841344
init. cond.
RK4
RK4
RK4
ABM
ABM
ABM
ABM
ABM
ABM
ABM
8. The first predictor for h = 0.2 is y4∗ = 2.61796154.
xn
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
h0.2
1.00000000 init. cond.
1.23369623 RK4
1.55308554 RK4
1.99610329 RK4
2.62136177 ABM
3.52079042 ABM
h0.1
1.00000000
1.10793839
1.23369772
1.38068454
1.55309381
1.75610064
1.99612995
2.28119129
2.62131818
3.02914333
3.52065536
init. cond.
RK4
RK4
RK4
ABM
ABM
ABM
ABM
ABM
ABM
ABM
329
Multistep Methods
6.3 Higher-Order
Multistep Methods
6.4
Equations and Systems
EXERCISES 6.4
Higher-Order Equations and Systems
1. The substitution y = u leads to the iteration formulas
un+1 = un + h(4un − 4yn ).
yn+1 = yn + hun ,
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.
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,
xn
u = 4u − 4y.
Using formula (4) in the text with x corresponding to
t, y corresponding to x, and u corresponding to y, we
obtain the table shown.
0.0
0.1
0.2
4. The substitution y = u leads to the system
y = u,
2
2
u = u − 2 y.
x
x
Using formula (4) in the text with x corresponding to
t, y corresponding to x, and u corresponding to y, we
obtain the table shown.
330
h0.2
yn
h0.2
un
2.0000
1.0000
1.4928
4.4731
h0.1
yn
2.0000
1.8321
1.4919
xn
h0.2
yn
h0.2
un
h0.1
yn
1.0
1.1
1.2
4.0000
9.0000
6.0001
11.0002
4.0000
4.9500
6.0000
h0.1
un
1.0000
2.4427
4.4753
h0.1
un
9.0000
10.0000
11.0000
6.4
5. The substitution y = u leads to the system
y = u,
u = 2u − 2y + et cos t.
Using formula (4) in the text with y corresponding to
x and u corresponding to y, we obtain the table shown.
Higher-Order Equations and Systems
xn
h0.2
yn
h0.2
un
h0.1
yn
h0.1
un
0.0
0.1
0.2
1.0000
2.0000
1.4640
2.6594
1.0000
1.2155
1.4640
2.0000
2.3150
2.6594
6. Using h = 0.1, the RK4 method for a system, and a numerical solver, we obtain
tn
0.0
0.1
0.2
0.3
0.4
0.5
7.
h0.2
i 1n
0.0000
2.5000
2.8125
2.0703
0.6104
1.5619
i1
7
h0.2
i 3n
0.0000
3.7500
5.7813
7.4023
9.1919
11.4877
i2
7
6
6
5
5
4
4
3
3
2
2
1
1
1
tn
h0.2
xn
h0.2
yn
h0.1
xn
h0.1
yn
0.0
0.1
0.2
6.0000
2.0000
8.3055
3.4199
6.0000
7.0731
8.3055
2.0000
2.6524
3.4199
2 3 4 5t
1
2 3 4 5t
x,y
20
xHtL
15
10
yHtL
5
0.5
8.
1
1.5
2t
1.5
2t
x,y
tn
h0.2
xn
h0.2
yn
h0.1
xn
h0.1
yn
50
0.0
0.1
0.2
1.0000
1.0000
1.0000
1.4006
2.0845
1.0000
1.8963
3.3502
40
2.0785
3.3382
yHtL
30
20
10
xHtL
0.5
331
1
6.4
Higher-Order Equations and Systems
x,y
9.
tn
0.0
0.1
0.2
h0.2
xn
h0.2
yn
3.0000
5.0000
3.9123
4.2857
h0.1
xn
h0.1
yn
3.0000
3.4790
3.9123
30
5.0000
4.6707
4.2857
25
20
15
xHtL
10
yHtL
5
5
10
15
20
25
30
t
-5
10.
tn
h0.2
xn
h0.2
yn
h0.1
xn
h0.1
yn
0.0
0.1
0.2
0.5000
0.2000
2.1589
2.3279
0.5000
1.0207
2.1904
0.2000
1.0115
2.3592
x,y
2
xHtL
1.5
1
0.5
yHtL
0.1
11. Solving for x and y we obtain the system
x = −2x + y + 5t
1
-5
h0.2
yn
tn
0.0
0.1
0.2
1.0000
2.0000
0.4179
2.1824
h0.1
xn
1.0000
0.6594
0.4173
h0.1
yn
-15
yHtL
-20
332
2
3
4t
xHtL
-10
2.0000
2.0476
2.1821
0.3t
x,y
y = 2x + y − 2t.
h0.2
xn
0.2
6.5
Second-Order Boundary-Value Problems
12. Solving for x and y we obtain the system
x,y
1
x = y − 3t2 + 2t − 5
2
1
y = − y + 3t2 + 2t + 5.
2
60
40
yHtL
20
tn
h0.2
xn
h0.2
yn
0.0
0.1
0.2
3.0000
1.0000
1.9867
0.0933
h0.1
xn
h0.1
yn
3.0000
2.4727
1.9867
2
1.0000
0.4527
0.0933
-20
4
6
8t
xHtL
-40
-60
EXERCISES 6.5
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
0.5
1.0
1.5
4.0000 -5.6774 -2.5807 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
0.25
0.50
0.75
1.00
0.0000 -0.0172 -0.0316 -0.0324 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
0.2
0.4
0.6
0.8
1.0
0.0000 -0.2259 -0.3356 -0.3308 -0.2167 0.0000
333
6.5
Second-Order Boundary-Value Problems
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
0.0
1.0000
0.2
1.9600
0.4
3.8800
0.6
0.8
1.0
7.7200 15.4000 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 1.1667 1.3333 1.5000 1.6667 1.8333 2.0000
1.0000 -0.5918 -1.1626 -1.3070 -1.2704 -1.1541 -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
1+
yi+1 + −2 +
yi + 1 −
yi−1 = 0.
xi
x2i
xi
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
1−
yi+1 + −2 +
yi + 1 +
yi−1 = 0.0156 ln xi .
xi
x2i
xi
The solution of the corresponding linear system gives
x
y
1.000
1.125
1.250
1.375
1.500
1.625
1.750
1.875
2.000
0.0000 -0.1988 -0.4168 -0.6510 -0.8992 -1.1594 -1.4304 -1.7109 -2.0000
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
0.4
0.9471
0.5
0.6
1.1465 1.3353
0.7
0.8
1.5149 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 .
334
6.5
Second-Order Boundary-Value Problems
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
0.6
0.4296 0.3216
0.7
0.8
0.2225 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
1+
ui+1 − 2ui + 1 −
ui−1 = 0.
ri
ri
The solution of the corresponding linear system gives
r
1.0
1.5
2.0
2.5
3.0
3.5
4.0
u 50.0000 72.2222 83.3333 90.0000 94.4444 97.6190 100.0000
13. (a) The difference equation
h
h
2
1 + Pi yi+1 + (−2 + h 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
2h
or y1 − y−1 = 2h.
The difference equation corresponding to i = 0,
h
h
1 + P0 y1 + (−2 + h2 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
0.0
0.2
0.4
0.6
0.8
1.0
y -2.2755 -2.0755 -1.8589 -1.6126 -1.3275 -1.0000
335
6.5
Second-Order Boundary-Value Problems
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
CHAPTER 6 REVIEW EXERCISES
1.
xn
1.00
1.05
1.10
1.15
1.20
1.25
1.30
1.35
1.40
1.45
1.50
2.
xn
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
Euler
h0.1
2.0000
2.1386
2.3097
2.5136
2.7504
3.0201
Euler
h0.1
0.0000
0.1000
0.2010
0.3049
0.4135
0.5279
Euler Imp . Euler Imp . Euler
h0.05
h0.1
h0.05
2.0000 2.0000
2.0000
2.0693
2.0735
2.1469 2.1549
2.1554
2.2328
2.2459
2.3272 2.3439
2.3450
2.4299
2.4527
2.5409 2.5672
2.5689
2.6604
2.6937
2.7883 2.8246
2.8269
2.9245
2.9686
3.0690 3.1157
3.1187
RK4
h0.1
2.0000
Euler Imp . Euler Imp . Euler
h0.05
h0.1
h0.05
0.0000 0.0000
0.0000
0.0500
0.0501
0.1001 0.1005
0.1004
0.1506
0.1512
0.2017 0.2030
0.2027
0.2537
0.2552
0.3067 0.3092
0.3088
0.3610
0.3638
0.4167 0.4207
0.4202
0.4739
0.4782
0.5327 0.5382
0.5378
RK4
h0.1
0.0000
2.1556
2.3454
2.5695
2.8278
3.1197
0.1003
0.2026
0.3087
0.4201
0.5376
336
RK4
h0.05
2.0000
2.0736
2.1556
2.2462
2.3454
2.4532
2.5695
2.6944
2.8278
2.9696
3.1197
RK4
h0.05
0.0000
0.0500
0.1003
0.1511
0.2026
0.2551
0.3087
0.3637
0.4201
0.4781
0.5376
1.0
1.50003
CHAPTER 6 REVIEW EXERCISES
3.
xn
0.50
0.55
0.60
0.65
0.70
0.75
0.80
0.85
0.90
0.95
1.00
4.
xn
1.00
1.05
1.10
1.15
1.20
1.25
1.30
1.35
1.40
1.45
1.50
Imp . Euler Imp . Euler
Euler
h0.05
h0.1
h0.05
0.5000
0.5000
0.5000
0.5500
0.5512
0.6024
0.6048
0.6049
0.6573
0.6609
0.7144
0.7191
0.7193
0.7739
0.7800
0.8356
0.8427
0.8430
0.8996
0.9082
0.9657
0.9752
0.9755
1.0340
1.0451
1.1044
1.1163
1.1168
Euler
h0.1
0.5000
0.6000
0.7095
0.8283
0.9559
1.0921
Euler
h0.1
1.0000
1.2000
1.4760
1.8710
2.4643
3.4165
Euler Imp . Euler Imp . Euler
h0.05
h0.1
h0.05
1.0000 1.0000
1.0000
1.1000
1.1091
1.2183 1.2380
1.2405
1.3595
1.4010
1.5300 1.5910
1.6001
1.7389
1.8523
1.9988 2.1524
2.1799
2.3284
2.6197
2.7567 3.1458
3.2360
3.3296
4.1528
4.1253 5.2510
5.6404
RK4
h0.1
0.5000
0.6049
0.7194
0.8431
0.9757
1.1169
RK4
h0.1
1.0000
1.2415
1.6036
2.1909
3.2745
5.8338
RK4
h0.05
0.5000
0.5512
0.6049
0.6610
0.7194
0.7801
0.8431
0.9083
0.9757
1.0452
1.1169
RK4
h0.05
1.0000
1.1095
1.2415
1.4029
1.6036
1.8586
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
0.0
0.1
0.2
0.3
0.4
yn
2.00000000
1.65620000
1.41097281
1.24645047
1.14796764
init. cond.
RK4
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
337
CHAPTER 6 REVIEW EXERCISES
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.
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
0.3
0.4
0.5
0.6
8.1049 11.3840 13.7038 14.7770 14.4083
0.7
12.5396
338
0.8
9.2847
0.9
4.9450
1.0
0.0000
Part II
7
Vectors, Matrices, and Vector Calculus
Vectors
EXERCISES 7.1
Vectors in 2-Space
√
1. (a) 6i + 12j
(b) i + 8j
(c) 3i
(d)
2. (a) 3, 3
(b) 3, 4
(c) −1,−2
(d) 5
3. (a) 12, 0
(b) 4, −5
(c) 4, 5
(d)
(c) − 13 i − j
√
(d) 2 2/3
(e)
4. (a)
1
2i
− 12 j
(b)
2
3i
+ 23 j
√
65
(e) 3
(e)
41
(e)
√
√
√
5
41
10/3
5. (a) −9i + 6j
(b) −3i + 9j
(c) −3i − 5j
√
(d) 3 10
(e)
6. (a) 3, 9
(b) −4,−12
(c) 6, 18
√
(d) 4 10
√
(e) 6 10
7. (a) −6i + 27j
(b) 0
(c) −4i + 18j
(d) 0
√
(e) 2 85
8. (a) 21, 30
(b) 8, 12
(c) 6, 8
√
(d) 4 13
(e) 10
9. (a) 4, −12 − −2, 2 = 6, −14
√
(b) −3, 9 − −5, 5 = 2, 4
10. (a) (4i + 4j) − (6i − 4j) = −2i + 8j
(b) (−3i − 3j) − (15i − 10j) = −18i + 7j
11. (a) (4i − 4j) − (−6i + 8j) = 10i − 12j
(b) (−3i + 3j) − (−15i + 20j) = 12i − 17j
12. (a) 8, 0 − 0, −6 = 8, 6
(b) −6, 0 − 0, −15 = −6, 15
13. (a) 16, 40 − −4, −12 = 20, 52
(b) −12, −30 − −10, −30 = −2, 0
14. (a) 8, 12 − 10, 6 = −2, 6
(b) −6, −9 − 25, 15 = −31, −24
339
34
7.1
Vectors in 2-Space
15.
16.
−−−→
P1 P2 = 2, 5
−−−→
P1 P2 = 6, −4
17.
18.
−−−→
P1 P2 = 2, 2
−−−→
P1 P2 = 2, −3
−−−→ −−→ −−→ −−→ −−−→ −−→
19. Since P1 P2 = OP2 − OP1 , OP2 = P1 P2 + OP1 = (4i + 8j) + (−3i + 10j) = i + 18j, and the terminal point is
(1, 18).
−−−→ −−→ −−→ −−→ −−→ −−−→
20. Since P1 P2 = OP2 − OP1 , OP1 = OP2 − P1 P2 = 4, 7 − −5, −1 = 9, 8, and the initial point is (9, 8).
21. a(= −a), b(= − 14 a), c(= 52 a), e(= 2a), and f (= − 12 a) are parallel to a.
22. We want −3b = a, so c = −3(9) = −27.
23. 6, 15
24. 5, 2
25. a =
26. a =
√
√
√
4 + 4 = 2 2 ; (a) u =
1
√
2, 2
2 2
= √12 , √12 ; (b) −u = − √12 , − √12 9 + 16 = 5; (a) u = 15 −3, 4 = − 35 , 45 ; (b) −u = 35 , − 45 27. a = 5; (a) u = 15 0, −5 = 0, −1; (b) −u = 0, 1
√
√
√
√
28. a = 1 + 3 = 2; (a) u = 12 1, − 3 = 12 , − 23 ; (b) −u = − 12 , 23 √
1
5
29. a + b = 5, 12 = 25 + 144 = 13; u = 13
5, 12 = 13
, 12
13 √
√
30. 2a − 3b = −5, 4 = 25 + 16 = 41 ; u = √141 −5, 4 = − √541 , √441 31. a =
32. a =
√
9 + 49 =
1
4
+
1
4
=
√
58 ; b = 2( √158 )(3i + 7j) =
√6 i
58
+
√14 j
58
; b = 3( 1/1√2 )( 12 i − 12 j) =
√
3 2
2 i
−
√
3 2
2 j
√1
2
33. − 34 a = −3, −15/2
34. 5(a + b) = 50, 1 = 0, 5
35.
36.
37. x = −(a + b) = −a − b
38. x = 2(a − b) = 2a − 2b
340
7.1
39.
Vectors in 2-Space
40.
b = (−c) − a; (b + c) + a = 0; a + b + c = 0
From Problem 39, e + c + d = 0. But b = e − a
and e = a + b, so (a + b) + c + d = 0.
41. From 2i + 3j = k1 b + k2 c = k1 (i + j) + k2 (i − j) = (k1 + k2 )i + (k1 − k2 )j we obtain the system of equations
k1 + k2 = 2, k1 − k2 = 3. Solving, we find k1 = 52 and k2 = − 12 . Then a = 52 b − 12 c.
42. From 2i + 3j = k1 b + k2 c = k1 (−2i + 4j) + k2 (5i + 7j) = (−2k1 + 5k2 )i + (4k1 + 7k2 )j we obtain the system of
1
7
equations −2k1 + 5k2 = 2, 4k1 + 7k2 = 3. Solving, we find k1 = 34
and k2 = 17
.
43. From y = 12 x we see that the slope of the tangent line at (2, 2) is 1. A vector with slope 1 is i + j. A unit vector
√
is (i + j)/i + j = (i + j)/ 2 = √12 i + √12 j. Another unit vector tangent to the curve is − √12 i − √12 j.
44. From y = −2x + 3 we see that the slope of the tangent line at (0, 0) is 3. A vector with slope 3 is i + 3j. A
√
unit vector is (i + 3j)/i + 3j = (i + 3j)/ 10 = √110 i + √110 j. Another unit vector is − √110 i − √110 j.
45. (a) Since Ff = −Fg , Fg = Ff = µFn and tan θ = Fg /Fn = µFn /Fn = µ.
(b) θ = tan−1 0.6 ≈ 31◦
46. Since w + F1 + F2 = 0,
−200j + F1 cos 20◦ i + F1 sin 20◦ j − F2 cos 15◦ i + F2 sin 15◦ j = 0
or
(F1 cos 20◦ − F2 cos 15◦ )i + (F1 sin 20◦ + F2 sin 15◦ − 200)j = 0.
Thus, F1 cos 20◦ − F2 cos 15◦ = 0; F1 sin 20◦ + F2 sin 15◦ − 200 = 0. Solving this system for F1 and
F2 , we obtain
F1 =
200 cos 15◦
200 cos 15◦
200 cos 15◦
=
≈ 336.8 lb
=
sin 15◦ cos 20◦ + cos 15◦ sin 20◦
sin(15◦ + 20◦ )
sin 35◦
and
F2 =
sin 15◦
200 cos 20◦
200 cos 20◦
=
≈ 327.7 lb.
◦
◦
◦
cos 20 + cos 15 sin 20
sin 35◦
47. Since y/2a(L2 + y 2 )3/2 is an odd function on [−a, a], Fy = 0. Now, using the fact that L/(L2 + y 2 )3/2 is an
even function, we have
a
L dy
dy
L a
=
y = L tan θ, dy = L sec2 θ dθ
2
2
3/2
2
a 0 (L + y 2 )3/2
−a 2a(L + y )
−1
tan−1 a/L
L sec2 θ dθ
sec2 θ dθ
L tan a/L
1
=
=
a 0
La 0
sec3 θ
L3 (1 + tan2 θ)3/2
−1
tan−1 a/L
tan a/L
1
1
=
cos θ dθ =
sin θ La 0
La
0
1
1
a
√
=
= √
.
La L2 + a2
L L2 + a2
√
√
Then Fx = qQ/4π0 L L2 + a2 and F = (qQ/4π0 L L2 + a2 )i.
48. Place one corner of the parallelogram at the origin and let two adja−−→
−−→
cent sides be OP1 and OP2 . Let M be the midpoint of the diagonal
connecting P1 and P2 and N be the midpoint of the other diagonal.
−−→
−−→ −−→
−−→ −−→
Then OM = 12 (OP1 + OP2 ). Since OP1 + OP2 is the main diagonal of the parallelogram and N is its midpoint,
−−→ 1 −−→ −−→
−−→ −−→
ON = 2 (OP1 + OP2 ). Thus, OM = ON and the diagonals bisect each other.
341
7.1
Vectors in 2-Space
−−→ −−→ −→
−−→ −−→ −−→ −→
49. By Problem 39, AB + BC + CA = 0 and AD + DE + EC + CA = 0. From the first equation,
−−→ −−→
−→
−−→
−−→
−−→
−−→
AB + BC = −CA. Since D and E are midpoints, AD = 12 AB and EC = 12 BC. Then,
−→ −−→ 1 −−→ −→
1−
2 AB + DE + 2 BC + CA = 0 and
1 −→
−−→
−→ 1 −−→ −−→
−→ 1 −→
DE = −CA − (AB + BC) = −CA − (−CA) = − CA.
2
2
2
Thus, the line segment joining the midpoints D and E is parallel to the side AC and half its length.
−→
−−→
−−→
50. We have OA = 150 cos 20◦ i + 150 sin 20◦ j, AB = 200 cos 113◦ i + 200 sin 113◦ j, BC = 240 cos 190◦ i + 240 sin 190◦ j.
Then
r = (150 cos 20◦ + 200 cos 113◦ + 240 cos 190◦ )i + (150 sin 20◦ + 200 sin 113◦ + 240 sin 190◦ )j
≈ −173.55i + 193.73j
and r ≈ 260.09 miles.
EXERCISES 7.2
Vectors in 3-Space
1. – 6.
7. A plane perpendicular to the z-axis, 5 units above the xy-plane
8. A plane perpendicular to the x-axis, 1 unit in front of the yz-plane
9. A line perpendicular to the xy-plane at (2, 3, 0)
10. A single point located at (4, −1, 7)
11. (2, 0, 0), (2, 5, 0), (2, 0, 8), (2, 5, 8), (0, 5, 0), (0, 5, 8), (0, 0, 8), (0, 0, 0)
12.
13. (a) xy-plane: (−2, 5, 0), xz-plane: (−2, 0, 4), yz-plane: (0, 5, 4); (b) (−2, 5, −2)
342
7.2
Vectors in 3-Space
(c) Since the shortest distance between a point and a plane is a perpendicular line, the point in the plane x = 3
is (3, 5, 4).
14. We find planes that are parallel to coordinate planes: (a) z = −5; (b) x = 1 and y = −1; (c) z = 2
15. The union of the planes x = 0, y = 0, and z = 0
16. The origin (0, 0, 0)
17. The point (−1, 2, −3)
18. The union of the planes x = 2 and z = 8
19. The union of the planes z = 5 and z = −5
20. The line through the points (1, 1, 1), (−1, −1, −1), and the origin
√
21. d = (3 − 6)2 + (−1 − 4)2 + (2 − 8)2 = 70
√
22. d = (−1 − 0)2 + (−3 − 4)2 + (5 − 3)2 = 3 6
23. (a) 7; (b) d = (−3)2 + (−4)2 = 5
24. (a) 2; (b) d = (−6)2 + 22 + (−3)2 = 7
√
25. d(P1 , P2 ) = 32 + 62 + (−6)2 = 9; d(P1 , P3 ) = 22 + 12 + 22 = 3
√
d(P2 , P3 ) = (2 − 3)2 + (1 − 6)2 + (2 − (−6))2 = 90 ; The triangle is a right triangle.
√
√
√
√
26. d(P1 , P2 ) = 12 + 22 + 42 = 21 ; d(P1 , P3 ) = 32 + 22 + (2 2)2 = 21
√
√
d(P2 , P3 ) = (3 − 1)2 + (2 − 2)2 + (2 2 − 4)2 = 28 − 16 2
The triangle is an isosceles triangle.
√
27. d(P1 , P2 ) = (4 − 1)2 + (1 − 2)2 + (3 − 3)2 = 10
√
d(P1 , P3 ) = (4 − 1)2 + (6 − 2)2 + (4 − 3)2 = 26
√
d(P2 , P3 ) = (4 − 4)2 + (6 − 1)2 + (4 − 3)2 = 26 ; The triangle is an isosceles triangle.
28. d(P1 , P2 ) = (1 − 1)2 + (1 − 1)2 + (1 − (−1))2 = 2
d(P1 , P3 ) = (0 − 1)2 + (−1 − 1)2 + (1 − (−1))2 = 3
√
d(P2 , P3 ) = (0 − 1)2 + (−1 − 1)2 + (1 − 1)2 = 5 ; The triangle is a right triangle.
√
29. d(P1 , P2 ) = (−2 − 1)2 + (−2 − 2)2 + (−3 − 0)2 = 34
√
d(P1 , P3 ) = (7 − 1)2 + (10 − 2)2 + (6 − 0)2 = 2 34
√
d(P2 , P3 ) = (7 − (−2))2 + (10 − (−2))2 + (6 − (−3))2 = 3 34
Since d(P1 , P2 ) + d(P1 , P3 ) = d(P2 , P3 ), the points P1 , P2 , and P3 are collinear.
√
30. d(P1 , P2 ) = (1 − 2)2 + (4 − 3)2 + (4 − 2)2 = 6
√
d(P1 , P3 ) = (5 − 2)2 + (0 − 3)2 + (−4 − 2)2 = 3 6
√
d(P2 , P3 ) = (5 − 1)2 + (0 − 4)2 + (−4 − 4)2 = 4 6
Since d(P1 , P2 ) + d(P1 , P3 ) = d(P2 , P3 ), the points P1 , P2 , and P3 are collinear.
√
(2 − x)2 + (1 − 2)2 + (1 − 3)2 = 21 =⇒ x2 − 4x + 9 = 21 =⇒ x2 − 4x + 4 = 16
31.
=⇒ (x − 2)2 = 16 =⇒ x = 2 ± 4 or x = 6, −2
343
7.2
Vectors in 3-Space
(0 − x)2 + (3 − x)2 + (5 − 1)2 = 5 =⇒ 2x2 − 6x + 25 = 25 =⇒ x2 − 3x = 0 =⇒ x = 0, 3
1 + 7 3 + (−2) 1/2 + 5/2
33.
,
,
= (4, 1/2, 3/2)
2
2
2
0 + 4 5 + 1 −8 + (−6)
34.
,
,
= (2, 3, −7)
2
2
2
32.
35. (x1 + 2)/2 = −1, x1 = −4;
(y1 + 3)/2 = −4, y1 = −11;
(z1 + 6)/2 = 8, z1 = 10
The coordinates of P1 are (−4, −11, 10).
36. (−3 + (−5))/2 = x3 = −4;
(4 + 8)/2 = y3 = 6;
(1 + 3)/2 = z3 = 2.
The coordinates of P3 are (−4, 6, 2).
−3 + (−4) 4 + 6 1 + 2
(a)
,
,
= (−7/2, 5, 3/2)
2
2
2
−4 + (−5) 6 + 8 2 + 3
(b)
,
,
= (−9/2, 7, 5/2)
2
2
2
−−−→
37. P1 P2 = −3, −6, 1
−−−→
38. P1 P2 = 8, −5/2, 8
−−−→
39. P1 P2 = 2, 1, 1
−−−→
40. P1 P2 −3, −3, 7
41. a + (b + c) = 2, 4, 12
42. 2a − (b − c) = 2, −6, 4 − −3, −5, −8 = 5, −1, 12
43. b + 2(a − 3c) = −1, 1, 1 + 2−5, −21, −25 = −11, −41, −49
44. 4(a + 2c) − 6b = 45, 9, 20 − −6, 6, 6 = 26, 30, 74
√
√
45. a + c = 3, 3, 11 = 9 + 9 + 121 = 139
√
√
√
46. c2b = ( 4 + 36 + 81 )(2)( 1 + 1 + 1 ) = 22 3
a b = 1 a + 5 1 b = 1 + 5 = 6
47. + 5
a
|b a
b
√
√
√
√
√
√
√
√
48. ba + ab = 1 + 1 + 1 1, −3, 2 + 1 + 9 + 4 −1, 1, 1 = 3 , −3 3 , 2 3 + − 14 , 14 , 14 √
√
√
√
√
√
= 3 − 14 , −3 3 + 14 , 2 3 + 14 49. a =
50. a =
√
√
100 + 25 + 100 = 15; u = −
1+9+4=
√
1
10, −5, 10 = −2/3, 1/3, −2/3
15
1
1
3
2
14 ; u = √ (i − 3j + 2k) = √ i − √ j + √ k
14
14
14
14
51. b = 4a = 4i − 4j + 4k
√
3
3 1
1 1
52. a = 36 + 9 + 4 = 7; b = −
−6, 3, −2 = , − , 2 7
7
14 7
53.
344
7.3 Dot Product
EXERCISES 7.3
Dot Product
√
1. a · b = 10(5) cos(π/4) = 25 2
√
2. a · b = 6(12) cos(π/6) = 36 3
3. a · b = 2(−1) + (−3)2 + 4(5) = 12
4. b · c = (−1)3 + 2(6) + 5(−1) = 4
5. a · c = 2(3) + (−3)6 + 4(−1) = −16
6. a · (b + c) = 2(2) + (−3)8 + 4(4) = −4
7. a · (4b) = 2(−4) + (−3)8 + 4(20) = 48
8. b · (a − c) = (−1)(−1) + 2(−9) + 5(5) = 8
9. a · a = 22 + (−3)2 + 42 = 29
10. (2b) · (3c) = (−2)9 + 4(18) + 10(−3) = 24
11. a · (a + b + c) = 2(4) + (−3)5 + 4(8) = 25
12. (2a) · (a − 2b) = 4(4) + (−6)(−7) + 8(−6) = 10
a·b
12
2(−1) + (−3)2 + 4(5)
13.
−1, 2, 5 =
b=
−1, 2, 5 = −2/5, 4/5, 2
b·b
(−1)2 + 22 + 52
30
14. (c · b)a = [3(−1) + 6(2) + (−1)5]2, −3, 4 = 42, −3, 4 = 8, −12, 16
15. a and f, b and e, c and d
16. (a) a · b = 2 · 3 + (−c)2 + 3(4) = 0 =⇒ c = 9
(b) a · b = c(−3) + 12 (4) + c2 = c2 − 3c + 2 = (c − 2)(c − 1) = 0 =⇒ c = 1, 2
17. Solving the system of equations 3x1 + y1 − 1 = 0, −3x1 + 2y1 + 2 = 0 gives x1 = 4/9 and y1 = −1/3. Thus,
v = 4/9, −1/3, 1.
18. If a and b represent adjacent sides of the rhombus, then a = b, the diagonals of the rhombus are a + b
and a − b, and
(a + b) · (a − b) = a · a − a · b + b · a − b · b = a · a − b · b = a2 − b2 = 0.
Thus, the diagonals are perpendicular.
19. Since
c·a=
a·b
a·b
a·b
b−
a ·a=b·a−
(a · a) = b · a −
a2 = b · a − a · b = 0,
a2
a2
a2
the vectors c and a are orthogonal.
√
√
20. a · b = 1(1) + c(1) = c + 1; a = 1 + c2 , b = 2
1
c+1
√
cos 45◦ = √ = √
1 + c2 = c + 1 =⇒ 1 + c2 = c2 + 2c + 1 =⇒ c = 0
=⇒
2
2
1+c 2
√
√
21. a · b = 3(2) + (−1)2 = 4; a = 10 , b = 2 2
4
1
1
√ =√
cos θ = √
=⇒ θ = cos−1 √ ≈ 1.11 rad ≈ 63.43◦
( 10)(2 2)
5
5
√
22. a · b = 2(−3) + 1(−4) = −10; a = 5 , b = 5
345
7.3
Dot Product
−10
2
cos θ = √
= −√
( 5 )5
5
√
=⇒ θ = cos−1 (−2/ 5 ) ≈ 2.68 rad ≈ 153.43◦
√
√
23. a · b = 2(−1) + 4(−1) + 0(4) = −6; a = 2 5 , b = 3 2
√
−6
1
√ = −√
cos θ = √
=⇒ θ = cos−1 (−1/ 10 ) ≈ 1.89 rad ≈ 108.43◦
(2 5)(3 2)
10
√
√
24. a · b = 12 (2) + 12 (−4) + 32 (6) = 8; a = 11/2, b = 2 14
√
8
8
√
cos θ = √
=√
=⇒ θ = cos−1 (8/ 154 ) ≈ 0.87 rad ≈ 49.86◦
( 11/2)(2 14 )
154
√
√
√
√
25. a = 14 ; cos α = 1/ 14 , α ≈ 74.50◦ ; cos β = 2/ 14 , β ≈ 57.69◦ ; cos γ = 3/ 14 , γ ≈ 36.70◦
26. a = 9; cos α = 2/3, α ≈ 48.19◦ ; cos β = 2/3, β ≈ 48.19◦ ; cos γ = −1/3, γ ≈ 109.47◦
√
27. a = 2; cos α = 1/2, α = 60◦ ; cos β = 0, β = 90◦ ; cos γ = − 3/2, γ = 150◦
√
√
√
√
28. a = 78 ; cos α = 5/ 78 , α ≈ 55.52◦ ; cos β = 7/ 78 , β ≈ 37.57◦ ; cos γ = 2/ 78 , γ ≈ 76.91◦
−−→
−−→
−−→
29. Let θ be the angle between AD and AB and a be the length of an edge of the cube. Then AD = ai + aj + ak,
−−→
AB = ai and
−−→ −−→
AD · AB
a2
1
√ =√
cos θ = −−→ −−→ = √
2
2
3
AD AB
3a a
−−→
−→
−→
◦
so θ ≈ 0.955317 radian or 54.7356 . Letting φ be the angle between AD and AC and noting that AC = ai + aj
we have
−−→ −→
AD · AC
2
a2 + a2
√
cos φ = −−→ −→ = √
=
2
2
3
AD AC
3a 2a
so φ ≈ 0.61548 radian or 35.2644◦ .
30. If a and b are orthogonal, then a · b = 0 and
a1 b1
a2 b2
a3 b3
+
+
a b a b a b
1
1
=
(a1 b1 + a2 b2 + a3 b3 ) =
(a · b) = 0.
a b
a b
√
√
√
√
31. a = 5, 7, 4; a = 3 10 ; cos α = 5/3 10 , α ≈ 58.19◦ ; cos β = 7/3 10 , β ≈ 42.45◦ ; cos γ = 4/3 10 ,
γ ≈ 65.06◦
cos α1 cos α2 + cos β1 cos β2 + cos γ1 cos γ2 =
32. We want cos α = cos β = cos γ or a1 = a2 = a3 . Letting a1 = a2 = a3 = 1 we obtain the vector i + j + k. A unit
vector in the same direction is √13 i + √13 j + √13 k.
33. compb a = a · b/b = 1, −1, 3 · 2, 6, 3/7 = 5/7
√
√
34. compa b = b · a/a = 2, 6, 3 · 1, −1, 3/ 11 = 5/ 11
√
√
35. b − a = 1, 7, 0; compa (b − a) = (b − a) · a/a = 1, 7, 0 · 1, −1, 3/ 11 = −6/ 11
36. a + b = 3, 5, 6; 2b = 4, 12, 6; comp2b (a + b) · 2b/|2b| = 3, 5, 6 · 4, 12, 6/14 = 54/7
√
√
√
−−→
−−→
−−→ −−→
37. OP = 3i + 10j; OP = 109 ; comp−
→ a = a · OP /OP = (4i + 6j) · (3i + 10j)/ 109 = 72/ 109
OP
√
√
−−→
−−→
−−→ −−→
38. OP = 1, −1, 1; OP = 3 ; comp−
→ a = a · OP /OP = 2, 1, −1 · 1, −1, 1/ 3 = 0
OP
39. compb a = a · b/b = (−5i + 5j) · (−3i + 4j)/5 = 7
28
projb a = (compb a)b/b = 7(−3i + 4j)/5 = − 21
5 i+ 5 j
√
√
40. compb a = a · b/b = (4i + 2j) · (−3i + j)/ 10 = − 10
√
√
projb a = (compb a)b/b = − 10(−3i + j)/ 10 = 3i − j
346
7.3 Dot Product
41. compb a = a · b/b = (−i − 2j + 7k) · (6i − 3j − 2k)/7 = −2
6
4
projb a = (compb a)b/b = −2(6i − 3j − 2k)/7 = − 12
7 i + 7j + 7k
42. compb a = a · b/b = 1, 1, 1 · −2, 2, −1/3 = −1/3
projb a = (compb a)b/b = − 13 −2, 2, −1/3 = 2/9, −2/9, 1/9
43. a + b = 3i + 4j; a + b = 5; comp(a+b) a = a · (a + b)/a + b = (4i + 3j) · (3i + 4j)/5 = 24/5
72
96
proj(a+b) a = (comp(a+b) a)(a + b)/a + b = 24
5 (3i + 4j)/5 = 25 i + 25 j
√
√
√
44. a − b = 5i + 2j; a − b = 29 ; comp(a−b) b = b · (a − b)/a − b = (−i + j) · (5i + 2j)/ 29 = −3/ 29
√
6
proj(a−b) b = (comp(a−b) b)(a − b)/a − b = − √329 (5i + 2j)/ 29 = − 15
29 i − 29 j
45. We identify F = 20, θ = 60◦ and d = 100. Then W = F d cos θ = 20(100)( 12 ) = 1000 ft-lb.
46. We identify d = −i + 3j + 8k. Then W = F · d = 4, 3, 5 · −1, 3, 8 = 45 N-m.
47. (a) Since w and d are orthogonal, W = w · d = 0.
√
(b) We identify θ = 0◦ . Then W = F d cos θ = 30( 42 + 32 ) = 150 N-m.
48. Using d = 6i + 2j and F = 3( 35 i + 45 j), W = F · d = 95 , 12
5 · 6, 2 =
78
5
ft-lb.
49. Let a and b be vectors from the center of the carbon atom to the centers of two distinct hydrogen atoms. The
distance between two hydrogen atoms is then
√
(b − a) · (b − a) = b · b − 2a · b + a · a
= b2 + a2 − 2a b cos θ = (1.1)2 + (1.1)2 − 2(1.1)(1.1) cos 109.5◦
= 1.21 + 1.21 − 2.42(−0.333807) ≈ 1.80 angstroms.
b − a =
50. Using the fact that | cos θ| ≤ 1, we have |a · b| = a b| cos θ| = a b| cos θ| ≤ a b.
51. a + b2 = (a + b) · (a + b) = a · a + 2a · b + b · b = a2 + 2a · b + b2
≤ a2 + 2|a · b| + b2
since x ≤ |x|
≤ a2 + 2a b + b2 = (a + b)2
by Problem 50
Thus, since a + b and a + b are positive, a + b ≤ a + b.
52. Let P1 (x1 , y1 ) and P2 (x2 , y2 ) be distinct points on the line ax + by = −c. Then
−−−→
n · P1 P2 = a, b · x2 − x1 , y2 − y1 = ax2 − ax1 + by2 − by1
= (ax2 + by2 ) − (ax1 + by1 ) = −c − (−c) = 0,
and the vectors are perpendicular. Thus, n is perpendicular to the line.
−−−→
53. Let θ be the angle between n and P2 P1 . Then
−−−→
|n · P2 P1 |
|ax1 − ax2 + by1 − by2 |
|a, b · x1 − x2 , y1 − y2 |
−−−→
√
√
d = P1 P2 | cos θ| =
=
=
n
a2 + b2
a2 + b2
|ax1 + by1 − (ax2 + by2 )|
|ax1 + by1 − (−c)|
|ax1 + by1 + c|
√
√
√
=
=
=
.
2
2
2
2
a +b
a +b
a2 + b2
347
7.4
Product
7.3 Cross
Dot Product
EXERCISES 7.4
Cross Product
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
i
j k 1 0
1 −1 −1 0 k = −5i − 5j + 3k
a × b = 1 −1 0 = i−
j+
3 5
0 5
0 3 0 3 5 i j k 1 0 2 0 i − j + 2 1 k = −i + 2j − 4k
a × b = 2 1 0 = 0 −1
4 −1
4 0
4 0 −1 i
j k 1 −3 1 1
−3 1 k = −12, −2, 6
j+
i−
a × b = 1 −3 1 = 2 0 2 4
0 4
2 0 4 i
j k 1 1
1 1
1 1 a×b= 1 1 1=
i − −5 3 j + −5 2 k = 1, −8, 7
2
3
−5 2 3 i
j
k 2
2 −1 2 −1 2 j + a × b = 2 −1 2 = i
−
−1 −1 −1 3 k = −5i + 5k
3 −1 −1 3 −1 i j k 1 −5 i − 4 −5 j + 4 1 k = 14i − 6j + 10k
a × b = 4 1 −5 = 3 −1
2 −1
2 3
2 3 −1 i
j
k 1/2 1/2 1/2 0 0 1/2 k = −3, 2, 3
a × b = 1/2 0 1/2 = i − 4
j + 4
6
0
0
6
4
6
0 i
j k 0 0
0 5 5 0 a × b = 0 5 0 = i − 2 4 j + 2 −3 k = 20, 0, −10
−3
4
2 −3 4 i
j
k 2 −4 2 2 −4 j + 2
a×b= 2
2 −4 = i
−
−3 6 −3 −3 k = 0, 0, 0
−3 6 −3 −3 6 i
j
k 8 −6 8 1 1 −6 a × b = 8 1 −6 = i − 1 10 j + 1 −2 k = −2, −86, −17
−2
10
1 −2 10 −−−→
−−−→
11. P1 P2 = (−2, 2, −4); P1 P3 = (−3, 1, 1)
i
j k −2
−−−→ −−−→ 2 −4 P1 P2 × P1 P3 = −2 2 −4 = i − 1 1
−3
−3 1 1 −2
−4 j + 1
−3
348
2 k = 6i + 14j + 4k
1
7.4 Cross Product
i j k 1 1 0
−−−→
−−−→
−−−→ −−−→ i − 12. P1 P2 = (0, 1, 1); P1 P3 = (1, 2, 2); P1 P2 × P1 P3 = 0 1 1 = 1
2 2
1 2 2 i j k 7 −4 2 −4 2 7
13. a × b = 2 7 −4 = i − 1 −1 j + 1 1 k = −3i − 2j − 5k
1
−1
1 1 −1 is perpendicular to
i
j
14. a × b = −1 −2
4 −1
both a and b.
k −1
−2 4 i − 4=
−1 0
4
0
−1
4 j + 0
4
−2 k = 4, 16, 9
−1 is perpendicular to both a and b.
i
j
k 5
−2 1 15. a × b = 5 −2 1 = i
−
2
0
−7
2 0 −7 5
1 j
+
2
−7
−2 k = 14, 37, 4
0 0
1
j + 1
2
1
k = j − k
2
a · (a × b) = 5, −2, −1 · 14, 37, 4 = 70 − 74 + 4 = 0; b · (a × b) = 2, 0, −7 · 14, 37, 4 = 28 + 0 − 28 = 0
i
j
k 1/2 0 −1/4 0 j + 1/2 −1/4 k = − 3 i − 3j − 1 k
16. a × b = 1/2 −1/4 0 = i
−
2
2
−2
6
2
6
2
−2 2
−2
6
a · (a × b) = ( 12 i − 14 j) · (− 32 i − 3j − 12 k) = − 34 +
b · (a × b) = (2i − 2j + 6k) ·
− 3j −
i j k 1 1 2
i − 17. (a) b × c = 2 1 1 = 1 1
3
3 1 1 i
j
k −1
a × (b × c) = 1 −1 2 = 1
0 1 −1 (− 32 i
1
2 k)
3
4
+0=0
= −3 + 6 − 3 = 0
2
1 j + 1
3
1
2 i − 0
−1
1 k=j−k
1
1
2 j + 0
−1
−1 k = −i + j + k
1 (b) a · c = (i − j + 2k) · (3i + j + k) = 4; (a · c)b = 4(2i + j + k) = 8i + 4j + 4k
a · b = (i − j + 2k) · (2i + j + k) = 3; (a · b)c = 3(3i + j + k) = 9i + 3j + 3k
a × (b × c) = (a · c)b − (a · b)c = (8i + 4j + 4k) − (9i + 3j + 3k) = −i + j + k
i
j k 1 −1 1 2
2 −1 k = 21i − 7j + 7k
i−
j+
18. (a) b × c = 1 2 −1 = 5 8 −1 8 −1 5 −1 5 8 i
j
k 3 −4 0 0 −4 j + 3
a × (b × c) = 3
0 −4 = i
−
21 7 21 −7 k = −28i − 105j − 21k
−7 7 21 −7 7 (b) a · c = (3i − 4k) · (−i + 5j + 8k) = −35; (a · c)b = −35(i + 2j − k) = −35i − 70j + 35k
a · b = (3i − 4k) · (i + 2j − k) = 7; (a · b)c = 7(−i + 5j + 8k) = −7i + 35j + 56k
a × (b × c) = (a · c)b − (a · b)c = (−35i − 70j + 35k) − (−7i + 35j + 56k) = −28i − 105j − 21k
19. (2i) × j = 2(i × j) = 2k
20. i × (−3k) = −3(i × k) = −3(−j) = 3j
349
7.4
Cross Product
21. k × (2i − j) = k × (2i) + k × (−j) = 2(k × i) − (k × j) = 2j − (−i) = i + 2j
22. i × (j × k) = i × i = 0
23. [(2k) × (3j)] × (4j) = [2 · 3(k × j) × (4j)] = 6(−i) × 4j = (−6)(4)(i × j) = −24k
24. (2i − j + 5k) × i = (2i × i) + (−j × i) + (5k × i) = 2(i × i) + (i × j) + 5(k × i) = 5j + k
25. (i + j) × (i + 5k) = [(i + j) × i] + [(i + j) × 5k] = (i × i) + (j × i) + (i × 5k) + (j × 5k)
= −k + 5(−j) + 5i = 5i − 5j − k
26. i × k − 2(j × i) = −j − 2(−k) = −j + 2k
27. k · (j × k) = k · i = 0
28. i · [j × (−k)] = i · [−(j × k)] = i · (−i) = −(i · i) = −1
√
29. 4j − 5(i × j) = 4j − 5k = 41
30. (i × j) · (3j × i) = k · (−3k) = −3(k · k) = −3
31. i × (i × j) = i × k = −j
32. (i × j) × i = k × i = j
33. (i × i) × j = 0 × j = 0
34. (i · i)(i × j) = 1(k) = k
35. 2j · [i × (j − 3k)] = 2j · [(i × j) + (i × (−3k)] = 2j · [k + 3(k × i)] = 2j · (k + 3j) = 2j · k + 2j · 3j
= 2(j · k) + 6(j · j) = 2(0) + 6(1) = 6
36. (i × k) × (j × i) = (−j) × (−k) = (−1)(−1)(j × k) = j × k = i
37. a × (3b) = 3(a × b) = 3(4i − 3j + 6k) = 12i − 9j + 18k
38. b × a = −a × b = −(a × b) = −4i + 3j − 6k
39. (−a) × b = −(a × b) = −4i + 3j − 6k
√
40. |a × b| = 42 + (−3)2 + 62 = 61
i
j
k 4
−3 6 41. (a × b) × c = 4 −3 6 = i
−
2
4 −1
2 4 −1 4
6 j
+
2
−1
−3 k = −21i + 16j + 22k
−4 42. (a × b) · c = 4(2) + (−3)4 + 6(−1) = −10
43. a · (b × c) = (a × b) · c = 4(2) + (−3)4 + 6(−1) = −10
44. (4a) · (b × c) = (4a × b) · c = 4(a × b) · c = 16(2) + (−12)4 + 24(−1) = −40
−−→
−→
45. (a) Let A = (1, 3, 0), B = (2, 0, 0), C = (0, 0, 4), and D = (1, −3, 4). Then AB = i − 3j, AC = −i − 3j + 4k,
−−→
−−→
−−→ −−→
−→ −−→
CD = i − 3j, and BD = −i − 3j + 4k. Since AB = CD and AC = BD, the quadrilateral is a parallelogram.
(b) Computing
i
−−→ −→ AB × AC = 1
−1
we find that the area is − 12i − 4j − 6k =
√
j k −3 0 = −12i − 4j − 6k
−3 4 144 + 16 + 36 = 14.
−−→
−→
46. (a) Let A = (3, 4, 1), B = (−1, 4, 2), C = (2, 0, 2) and D = (−2, 0, 3). Then AB = −4i + k, AC = −i − 4j + k,
−−→
−−→
−−→ −−→
−→ −−→
CD = −4i+k, and BD = −i−4j+k. Since AB = CD and AC = BD, the quadrilateral is a parallelogram.
(b) Computing
i
−−→ −→ AB × AC = −4
−1
j k 0 1 = 4i + 3j + 16k
−4 1 350
7.4 Cross Product
√
√
we find that the area is 4i + 3j + 16k = 16 + 9 + 256 = 281 ≈ 16.76.
−−−→
−−−→
47. P1 P2 = j; P2 P3 = −j + k
i
j k 0 0
0 1 −−−→ −−−→ 1 0 1
j + P1 P2 × P2 P3 = 0 1 0 = i
−
0 1
0 −1 k = i; A = 2 i =
−1 1 0 −1 1 −−−→
−−−→
48. P1 P2 = j + 2k; P2 P3 = 2i + j − 2k
i j k 1 2 0 2 0 1
−−−→ −−−→ i − j + P1 P2 × P2 P3 = 0 1 2 = 2 −2 2 1 k = −4i + 4j − 2k
1 −2 2 1 −2 A = 12 − 4i + 4j − 2k = 3 sq. units
−−−→
−−−→
49. P1 P2 = −3j − k; P2 P3 = −2i − k
i
j
k −−−→ −−−→ −3
P1 P2 × P2 P3 = 0 −3 −1 = 0
−2 0 −1 0
−1 i
−
−2
−1
A = 12 3i + 2j − 6k = 72 sq. units
−−−→
−−−→
50. P1 P2 = −i + 3k; P2 P3 = 2i + 4j − k
i
j k −1
−−−→ −−−→ 0 3 P1 P2 × P2 P3 = −1 0 3 = i
−
2
4 −1
2 4 −1 0
−1 j
+
−2
−1
−1
3 j + 2
−1
1
2
sq. unit
−3 k = 3i + 2j − 6k
0 0
k = −12i + 5j − 4k
4
√
A = 12 − 12i + 5j − 4k = 185
sq. units
2
i
j k −1 0 −1
4 0 51. b × c = −1 4 0 = i−
j + 2 2
2 2
2
2 2 2
4
k = 8i + 2j − 10k
2
v = |a · (b × c)| = |(i + j) · (8i + 2j − 10k)| = |8 + 2 + 0| = 10 cu. units
i j k 4 1 1 1
1 4
52. b × c = 1 4 1 = i − 1 5 j + 1 1 k = 19i − 4j − 3k
1
5
1 1 5 v = |a · (b × c)| = |(3i + j + k) · (19i − 4j − 3k)| = |57 − 4 − 3| = 50 cu. units
i
j
k −2 −6 6 −6 j + −2 6 k = 21i − 14j − 21k
53. b × c = −2 6 −6 = i
−
3 2/2
5/2 1/2
5/2 3 5/2 3 1/2 a · (b × c) = (4i + 6j) · (21i − 14j − 21k) = 84 − 84 + 0 = 0. The vectors are coplanar.
−−−→
−−−→
−−−→
54. The four points will be coplanar if the three vectors P1 P2 = 3, −1, −1, P2 P3 = −3, −5, 13, and P3 P4 =
−8, 7, −6 are coplanar.
i
j
k −3 13 −3 −5 −−−→ −−−→ −5 13 P2 P3 × P3 P4 = −3 −5 13 = i − −8 −6 j + −8 7 k = −61, −122, −61
7
−6
−8 7 −6 −−−→ −−−→ −−−→
P1 P2 · (P2 P3 × P3 P4 ) = 3, −1, −1 · −61, −122, −61 = −183 + 122 + 61 = 0
The four points are coplanar.
55. (a) Since θ = 90◦ , a × b = a b | sin 90◦ | = 6.4(5) = 32.
351
7.4
Cross Product
(b) The direction of a × b is into the fourth quadrant of the xy-plane or to the left of the plane determined by
a and b as shown in Figure 7.54 in the text. It makes an angle of 30◦ with the positive x-axis.
√
√
(c) We identify n = ( 3 i − j)/2. Then a × b = 32n = 16 3 i − 16j.
√
√ √
56. Using Definition 7.4, a × b = 27 (8) sin 120◦ n = 24 3 ( 3/2)n = 36n. By the right-hand rule, n = j or
n = −j. Thus, a × b = 36j or −36j.
57. (a) We note first that a × b = k, b × c = 12 (i − k), c × a = 12 (j − k), a · (b × c) =
c · (a × b) = 12 . Then
A=
1
2 (i
− k)
1
2
= i − k,
B=
1
2 (j
− k)
1
2
= j − k,
and C =
k
1
2
1
2
, b · (c × a) =
1
2
, and
= 2k.
(b) We need to compute A · (B × C). Using formula (10) in the text we have
(c × a) × (a × b)
[(c × a) · b]a − [(c × a) · a]b
=
[b · (c × a)][c · (a × b)]
[b · (c × a)][c · (a × b)]
a
=
since (c × a) · a = 0.
c · (a × b)
B×C=
Then
A · (B × C) =
b×c
a
1
·
=
a · (b × c) c · (a × b)
c · (a × b)
and the volume of the unit cell of the reciprocal latrice is the reciprocal of the volume of the unit cell of
the original lattice.
a2
a1
a1
a3 a3 a2 k
58. a × (b + c) = i−
j+
b2 + c2 b3 + c3 b1 + c1 b3 + c3 b1 + c1 b2 + c2 = (a2 b3 − a3 b2 )i + (a2 c3 − a3 c2 )i − [(a1 b3 − a3 b1 )j + (a1 c3 − a3 c1 )j] + (a1 b2 − a2 b1 )k + (a1 c2 − a2 c1 )k
= (a2 b3 − a3 b2 )i − (a1 b3 − a3 b1 )j + (a1 b2 − a2 b1 )k + (a2 c3 − a3 c2 )i − (a1 c3 − a3 c1 )j + (a1 c2 − a2 c1 )k
=a×b+a×c
59. b × c = (b2 c3 − b3 c2 )i − (b1 c3 − b3 c1 )j + (b1 c2 − b2 c1 )k
a × (b × c) = [a2 (b1 c2 − b2 c1 ) + a3 (b1 c3 − b3 c1 )]i − [a1 (b1 c2 − b2 c1 ) − a3 (b2 c3 − b3 c2 )]j
+ [−a1 (b1 c3 − b3 c1 ) − a2 (b2 c3 − b3 c2 )]k
= (a2 b1 c2 − a2 b2 c1 + a3 b1 c3 − a3 b3 c1 )i − (a1 b1 c2 − a1 b2 c1 − a3 b2 c3 + a3 b3 c2 )j
− (a1 b1 c3 − a1 b3 c1 + a2 b2 c3 − a2 b3 c2 )k
(a · c)b − (a · b)c = (a1 c1 + a2 c2 + a3 c3 )(b1 i + b2 j + b3 k) − (a1 b1 + a2 b2 + a3 b3 )(c1 i + c2 j + c3 k)
= (a2 b1 c2 − a2 b2 c1 + a3 b1 c3 − a3 b3 c1 )i − (a1 b1 c2 − a1 b2 c1 − a3 b2 c3 + a3 b3 c2 )j
− (a1 b1 c3 − a1 b3 c1 + a2 b2 c3 − a2 b3 c2 )k
60. The statement is false since i × (i × j) = i × k = −j and (i × i) × j = 0 × j = 0.
61. Using equation 9 in the text,
a1
a · (b × c) = b1
c1
a2
b2
c2
a3 b3 c3 and
c1
(a × b) · c = c · (a × b) = a1
b1
c2
a2
b2
c3 a3 .
b3 Expanding these determinants out we obtain a · (b × c) = a1 b2 c3 + a2 b3 c1 + a3 b1 c2 − a3 b2 c1 − a1 b3 c2 − a2 b1 c3
and c · (a × b) = a2 b3 c1 + a3 b1 c2 + a1 b2 c3 − a2 b1 c3 − a3 b2 c1 − a1 b3 c2 . These are equal so a · (b × c) = (a × b) · c.
352
7.5 Lines and Planes in 3-Space
62. a × (b × c) + b × (c × a) + c × (a × b)
= (a · c)b − (a · b)c + (b · a)c − (b · c)a + (c · b)a − (c · a)b
= [(a · c)b − (c · a)b] + [(b · a)c − (a · b)c] + [(c · b)a − (b · c)a] = 0
63. Since
a × b2 = (a2 b3 − a3 b2 )2 + (a1 b3 − a3 b1 )2 + (a1 b2 − a2 b1 )2
= a22 b23 − 2a2 b3 a3 b2 + a23 b22 + a21 b23 − 2a1 b3 a3 b1 + a23 b21 + a21 b22 − 2a1 b2 a2 b1 + a22 b21
and
a2 b2 − (a · b)2 = (a21 + a22 + a23 )(b21 + b22 + b23 ) − (a1 b1 + a2 b2 + a3 b3 )2
= a21 a22 + a21 b22 + a21 b23 + a22 b21 + a22 b22 + a22 b23 + a23 b21 + a23 b22 + a23 b23
− a21 b21 − a22 b22 − a23 b23 − 2a1 b1 a2 b2 − 2a1 b1 a3 b3 − 2a2 b2 a3 b3
= a21 b22 + a21 b23 + a22 b21 + a22 b23 + a23 b21 + a23 b22 − 2a1 a2 b1 b2 − 2a1 a3 b1 b3 − 2a2 a3 b2 b3
we see that a × b2 = a2 b2 − (a · b)2 .
64. No. For example i × (i + j) = i × j by the distributive law (iii) in the text, and the fact that i × i = 0. But i + j
does not equal j.
65. By the distributive law (iii) in the text:
(a + b) × (a − b) = (a + b) × a − (a + b) × b = a × a + b × a − a × b − b × b = 2b × a
since a × a = 0, b × b = 0, and −a × b = b × a.
EXERCISES 7.5
Lines and Planes in 3-Space
−−→
−−→
The equation of a line through P1 and P2 in 3-space with r1 = OP1 and r2 = OP2 can be expressed as r = r1 + t(ka)
or r = r2 + t(ka) where a = r2 − r1 and k is any non-zero scalar. Thus, the form of the equation of a line is not
unique. (See the alternate solution to Problem 1.)
1. a = 1 − 3, 2 − 5, 1 − (−2) = −2, −3, 3; x, y, z = 1, 2, 1 + t−2, −3, 3
Alternate Solution: a = 3 − 1, 5 − 2, −2 − 1 = 2, 3, −3; x, y, z = 3, 5, −2 + t2, 3, −3
2. a = 0 − (−2), 4 − 6, 5 − 3 = 2, −2, 2; x, y, z = 0, 4, 5 + t2, −2, 2
3. a = 1/2 − (−3/2), −1/2 − 5/2, 1 − (−1/2) = 2, −3, 3/2; x, y, z = 1/2, −1/2, 1 + t2, −3, 3/2
4. a = 10 − 5, 2 − (−3), −10 − 5 = 5, 5, −15; x, y, z = 10, 2, −10 + t5, 5, −15
5. a = 1 − (−4), 1 − 1, −1 − (−1) = 5, 0, 0; x, y, z = 1, 1, −1 + t5, 0, 0
6. a = 3 − 5/2, 2 − 1, 1 − (−2) = 1/2, 1, 3; x, y, z = 3, 2, 1 + t1/2, 1, 3
7. a = 2 − 6, 3 − (−1), 5 − 8 = −4, 4, −3; x = 2 − 4t, y = 3 + 4t, z = 5 − 3t
8. a = 2 − 0, 0 − 4, 0 − 9 = 2, −4, −9; x = 2 + 2t, y = −4t, z = −9t
9. a = 1 − 3, 0 − (−2), 0 − (−7) = −2, 2, 7; x = 1 − 2t, y = 2t, z = 7t
10. a = 0 − (−2), 0 − 4, 5 − 0 = 2, −4, 5; x = 2t, y = −4t, z = 5 + 5t
353
7.5
Lines and Planes in 3-Space
11. a = 4 − (−6), 1/2 − (−1/4), 1/3 − 1/6 = 10, 3/4, 1/6; x = 4 + 10t, y =
1 3
1 1
+ t, z = + t
2 4
3 6
12. a = −3 − 4, 7 − (−8), 9 − (−1) = −7, 15, 10; x = −3 − 7t, y = 7 + 15t, z = 9 + 10t
13. a1 = 10 − 1 = 9, a2 = 14 − 4 = 10, a3 = −2 − (−9) = 7;
x − 10
y − 14
z+2
=
=
9
10
7
14. a1 = 1 − 2/3 = 1/3, a2 = 3 − 0 = 3, a3 = 1/4 − (−1/4) = 1/2;
15. a1 = −7 − 4 = −11, a2 = 2 − 2 = 0, a3 = 5 − 1 = 4;
x−1
y−3
z − 1/4
=
=
1/3
3
1/2
z−5
x+7
=
, y=2
−11
4
16. a1 = 1 − (−5) = 6, a2 = 1 − (−2) = 3, a3 = 2 − (−4) = 6;
x−1
y−1
z−2
=
=
6
3
6
17. a1 = 5 − 5 = 0, a2 = 10 − 1 = 9, a3 = −2 − (−14) = 12; x = 5,
z+2
y − 10
=
9
12
18. a1 = 5/6 − 1/3 = 1/2; a2 = −1/4 − 3/8 = −5/8; a3 = 1/5 − 1/10 = 1/10
x − 5/6
y + 1/4
z − 1/5
=
=
1/2
−5/8
1/10
19. parametric: x = 4 + 3t, y = 6 + t/2, z = −7 − 3t/2; symmetric:
x−4
y−6
z+7
=
=
3
1/2
−3/2
x−1
y−8
=
, z = −2
−7
−8
y
z
x
= =
21. parametric: x = 5t, y = 9t, z = 4t; symmetric:
5
9
4
x
y+3
z − 10
22. parametric: x = 12t, y = −3 − 5t, z = 10 − 6t; symmetric:
=
=
12
−5
−6
20. parametric: x = 1 − 7t, y = 8 − 8t, z = −2; symmetric:
23. Writing the given line in the form x/2 = (y − 1)/(−3) = (z − 5)/6, we see that a direction vector is 2, −3, 6.
Parametric equations for the line are x = 6 + 2t, y = 4 − 3t, z = −2 + 6t.
24. A direction vector is 5, 1/3, −2. Symmetric equations for the line are (x−4)/5 = (y +11)/(1/3) = (z +7)/(−2).
25. A direction vector parallel to both the xz- and xy-planes is i = 1, 0, 0. Parametric equations for the line are
x = 2 + t, y = −2, z = 15.
26. (a) Since the unit vector j = 0, 1, 0 lies along the y-axis, we have x = 1, y = 2 + t, z = 8.
(b) since the unit vector k = 0, 0, 1 is perpendicular to the xy-plane, we have x = 1, y = 2, z = 8 + t.
27. Both lines go through the points (0, 0, 0) and (6, 6, 6). Since two points determine a line, the lines are the same.
28. a and f are parallel since 9, −12, 6 = −3−3, 4, −2. c and d are orthogonal since 2, −3, 4 · 1, 4, 5/2 = 0.
29. In the xy-plane, z = 9 + 3t = 0 and t = −3. Then x = 4 − 2(−3) = 10 and y = 1 + 2(−3) = −5. The point is
(10, −5, 0). In the xz-plane, y = 1+2t = 0 and t = −1/2. Then x = 4−2(−1/2) = 5 and z = 9+3(−1/2) = 15/2.
The point is (5, 0, 15/2). In the yz-plane, x = 4−2t = 0 and t = 2. Then y = 1+2(2) = 5 and z = 9+3(2) = 15.
The point is (0, 5, 15).
30. The parametric equations for the line are x = 1 + 2t, y = −2 + 3t, z = 4 + 2t. In the xy-plane, z = 4 + 2t = 0
and t = −2. Then x = 1 + 2(−2) = −3 and y = −2 + 3(−2) = −8. The point is (−3, −8, 0). In the xz-plane,
y = −2 + 3t = 0 and t = 2/3. Then x = 1 + 2(2/3) = 7/3 and z = 4 + 2(2/3) = 16/3. The point is (7/3, 0, 16/3).
In the yz-plane, x = 1 + 2t = 0 and t = −1/2. Then y = −2 + 3(−1/2) = −7/2 and z = 4 + 2(−1/2) = 3. The
point is (0, −7/2, 3).
354
7.5 Lines and Planes in 3-Space
31. Solving the system 4 + t = 6 + 2s, 5 + t = 11 + 4s, −1 + 2t = −3 + s, or t − 2s = 2, t − 4s = 6, 2t − s = −2
yields s = −2 and t = −2 in all three equations. Thus, the lines intersect at the point x = 4 + (−2) = 2,
y = 5 + (−2) = 3, z = −1 + 2(−2) = −5, or (2, 3, −5).
32. Solving the system 1 + t = 2 − s, 2 − t = 1 + s, 3t = 6s, or t + s = 1, t + s = 1, t − 2s = 0 yields s = 1/3 and
t = 2/3 in all three equations. Thus, the lines intersect at the point x = 1 + 2/3 = 5/3, y = 2 − 2/3 = 4/3,
z = 3(2/3) = 2, or (5/3, 4/3, 2).
33. The system of equations 2 − t = 4 + s, 3 + t = 1 + s, 1 + t = 1 − s, or t + s = −2, t − s = −2, t + s = 0 has no
solution since −2 = 0. Thus, the lines do not intersect.
34. Solving the system 3 − t = 2 + 2s, 2 + t = −2 + 3s, 8 + 2t = −2 + 8s, or t + 2s = 1, t − 3s = −4, 2t − 8s = −10
yields s = 1 and t = −1 in all three equations. Thus, the lines intersect at the point x = 3 − (−1) = 4,
y = 2 + (−1) = 1, z = 8 + 2(−1) = 6, or (4, 1, 6).
35. a = −1, 2, −2, b = 2, 3, −6, a · b = 16, a = 3, b = 7; cos θ =
a·b
16
=
;
a b
3·7
16
≈ 40.37◦
21
√
√
36. a = 2, 7, −1, b = −2, 1, 4, a · b = −1, a = 3 6 , b = 21 ;
1
a·b
1
−1
cos θ =
= − √ ; θ = cos−1 (− √ ) ≈ 91.70◦
= √ √
a b
(3 6 )( 21 )
9 14
9 14
θ = cos−1
37. A direction vector perpendicular to the given lines will be 1, 1, 1 × −2, 1, −5 = −6, 3, 3. Equations of the
line are x = 4 − 6t, y = 1 + 3t, z = 6 + 3t.
38. The direction vectors of the given lines are 3, 2, 4 and 6, 4, 8 = 23, 2, 4. These are parallel, so we need a
third vector parallel to the plane containing the lines which is not parallel to them. The point (1, −1, 0) is on
the first line and (−4, 6, 10) is on the second line. A third vector is then 1, −1, 0 − −4, 6, 10 = 5, −7, −10.
Now a direction vector perpendicular to the plane is 3, 2, 4 × 5, −7, −10 = 8, 50, −31. Equations of the line
through (1, −1, 0) and perpendicular to the plane are x = 1 + 8t, y = −1 + 50t, z = −31t.
39. 2(x − 5) − 3(y − 1) + 4(z − 3) = 0; 2x − 3y + 4z = 19
40. 4(x − 1) − 2(y − 2) + 0(z − 5) = 0; 4x − 2y = 0
41. −5(x − 6) + 0(y − 10) + 3(z + 7) = 0; −5x + 3z = −51
42. 6x − y + 3z = 0
43. 6(x − 1/2) + 8(y − 3/4) − 4(z + 1/2) = 0; 6x + 8y − 4z = 11
44. −(x + 1) + (y − 1) − (z − 0) = 0; −x + y − z = 2
45. From the points (3, 5, 2) and (2, 3, 1) we obtain the vector u = i + 2j + k. From the points (2, 3, 1) and
(−1, −1, 4) we obtain the vector v = 3i + 4j − 3k. From the points (−1, −1, 4) and (x, y, z) we obtain the vector
w = (x + 1)i + (y + 1)j + (z − 4)k. Then, a normal vector is
i j k u × v = 1 2 1 = −10i + 6j − 2k.
3 4 −3 A vector equation of the plane is −10(x + 1) + 6(y + 1) − 2(z − 4) = 0 or 5x − 3y + z = 2.
46. From the points (0, 1, 0) and (0, 1, 1) we obtain the vector u = k. From the points (0, 1, 1) and (1, 3, −1) we obtain
the vector v = i + 2j − 2k.
From the points (1, 3, −1) and (x, y, z) we obtain the vector
355
7.5
Lines and Planes in 3-Space
w = (x − 1)i + (y − 3)j + (z + 1)k. Then, a normal
i
u × v = 0
1
vector is
j k 0 1 = −2i + j.
2 −2 A vector equation of the plane is −2(x − 1) + (y − 3) + 0(z + 1) = 0 or −2x + y = 1.
47. From the points (0, 0, 0) and (1, 1, 1) we obtain the vector u = i + j + k. From the points (1, 1, 1) and
(3, 2, −1) we obtain the vector v = 2i + j − 2k. From the points (3, 2, −1) and (x, y, z) we obtain the vector
w = (x − 3)i + (y − 2)j + (z + 1)k. Then, a normal vector is
i j k u × v = 1 1 1 = −3i + 4j − k.
2 1 −2 A vector equation of the plane is −3(x − 3) + 4(y − 2) − (z + 1) = 0 or −3x + 4y − z = 0.
48. The three points are not colinear and all satisfy x = 0, which is the equation of the plane.
49. From the points (1, 2, −1) and (4, 3, 1) we obtain the vector u = 3i + j + 2k. From the points (4, 3, 1) and
(7, 4, 3) we obtain the vector v = 3i + j + 2k. From the points (7, 4, 3) and (x, y, z) we obtain the vector
w = (x − 7)i + (y − 4)j + (z − 3)k. Since u × v = 0, the points are colinear.
50. From the points (2, 1, 2) and (4, 1, 0) we obtain the vector u = 2i − 2k.
From the points (4, 1, 0) and
(5, 0, −5) we obtain the vector v = i − j − 5k. From the points (5, 0, −5) and (x, y, z) we obtain the vector
w = (x − 5)i + yj + (z + 5)k. Then, a normal vector is
i
j
k u × v = 2 0 −2 = −2i + 8j − 2k.
1 −1 −5 A vector equation of the plane is −2(x − 5) + 8y − 2(z + 5) = 0 or x − 4y + z = 0.
51. A normal vector to x + y − 4z = 1 is 1, 1, −4. The equation of the parallel plane is
(x − 2) + (y − 3) − 4(z + 5) = 0 or x + y − 4z = 25.
52. A normal vector to 5x−y +z = 6 is 5, −1, 1, . The equation of the parallel plane is 5(x−0)−(y −0)+(z −0) = 0
or 5x − y + z = 0.
53. A normal vector to the xy-plane is 0, 0, 1. The equation of the parallel plane is z − 12 = 0 or z = 12.
54. A normal vector is 0, 1, 0. The equation of the plane is y + 5 = 0 or y = −5.
55. Direction vectors of the lines are 3, −1, 1. and 4, 2, 1. A normal vector to the plane is 3, −1, 1 × 4, 2, 1 =
−3, 1, 10. A point on the first line, and thus in the plane, is 1, 1, 2. The equation of the plane is
−3(x − 1) + (y − 1) + 10(z − 2) = 0 or −3x + y + 10z = 18.
56. Direction vectors of the lines are 2, −1, 6 and 1, 1, −3. A normal vector to the plane is 2, −1, 6×1, 1, −3 =
−3, 12, 3. A point on the first line, and thus in the plane, is (1, −1, 5). The equation of the plane is
−3(x − 1) + 12(y + 1) + 3(z − 5) = 0 or −x + 4y + z = 0.
57. A direction vector for the two lines is 1, 2, 1. Points on the lines are (1, 1, 3) and (3, 0, −2). Thus, another
vector parallel to the plane is 1−3, 1−0, 3+2 = −2, 1, 5. A normal vector to the plane is 1, 2, 1×−2, 1, 5 =
9, −7, 5. Using the point (3, 0, −2) in the plane, the equation of the plane is 9(x − 3) − 7(y − 0) + 5(z + 2) = 0
or 9x − 7y + 5z = 17.
356
7.5 Lines and Planes in 3-Space
58. A direction vector for the line is 3, 2, −2. Letting t = 0, we see that the origin is on the line and hence in the
plane. Thus, another vector parallel to the plane is 4 − 0, 0 − 0, −6 − 0 = 4, 0, −6. A normal vector to the
plane is 3, 2, −2 × 4, 0, −6 = −12, 10, −8. The equation of the plane is −12(x − 0) + 10(y − 0) − 8(z − 0) = 0
or 6x − 5y + 4z = 0.
59. A direction vector for the line, and hence a normal vector to the plane, is −3, 1, −1/2. The equation of the
plane is −3(x − 2) + (y − 4) − 12 (z − 8) = 0 or −3x + y − 12 z = −6.
60. A normal vector to the plane is 2 − 1, 6 − 0, −3 + 2 = 1, 6, −1. The equation of the plane is
(x − 1) + 6(y − 1) − (z − 1) = 0 or x + 6y − z = 6.
61. Normal vectors to the planes are (a) 2, −1, 3, (b) 1, 2, 2, (c) 1, 1, −3/2, (d) −5, 2, 4,
(e) −8, −8, 12, (f ) −2, 1, −3. Parallel planes are (c) and (e), and (a) and (f ). Perpendicular planes
are (a) and (d), (b) and (c), (b) and (e), and (d) and (f ).
62. A normal vector to the plane is −7, 2, 3. This is a direction vector for the line and the equations of the line
are x = −4 − 7t, y = 1 + 2t, z = 7 + 3t.
63. A direction vector of the line is −6, 9, 3, and the normal vectors of the planes are (a) 4, 1, 2, (b) 2, −3, 1,
(c) 10, −15, −5, (d) −4, 6, 2. Vectors (c) and (d) are multiples of the direction vector and hence the
corresponding planes are perpendicular to the line.
64. A direction vector of the line is −2, 4, 1, and normal vectors to the planes are (a) 1, −1, 3,
(b) 6, −3, 0, (c) 1, −2, 5, (d) −2, 1, −2. Since the dot product of each normal vector with the direction vector is non-zero, none of the planes are parallel to the line.
65. Letting z = t in both equations and solving 5x − 4y = 8 + 9t, x + 4y = 4 − 3t, we obtain x = 2 + t, y =
1
2
− t,
z = t.
66. Letting y = t in both equations and solving x − z = 2 − 2t, 3x + 2z = 1 + t, we obtain x = 1 − 35 t, y = t,
z = −1 + 75 t or, letting t = 5s, x = 1 − 3s, y = 5s, z = −1 + 7s.
67. Letting z = t in both equations and solving 4x − 2y = 1 + t, x + y = 1 − 2t, we obtain x =
1
2
− 12 t, y =
1
2
− 32 t,
z = t.
68. Letting z = t and using y = 0 in the first equation, we obtain x = − 12 t, y = 0, z = t.
69. Substituting the parametric equations into the equation of the plane, we obtain 2(1+2t)−3(2−t)+2(−3t) = −7
or t = −3. Letting t = −3 in the equation of the line, we obtain the point of intersection (−5, 5, 9).
70. Substituting the parametric equations into the equation of the plane, we obtain (3−2t)+(1+6t)+4(2− 12 t) = 12
or 2t = 0. Letting t = 0 in the equation of the line, we obtain the point of intersection (3, 1, 2).
71. Substituting the parametric equations into the equation of the plane, we obtain 1 + 2 − (1 + t) = 8 or t = −6.
Letting t = −6 in the equation of the line, we obtain the point of intersection (1, 2, −5).
72. Substituting the parametric equations into the equation of the plane, we obtain 4 + t − 3(2 + t) + 2(1 + 5t) = 0
or t = 0. Letting t = 0 in the equation of the line, we obtain the point of intersection (4, 2, 1).
357
7.5
Lines and Planes in 3-Space
In Problems 73 and 74, the cross product of the normal vectors to the two planes will be a vector parallel to both
planes, and hence a direction vector for a line parallel to the two planes.
73. Normal vectors are 1, 1, −4 and 2, −1, 1. A direction vector is
1, 1, −4 × 2, −1, 1 = −3, −9, −3 = −31, 3, 1.
Equations of the line are x = 5 + t, y = 6 + 3t, z = −12 + t.
74. Normal vectors are 2, 0, 1 and −1, 3, 1. A direction vector is
2, 0, 1 × −1, 3, 1 = −3, −3, 6 = −31, 1, −2.
Equations of the line are x = −3 + t, y = 5 + t, z = −1 − 2t.
In Problems 75 and 76, the cross product of the direction vector of the line with the normal vector of the given plane
will be a normal vector to the desired plane.
75. A direction vector of the line is 3, −1, 5 and a normal vector to the given plane is 1, 1, 1. A normal vector
to the desired plane is 3, −1, 5 × 1, 1, 1 = −6, 2, 4. A point on the line, and hence in the plane, is (4, 0, 1).
The equation of the plane is −6(x − 4) + 2(y − 0) + 4(z − 1) = 0 or 3x − y − 2z = 10.
76. A direction vector of the line is 3, 5, 2 and a normal vector to the given plane is 2, −4, −1. A normal vector to
the desired plane is −3, 5, 2 × 2, −4, −1 = 3, 1, 2. A point on the line, and hence in the plane, is (2, −2, 8).
The equation of the plane is 3(x − 2) + (y + 2) + 2(z − 8) = 0 or 3x + y + 2z = 20.
77.
78.
79.
80.
81.
82.
358
7.6
Vector Spaces
EXERCISES 7.6
Vector Spaces
1. Not a vector space. Axiom (vi) is not satisfied.
2. Not a vector space. Axiom (i) is not satisfied.
3. Not a vector space. Axiom (x) is not satisfied.
4. A vector space
5. A vector space
6. A vector space
7. Not a vector space. Axiom (ii) is not satisfied.
8. A vector space
9. A vector space
10. Not a vector space. Axiom (i) is not satisfied.
11. A subspace
12. Not a subspace. Axiom (i) is not satisfied.
13. Not a subspace. Axiom (ii) is not satisfied.
14. A subspace
15. A subspace
16. A subspace
17. A subspace
18. A subspace
19. Not a subspace. Neither axioms (i) nor (ii) are satisfied.
20. A subspace
21. Let (x1 , y1 , z1 ) and (x2 , y2 , z2 ) be in S. Then
(x1 , y1 , z1 ) + (x2 , y2 , z2 ) = (at1 , bt1 , ct1 ) + (at2 , bt2 , ct2 ) = (a(t1 + t2 ), b(t1 + t2 ), c(t1 + t2 ))
is in S. Also, for (x, y, z) in S then k(x, y, z) = (kx, ky, kz) = (a(kt), b(kt), c(kt)) is also in S.
22. Let (x1 , y1 , z1 ) and (x2 , y2 , z2 ) be in S. Then ax1 + by1 + cz1 = 0 and ax2 + by2 + cz2 = 0. Adding gives
a(x1 + x2 ) + b(y1 + y2 ) + c(z1 + z2 ) = 0 and so (x1 , y1 , z1 ) + (x2 , y2 , z2 ) = (x1 + x2 , y1 + y2 , z1 + z2 ) is in S. Also,
for (x, y, z) then ax + by + cz = 0 implies k(ax + by + cz) = k · 0 = 0 and a(kx) + b(ky) + c(kz) = 0. this means
k(x, y, z) = (kx, ky, kz) is in S.
23. (a) c1 u1 + c2 u2 + c3 u3 = 0 if and only if c1 + c2 + c3 = 0, c2 + c3 = 0, c3 = 0. The only solution of this system
is c1 = 0, c2 = 0, c3 = 0.
(b) Solving the system c1 + c2 + c3 = 3, c2 + c3 = −4, c3 = 8 gives c1 = 7, c2 = −12, c3 = 8. Thus
a = 7u1 − 12u2 + 8u3 .
24. (a) The assumption c1 p1 + c2 p2 = 0 is equivalent to (c1 + c2 )x + (c1 − c2 ) = 0. Thus c1 + c2 = 0, c1 − c2 = 0.
The only solution of this system is c1 = 0, c2 = 0.
(b) Solving the system c1 + c2 = 5, c1 − c2 = 2 gives c1 =
7
2
, c2 =
25. Linearly dependent since −6, 12 = − 32 4, −8
26. Linearly dependent since 21, 1 + 30, 1 + (−1)2, 5 = 0, 0
359
3
2
. Thus p(x) = 72 p1 (x) + 32 p2 (x)
7.6
Vector Spaces
27. Linearly independent
28. Linearly dependent since for all x (1) · 1 + (−2)(x + 1) + (1)(x + 1)2 + (−1)x2 = 0.
29. f is discontinuous at x = −1 and at x = −3.
2π
2π
30. (x, sin x) =
x sin x dx = (−x cos x + sin x) = −2π
0
0
2π
2π 3
1 3 8
2
2
31. x =
x dx = x = π 3 and so x = 2
. Now
3
3
3
0
0
2π
1 2π
1
1
2π
sin x2 =
sin2 x dx =
(1 − cos 2x) dx =
x − sin 2x = π
2 0
2
2
0
0
√
and so sin x = π .
2π
32. A basis could be 1, x, ex cos 3x, ex sin 3x.
33. We need to show that Span{x1 , x2 , . . . , xn } is closed under vector addition and scalar multiplication. Suppose
u and v are in Span{x1 , x2 , . . . , xn }. Then u = a1 x1 + a2 x2 + · · · + an xn and v = b1 x1 + b2 x2 + · · · + bn xn , so
that
u + v = (a1 + b1 )x1 + (a2 + b2 )x2 + · · · + (an + bn )xn ,
which is in Span{x1 , x2 , . . . , xn }. Also, for any real number k,
ku = k(a1 x1 + a2 x2 + · · · + an xn ) = ka1 x1 + ka2 x2 + · · · + kan xn ,
which is in Span{x1 , x2 , . . . , xn }. Thus, Span{x1 , x2 , . . . , xn } is a subspace of V.
34. R2 is not a subspace of either R3 or R4 and R3 is not a subspace of R4 . The vectors in R2 are ordered pairs,
while the vectors in R3 are ordered triples.
35. Since a basis for M22 is
B=
1
0
0
0
,
0
1
0
0
,
0
0
1
0
,
0
0
0
1
,
the dimension of M22 is 4.
36. To show that the set of nonzero orthogonal vectors is linearly independent we set c1 v1 + c2 v2 + · · · + cn vn = 0.
For 0 ≤ i ≤ n,
(c1 v1 + c2 v2 + · · · + ci vi · · · + cn vn ) · vi = c1 v1 · vi + c2 v2 · vi + · · · + ci vi · vi · · · + cn vn · vi = ci ||vi ||2 ,
so ci ||vi ||2 = 0 because
(c1 v1 + c2 v2 + · · · + ci vi · · · + cn vn ) · vi = 0 · vi = 0.
Since vi is a nonzero vector, ci = 0. Thus, the assumption that c1 v1 + c2 v2 + · · · + cn vn = 0 leads to
c1 = c2 = · · · = cn = 0, and the set is linearly independent.
37. We verify the four properties:
(i) (u, v) = u1 v1 + 4u2 v2 = v1 u1 + 4v2 u2 = (v, u)
(ii) (ku, v) = (ku1 )v1 + 4(ku2 )v2 = k(u1 v1 + 4u2 v2 ) = k(u, v)
(iii) (u, u) = u21 + 4ku22 > 0 for u = 0. Furthermore, u21 + 4ku22 = 0 if and only if u1 = 0 and u2 = 0, or
equivalently, u = 0.
(iv) (u, v + w) = u1 (v1 + w1 ) + 4u2 (v2 + w2 ) = (u1 v1 + 4u2 v2 ) + (u1 w1 + 4u2 w2 ) = (u, v) + (u, w)
360
7.7
Gram-Schmidt Orthogonalization Process
38. (a) Let u = 2, 1 and v = 2, −1 be nonzero vectors in R2 . With respect to the standard inner or dot product
on R2 ,
u · v = 2, 1 · 2, −1 = 2 · 2 + 1 · (−1) = 3.
We see that u and v are not orthogonal with respect to that inner product. But using the inner product
in Problem 37, we have
(u, v) = 2 · 2 + 4(1) · (−1) = 0,
and so u and v are orthogonal with respect to that inner product.
(b) Consider f (x) = sin x and g(x) = cos x in C[0, 2π]. Since
2π
2π
1 2π
1
1
sin x cos x dx =
sin 2x dx = − cos 2x = − (1 − 1) = 0,
2
4
4
0
0
0
these functions are orthogonal in C[0, 2π].
EXERCISES 7.7
Gram-Schmidt Orthogonalization Process
5
5
12
1. Letting w1 = 12
13 , 13 and w2 = 13 , − 13 , we have
12
5
5
12
w1 · w2 =
+
−
= 0,
13
13
13
13
so the vectors are orthogonal. Also,
||w1 || =
12
13
2
+
5
13
2
=1
and ||w2 || =
5
13
2
2
12
+ −
= 1,
13
so the basis is orthonormal. To express u = 4, 2 in terms of w1 and w2 we compute
12
5
58
12 5
u · w1 = 4, 2 ·
,
= (4)
+ (2)
=
13 13
13
13
13
12
5
12
4
5
u · w2 = 4, 2 ·
,−
= (4)
+ (2) −
=− ,
13 13
13
13
13
so
58
4
w1 −
w2 .
13
13
√
√
√
√
√
√
√
√
2. Letting w1 = 1/ 3, 1/ 3, −1/ 3, w2 = 0, −1/ 2, −1/ 2, and w3 = −2/ 6, 1/ 6, −1/ 6, we have
1
1
1
1
1
w1 · w2 = √
(0) + √
−√
+ −√
−√
=0
3
3
2
3
2
1
2
1
1
1
1
√
w1 · w3 = √
−√
+ √
+ −√
−√
=0
3
6
3
6
3
6
2
1
1
1
1
√
w2 · w3 = (0) − √
+ −√
+ −√
−√
= 0,
6
2
6
2
6
u=
361
7.7
Gram-Schmidt Orthogonalization Process
so the vectors are orthogonal. Also,
||w1 || =
1
√
3
2
2
2 2
1
1
1
2
+
+ −√
= 1, ||w2 || = 0 + − √
+ −√
= 1,
3
2
2
2 2 2
2
1
1
and ||w3 || =
−√
+ √
+ −√
= 1,
6
6
6
1
√
3
2
so the basis is orthonormal. To express u = 5, −1, 6 in terms of w1 , w2 , and w3 we compute
1
1
1
1
1
1
2
√ , √ , −√
= (5) √
+ (−1) √
+ (6) − √
= −√
3
3
3
3
3
3
3
1
1
1
1
5
u · w2 = 5, −1, 6 · 0, − √ , − √
= (5)(0) + (−1) − √
+ (6) − √
= −√
2
2
2
2
2
2
1
1
2
1
1
17
u · w3 = 5, −1, 6 · − √ , √ , − √
= (5) − √
+ (−1) √
+ (6) − √
= −√
6
6
6
6
6
6
6
u · w1 = 5, −1, 6 ·
so
2
5
17
u = − √ w1 − √ w2 − √ w3 .
3
2
6
Since the basis vectors in Problems 3 and 4 are orthogonal but not orthonormal, the result of Theorem 7.5 must be
slightly modified to read
u=
u · w1
u · w2
u · wn
w1 +
w2 + · · · +
wn .
2
2
||w1 ||
||w2 ||
||wn ||2
The proof is very similar to that given in the text for Theorem 7.5.
3. Letting w1 = 1, 0, 1, w2 = 0, 1, 0, and w3 = −1, 0, 1 we have
w1 · w2 = (1)(0) + (0)(1) + (1)(0) = 0
w1 · w3 = (1)(−1) + (0)(0) + (1)(1) = 0
w2 · w3 = (0)(−1) + (1)(0) + (0)(1) = 0
so the vectors are orthogonal. We also compute
||w1 ||2 = 12 + 02 + 12 = 2
||w2 ||2 = 02 + 12 + 02 = 1
||w3 ||2 = (−1)2 + 02 + 12 = 2
and, with u = 10, 7, −13,
u · w1 = (10)(1) + (7)(0) + (−13)(1) = −3
u · w2 = (10)(0) + (7)(1) + (−13)(0) = 7
u · w3 = (10)(−1) + (7)(0) + (−13)(1) = −23.
Then, using the result given before the solution to this problem, we have
3
23
u = − w1 + 7w2 −
w3 .
2
2
362
7.7
Gram-Schmidt Orthogonalization Process
4. Letting w1 = 2, 1, −2, 0, w2 = 1, 2, 2, 1, w3 = 3, −4, 1, 3, and w4 = 5, −2, 4, −9 we have
w1 · w2 = (2)(1) + (1)(2) + (−2)(2) + (0)(1) = 0
w1 · w3 = (2)(3) + (1)(−4) + (−2)(1) + (0)(3) = 0
w1 · w4 = (2)(5) + (1)(−2) + (−2)(4) + (0)(−9) = 0
w2 · w3 = (1)(3) + (2)(−4) + (2)(1) + (1)(3) = 0
w2 · w4 = (1)(5) + (2)(−2) + (2)(4) + (1)(−9) = 0
w3 · w4 = (3)(5) + (−4)(−2) + (1)(4) + (3)(−9) = 0
so the vectors are orthogonal. We also compute
||w1 ||2 = 22 + 12 + (−2)2 + 02 = 9
||w2 ||2 = 12 + 22 + 22 + 12 = 10
||w3 ||2 = 32 + (−4)2 + 12 + 32 = 35
||w4 ||2 = 52 + (−2)2 + 42 + (−9)2 = 126
and, with u = 1, 2, 4, 3,
u · w1 = (1)(2) + (2)(1) + (4)(−2) + (3)(0) = −4
u · w2 = (1)(1) + (2)(2) + (4)(2) + (3)(1) = 16
u · w3 = (1)(3) + (2)(−4) + (4)(1) + (3)(3) = 8
u · w4 = (1)(5) + (2)(−2) + (4)(4) + (3)(−9) = −10.
Then, using the result given before the solution to this problem, we have
4
8
8
5
u = − w1 + w2 +
w3 −
w4 .
9
5
35
63
5. (a) We have u1 = −3, 2 and u2 = −1, −1. Taking v1 = u1 = −3, 2, and using u2 · v1 = 1 and v1 · v1 = 13
we obtain
u2 · v1
1
10 15
v2 = u2 −
v1 = −1, −1 − −3, 2 = − , −
.
v1 · v1
13
13 13
15
Thus, an orthogonal basis is {−3, 2, − 10
13 , − 13 } and an orthonormal basis is {w1 , w2 }, where
1
3
2
1
w1 =
−3, 2 = √ −3, 2 = − √ , √
||−3, 2||
13
13
13
and
w2
1
=
10
||− 13 , − 15
13 ||
10 15
− ,−
13 13
=
1
√
5/ 13
10 15
− ,−
13 13
=
2
3
−√ , −√
13
13
.
(b) We have u1 = −3, 2 and u2 = −1, −1. Taking v1 = u2 = −1, −1, and using u1 · v1 = 1 and v1 · v1 = 2
we obtain
v2 = u1 −
u1 · v1
1
v1 = −3, 2 − −1, −1 =
v1 · v1
2
5 5
− ,
2 2
.
Thus, an orthogonal basis is {−1, −1, − 52 , 52 } and an orthonormal basis is {w3 , w4 }, where
1
1
1
1
w3 =
−1, −1 = √ −1, −1 = − √ , − √
||−1, −1||
2
2
2
and
w4
1
=
5 5
||− 2 , 2 ||
5 5
− ,
2 2
1
= √
5/ 2
363
5 5
− ,
2 2
=
1
1
−√ , √
2
2
.
7.7
Gram-Schmidt Orthogonalization Process
(c)
1
4
u
-4
w1
2
-2
2
4
-1
1
w4
0.5
-0.5
v-2
0.5
0.5
1
-1
-0.5
-0.5
w2
-4
0.5
1
-0.5
w3
-1
-1
6. (a) We have u1 = −3, 4 and u2 = −1, 0. Taking v1 = u1 = −3, 4, and using u2 · v1 = 3 and v1 · v1 = 25
we obtain
u2 · v1
3
16 12
v2 = u2 −
v1 = −1, 0 − −3, 4 = − , −
.
v1 · v1
25
25 25
12
Thus, an orthogonal basis is {−3, 4, − 16
25 , − 25 } and an orthonormal basis is {w1 , w2 }, where
1
1
3 4
w1 =
−3, 4 = −3, 4 = − ,
||−3, 4||
5
5 5
and
w2
1
=
16
||− 25 , − 12
25 ||
16 12
− ,−
25 25
1
=
4/5
16 12
− ,−
25 25
=
4 3
− ,−
5 5
.
(b) We have u1 = −3, 4 and u2 = −1, 0. Taking v1 = u2 = −1, 0, and using u1 · v1 = 3 and v1 · v1 = 1
we obtain
u1 · v1
3
v2 = u1 −
v1 = −3, 4 − −1, 0 = 0, 4 .
v1 · v1
1
Thus, an orthogonal basis is {−1, 0, 0, 4} and an orthonormal basis is {w3 , w4 }, where
w3 =
1
1
−1, 0 = −1, 0 = −1, 0
||−1, 0||
1
and
w4 =
(c)
u
1
1
0, 4 = 0, 4 = 0, 1 .
||0, 4||
4
1
4
1
w1
0.5
2
0.5
w3
v
-4
w4
-2
2
4
-1
-0.5
-2
w2
-4
1
0.5
-1
-0.5
0.5
-0.5
-0.5
-1
-1
1
7. (a) We have u1 = 1, 1 and u2 = 1, 0. Taking v1 = u1 = 1, 1, and using u2 · v1 = 1 and v1 · v1 = 2 we
obtain
1 1
u2 · v1
1
v2 = u2 −
,−
.
v1 = 1, 0 − 1, 1 =
v1 · v1
2
2 2
Thus, an orthogonal basis is {1, 1, 12 , − 12 } and an orthonormal basis is {w1 , w2 }, where
1
1
1
1
w1 =
1, 1 = √ 1, 1 = √ , √
||1, 1||
2
2
2
and
w2 =
1
|| √12 , − √12 ||
1
1
√ , −√
2
2
=
1
1
364
1
1
√ , −√
2
2
=
1
1
√ , −√
2
2
.
7.7
Gram-Schmidt Orthogonalization Process
(b) We have u1 = 1, 1 and u2 = 1, 0. Taking v1 = u2 = 1, 0, and using u1 · v1 = 1 and v1 · v1 = 1 we
obtain
u1 · v1
1
v2 = u1 −
v1 = 1, 1 − 1, 0 = 0, 1 .
v1 · v1
1
Thus, an orthogonal basis is {1, 0, 0, 1}, which is also an orthonormal basis.
(c)
2
1
1
u
0.5
-2
1
w1
1.5
0.5
w4
0.5
w3
v
-1
1
-1
2
-0.5
0.5
-0.5
-1
-0.5
w2
-0.5
-1
1
1
0.5
-0.5
-1.5
-1
-1
-2
8. (a) We have u1 = 5, 7 and u2 = 1, −2. Taking v1 = u1 = 5, 7, and using u2 · v1 = −9 and v1 · v1 = 74
we obtain
u2 · v1
9
119 85
v2 = u2 −
v1 = 1, −2 − 5, 7 =
,−
.
v1 · v1
74
74
74
85
Thus, an orthogonal basis is {5, 7, 119
74 , − 74 } and an orthonormal basis is {w1 , w2 }, where
1
5
7
1
w1 =
5, 7 = √ 5, 7 = √ , √
||5, 7||
74
74
74
and
w2 =
1
85
|| 119
,
74 − 74 ||
119 85
,−
74
74
=
1
√
17/ 74
119 85
,−
74
74
=
7
5
√ , −√
74
74
.
(b) We have u1 = 5, 7 and u2 = 1, −2. Taking v1 = u2 = 1, −2, and using u1 · v1 = −9 and v1 · v1 = 5
we obtain
u1 · v1
9
34 17
v2 = u1 −
v1 = 5, 7 − 1, −2 =
,
.
v1 · v1
5
5 5
17
Thus, an orthogonal basis is {1, −2, 34
5 , 5 } and an orthonormal basis is {w3 , w4 }, where
1
1
1
2
w3 =
1, −2 = √ 1, −2 = √ , − √
||1, −2||
5
5
5
and
w4
(c)
1
= 34 17
|| 5 , 5 ||
34 17
,
5 5
8
1
√
=
17/ 5
34 17
,
5 5
=
1
2
√ ,√
5
5
1
u
6
4
0.5
.
1
w1
0.5
w4
2
-7.5 -5 -2.5
-2
-4
2.5 5 7.5
v
-1
-0.5
0.5
-0.5
-6
-8
w2
1
-1
-0.5
0.5
-0.5
-1
-1
1
w3
9. We have u1 = 1, 1, 0, u2 = 1, 2, 2, and u3 = 2, 2, 1. Taking v1 = u1 = 1, 1, 0 and using u2 · v1 = 3 and
v1 · v1 = 2 we obtain
u2 · v1
3
1 1
v1 = 1, 2, 2 − 1, 1, 0 = − , , 2 .
v2 = u2 −
v1 · v1
2
2 2
365
7.7
Gram-Schmidt Orthogonalization Process
Next, using u3 · v1 = 4, u3 · v2 = 2, and v2 · v2 = 92 , we obtain
u3 · v1
u3 · v2
4
2
v3 = u3 −
v1 −
v2 = 2, 2, 1 − 1, 1, 0 −
v1 · v1
v2 · v2
2
9/2
Thus, an orthogonal basis is
B =
and an orthonormal basis is
B =
1 1
− , ,2
2 2
=
2
2 1
,− ,
9
9 9
.
1 1
2
2 1
1, 1, 0 , − , , 2 ,
,− ,
,
2 2
9
9 9
1
1
1
1
4
2
2 1
√ , √ ,0 , − √ , √ , √
,
,− ,
.
3
3 3
2
2
3 2 3 2 3 2
10. We have u1 = −3, 1, 1, u2 = 1, 1, 0, and u3 = −1, 4, 1. Taking v1 = u1 = −3, 1, 1 and using u2 · v1 = −2
and v1 · v1 = 11 we obtain
u2 · v1
−2
5 13 2
v2 = u2 −
v1 = 1, 1, 0 −
−3, 1, 1 =
, ,
.
v1 · v1
11
11 11 11
Next, using u3 · v1 = 8, u3 · v2 =
v3 = u3 −
49
11 ,
and v2 · v2 =
18
11 ,
we obtain
49/11
u3 · v1
u3 · v2
8
−3, 1, 1 −
v1 −
v2 = −1, 4, 1 −
v1 · v1
v2 · v2
11
18/11
Thus, an orthogonal basis is
B =
−3, 1, 1 ,
5 13 2
,
,
11 11 11
5 13 2
, ,
11 11 11
−
=
1 1
2
,
,−
18 18
9
.
1 1
2
, − ,
,−
,
18 18
9
and an orthonormal basis is
3
1
1
5
13
2
1
1
4
√ , √ , √
B =
−√ , √ , √
,
, − √ , √ , √
.
11
11
11
3 22 3 22 3 22
3 2 3 2 3 2
11. We have u1 = 12 , 12 , 1, u2 = −1, 1, − 12 , and u3 = −1, 12 , 1. Taking v1 = u1 = 12 , 12 , 1 and using u2 ·v1 = − 12
and v1 · v1 =
3
2
we obtain
v2 = u2 −
u2 · v1
v1 =
v1 · v1
−1, 1, −
1
2
−
−1/2
3/2
1 1
, ,1
2 2
=
5 7 1
− , ,−
6 6 6
.
Next, using u3 · v1 = 34 , u3 · v2 = 54 , and v2 · v2 = 25
12 , we obtain
3/4 1 1
5/4
5 7 1
3
9 3
u3 · v1
u3 · v2
1
v3 = u3 −
, ,1 −
− , ,−
= − ,− ,
.
v1 −
v2 = −1, , 1 −
v1 · v1
v2 · v2
2
3/2 2 2
25/12
6 6 6
4
20 5
Thus, an orthogonal basis is
B =
5 7
1
3
9 3
1 1
, ,1 , − , ,−
, − ,− ,
,
2 2
6 6
6
4
20 5
and an orthonormal basis is
1
1
2
1
7
1
1
3
4
√ ,√ ,√
B =
, −√ , √ , − √
, −√ , − √ , √
.
6
6
6
3 5 3
5 3
2
5 2 5 2
12. We have u1 = 1, 1, 1, u2 = 9, −1, 1, and u3 = −1, 4, −2. Taking v1 = u1 = 1, 1, 1 and using u2 · v1 = 9
and v1 · v1 = 3 we obtain
v2 = u2 −
u2 · v1
9
v1 = 9, −1, 1 − 1, 1, 1 = 6, −4, −2 .
v1 · v1
3
366
7.7
Gram-Schmidt Orthogonalization Process
Next, using u3 · v1 = 1, u3 · v2 = −18, and v2 · v2 = 56, we obtain
u3 · v1
u3 · v2
1
−18
v1 −
v2 = −1, 4, −2 − 1, 1, 1 −
v3 = u3 −
6, −4, −2 =
v1 · v1
v2 · v2
3
56
Thus, an orthogonal basis is
B =
1, 1, 1 , 6, −4, −2 ,
125
25 50
,
,−
42 21
42
25 50
125
,
,−
42 21
42
.
,
and an orthonormal basis is
1
1
1
3
2
1
1
4
5
√ ,√ ,√
B =
, √ , −√ , −√
, √ , √ , −√
.
3
3
3
14
14
14
42
42
42
13. We have u1 = 1, 5, 2, and u2 = −2, 1, 1. Taking v1 = u1 = 1, 5, 2 and using u2 · v1 = 5 and v1 · v1 = 30
we obtain
u2 · v1
5
13 1 2
v2 = u2 −
v1 = −2, 1, 1 −
1, 5, 2 = − , ,
.
v1 · v1
30
6 6 3
1 2
Thus, an orthogonal basis is B = 1, 5, 2 , − 13
, and an orthonormal basis is
6 , 6 , 3
1
5
2
13
1
4
√ ,√ ,√
B =
, −√
,√
,√
.
30
30
30
186
186
186
14. We have u1 = 1, 2, 3, and u2 = 3, 4, 1. Taking v1 = u1 = 1, 2, 3 and using u2 · v1 = 14 and v1 · v1 = 14 we
obtain
u2 · v1
14
v2 = u2 −
v1 = 3, 4, 1 −
1, 2, 3 = 2, 2, −2 .
v1 · v1
14
Thus, an orthogonal basis is B = {1, 2, 3 , 2, 2, −2} , and an orthonormal basis is
1
2
3
1
1
1
√ ,√ ,√
B =
, √ , √ , −√
.
14
14
14
3
3
3
15. We have u1 = 1, −1, 1, −1, and u2 = 1, 3, 0, 1. Taking v1 = u1 = 1, −1, 1, −1 and using u2 · v1 = −3 and
v1 · v1 = 4 we obtain
7 9 3 1
u2 · v1
−3
v2 = u2 −
1, −1, 1, −1 =
, , ,
.
v1 = 1, 3, 0, 1 −
v1 · v1
4
4 4 4 4
Thus, an orthogonal basis is B = 1, −1, 1, −1 , 74 , 94 , 34 , 14 , and an orthonormal basis is
1
9
3
1
1 1
1
7
√
√
√
√
B =
,
,
,
.
,− , ,−
,
2
2 2
2
2 35 2 35 2 35 2 35
16. We have u1 = 4, 0, 2, −1, u2 = 2, 1, −1, 1, and u3 = 1, 1, −1, 0. Taking v1 = u1 = 4, 0, 2, −1 and using
u2 · v1 = 5 and v1 · v1 = 21 we obtain
22
31 26
u2 · v1
5
v2 = u2 −
4, 0, 2, −1 =
, 1, − ,
.
v1 = 2, 1, −1, 1 −
v1 · v1
21
21
21 21
Next, using u3 · v1 = 2, u3 · v2 =
74
21
, and v2 · v2 =
122
21
, we obtain
u3 · v1
u3 · v2
v1 −
v2
v1 · v1
v2 · v2
74/21
22
31 26
1 24
18
40
2
4, 0, 2, −1 −
, 1, − ,
= − ,
,− ,−
.
= 1, 1, −1, 0 −
21
122/21 21
21 21
61 61
61
61
v3 = u3 −
Thus, an orthogonal basis is
B =
4, 0, 2, −1 ,
22
31 26
, 1, − ,
21
21 21
367
1 24
18
40
, − ,
,− ,−
,
61 61
61
61
7.7
Gram-Schmidt Orthogonalization Process
and an orthonormal basis is
4
2
1
22
21
31
26
√ , 0, √ , − √
B =
, √
,√
, −√
,√
,
21
21
21
2562
2562
2562
2562
1
24
18
40
−√
,√
, −√
, −√
.
2501
2501
2501
2501
17. We have u1 = 1, u2 = x, and u3 = x2 . Taking v1 = u1 = 1 and using
1
(u2 , v1 ) =
1 · x2 dx = 0 and (v1 , v1 ) =
−1
1
−1
x · x dx = 2
we obtain
v2 = u2 −
(u2 , v1 )
0
v1 = x − x = x.
(v1 , v1 )
2
Next, using
1
2
(u3 , v1 ) =
x · 1 dx = ,
3
−1
2
1
x · x dx = 0,
2
(u3 , v2 ) =
−1
and
(v2 , v2 ) =
1
−1
x · x dx =
2
,
3
we obtain
0
(u3 , v1 )
(u3 , v2 )
2/3
1
v1 −
v2 = x2 −
1−
x = x2 − .
(v1 , v1 )
(v2 , v2 )
2
2/3
3
Thus, an orthogonal basis is B = 1, x, x2 − 13 .
v3 = u3 −
18. We have u1 = x2 − x, u2 = x2 + 1, and u3 = 1 − x2 . Taking v1 = u1 = x2 − x and using
1
1
16
16
2
2
(u2 , v1 ) =
(x + 1)(x − x)dx =
(x2 − x)(x2 − x)dx =
and (v1 , v1 ) =
15
15
−1
−1
we obtain
v2 = u2 −
(u2 , v1 )
16/15 2
v1 = x2 + 1 −
(x − x) = x + 1.
(v1 , v1 )
16/15
Next, using
(u3 , v1 ) =
1
−1
(1 − x2 )(x2 − x)dx =
and
(v2 , v2 ) =
4
,
15
(u3 , v2 ) =
1
−1
1
(x + 1)(x + 1)dx =
−1
(1 − x2 )(x + 1)dx =
4
,
3
8
,
3
we obtain
(u3 , v1 )
(u3 , v2 )
4/15 2
4/3
1
5
1
v1 −
v2 = 1 − x2 −
(x − x) −
(x + 1) = − x3 − x + .
(v1 , v1 )
(v2 , v2 )
16/15
8/3
4
4
2
Thus, an orthogonal basis is B = x2 − x, x + 1, − 54 x3 − 14 x + 12 .
v3 = u3 −
19. Using the solution of Problem 17 and computing
1
||v1 ||2 = (v1 , v1 ) =
1 · 1 dx = 2,
−1
and
||v3 ||2 = (v3 , v3 ) =
1
−1
||v2 ||2 = (v2 , v2 ) =
x2 −
368
1
3
x2 −
1
3
1
−1
dx =
x · x dx =
8
,
45
2
,
3
7.7
we see that an orthonormal basis is
1
x2 − 1/3
x
B = √ ,
, =
2
2/3
8/45
20. Using the solution of Problem 18 and computing
1
16
2
||v1 || = (v1 , v1 ) =
(x2 − x)(x2 − x)dx =
,
15
−1
and
||v3 || = (v3 , v3 ) =
1
2
−1
Gram-Schmidt Orthogonalization Process
1
3
15
√ , √ x, √
2
6 2 10
x2 −
||v2 || = (v2 , v2 ) =
5
1
1
− x3 − x +
4
4
2
.
1
2
1
3
(x + 1)(x + 1)dx =
−1
1
5
1
− x3 − x +
4
4
2
dx =
8
,
3
1
,
3
we see that an orthonormal basis is
√
B =
√
15 2
3
3
(x − x), √ (x + 1),
(−5x2 − x + 2) .
4
4
2 6
√
√
√
21. Using w1 = 1/ 2, w2 = 3x/ 6, and w3 = (15/2 10)(x2 − 1/3), and computing
1
√
1
(p, w1 ) =
(9x2 − 6x + 5) √ dx = 8 2,
2
−1
1
√
3
(p, w2 ) =
(9x2 − 6x + 5) √ x dx = −2 6
6
−1
1
15
1
12
2
2
(p, w3 ) =
(9x − 6x + 5) √
x −
dx = √ ,
3
2 10
10
−1
we find from Theorem 7.5
√
√
12
p(x) = 9x2 − 6x + 5 = (p, w1 )w1 + (p, w2 )w2 + (p, w3 )w3 = 8 2 w1 − 2 6 w2 + √ w3 .
10
√
√
√
22. Using w1 = ( 15/4)(x2 − x), w2 = (3/2 6)(x + 1), and w3 = −( 3/4)(5x2 + x − 2), and computing
√
1
15
41
(p, w1 ) =
(9x2 − 6x + 5)
x2 − x dx = √ ,
4
15
−1
1
√
3
(p, w2 ) =
(9x2 − 6x + 5) √ (x + 1) dx = 3 6
2 6
−1
√
1
3
1
2
2
(p, w3 ) =
(5x + x − 2) dx = √ ,
(9x − 6x + 5) −
4
3
−1
we find from Theorem 7.5
√
41
1
p(x) = 9x2 − 6x + 5 = (p, w1 )w1 + (p, w2 )w2 + (p, w3 )w3 = √ w1 + 3 6 w2 + √ w3 .
15
3
23. Since u3 depends on u1 and u2 we would expect the Gram-Schmidt process to yield a pair of orthogonal vectors
v1 and v2 , with a third vector v3 that is 0. This is because u3 lies in the subspace W2 of R3 spanned by u1
and u2 , and hence the projection of u3 onto W2 is u3 itself. In other words,
u3 · v1
u3 · v2
u3 · v1
u3 · v2
u3 = projW3 u3 =
v1 +
v2 so v3 = u3 −
v1 +
v2 = 0.
v1 · v1
v2 · v 2
v1 · v1
v2 · v2
To carry out the orthogonalization process we take v1 = u1 = 1, 1, 3. Then, using u2 · v1 = 8 and v1 · v1 = 11
we obtain
u2 · v1
8
3 36 13
v2 = u2 −
v1 = 1, 4, 1 − 1, 1, 3 =
, ,−
.
v1 · v1
11
11 11 11
369
7.7
Gram-Schmidt Orthogonalization Process
Next, using u3 · v1 = 2, u3 · v2 =
402
11 ,
and v2 · v2 =
134
11 ,
we obtain
402/11
u3 · v1
u3 · v2
2
1, 1, 3 −
v3 = u3 −
v1 −
v2 = 1, 10, −3 −
v1 · v1
v2 · v2
11
134/11
3 36 13
, ,−
11 11 11
= 0, 0, 0 .
3 36
3
In this case {v1 , v2 } = {1, 1, 3}, 11
, 11 , − 13
11 } is an orthogonal subset of R containing the third vector
u3 = 1, 10, −3.
CHAPTER 7 REVIEW EXERCISES
1. True
2. False; the points must be non-collinear.
3. False; since a normal to the plane is 2, 3, −4 which is not a multiple of the direction vector 5, −2, 1 of the
line.
4. True
5. True
6. True
7. True
8. True
9. True
10. True; since a × b and c × d are both normal to the plane and hence parallel (unless a × b = 0 or c × d = 0.)
11. 9i + 2j + 2k
12. orthogonal
13. −5(k × j) = −5(−i) = 5i
15.
(−12)2 + 42 + 62 = 14
14. i · (i × j) = i × k = 0
16. (−1 − 20)i − (−2 − 0)j + (8 − 0)k = −21i + 2j + 8k
17. −6i + j − 7k
18. The coordinates of (1, −2, −10) satisfy the given equation.
19. Writing the line in parametric form, we have x = 1 + t, y = −2 + 3t, z = −1 + 2t. Substituting into the equation
of the plane yields (1 + t) + 2(−2 + 3t) − (−1 + 2t) = 13 or t = 3. Thus, the point of intersection is x = 1 + 3 = 4,
y = −2 + 3(3) = 7, z = −1 + 2(3) = 5, or (4, 7, 5).
20. |a| =
√
1
4
3
1
42 + 32 + (−5)2 = 5 2 ; u = − √ (4i + 3j − 5k) = − √ i − √ j + √ k
5 2
5 2
5 2
2
21. x2 − 2 = 3, x2 = 5; y2 − 1 = 5, y2 = 6; z2 − 7 = −4, z2 = 3; P2 = (5, 6, 3)
22. (5, 1/2, 5/2)
√
23. (7.2)(10) cos 135◦ = −36 2
24. 2b = −2, 4, 2; 4c = 0, −8, 8; a · (2b + 4c) = 3, 1, 0 · −2, −4, 10 = −10
25. 12, −8, 6
a·b
1
1
26. cos θ =
= √ √ = ; θ = 60◦
|a||b|
2
2 2
√
1
3 10
27. A = |5i − 4j − 7k| =
2
2
370
CHAPTER 7 REVIEW EXERCISES
28. From 3(x − 3) + 0(y − 6) + (1)(z − (−2)) = 0 we obtain 3x + z = 7.
29. | − 5 − (−3)| = 2
30. parallel: −2c = 5,
i
j
31. a × b = 1 1
1 −2
and b is
c = −5/2; orthogonal: 1(−2) + 3(−6) + c(5) = 0, c = 4
k 1 0
1 1 1 0 k = i − j − 3k A unit vector perpendicular to both a
0 = i−
j+
−2 1 1 1
1 −2 1
a×b
1
1
1
3
=√
(i − j − 3k) = √ i − √ j − √ k.
a × b
1+1+9
11
11
11
3
1/2
1/2
2
2
1/4 + 1/4 + 1/6 = ; cos α =
= , α ≈ 48.19◦ ; cos β =
= , β ≈ 48.19◦ ;
4
3/4
3
3/4
3
−1/4
1
cos γ =
= − , γ ≈ 109.47◦
3/4
3
32. a =
33. compb a = a · b/b = 1, 2, −2 · 4, 3, 0/5 = 2
34. compa b = b · a/a = 4, 3, 0 · 1, 2, −2/3 = 10/3
proja b = (compa b)a/a = (10/3)1, 2, −2/3 = 10/9, 20/9, −20/9
35. a + b = 1, 2, −2 + 4, 3, 0 = 5, 5, −2
√
compa (a + b) = (a + b) · a/ 1 + 4 + 4 = 13 (a · a + b · a) = 13 [(1 + 4 + 4) + (4 + 6 + 0)] =
1 2
2
19 38
38
proja (a + b) = [compa (a + b)](a/a) = 19
3 3, 3, −3 = 9 , 9 , − 9 19
3
36. a − b = 1, 2, −2 − 4, 3, 0 = −3, −1, −2
√
compb (a − b) = (a − b) · b/ 16 + 9 = 15 (a · b − b · b) = 15 [(4 + 6 + 0) − (16 + 9)] = −3
9
projb (a − b) = [compb (a − b)](b/b) = −3 45 , 35 , 0 = − 12
5 , − 5 , 0
37. Let a = a, b, c and r = x, y, z. Then
(a) (r − a) · r = x − a, y − b, z − c · x, y, z = x2 − ax + y 2 − by + z 2 − zc = 0 implies
a
b
c
a2 + b2 + c2
(x − )2 + (y − )2 + (z − )2 =
. The surface is a sphere.
2
2
2
4
(b) (r − a) · a = x − a, y − b, z − c · a, b, c = a(x − a) + b(y − b) + c(z − c) = 0
The surface is a plane.
38. 4, 2, −2 − 2, 4, −3 = 2, −2, 1; 2, 4, −3 − 6, 7, −5 = −4, −3, 2; 2, −2, 1 · −4, −3, 2 = 0
The points are the vertices of a right triangle.
39. A direction vector of the given line is 4, −2, 6. A parallel line containing (7, 3, −5) is (x−7)/4 = (y −3)/(−2) =
(z + 5)/6.
40. A normal to the plane is 8, 3, −4. The line with this direction vector and through (5, −9, 3) is x = 5 + 8t,
y = −9 + 3t, z = 3 − 4t.
41. The direction vectors are −2, 3, 1 and 2, 1, 1. Since −2, 3, 1 · 2, 1, 1 = 0, the lines are orthogonal. Solving
1 − 2t = x = 1 + 2s, 3t = y = −4 + s, we obtain t = −1 and s = 1. The point (3, −3, 0) obtained by letting
t = −1 and s = 1 is common to the two lines, so they do intersect.
42. Vectors in the plane are 2, 3, 1 and 1, 0, 2. A normal vector is 2, 3, 1 × 1, 0, 2 = 6, −3, −3 = 32, −1, −1.
An equation of the plane is 2x − y − z = 0
43. The lines are parallel with direction vector 1, 4, −2. Since (0, 0, 0) is on the first line and (1, 1, 3) is on the second
line, the vector 1, 1, 3 is in the plane. A normal vector to the plane is thus 1, 4, −2 × 1, 1, 3 = 14, −5, −3.
An equation of the plane is 14x − 5y − 3z = 0.
371
CHAPTER 7 REVIEW EXERCISES
44. Letting z = t in the equations of the plane and solving −x + y = 4 + 8t, 3x − y = −2t, we obtain x = 2 + 3t,
y = 6 + 11t, z = t. Thus, a normal to the plane is 3, 11, 1 and an equation of the plane is
3(x − 1) + 11(y − 7) + (z + 1) = 0
or
3x + 11y + z = 79.
√
√
a
10
= √ (i + j) = 5 2 i + 5 2 j; d = 7, 4, 0 − 4, 1, 0 = 3i + 3j
a
2
√
√
√
W = F · d = 15 2 + 15 2 = 30 2 N-m
√
√
√
√
46. F = 5 2 i + 5 2 j + 50i = (5 2 + 50)i + 5 2 j; d = 3i + 3j
√
√
√
W = 15 2 + 150 + 15 2 = 30 2 + 150 N-m ≈ 192.4 N-m
√
√
√
√
√
47. Since F2 = 200(i + j)/ 2 = 100 2 i + 100 2 j, F3 = F2 − F1 = (100 2 − 200)i + 100 2 j and
√
√
√
F3 = (100 2 − 200)2 + (100 2)2 = 200 2 − 2 ≈ 153 lb.
45. F = 10
48. Let F1 = F1 and F2 = F2 . Then F1 = F1 [(cos 45◦ )i + (sin 45◦ )j] and F2 = F2 [(cos 120◦ )i + (sin 120◦ )j], or
√
F1 = F1 ( √12 i + √12 j) and F2 = F2 (− 12 i + 23 j). Since w + F1 + F2 = 0,
√
√
3
3
1
1
1
1
1
1
F1 ( √ i + √ j) + F2 (− i +
j) = 50j,
( √ F1 − F2 )i + ( √ F1 +
F2 )j = 50j
2
2
2
2
2
2
2
2
and
√
1
3
1
1
√ F1 − F2 = 0,
√ F1 +
F2 = 50.
2
2
2
2
√
√
√
Solving, we obtain F1 = 25( 6 − 2 ) ≈ 25.9 lb and F2 = 50( 3 − 1) ≈ 36.6 lb.
49. Not a vector space. Axiom (viii) is not satisfied.
50. The vectors are linearly independent. The only solution of the system
c1 = 0,
c1 + 2c2 + c3 = 0,
2c1 + 3c2 − c3 = 0
is c1 = 0, c2 = 0, c3 = 0.
51. Let p1 and p2 be in Pn such that
d 2 p1
d 2 p2
=
0
and
= 0. Since
dx2
dx2
d 2 p1
d 2 p2
d2
d 2 p1
d2
+
= 2 (p1 + p2 ) and 0 = k
= 2 (kp1 )
2
2
2
dx
dx
dx
dx
dx
we conclude that the set of polynomials with the given property is a subspace of Pn . A basis for the subspace
is 1, x.
0=
52. The intersection W1 ∩ W2 is a subspace of V . If x and y are in W1 ∩ W2 then x and y are in each subspace
and so x + y is in each subspace. That is, x + y is in W1 ∩ W2 . Similarly, if x is in W1 ∩ W2 then x is in each
subspace and so kx is in each subspace. That is, kx is in W1 ∩ W2 for any scalar k.
The union W1 ∪ W2 is generally not a subspace. For example, W1 = {x, y y = x} and W2 = {x, y y = 2x}
are subspaces of R2 . Now 1, 1 is in W1 and 1, 2 is in W2 but 1, 1 + 1, 2 = 2, 3 is not in W1 ∪ W2 .
372
8
Matrices
EXERCISES 8.1
Matrix Algebra
1. 2 × 4
2. 3 × 2
3. 3 × 3
4. 1 × 3
6. 8 × 1
7. Not equal
8. Not equal
9. Not equal
11. Solving x = y − 2, y = 3x − 2 we obtain x = 2, y = 4.
12. Solving x2 = 9, y = 4x we obtain x = 3, y = 12 and x = −3, t = −12.
13. c23 = 2(0) − 3(−3) = 9; c12 = 2(3) − 3(−2) = 12
14. c23 = 2(1) − 3(0) = 2; c12 = 2(−1) − 3(0) = −2
4−2 5+6
2 11
15. (a) A + B =
=
−6 + 8 9 − 10
2 −1
−2 − 4
6−5
−6
1
(b) B − A =
=
8 + 6 −10 − 9
14 −19
8 10
−6
18
2
(c) 2A + 3B =
+
=
−12 18
24 −30
12

 

−2 − 3 0 + 1
−5
1

 

16. (a) A − B =  4 − 0 1 − 2  =  4 −1 
7+4 3+2
11
5

 

3 + 2 −1 − 0
5 −1

 

(b) B − A =  0 − 4
2 − 1  =  −4
1
−4 − 7 −2 − 3
−11 −5

 

1 −1
2 −2

 

(c) 2(A + B) = 2  4
3 = 8
6
3
1
6
2
−2 − 9
12 − 6
−11
6
17. (a) AB =
=
5 + 12 −30 + 8
17 −22
−2 − 30 3 + 24
−32 27
(b) BA =
=
6 − 10 −9 + 8
−4 −1
4
+
15
−6
−
12
19 −18
(c) A2 =
=
−10 − 20 15 + 16
−30
31
28
−12
373
5. 3 × 4
10. Not equal
8.1
Matrix Algebra
1 + 18 −6 + 12
19
6
=
−3 + 6
18 + 4
3 22

 
−4 + 4 6 − 12
−3 + 8
0

 
18. (a) AB =  −20 + 10 30 − 30 −15 + 20  =  −10
−32 + 12 48 − 36 −24 + 24
−20
(d) B2 =

−6 5

0 5
12 0
(b)
19. (a)
(b)
(c)
(d)
20. (a)
−4 + 30 − 24 −16 + 60 − 36
2
8
BA =
=
1 − 15 + 16
4 − 30 + 24
2 −2
9 24
BC =
3 8
1 −2
9 24
3
8
A(BC) =
=
−2
4
3
8
−6 −16
0 2
0 0
0 0
C(BA) =
=
3 4
0 0
0 0
1 −2
6 5
−4 −5
A(B + C) =
=
−2
4
5 5
8 10


3


AB = ( 5 −6 7 )  4  = (−16)

3
−1



(b) BA =  4  ( 5 −6
−1

15

(c) (BA)C =  20
−5

15 −18

7 ) =  20 −24
−5
6

−18
21
−24
6
1

28   0
2
1
−7
2
3

21

28 
−7
 
78
4
 
−1  =  104
−26
1
54
72
−18

99

132 
−33
(d) Since AB is 1 × 1 and C is 3 × 3 the product (AB)C is not defined.


4


21. (a) AT A = ( 4 8 −10 )  8  = (180)
−10
 
2
 
T
(b) B B =  4  ( 2
5

4

4
8 10


5 ) =  8 16 20 
10 20 25

   

4
2
6

   

(c) A + BT =  8  +  4  =  12 
−10
5
−5
1 2
−2 5
−1
7
22. (a) A + BT =
+
=
2 4
3 7
5 11
2 4
−2 5
4 −1
(b) 2AT − BT =
−
=
4 8
3 7
1
1
374
8.1 Matrix Algebra
2
3 −1
−3
−7
=
4
−3 −3
−6 −14
T 7 10
7 38
(AB)T =
=
38 75
10 75
5 −2
3 8
7 38
T T
B A =
=
10 −5
4 1
10 75
5 −4
−3 11
2 7
AT + B =
+
=
9
6
−7
2
2 8
10 18
−3 −7
7 11
2A + BT =
+
=
−8 12
11
2
3 14
(c) AT (A − B) =
23. (a)
(b)
24. (a)
(b)
25.
27.
−4
8
−19
18
−
4
16
−
19
20
+
1
2
−6
9
=
−38
−2
=
−14
1
29. 4 × 5



 
 

−5
−6
−5

 
 
 

26.  3  +  −5  +  −8  =  −10 
−3
15
10
22
6

 
 
 

−7
−1
2
−10

 
 
 

28.  17  +  1  −  8  =  10 
−6
4
−6
4
30. 3 × 2
2 −3
2 4
6 −6
;
(AT )T =
=A
32. (A + B)T =
= A T + BT
4
2
−3 2
14 10
T 16
40
16
−8
4 2
2 −3
16 −8
T
T T
33. (AB) =
=
;
B A =
=
−8 −20
40 −20
10 5
4
2
40 −20
12 −18
34. (6A)T =
= 6AT
24
12




2 1 5 15 9

 2 6 2


35. B = AAT =  6 3 
=  15 39 27  = BT
1 3 5
2 5
9 27 29
31. AT =
36. Using Problem 33 we have (AAT )T = (AT )T AT = AAT , so that AAT is symmetric.
1 0
0 0
37. Let A =
and B =
. Then AB = 0.
0 0
0 1


2 3 4


38. We see that A = B, but AC =  4 6 8  = BC.
6 9 12
39. Since (A+B)2 = (A+B)(A+B) = A2 +AB+BA+B2 , and AB = BA in general, (A+B)2 = A2 +2AB+B2 .
40. Since (A + B)(A − B) = A2 − AB + BA − B2 , and AB = BA in general, (A + B)(A − B) = A2 − B2 .
41. a11 x1 + a12 x2 = b1 ;
a21 x1 + a22 x2 = b2


  
7
2 6
1
x1


  
42.  1 2 −1   x2  =  −1 
9
5 7 −4
x3
375
8.1
Matrix Algebra
43. ( x

y)
a
b/2
b/2
c
x
y
=( ax + by/2
bx/2 + cy )
x
y
=( ax2 + bxy/2 + bxy/2 + cy 2 )=( ax2 + bxy + cy 2 )
  

0
−∂/∂z
∂/∂y
P
−∂Q/∂z + ∂R/∂y

  

44.  ∂/∂z
0
−∂/∂x   Q  =  ∂P/∂z − ∂R/∂x  = curl F
−∂/∂y
∂/∂x
0
R
−∂P/∂y + ∂Q/∂x
  
  
 

x
cos γ sin γ 0
x
x cos γ + y sin γ
xY
  
  
 

45. (a) MY  y  =  − sin γ cos γ 0   y  =  −x sin γ + y cos γ  =  yY 
z
0
0
1
z
z
zY




1
0
0
cos β 0 − sin β




(b) MR =  0
cos α sin α 
1
0 ;
MP  0
0 − sin α cos α
sin β 0
cos β
  
  
  

1 0
0
1
1
1
1
0
0
1
√
  
  
1 √

3
1  
(c) MP  1  =  0
cos 30◦ sin 30◦   1  =  0
2
2   1  =  2 ( 3 + 1) 
√
√
1
3
1
1
1
0 − sin 30◦ cos 30◦
0 − 12
2 ( 3 − 1)
2
√
√
  






2
1
cos 45◦ 0 − sin 45◦
1
1
0 − 22
2
  
 1 √
 √
 

MR MP  1  =  0
1
0
0   12 ( 3 + 1) 
  2 ( 3 + 1)  =  √0 1
√
√
√
1
2
2
1
1
sin 45◦ 0
cos 45◦
0
2 ( 3 − 1)
2
2
2 ( 3 − 1)
√ 
1 √
6)
4 (3 2 −
 1 √

=  2 ( 3 + 1) 
√
√
1
6)
4( 2 +
√ 
  
 1 √
6)
1
cos 60◦ sin 60◦ 0
4 (3 √2 −
  

 1
◦
◦
MY MR MP  1  =  − sin 60 cos 60 0   2 ( 3 + 1) 
√
√
1
1
0
0
1
6)
4( 2 +
√
√
√
√
√
 1


 1 √


3
1
0
(3 2 − 6 )
(3 2 − 6 + 6 + 2 3 )
2
2
4
8
√
√
√
 √
 1

 1 √
1
=  − 23
0   2 ( 3 + 1)  =  8 (−3 6 + 3 2 + 2 3 + 2) 
2
√
√
√
√
1
1
6)
6)
0
0 1
4( 2 +
4( 2 +
1 0
2 −2
2 −2
46. (a) LU = 1
=
=A
1
0
3
1
2
2
1 0
6 2
6
2
(b) LU = 2
=
=A
4 1
1
0 − 13
3


 

1
0 0
1 −2
1
1 −2 1


 

(c) LU =  0
1 00
1
2 = 0
1 2 = A
2 10 1
0
0 −21
2
6 1


 

1 0 0
1
1
1
1
1 1


 

(d) LU =  3 1 0   0 −2 −1  =  3
1 2 = A
1 1 1
0
0
1
1 −1 1


17 43
A11 A12
B1
A11 B1 + A12 B2


47. (a) AB =
=
=  3 75 
A21 A22
B2
A21 B1 + A22 B2
−14 51
376
8.2
since
A11 B1 + A12 B2 =
13
−9
25
49
+
4 18
12 26
Systems of Linear Algebraic Equations
=
17 43
3 75
and
A21 B1 + A22 B2 = ( −24
17 ) = ( −14
34 ) + ( 10
51 ) .
(b) It is easier to enter smaller strings of numbers and the chance of error is decreased. Also, if the large matrix
has submatrices consisting of all zeros or diagonal matrices, these are easily entered without listing all of
the entries.
EXERCISES 8.2
Systems of Linear Algebraic Equations
1.
1 −1
4
3
11
−5
−4R1 +R2
−−−−−−→
1
−1
0
7
The solution is x1 = 4, x2 = −7.
4
3 −2
1 −1
R12
2.
−−−−−
−→
1 −1 −2
3 −2
The solution is x1 = 8, x2 = 10.
1
1 13
9
3 −5
R1
3.
−−−9−−−→
2 −1 −1
2 1
− 1 R2 +R1
−−3−−−−→
11
−49
−2
− 59
−1
0
− 23
0
1
1
3
1
7 R2
−−−−−−→
4
1
−3R +R
2
−−−−1−−→
−2R +R
2
−−−−1−−→
1
−1
0
1
1 −1
0
1
1
0
1
3
1
3
11
−7
−2
1
9
R3 +R1
−−−−−−→
R +R
1
−−−2−−−→
10
− 59
3R
−−−−−2−→
1
0
0
1
4
−7
8
1 0
0 1
10
1
1
3
− 59
0
1
1
3
The solution is x1 = − 23 , x2 = 13 .
4.
10
15
3
2
1
−1
1
R1
10
−−−
−−−→
− 3 R2 +R1
−−2−−−−→
1
3
2
1
10
3
2
−1
1
0
− 17
25
0
1
13
25
−3R +R
2
−−−−1−−→
1
0
3
2
− 52
1
10
− 13
10
− 2 R2
5
−−−−
−−→
1
3
2
0
1
1
10
13
25
13
The solution is x1 = − 17
25 , x2 = 25 .






−3
1 −1 −1 −3
1 −1 −1
1 −1 −1 −3
1
R2


 −2R +R2 

7
13 
3
5
5
7 13  −−−5−−−→  0
1
5.  2
7  −−−−1−−→
0
5
5 
−R1 +R3
1 −2
3 −11
0 −1
4 −8
0 −1
4 −8





1 0 25
1 0 0
1 0 25 − 25
− 25
5
2
R
−
R
+R
1 
R +R1 

27 3
13 
7
13  5 3
−−−2−−−→
−
−
−
−
−
−→
0
1
−
−
−
−
−
−→
0 1 0
 0 1 75




5
5
5
R2 +R3
− 75 R3 +R2
27
27
0 0 1
0 0 5 −5
0 0 1 −1
The solution is x1 = 0, x2 = 4, x3 = −1.
377

0

4
−1
8.2
Systems of Linear Algebraic Equations

1

6.  2
1



0
1
2 −1
1
2 −1 0
1
 −2R +R2 
 − 3 R2 

1
2 9  −−−−1−−→
4 9  −−−−
1 − 43 −3 
−−→  0
 0 −3
−R1 +R3
−1
1 3
0 −3
2 3
0 −3
2
3





5
5
6
6
1 0
1 0 0
1 0
3
3
1
5
−2R2 +R1 
 − 2 R3 
 − 3 R3 +R1 
4
4
−−−−−−→  0 1 − 3 −3  −−−−−−→  0 1 − 3 −3  −−
−−−−→  0 1 0
4
3R2 +R3
3 R3 +R2
0 0 −2 −6
0 0
1
3
0 0 1
2
−1
0


1


1
3
The solution is x1 = 1, x2 = 1, x3 = 3.
1 1 1 0 −R1 +R2
1 1 1 0
7.
−−−−−−→
1 1 3 0
0 0 2 0
Since x3 = 0, setting x2 = t we obtain x1 = −t, x2 = t, x3 = 0.
1
1
2 −4 9 −5R1 +R2 1
9
2 −4
1 2 −4
− 11
R2
8.
−−−−−−→
−−−−
−−→
5 −1
2 1
0 −11 22 −44
0 1 −2
If x3 = t, the solution is x1 = 1, x2 = 4 + 2t, x3 = t




1 −1 −1 8
8
1 −1 −1
row




9.  1 −1
1 3  −−−−−−→  0
0
2 −5 
operations
−1
1
1 4
0
0
0 12
Since the bottom row implies 0 = 12, the system is inconsistent.



4 
1 13
3
1
4
3
row




10.  4
3 −3  −−−−−−→  0 1 −5 
operations
2 −1 11
0 0
0
The solution

2 2 0

11.  −2 1 1
3 0 1
is x1 = 3, x2 = −5.


1 1
0
row


0  −−−−−−→  0 1
operations
0
0 0
0
0
1
3

0
1
0
The solution is x1 = x2 = x3 = 0.



1 −1
1 −1 −2 0
row



4
5 0  −−−−−−→  0
12.  2
1
operations
6
0 −3 0
0
0
The solution is x1 = 12 t, x2 = − 32 t,



1
2
2 2
1
row



13.  1
1
1 0  −−−−−−→  0
operations
1 −3 −1 0
0

−2
0

3
2

0
0
0
x3 = t.
2 2
1 1
2
0
4
1
The solution is x1 = −2, x2 − 2, x3 = 4.



1 −2
1
1 −2 1 2
row



1
14.  3 −1 2 5  −−−−−−→  0
1 −5
operations
2
1 1 1
0
0
0


2
2


− 15 
−2
Since the bottom row implies 0 = −2, the system is inconsistent.
378
9
4
−2R +R
1
−−−−2−−→
1 0
0 1
0
−2
1
4
8.2

1
1
1

15.  1 −1 −1
3
1
1
3


3
1 1 1
row


−1  −−−−−−→  0 1 1
operations
0 0 0
5
Systems of Linear Algebraic Equations


2
0
If x3 = t the solution is x1 = 1, x2 = 2 − t, x3 = t.




1 −1 −2 −1
1 −1 −2 −1
row




16.  −3 −2
1 −7  −−−−−−→  0
1
1
2
operations
2
3
1
8
0
0
0
0
If x3 = t the solution is x1 = 1 + t, x2 = 2 − t, x3 = t.


 1 0 1 −1 1 
1
0 1 −1
1
0
0 1 1
3 
1
3
2 1
1
row



2
2
2 
17. 
 −−−−−−→ 

 1 −1 0


1 −1 operations 0 0 1 −5 1 
1
1 1
1
2
0 0 0
1 0
The

2
3

18. 
1
4
solution is x1 = 0, x2 = 1, x3 = 1, x4


1 12
1
1 0
3
0 1
1
1 1
4
row


 −−−−−−→ 

2
2 3
3 operations  0 0
5 −2 1 16
0 0
The

1
0

19. 
1
1
solution is x1 = 1, x2 = 2, x3 = −1, x4 = 0.


3 5 −1
1
1 3 5 −1
 0 1 1 −1
1 1 −1
4
row


 −−−−−−→ 
2 5 −4 −2  operations  0 0 1 −4
4 6
−2
0
6
Since the bottom row implies 0 = 1,



1 2 0
1 0
1
 4 9 1 12 0 
0
row



20. 
 −−−−−−→ 
 3 9 6 21 0  operations  0
1
3 1
9
0
0
0
= 0.
1
2
0
1
1
−2
−1
0
1
0
3
2

1


−1 
0
1

4


1
0
1
the system is inconsistent.

2 0
1 0
1 1
8 0


0 1 −2 0 
0
0
0
If x4 = t the solution is x1 = 19t, x2 = −10t, x3



1
1
1
4.280
1
row



21.  0.2 −0.1 −0.5 −1.978  −−−−−−→  0
operations
4.1
0.3
0.12
1.686
0
0
= 2t, x4 = t.
1
1
1
2.333
0
1
The solution is x1 = 0.3, x2 = −0.12, x3 = 4.1.



2.5 1.4
4.5
2.6170
1
row



22.  1.35 0.95
1.2
0.7545  −−−−−−→  0
operations
2.7 3.05 −1.44 −1.4292
0
0.56
1
4.28

9.447 
4.1
1.8
−6.3402
0

1
1.0468


−3.3953 
0.28
The solution is x1 = 1.45, x2 = −1.62, x3 = 0.28.
23. From x1 Na + x2 H2 O → x3 NaOH + x4 H2 we obtain the system x1 = x3 , 2x2 = x3 + 2x4 , x2 = x3 . We see
that x1 = x2 = x3 , so the second equation becomes 2x1 = x1 + 2x4 or x1 = 2x4 . A solution of the system is
x1 = x2 = x3 = 2t, x4 = t. Letting t = 1 we obtain the balanced equation 2Na + 2H2 O → 2NaOH + H2 .
379
8.2
Systems of Linear Algebraic Equations
24. From x1 KClO3 → x2 KCl + x3 O2 we obtain the system x1 = x2 , x1 = x2 , 3x1 = 2x3 . Letting x3 = t we see that
a solution of the system is x1 = x2 = 23 t, x3 = t. Taking t = 3 we obtain the balanced equation
2KClO3 → 2KCl + 3O2 .
25. From x1 Fe3 O4 + x2 C → x3 Fe + x4 CO we obtain the system 3x1 = x3 , 4x1 = x4 , x2 = x4 . Letting x1 = t we
see that x3 = 3t and x2 = x4 = 4t. Taking t = 1 we obtain the balanced equation
Fe3 O4 + 4C → 3Fe + 4CO.
26. From x1 C5 H8 + x2 O2 → x3 CO2 + x4 H2 O we obtain the system 5x1 = x3 , 8x1 = 2x4 , 2x2 = 2x3 + x4 . Letting
x1 = t we see that x3 = 5t, x4 = 4t, and x2 = 7t. Taking t = 1 we obtain the balanced equation
C5 H8 + 7O2 → 5CO2 + 4H2 O.
27. From x1 Cu + x2 HNO3 → x3 Cu(NO3 )2 + x4 H2 O + x5 NO we obtain the system
x1 = 3,
x2 = 2x4 ,
x2 = 2x3 + x5 ,
3x2 = 6x3 + x4 + x5 .
Letting x4 = t we see that x2 = 2t and
2t = 2x3 + x5
6t = 6x3 + t + x5
Then x3 =
3
4t
and x5 =
1
2 t.
Finally, x1 = x3 =
3
4 t.
2x3 + x5 = 2t
or
6x3 + x5 = 5t.
Taking t = 4 we obtain the balanced equation
3Cu + 8HNO3 → 3Cu(NO3 )2 + 4H2 O + 2NO.
28. From x1 Ca3 (PO4 )2 + x2 H3 PO4 → x3 Ca(H2 PO4 )2 we obtain the system
3x1 = x3 ,
2x1 + x2 = 2x3 ,
8x1 + 4x2 = 8x3 ,
3x2 = 4x3 .
Letting x1 = t we see from the first equation that x3 = 3t and from the fourth equation that x2 = 4t. These
choices also satisfy the second and third equations. Taking t = 1 we obtain the balanced equation
Ca3 (PO4 )2 + 4H3 PO4 → 3Ca(H2 PO4 )2 .
29. The system of equations is
−i1 + i2 − i3 = 0
10 − 3i1 + 5i3 = 0
−i1 + i2 − i3 = 0
3i1 − 5i3 = 10
or
27 − 6i2 − 5i3 = 0
Gaussian elimination gives

−1
1 −1

 3 0
0 6
−5
5
6i2 + 5i3 = 27


1 −1
row


1
10  −−−−−−→  0
operations
0
0
27
0
35
38
1
, i2 =
, i3 = .
9
9
3
30. The system of equations is
i1 − i2 − i3 = 0
1
−8/3
0

10/3  .
1
1/3
The solution is i1 =
52 − i1 − 5i2 = 0
i1 − i2 − i3 = 0
or
−10i3 + 5i2 = 0
i1 + 5i2 = 52
5i2 − 10i3 = 0
380

8.2
Gaussian elimination gives

−1
1

1
0


0
1 −1
row


52  −−−−−−→  0
1
operations
0
0
0
−1
5
0
5 −10
Systems of Linear Algebraic Equations
−1
1/6
1

0

26/3  .
4
The solution is i1 = 12, i2 = 8, i3 = 4.
31. Interchange row 1 and row in I3 .
32. Multiply row 3 by c in I3 .
33. Add c times row 2 to row 3 in I3 .
34. Add row 4 to row 1 in I4 .

a21

35. EA =  a11
a22
a12

a23

a13 
a31
a32
a33


37. EA = 
a11

38. E1 E2 A = E1 
a11

36. EA =  a21
ca31

a12
a22
a13
a23
ca22 + a32
ca23 + a33
a21
ca21 + a31


ca32

a13

a23 
ca33



a11
a21
a12
a22
a13
a23
ca21 + a31
ca22 + a32
ca23 + a33
39. The system is equivalent to
a12
a22
1
0
1
2
1
Letting
Y=
a21
a11
a22
a12
a23
a13
ca21 + a31
ca22 + a32
ca23 + a33
 
=
we have

y1
0
1
2
1
=
y2
1
−2
3
2
0
y1
y2
X=
2
−2
0
3
=



2
.
6
X
2
6
.
This implies y1 = 2 and 12 y1 + y2 = 1 + y2 = 6 or y2 = 5. Then
2
2 −2
x1
=
,
5
0
3
x2
which implies 3x2 = 5 or x2 =
40. The system is equivalent to
5
3
and 2x1 − 2x2 = 2x1 −
1
0
2
3
1
Letting
Y=
we have
6
2
0
− 13
y1
y2
1
0
2
3
1
10
3
=
y1
y2
X=
6
0
381
8
3
= 2 or x1 =
2
− 13
=
1
−1
1
−1
X
.
.
. The solution is X =
8 5
3, 3
.
8.2
Systems of Linear Algebraic Equations
This implies y1 = 1 and 23 y1 + y2 =
+ y2 = −1 or y2 = − 53 . Then
6
2
x1
1
=
,
0 − 13
− 53
x2
2
3
which implies − 13 x2 = − 53 or x2 = 5 and 6x1 + 2x2 = 6x1 + 10 = 1 or x1 = − 32 . The solution is X = − 32 , 5 .
41. The system is equivalent to

1

0
2
0
1
10
Letting

0
1

00
1
0
−2
1
0






1
2



2  X =  −1  .
−21
1
1
  
Y =  y2  =  0
−2
1
0
0
y1
y3
we have

1

0
0
1
2
10

1

2X
−21
  

0
y1
2
  

0   y2  =  −1  .
1
1
y3
This implies y1 = 2, y2 = −1, and 2y1 + 10y2 + y3 = 4 − 10 + y3 = 1 or y3 = 7. Then

  

1 −2
1
x1
2

  

1
2   x2  =  −1  ,
0
0
0 −21
7
x3
which implies −21x3 = 7 or x3 = − 13 , x2 + 2x3 = x2 − 23 = −1 or x2 = − 13 , and x1 − 2x2 + x3 = x1 + 23 − 13 = 2
or x1 =
5
3
. The solution is X =
42. The system is equivalent to
5
1
1
3, −3, −3

1

3
1
0
1
1
Letting
.

1

00
1
0

 
1
1
0

 
−2 −1  X =  1  .
0
1
4


0
y1

1
  
Y =  y2  =  0
0
y3
we have

1

3
1
0
1
1
1
−2
0

1

−1  X
1
   
0
y1
0
   
0   y2  =  1  .
1
4
y3
This implies y1 = 0, 3y1 + y2 = y2 = 1, and y1 + y2 + y3 = 0 + 1 + y3 = 4 or y3 = 3. Then

   
1
1
1
x1
0

   
0
−2
−1
=
x

 2  1,
0
0
1
3
x3
which implies x3 = 3, −2x2 − x3 = −2x2 − 3 = 1 or x2 = −2, and x1 + x2 + x3 = x1 − 2 + 3 = 0 or x1 = −1.
The solution is X = (−1, −2, 3).
382
8.3 Rank of a Matrix
43. Using the Solve function in Mathematica we find x1 = −0.0717393 − 1.43084c, x2 = −0.332591 + 0.855709c,
x3 = c, where c is any real number
44. Using the Solve function in Mathematica we find x1 = c/3, x2 = 5c/6, x3 = c, where c is any real number
45. Using the Solve function in Mathematica we find x1 = −3.76993, x2 = −1.09071, x3 = −4.50461, x4 = −3.12221
46. Using the Solve function in Mathematica we find x1 =
where b and c are any real numbers.
8
3
− 73 b + 23 c, x2 =
EXERCISES 8.3
Rank of a Matrix
1.
2.
3 −1
1
3
2 −2
0
2

3.  6
−1
1
1
2
0
1
3

7. 
7
1
0
3
; The rank is 2.
1
1
−1
0
0
2

5
1
1

8. 
0
2
−2
4
1
5
;

1
row


4  −−−−−−→  0
operations
3
0
0
3
6
0
6
1
1
0
The rank is 1.
1
2
0
0
3
2


0 ;
0

2

5 ;
1
The rank is 3.
1
− 13
2
3
0
0
1
0
5
4
−2
The rank is 1.
The rank is 3.

1
1

1 −3 ;
0
1
row
−−−−−−→
operations


−2
1
0
−6 
row


 −−−−−−→ 
−1  operations  0
4


1
3
row


9  −−−−−−→  0
operations
− 32
0
3 −1 2 0
6
2 4 5



1
1
row


4  −−−−−−→  0
operations
1
0
1 1

5.  1 0
1 4

row
−−−−−−→
operations
− 12

row
−−−−−−→
operations
1
3

4.  −1
−1
6.
0


;
The rank is 2.

1

;
0
0
The rank is 2.

 1 −2 3
4
0
1 0
8
row


 −−−−−−→ 
0  operations  0
0 1
8
0
0 0
4
0

;
4 

3
The rank is 3.
0
383
2
3
− 13 b − 13 c, x3 = −3, x4 = b, x5 = c,
8.3
Rank of a Matrix
0
2
4
2
4

9. 
2
1
0
5
1
2
3
3
6
6
6
12
1
2
0
1
row


−−−−−−→ 
1 

operations  0
3
0

1 −2 1 8 −1 1
1
0
0 1 3 −1 1
1


10.  0
0 1 3 −1 2 10

0
0 0 0
0 1
1
1 −2 1 8 −1 1
2



1
2 3
1
row



11.  1
0 1  −−−−−−→  0
operations
1 −1 5
0
1
2
0
1
1
3
4
3
3
2
1
1
6
− 13
0
0
1
0
0
0
0


;
2


6
1


5
0
row


8  −−−−−−→  0
 operations 
0
3
6

0
−2
0
0
0
0
The rank is 3.
1
1
0
0
0
8
3
0
0
0

−1 1 1 6
−1 1 1 5 


0 1 9 3 ;

0 0 1 0
0 0 0 0
The rank is 4.

2
1
0
3

1 ;
1
Since the rank of the matrix is 3 and there are 3 vectors, the vectors are linearly independent.


1
2
6 3
 1 −1 4 
0
row



12. 
 −−−−−−→ 
3
2 1  operations  0
2
5 4
0
Since the

1 −1

13.  1 −1
1
−1
Since the

2
1
2
2

14. 
 3 −1
−1
1
4
− 58 


1
0
0
0
rank of the matrix is 3 and there are 4 vectors, the vectors are linearly dependent.



3 −1
1 −1 3 −1
row



4
2  −−−−−−→  0
0 1
3
operations
5
7
0
0 0
1
rank of the matrix is 3 and there are 3 vectors, the vectors are linearly independent.



1
5
1 1 1 −1
 0 1 1 −7 
1
1
row



 −−−−−−→ 

6
1  operations  0 0 1 −3 
1
1 1 −1
0 0 0
1
Since the rank of the matrix is 4 and there are 4 vectors, the vectors are linearly independent.
15. Since the number of unknowns is n = 8 and the rank of the coefficient matrix is r = 3, the solution of the
system has n − r = 5 parameters.
16. (a) The maximum possible rank of A is the number of rows in A, which is 4.
(b) The system is inconsistent if rank(A) < rank(A/B) = 2 and consistent if rank(A) = rank(A/B) = 2.
(c) The system has n = 6 unknowns and the rank of A is r = 3, so the solution of the system has n − r = 3
parameters.
17. Since 2v1 + 3v2 − v3 = 0 we conclude that v1 , v2 , and v3 are linearly dependent. Thus, the rank of A is at
most 2.
18. Since the rank of A is r = 3 and the number of equations is n = 6, the solution of the system has n − r = 3
parameters. Thus, the solution of the system is not unique.
19. The system consists of 4 equations, so the rank of the coefficient matrix is at most 4, and the maximum number
of linearly independent rows is 4. However, the maximum number of linearly independent columns is the same
384
8.4 Determinants
as the maximum number of linearly independent rows. Thus, the coefficient matrix has at most 4 linearly
independent columns. Since there are 5 column vectors, they must be linearly dependent.
20. Using the RowReduce in Mathematica we find

1 0 0

0 1 0


0 0 1


0 0 0
0
0
that the reduced row-echelon form of the augmented matrix is

834
0 0
− 261
2215
443
1818
282 
0 0
2215
443 

13
6 
.
0 0
−
443
443 


4214
1 0
− 130
2215
443 
0
0
− 6079
2215
1
677
443
We conclude that the system is consistent and the solution is x1 = − 226
443 −
6
x3 = − 443
−
13
443 c,
x4 = − 130
443 −
4214
2215 c,
x5 =
677
433
+
6079
2215 c,
834
2215 c,
x2 =
282
443
−
x6 = c.
EXERCISES 8.4
Determinants
1. M12 =
1
−2
2
=9
5
3. C13 = (−1)1+3
5. M33
0
= 1
1
2
2
1
3+4
7. C34 = (−1)
2. M32 =
1
−1
−2
3
6. M41
0
2
1
1
2 −2 = 10
1
1
13. (1 − λ)(2 − λ) − 6 = λ2 − 3λ − 4
0 2
3 0
0 5
0
2
1 = −3
5
8
17.
3 0
2 7
2 6
2
7
1 =3
6
4
2
= 2
1
4
−2
0
4
10. 2
15.
4
2
4. C22 = (−1)2+2
=1
0
3 =2
2
9. −7
2
1
0
= −48
8
1
2
+2
4
2
2+3
8. C23 = (−1)
2
4
−2
5
= 18
0
3 = 24
−1
0
2
5
1
1
1
0
−1 = 22
2
12. −1/2
11. 17
14. (−3 − λ)(5 − λ) − 8 = λ2 − 2λ − 23
16.
7
= 3(22) + 2(−2) = 62
6
385
5
0
0
0
−3
0
0
−3
0 =5
0
2
0
= 5(−3)(2) = −30
2
1818
2215 c,
8.4
Determinants
18.
1 −1 −1
2
2
2 −2 =
1
1
1
9
19.
4 5
1 2
1 2
3
2
3 =4
2
3
6
0
20.
1
4
1
3
1
2
8
0 = 0, expanding along the third column.
9
0
−2
3
1
−5
3
1
21.
−2 −1
−3
6
−3
4
4
6
1 = −2
4
8
22.
3
−1
7
1
2
5 =3
−4
10
23.
5
2
−4
1
x
1
y
2
3
1
y
z =
3
4
1
24.
1
−2
9
−1 −1
1
3
1
+3
3
1
1
−1
+3
8
4
5
−5
10
z
x
−
4
2
y
z
=
3+y
4+z
−1
−1
2
−2
= 20 − 2(−8) + 4 = 40
2
=0
2
4
−1
−3
8
6
−1
5
7
10
z
x
+
4
2
1
x
y
2+x 3+y
+
9
+
4
= −2(44) + 3(−24) − 3(−25) = −85
1
−1
2
7
−4
= 3(40) − 5(−45) + (−10) = 335
y
= (4y − 3z) − (4x − 2z) + (3x − 2y) = −x + 2y − z
3
z
4+z
−
x
z
x
y
+
2+x 4+z
2+x 3+y
= (4y + yz − 3z − yz) − (4x + xz − 2z − xz) + (3x + xy − 2y − xy) = −x + 2y − z
25.
26.
1
1 −3
1
5
3
1 −2
1
4
8
0
2
1
−2
0
1
5
5
6
−1
0
1
1
0
1
2
=2 1
0
4
0
1
2
4
=5 1
0
5
1
1
−2
8
−2
1
1
−3
1
1 = 2(4)
−2
0
1
2
0 +4 1
1
5
1
6
−1
−3
1
− 2(8)
1
−3
1
1
= 8(−5) − 16(4) = −104
−2
1 = 5(0) + 4(80) = 320
1
27. Expanding along the first column in the original matrix and each succeeding minor, we obtain 3(1)(2)(4)(2) = 48.
28. Expanding along the bottom row we obtain
2 0
0 −2
2
1 6
0
5
1
−1
+
1 2 −1
1
1
2 1 −2
3
2
2
1
0
0
0
6
2
1
0
0
= −1(−48) + 0 = 48.
−1
−2
29. Solving λ2 − 2λ − 15 − 20 = λ2 − 2λ − 35 = (λ − 7)(λ + 5) = 0 we obtain λ = 7 and −5.
30. Solving −λ3 + 3λ2 − 2λ = −λ(λ − 2)(λ − 1) = 0 we obtain λ = 0, 1, and 2.
386
8.5
Properties of Determinants
EXERCISES 8.5
Properties of Determinants
1. Theorem 8.11
2. Theorem 8.14
3. Theorem 8.14
4. Theorem 8.12 and 8.11
5. Theorem 8.12 (twice)
6. Theorem 8.11 (twice)
7. Theorem 8.10
8. Theorem 8.12 and 8.9
9. Theorem 8.8
10. Theorem 8.11 (twice)
11. det A = −5
12. det B = 2(3)(5) = 30
13. det C = −5
14. det D = 5
15. det A = 6( 23 )(−4)(−5) = 80
16. det B = −a13 a22 a31
17. det C = (−5)(7)(3) = −105
18. det D = 4(7)(−2) = −56
19. det A = 14 = det AT
20. det A = 96 = det T
0
21. det AB = 10
8
−2
2
7 23 = −80 = 20(−4) = det A det B
4 16
22. From Problem 21, (det A)2 = det A2 = det I = 1, so det A = ±1.
a 1
23. Using Theorems 8.14, 8.12, and 8.9, det A = b 1
c
1
2
a 1
2 =2 b 1
1
1 = 0.
2
1
c
1
24. Using Theorems 8.14 and 8.9,
1
det A =
x
x+y+z
25.
1
4
0
26.
2 4
4 2
8 7
1
y
x+y+z
1 5
1
1
5
1
1
3 6 = 0 −1 −14 = 0 −1
−1 1
0 −1
1
0
0
5
2
4
5
2
0 = 0 −6 −10 = −2 0
−2
0 −9 −22
0
1 1
1
z
= (x + y + z) x y
x+y+z
1 1
1
z = 0.
1
5
−14 = 1(−1)(15) = −15
15
4
3
−9
5
2
5 = −2 0
−22
0
387
4
3
0
5
5 = −2(2)(3)(−7) = 84
−7
8.5
Properties of Determinants
27.
−1
4
2
−5
9
−9
−2
5
1
2
0
−2
1
2
−2
2 1
1 −2 3
28.
29.
30.
31.
32.
3
3
4
−11
0
2
1
3
1
5
2
1
4
0
2
3
1
2 3
1
2
1
3 5
3 6
5 8
2 9
1 3
0 1
1
7
6
3 1
4
3
−1
−2 =
0
6
0
2
3
3
−1
10 =
0
2
3
9
33
0
0
1 −2
−6
1 =− 5
0
2
−2
2
−8
12
3
10 = −1(3)(3) = −9
3
−2
10
−2
2
1
1 =− 0
0
−6
1 −2
1
1
2
2
0
2
0
3
5
1
1
0
4
3
1
=−
0
5
0
2
= −(1)(1)(8)(
37
) = −148
2
5
1
=−
0
2
4
1
0
2
1
2
−4
1
−5
4
−3
1
2
3
0
1
1
2
2
0
0
0
0
1
0
0
−4
8
−23
1
4
1
0
=−
0
5
0
2
2
1
0
0
2 0
−4 1
=−
8 4
−23 7
1
2
3
4
2 3
0
=
2 2
0
−1 7
0
1
0
0
2
2
0
3
= 1(1)(2)(8) = 16
2
8
1
2
3
7
0
=
7
0
20
0
1
−1
3
2
0
5
3
0
=
−1
0
16
0
1
0
0
3
9
7
1
4
8
1
1
6
4
1
0
=−
0
5
0
2
3
3
1
−8
7
−13
6
−17
1
4
0
0
=
5
0
−10
0
4
5
3
1
0
0
7
6
−31
0
4
5
= 1(1)(−31)(15) = −465
−15
15
1
8
2
4
=−
5
0
3
2
=
1
0
0
0
3
1
7
6
4
2
19
−9 = −1(10)(− ) = 38
5
− 19
5
−2
2
1
5 −6
1
= 1(5)(−2)(0) = 0
0 −2 −4
0
0
0
1
0
0
5 −6
1
=
=
0
0 10 −14 −2
1
0
0 −5
6 −1
2
2
−2
10
0
2
1
−9 = − 0
−2
0
1
0
=
0 −31 −15
0
0
31
30
0
4
3
1
7
6
4
5
3
−8
−13
−17
0
−10
37
2
33. We first use the second row to reduce the third row. Then we use the first row to reduce the second row.
1
1
1
1
a
b
c
= 0
0 b2 − ab c2 − ac
0
1
1
1
b−a
c − a = (b − a)(c − a) 0
b(b − a) c(c − a)
0
1
1
b
1
1 .
c
Expanding along the first row gives (b − a)(c − a)(c − b).
34. In order, we use the third row to reduce the fourth row, the second row to reduce the third row, and the first
row to reduce the second row. We then pull out a common factor from each column.
1
a
1
b
1
c
1
d
a2
b2
c2
d2
a3
b3
c3
d3
1
0
=
0
1
b−a
1
c−a
1
d−a
b2 − ab
c2 − ac
d2 − ac
0 b3 − ab2
c3 − ac2
d3 − ad2
1
0
= (b − a)(c − a)(d − a)
0
0
1
1
b
b2
1
1
c
1
1
.
d
c2
d2
Expanding along the first column and using Problem 33 we obtain (b − a)(c − a)(d − a)(c − b)(d − b)(d − c).
388
8.6
Inverse of a Matrix
35. Since C11 = 4, C12 = 5, and C13 = −6, we have a21 C11 + a22 C12 + a23 C13 = (−1)(4) + 2(5) + 1(−6) = 0. Since
C12 = 5, C22 = −7, and C23 = −3, we have a13 C12 + a23 C22 + a33 C32 = 2(5) + 1(−7) + 1(−3) = 0.
36. Since C11 + −7, C12 = −8, and C13 = −10 we have a21 C11 + a22 C12 + a23 C13 = −2(−7) + 3(−8) − 1(−10) = 0.
Since C12 = −8, C22 = −19, and C32 = −7 we have a13 C12 + a23 C22 + a33 C32 = 5(−8) − 1(−19) − 3(−7) = 0.
37. det(A + B) =
10
0
= −30;
0 −3
det A + det B = 10 − 31 = −21
38. det(2A) = 25 det A = 32(−7) = −224
39. Factoring −1 out of each row we see that det(−A) = (−1)5 det A = − det A. Then − det A = det(−A) =
det AT = det A and det A = 0.
Row reduction: 253 /3 ≈ 5.2(103 )
40. (a) Cofactors: 25! ≈ 1.55(1025 );
(b) Cofactors: about 90 billion centuries;
Row reduction: about
1
10
second
EXERCISES 8.6
Inverse of a Matrix
1. AB =

3 − 2 −1 + 1
6 − 6 −2 + 3
2−1
=
1
0
0
1
−1 + 1
−2 + 2


1

 
2. AB =  6 − 6
−3 + 4
6 − 6  = 0
2 + 1 − 3 −1 − 1 + 2 2 + 2 − 3
0
3. det A = 9. A is nonsingular. A−1 =
4. det A = 5. A is nonsingular. A
−1
1
=
5
5. det A = 12. A is nonsingular. A−1 =
6. det A = −3π . A is nonsingular. A
2
1
9
−1
7. det A = −16. A is nonsingular. A−1
8. det A = 0. A is singular.
9. det A = −30. A is nonsingular. A−1
1
12
1
−4
1
5
3
−4
1
2
3
=
1
9
− 49
1
9
5
9
=
3
5
− 45
1
5
1
15
=
1
6
1
4
1
3
0
6
1
=− 2
3π

π
π
8
1 
=−  2
16
−6


0

0
1
0
1
0
−14
1 
= −  −2
30
−4
π
−2π
−8
−4
4
13
4
−7
389
0
1
2
=
1
− 3π
1
− 3π
1
− 3π
  1
−8
−2
  1
6  =  −8
3
−2
8
  7
16
15
  1
−2  =  15
2
−4
15
2
3π
1
2
1
4
− 14
− 13
30
2
− 15
7
30
1
2
− 38
1
8



8
− 15
1
15
2
15



8.6
Inverse of a Matrix

 4
10
8 20
2
39
39
1 
 1
−1
5
10. det A = 78. A is nonsingular. A =
 −2 −5 19   − 39 − 78
78
3
2
12 −9
3
− 26
13

 1
0
−12
0 0
3
1 
 
−1
11. det A = −36. A is nonsingular. A = −  0 −6
0  =  0 16
36
0
0 18
0 0


 
0 0 2
0 0 18
1 

 
12. det A = 16. A is nonsingular. A−1 =
 8 0 0  =  21 0 0 
16
0 1 0
0 16 0

13. det A = 27. A is nonsingular. A−1
−4
4

15.
16.
17.
1
0
0
4
0
1
−−−1−−−→
8
0
1
0
0
1
2
0
1
1 3
1
0
0
1
3
A−1 =
− 14
A−1
4 R2
1
3
2R2
−5R +R
4

21.  2
−1
2
−−−−1−−→
1
0
1
0

19.  4 5 6 0 1
7 8 9 0 0

1
0 −1 1

20.  0 −2
1 0
2 −1
3 0

R1
−−−8−−−→
4 0 1
2 32
=
1 1
1 2
1
6 R1
2
3
1
−2
0
0
1
0
0
1
R1

1
6
0
0
1
4
0
1
8
0
0
1
0
2
1
3
0
−12
−−−2−−−→
0
1 − 13
0
1
1
1
4
1
− 12
5
12
2 −3
−2

−2
5
18.
6
−6
−1
−3

−−−−−−→
; A−1 =
1 0
−5 1
1 − 32
−2
4
1 2
row


0  −−−−−−→  0 1
operations
1
0 0


1
0 0
row


1 0  −−−−−−→  0
operations
0 1
0


−1
0 0
R13


1 0  −−−−−−→  2
4
0 1
1
3 R2 +R1
−
1
1
0
12
−−−−
−−→
1
2
0
0
1
1 0
0 1
2R +R
1
0
4
3
0
1
− 13
−2
0
0
1
0
0
1
5
9
− 29
− 49
−2
0
1
2
0
3
0
0
1
12
1
4
0
1
0
390
− 52
1
12
1
4
1
6
; A−1 =
3
1
0
0
1
5
12
1
− 12
1
0
− 32
1
0
−3R +R
1
−−−−2−−→
1
2
0
1
1
3
R2 +R1
1
0
− 14
0
1
5
12
2
−−
−−−−→
1
4
1
− 12
1
0
2
3
2
0
1
1
1

− 19
0
1



0
1

9
4
9
3
2
1
2

0 ; A is singular.
1
 5

− 19 92
9
 2
5
1 
−1
− 9 9 ; A =  − 9
0
0
− 12
1
2
−−−−
−−→
3
2
1
6
2
9
− 29
1
9
1
2
− 12
1
6
1
− 43 
− 19 


− 17 
− 13
7
9
1
27
17
27
4
− 27
27
4
27
15
0
2
8
R2
− 12
 
0 − 16
3

−9 
  0 − 16
=
1
0 0
3
−3
3
0


0
12
1
1
−2


3
0
0
0
1

14. det A = −6. A is nonsingular. A−1 = − 
6

0
  2
−36
9
− 1
−3 
  27
=
−51   10
21 −9
1
6
17 −6
6

1  −1
=

27  10
1
39
19
78
1
26
2
9
− 49
− 19
2
9
1
9
2
9
− 59
− 19


1 0 0
1
row


0  −−−−−−→  0 1 0
operations
0
0 0 1
0
0
1
3



2
3
− 13
− 23
1
3
− 23
0


;
;
;
8.6

2
3
− 13
− 23
0

A−1 =  0
1
3

2

22.  4
−6
0
A
−1

0
1
−2
1
1

2
0
0
2 3
1
5

= 2
−1
1
0
3

23.  3
0
6
2
−1
0

24.  0 1 4 0
0 0 8 0

1 2
3
 −1 0
2

25. 
 2 1 −3
1
1
1
0
1
1
0
2


0
1
−1

1
3
− 23
4 −2
2 −2
8 10


1 0

1
0
1 2 −1
2
row


1
1
1
1 0  −−−−−−→  0 1 − 3
−6
3
operations
0 0
0 −2 −1
0 1


0 0
0
1 −3 0 −1
row


1 1
1 0  −−−−−−→  0
1
1
operations
0 1
0
0 1 −1 −1

−3

−1 
1



5
0
1 0 0 1 −2
8
row



0  −−−−−−→  0 1 0 0
1 − 12 ;
operations
1
1
0 0 1 0
0
8


1 0 0 0
1 2 3
1


5
0 1 0 0
1
row
0 1 2
 −−−−−−→ 
0 0 1 0  operations  0 0 1 − 23
0
0
1
0 0 0 1

1 0 0 0

0 1 0 0
row
−−−−−−→ 
operations 
0 0 1 0
0

0

26. 
0
0
0
0 1
0 0
0
1
0
1 0
0
Inverse of a Matrix
0
0

− 12
− 23
− 16
1
1
3
− 13
1
3
− 13
1
2
0
− 12
0
0
7
6
− 43
1
3
1
2
1
0
0
1
0
0
0
0
1
1

0
1 0 0 0
0 1 0 0
0
row


 −−−−−−→ 
0  interchange  0 0 1 0
0
0
0
1
− 13
27. (AB)−1 = B−1 A−1 =
−1
1
0
0
0
0
0


0 ; A is singular.
1


0
1 0 0
row


0  −−−−−−→  0 1 0
operations
1
0 0 1
1


A−1
0
0
1
0
1
2
1
3
− 12
1
2
0
0
−1
− 23


; A−1


1
1 −2

= 0
1
2
−1

0
0


0
1
2
1
1
0
0
0
0
0
0
1
0

0
1
0
1


; A−1 = 
0
0
0
0
1
0
7
6
− 43
1
3
1
2
0
29. A = (A−1 )−1 =
T
30. A =
1
4
2
;
10
31. Multiplying
−2
3
3
−4

0
1


0
0
1
0
6 −32
T −1
(A )
4 −3
x −4
=
5 −1
5
−1
; A =
1
−2
−1
2
4 −3
x −4
=
16 − 3x
0
0
16 − 3x
−2
1
2
; (A
−1 T
) =
391
we see that x = 5.
5
−2
−1
1
2





0
0
0

−1 −4
20


= 2
6 −30 
3

0
0
1
1
3
10
3
−3


2 −1 ;
−1
1



28. (AB)−1 = B−1 A−1
6

5
8
− 12
1
8
1
1
1
2
2
 1
− 2 − 23 − 16

1
1
 1
3
3
=

1
1
0
−
−

3
3
− 12

5
8.6
Inverse of a Matrix
32. A−1 =
− cos θ
sin θ
sin θ
cos θ
33. (a) AT =
− cos θ
sin θ
sin θ
cos θ


(b) AT = 
= A−1
√1
3
0
− √26
√1
3
√1
2
√1
6
√1
3
− √12
√1
6


 = A−1
34. Since det A · det A−1 = det AA−1 = det I = 1, we see that det A−1 = 1/ det A. If A is orthogonal, det A =
det AT = det A−1 = 1/ det A and (det A)2 = 1, so det A = ±1.
35. Since A and B are nonsingular, det AB = det A · det B = 0, and AB is nonsingular.
36. Suppose A is singular. Then det A = 0, det AB = det A · det B = 0, and AB is singular.
37. Since det A · det A−1 = det AA−1 = det I = 1, det A−1 = 1/ det A.
38. Suppose A2 = A and A is nonsingular. Then A2 A−1 = AA−1 , and A = I. Thus, if A2 = A, either A is
singular or A = I.
39. If A is nonsingular, then A−1 exists, and AB = 0 implies A−1 AB = A−1 0, so B = 0.
40. If A is nonsingular, A−1 exists, and AB = AC implies A−1 AB = A−1 AC, so B = C.
1 0
0 0
41. No, consider A =
and B =
.
0 0
0 1
42. A is nonsingular if a11 a22 a33 = 0 or a11 , a22 , and a33 are all nonzero.


0
0
1/a11


A−1 =  0
1/a22
0 
0
0
1/a33
For any diagonal matrix, the inverse matrix is obtaining by taking the reciprocals of the diagonal entries and
leaving all other entries 0.
43. A−1 =
1
3
2
3
44. A−1 =
45. A
−1
=
46. A−1 =

47. A−1
; A−1
2
3
− 13
1
6
1
6
; A−1
1
16
− 18
3
8
1
4
; A
−2
1
3
2
− 12
1
5
6
5
− 15
2
−5
; A−1
1
1
− 12
1
− 12
1
3
5
12
=
−2
4
−3
6
=
1
2
− 32
3
4
− 12
=

; x1 = 6, x2 = −2
; x1 =
; x1 =
−11
15
2
1
3
, x2 = −
2
2
3
1
, x2 = −
4
2
; x1 = −11, x2 =
15
2
 

−4
2


 

0 ; A−1  0  =  4 ; x1 = 2, x2 = 4, x3 = −6
6
−6
− 15





1
1
− 12
4
1
3


 

0 ; A−1  2  =  0 ; x1 = − , x2 = 0, x3 =
2
2
3
−3
−1
2
1
5

6
=
1
5
12

=  − 23
−1
4
14
− 15

=  −1

48. A−1
1
3
− 13
4
392
8.6


49. A−1 = 

50.
51.
52.
53.
−2
−3
1
4
5
4
− 14
7
4
Inverse of a Matrix
 

1
21

 

0
; A−1  −3  =  1 ; x1 = 21, x2 = 1, x3 = −11
7
−11
−1

 


2
1
1
1
 1  2
−1 −1 

 


=
; x1 = 1, x2 = 2, x3 = −1, x4 = −4
; A−1 
 −5   −1 
1
1
2


2 −1
 −1
2

A−1 = 
 1 −1
3
−4
1 −1
1
0
1
1
9
7 −2
x1
b1
5
10
6
10
10
10
−1
−1
−1
; X=A
=
; A =
= 13 ; X = A
=
;
3
7
3
2
4
50
16
x2
b2
− 20 20
20
0
−2
−1
X=A
=
−20
−7

   

 



1 2 5
x1
b1
−1
−12
2 −1 −1

   

 



 2 3 8   x2  =  b2 ; A−1 =  12 −7 −2 ; X = A−1  4  =  −52 ;
−1 1 2
6
23
−5
3
1
x3
b3
  


 

3
0
0
1


 

−1  
−1 
X = A  3  =  9 ; = A  −5  =  27 
3
4
−3
−11
det A = 18 = 0, so the system has only the trivial solution.
54. det A = 0, so the system has a nontrivial solution.
55. det A = 0, so the system has a nontrivial solution.
56. det A = 12 = 0, so the system has only the trivial solution.

  

1
1
1
i1
0

  

57. (a)  −R1
R2 0   i2  =  E2 − E1 
0
−R2 R3
i3
E3 − E2
(b) det A = R1 R2 + R1 R3 + R2 R3 > 0, so A is nonsingular.


−R2
R2 R3 −R2 − R3
1


(c) A−1 =
R3
−R1 ;
 R1 R3
R1 R2 + R1 R3 + R2 R3
R1 R2
R2
R1 + R 2




0
R 2 E1 − R 2 E 3 + R 3 E 1 − R 3 E 2
1




A−1  E2 − E1  =
 R 1 E2 − R 1 E 3 − R 3 E 1 + R 3 E 2 
R1 R2 + R1 R3 + R2 R3
E3 − E2
−R1 E2 + R1 E3 − R2 E1 + R2 E3
58. (a) We write the equations in the form
−4u1 + u2 + u4 = −200
u1 − 4u2 + u3 = −300
u2 − 4u3 + u4 = −300

In matrix form this becomes
−4
 1


 0
1
u1 + u3 − 4u4 = −200.

  
−200
1
0
1
u1

  
−4
1
0
  u2   −300 
.
  = 
1 −4
1   u3   −300 
0
1
−4
393
u4
−200
8.6
Inverse of a Matrix

(b) A
−1
7
− 24
 1
 − 12
=
 1
 − 24
1
− 12
1
− 12
1
− 24
7
− 24
1
− 12
1
− 12
7
− 24
1
− 24
1
− 12
  225 
−200
2
 275 
1 


− 24 
−300   2 
; A−1 
; u1 = u4 = 225 , u2 = u3 = 275

=


275
1 
2
2


−300
− 12 
 2 
225
−200
−7
1
− 12


24
2
EXERCISES 8.7
Cramer’s Rule
1. det A = 10, det A1 = −6, det A2 = 12; x1 =
−6
10
2. det A = −3, det A1 = −6, det A2 = −6; x1 =
= − 35 , x2 =
−6
−3
−6
−3
= 2, x2 =
3. det A = 0.3, det A1 = 0.03, det A2 = −0.09; x1 =
0.03
0.3
12
10
=
=2
= 0.1 , x2 =
4. det A = −0.015, det A1 = −0.00315, det A2 = −0.00855; x1 =
6
5
−0.09
0.3
−0.00315
−0.015
= −0.3
= 0.21, x2 =
−0.00855
−0.015
= 0.57
5. det A = 1, det A1 = 4, det A2 = −7; x = 4, y = −7
6. det A = −70, det A1 = −14, det A2 = 35; r =
−14
−70
, s=
35
−70
= − 12
7. det A = 11, det A1 = −44, det A2 = 44, det A3 = −55; x1 =
−44
11
= −4, x2 =
=
1
5
44
11
173
8. det A = −63, det A1 = 173, det A2 = −136, det A3 = − 61
2 ; x1 = − 63 , x2 =
9. det A = −12, det A1 = −48, det A2 = −18, det A3 = −12; u =
48
12
= 4, v =
18
12
= 4, x3 =
136
63
=
, x3 =
3
2
−55
11
= −5
61
126
, w=1
10. det A = 1, det A1 = −2, det A2 = 2, det A3 = 5; x = −2, y = 2, z = 5
11. det A = 6 − 5k, det A1 = 12 − 7k, det A2 = 6 − 7k; x1 =
for k = 6/5.
12 − 7k
6 − 7k
, x2 =
. The system is inconsistent
6 − 5k
6 − 5k
−1−1
1
−2
1
=
=1−
, x2 =
−1
−1
−1
−1
(b) When = 1.01, x1 = −99 and x2 = 100. When = 0.99, x1 = 101 and x2 = −100.
12. (a) det A = − 1, det A1 = − 2, det A2 = 1; x1 =
13. det A ≈ 0.6428, det A1 ≈ 289.8, det A2 ≈ 271.9; x1 ≈
289.8
0.6428
≈ 450.8, x2 ≈
271.9
0.6428
≈ 423
14. We have (sin 30◦ )F + (sin 30◦ )(0.5N ) + N sin 60◦ = 400 and (cos 30◦ )F + (cos 30◦ )(0.5N ) − N cos 60◦ = 0. The
system is
(sin 30◦ )F + (0.5 sin 30◦ + sin 60◦ )N = 400
(cos 30◦ )F + (0.5 cos 30◦ − cos 60◦ )N = 0.
det A ≈ −1, det A1 ≈ −26.795, det A2 ≈ −346.41; F ≈ 26.795, N ≈ 346.41
15. The system is
i1 + i2 − i3 = 0
r1 i1 − r2 i2 = E1 − E2
r2 i2 + Ri3 = E2
det A = −r1 R − r2 R − r1 r2 , det A3 = −r1 E2 , −r2 E1 ; i3 =
394
r1 E2 + r2 E1
r1 R + r2 R + r1 r2
8.8 The Eigenvalue Problem
EXERCISES 8.8
The Eigenvalue Problem
−2
; λ = −1
5
√
√
√
2 −1
1
1
2
√
√ , λ= 2
√
and K2 since
=
= 2
2 −2
2− 2
2− 2
−2 + 2 2
√ √ √ √
√
2 −1
2+ 2
2+2 2
2+ 2
√
=
; λ= 2
= 2
2 −2
2
2 2
2
6 3
−5
0
−5
since
=
=0
; λ=0
2 1
10
0
10
2
8
2 + 2i
−4 + 4i
2 + 2i
since
=
= 2i
; λ = 2i
−1 −2
−1
−2i
−1


 



1 −2
2
4
12
4


 



and K3 since  −2
1 −2   −4  =  −12  = 3  −4 ; λ = 3
2
2
1
0
0
0


 

1 −2
2
−1
−1


 

1 −2   1  =  1 ; λ = 1
 −2
2
2
1
1
1

   
 
−1 1
0
1
3
1

   
 
since  1 2
1   4  =  12  = 3  4 ; λ = 3
0 3 −1
3
9
3
1. K3 since
2. K1
3. K3
4. K2
5. K2
6. K2
4
5
2
1
7. We solve
−2
5
=
2
−5
= (−1)
det(A − λI) =
For λ1 = 6 we have
−1 − λ
2
= (λ − 6)(λ − 1) = 0.
−7
8−λ
1 −2/7 0
−7 2 0
=⇒
−7 2 0
0
0
0
2
. For λ2 = 1 we have
7
−2 2 0
1 −1
=⇒
−7 7 0
0
0
1
so that k1 = k2 . If k2 = 1 then K2 =
.
1
so that k1 = 27 k2 . If k2 = 7 then K1 =
8. We solve
For λ1 = 0 we have
0
0
2−λ
1
= λ(λ − 3) = 0.
2
1−λ
2 1 0
1 1/2 0
=⇒
2 1 0
0
0
0
det(A − λI) =
395
8.8
The Eigenvalue Problem
−1
. For λ2 = 3 we have
2
−1
1 0
1 −1 0
=⇒
2 −2 0
0
0 0
1
so that k1 = k2 . If k2 = 1 then K2 =
.
1
so that k1 = − 12 k2 . If k2 = 2 then K1 =
−8 − λ −1
= (λ + 4)2 = 0.
16
−λ
1 1/4 0
−4 −1 0
For λ1 = λ2 = −4 we have
=⇒
16
4 0
0
0
0
−1
so that k1 = − 14 k2 . If k2 = 4 then K1 =
.
4
9. We solve
10. We solve
det(A − λI) =
det(A − λI) =
1−λ
1/4
1
= (λ − 3/2)(λ − 1/2) = 0.
1−λ
1 −2 0
0
1
=⇒
−1/2 0
0
0 0
−1/2
1/4
2
so that k1 = 2k2 . If k2 = 1 then K1 =
. If λ2 = 1/2 then
1
1/2
1
0
1 2
=⇒
1/4 1/2 0
0 0
−2
so that k1 = −2k2 . If k2 = 1 then K2 =
.
1
For λ1 = 3/2 we have
0
0
−1 − λ
−5
2
= λ2 + 9 = (λ − 3i)(λ + 3i) = 0.
1−λ
0
1 −(1/5) + (3/5)i 0
−1 − 3i
2
For λ1 = 3i we have
=⇒
−5
1 − 3i 0
0
0
0
1 − 3i
so that k1 = 15 − 35 i k2 . If k2 = 5 then K1 =
. For λ2 = −3i we have
5
−1 + 3i
2
0
1 − 15 − 35 i 0
=⇒
0
−5
1 + 3i 0
0
0
1 + 3i
so that k1 = 15 + 35 i k2 . If k2 = 5 then K2 =
.
5
11. We solve
det(A − λI) =
−1
= λ2 − 2λ + 2 = 0.
1−λ
i −1 0
i −1 0
For λ1 = 1 − i we have
=⇒
1
i 0
0
0 0
−i
i
so that k1 = −ik2 . If k2 = 1 then K1 =
and K2 = K1 =
.
1
1
12. We solve
det(A − λI) =
1−λ
1
396
8.8 The Eigenvalue Problem
det(A − λI) =
13. We solve
For λ1 = 4 we have
4−λ
0
0
8
8
= (λ − 4)(λ + 5) = 0.
−5 − λ
0
0 1 0
=⇒
0
0 0 0
0 −9
1
so that k2 = 0. If k1 = 1 then K1 =
. For λ2 = −5 we have
0
9 8 0
1 89
=⇒
0 0 0
0 0
−8
.
so that k1 = − 89 k2 . If k2 = 9 then K2 =
9
det(A − λI) =
14. We solve
For λ1 = 7 we have
7−λ
0
0 0
0 6
5−λ
−1
0
0
0
= (λ − 7)(λ − 13) = 0.
13 − λ
0
0 1 0
=⇒
0 0 0
0
1
so that k2 = 0. If k1 = 1 then K1 =
. For λ2 = 13 we have
0
−6 0 0
1 0
=⇒
0 0 0
0 0
0
so that k1 = 0. If k2 = 1 then K2 =
.
1
4−λ
0
0
−1
0
−5 − λ
9 = 4 − λ −5 − λ
9 = λ(4 − λ)(λ + 4) = 0.
−1
−λ
4−λ
−1
−λ




5 −1 0 0
1 0 −9/25 0




For λ1 = 0 we have
 0 −5 9 0  =⇒  0 1 −9/5 0 
5 −1 0 0
0 0
0
0
 
9
 
9
9
so that k1 = 25 k3 and k2 = 5 k3 . If k3 = 25 then K1 =  45 . If λ2 = 4 then
25




1 −1
0 0
1 0 −1 0




9 0  =⇒  0 1 −1 0 
 0 −9
15. We solve
det(A − λI) =
0
0
5
5
−1
−4
0
0
 
1
 
so that k1 = k3 and k2 = k3 . If k3 = 1 then K2 =  1 . If λ3
1



9 −1 0 0
1



 0 −1 9 0  =⇒  0
5 −1 4 0
0
 
1
 
so that k1 = k3 and k2 = 9k3 . If k3 = 1 then K3 =  9 .
1
397
0
0
0
= −4 then
0 −1
1 −9
0
0

0

0
0
8.8
The Eigenvalue Problem
det(A − λI) =
16. We solve
3−λ
0
0
2−λ
4

0
0
0
1−λ


2 0 0

0 1 0
4 0 0

For λ1 = 1 we have
= (3 − λ)(2 − λ)(1 − λ) = 0.
0
1 0 0


0  =⇒  0 1 0
0 0 0
0


0

0
0
0
 
so that k1 = 0 and k2 = 0. If k3 = 1 then K1 =  0 . If λ2 = 2 then
1




1 0
0 0
1 0 0 0




0 0  =⇒  0 0 1 0 
0 0
4 0 −1 0
0 0 0 0
 
0
 
so that k1 = 0 and k3 = 0. If k2 = 1 then K2 =  1 . If λ3 = 3 then
0




0
0
0 0
1 0 −1/2 0




0 0  =⇒  0 1
0
0
 0 −1
4
0 −2 0
0 0
0
0
 
1
 
1
so that k1 = 2 k3 and k2 = 0. If k3 = 2 then K3 =  0 .
2
−λ
det(A − λI) = −1
17. We solve
0

4
−4 − λ
2

 −1
0
For λ1 = λ2 = λ3 = −2 we have
so that k1 = −2k2 . If k2 = 1 and k3 = 1 then

18. We solve
det(A − λI) =
For λ1 = 3 we have
0
0
= −(λ + 2)3 = 0.
−2 − λ


4 0 0
1 2 0


−2 0 0  =⇒  0 0 0
0 0 0
0 0 0

−2


K1 =  1 
0
1−λ
0
0
0

0

0
0
 
0
 
and K2 =  0  .
1
1−λ
6
0
=
0
3 − λ 3 − λ = (3 − λ)(1 − λ)2 = 0.
2−λ
1
0
1
2−λ
1
2−λ




−2 6
0 0
1 0 −3 0




0 0  =⇒  0 1 −1 0 
 0 0
0 1 −1 0
0 0
0 0
6
0
398
8.8 The Eigenvalue Problem
 
3
 
so that k1 = 3k3 and k2 = k3 . If k3 = 1 then K1 =  1 . For λ2 = λ3 = 1 we have
1




0 6 0 0
0 1 0 0




 0 1 1 0  =⇒  0 0 1 0 
0
1
1
0
 
1
 
so that k2 = 0 and k3 = 0. If k1 = 1 then K2 =  0 .
0
−λ
det(A − λI) =
0
0
0
0
−1
0
0
= −(λ + 1)(λ2 + 1) = 0.
−1 − λ




1 0 −1 0
1 0 −1 0




For λ1 = −1 we have
0 0  =⇒  0 1
1 0
1 1
1 1
0 0
0 0
0 0


1


so that k1 = k3 and k2 = −k3 . If k3 = 1 then K1 =  −1 . For λ2 = i we have
1




−i
0
−1
0
1 0 −i 0




0
0  =⇒  0 1 −1 0 
 1 −i
19. We solve
1
−λ
1
1
1
1
−1 − i
0 0
0 0
0
 
 
i
−i
 
 
so that k1 = ik3 and k2 = k3 . If k3 = 1 then K2 =  1  and K3 = K2 =  1 .
1
1
20. We solve
det(A − λI) =
2−λ
−1
5
0
2−λ
1
0
4
= −λ3 + 6λ2 − 13λ + 10 = (λ − 2)(−λ2 + 4λ − 5)
2−λ
= (λ − 2)(λ − (2 + i))(λ − (2 − i)) = 0.

0 −1 0

0 4
5
0
1 0
For λ1 = 2 we have


0
1 0 4/5


0  =⇒  0 1 0
0 0 0
0

so that k1 = − 45 k3 and k2 = 0. If k3 = 5 then K1

−i −1
0

4
 5 −i
0
1 −i

0

0
0

−4


=  0 . For λ2 = 2 + i we have
5



0
1 −i
0 0



0  =⇒  0
1 −i 0 
0
0
0 0
0
399
8.8
The Eigenvalue Problem


−i


so that k1 = ik2 and k2 = ik3 . If k3 = i then K2 =  −1 . For λ3 = 2 − i we have
i




i −1 0 0
1 i 0 0




i 4 0  =⇒  0 1 i 0 
5
0
1 i 0
0 0 0 0


−1


so that k1 = −ik2 and k2 = −ik3 . If k3 = i then K3 =  1 .
i
1−λ
2
3
0
5−λ
6
= −(λ − 1)(λ − 5)(λ + 7) = 0.
0
0
−7 − λ




0 2
3 0
0 1 0 0




For λ1 = 1 we have
6 0  =⇒  0 0 1 0 
0 4
0 0 −6 0
0 0 0 0
 
1
 
so that k2 = k3 = 0. If k1 = 1 then K1 =  0 . For λ2 = 5 we have
0




−4 2
3 0
1 − 12 0 0




6 0  =⇒  0
0 1 0
 0 0
0 0 −12 0
0
0 0 0
 
1
 
so that k3 = 0 and k2 = 2k1 . If k1 = 1 then K2 =  2 . For λ3 = −7 we have
0




8 2 3 0
1 0 14 0




 0 12 6 0  =⇒  0 1 12 0 
det(A − λI) =
21. We solve
0
0
0
0
0

so that k1 =
− 14
k3 and k2 =
− 12
0

0
0
−1


k3 . If k3 = 4 then K3 =  −2 .
4
−λ 0
0
22. We solve
det(A − λI) = 0 −λ
0
= −λ2 (λ − 1) = 0.
0
0 1−λ




0 0 0 0
0 0 1 0




For λ1 = λ2 = 0 we have
 0 0 0 0  =⇒  0 0 0 0 
0 0 1 0
0 0 0 0
 
 
1
0
 
 
so that k3 = 0. If k1 = 1 and k2 = 0 then K1 =  0  and if k1 = 0 and k2 = 1 then K2 =  1 . For λ3 = 1
0
0
400
8.8 The Eigenvalue Problem
we have

−1

 0
0
0
0
−1
0
0


1 0 0


0  =⇒  0 1 0
0 0 0
 
0
 
so that k1 = k2 = 0. If k3 = 1 then K3 =  0 .
1
5 1
23. The eigenvalues and eigenvectors of A =
are
1 5
0
0
0
0


0
0
1
1
, K2 =
−1
1
5 −1
1
and the eigenvalues and eigenvectors of A−1 =
are
24 −1
5
1
1
1
1
λ 1 = , λ2 = , K 1 =
, K2 =
.
4
6
−1
1


1
2 −1


24. The eigenvalues and eigenvectors of A =  1
0
1  are
4 −4
5






−1
−2
−1






λ1 = 1, λ2 = 2, λ3 = 3, K1 =  1  , K2 =  1  , K3 =  1  .
2
4
4


4 −6
2
1

−1
and the eigenvalues and eigenvectors of A =  −1
9 −2  are
6
−4 12 −2






−1
−2
−1
1
1






λ1 = 1, λ2 = , λ3 = , K1 =  1  , K2 =  1  , K3 =  1  .
2
3
2
4
4
λ1 = 4,
25. Since det A =
6
0
3
0
λ2 = 6,
K1 =
= 0 the matrix is singular. Now from
det(A − λI) =
6−λ
0
3
−λ
= λ(λ − 6)
we see λ = 0 is an eigenvalue.
1
0
26. Since det A = 4 −4
7 −4
1
5 = 0 the matrix is singular. Now from
8
det(A − λI) =
1−λ
4
7
0
−4 − λ
−4
we see λ = 0 is an eigenvalue.
401
1
5
= −λ(λ2 − 5λ − 15)
8−λ
8.8
The Eigenvalue Problem
27. (a) Since p + 1 − p = 1 and q + 1 − q = 1, the first matrix A is stochastic. Since
and
1
6
+
1
3
+
1
2
1
2
+
1
4
+
1
4
= 1,
1
3
+
1
3
+
1
3
= 1,
= 1, the second matrix A is stochastic.
(b) The matrix from part (a) is shown with its eigenvalues and corresponding eigenvectors.
1 1 1
2
4
4
√ 1
√
1 1 1
1
1
2 , 6 + 12
2;
 3 3 3 ; eigenvalues: 1, 16 − 12
1
6
1
3
1
2
√
√
√
√
2) 2(2+ 2)
3(1+ 2) 2(−2+ 2)
√ ,
√ ,1 , −
√ ,
√
eigenvectors: (1, 1, 1), − 3(−1+
,1
−6+ 2
−6+ 2
6+ 2
6+ 2
Further examples indicate that 1 is always an eigenvalue with corresponding eigenvector (1, 1, 1). To prove
this, let A be a stochastic matrix and K = (1, 1, 1). Then

  
  
a11 · · · a1n
a11 + · · · + a1n
1
1
..   ..  
..
 ..
  .. 
AK =  .
 =  .  = 1K,
.  .  = 
.
· · · ann
an1
an1 + · · · + ann
1
1
and 1 is an eigenvalue of A with corresponding eigenvector (1, 1, 1).
(c) For the 3 × 3 matrix in part (a) we have
 3
7

A2 = 
8
1
3
5
18
24
11
36
23
72
1
3
13
36
29
72



,

A3 = 
49
144
71
216
5
16
29
96
11
36
67
216
103
288
79
216
163
432


.
These powers of A are also stochastic matrices. To prove that this is true in general for 2 × 2 matrices, we
prove the more general theorem that any product of 2 × 2 stochastic matrices is stochastic. Let
a11 a12
b11 b12
A=
and B =
a21 a22
b21 b22
be stochastic matrices. Then
AB =
a11 b11 + a12 b21
a11 b12 + a12 b22
a21 b11 + a22 b21
a21 b12 + a22 b22
.
The sums of the rows are
a11 b11 + a12 b21 + a11 b12 + a12 b22 = a11 (b11 + b12 ) + a12 (b21 + b22 )
= a11 (1) + a12 (1) = a11 + a12 = 1
a21 b11 + a22 b21 + a21 b12 + a22 b22 = a21 (b11 + b12 ) + a22 (b21 + b22 )
= a21 (1) + a22 (1) = a21 + a22 = 1.
Thus, the product matrix AB is stochastic. It follows that any power of a 2 × 2 matrix is stochastic. The
proof in the case of an n × n matrix is very similar.
402
8.9
Powers of Matrices
EXERCISES 8.9
Powers of Matrices
1. The characteristic equation is λ2 − 6λ + 13 = 0. Then
−7 −12
6
A2 − 6A + 13I =
−
24
17
24
2. The characteristic equation is −λ3 + λ2 + 4λ − 1.

 
2 6 13
1

 
3
2
−A + A + A − I = −  4 5 17  +  0
1 5
9
1
−12
30
+
Then


2 5
0


4 5 + 41
1
4
1
0
1
0
13
0
2

0
13
=

1
 
3 + 0
1
0
0
0
0
1
0
0
0
0

.

0
 
0 = 0
1
0
0
0

0

0.
0
0
3. The characteristic equation is λ2 − 3λ − 10 = 0, with eigenvalues −2 and 5. Substituting the eigenvalues into
λm = c0 + c1 λ generates
(−2)m = c0 − 2c1
5m = c0 + 5c1 .
Solving the system gives
c0 =
1
[5(−2)m + 2(5)m ],
7
1
[−(−2)m + 5m ].
7
c1 =
Thus
Am = c0 I + c1 A =
1
m m+1
7 [3(−1) 2
2
m
7 [−(−2)
and
3
A =
11
38
3
m
m
7 [−(−2) + 5 ]
1
m
m
7 [(−2) + 6(5) ]
+ 5m ]
+ 5m ]
57
106
.
4. The characteristic equation is λ2 − 10λ + 16 = 0, with eigenvalues 2 and 8. Substituting the eigenvalues into
λm = c0 + c1 λ generates
2m = c0 + 2c1
8m = c0 + 8c1 .
Solving the system gives
c0 =
1 m+2
− 8m ),
(2
3
c1 =
1
(−2m + 8m ).
6
Thus
Am = c0 I + c1 A =
and
A4 =
1 m
2 (2
1 m
2 (2
2056
−2040
403
+ 8m )
− 8m )
−2040
2056
1 m
2 (2
1 m
2 (2
.
− 8m )
+ 8m )
8.9
Powers of Matrices
5. The characteristic equation is λ2 − 8λ − 20 = 0, with eigenvalues −2 and 10. Substituting the eigenvalues into
λm = c0 + c1 λ generates
(−2)m = c0 − 2c1
10m = c0 + 10c1 .
Solving the system gives
c0 =
1
[5(−2)m + 10m ],
6
c1 =
1
[−(−2)m + 10m ].
12
Thus
Am = c0 I + c1 A =
1
m
m m+1
]
6 [(−2) + 2 5
1
m
m
3 [−(−2) + 10 ]
and
5
A =
83328
33344
41680
16640
5
m
m
12 [−(−2) + 10 ]
1
m
m
6 [5(−2) + 10 ]
.
6. The characteristic equation is λ2 + 4λ + 3 = 0, with eigenvalues −3 and −1. Substituting the eigenvalues into
λm = c0 + c1 λ generates
(−3)m = c0 − 3c1
(−1)m = c0 − c1 .
Solving the system gives
c0 =
1
[−(−3)m + 3(−1)m ],
2
c1 =
1
[−(−3)m + (−1)m ].
2
Thus
Am = c0 I + c1 A =
and
(−1)m
−(−3)m + (−1)m
0
(−3)m
A6 =
1
0
−728
729
.
7. The characteristic equation is −λ3 + 2λ2 + λ − 2 = 0, with eigenvalues −1, 1, and 2. Substituting the eigenvalues
into λm = c0 + c1 λ + c2 λ2 generates
(−1)m = c0 − c1 + c2
1 = c0 + c1 + c2
2
m
= c0 + 2c1 + 4c2 .
Solving the system gives
1
[3 + (−1)m − 2m ],
3
1
c1 = [1 − (−1)m ],
2
1
c2 = [−3 + (−1)m + 2m+1 ].
6
c0 =
Thus
404
8.9

1
−1 + 2m
0
1
m
m+1
]
3 [(−1) + 2
1
m
m
3 [−(−1) + 2 ]

Am = c0 I + c1 A + c2 A2 =  0
and

A10
1

= 0
0
1023
683
341
−1 + 2m
Powers of Matrices


− 23 [(−1)m − 2m ] 
1
m
3 [2(−1)
+ 2m ]

1023

682  .
342
√
√
8. The characteristic equation is −λ3 − λ2 + 2λ + 2 = 0, with eigenvalues −1, − 2 , and 2 . Substituting the
eigenvalues into λm = c0 + c1 λ + c2 λ2 generates
(−1)m = c0 − c1 + c2
√
√
(− 2 )m = c0 − 2c1 + 2c2
√
√
( 2 )m = c0 + 2c1 + 2c2 .
Solving the system gives
√
√
√
√
c0 = [2 − ( 2 )m−1 − ( 2 )m−2 ](−1)m + ( 2 − 1)( 2 )m−2 ,
√
1
c1 = [1 − (−1)m ]( 2 )m−1 ,
2
√
√
√
1
1 √
c2 = (−1)m+1 + (1 + 2 )(−1)m ( 2 )m−1 + ( 2 − 1)( 2 )m−1 .
2
2
Thus Am = c0 I + c1 A + c2 A2 and

1

6
A = 7
0
0
8
0

7

−7  .
8
9. The characteristic equation is −λ3 +3λ2 +6λ−8 = 0, with eigenvalues −2, 1, and 4. Substituting the eigenvalues
into λm = c0 + c1 λ + c2 λ2 generates
(−2)m = c0 − 2c1 + 4c2
1 = c0 + c1 + c2
4
m
= c0 + 4c1 + 16c2 .
Solving the system gives
1
[8 + (−1)m 2m+1 − 4m ],
9
1
c1 =
[4 − 5(−2)m + 4m ],
18
1
c2 =
[−2 + (−2)m + 4m ].
18
c0 =
Thus
1

Am = c0 I + c1 A + c2 A2 = 
m
9 [(−2)
+ (−1)m 2m+1 + 3 · 22m+1 ]
− 23 [(−2)m − 4m ]
1
3 [−3
+ (−2)m + 22m+1 ]
405
1
m
m
3 [−(−2) + 4 ]
1
m m+1
+ 4m ]
3 [(−1) 2
1
m
m
3 [−(−2) + 4 ]
0


0
1
8.9
Powers of Matrices
and

A10
699392

=  698368
699391

0

0.
1
349184
350208
349184
10. The characteristic equation is −λ3 − 32 λ2 + 32 λ + 1 = 0, with eigenvalues −2, − 12 , and 1. Substituting the
eigenvalues into λm = c0 + c1 λ + c2 λ2 generates
(−2)m = c0 − 2c1 + 4c2
1 m
1
1
−
= c0 − c1 + c2
2
2
4
1 = c0 + c1 + c2 .
Solving the system gives
1 −m
[2 [(−4)m + 8(−1)m + 2m+1 − (−1)m 22m+1 ],
9
1
c1 = − 2−m [(−4)m + 4(−1)m − 5 · 2m ],
9
2
c2 = [1 + (−2)m − (−1)m 2m−1 ].
9
c0 =
Thus
Am = c0 I + c1 A + c2 A2

1

=
−m
[2(−1)m + 2m ]
32
2
1 m
3 −1 + − 2
m
m
m
− 19 2−m [7(−4) − 6(−1) − 3 · 2 + (−1)m 22m+1 ]
and

43
128

85
A8 =  − 128
− 32725
128
85
− 256
171
256
85
− 256
1
1 m
3 [ −1 + − 2
1
1 m
3 2 + −2
1
1 m
3 −1 + − 2
0
0
1
m
3 [(−2)



+ (−1)m 2m+1 ]

0

0 .
256
11. The characteristic equation is λ2 − 8λ + 16 = 0, with eigenvalues 4 and 4. Substituting the eigenvalues into
λm = c0 + c1 λ generates
4m = c0 + 4c1
4m−1 m = c1 .
Solving the system gives
c0 = −4m (m − 1),
c1 = 4m−1 m.
Thus
Am = c0 I + c1 A =
and
A6 =
4m−1 (3m + 4)
3 · 4m−1 m
−3 · 4m−1 m
4m−1 (−3m + 4)
22528
−18432
406
18432
−14336
.
8.9
Powers of Matrices
12. The characteristic equation is −λ3 − λ2 + 21λ + 45 = 0, with eigenvalues −3, −3, and 5. Substituting the
eigenvalues into λm = c0 + c1 λ + c2 λ2 generates
(−3)m = c0 − 3c1 + 9c2
(−3)m−1 m = c1 − 6c2
5m = c0 + 5c1 + 25c2 .
Solving the system gives
1
[73(−3)m − 2(−1)m 3m+2 + 9 · 5m − 40(−3)m m],
64
1
c1 =
[−(−1)m 3m+2 + 9 · 5m − 8(−3)m m],
96
1
c2 =
[−(−3)m + 5m − 8(−3)m−1 m].
64
c0 =
Thus
Am = c0 I + c1 A + c2 A2
 1
m
m

=


[31(−3)
− (−1) 3m+1 + 4 · 5m ]
1
[−(−3)m
16
− (−1)m 3m+1 + 4 · 5m ]
1
[(−3)m
32
+ (−1)m 3m+1 − 4 · 5m ]
1
[−(−3)m
16
− (−1)m 3m+1 + 4 · 5m ]
1
[7(−3)m
8
− (−1)m 3m+1 + 4 · 5m ]
1
[(−3)m
16
+ (−1)m 3m+1 − 4 · 5m ] 
3
[(−3)m
32
+ (−1)m 3m+1 − 4 · 5m ]
3
[(−3)m
16
+ (−1)m 3m+1 − 4 · 5m ]
32
and

178

5
A =  842
−1263
1
[29(−3)m
32


− (−1)m 3m+2 + 12 · 5m ]

−421

−842  .
1020
842
1441
−2526
13. (a) The characteristic equation is λ2 − 4λ = λ(λ − 4) = 0, so 0 is an eigenvalue. Since the matrix satisfies the
characteristic equation, A2 = 4A, A3 = 4A2 = 42 A, A4 = 42 A2 = 43 A, and, in general,
4m
4m
m
m
A =4 A=
.
3(4)m 3(4)m
(b) The characteristic equation is λ2 = 0, so 0 is an eigenvalue. Since the matrix satisfies the characteristic
equation, A2 = 0, A3 = AA2 = 0, and, in general, Am = 0.
(c) The characteristic equation is −λ3 + 5λ2 − 6λ = 0, with eigenvalues 0, 2, and 3. Substituting λ = 0 into
λm = c0 + c1 λ + c2 λ2 we find that c0 = 0. Using the nonzero eigenvalues, we find
2m = 2c1 + 4c2
3m = 3c1 + 9c2 .
Solving the system gives
c1 =
1
[9(2)m − 4(3)m ],
6
c2 =
1
[−3(2)m + 2(3)m ].
6
3m−1
3m−1
Thus Am = c1 A + c2 A2 and


Am = 
2(3)m−1
1
m
6 [9(2)
− 4(3)m ]
1
m
6 [−9(2)
+ 8(3)m ]
1
m
6 [3(2)
− 2(3)m ]
1
m
6 [−3(2)
407
+ 4(3)m ]
1
m
6 [−3(2)
1
m
6 [3(2)

− 2(3)m ] 
.
+ 4(3)m ]
8.9
Powers of Matrices
14. (a) Let
Xn−1 =
xn−1
and A =
yn−1
Then
Xn = AXn−1 =
1
1
1
0
xn−1
=
yn−1
1
0
1
1
.
xn−1 + yn−1
.
xn−1
√
√
(b) The characteristic equation of A is λ2 − λ − 1 = 0, with eigenvalues λ1 = 12 (1 − 5 ) and λ2 = 12 (1 + 5 ).
m
From λm = c0 + c1 λ we get λm
1 = c0 + c1 λ1 and λ2 = c0 + c1 λ2 . Solving this system gives
m
m
m
c0 = (λ2 λm
1 − λ1 λ2 )/(λ2 − λ1 ) and c1 = (λ2 − λ1 )/(λ2 − λ1 ).
Thus
Am = c0 I + c1 A
√
√
(1 + 5 )m+1 − (1 − 5 )m+1
1
√
√
√
=
2(1 + 5 )m − 2(1 − 5 )m
2m+1 5
√
√
2(1 + 5 )m − 2(1 − 5 )m
√
√ m
√
√ m .
(1 + 5 )(1 − 5 ) − (1 − 5 )(1 + 5 )
(c) From part (a), X2 = AX1 , X3 = AX2 = A2 X1 , X4 = AX3 = A3 X1 , and, in general, Xn = An−1 X1 .
With
1
144 89
1
233
X1 =
we have X12 = A11 X1 =
=
,
1
89 55
1
144
so the number of adult pairs is 233. With
1
144 89
1
144
X1 =
we have A11 X1 =
=
,
0
89 55
0
89
so the number of baby pairs is 144. With
2
144 89
2
377
X1 =
we have A11 X1 =
=
,
1
89 55
1
233
so the total number of pairs is 377.
1
15. The characteristic equation of A is λ2 − 5λ + 10 = 0, so A2 − 5A + 10I = 0 and I = − 10
A2 + 12 A. Multiplying
by A−1 we find
3
2
1
1
1 2 −4
1 1 0
10
5
.
A−1 = − A + I = −
+
=
10
2
10 1
2 0 1
3
−1 1
10
5
16. The characteristic equation of A is −λ3 +2λ2 +λ−2 = 0, so −A3 +2A2 +A−2I = 0 and I = − 12 A3 +A2 + 12 A.
Multiplying by A−1 we find
3
1
2
2
1
1

A−1 = − A2 + A + I =  21
2
2
1
2
17. (a) Since
2
A =
1
−1
0
0
− 52
1
2

− 12  .
1
2
− 32
we see that
for all integers m ≥ 2. Thus A is not nilpotent.
(b) Since A2 = 0, the matrix is nilpotent with index 2.
(c) Since A3 = 0, the matrix is nilpotent with index 3.
408
A
m

=
1
−1
0
0
8.10 Orthogonal Matrices
(d) Since A2 = 0, the matrix is nilpotent with index 2.
(e) Since A4 = 0, the matrix is nilpotent with index 4.
(f ) Since A4 = 0, the matrix is nilpotent with index 4.
18. (a) If Am = 0 for some m, then (det A)m = det Am = det 0 = 0, and A is a singular matrix.
(b) By (1) of Section 8.8 we have AK = λK, A2 K = λAK = λ2 K, A3 K = λ2 AK = λ3 K, and, in general,
Am K = λm K. If A is nilpotent with index m, then Am = 0 and λm = 0.
EXERCISES 8.10
Orthogonal Matrices

  

 
0
0 −4
0
0
0

  

 
1. (a)–(b)  0 −4
0   1  =  −4  = −4  1 ; λ1 = −4
−4
0 15
0
0
0

  

 
0
0 −4
4
−4
4

  

 
0   0  =  0  = (−1)  0 ; λ2 = −1
 0 −4
−4
0 15
1
1
1


 



0
0 −4
1
16
1


 



0   0  =  0  = 16  0 ; λ3 = 16
 0 −4
−4
0 15
−4
−64
−4
(c) KT1 K2 = ( 0
 
4
 
0 )  0  = 0; KT1 K3 = ( 0
1
1

1

1


0 )  0  = 0; KT2 K3 = ( 4
−4

0

1


1) 0 = 0
−4


 



1 −1 −1
−2
−4
−2


 



2. (a)–(b)  −1
1 −1   1  =  2  = 2  1 ; λ1 = 2
−1 −1
1
1
2
1


 



1 −1 −1
0
0
0


 



1 −1   1  =  2  = 2  1 ; λ2 = 2
 −1
−1 −1
1
−1
−2
−1

  

 
1 −1 −1
1
−1
1

  

 
−1
1
−1
1
=
−1
=
(−1)

  

 1 ; λ3 = −1
−1 −1
1
1
−1
1

(c) KT1 K2 = ( −2
1

0


1 )  1  = 1 − 1 = 0; KT1 K3 = ( −2
−1
409
1
 
1
 
1 )  1  = −2 + 1 + 1 = 0
1
8.10 Orthogonal Matrices
 
1
 
KT2 K3 = ( 0 1 −1 )  1  = 1 − 1 = 0
1
  √2   √ 
 √2 
13
0
2
9
2
2
 √   √ 
√ 
5
0   22  =  9 2  = 18  22 ; λ1 = 18
0 −8
0
0
0

5

3. (a)–(b)  13
0
  √ 3   8√ 2 
 √3 
− 3
5 13
0
3
3

  √ 3   8√ 3 
 √3 
13
5
0
=
=
(−8)
−
−

 3  
 3 ; λ2 = −8
3 
√
√
√
3
3
0 0 −8
−8 3


5 13

5
 13
0
0
3
√
6
0
6
√

0   − 66
√
−8
− 36


(c)
KT1 K2
√
√
√
√
2
2
=(
KT1 K3 = (
2
2


=
√
3
3
√

0 )  − 33
√
3
3
√

6
6
√

0 )  − 66
√
− 36
2
2
2
2

KT2 K3

√
=(
3
2
2
0
3

2
2
2
2
0

4. (a)–(b)  2 2
2 0

3 2

2 2

3
3
−


=


=
√
6
6
√
3 
6
3 )− 6
√
− 36
√
√
3
3

3
√
8 6
− 6
 8√ 6

6
√
8 6
3



=
3
√
6
6
√

(−8)  − 66
√
− 36



; λ3 = −8
√
√
6
6
−
= 0;
6
6
√
√
12
12
−
=0
12
12


=
√
√
√
18
18
18
+
−
=0
18
18
9
 


0
−2

  


0   2  =  0  = 0  2 ; λ 1 = 0
4
1
0
1

 



2
1
3
1

 



0   2  =  6  = 3  2 ; λ2 = 3
4
−2
−6
−2
   
 
2
2
12
2
   
 
0   1  =  6  = 6  1 ; λ3 = 6
4
2
12
2
2
−2



1


(c) KT1 K2 = ( −2 2 1 )  2  = −2 + 4 − 2 = 0; KT1 K3 = ( −2
−2
 
2
 
KT2 K3 = ( 1 2 −2 )  1  = 2 + 2 − 4 = 0
2
5. Orthogonal. Columns form an orthonormal set.
410
2
 
2
 
1 )  1  = −4 + 2 + 2 = 0
2
8.10 Orthogonal Matrices
6. Not orthogonal. Columns one and three are not unit vectors.
7. Orthogonal. Columns form an orthonormal set.
8. Not orthogonal. The matrix is singular.
9. Not orthogonal. Columns are not unit vectors.
10. Orthogonal. Columns form an orthogonal set.
1
1
11. λ1 = −8, λ2 = 10, K1 =
, K2 =
, P=
−1
1
12. λ1 = 7, λ2 = 4, K1 =
1
0
, K2 =
, P=
0
1
13. λ1 = 0, λ2 = 10, K1 =
√
1
0
0
1
3
1
, K2 =
, P=
−1
3
√
5
5
1
1
14. λ1 = +
, λ2 = −
, K1 =
2
2
2
2

√1
2
− √12
√3
10
1
√
− 10
√1
2
√1
2
√1
10
√3
10
 1+ 5
√ √ √
√
1+ 5
1− 5
10+2 5

, K2 =
, P=
2
2
√ 2 √
√
10+2 5

√
√1−
5
√
10−2 5
√ 2 √
10−2 5
 
 
 1

− √2 √12 0
−1
1
0


 
 


15. λ1 = 0, λ2 = 2, λ3 = 1, K1 =  0 , K2 =  0 , K3 =  1 , P = 
0 0 1
√1
√1
1
1
0
0
2
2







− √12
−1
1
1

√
√
√
√






16. λ1 = −1, λ2 = 1− 2 , λ3 = 1+ 2 , K1 =  0 , K2 =  − 2 , K3 =  2 , P = 
 0
1
1
1
√1


 
 √3
− 11
−3
1
1




 
 √1
17. λ1 = −11, λ2 = 0, λ3 = 6, K1 =  1 , K2 =  −4 , K3 =  2 , P = 
11
1
7
1
√1


 
 1
1
−2
2
3




 

18. λ1 = −18, λ2 = 0, λ3 = 9, K1 =  −2 , K2 =  1 , K3 =  2 , P =  − 23
2
2
2
1
19.
3
5
4
5
But
a
3
5
4
5
b
a
b
12
25
=
1
0
0
1


3
implies
9
25
+ a2 = 1 and
16
25
11
−
2
√1
66
− √466
√7
66
− 23
1
3
2
3
2
3
2
3
1
3


1
2
√
2
2
1
2
√1
6
√2
6
√1
6

√1
5
b
a
√1
5
− 45
3
5
4
5
√1
5
b
a
√1
5
=
1
0
0
1
1
5
+ b2 = 1 and a2 +
1
5
= 1. These give a = ± √25 , b = ± √25 .
a
b
But √ + √ = 0 indicates a and b must have opposite signs. Therefore choose a = − √25 , b =
5
5
The matrix
√1
5
2
√
− 5
√2
5
√1
5
is orthogonal.
411



+ b2 = 1. These equations give a = ± 45 , b = ± 35 .
implies



is orthogonal.
3
5
1
2
√
2
2
1
2


+ ab = 0 indicates a and b must have opposite signs. Therefore choose a = − 45 , b = 35 .
The matrix
20.



√2
2
.
8.10 Orthogonal Matrices
21. (a)–(b) We compute

0

AK1 =  2
2

0

AK2 =  2
2

0

AK3 =  2
2

 



2
1
−2
1

 



2   −1  =  2  = −2  −1  = −2K1
0
0
0
0

 



2
1
−2
1

 



2   0  =  0  = −2  0  = −2K2
0
−1
2
−1
   
 
2
1
4
1
   
 
2   1  =  4  = 4  1  = 4K3 ,
0
1
4
1
2
0
2
2
0
2
2
0
2
and observe that K1 is an eigenvector with corresponding eigenvalue −2, K2 is an eigenvector with corresponding eigenvalue −2, and K3 is an eigenvector with corresponding eigenvalue 4.
(c) Since K1 · K2 = 1 = 0, K1 and K2 are not orthogonal, while K1 · K3 = 0 and K2 · K3 = 0 so K3
is orthogonal to both K1 and K2 , To transform {K1 , K2 } into an orthogonal set we let V1 = K1 and
compute K2 · V1 = 1 and V1 · V1 = 2. Then



  1
1
1
2
K2 · V1

 1
 

V2 = K2 −
V1 =  0  −  −1  =  12  .
V1 · V1
2
−1
0
−1
Now, {V1 , V2 , K3 } is an orthogonal set of eigenvectors with
||V1 || =
√
2,
3
||V2 || = √ ,
6
and ||K3 || =
An orthonormal set of vectors is

√1
2
 √1
− 2


,
0

√1
6
 √1

6
− √26


,

and


and so the matrix

P=
√1
2
 √1
− 2
0
√1
6
√1
6
− √26
is orthogonal.
412
√1
3
√1
3
√1
3



√1
3
√1
3
√1
3


,
√
3.
8.10 Orthogonal Matrices
22. (a)–(b) We compute

AK1
AK2
AK3
AK4

  


1
−1
0
−1

  
 0
1
 0 0



 =   = 0
 = 0K1
 0
1 0 0
1
1

=
1
1

1
1

=
1
1

1
1

=
1
1

1
1

=
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
0
0
0
   
 
1
1
4
1







11 4
1
   =   = 4   = 4K4 ,
1
11 4
1
1
1
1
1
1
0
1

  


1
−1
0
−1

  
 0
1
 0 0



 =   = 0
 = 0K2
 1
1 1 0
1
0
0
0

  


1
−1
0
−1

  
 1
1
 1 0



 =   = 0
 = 0K3
 0
1 0 0
1
1
4
1
and observe that K1 is an eigenvector with corresponding eigenvalue 0, K2 is an eigenvector with corresponding eigenvalue 0, K3 is an eigenvector with corresponding eigenvalue 0, and K4 is an eigenvector with
corresponding eigenvalue 4.
(c) Since K1 · K2 = 1 = 0, K1 and K2 are not orthogonal. Similarly , K1 · K3 = 1 = 0 and K2 · K3 = 1 = 0
so K1 and K3 and K2 and K3 are not orthogonal. However, K1 · K4 = 0, K2 · K4 = 0, and K3 · K4 = 0,
so each of K1 , K2 , and K3 is orthogonal to K4 . To transform {K1 , K2 , K3 } into an orthogonal set we let
V1 = K1 and compute K2 · V1 = 1 and V1 · V1 = 2. Then



  1
−1
−1
−2






K2 · V1
 0 1  0  0
V2 = K2 −
V1 = 
.
− 
=
V1 · V1
 1 2  0  1
0
1
− 12
Next, using K3 · V1 = 1, K3 · V2 = 12 , and V2 · V2 = 32 , we obtain




 1  1
−3
−1
−1
−2








K3 · V 1
K3 · V2
 1  1  0  1/2  0   1 
V3 = K 3 =
V1 −
V2 = 
 =  1 .
− 
−

V1 · V1
V2 · V2
 0  2  0  3/2  1   − 3 
0
− 12
1
Now, {V1 , V2 , V3 , K4 } is an orthogonal set of eigenvectors with
√
3
2
||V1 || = 2, ||V2 || = √ , ||K3 || = √
and ||K4 = 2.
6
3
An orthonormal set of vectors is
 1 
− √2
 0 



,
 0 
√1
2





− √16

0 

,
√2 
6
− √16

1
− 2√
3

3 
 √
 2 3

,
1 
− √
2 3
1
− 2√
3
413
1
2
and
1
2
 1 ,
2
1
2
− 13
8.10 Orthogonal Matrices
and so the matrix



P=

− √12
− √16
0
0
√2
6
− √16
0
√1
2
1
− 2√
3
3
√
2 3
1
− 2√
3
1
− 2√
3
1
2
1
2
1
2
1
2





is orthogonal.
 
 
0
a
 
 
23. If we take K1 =  1  as in Example 4 in the text then we look for a vector K2 =  b  such that 1(a) +
1
c
1
1
1
1
1
4 b − 4 c = 0 and K1 · K2 = 0 or b + c = 0. The last equation implies c = −b so a + 4 b − 4 (−b) = a + 2 b = 0.
If we let
 b =−2, then a = 1 and c = 2, so a second eigenvector with eigenvalue −9 and orthogonal to K1 is
1


K2 =  −2 .
2
24. The eigenvalues and corresponding eigenvectors of A are


−1
 1


λ1 = λ2 = −1, λ3 = λ4 = 3, and K1 = 
,
 0


0
 0


K2 = 
,
 −1 
0
 
0
0
 
K3 =   ,
1
1
1
 
1
1
 
K4 =   .
0
0
Since K1 · K2 = K1 · K3 = K1 · K4 = K2 · K3 = K2 · K4 = K3 · K4 = 0, the vectors are orthogonal. Using
√
K1 = K2 = K3 = K4 = 2 , we construct the orthogonal matrix
 √1

√1
− 2
0
0
2
 √1

√1 
0
0

2
2 
P=

.
0 
 0 − √12 √12
0
√1
2
√1
2
0
25. Suppose A and B are orthogonal matrices. Then A−1 = AT and B−1 = BT and
(AB)−1 = B−1 A−1 = BT AT = (AB)T .
Thus AB is an orthogonal matrix.
EXERCISES 8.11
Approximation of Eigenvalues
1
and computing Xi = AXi−1 for i = 1, 2, 3, 4 we obtain
1
2
4
8
16
X1 =
, X2 =
, X3 =
, X4 =
.
2
4
8
16
1
AK · K
4
We conclude that a dominant eigenvector is K =
with corresponding eigenvalue λ =
= = 2.
K·K
2
1
1. Taking X0 =
414
8.11 Approximation of Eigenvalues
1
and computing Xi = AXi−1 for i = 1, 2, 3, 4, 5 we obtain
1
−5
49
−437
3937
−35429
X1 =
, X2 =
, X3 =
, X4 =
, X5 =
.
7
−47
439
−3935
35431
−35429
−0.99994
1
We conclude that a dominant eigenvector is K =
≈
with corresponding
35439
1
35431
AK · K
eigenvalue λ =
= −8.9998.
K·K
1
6
6
0.375
1
3. Taking X0 =
and computing AX0 =
, we define X1 =
=
. Continuing in this
16 16
1
16
1
2. Taking X0 =
manner we obtain
X2 =
0.3363
1
,
0.3335
1
0.3333
X3 =
We conclude that a dominant eigenvector is K =
1
,
X4 =
0.3333
1
.
with corresponding eigenvalue λ = 14.
1
1
0.2
1 1
4. Taking X0 =
and computing AX0 =
, we define X1 =
=
. Continuing in this manner
5 5
1
5
1
we obtain
0.2727
0.2676
0.2680
0.2679
, X3 =
, X4 =
, X5 =
.
X2 =
1
1
1
1
0.2679
We conclude that a dominant eigenvector is K =
with corresponding eigenvalue λ = 6.4641.
1
 
 
  

1
11
11
1
1   
 
 

5. Taking X0 =  1  and computing AX0 =  11 , we define X1 =
 11  =  1 . Continuing in
11
1
6
6
0.5455
this manner we obtain


1


X2 =  1  ,
0.5045




1
1




X3 =  1  , X4 =  1  .
0.5005
0.5


1


We conclude that a dominant eigenvector is K =  1  with corresponding eigenvalue λ = 10.
0.5
 
 
 


1
5
5
1
1 
 
 


6. Taking X0 =  1  and computing AX0 =  2 , we define X1 =  2  =  0.4 . Continuing in this
5
1
2
2
0.4
manner we obtain








1
1
1
1








X2 =  0.2105  , X3 =  0.1231  , X4 =  0.0758  , X5 =  0.0481  .
0.2105
0.1231
0.0758
0.0481
 
 
1
1
 
 
At this point if we restart with X0 =  0  we see that K =  0  is a dominant eigenvector with corresponding
0
0
eigenvalue λ = 3.
415
8.11 Approximation of Eigenvalues
1
and using scaling we obtain
1
0.625
0.5345
0.5098
0.5028
0.5008
X1 =
, X2 =
, X3 =
, X4 =
, X5 =
.
1
1
1
1
1
0.5
Taking K =
as the dominant eigenvector we find λ1 = 7. Now the normalized eigenvector is
1
0.4472
1.6 −0.8
1
K1 =
and B =
. Taking X0 =
and using scaling again we obtain X1 =
0.8944
−0.8
0.4
1
1
1
1
, X2 =
. Taking K =
we find λ2 = 2. The eigenvalues are 7 and 2.
−0.5
−0.5
−0.5
7. Taking X0 =
8. Taking X0 =
1
0.3333
0.3333
1/3
and using scaling we obtain X1 =
, X2 =
. Taking K =
as the
1
1
1
1
dominant eigenvector we find λ1 = 10. Now the normalized eigenvector is K1 =
0.3162
0.9486
and B =
0
0
0
.
0
An eigenvector for the zero matrix is λ2 = 0. The eigenvalues are 10 and 0.
 
1
 
9. Taking X0 =  1  and using scaling we obtain
1
 








1
1
1
1
1
 








X1 =  0  , X2 =  −0.6667  , X3 =  −0.9091  , X4 =  −0.9767  , X5 =  −0.9942  .
1
1
1
1
1


1


Taking K =  −1  as the dominant eigenvector we find λ1 = 4. Now the normalized eigenvector is

1



 
0.5774
1.6667 0.3333 −1.3333
1




 
K1 =  −0.5774 and B =  0.3333 0.6667
0.3333 . If X0 =  1  is now chosen only one more
0.5774
1
−1.3333 0.3333
1.6667
 
1
 
eigenvalue is found. Thus, try X0 =  1 . Using scaling we obtain
0










1
1
1
1
1










X1 =  0.5  , X2 =  0.2  , X3 =  0.0714  , X4 =  0.0244  , X5 =  0.0082  .
−0.5
−0.8
−0.9286
−0.9756
−0.9918


1


Taking K =  0  as the eigenvector we find λ2 = 3. The normalized eigenvector in this case is
−1




 
0.7071
0.1667 0.3333 0.1667
1




 
K2 = 
0
 and C =  0.3333 0.6667 0.3333 . If X0 =  1  is chosen, and scaling is used we
−0.7071
0.1667 0.3333 0.1667
1
416
8.11 Approximation of Eigenvalues






0.5
0.5
0.5






obtain X1 =  1 , X2 =  1 . Taking K =  1  we find λ3 = 1. The eigenvalues are 4, 3, and 1.
0.5
0.5
0.5
 
1
 
The difficulty in choosing X0 =  1  to find the second eigenvector results from the fact that this vector is a
1
linear combination of the eigenvectors corresponding to the other two eigenvalues, with 0 contribution from the
second eigenvector. When this occurs the development of the power method, shown in the text, breaks down.
 
1
 
10. Taking X0 =  1  and using scaling we obtain
1








−0.3636
−0.2431
−0.2504
−0.2499








X1 =  −0.3636  , X2 =  0.0884  , X3 =  −0.0221  , X4 =  −0.0055  .
1
1
1
1


−0.25


Taking K =  0  as the dominant eigenvector we find λ1 = 16. The normalized eigenvector is
1




 
−0.2425
−0.9412
0 −0.2353
1




 
K1 = 
0
0
−4
0
 and B = 
. Taking X0 =  1  and using scaling we obtain
0.9701
1
−0.2353
0 −0.0588








−0.2941
0.0735
−0.0184
0.0046








X1 =  −1  , X2 =  1  , X3 =  −1  , X4 =  1  .
−0.0735
0.0184
−0.0046
0.0011
 
0
 
Taking K =  1  as the eigenvector we find λ2 = −4. The normalized eigenvector in this case is K2 = K =
0
 


 


0
−0.9412 0 −0.2353
1
−1
 


 


0
0
0
 1  and C = 
. Taking X0 =  1  and using scaling we obtain X1 =  0 ,
0
−0.2353 0 −0.0588
1
−0.25




1
1




X2 =  0 . Using K =  0  we find λ3 = −1. The eigenvalues are 16, −4, and −1.
0.25
0.25
4 −1
1
11. The inverse matrix is
. Taking X0 =
and using scaling we obtain
−3
1
1
1
1
1
1
X1 =
, X2 =
, X3 =
, X4 =
.
−0.6667
−0.7857
−0.7910
−0.7913
1
1 1
Using K =
we find λ = 4.7913. The minimum eigenvalue of
is 1/4.7913 ≈ 0.2087.
−0.7913
3 4
417
8.11 Approximation of Eigenvalues
1
and using scaling we obtain
4 2
1
0.6667
0.7857
0.75
X1 =
, X2 =
, . . . , X10 =
.
1
1
1
0.75
−0.2
0.3
Using K =
we find λ = 5. The minimum eigenvalue of
is 1/5 = 0.2
1
0.4 −0.1
12. The inverse matrix is
1
3
. Taking X0 =
13. (a) Replacing the second derivative with the difference expression we obtain
EI
yi+1 − 2yi + yi−1
+ P yi = 0
h2
or EI(yi+1 − 2yi + yi−1 ) + P h2 yi = 0.
(b) Expanding the difference equation for i = 1, 2, 3 and using h = L/4, y0 = 0, and y4 = 0 we obtain
EI(y2 − 2y1 + y0 ) +
P L2
y1 = 0
16
EI(y3 − 2y2 + y1 ) +
P L2
y2 = 0
16
EI(y4 − 2y3 + y2 ) +
P L2
y3 = 0
16
In matrix form this becomes

2

 −1
0
−1
2
−1
or
2y1 − y2 =
P L2
y1
16EI
−y1 + 2y2 − y3 =
P L2
y2
16EI
−y2 + 2y3 =
P L2
y3 .
16EI
 
 
0
y1
y1
P L2  
 
−1   y2  =
 y2  .
16EI
2
y3
y3


0.75 0.5 0.25


(c) A−1 =  0.5 1
0.5 
0.25 0.5 0.75
 
1
 
(d) Taking X0 =  1  and using scaling we obtain
1






0.75
0.7143
0.7083






X1 =  1  , X2 =  1  , X3 =  1  ,
0.75


Using K = 
(e) Solving
0.7143
0.7071

0.7073


X4 =  1  ,
0.7083

0.7073


0.7071


X5 =  1  .
0.7071


1  we find λ = 1.7071. Then 1/λ = 0.5859 is the minimum eigenvalue of A.
0.7071
P L2
EI
= 0.5859 for P we obtain P = 9.3726 2 . In Example 3 of Section 3.9 we saw
16EI
L
P = π2
EI
EI
≈ 9.8696 2 .
2
L
L
14. (a) The difference equation is
EIi (yi+1 − 2yi + yi−1 ) + P h2 yi = 0, i = 1, 2, 3,
418
8.11 Approximation of Eigenvalues
where I0 = 0.00200, I1 = 0.00175, I2 = 0.00150, I3 = 0.00125, and I4 = 0.00100. The system of equations
is
P L2
y1 = 0
16
P L2
0.00150E(y3 − 2y2 + y1 ) +
y2 = 0
16
P L2
0.00125E(y4 − 2y3 + y2 ) +
y2 = 0
16
In matrix form this becomes

0.0035 −0.00175

0.003
 −0.0015
0 −0.00125
P L2
y1
16E
P L2
−0.0015y1 + 0.003y2 − 0.0015y3 =
y2
16E
P L2
−0.00125y2 + 0.0025y3 =
y3 .
16E
0.00175E(y2 − 2y1 + y0 ) +
(b) The inverse of A is

A−1
0.0035y1 − 0.00175y2 =
or
 
 
0
y1
y1
   P L2  
−0.0015   y2  =
 y2  .
16E
0.0025
y3
y3
428.571

=  285.714
142.857
333.333
666.667
333.333
 
1
 
Taking X0 =  1  and using scaling we obtain
1






0.7113
0.6710
0.6645






X1 =  1  , X2 =  1  , X3 =  1  ,
0.7958
0.7679
0.7635

200

400  .
600


0.6634


X4 =  1  ,
0.7628


0.6632


X5 =  1  .
0.7627
This yields the eigenvalue λ = 1161.23. The smallest eigenvalue of A is then 1/λ = 0.0008612. The lowest
critical load is
16E
E
P =
(0.0008612) − 0.01378 2 .
L2
L

15. (a) A10
(b) X10
X12

67,745,349 −43,691,832
8,258,598


=  −43,691,832
28,182,816 −5,328,720 
8,258,598 −5,328,720
1,008,180
  



1
67,745,349
1
  



= A10  0  =  −43,691,832  ≈ 67,745,349  −0.644942 
0
8,258,598
0.121906
  



1
2,680,201,629
1
  



= A12  0  =  −1,728,645,624  ≈ 2,680,201,629  −0.644968  .
0
326,775,222
0.121922
The vectors appear to be approaching scalar multiples of K = (1, −0.644968, 0.121922), which approximates
the dominant eigenvector.
(c) The dominant eigenvalue is λ1 = (AK · K)/(K · K) = 6.28995.
419
8.12
8.11 Diagonalization
Approximation of Eigenvalues
EXERCISES 8.12
Diagonalization
1. Distinct eigenvalues λ1 = 1, λ2 = 5 imply A is diagonalizable.
−3 1
1
, D=
P=
1 1
0
0
5
2. Distinct eigenvalues λ1 = 0, λ2 = 6 imply A is diagonalizable.
−5 −1
0
P=
, D=
4
2
0
1
3. For λ1 = λ2 = 1 we obtain the single eigenvector K1 =
. Hence
1
√
√
4. Distinct eigenvalues λ1 = 5 , λ2 = − 5 imply A is diagonalizable.
√ √
√
5 − 5
5
P=
, D=
1
1
0
A is not diagonalizable.
5. Distinct eigenvalues λ1 = −7, λ2 = 4 imply A is diagonalizable.
13 1
−7
P=
, D=
2 1
0
0
6. Distinct eigenvalues λ1 = −4, λ2 = 10 imply A is diagonalizable.
−3
1
−4
P=
, D=
1 −5
0
0
6
0
√
− 5
4
0
10
1
2
, λ2 =
imply A is diagonalizable.
3
3
1
0
1 1
3
P=
, D=
0 23
−1 1
1
8. For λ1 = λ2 = −3 we obtain the single eigenvector K1 =
. Hence A is not diagonalizable.
1
9. Distinct eigenvalues λ1 = −i, λ2 = i imply A is diagonalizable.
1 1
−i 0
P=
, D=
−i i
0 i
7. Distinct eigenvalues λ1 =
10. Distinct eigenvalues λ1 = 1 + i, λ2 = 1 − i imply A is diagonalizable.
2
2
1+i
0
P=
, D=
i −i
0
1−i
11. Distinct eigenvalues λ1 = 1, λ2 = −1, λ3 = 2

1

P = 0
0
imply A is diagonalizable.



0 1
1
0 0



1 1  , D =  0 −1 0 
0 1
0
0 2
420
8.12 Diagonalization
12. Distinct eigenvalues λ1 = 3, λ2 = 4, λ3 = 5 imply A is diagonalizable.



1 2
0
3 0



P = 0 2
1, D = 0 4
1 1 −1
0 0
13. Distinct eigenvalues λ1 = 0, λ2 = 1, λ3 = 2 imply A is diagonalizable.



1 1 1
0 0



P =  0 1 0, D = 0 1
−1 1 1
0 0
0


0
5
0


0
2
14. Distinct eigenvalues λ1 = 1, λ2 = −3i, λ3 = 3i imply A is diagonalizable.




0 −3i 3i
1
0 0




P = 0
1 1  , D =  0 −3i 0 
1
0 0
0
0 3i
 
1
 
15. The eigenvalues are λ1 = λ2 = 1, λ3 = 2. For λ1 = λ2 = 1 we obtain the single eigenvector K1 =  0 .
0
Hence A is not diagonalizable.
16. Distinct eigenvalues λ1 = 1, λ2 = 2, λ3 = 3 imply A is diagonalizable.




1 1 0
1 0 0




P = 0 1 0, D = 0 2 0
0 0 1
0 0 3
√
√
17. Distinct eigenvalues λ1 = 1, λ2 = 5 , λ3 = − 5 imply A is diagonalizable.
√
√ 



0 1+ 5 1− 5
1 0
0
√




P = 0
2
2 , D = 0
5
0
√
0 0 − 5
1
0
0


1


18. For λ1 = λ2 = λ3 = 1 we obtain the single eigenvector K1 =  −2 . Hence A is not diagonalizable.
1
19. For the eigenvalues λ1 = λ2 = 2, λ3 = 1, λ4 = −1 we obtain four linearly independent eigenvectors. Hence A
is diagonalizable and




−3 −1 −1 1
2 0 0
0
 0
0 2 0
1
0 0
0




P=
, D = 
.
 −3
0 0 1
0
0 1
0
1
0
1 0
0 0 0 −1
 
1
0
 
20. The eigenvalues are λ1 = λ2 = 2, λ3 = λ4 = 3. For λ3 = λ4 = 3 we obtain the single eigenvector K1 =  .
1
0
Hence A is not diagonalizable.
1
1
21. λ1 = 0, λ2 = 2, K1 =
, K2 =
, P=
−1
1
√1
2
1
√
− 2
421
√1
2
√1
2
, D=
0
0
0
2
8.12 Diagonalization
22. λ1 = −1, λ2 = 4, K1 =
1
2
, K2 =
, P=
−2
1
√ √ 10
− 10
23. λ1 = 3, λ2 = 10, K1 =
, K2 =
, P=
2
5
24. λ1 = −1, λ2 = 3, K1 =
√2
5
√1
5
√1
5
− √25
1
1
, K2 =
, P=
1
−1
√1
2
√1
2
√
− √10
14
√2
14
√1
2
− √12
, D=
√
√10
35
√15
35
−1
0
0
4
, D=
, D=
−1
0


 
 
 1
− √2
−1
1
0


 
 
 1
25. λ1 = −1, λ2 = λ3 = 1, K1 =  1 , K2 =  1 , K3 =  0 , P =  √
2
0
0
1
0
0
3
√1
2
√1
2
0


 


 √1
− 2
−1
1
1


 



26. λ1 = λ2 = −1, λ3 = 5, K1 =  0 , K2 =  1 , K3 =  −1 , P = 
0
1
0
1
√1
2


−1
0 0


D =  0 −1 0 
0
0
3
0
0
10


−1


0 , D =  0
0
1
0
√1
2
√1
2
√1
3
− √13
√1
3
0
0
1
0

0

0
1
0
0


,
5
 




2
2
2
1
3
 





27. λ1 = 3, λ2 = 6, λ3 = 9, K1 =  2 , K2 =  −1 , K3 =  −2 , P =  32
1
1
−2
2
2
3
− 13
− 23
3
1
3
− 23
2
3


3


, D =  0
0

6
0

0
9
0
−6
0

0

0
8
√ 
√ 
 


0
1− 2
1+ 2
√
√


 


28. λ1 = 1, λ2 = 2 − 2 , λ3 = 2 + 2 , K1 =  1 , K2 =  0 , K3 =  0 ,
1
1
0
√ √
√ √ 



2
2+ 2
0 − 2−
1
0
0
2
2


√


P=
, D = 0 2 − 2 2
0 
0
0 
1

√ √
√ √
√
2+ 2
2− 2
0
0
2+ 2
0
2
2
 

 


0
0
1
1
 


 

29. λ1 = 1, λ2 = −6, λ3 = 8, K1 =  1 , K2 =  0 , K3 =  0 , P =  1
0
−1
1
0



30. λ1 = λ2 = 0, λ3 = −2, λ4 = 2, K1 = 


0



P=


− √12
0
0
− √12
√1
2
0
0
√1
2
1 
1
2
2
− 12
1
2
1
2
− 12


,
1 
2 
1
2

0
0

D=
0
0
0
0
0
0
0
1
0
0
−2
0

0
0


0
2
422


0
− √12
√1
2


1


0 , D =  0
√1
0
2
 
1







0
 −1 
 −1 
1
, K2 = 
, K 3 = 
, K4 =  
 0
 1
1
1
−1

√1
2
1
−1

1
8.12 Diagonalization
T
31. The given equation can be written as X AX = 24: ( x
λ1 = 6, λ2 = 4, K1 =
y)
√1
1
1
2
, K2 =
, P=
−1
1
− √12
6 0
X
(X Y )
= 24
0 4
Y
−1
5
5
−1
√1
2
√1
2
or
x
= 24. Using
y
and X = PX we find
6X 2 + 4Y 2 = 24.
The conic section is an ellipse. Now from X = PT X we see that the XY -coordinates of (1, −1) and (1, 1)
√
√
are ( 2 , 0) and (0, 2 ), respectively. From this we conclude that the X-axis and Y -axis are as shown in the
accompanying figure.
13 −5
x
32. The given equation can be written as XT AX = 288: ( x y )
= 288.
−5 13
y
√1
− √12
1
−1
2
and X= PX we find
Using λ1 = 8, λ2 = 18, K1 =
, K2 =
, P=
1
1
√1
√1
2
2
8 0
X
(X Y )
= 288 or 8X 2 + 18Y 2 = 288.
0 18
Y
The conic section is an ellipse. Now from X = PT X we see that the XY -coordinates of (1, 1) and (1, −1) are
√
√
( 2 , 0) and (0, − 2 ), respectively. From this we conclude that the X-axis and Y -axis are as shown in the
accompanying figure.
−3 4
x
T
33. The given equation can be written as X AX = 20: ( x y )
= 20. Using
4 3
y
√1
− √25
1
−2
5
λ1 = 5, λ2 = −5, K1 =
and X = PX we find
, K2 =
, P=
2
1
√2
√1
5
5
5
0
X
(X Y )
= 20 or 5X 2 − 5Y 2 = 20.
0 −5
Y
The conic section is a hyperbola. Now from X = PT X we see that the XY -coordinates of (1, 2) and (−2, 1)
√
√
are ( 5 , 0) and (0, 5 ), respectively. From this we conclude that the X-axis and Y -axis are as shown in the
accompanying figure.
34. The given equation can be written as XT AX = 288:
16 12
x
x
(x y)
+ ( −3 4 )
= 0.
12 9
y
y
4
3
4
−3
5
5
Using λ1 = 25, λ2 = 0, K1 =
and X = PX we find
, K2 =
, P= 3
4
3
4
−
5
5
25 0
X
X
(X Y )
+ (0 5)
= 0 or 25X 2 + 5Y = 0.
0 0
Y
Y
The conic section is a parabola. Now from X = PT X we see that the XY -coordinates of (4, 3) and (3, −4)
are (5, 0) and (0, −5), respectively. From this we conclude that the X-axis and Y -axis are as shown in the
accompanying figure.
35. Since D = P−1 AP we have A = PDP−1 . Hence
1 1
2 0
−1
A=
2 1
0 3
2
423
1
−1
=
4
−1
2
1
.
8.12 Diagonalization
36. Since eigenvectors are mutually orthogonal we use an orthogonal matrix P and A = PDPT .
 √1

  √1
  8 4

√1
√1
√1
− √13
− 13
1 0 0
3
3
3
2
6
3
3



 
4 
A =  − √13
0 √26   0 3 0   √12
0 − √12  =  34 11
3
3 
1
4
8
1
1
1
1
2
1
0
0
5
√
√
√
√
√
−
−√
3
2
6
6
6
3
6
3
3
37. Since D = P−1 AP we have
A = PDP−1
A2 = PDP−1 PDP−1 = PDDP−1 = PD2 P−1
A3 = A2 A = PD2 P−1 PDP−1 = PD2 DP−1 = PD3 P−1
and so on.
 4
2
0
0
 0 34
0

38. 
 0 0 (−1)4

16 0 0
0

0 
0 
  0 81 0

=

0 
0   0 0 1
0
0 0 625
0 0
0
(5)4
1
−1
1 −1
39. λ1 = 2, λ2 = −1, K1 =
, K2 =
, P=
, P−1 =
1
2
1
2
2
1
21 11
1 −1
32
0
3
3
A5 =
=
22 10
1
2
0 −1
− 13 31
5
2
5 2
−1
40. λ1 = 0, λ2 = 1, K1 =
, K2 =
, P=
, P−1 =
3
1
3 1
3
5 2
0 0
−1
2
6 −10
=
A10 =
3 1
0 1
3 −5
3
−5
0


2
3
− 13
2
−5
1
3
1
3
EXERCISES 8.13
Cryptography
5 14 4 0
. The encoded message is
5 12 16 0
1 2
19 5 14 4 0
35 15
B = AM =
=
1 1
8 5 12 16 0
27 10
1. (a) The message is M =
19
8
(b) The decoded message is
M=A
−1
B=
2. (a) The message is M =
B = AM =
3
5
1
2
−1
2
1 −1
20 8 5
0 9 19
0
0
35
27
13
8
20
8
5
0
13
15
0
9
19
0
8
5
15
10
38
26
36
20
0
0
=
19
8
38
26
36
20
5 14
5 12
0
0
4
16
.
0
0
.
25
. The encoded message is
0
14 5 25
60 69 110 0 79 70 132
=
18 5 0
20 26 43 0 29 25 50
15
5
14 5
18 5
424
40
75
15
25
.
8.13
(b) The decoded message is
M = A−1 B =
=
2
−1
−5
3
20
0
8
9
5
19
60
20
69
26
0 13
0 8
110
43
15
5
0 79 70 132
0 29 25 50
14 5 25
.
18 5 0
16 8 15 14 5
. The encoded message is
0 8 15 13 5
3 5
16 8 15 14 5
48 64
B = AM =
=
2 3
0 8 15 13 5
32 40
40 75
15 25
Cryptography
3. (a) The message is M =
(b) The decoded message is
M=A
−1
B=

7

4. (a) The message is M =  8
−3
2
15
5
−3
0
=
48
32
64
40
120
75
14
15
18
20
1

B = AM =  1
0
(b) The decoded message is


0
1 −1
50


−1
M = A B =  2 −2 −1   33
−1
1
1
26

7

5. (a) The message is M =  8

2

B = AM =  1
−1
1
1
1
15
0
40
25
=
16
0
20


50
 
1  =  33
0
26
14
13
.
75
55
44
31

22

21  .
28
15
52
40
13
1
15
0
14
15
18
0
14
15
0
14
19
0
20
13
0
44
31
 
22
7
 
21  =  8
28
15
52
40
13
1
9
20


1 .
0


0 15 14 0 13 1 . The encoded message is
9 14 0 19 20 0
0

 
1
7 15 0 14 15 18 20
31
44 15 61

 
1   8 0 15 14 0 13 1  =  24
29 15 47
0
9 14 0 19 20 0
0
1 −15 15 0
(b) The decoded message is


1 −1
0
31


M = A−1 B =  1 −1
1   24
−2
3 −1
1
5
5
99
66
75
55
20
.
30
15
99
66
18
15
15
57
43
30
15
15
8
8
40
25
The encoded message is
57
43
14
107
67


0 15 14 0 13 1 .
9 14 0 19 20 0
0

2 3
7 15 0 14 15 18

1 2   8 0 15 14 0 13
1 2
9 14 0 19 20 0

107
67
120
75
44
15
61
50
49
29
−15
15
15
47
0
35
−15
31
−5
41

50
35
−15
49
31
−5

41

21  .
−19
14
14
19
15
0
20
18
13
0
107
75
100
27 78
18 126
44
107

7
 
21  =  8
−19
9

4 18 0 10 15 8


6. (a) The message is M =  14 0 9 19 0 20 . The encoded message
8
5 0 19 16 25


 
5 3
0
4 18 0 10 15 8
62 90


 
B = AM =  4 3 −1   14 0 9 19 0 20  =  50 67
5 2
2
8 5 0 19 16 25
64 100
15
0
14
0
15
0

425
is
27


67  .
130

20

1 .
0
8.13 Cryptography
(b) The decoded message is

8 −6

−1
M = A B =  −13 10
−7
5
7. The decoded message is
M = A−1 B =

−3
62

5   50
3
64
2 −3
−5
8
90
67
100
152
95
 
100
4
 
67  =  14
130
8
27 107 75
27 78 44
18 126 107
184
116
171
107
86
56
212
133
=
19
0
20
8
18
0
5
0 10 15
9 19 0
0 19 16
21
1
4 25
18 4

8

20  .
25
.
From correspondence (1) we obtain: STUDY HARD.
8. The decoded message is
1 −1
46
−1
M=A B=
1 −2
23
−7
−15
−13
−14
−18
−18
22
2
1
−12
10
5
=
23
0
8
23
1
15
20 0
18 18
13 5
25 0
From correspondence (1) we obtain: WHAT ME WORRY .
9. The decoded message is

0

−1
M = A B = 0
1

0
1
31

1
0   19
0 −1
13
21
0
21
9
22 20
13 16
 
9
13
 
15  =  19
1
20
8
9
0
18
1
20
8
0
0
20
9
1
13
14
16
20
9

15  .
0
From correspondence (1) we obtain: MATH IS IMPORTANT.
10. The decoded message is


1
0 −1
36 32
28 61


−1
M = A B =  −1
1
2   −9 −2 −18 −1
0 −1
0
23 27
23 41


13 5 5 20 0 13 5 0


=  1 20 0 20 8
5 0 12  .
9
2
18
1
18
25
0
26
−18
56 10
−25 0
26
43
5

12

0 
12
0
From correspondence (1) we obtain: MEET ME AT THE LIBRARY .
u v
−1
11. Let A =
. Then
x y
u v
17
16
18
5
34 0
34
20
−1
A B=
x y
−30 −31 −32 −10 −59 0 −54 −35
9
5
25
,
−13 −6 −50
2
1
so 17u − 30v = 4, 16u − 31v = 1 and 5x − 6y = 1, 25x − 50y = 25. Then A−1 =
and
−1 −1
4 1
4 0 9 0 14 5 5 4 0
A−1 B
.
13 15 14 5 25 0 20 15 4 1 25
From correspondence

22 8

T
12. (a) M =  13 3
2 27

1

T
(b) B = M =  1
1
(1) we obtain: DAD I NEED MONEY TODAY.

19 27 21 3 3 27 21 18 21

21 22 3 25 27 6
7 14 23 
21 7 27 5 21 17 2 25 7
 
1
0
37
38 61 56
51 33 51
50 30
 
0
1  =  24
35 40 34
48 8 24
44 23
1 −1
11 −24 0 15 −24 20 6 −11 5
426

57 51

43 28 
−11 16

.
8.14
An Error-Correcting Code


−1
1
1


(c) BA−1 = B  2 −1 −1  = M
1
0 −1


15 22 20 8 23 6 21 22


13. (a) B =  10 22 18 23 25 2 23 25 
3 26 26 14 23 16 26 12
(b) Using correspondence (1) the encoded message is: OVTHWFUVJVRWYBWYCZZNWPZL.




1
4 −3
46
32
14
58
54 −34
35
86




(c) M A−1 B =  2
3 −2  B =  54
58
42
57
75 −14
59
95 
−2 −4
3
−61 −54 −34 −66 −77
28 −56 −108


19 5 14 4
0 20 8
5


M = M mod 27 =  0 4 15 3 21 13 5 14 .
20 0 20 15 4
1 25 0
Using correspondence (1) the encoded message is: SEND THE DOCUMENT TODAY.
EXERCISES 8.14
An Error-Correcting Code
1. ( 0 1
3. ( 0
0 0
5. ( 1 0
7. ( 1
1
1
0)
1
1)
0
1
0
0
1)
0 0)
9. Parity error
11. ( 1 0
0
1
In Problems 13-18, D = ( c1
T
14. D = P ( 0
15. DT = P ( 0
T
16. D = P ( 0
17. DT = P ( 0
1
0
1
0
1
1
1
0
0
1
1
1
1)
4. ( 1
0
1
0
0)
6. ( 0
1
1
0
1
8. ( 0
0
1)
10. ( 1
0
1
0)
12. Parity error
1)

13. DT = P ( 1
2. ( 1
c2
T
0) = (0
T
1) = (1
T
1) = (0
T
1) = (1
T
0) = (1
1

c3 ) and P =  1
0
0
0
1
1
1
1
0
1

1

1 .
1
0
1
1
T
0
1
0
1
1
0)
T
0
0
0
0
1
1)
T
1
0
0
1
0
1)
T
1
0
1
0
0
1)
T
1
0
0
1
1
0)
0) ; C = (0
0) ; C = (1
0) ; C = (0
1) ; C = (1
0) ; C = (1
427
0
1
0)
8.14 An Error-Correcting Code
18. DT = P ( 1
1
T
0
0) = (0
T
1
1) ; C = (0
1
1
1
1
0
0)
In Problems 19-28, W represents the correctly decoded message.
19. S = HRT = H ( 0
0
20. S = HRT = H ( 1
W = (1 0 0 0)
0
0
0
0
0) = (0
T
0
0) ;
a code word. W = ( 0
T
0
0
0)
1
0
0
0
0
0) = (0
1
1) ;
not a code word. The error is in the third bit.
21. S = HRT = H ( 1 1
W = (0 0 0 1)
0
1
1
0
1) = (1
0
1 ) ; not a code word. The error is in the fifth bit.
T
T
a code word. W = ( 0
0
1
0)
T
a code word. W = ( 1
1
1
1)
T
a code word. W = ( 0
1
1
0)
22. S = HRT = H ( 0
1
0
1
0
1
0) = (0
0
0) ;
23. S = HRT = H ( 1
1
1
1
1
1
1) = (0
0
0) ;
24. S = HRT = H ( 1
1
0
0
1
1
0) = (0
0
0) ;
T
not a code word. The error is in the second bit.
0) ;
T
not a code word. The error is in the second bit.
T
not a code word. The error is in the seventh bit.
T
not a code word. The error is in the second bit.
25. S = HRT = H ( 0 1
W = (1 0 0 1)
1
1
0
0
1) = (0
1
0) ;
26. S = HRT = H ( 1 0
W = (0 0 0 1)
0
1
0
0
1) = (0
1
27. S = HRT = H ( 1 0
W = (1 0 1 0)
1
1
0
1
1) = (1
1
1) ;
28. S = HRT = H ( 0 0
W = (1 0 1 1)
1
0
0
1
1) = (0
1
0) ;
29. (a) 27 = 128
(b) 24 = 16
(c) ( 0 0
0
0
0
0
0 ), ( 1
1
0
1
0
0
1 ), ( 0
1
0
1
0
1
0 ), ( 1
0
0
0
0
1
1 ),
(1 0
0
1
1
0
0 ), ( 0
1
0
0
1
0
1 ), ( 1
1
0
0
1
1
0 ), ( 0
0
0
1
1
1
1 ),
(1 1
1
0
0
0
0 ), ( 0
0
1
1
0
0
1 ), ( 1
0
1
1
0
1
0 ), ( 0
1
1
0
0
1
1 ),
(0 1
1
1
1
0
0 ), ( 1
0
1
0
1
0
1 ), ( 0
0
1
0
1
1
0 ), ( 1
1
1
1
1
1
1)
0) = (0
0
0
0)
30. (a) c4 = 0, c3

0
0

(b) H = 
0
1
= 1, c2 = 1, c1 = 0; ( 0 1 1 0 0 1 1 0 )

0 0 0 1 1 1 1
0 1 1 0 0 1 1

 (c) S = HRT = H ( 0
1 0 1 0 1 0 1
1
1
1
1
1
1
1
428
0
1
1
1
1
0
T
T
8.15 Method of Least Squares
EXERCISES 8.15
Method of Least Squares
YT = ( 1
1. We have
AT A =
Now
2
5
3
5
so X = (AT A)−1 AT Y =
54
14
T
Now
A A=
so X = (AT A)−1 AT Y =
13
5
− 25
A A=
so X = (A A)
4. We have
T
A Y=
(AT A)−1
1.1
−0.3
5
6
4
7)
and
T
and A =
0
1
1
1
1
20
(AT A)−1 =
T
−1
and
3 4 5
.
1 1 1
4 −14
1
=
20 −14
54
2
1
3
.
1
−6
14
2
1
4
−6
and the least squares line is y = 2.6x − 0.4.
Y = ( 1 1.5
T
3
14
6
T
Now
14
4
and AT =
2)
Y = ( −1
3. We have
3
and the least squares line is y = 0.4x + 0.6.
T
2. We have
2
55
15
3
15
5
4.5
5)
T
and A =
and
T
−1
(A A)
1
1
1
=
50
2
1
3
1
4
1
−15
55
5
−15
5
.
1
and the least squares line is y = 1.1x − 0.3.
T
Y = ( 0 1.5
3
4.5
5)
T
and A =
0
1
2
1
3
1
4
1
5
.
1
5 −14
54 14
1
T
−1
Now
A A=
and (A A) =
74 −14
54
14 5
1.06757
so X = (AT A)−1 AT Y =
and the least squares line is y = 1.06757x − 0.189189.
−0.189189
0 1 2 3 4 5 6
T
T
5. We have
Y = ( 2 3 5 5 9 8 10 ) and A =
.
1 1 1 1 1 1 1
7 −21
91 21
1
T
T
−1
Now
A A=
and (A A) =
196 −21
91
21 7
T
so X = (AT A)−1 AT Y =
6. We have
19
14
27
14
YT = ( 2 2.5
and the least squares line is y = 1.35714x + 1.92857.
1
Now
AT A =
1.5
2
140
28
28
7
3.2
5)
and
and AT =
(AT A)−1 =
429
1
2
3
4
1
1
1
1
1
196
7
−28
5
6
7
1 1
−28
140
1
.
8.15 Method of Least Squares
so X = (AT A)−1 AT Y =
7. We have
YT = ( 220
0.407143
0.828571
200
180
T
Now
A A=
so X = (AT A)−1 AT Y =
and the least squares line is y = 0.407143x + 0.828571.
170
36400
420
− 117
140
150
420
6
and AT =
135 )
and
(AT A)−1 =
20
1
1
42000
40
1
60
1
80
1
100
1
6 −420
−420 36400
120
.
1
and the least squares line is v = −0.835714T + 234.333. At T = 140,
703
3
v ≈ 117.333 and at T = 160, v ≈ 100.619.
T
8. We have Y = ( 0.47
0.90
2.0
3.7
7.5
15 )
T
and A =
600 650
.
1
1
6
−3150
−3150 1697500
400
1
450
1
500
1
550
1
1697500 3150
1
Now
A A=
and (AT A)−1 =
262500
3150
6
0.0538
so X = (AT A)−1 AT Y =
and the least squares line is R = 0.0538T − 23.3167. At T = 700,
−23.3167
R ≈ 14.3433.
T
EXERCISES 8.16
Discrete Compartmental Models
In Problems 1-5 we use the fact that the element τij in the transfer matrix T is the rate of transfer from compartment
j to compartment i, and the fact that the sum of each column in T is 1.
1. (a) The initial state and the transfer matrix are
90
0.8
X0 =
and T =
60
0.2
(b) We have
X1 = TX0 =
and
X2 = TX1 =
0.8
0.2
0.4
0.6
0.8
0.2
0.4
0.6
90
60
96
54
0.4
0.6
=
=
96
54
.
98.4
51.6
.
(c) From TX̂ − X̂ = (T − I)X̂ = 0 and the fact that the system is closed we obtain
−0.2x1 + 0.4x2 = 0
x1 +
x2 = 150.
The solution is x1 = 100, x2 = 50, so the equilibrium state is X̂ =
430
100
.
50
8.16 Discrete Compartmental Models
2. (a) The initial state and the transfer matrix are



100
0.7 0



X0 =  200  and T =  0.3 0.8
150
0 0.2
(b) We have


145


X1 = TX0 =  190 
115

0.5

0 .
0.5


159


and X2 = TX1 =  195.5  .
95.5
(c) From TX̂ − X̂ = (T − I)X̂ = 0 and the fact that the system is closed we obtain
−0.8x1 + 0.5x2
=0
0.3x1 − 0.9x2
=0
x1 +
x2 + x3 = 450.


145.161


The solution is x1 = 145.161, x2 = 217.742, x3 = 87.0968, so the equilibrium state is X̂ =  217.742 .
87.097
3. (a) The initial state and the transfer matrix are



100
0.2



X0 =  0  and T =  0.3
0
0.5
(b) We have


20
 
X1 = TX0 =  30 
50
0.5
0.1

0

0.
0.4
1


19
 
and X2 = TX1 =  9  .
72
(c) From TX̂ − X̂ = (T − I)X̂ = 0 and the fact that the system is closed we obtain
−0.8x1 + 0.5x2
=0
0.3x1 − 0.9x2
=0
x1 +
x2 + x3 = 100.


0


The solution is x1 = x2 = 0, x3 = 100, so the equilibrium state is X̂ =  0 .
100
4. (a) The transfer matrix is

0.7
0.05
0
0.75
0.2

T =  0.3
431
0.15


0 .
0.85
8.16 Discrete Compartmental Models
(b)
Year
Bare Space
Grasses
Small Shrubs
0
10.00
0.00
0.00
1
2
7.00
5.05
3.00
4.35
0.00
0.60
3
3.84
4.78
1.38
4
5
3.14
2.75
4.74
4.49
2.13
2.76
6
2.56
4.19
3.24
5. From TX̂ = 1X̂ we see that the equilibrium state vector X̂ is the eigenvector of the transfer matrix T corresponding to the eigenvalue 1. It has the properties that its components add up to the sum of the components
of the initial state vector.
6. (a) The initial state and the transfer matrix are



0
0.88



X0 =  100  and T =  0.06
0
0.06
(b)
Year
Phytoplankton
Water
Zooplankton
0
1
0.00
2.00
100.00
97.00
0.00
1.00
2
3.70
94.26
2.04
3
4
5.14
6.36
91.76
89.47
3.10
4.17
5
6
7.39
8.25
87.37
85.46
5.24
6.30
7
8.97
83.70
7.33
8
9
9.56
10.06
82.10
80.62
8.34
9.32
10
11
10.46
10.79
79.28
78.04
10.26
11.17
12
11.06
76.90
12.04
0.02
0.97

0

0.05  .
0.01
0.95
CHAPTER 8 REVIEW EXERCISES

2
3

1. 
4
5
3
4
5
6
3. AB =

4
5


6
2. 4 × 3
7
3
6
4
; BA = ( 11 )
8
4. A−1 = −
432
1
2
4
−3
−2
1
=
−2
1
3
2
− 12
CHAPTER 8 REVIEW EXERCISES
5. False; consider A =
7. det
1
2A
=
1 3
2
1
0
(5) =
5
8
0
1
and B =
0
1
1
0
6. True
; det(−AT ) = (−1)3 (5) = −5
8. det AB−1 = det A/ det B = 6/2 = 3
10. det C = (−1)3 / det B = −1/103 (2) = −1/2000
9. 0
12. True
11. False; an eigenvalue can be 0.
13. True
14. True, since complex roots of real polynomials occur in conjugate pairs.
15. False; if the characteristic equation of an n×n matrix has repeated roots, there may not be n linearly independent
eigenvectors.
16. True
17. True
18. True
19. False; A is singular and thus not orthogonal.
20. True
21. A = 12 (A + AT ) + 12 (A − AT ) where 12 (A + AT ) is symmetric and 12 (A − AT ) is skew-symmetric.
0 1
0 1
22. Since det A2 = (det A)2 ≥ 0 and det
= −1, there is no A such that A2 =
.
1 0
1 0
1
1
23. (a)
is nilpotent.
−1 −1
(b) Since det An = (det A)n = 0 we see that det A = 0 and A is singular.
i
0
0 −1
0
24. (a) σx σy =
= −σy σx ; σx σz =
= −σz σx ; σy σz =
0 −i
1
0
i
i
0
= −σz σy
(b) We first note that for anticommuting matrices AB = −BA, so C = 2AB. Then Cxy
0 2i
0 2
Cyz =
, and Czx =
.
2i 0
−2 0







5 −1 1 −9
9
1
1
1 5
1 1
5 9
R13
row
row







25.  2
4 0 27  −−−−−
4 0 27  −−−−−−→  0 1 −5 92  −−−−−−→  0
−→  2
operations
operations
1
1 5
9
5 −1 1 −9
0 0
1 12
0
=
2i
0
,
0 −2i
0
− 12
1 0
0 1
1
2
0


7 .
T
The solution is X = ( − 12 7 12 ) .



1
1
1 6
1 1
row



26.  1 −2
3 2  −−−−−−→  0 1
operations
0 0
2
0 −3 3
1
− 23
1
6


1 0 0
row

4 
3  −−−−−−→  0 1 0
operations
1
0 0 1
3


2 .
1
The solution is x1 = 3, x2 = 2, x3 = 1.
27. Multiplying the second row by abc we obtain the third row. Thus the determinant is 0.
28. Expanding along the first row we see that the result is an expression of the form ay + bx2 + cx + d = 0, which
is a parabola since, in this case a = 0 and b = 0. Letting x = 1 and y = 2 we note that the first and second
rows are the same. Similarly, when x = 2 and y = 3, the first and third rows are the same; and when x = 3
433
CHAPTER 8 REVIEW EXERCISES
and y = 5, the first and fourth rows are the same. In each case the determinant is 0 and the points lie on the
parabola.
29. 4(−2)(3)(−1)(2)(5) = 240
30. (−3)(6)(9)(1) = −162


1 −1
1


31. Since  5
1 −1  = 18 = 0, the system has only the trivial solution.
1
2
1


1 −1 −1


32. Since  5
1 −1  = 0, the system has infinitely many solutions.
1
2
1
33. From x1 I2 + x2 HNO3 → x3 HIO3 + x4 NO2 + x5 H2 O we obtain the system 2x1 = x3 , x2 = x3 + 2x5 , x2 = x4 ,
3x2 = 3x3 +2x4 +x5 . Letting x4 = x2 in the fourth equation we obtain x2 = 3x3 +x5 . Taking x1 = t we see that
x3 = 2t, x2 = 2t + 2x5 , and x2 = 6t + x5 . From the latter two equations we get x5 = 4t. Taking t = 1 we have
x1 = 1, x2 = 10, x3 = 2, x4 = 10, and x5 = 4. The balanced equation is I2 +10HNO3 → 2HIO3 +10NO2 +4H2 O.
34. From x1 Ca + x2 H3 PO4 → x3 Ca3 P2 O8 + x4 H2 we obtain the system x1 = 3x3 , 3x2 = 2x4 , x2 = 2x3 , 4x2 = 8x3 .
Letting x3 = t we see that x1 = 3t, x2 = 2t, and x4 = 3t. Taking t = 1 we obtain the balanced equation
3Ca + 2H3 PO4 → Ca3 P2 O8 + 3H2 .
42
−21
−56
1
1
2
= − , x2 =
= , x3 =
=
−84
2
−84
4
−84
3
16
−4
0
36. det = 4, det A1 = 16, det A2 = −4, det A3 = 0; x1 =
= 4, x2 =
= −1, x3 = = 0
4
4
4
35. det A = −84, det A1 = 42, det A2 = −21, det A3 = −56; x1 =
37. det A = cos2 θ + sin2 θ, det A1 = X cos θ − Y sin θ, det A2 = Y cos θ + X sin θ;
x1 = X cos θ − Y sin θ, y = Y cos θ + X sin θ
38. (a) i1 − i2 − i3 − i4 = 0, i2 R1 = E, i2 R1 − i3 R2 = 0, i3 R2 − i4 R3 = 0


1 −1
−1
−1
0 R
0
0
1


(b) det A = 
 = R1 R2 R3 ;
 0 R1 −R2
0
0

0
0
E

det A1 = 
0
0
R2
−1
−1
R1
R1
0
−R2
0
R2
−R3
−1

0

 = −E[−R2 R3 − R1 (R3 + R2 )] = E(R2 R3 + R1 R3 + R1 R2 );
0
−R3
det A1
1
1
1
E(R2 R3 + R1 R3 + R1 R2 )
=E
+
+
=
det A
R1 R2 R 3
R1
R2
R3

  


2
3 −1
x1
6
−2
1

  

−1
39. AX = B is  1 −2
0   x2  =  −3 . Since A = −  −1
3
−2
0
1
9
−4
x3
 
7
 
X = A−1 B =  5 .
23
i1 =
434
−3
0

−2

−1 , we have
−6
−7
CHAPTER 8 REVIEW EXERCISES

− 14
− 94
1
2
1
2
− 14
4

40. (a) A−1 B =  −1

  

1
−1


3  
2 1 =  1
1
0
−1
3
2
3
2

(b) A−1 B =  −1
1
2
− 14
1
2
− 14
 

−2
−10
 

3 
7
2  1 = 
3
−2
−1
− 94

4
41. From the characteristic equation λ2 − 4λ − 5 = 0 wesee that
the eigenvalues are λ1 = −1 and λ2 = 5. For
−1
λ1 = −1 we have 2k1 +2k2 = 0, 4k1 4k2 = 0 and K1 =
. For λ2 = 5 we have −4k1 +2k2 = 0, 4k1 −2k2 = 0
1
1
and K2 =
.
2
2
42. From the characteristic
equation λ = 0 we see that the eigenvalues are λ1 = λ2 = 0. For λ1 = λ2 = 0 we have
0
4k1 = 0 and K1 =
is a single eigenvector.
1
43. From the characteristic equation −λ3 + 6λ2 + 15λ + 8 = −(λ + 1)2 (λ − 8) = 0 we see that the eigenvalues are
λ1 = λ2 = −1 and λ3 = 8. For λ1 = λ2 = −1 we have




4 2 4 0
1 12 1 0
row




 2 1 2 0  −−−−−−→  0 0 0 0  .
operations
4 2 4 0
0 0 0 0
Thus K1 = ( 1 −2
Thus K3 = ( 2
T
T
0 ) and K2 = ( 1 0 −1 ) . For λ3 = 8 we have



−5
2
4 0
1 − 25 − 45
row



2 0  −−−−−−→  0
1 − 12
 2 −8
operations
4
2 −5 0
0
0
0
0


0.
0
T
1
2) .
44. From the characteristic equation −λ3 +18λ2 −99λ+162 = −(λ−9)(λ−6)(λ−3) = 0 we see that the eigenvalues
are λ1 = 9, λ2 = 6, and λ3 = 3. For λ1 = 9 we have



−2 −2
0 0
1 1
row



2 0  −−−−−−→  0 1
 −2 −3
operations
0
2 −4 0
0 0
Thus K1 = ( −2
Thus K2 = ( 2
2
1
0
−2
0
0


0.
0
T
1 ) . For λ2 = 6 we have



1 −2
0 0
1 −2
row



0
2 0  −−−−−−→  0
1
 −2
operations
0
2 −1 0
0
0

0

0.
0
− 12
0
0
T
2 ) . For λ3 = 3 we have



4 −2 0 0
1 − 12 0
row



3 2 0  −−−−−−→  0
1 1
 −2
operations
0
2 2 0
0
0 0
T
Thus K3 = ( 1 2 −2 ) .
435

0

0.
0
CHAPTER 8 REVIEW EXERCISES
45. From the characteristic equation −λ3 − λ2 + 21λ + 45 = −(λ + 3)2 (λ − 5)
λ1 = λ2 = −3 and λ3 = 5. For λ1 = λ2 = −3 we have



1
2 −3 0
1 2 −3
row



4 −6 0  −−−−−−→  0 0
0
 2
operations
−1 −2
3 0
0 0
0
Thus K1 = ( −2
T
1
Thus K3 = ( −1
= 0 we see that the eigenvalues are

0

0.
0
T
0 ) and K2 = ( 3 0 1 ) . For λ3 = 5 we have



−7
2 −3 0
1 − 27
row



1
 2 −4 −6 0  −−−−−−→  0
operations
−1 −2 −5 0
0
0
3
7

0
2

0.
0
0
T
−2
1) .
46. From the characteristic equation −λ3 + λ2 + 2λ = −λ(λ + 1)(λ − 2) = 0 we see that the eigenvalues are λ1 = 0,
T
λ2 = −1, and λ3 = 2. For λ1 = 0 we have k3 = 0, 2k1 + 2k2 + k3 = 0 and K1 = ( 1 −1 0 ) . For λ2 = −1 we
have




1 0 0 0
1 0 0 0
row




 0 1 1 0  −−−−−−→  0 1 1 0  .
operations
2 2 2 0
0 0 0 0
T
Thus K2 = ( 0 1 −1 ) . For λ3 = 2 we have



−2
0
0 0
1 0
row



0
−2
1
0
−
−
−
−
−
−→


0 1
operations
2
2 −1 0
0 0
Thus K3 = ( 0
1
47. Let X1 = ( a b
XT1 ( √13
√1
3
√1
3

0

0.
0
− 12
0
0
T
2) .
T
c)
T
be the first column of the matrix. Then XT1 ( − √12
) =
√1 (a
3
T
√1 )
2
0
=
√1 (c
2
− a) = 0 and
+ b + c) = 0. Also XT1 X1 = a2 + b2 + c2 = 1. We see that c = a and b = −2a
from the first two equations. Then a2 + 4a2 + a2 = 6a2 = 1 and a =
√1
6
. Thus X1 = ( √16
− √26
T
√1 ) .
6
T
48. (a) Eigenvalues are λ1 = λ2 = 0 and λ3 = 5 with corresponding eigenvectors K1 = ( 0 1 0 ) , K2 =
√
√
T
T
( 2 0 1 ) , and K3 = ( −1 0 2 ) . Since K1 = 1, K2 = 5 , and K3 = 5 , we have




0
1 0
0 √25 − √15


 2
1 
−1
T
P = 1 0
0  and P = P =  √5 0 √5  .
√2
0 √15
− √1 0 √2
5
5

0 0

−1
(b) P AP =  0 0
0 0
1
49. We identify A = 3
(X
Y )D
X
Y
2

0

0
5
3 2
. Eigenvalues are λ1 = − 12 and λ2 =
1
5
2
so D =
= − 12 X 2 + 52 Y 2 = 1. The graph is a hyperbola.
436
− 12
0
5
0
5
2
and the equation becomes
CHAPTER 8 REVIEW EXERCISES
50. We measure years in units of 10, with 0 corresponding to 1890. Then Y = ( 63 76
T
0 1 2 3 4
30 10
A=
, so AT A =
. Thus
1 1 1 1 1
10 5
5 −10
15
1
T
−1 T
T
X = (A A) A Y =
A Y=
,
50 −10
30
62
92
106
T
123 )
and
and the least squares line is y = 15t + 62. At t = 5 (corresponding to 1940) we have y = 137. The error in the
predicted population is 5 million or 3.7%.
51. The encoded message is
B = AM =
=
10 1
9 1
204
185
52. The encoded message is
13
12
B = AM =
=
53. The decoded message is

−3
2

−1
M=A B= 1
0
2 −1
19
14
1
3
208
188
10 1
9 1
20
8
5 12
5 4
55
50
124
112
19
18
5
22
12
0
120
108
3
19
9 20 5 0 12 1 21
15 14 0 6 18 9 0
105
96
214 50 6 138 19
194 45 6 126 18
0 1
0 20
7
21
72
49
0
30
91
145
189
67
46
0
29
84
131

−1
19

0   35
1
5
0
10
 
14 0 20
8
 
53 1 54  =  19
−3 48 2 39
8
15
27
15
210
189
14 20 0 1 18
5 19 0 1 13
219 0 11 193
.
199 0 10 175
208
5
12
.
16
0

−3
2

−1
M=A B= 1
0
2 −1

−1
5

0   27
1
21
From correspondence (1) we obtain: ROSEBUD
2
17
13
 
21
18
 
40  =  5
−2
4
15
19


21  .
0
2
0
.
55. (a) The parity is even so the decoded message is ( 1
1
0
0
1)
(b) The parity is odd; there is a parity error.
 
 
 1
 
c1
1 1 0 1  
0
 
0  
56. From  c2   1 0 1 1    =  0  we obtain the codeword ( 0
0
0 1 1 1
1
c3
1
437
0
1
1


14 0 20  .
23 1 25
0 15
5 0
From correspondence (1) we obtain: HELP IS ON THE WAY.
54. The decoded message is
9
0
0
1 ).
9
Vector Calculus
EXERCISES 9.1
Vector Functions
1.
2.
3.
4.
5.
6.
7.
8.
9.
Note: the scale is distorted in this graph. For t = 0, the graph starts at (1, 0, 1). The upper loop shown
intersects the xz-plane at about (286751, 0, 286751).
438
9.1
10.
11. x = t, y = t, z = t2 + t2 = 2t2 ; r(t) = ti + tj + 2t2 k
√
√
√
12. x = t, y = 2t, z = ± t2 + 4t2 + 1 = ± 5t2 − 1 ; r(t) = ti + 2tj ± 5t2 − 1 k
13. x = 3 cos t, z = 9 − 9 cos2 t = 9 sin2 t, y = 3 sin t; r(t) = 3 cos ti + 3 sin tj + 9 sin2 tk
14. x = sin t, z = 1, y = cos t; r(t) = sin ti + cos tj + k
439
Vector Functions
9.1
Vector Functions
15. r(t) =
sin 2t
ln t
i + (t − 2)5 j +
k. Using L’Hôpital’s Rule,
t
1/t
2 cos 2t
1/t
5
lim r(t) =
k = 2i − 32j.
i + (t − 2) j +
1
−1/t2
t→0+
16. (a) limt→α [−4r1 (t) + 3r2 (t)] = −4(i − 2j + k) + 3(2i + 5j + 7k) = 2i + 23j + 17k
(b) limt→α r1 (t) · r2 (t) = (i − 2j + k) · (2i + 5j + 7k) = −1
17. r (t) =
1
1
1
2
i − 2 j; r (t) = − 2 i + 3 j
t
t
t
t
18. r (t) = −t sin t, 1 − sin t; r (t) = −t cos t − sin t, − cos t
19. r (t) = 2te2t + e2t , 3t2 , 8t − 1; r (t) = 4te2t + 4e2t , 6t, 8
1
2t
20. r (t) = 2ti + 3t2 j +
k; r (t) = 2i + 6tj −
k
1 + t2
(1 + t2 )2
21. r (t) = −2 sin ti + 6 cos tj
√
r (π/6) = −i + 3 3 j
23. r (t) = j −
8t
k
(1 + t2 )2
r (1) = j − 2k
22. r (t) = 3t2 i + 2tj
r (−1) = 3i − 2j
24. r (t) = −3 sin ti + 3 cos tj + 2k
√
√
3 2
−3 2
r (π/4) =
i+
j + 2k
2
2
1
1
8
25. r(t) = ti + t2 j + t3 k; r(2) = 2i + 2j + k; r (t) = i + tj + t2 k; r (2) = i + 2j + 4k
2
3
3
Using the point (2, 2, 8/3) and the direction vector r (2), we have x = 2 + t, y = 2 + 2t, z = 8/3 + 4t.
6t
6
3
26. r(t) = (t3 −t)i+
j+(2t+1)2 k; r(1) = 3j+9k; r (t) = (3t2 −1)i+
j+(8t+4)k; r (1) = 2i+ j+12k.
t+1
(t + 1)2
2
Using the point (0, 3, 9) and the direction vector r (1), we have x = 2t, y = 3 + 32 t, z = 9 + 12t.
d
[r(t) × r (t)] = r(t) × r (t) + r (t) × r (t) = r(t) × r (t)
dt
d
d
28.
[r(t) · (tr(t))] = r(t) · (tr(t)) + r (t) · (tr(t)) = r(t) · (tr (t) + r(t)) + r (t) · (tr(t))
dt
dt
= r(t) · (tr (t)) + r(t) · r(t) + r (t) · (tr(t)) = 2t(r(t) · r (t)) + r(t) · r(t)
27.
440
9.1
29.
Vector Functions
d
d
[r(t) · (r (t) × r (t))] = r(t) · (r (t) × r (t)) + r (t) · (r (t) × r (t))
dt
dt
= r(t) · (r (t) × r (t) + r (t) × r (t)) + r (t) · (r (t) × r (t))
= r(t) · (r (t) × r (t))
30.
d
d
[r1 (t) × (r2 (t) × r3 (t))] = r1 (t) × (r2 (t) × r3 (t)) + r1 (t) × (r2 (t) × r3 (t))
dt
dt
= r1 (t) × (r2 (t) × r3 (t) + r2 (t) × r3 (t)) + r1 (t) × (r2 (t) × r3 (t))
= r1 (t) × (r2 (t) × r3 (t)) + r1 (t) × (r2 (t) × r3 (t)) + r1 (t) × (r2 (t) × r3 (t))
d
1
1 1
31.
r1 (2t) + r2
= 2r1 (2t) − 2 r2
dt
t
t
t
d 3 2
[t r(t )] = t3 (2t)r (t2 ) + 3t2 r(t2 ) = 2t4 r (t2 ) + 3t2 r(t2 )
dt
2
2
2
2
2
2
1 2
3
33.
r(t) dt =
t dt i +
3t2 dt j +
4t3 dt k = t2 i + t3 j + t4 k = i + 9j + 15k
2
2
−1
−1
−1
−1
−1
−1
−1
4
4
4
4
√
√
34.
r(t) dt =
2t + 1 dt i +
− t dt j +
sin πt dt k
32.
0
0
0
0
4
4
4
1
16
2
1
26
= (2t + 1)3/2 i − t3/2 j − cos πt k =
i− j
3
3
π
3
3
0
0
0
2
35.
r(t) dt =
tet dt i +
−e−2t dt j +
tet dt k
= [tet − et + c1 ]i +
1 2
1
1 2
e−2t + c2 j + et + c3 k = et (t − 1)i + e−2t j + et k + c,
2
2
2
2
1
where c = c1 i + c2 j + c3 k.
1
t
t2
36.
r(t) dt =
dt
i
+
dt
j
+
dt
k
1 + t2
1 + t2
1 + t2
1
1
= [tan−1 t + c1 ]i +
dt
k
1−
ln(1 + t2 ) + c2 j +
2
1 + t2
1
= [tan−1 t + c1 ]i +
ln(1 + t2 ) + c2 j + [t − tan−1 t + c3 ]k
2
1
−1
= tan ti + ln(1 + t2 )j + (t − tan−1 t)k + c,
2
where c = c1 i + c2 j + c3 k.
37. r(t) = r (t) dt =
6 dt i +
6t dt j +
3t2 dt k = [6t + c1 ]i + [3t2 + c2 ]j + [t3 + c3 ]k
Since r(0) = i − 2j + k = c1 i + c2 j + c3 k, c1 = 1, c2 = −2, and c3 = 1. Thus,
r(t) = (6t + 1)i + (3t2 − 2)j + (t3 + 1)k.
1
1
38. r(t) = r (t) dt =
t sin t2 dt i +
− cos 2t dt j = −[ cos t2 + c1 ]i + [− sin 2t + c2 ]j
2
2
Since r(0) = 32 i = (− 12 + c1 )i + c2 j, c1 = 2 and c2 = 0. Thus,
1
1
r(t) = − cos t2 + 2 i − sin 2tj.
2
2
39. r (t) = r (t) dt =
12t dt i +
−3t−1/2 dt j +
2 dt k = [6t2 + c1 ]i + [−6t1/2 + c2 ]j + [2t + c3 ]k
441
9.1
Vector Functions
Since r (1) = j = (6 + c1 )i + (−6 + c2 )j + (2 + c3 )k, c1 = −6, c2 = 7, and c3 = −2. Thus,
r (t) = (6t2 − 6)i + (−6t1/2 + 7)j + (2t − 2)k.
r(t) = r (t) dt =
(6t2 − 6) dt i +
(−6t1/2 + 7) dt j +
(2t − 2) dt k
= [2t3 − 6t + c4 ]i + [−4t3/2 + 7t + c5 ]j + [t2 − 2t + c6 ]k.
Since
r(1) = 2i − k = (−4 + c4 )i + (3 + c5 )j + (−1 + c6 )k,
c4 = 6, c5 = −3, and c6 = 0. Thus,
r(t) = (2t3 − 6t + 6)i + (−4t3/2 + 7t − 3)j + (t2 − 2t)k.
40. r (t) = r (t) dt =
sec2 t dt i +
cos t dt j +
− sin t dt k
= [tan t + c1 ]i + [sin t + c2 ]j + [cos t + c3 ]k
Since r (0) = i + j + k = c1 i + c2 j + (1 + c3 )k, c1 = 1, c2 = 1, and c3 = 0. Thus,
r (t) = (tan t + 1)i + (sin t + 1)j + cos tk.
r(t) = r (t) dt =
(tan t + 1) dt i +
(sin t + 1) dt j +
cos t dt k
= [ln | sec t| + t + c4 ]i + [− cos t + t + c5 ]j + [sin t + c6 ]k.
Since r(0) = −j + 5k = c4 i + (−1 + c5 )j + c6 k, c4 = 0, c5 = 0, and c6 = 5. Thus,
r(t) = (ln | sec t| + t)i + (− cos t + t)j + (sin t + 5)k.
41. r (t) = −a sin ti + a cos tj + ck; r (t) = (−a sin t)2 + (a cos t)2 + c2 =
2π
2π
s=
a2 + c2 dt = a2 + c2 t = 2π a2 + c2
a2 + c2
0
0
42. r (t) = i + (cos t − t sin t)j + (sin t + t cos t)k
r (t) = 12 + (cos t − t sin t)2 + (sin t + t cos t)2 =
π
t
s=
2 + t2 dt =
2 + t2 + ln t + 2 + t2 2
0
√
t
t
2 + t2
π
= π 2 + π 2 + ln(π +
2
√
2 + π 2 ) − ln 2
0
43. r (t) = (−2e sin 2t + e cos 2t)i + (2e cos 2t + e sin 2t)j + et k
√
√
r (t) = 5e2t cos2 2t + 5e2t sin2 2t + e2t = 6e2t = 6 et
3π √
√ 3π √
s=
6 et dt = 6 et = 6 (e3π − 1)
0
t
√
t
0
√
√
√
44. r (t) = 3i + 2 3 tj + 2t2 k; r (t) = 32 + (2 3 t)2 + (2t2 )2 = 9 + 12t2 + 4t4 = 3 + 2t2
1
1
2
2
11
s=
(3 + 2t2 ) dt = (3t + t3 ) = 3 + =
3
3
3
0
0
t
2
2
2
2
45. r (t) = −a sin ti + a cos tj; r (t) = a sin t + a cos t = a, a > 0; s =
a du = at
0
r(s) = a cos(s/a)i + a sin(s/a)j; r (s) = − sin(s/a)i + cos(s/a)j
r (s) =
sin2 (s/a) + cos2 (s/a) = 1
442
9.2
Motion on a Curve
√
√
2
2
1
46. r (s) = − √ sin(s/ 5 )i + √ cos(s/ 5 )j + √ k
5
5
5
√
√
4
4 1
4
1
r (s) =
sin2 (s/ 5 ) + cos2 (s/ 5 ) + =
+ =1
5
5
5
5 5
d
d 2
d
d
(r · r) =
r2 =
c = 0 and
(r · r) = r · r + r · r = 2r · r , we have r · r = 0. Thus, r is
dt
dt
dt
dt
perpendicular to r.
47. Since
48. Since r(t) is the length of r(t), r(t) = c represents a curve lying on a sphere of radius c centered at the
origin.
49. Let r1 (t) = x(t)i + y(t)j. Then
d
d
[u(t)r1 (t)] = [u(t)x(t)i + u(t)y(t)j] = [u(t)x (t) + u (t)x(t)]i + [u(t)y (t) + u (t)y(t)]j
dt
dt
= u(t)[x (t)i + y (t)j] + u (t)[x(t)i + y(t)j] = u(t)r1 (t) + u (t)r1 (t).
50. Let r1 (t) = x1 (t)i + y1 (t)j and r2 (t) = x2 (t)i + y2 (t)j. Then
d
d
[r1 (t) · r2 (t)] = [x1 (t)x2 (t) + y1 (t)y2 (t)] = x1 (t)x2 (t) + x1 (t)x2 (t) + y1 (t)y2 (t) + y1 (t)y2 (t)
dt
dt
= [x1 (t)x2 (t) + y1 (t)y2 (t)] + [x1 (t)x2 (t) + y1 (t)y2 (t)] = r1 (t) · r2 (t) + r1 (t) · r2 (t).
51.
d
r1 (t + h) × r2 (t + h) − r1 (t) × r2 (t)
[r1 (t) × r2 (t)] = lim
h→0
dt
h
r1 (t + h) × r2 (t + h) − r1 (t + h) × r2 (t) + r1 (t + h) × r2 (t) − r1 (t) × r2 (t)
= lim
h→0
h
r1 (t + h) × [r2 (t + h) − r2 (t)]
[r1 (t + h) − r1 (t)] × r2 (t)
= lim
+ lim
h→0
h→0
h
h
r2 (t + h) − r2 (t)
r1 (t + h) − r1 (t)
+ lim
× r2 (t)
= r1 (t) × lim
h→0
h→0
h
h
= r1 (t) × r2 (t) + r1 (t) × r2 (t)
52. Let v = ai + bj and r(t) = x(t)i + y(t)j. Then
b
b
v · r(t) dt =
[ax(t) + by(t)] dt = a
a
a
b
EXERCISES 9.2
Motion on a Curve
1. v(t) = 2ti + t3 j; v(1) = 2i + j; v(1) =
a(t) = 2i + 3t2 j; a(1) = 2i + 3j
√
4+1=
√
5;
443
y(t) dt = v ·
x(t) dt + b
a
b
a
b
r(t) dt.
a
9.2
Motion on a Curve
√
√
2
j; v(1) = 2i − 2j; v(1) = 4 + 4 = 2 2 ;
3
t
6
a(t) = 2i + 4 j; a(1) = 2i + 6j
t
2. v(t) = 2ti −
3. v(t) = −2 sinh 2ti+2 cosh 2tj; v(0) = 2j; v(0) = 2;
a(t) = −4 cosh 2ti + 4 sinh 2tj; a(0) = −4i
√
√
1
4. v(t) = −2 sin ti + cos tj; v(π/3) = − 3 i + j; v(π/3) = 3 + 1/4 = 13/2;
2
√
3
a(t) = −2 cos ti − sin tj; a(π/3) = −i −
j
2
√
√
5. v(t) = (2t − 2)j + k; v(2) = 2j + k v(2) = 4 + 1 = 5 ;
a(t) = 2j; a(2) = 2j
6. v(t) = i+j+3t2 k; v(2) = i+j+12k; v(2) =
a(2) = 12k
√
1 + 1 + 144 =
7. v(t) = i + 2tj + 3t2 k; v(1) = i + 2j + 3k; v(1) =
a(t) = 2j + 6tk; a(1) = 2j + 6k
8. v(t) = i + 3t2 j + k; v(1) = i + 3j + k; v(1) =
a(t) = 6tj; a(1) = 6j
√
√
√
146 ; a(t) = 6tk;
1+4+9=
1+9+1=
√
√
14 ;
11 ;
9. The particle passes through the xy-plane when z(t) = t2 − 5t = 0 or t = 0, 5 which gives us the points (0, 0, 0)
and (25, 115, 0). v(t) = 2ti + (3t2 − 2)j + (2t − 5)k; v(0) = −2j − 5k, v(5) = 10i + 73j + 5k; a(t) = 2i + 6tj + 2k;
a(0) = 2i + 2k, a(5) = 2i + 30j + 2k
10. If a(t) = 0, then v(t) = c1 and r(t) = c1 t + c2 . The graph of this equation is a straight line.
444
9.2
Motion on a Curve
√
11. Initially we are given s0 = 0 and v0 = (480 cos 30◦ )i + (480 sin 30◦ )j = 240 3 i + 240j. Using a(t) = −32j we
find
v(t) = a(t) dt = −32tj + c
√
240 3 i + 240j = v(0) = c
√
√
v(t) = −32tj + 240 3 i + 240j = 240 3 i + (240 − 32t)j
√
r(t) = v(t) dt = 240 3 ti + (240t − 16t2 )j + b
0 = r(0) = b.
√
√
(a) The shell’s trajectory is given by r(t) = 240 3 ti + (240t − 16t2 )j or x = 240 3 t, y = 240t − 16t2 .
(b) Solving dy/dt = 240 − 32t = 0, we see that y is maximum when t = 15/2. The maximum
altitude is y(15/2) = 900 ft.
(c) Solving y(t) = 240t − 16t2 = 16t(15 − t) = 0, we see that the shell is at ground level when
√
t = 0 and t = 15. The range of the shell is x(15) = 3600 3 ≈ 6235 ft.
(d) From (c), impact is when t = 15. The speed at impact is
√
v(15) = |240 3 i + (240 − 32 · 15)j| = 2402 · 3 + (−240)2 = 480 ft/s.
√
12. Initially we are given s0 = 1600j and v0 = (480 cos 30◦ )i + (480 sin 30◦ )j = 240 3 i + 240j. Using a(t) = −32j
we find
v(t) = a(t) dt = −32tj + c
√
240 3 i + 240j = v(0) = c
√
√
v(t) = −32tj + 240 3 i + 240j = 240 3 i + (240 − 32t)j
√
r(t) = v(t) dt = 240 3 ti + (240t − 16t2 )j + b
1600j = r(0) = b.
√
√
(a) The shell’s trajectory is given by r(t) = 240 3 ti + (240t − 16t2 + 1600)j or x = 240 3 t,
y = 240t − 16t2 + 1600.
(b) Solving dy/dt = 240 − 32t = 0, we see that y is maximum when t = 15/2. The maximum
altitude is y(15/2) = 2500 ft.
(c) Solving y(t) = −16t2 + 240t + 1600 = −16(t − 20)(t + 5) = 0, we see that the shell hits the ground
√
when t = 20. The range of the shell is x(20) = 4800 3 ≈ 8314 ft.
(d) From (c), impact is when t = 20. The speed at impact is
√
√
|v(20) = |240 3 i + (240 − 32 · 20)j| = 2402 · 3 + (−400)2 = 160 13 ≈ 577 ft/s.
13. We are given s0 = 81j and v0 = 4i. Using a(t) = −32j, we have
v(t) = a(t) dt = −32tj + c
4i = v(0) = c
r(t) =
v(t) = 4i − 32tj
v(t) dt = 4ti − 16t2 j + b
81j = r(0) = b
r(t) = 4ti + (81 − 16t2 )j.
445
9.2
Motion on a Curve
Solving y(t) = 81 − 16t2 = 0, we see that the car hits the water when t = 9/4. Then
√
v(9/4) = |4i − 32(9/4)j| = 42 + 722 = 20 13 ≈ 72.11 ft/s.
14. Let θ be the angle of elevation. Then v(0) = 98 cos θi + 98 sin θj. Using a(t) = −9.8j, we have
v(t) = a(t) dt = −9.8tj + c
98 cos θi + 98 sin θj = v(0) = c
v(t) = 98 cos θi + (98 sin θ − 9.8t)j
r(t) = 98t cos θi + (98t sin θ − 4.9t2 )j + b.
Since r(0) = 0, b = 0 and r(t) = 98t cos θi + (98t sin θ − 4.9t2 )j.
Setting y(t) = 98t sin θ − 4.9t2 =
t(98 sin θ − 4.9t) = 0, we see that the projectile hits the ground when t = 20 sin θ. Thus, using x(t) = 98t cos θ,
490 = x(t) = 98(20 sin θ) cos θ or sin 2θ = 0.5. Then 2θ = 30◦ or 150◦ . The angles of elevation are 15◦ and 75◦ .
√
√
s 2
s 2
15. Let s be the initial speed. Then v(0) = s cos 45◦ i + s sin 45◦ j =
i+
j. Using a(t) = −32j, we have
2
2
v(t) = a(t) dt = −32tj + c
√
√
s 2
s 2
i+
j = v(0) = c
2
2
√
√
s 2
s 2
v(t) =
i+
− 32t j
2
2
√
√
s 2
s 2
2
r(t) =
ti +
t − 16t j + b.
2
2
Since r(0) = 0, b = 0 and
√
√
s 2
s 2
2
r(t) =
ti +
t − 16t j.
2
2
√
√
Setting y(t) = s 2 t/2 − 16t2 = t(s 2/2 − 16t) = 0 we see that the ball hits the
√
Thus, using x(t) = s 2 t/2 and the fact that 100 yd = 300 ft, 300 = x(t) =
√
s = 9600 ≈ 97.98 ft/s.
√
ground when t = 2 s/32.
√
s 2 √
s2
( 2 s/32) =
and
2
32
16. Let s be the initial speed and θ the initial angle. Then v(0) = s cos θi + s sin θj. Using a(t) = −32j, we have
v(t) = a(t) dt = −32tj + c
s cos θi + s sin θj = v(0) = c
v(t) = s cos θi + (s sin θ − 32t)j
r(t) = st cos θi + (st sin θ − 16t2 )j + b.
Since r(0) = 0, b = 0 and r(t) = st cos θi + (st sin θ − 16t2 )j. Setting y(t) = st sin θ − 16t2 = t(s sin θ − 16t) = 0,
we see that the ball hits the ground when t = (s sin θ)/16. Using x(t) = st cos θi, we see that the range of the
ball is
s sin θ
s2 sin θ cos θ
s2 sin 2θ
x
=
=
.
16
16
32
446
9.2
Motion on a Curve
√
√
For θ = 30◦ , the range is s2 sin 60◦ /32 = 3 s2 /64 and for θ = 60◦ the range is s2 sin 120◦ /32 = 3 s2 /64. In
general, when the angle is 90◦ − θ the range is
[s2 sin 2(90◦ − θ)]/32 = s2 [sin(180◦ − 2θ)]/32 = s2 (sin 2θ)/32.
Thus, for angles θ and 90◦ − θ, the range is the same.
17. Let the initial speed of the projectile be s and let the target be at (x0 , y0 ). Then
vp (0) = s cos θi + s sin θj and vt (0) = 0. Using a(t) = −32j, we have
vp (t) = a(t) dt = −32tj + c
s cos θi + s sin θj = vp (0) = c
vp (t) = s cos θi + (s sin θ − 32t)j
rp (t) = st cos θi + (st sin θ − 16t2 )j + b.
Since rp (0) = 0, b = 0 and rp (t) = st cos θi+(st sin θ−16t2 )j. Also, vt (t) = −32tj+c and since vt (0) = 0, c = 0
and vt (t) = −32tj. Then rt (t) = −16t2 j + b. Since rt (0) = x0 i + y0 j, b = x0 i + y0 j and rt (t) = x0 i + (y0 − 16t2 )j.
Now, the horizontal component of rp (t) will be x0 when t = x0 /s cos θ at which time the vertical component of
rp (t) will be
(sx0 /s cos θ) sin θ − 16(x0 /s cos θ)2 = x0 tan θ − 16(x0 /s cos θ)2 = y0 − 16(x0 /s cos θ)2 .
Thus, rp (x0 /s cos θ) = rt (x0 /s cos θ) and the projectile will strike the target as it falls.
18. The initial angle is θ = 0, the initial height is 1024 ft, and the initial speed is s = 180(5280)/3600 = 264 ft/s.
Then x(t) = 264t and y(t) = −16t2 +1024. Solving y(t) = 0 we see that the pack hits the ground at t = 8 seconds
The horizontal distance travelled is x(8) = 2112 feet. From the figure in the text, tan α = 1024/2112 = 16/33
and α ≈ 0.45 radian or 25.87◦ .
19. r (t) = v(t) = −r0 ω sin ωti + r0 ω cos ωtj; v = v(t) =
r02 ω 2 sin2 ωt + r02 ω 2 cos2 ωt = r0 ω
ω = v/r0 ; a(t) = r (t) = −r0 ω 2 cos ωti − r0 ω 2 sin ωtj
a = a(t) =
r02 ω 4 cos2 ωt + r02 ω 4 sin2 ωt = r0 ω 2 = r0 (v/r0 )2 = v 2 /r0 .
20. (a) v(t) = −b sin ti + b cos tj + ck; v(t) = b2 sin2 t + b2 cos2 t + c2 =
t
t
ds
(b) s =
v(u) du
b2 + c2 du = t b2 + c2 ;
= b2 + c2
dt
0
0
√
b2 + c2
d2 s
= 0; a(t) = −b cos ti − b sin tj; a(t) = b2 cos2 t + b2 sin2 t = |b|. Thus, d2 s/dt2 = a(t).
dt2
21. By Problem 19, a = v 2 /r0 = 15302 /(4000 · 5280) ≈ 0.1108. We are given mg = 192, so m = 192/32 and
we = 192 − (192/32)(0.1108) ≈ 191.33 lb.
(c)
22. By Problem 19, the centripetal acceleration is v 2 /r0 . Then the horizontal force is
mv 2 /r0 . The vertical force is 32m. The resultant force is U = (mv 2 /r0 )i + 32mj.
From the figure, we see that tan φ = (mv 2 /r0 )/32m = v 2 /32r0 . Using r0 = 60 and
v = 44 we obtain tan φ = 442 /32(60) ≈ 1.0083 and φ ≈ 45.24◦ .
23. Solving x(t) = (v0 cos θ)t for t and substituting into y(t) = − 12 gt2 + (v0 sin θ)t + s0 we obtain
1
y=− g
2
x
v0 cos θ
2
+ (v0 sin θ)
x
g
+ s0 = − 2
x2 + (tan θ)x + s0 ,
v0 cos θ
2v0 cos2 θ
447
9.2
Motion on a Curve
which is the equation of a parabola.
24. Since the projectile is launched from ground level, s0 = 0. To find the maximum height we maximize
y(t) = − 12 gt2 + (v0 sin θ)t. Solving y (t) = −gt + v0 sin θ = 0, we see that t = (v0 /g) sin θ is a critical point.
Since y (t) = −g < 0,
H=y
v0 sin θ
g
v0 sin θ
1 v 2 sin2 θ
v 2 sin2 θ
+ v0 sin θ
=− g 0 2
= 0
2
g
g
2g
is the maximum height. To find the range we solve y(t) = − 12 gt2 + (v0 sin θ)t = t(v0 sin θ − 12 gt) = 0. The
positive solution of this equation is t = (2v0 sin θ)/g. The range is thus
x(t) = (v0 cos θ)
v 2 sin 2θ
2v0 sin θ
= 0
.
g
g
25. Letting r(t) = x(t)i + y(t)j + z(t)k, the equation dr/dt = v is equivalent to dx/dt = 6t2 x, dy/dt = −4ty 2 ,
dz/dt = 2t(z + 1). Separating variables and integrating, we obtain dx/x = 6t2 dt, dy/y 2 = −4t dt, dz/(z + 1) =
2t dt, and ln x = 2t3 + c1 , −1/y = −2t2 + c2 , ln(z + 1) = t2 + c3 . Thus,
3
r(t) = k1 e2t i +
2
1
j + (k3 et − 1)k.
2t2 + k2
26. We require the fact that dr/dt = v. Then
dL
d
dp dr
= (r × p) = r ×
+
× p = τ + v × p = τ + v × mv = τ + m(v × v) = τ + 0 = τ .
dt
dt
dt
dt
27. (a) Since F is directed along r we have F = cr for some constant c. Then
τ = r × F = r × (cr) = c(r × r) = 0.
(b) If τ = 0 then dL/dt = 0 and L is constant.
28. (a) Using Problem 27, F = −k(M m/r2 )u = ma. Then a = d2 r/dt = −k(M/r2 )u.
(b) Using u = r/r we have
M
r × r = r × −k 2 u
r
1
kM
kM
= − 2 r × ( r) = − 3 (r × r) = 0.
r
r
r
(c) From Theorem 9.4 (iv) we have
d
dv dr
(r × v) = r ×
+
× v = r × r + v × v = 0 + 0 = 0.
dt
dt
dt
(d) Since r = ru we have c = r × v = ru × ru = r2 (u × u ).
(e) Since u = (1/r)r is a unit vector, u · u = 1 and
d
du du
du
d
(u · u) = u ·
+
· u = 2u ·
= (1) = 0.
dt
dt
dt
dt
dt
Thus, u · u = 0.
(f )
d
dc dv
kM
kM
(v × c) = v ×
+
× c = v × 0 + a × c = − 2 u × c = − 2 u × [r2 (u × u )]
dt
dt
dt
r
r
= −kM [u × (u × u )] = −kM = −kM [(u · u )u − (u · u)u ]
by (10) of 7.4
= −kM [0 − u ] = kM u = kM
du
dt
448
9.3
Curvature and Components of Acceleration
(g) Since
r · (v × c) = (r × v) · c
= c · c = c2
and
by Problem 61 in 7.4
where c = c
(kM u + d) · r = (kM u + d) · ru = kM ru · u + rd · u
= kM r + rd cos θ
we have c2 = kM r + rd cos θ or r =
where d = d
c2
c2 /kM
=
.
kM + d cos θ
1 + (d/kM ) cos θ
(h) First note that c > 0 (otherwise there is no orbit) and d > 0 (since the orbit is not a circle). We recognize
the equation in (g) to be that of a conic section with eccentricity e = d/kM . Since the orbit of the planet
is closed it must be an ellipse.
(i) At perihelion c = c = r × v = r0 v0 sin(π/r) = r0 v0 . Since r is minimum at this point, we want the
denominator in the equation r0 = [c2 /kM ]/[1 + (d/kM ) cos θ] to be maximum. This occurs when θ = 0. In
this case
r2 v 2 /kM
r0 = 0 0
and
d = r0 v02 − kM.
1 + d/kM
EXERCISES 9.3
Curvature and Components of Acceleration
1. r (t) = −t sin ti + t cos tj + 2tk; |r (t)| =
sin t
cos t
2
T(t) = − √ i + √ j + √ k
5
5
5
t2 sin2 t + t2 cos2 t + 4t2 =
2. r (t) = et (− sin t + cos t)i + et (cos t + sin t)j +
√
√
5 t;
2 et k,
|r (t)| = [e2t (sin2 t − 2 sin t cos t + cos2 t) + e2t (cos2 t + 2 sin t cos t + sin2 t) + 2e2t ]1/2 =
√
2
1
1
T(t) = (− sin t + cos t)i + (cos t + sin t)j +
k
2
2
2
√
4e2t = 2et ;
√
3. We assume a > 0. r (t) = −a sin ti + a cos tj + ck; |r (t)| = a2 sin2 t + a2 cos2 t + c2 = a2 + c2 ;
a sin t
a cos t
c
dT
a sin t
a cos t
T(t) = − √
i+ √
j+ √
k;
i− √
j,
= −√
2
2
2
2
2
2
2
2
dt
a +c
a +c
a +c
a +c
a2 + c2
2
dT a cos2 t a2 sin2 t
a
+ 2
=√
; N = − cos ti − sin tj;
dt =
a2 + c2
a + c2
a2 + c2
i
j
k
a sin t
a cos t
c
= √c sin t i − √c cos t j + √ a
√
√
B = T × N = − √ 2
k;
2
2
2
2
2
a
+
c
a
+
c
a
+
c
a2 + c2
a2 + c2
a2 + c2
− cos t
− sin t
0
√
|dT/dt|
a/ a2 + c2
a
√
κ=
=
= 2
|r (t)|
a + c2
a2 + c2
449
9.3
Curvature and Components of Acceleration
4. r (t) = i + tj + t2 k, r (1) = i + j + k; |r (t)| =
√
1 + t2 + t4 , |r (1)| =
√
3;
1
T(t) = (1 + t2 + t4 )−1/2 (i + tj + t2 k), T(1) = √ (i + j + k);
3
dT
t
1
= − (1 + t2 + t4 )−3/2 (2t + 4t3 )i + [(1 + t2 + t4 )−1/2 − (1 + t2 + t4 )−3/2 (2t + 4t3 )]j
dt
2
2
t2
+ [2t(1 + t2 + t4 )−1/2 − (1 + t2 + t4 )−3/2 (2t + 4t3 )]k;
2
√
d
1 1
2
1
1
d
1
T(1) = − √ i + √ k, T(1) =
+ = √ ; N(1) = − √ (i − k);
dt
dt
3 3
3
3
3
2
√ √
√
i
j
k d
√
√
√ 1
2/ 3
2
B(1) = 1/ 3 1/ 3 1/ 3 = √ (i − 2j + k); κ = T(1) /|r (1)| = √
=
√
√ dt
3
6
3
−1/ 2
0
1/ 2 √
5. From Example 2 in the text, a normal to the osculating plane is B(π/4) = √126 (3i−3j+2 2 k). The point on the
√ √
√
√
√
curve when t = π/4 is ( 2 , 2 , 3π/4). An equation of the plane is 3(x − 2 ) − 3(y − 2 ) + 2 2(z − 3π/4) = 0,
√
√
√
√
3x − 3y + 2 2 z = 3 2 π/2, or 3 2 x − 3 2 y + 4z = 3π.
6. From Problem 4, a normal to the osculating plane is B(1) =
√1 (i − 2j + k).
6
The point on the curve when t = 1
is (1, 1/2, 1/3). An equation of the plane is (x − 1) − 2(y − 1/2) + (z − 1/3) = 0 or x − 2y + z = 1/3.
√
7. v(t) = j + 2tk, |v(t)| = 1 + 4t2 ; a(t) = 2k; v · a = 4t, v × a = 2i, |v × a| = 2;
4t
2
aT = √
, aN = √
1 + 4t2
1 + 4t2
8. v(t) = −3 sin ti + 2 cos tj + k,
|v(t)| =
9 sin2 t + 4 cos2 t + 1 = 5 sin2 t + 4 sin2 t + 4 cos2 t + 1 = 5
a(t) = −3 cos ti − 2 sin tj; v · a = 9 sin t cos t − 4 sin t cos t = 5 sin t cos t,
9.
10.
11.
12.
sin2 t + 1 ;
v × a = 2 sin ti − 3 cos tj + 6k, |v × a| = 4 sin2 t + 9 cos2 t + 36 = 5 cos2 t + 8 ;
√
5 sin t cos t
cos2 t + 8
aT =
, aN =
sin2 t + 1
sin2 t + 1
√
v(t) = 2ti + 2tj + 4tk, |v(t)| = 2 6 t, t > 0; a(t) = 2i + 2j + 4k; v · a = 24t, v × a = 0;
√
24t
aT = √ = 2 6 , aN = 0, t > 0
2 6t
√
v(t) = 2ti − 3t2 j + 4t3 k, |v(t)| = t 4 + 9t2 + 16t4 , t > 0; a(t) = 2i − 6tj + 12t2 k;
√
v · a = 4t + 18t3 + 48t5 ; v × a = −12t4 i − 16t3 j − 6t2 k, |v × a| = 2t2 36t4 + 64t2 + 9 ;
√
4 + 18t2 + 48t4
2t 36t4 + 64t2 + 9
aT = √
, aN = √
,t>0
4 + 9t2 + 16t4
4 + 9t2 + 16t4
√
v(t) = 2i + 2tj, |v(t)| = 2 1 + t2 ; a(t) = 2j; v · a = 4t; v × a = 4k, |v × a| = 4;
2t
2
, aN = √
aT = √
1 + t2
1 + t2
√
1 + t2
1
t
2t
1 − t2
v(t) =
i
+
j,
|v(t)|
=
; a(t) = −
i+
j;
2
2
2
2
2
1+t
1+t
1+t
(1 + t )
(1 + t2 )2
v·a=−
2t
t − t3
t
1
1
+
=−
; v×a=
k, |v × a| =
;
(1 + t2 )3
(1 + t2 )3
(1 + t2 )2
(1 + t2 )2
(1 + t2 )2
aT = − √
t/(1 + t2 )2
1/(1 + t2 )2
t
1
√
,
a
=
=−
=
N
2
3/2
2
2
2
2
(1 + t )
(1 + t2 )3/2
1 + t /(1 + t )
1 + t /(1 + t )
450
9.3
Curvature and Components of Acceleration
13. v(t) = −5 sin ti + 5 cos tj, |v(t)| = 5; a(t) = −5 cos ti − 5 sin tj; v · a = 0, v × a = 25k, |v × a| = 25;
aT = 0, aN = 5
sinh2 t + cosh2 t ; a(t) = cosh ti + sinh tj v · a = 2 sinh t cosh t;
2 sinh t cosh t
1
v × a = (sinh2 t − cosh2 t)k = −k, |v × a| = 1; aT =
, aN =
2
2
2
sinh t + cosh t
sinh t + cosh2 t
√
15. v(t) = −e−t (i + j + k), |v(t)| = 3 e−t ; a(t) = e−t (i + j + k); v · a = −3e−2t ; v × a = 0, |v × a| = 0;
√ −t
aT = − 3 e , aN = 0
√
16. v(t) = i + 2j + 4k, |v(t)| = 21 ; a(t) = 0; v · a = 0, v × a = 0, |v × a| = 0; aT = 0, aN = 0
14. v(t) = sinh ti + cosh tj, |v(t)| =
17. v(t) = −a sin ti + b cos tj + ck, |v(t)| =
a2 sin2 t + b2 cos2 t + c2 ; a(t) = −a cos ti − b sin tj;
v × a = bc sin ti − ac cos tj + abk, |v × a| =
κ=
|v × a|
=
|v|3
b2 c2 sin2 t + a2 c2 cos2 t + a2 b2
b2 c2 sin2 t + a2 c2 cos2 t + a2 b2
;
(a2 sin2 t + b2 cos2 t + c2 )3/2
18. (a) v(t) = −a sin ti + b cos tj, |v(t)| =
a2 sin2 t + b2 cos2 t ; a(t) = −a cos ti − b sin tj;
ab
v × a = abk; |v × a| = ab; κ = 2 2
(a sin t + b2 cos2 t)3/2
(b) When a = b, |v(t)| = a, |v × a| = a2 , and κ = a2 /a3 = 1/a.
19. The equation of a line is r(t) = b + tc, where b and c are constant vectors.
v(t) = c, |v(t)| = |c|; a(t) = 0; v × a = 0, |v × a| = 0; κ = |v × a|/|v|3 = 0
20. v(t) = a(1 − cos t)i + a sin tj; v(π) = 2ai, |v(π)| = 2a; a(t) = a sin ti + a cos tj, a(π) = −aj;
i
j k |v × a|
2a2
1
|v × a| = 2a 0 0 = −2a2 k; |v × a| = 2a2 ; κ =
=
=
3
3
|v|
8a
4a
0 −a 0 21. v(t) = f (t)i + g (t)j, |v(t)| =
[f (t)]2 + [g (t)]2 ; a(t) = f (t)i + g (t)j;
v × a = [f (t)g (t) − g (t)f (t)]k, |v × a| = |f (t)g (t) − g (t)f (t)|;
κ=
|v × a|
|f (t)g (t) − g (t)f (t)|
=
3
|v|
([f (t)]2 + [g (t)]2 )3/2
22. For y = F (x), r(x) = xi + F (x)j. We identify f (x) = x and g(x) = F (x) in Problem 21. Then f (x) = 1,
f (x) = 0, g (x) = F (x), g (x) = F (x), and κ = |F (x)|/(1 + [F (x)]2 )3/2 .
23. F (x) = x2 , F (0) = 0, F (1) = 1; F (x) = 2x, F (0) = 0, F (1) = 2; F (x) = 2, F (0) = 2, F (1) = 2;
2
2
1
2
κ(0) =
= 2; ρ(0) = ; κ(1) =
= √ ≈ 0.18;
2
3/2
2
3/2
2
(1 + 0 )
(1 + 2 )
5 5
√
√
5 5
ρ(1) =
≈ 5.59; Since 2 > 2/5 5 , the curve is “sharper” at (0, 0).
2
24. F (x) = x3 , F (−1) = −1, F (1/2) = 1/8; F (x) = 3x2 , F (−1) = 3, F (1/2) = 3/4; F (x) = 6x,
F (−1) = −6, F (1/2) = 3; κ(−1) =
| − 6|
6
3
= √ = √ ≈ 0.19;
(1 + 32 )3/2
10 10
5 10
√
5 10
3
1
3
192
1
125
ρ(−1) =
=
≈ 5.27; κ( ) =
=
≈ 1.54; ρ( ) =
≈ 0.65
2
3/2
3
2
125/64
125
2
192
[1 + (3/4) ]
Since 1.54 > 0.19, the curve is “sharper” at (1/2, 1/8).
25. At a point of inflection (x0 , F (x0 )), if F (x0 ) exists then F (x0 ) = 0. Thus, assuming that limx→x0 F (x)
exists, F (x) and hence κ is near 0 for x near x0 .
451
9.3
Curvature and Components of Acceleration
26. We use the fact that T · N = 0 and T · T = N · N = 1. Then
|a(t)|2 = a · a = (aN N + aT T) · (aN N + aT T) = a2N N · N + 2aN aT N · T + a2T T · T = a2N + a2T .
EXERCISES 9.4
Partial Derivatives
1. y = − 12 x + C
2. x = y 2 − c
3. x2 − y 2 = 1 + c2
4. 4x2 + 9y 2 = 36 − c2 , −6 ≤ c ≤ 6
5. y = x2 + ln c, c > 0
6. y = x + tan c, −π/x < c < π/2
7. x2 /9 + z 2 /4 = c; elliptical cylinder
8. x2 + y 2 + z 2 = c; sphere
9. x2 + 3y 2 + 6z 2 = c; ellipsoid
10. 4y − 2z + 1 = c; plane
11.
452
9.4
Partial Derivatives
12. Setting x = −4, y = 2, and z = −3 in x2 /16 + y 2 /4 + z 2 /9 = c we obtain c = 3. The equation of the surface is
√
x2 /16 + y 2 /4 + z 2 /9 = 3. Setting y = z = 0 we find the x-intercepts are ±4 3 . Similarly, the y-intercepts are
√
√
±2 3 and the z-intercepts are ±3 3 .
13. zx = 2x − y 2 ; zy = −2xy + 20y 4
14. zx = −3x2 + 12xy 3 ; zy = 18x2 y 2 + 10y
15. zx = 20x3 y 3 − 2xy 6 + 30x4 ; zy = 15x4 y 2 − 6x2 y 5 − 4
16. zx = 3x2 y 2 sec2 (x3 y 2 ); zy = 2x3 y sec2 (x3 y 2 )
√
2
24y x
17. zx = √
;
z
=
−
y
(3y 2 + 1)2
x (3y 2 + 1)
18. zx = 12x2 − 10x + 8; zy = 0
19. zx = −(x3 − y 2 )−2 (3x2 ) = −3x2 (x3 − y 2 )−2 ; zy = −(x3 − y 2 )−2 (−2y) = 2y(x3 − y 2 )−2
20. zx = 6(−x4 + 7y 2 + 3y)5 (−4x3 ) = −24x3 (−x4 + 7y 2 + 3y)5 ; zy = 6(−x4 + 7y 2 + 3y)5 (14y + 3)
21. zx = 2(cos 5x)(− sin 5x)(5) = −10 sin 5x cos 5x; zy = 2(sin 5y)(cos 5y)(5) = 10 sin 5y cos 5y
tan−1 y 2
22. zx = (2x tan−1 y 2 )ex
2
3
23. fx = x(3x2 yex
24. fθ = φ2 cos
y
3
+ ex
θ
φ
y
1
φ
; zy =
2x2 y x2 tan−1 y2
e
1 + y4
3
3
= (3x3 y + 1)ex y ; fy = x4 ex
= φ cos
θ
θ
; fφ = φ2 cos
φ
φ
y
−
θ
φ2
+ 2φ sin
θ
θ
θ
= −θ cos + 2φ sin
φ
φ
φ
25. fx =
(x + 2y)3 − (3x − y)
7y
(x + 2y)(−1) − (3x − y)(2)
−7x
=
; fy =
=
(x + 2y)2
(x + 2y)2
(x + 2y)2
(x + 2y)2
26. fx =
(x2 − y 2 )2 y − xy[2(x2 − y 2 )2x]
−3x2 y − y 3
=
;
2
2
4
(x − y )
(x2 − y 2 )3
fy =
(x2 − y 2 )2 x − xy[2(x2 − y 2 )(−2y)]
3xy 2 + x3
=
(x2 − y 2 )4
(x2 − y 2 )3
8u
15v 2
;
g
=
v
4u2 + 5v 3
4u2 + 5v 3
√
√
s
r
1
1
28. hr = √ + 2 ; hs = − 2 − √
r
s
2s r
2r s
27. gu =
√
y
29. wx = √ ; wy = 2 x − y
x
30. wx = xy
1
x
1 y/z
e
z
y y2
y
√
− ey/z = 2 x −
+ 1 ey/z ; wz = −yey/z − 2 = 2 ey/z
z
z
z
+ (ln xz)y = y + y ln xz; wy = x ln xz; wz =
xy
z
31. Fu = 2uw2 − v 3 − vwt2 sin(ut2 ); Fv = −3uv 2 + w cos(ut2 );
Fx = 4(2x2 t)3 (4xt) = 16xt(2x2 t)3 = 128x7 t4 ; Ft = −2uvwt sin(ut2 ) + 64x8 t3
s −1
32. Gp = r4 s5 (p2 q 3 )r
4 5
Gq = r4 s5 (p2 q 3 )r
4 5
Gs = (p2 q 3 )r
33.
4 5
s
s −1
(2pq 3 ) = 2pq 3 r4 s5 (p2 q 3 )r
s −1
4 5
(3p2 q 2 ) = 3p2 q 2 r4 s5 (p2 q 3 )r
;
s −1
4 5
; Gr = (p2 q 3 )r
4 5
s
(4r3 s5 ) ln(p2 q 3 );
(5r4 s4 ) ln(p2 q 3 )
∂z
∂2z
(x2 + y 2 )2 − 2x(2x)
2y 2 − 2x2
∂z
2x
2y
,
=
=
;
,
= 2
= 2
2
2
2
2
2
2
2
2
∂x
x +y
∂x
(x + y )
(x + y )
∂y
x + y2
∂2z
(x2 + y 2 )2 − 2y(2y)
2x2 − 2y 2
∂2z
∂2z
2y 2 − 2x2 + 2x2 − 2y 2
=
= 2
;
+ 2 =
=0
2
2
2
2
2
2
2
∂y
(x + y )
(x + y )
∂x
∂y
(x2 + y 2 )2
453
9.4
34.
Partial Derivatives
2
2
2
2
∂z
= ex −y (−2y sin 2xy) + 2xex −y cos 2xy
∂x
2
2
∂2z
= ex −y (−4y 2 cos 2xy − 8xy sin 2xy + 4x2 cos 2xy + 2 cos 2xy)
2
∂x
2
2
2
2
∂z
= ex −y (−2x sin 2xy) − 2yex −y cos 2xy
∂y
2
2
∂2z
= ex −y (−4x2 cos 2xy + 8xy sin 2xy + 4y 2 cos 2xy − 2 cos 2xy)
∂y 2
Adding the second partial derivatives gives
∂2z
∂2z
+
= [−4(y 2 + x2 ) cos 2xy + 4(x2 + y 2 ) cos 2xy] = 0.
∂x2
∂y 2
35.
36.
37.
∂u
∂u
∂2u
∂2u
=
−
cos
at
sin
x;
= −a2 cos at sin x;
= cos at cos x,
=
−a
sin
at
sin
x,
∂x
∂x2
∂t
∂t2
∂2u
∂2u
a2 2 = a2 (− cos at sin x) = 2
∂x
∂t
∂u
∂2u
= − cos(x + at) − sin(x − at);
= − sin(x + at) + cos(x − at),
∂x
∂x2
∂u
∂2u
= −a2 cos(x + at) − a2 sin(x − at);
= −a sin(x + at) − a cos(x − at),
∂t
∂t2
∂2u
∂2u
a2 2 = −a2 cos(x + at) − a2 sin(x − at) = 2
∂x
∂t
2
∂C
∂2C
4x2 −1/2 −x2 /kt
2 −1/2 −x2 /kt
2x
=
t
e
−
e
;
= − t−1/2 e−x /kt ,
t
2
2
2
∂x
kt
∂x
k t
kt
2
∂C
x2
t−3/2 −x2 /kt k ∂ 2 C
x2 −1/2 −x2 /kt t−1/2 −x2 /kt
∂C
;
=
t
e
−
=
= t−1/2 2 e−x /kt −
e
e
∂t
kt
2
4 ∂x2
kt2
2t
∂t
38. (a) Pv = −k(T /V 2 )
(b) P V = kt, P VT = k, VT = k/P
(c) P V = kT , V = kTp , Tp = V /k
2
2
2
2
2
2
39. zx = v 2 euv (3x2 ) + 2uveuv (1) = 3x2 v 2 euv + 2uveuv ; zy = v 2 euv (0) + 2uveuv (−2y) = −4yuveuv
40. zx = (2u cos 4v)(2xy 3 ) − (4u2 sin 4v)(3x2 ) = 4xy 3 u cos 4v − 12x2 u2 sin 4v
zy = (2u cos 4v)(3x2 y 2 ) − (4v 2 sin 4v)(3y 2 ) = 6x2 y 2 u cos 4v − 12y 2 u2 sin 4v
41. zu = 4(4u3 ) − 10y[2(2u − v)(2)] = 16u3 − 40(2u − v)y
zv = 4(−24v 2 ) − 10y[2(2u − v)(−1)] = −96v 2 + 20(2u − v)y
42. zu =
zv =
2y
(x + y)2
1
v
+
−2x
(x + y)2
u
−2x
2y
− 2 +
2
(x + y)
v
(x + y)2
−
v2
u2
=
2y
2xv 2
+ 2
2
v(x + y)
u (x + y)2
2v
u
=−
4xv
2yu
−
v 2 (x + y)2
u(x + y)2
3 2
3
(u + v 2 )1/2 (2u)(−e−t sin θ) + (u2 + v 2 )1/2 (2v)(−e−t cos θ)
2
2
2
2 1/2 −t
2
= −3u(u + v ) e sin θ − 3v(u + v 2 )1/2 e−t cos θ
3
3
wθ = (u2 + v 2 )1/2 (2u)e−t cos θ + (u2 + v 2 )1/2 (2v)(−e−t sin θ)
2
2
2
2 1/2 −t
= 3u(u + v ) e cos θ − 3v(u2 + v 2 )1/2 e−t sin θ
43. wt =
454
2
9.4
Partial Derivatives
√
√
u/2 uv
v/2 uv
rv
rs2 u
(2r) +
(2rs2 ) = √
+√
1 + uv
1 + uv
uv (1 + uv)
uv (1 + uv)
√
√
v/2 uv
−sv
r2 su
u/2 uv
ws =
+√
(−2s) +
(2r2 s) = √
1 + uv
1 + uv
uv (1 + uv)
uv (1 + uv)
44. wr =
45. Ru = s2 t4 (ev ) + 2rst4 (−2uve−u ) + 4rs2 t3 (2uv 2 eu
2
2
Rv = s2 t4 (2uvev ) + 2rst4 (e−u ) + 4rs2 t3 (2u2 veu
2
46. Qx =
Qt =
47. wt =
wr =
wu =
1
P
√
t2
1 − x2
2
+
1
1
(2t sin−1 x) +
p
q
1
q
−
2 2
2 2
v
v
) = s2 t4 ev − 4uvrst4 e−u + 8uv 2 rs2 t3 eu
2
2
) = 2s2 t4 uvev + 2rst4 e−u + 8rs2 t3 u2 veu
2
2
1
t2
+
1
r
1/t
1 + (x/t)2
t2
1
t
= √
+ 2+
2 + x2 )
2
qt
r(t
p 1−x
2x
t3
+
1
r
−x/t2
1 + (x/t)2
=
2
u
2x
+
x2 + y 2 rs + tu 2
x2 + y 2
2
s
2x
+
x2 + y 2 rs + tu 2
x2 + y 2
2
t
+
x2 + y 2 rs + tu 2
−t cosh rs
=
u2
x2 + y 2
2y
2y
2x
cosh rs
=
u
v
2 2
v
2t sin−1 x
x
2x
− 3−
p
qt
r(t2 + x2 )
xu
x2 + y 2 (rs + tu)
st sinh rs
=
u
2 2
+
xs
x2 + y 2 (rs + tu)
2y
y cosh rs
x2 + y 2
u
+
xt
x2 + y 2 (rs + tu)
yst sinh rs
x2 + y 2
u
−
yt cosh rs
u2
x2 + y 2
48. sφ = 2pe3θ + 2q[− sin(φ + θ)] − 2rθ2 + 4(2) = 2pe3θ − 2q sin(φ + θ) − 2rθ2 + 8
sθ = 2p(3φe3θ ) + 2q[− sin(φ + θ)] − 2r(2φθ) + 4(8) = 6pφe3θ − 2q sin(φ + θ) − 4rφθ + 32
49.
dz
2u
2v
4ut − 4vt−3
= 2
(2t) + 2
(−2t−3 ) =
2
2
dt
u +v
u +v
u2 + v 2
50.
dz
= (3u2 v − v 4 )(−5e−5t ) + (u3 − 4uv 3 )(5 sec 5t tan 5t) = −5(3u2 v − v 4 )e−5t + 5(u3 − 4uv 3 ) sec 5t tan 5t
dt
dw
dt
dw
dt
= −3 sin(3u + 4v)(2) − 4 sin(3u + 4v)(−1); u(π) = 5π/2, v(π) = −5π/4
= −6 sin 15π − 5π + 4 sin 15π − 5π = −2 sin 5π = −2
2
2
2
π
dw
−8
dw xy
52.
= yexy
+
xe
(3);
x(0)
=
4,
y(0)
=
5;
= 5e20 (−8) + 4e20 (3) = −28e20
dt
(2t + 1)2
dt 0
51.
53. With x = r cos θ and y = r sin θ
∂u ∂x ∂u ∂y
∂u
∂u
∂u
=
+
=
cos θ +
sin θ
∂r
∂x ∂r
∂y ∂r
∂x
∂y
∂2u
∂ 2 u ∂x
∂2u
∂ 2 u ∂y
∂2u
2
=
cos
θ
+
sin2 θ
cos
θ
+
sin
θ
=
∂r2
∂x2 ∂r
∂y 2 ∂r
∂x2
∂y 2
∂u
∂u ∂x ∂u ∂y
∂u
∂u
=
+
=
(−r sin θ) +
(r cos θ)
∂θ
∂x ∂θ
∂y ∂θ
∂x
∂y
∂2u
∂u
∂ 2 u ∂x
∂u
∂ 2 u ∂y
=
(−r cos θ) +
(−r sin θ) +
(−r sin θ) + 2
(r cos θ)
2
2
∂θ
∂x
∂x ∂θ
∂y
∂y ∂θ
= −r
∂u
∂2u
∂2u
∂u
cos θ + r2 2 sin2 θ − r
sin θ + r2 2 cos2 θ.
∂x
∂x
∂y
∂y
455
9.4
Partial Derivatives
Using
∂2u ∂2u
+ 2 = 0, we have
∂x2
∂y
∂ 2 u 1 ∂u
∂2u
∂2u
1
1 ∂2u
+
=
cos2 θ + 2 sin2 θ +
+ 2
2
2
2
∂r
r ∂r
r ∂θ
∂x
∂y
r
+
=
54.
−r
∂u
∂2u
∂2u
∂u
cos θ + r2 2 sin2 θ − r
sin θ + r2 2 cos2 θ
∂x
∂x
∂y
∂y
∂2u
∂2u
∂u
2
2
(cos
θ
+
sin
θ)
+
(sin2 θ + cos2 θ) +
2
2
∂x
∂y
∂x
+
=
1
r2
∂u
∂u
cos θ +
sin θ
∂x
∂y
∂u
∂y
1
1
cos θ − cos θ
r
r
1
1
sin θ − sin θ
r
r
∂2u ∂2u
+ 2 = 0.
∂x2
∂y
dP
0.08T (dV /dt)
3.6 dV
(V − 0.0427)(0.08)dT /dt
−
+ 3
=
dt
(V − 0.0427)2
(V − 0.0427)2
V dt
=
0.08
dT
+
V − 0.0427 dt
0.08T
3.6
−
V3
(V − 0.0427)2
dV
dt
55. Since dT /dT = 1 and ∂P/∂T = 0,
0 = FT =
∂F ∂P
∂F ∂V
∂F dT
∂V
∂F/∂T
1
+
+
=⇒
=−
=−
.
∂P ∂T
∂V ∂T
∂T dT
∂T
∂F/∂V
∂T /∂V
56. We are given dE/dt = 2 and dR/dt = −1. Then
when E = 60 and R = 50,
dI
∂I dE
∂I dR
1
E
=
+
=
(2) − 2 (−1),
dt
∂E dt
∂R dt
R
R
dI
1
2
60
3/5
8
=
=
+
+
=
amp/min.
dt
50 502
25
25
125
57. Since the height of the triangle is x sin θ, the area is given by A = 12 xy sin θ. Then
dA ∂A dx ∂A dy ∂A dθ
1
dx 1
dy 1
dθ
+
+
= y sin θ
+ x sin θ
+ xy cos θ
.
dt ∂x dt
∂y dt
∂θ dt
2
dt
2
dt
2
dt
When x = 10, y = 8, θ = π/6, dx/dt = 0.3, dy/dt = 0.5, and dθ/dt = 0.1,
dA
1
= (8)
dt
2
1
1
1
(0.3) + (10)
(0.5) + (10)(8)
2
2
2
√
√
= 0.6 + 1.25 + 2 3 = 1.85 + 2 3 ≈ 5.31 cm2 /s.
58.
1
2
dw
∂w dx ∂w dy ∂w dz
x dx/dt + y dy/dt + z dz/dt
=
=
+
+
=
dt
∂x dt
∂y dt
∂z dt
x2 + y 2 + z 2
=
−16 sin t cos t + 16 sin t cos t + 25t
25t
√
=√
2
16 + 25t
16 + 25t2
dw =
dt t=5π/2
125π/2
16 +
625π 2 /4
=√
125π
≈ 4.9743
64 + 625π 2
456
√ 3
(0.1)
2
−4x sin t + 4y cos t + 5z
16 cos2 t + 16 sin2 t + 25t2
and
9.5
Directional Derivative
EXERCISES 9.5
Directional Derivative
1. ∇f = (2x − 3x2 y 2 )i + (4y 3 − 2x3 y)j
2. ∇f = 4xye−2x y i + (1 + 2x2 e−2x y )j
2
2
y2
2xy
3xy 2
i
+
j
−
k
z3
z3
z4
4. ∇F = y cos yzi + (x cos yz − xyz sin yz)j − xy 2 sin yzk
3. ∇F =
5. ∇f = 2xi − 8yj; ∇f (2, 4) = 4i − 32j
6. ∇f =
3x2
2
x3 y
−
y4
i+
x3 − 4y 3
2
x3 y
−
y4
27
5
j; ∇f (3, 2) = √ i − √ j
38
2 38
7. ∇F = 2xz sin 4yi + 4x z cos 4yj + 2x2 z sin 4yk
√
√
4π
4π
4π
∇F (−2, π/3, 1) = −4 sin
i + 16 cos
j + 8 sin
k = 2 3 i − 8j − 4 3 k
3
3
3
2x
3
1
2y
2z
4
8. ∇F = 2
j+ k
i+ 2
j+ 2
k; ∇F (−4, 3, 5) = − i +
x + y2 + z2
x + y2 + z2
x + y2 + z2
25
25
5
√
√
f (x + h 3/2, y + h/2) − f (x, y)
(x + h 3/2)2 + (y + h/2)2 − x2 − y 2
9. Du f (x, y) = lim
= lim
h→0
h→0
h
h
√
2
2
√
√
h 3 x + 3h /4 + hy + h /4
= lim
= lim ( 3 x + 3h/4 + y + h/4) = 3 x + y
h→0
h→0
h
√
√
√
√
f (x + h 2/2, y + h 2/2) − f (x, y)
3x + 3h 2/2 − (y + h 2/2)2 − 3x + y 2
10. Du f (x, y) = lim
= lim
h→0
h→0
h
h
√
√
2
√
√
√
√
3h 2/2 − h 2 y − h /2
= lim
= lim (3 2/2 − 2 y − h/2) = 3 2/2 − 2 y
h→0
h→0
h
√
3
1
11. u =
i + j; ∇f = 15x2 y 6 i + 30x3 y 5 j; ∇f (−1, 1) = 15i − 30j;
2
2
√
15 3
15 √
Du f (−1, 1) =
− 15 =
( 3 − 2)
2
2
√
√
2
2
12. u =
i+
j; ∇f = (4 + y 2 )i + (2xy − 5)j; ∇f (3, −1) = 5i − 11j;
2
2
√
√
√
5 2 11 2
Du f (3, −1) =
−
= −3 2
2
2
√
√
10
x
1
3 10
−y
1
13. u =
i+ 2
j; ∇f (2, −2) = i + j
i−
j; ∇f = 2
10
10
x + y2
x + y2
4
4
√
√
√
10 3 10
10
Du f (2, −2) =
−
=−
40
40
20
2
2 2
6
8
3
4
y2
x2
i+
j = i + j; ∇f =
i
+
j; ∇f (2, −1) = i + 4j
10
10
5
5
(x + y)2
(x + y)2
3 16
19
Du f (2, −1) = +
=
5
5
5
14. u =
457
9.5
Directional Derivative
√
15. u = (2i + j)/ 5 ; ∇f = 2y(xy + 1)i + 2x(xy + 1)j; ∇f (3, 2) = 28i + 42j
2(28)
42
98
Du f (3, 2) = √ + √ = √
5
5
5
16. u = −i; ∇f = 2x tan yi + x2 sec2 yj; ∇f (1/2, π/3) =
√
√
3 i + j; Du f (1/2, π/3) = − 3
1
1
17. u = √ j + √ k; ∇F = 2xy 2 (2z + 1)2 i + 2x2 y(2z + 1)2 j + 4x2 y 2 (2z + 1)k
2
2
√
18
12
6
∇F (1, −1, 1) = 18i − 18j + 12k; Du F (1, −1, 1) = − √ + √ = − √ = −3 2
2
2
2
1
2y
2y 2 − 2x2
2
1
2x
18. u = √ i − √ j + √ k; ∇F = 2 i − 2 j +
k; ∇F (2, 4, −1) = 4i − 8j − 24k
z
z
z3
6
6
6
√
4
16
24
Du F (2, 4, −1) = √ − √ − √ = −6 6
6
6
6
19. u = −k; ∇F =
xy
x2 y + 2y 2 z
i+
x2 + 4z
2
x2 y + 2y 2 z
j+
y2
x2 y + 2y 2 z
k
∇F (−2, 2, 1) = −i + j + k; Du F (−2, 2, 1) = −1
√
2
1
2
20. u = −(4i − 4j + 2k)/ 36 = − i + j − k; ∇F = 2i − 2yj + 2zk; ∇F (4, −4, 2) = 2i + 8j + 4k
3
3
3
8
4 16 4
Du F (4, −4, 2) = − +
− =
3
3
3
3
√
12
16 4
21. u = (−4i − j)/ 17 ; ∇f = 2(x − y)i − 2(x − y)j; ∇f (4, 2) = 4i − 4j; Du F (4, 2) = − √ √ = − √
17 17
17
√
22. u = (−2i + 5j)/ 29 ; ∇f = (3x2 − 5y)i − (5x − 2y)j; ∇f (1, 1) = −2i − 3j;
4
15
11
Du f (1, 1) = √ − √ = − √
29
29
29
√
√
2
2x
2x
23. ∇f = 2e sin yi + e cos yj; ∇f (0, π/4) = 2 i +
j
2
√ 2
√
√
√
The maximum Du is [( 2 ) + ( 2/2)2 ]1/2 = 5/2 in the direction 2 i + ( 2/2)j.
24. ∇f = (xyex−y + yex−y )i + (−xyex−y + xex−y )j; ∇f (5, 5) = 30i − 20j
√
The maximum Du is [302 + (−20)2 ]1/2 = 10 13 in the direction 30i − 20j.
25. ∇F = (2x + 4z)i + 2z 2 j + (4x + 4yz)k; ∇F (1, 2, −1) = −2i + 2j − 4k
√
The maximum Du is [(−2)2 + 22 + (−4)2 ]1/2 = 2 6 in the direction −2i + 2j − 4k.
26. ∇F = yzi + xzj + xyk; ∇F (3, 1, −5) = −5i − 15j + 3k
√
The maximum Du is [(−5)2 + (−15)2 + 32 ]1/2 = 259 in the direction −5i − 15j + 3k.
27. ∇f = 2x sec2 (x2 + y 2 )i + 2y sec2 (x2 + y 2 )j;
∇f ( π/6 ,
π/6 ) = 2 π/6 sec2 (π/3)(i + j) = 8 π/6 (i + j)
The minimum Du is −8
π/6 (12 + 12 )1/2 = −8
π/3 in the direction −(i + j).
28. ∇f = 3x2 i − 3y 2 j; ∇f (2, −2) = 12i − 12j = 12(i − j)
√
The minimum Du is −12[12 + (−1)2 ]1/2 = −12 2 in the direction −(i − j) = −i + j.
√ y
√
√
ze
x
2
3
y
29. ∇F = √ i + xz e j + √ k; ∇F (16, 0, 9) = i + 12j + k. The minimum Du is
8
3
2 x
2 z
√
3
2
−[(3/8)2 + 122 + (2/3)2 ]1/2 = − 83,281/24 in the direction − i − 12j − k.
8
3
458
9.5
Directional Derivative
1
1
1
i + j − k; ∇F (1/2, 1/6, 1/3) = 2i + 6j − 3k
x
y
z
The minimum Du is −[22 + 62 + (−3)2 ]1/2 = −7 in the direction −2i − 6j + 3k.
30. ∇F =
31. Using implicit differentiation on 2x2 + y 2 = 9 we find y = −2x/y. At (2, 1) the slope of the tangent line is
√
−2(2)/1 = −4. Thus, u = ±(i − 4j)/ 17 . Now, ∇f = i + 2yj and ∇f (3, 4) = i + 8j. Thus,
√
√
√
Du = ±(1/ 17 − 32 17 ) = ±31/ 17 .
2x + y − 1 x + 2y
3x + 3y − 1
√
√
32. ∇f = (2x + y − 1)i + (x + 2y)j; Du f (x, y) =
+ √
=
2
2
2
√
Solving (3x + 3y − 1)/ 2 = 0 we see that Du is 0 for all points on the line 3x + 3y = 1.
33. (a) Vectors perpendicular to 4i + 3j are ±(3i − 4j). Take u = ±
3
4
i− j .
5
5
√
4
3
(b) u = (4i + 3j)/ 16 + 9 = i + j
5
5
4
3
(c) u = − i − j
5
5
34. D−u f (a, b) = ∇f (a, b) · (−u) = −∇f (a, b) · u = −Du f (a, b) = −6
35. (a) ∇f = (3x2 − 6xy 2 )i + (−6x2 y + 3y 2 )j
Du f (x, y) =
9x2 − 18xy 2 − 6x2 y + 3y 2
3(3x2 − 6xy 2 ) −6x2 y + 3y 2
√
√
√
+
=
10
10
10
3
3
(b) F (x, y) = √ (3x2 − 6xy 2 − 2x2 y + y 2 ); ∇F = √ [(6x − 6y 2 − 4xy)i + (−12xy − 2x2 + 2y)j]
10
10
3
1
3
3
√
√
Du F (x, y) = √
(6x − 6y 2 − 4xy) + √
(−12xy − 2x2 + 2y)
10
10
10
10
9
3
1
2
2
= (3x − 3y − 2xy) + (−6xy − x + y) = (27x − 27y 2 − 36xy − 3x2 + 3y)
5
5
5
Gmx
Gmy
Gm
36. ∇U = 2
i+ 2
j= 2
(xi + yj)
(x + y 2 )3/2
(x + y 2 )3/2
(x + y 2 )3/2
The maximum and minimum values of Du U (x, y) are obtained when u is in the directions ∇U and −∇U ,
respectively. Thus, at a point (x, y), not (0, 0), the directions of maximum and minimum increase in U are
xi + yj and −xi − yj, respectively. A vector at (x, y) in the direction ±(xi + yj) lies on a line through the origin.
37. ∇f = (3x2 − 12)i + (2y − 10)j. Setting |∇f | = [(3x2 − 12)2 + (2y − 10)2 ]1/2 = 0, we obtain 3x2 − 12 = 0 and
2y − 10 = 0. The points where |∇f | = 0 are (2, 5) and (−2, 5).
38. Let ∇f (a, b) = αi + βj. Then
Du f (a, b) = ∇f (a, b) · u =
5
12
α− β =7
13
13
and Dv f (a, b) = ∇f (a, b) · v =
5
12
α − β = 3.
13
13
Solving for α and β, we obtain α = 13 and β = −13/6. Thus, ∇f (a, b) = 13i − (13/6)j.
39. ∇T = 4xi + 2yj; ∇T (4, 2) = 16i + 4j. The minimum change in temperature (that is, the maximum decrease
in temperature) is in the direction −∇T (4, 3) = −16i − 4j.
40. Let x(t)i + y(t)j be the vector equation of the path. At (x, y) on this curve, the direction of a tangent vector is
x (t)i + y (t)j. Since we want the direction of motion to be −∇T (x, y), we have x (t)i + y (t)j = −∇T (x, y) =
4xi + 2yj. Separating variables in dx/dt = 4x, we obtain dx/x = 4 dt, ln x = 4t + c1 , and x = C1 e4t . Separating
variables in dy/dt = 2y, we obtain dy/y = 2 dt, ln y = 2t + c2 , and y = C2 e2t . Since x(0) = 4 and y(0) = 2, we
459
9.5
Directional Derivative
have x = 4e4t and y = 2e2t . The equation of the path is 4e4t i + 2e2t j for t ≥ 0, or eliminating the parameter,
x = y 2 , y ≥ 0.
41. Let x(t)i + y(t)j be the vector equation of the path. At (x, y) on this curve, the direction of a tangent vector
is x (t)i + y (t)j. Since we want the direction of motion to be ∇T (x, y), we have x (t)i + y (t)j = ∇T (x, y) =
−4xi − 2yj. Separating variables in dx/dt = −4x we obtain dx/x = −4 dt, ln x = −4t + c1 and x = C1 e−4t .
Separating variables in dy/dt = −2y we obtain dy/y = −2 dt, ln y = −2t + c2 and y = C2 e−2t . Since x(0) = 3
and y(0) = 4, we have x = 3e−4t and y = 4e−2t . The equation of the path is 3e−4t i + 4e−2t j, or eliminating the
parameter, 16x = 3y 2 , y ≥ 0.
42. Substituting x = 0, y = 0, z = 1, and T = 500 into T = k/(x2 + y 2 + z 2 ) we see that k = 500 and
T (x, y, z) = 500/(x2 + y 2 + z 2 ) . To find the rate of change of T at 2, 3, 3 in the direction of 3, 1, 1 we first
compute 3, 1, 1 − 2, 3, 3 = 1, −2, −2. Then u = 13 1, −2, −2 = 13 i − 23 j − 23 k. Now
∇T = −
(x2
1000x
1000y
1000z
500
750
750
i− 2
j− 2
k and ∇T (2, 3, 3) = −
i−
j−
k,
2
2
2
2
2
2
2
2
2
+y +z )
(x + y + z )
(x + y + z )
121
121
121
so
Du T (2, 3, 3) =
1
3
−
500
121
−
2
3
−
750
121
−
2
3
−
750
121
750
750
The direction of maximum increase is ∇T (2, 3, 3) = − 500
121 i− 121 j− 121 k =
√
√
250
rate of change of T is |∇T (2, 3, 3)| = 250
22.
121 4 + 9 + 9 = 121
=
2500
.
363
250
121 (−2i−3j−3k),
and the maximum
43. Since ∇f = fx (x, y)i + fy (x, y)j, we have ∂f /∂x = 3x2 + y 3 + yexy . Integrating, we obtain f (x, y) = x3 + xy 3 +
exy + g(y). Then fy = 3xy 2 + xexy + g (y) = −2y 2 + 3xy 2 + xexy . Thus, g (y) = −2y 2 , g(y) = − 23 y 3 + c, and
f (x, y) = x3 + xy 3 + exy − 23 y 3 + C.
44. Let u = u1 i + u2 j and v = v1 i + v2 j. Dv f = (fx i + fy j) · v = v1 fx + v2 fy
∂
∂
(v1 fx + v2 fy )i +
(v1 fx + v2 fy )j · u = [(v1 fxx + v2 fyx )i + (v1 fxy + v2 fyy )j] · u
Du Dv f =
∂x
∂y
= u1 v1 fxx + u1 v2 fyx + u2 v1 fxy + u2 v2 fyy
Du f = (fx i + fy j) · u = u1 fx + u2 fy
∂
∂
Dv Du f =
(u1 fx + u2 fy )i +
(u1 fx + u2 fy )j · v = [(u1 fxx + u2 fyx )i + (u1 fxy + u2 fyy )j] · v
∂x
∂y
= u1 v1 fxx + u2 v1 fyx + u1 v2 fxy + u2 v2 fyy
Since the second partial derivatives are continuous, fxy = fyx and Du Dv f = Dv Du f . [Note that this result is
a generalization of fxy = fyx since Di Dj f = fyx and Dj Di f = fxy .]
∂
∂
(cf )i +
(cf )j = cfx i + cfy j = c(fx i + fy j) = c∇f
∂x
∂y
46. ∇(f + g) = (fx + gx )i + (fy + gy )j = (fx i + fy j) + (gx i + gy j) = ∇f + ∇g
45. ∇(cf ) =
47. ∇(f g) = (f gx + fx g)i + (f gy + fy g)j = f (gx i + gy j) + g(fx i + fy j) = f ∇g + g∇f
48. ∇(f /g) = [(gfx − f gx )/g 2 ]i + [(gfy − f gy )/g 2 ]j = g(fx i + fy j)/g 2 − f (gx i + gy j)/g 2
= g∇f /g 2 − f ∇g/g 2 = (g∇f − f ∇g)/g 2
i
j
k ∂f3
∂f2
−
i+
49. ∇ × F = ∂/∂x ∂/∂y ∂/∂z =
∂y
∂z
f1
f2
f3
∂f3
∂f1
−
∂z
∂x
460
j+
∂f1
∂f2
−
∂x
∂y
k
9.6 Tangent Planes and Normal Lines
EXERCISES 9.6
Tangent Planes and Normal Lines
1. Since f (6, 1) = 4, the level curve is x − 2y = 4.
∇f = i − 2j; ∇f (6, 1) = i − 2j
2. Since f (1, 3) = 5, the level curve is y + 2x = 5x or y = 3x, x = 0.
y
1
∇f = − 2 i + j; ∇f (1, 3) = −3i + j
x
x
3. Since f (2, 5) = 1, the level curve is y = x2 + 1.
∇f = −2xi + j; ∇f (2, 5) = −10i + j
4. Since f (−1, 3) = 10, the level curve is x2 + y 2 = 10.
∇f = 2xi + 2yj; ∇f (−1, 3) = −2i + 6j
5. Since f (−2, −3) = 2, the level curve is x2 /4 + y 2 /9 = 2
2y
2
x
or x2 /8 + y 2 /18 = 1. ∇f = i +
j; ∇f (−2, −3) = −i − j
2
9
3
6. Since f (2, 2) = 2, the level curve is y 2 = 2x, x = 0. ∇f = −
∇f (2, 2) = −i + 2j
y2
2y
i + j;
x2
x
7. Since f (1, 1) = −1, the level curve is (x − 1)2 − y 2 = −1 or y 2 − (x − 1)2 = 1.
∇f = 2(x − 1)i − 2yj; ∇f (1, 1) = −2j
8. Since f (π/6, 3/2) = 1, the level curve is y − 1 = sin x or y = 1 + sin x, sin x = 0.
√
−(y − 1) cos x
1
∇f =
j; ∇f (π/6, 3/2) = − 3 i + 2j
i+
2
sin x
sin x
9. Since F (3, 1, 1) = 2, the level surface is y + z = 2. ∇F = j + k;
∇F (3, 1, 1) = j + k
461
9.6
Tangent Planes and Normal Lines
10. Since F (1, 1, 3) = −1, the level surface is x2 + y 2 − z = −1 or z = 1 + x2 + y 2 .
∇F = 2xi + 2yj − k; ∇F (1, 1, 3) = 2i + 2j − k
11. Since F (3, 4, 0) = 5, the level surface is x2 + y 2 + z 2 = 25.
x
y
z
∇F =
i+
j+
k;
2
2
2
2
2
2
2
x +y +z
x +y +z
x + y2 + z2
3
4
∇F (3, 4, 0) = i + j
5
5
12. Since F (0, −1, 1) = 0, the level surface is x2 − y 2 + z = 0 or z = y 2 − x2 .
∇F = 2xi − 2yj + k; ∇F (0, −1, 1) = 2j + k
13. F (x, y, z) = x2 + y 2 − z; ∇F = 2xi + 2yj − k. We want ∇F = c(4i + j + 12 k) or 2x = 4c, 2y = c, −1 = c/2.
From the third equation c = −2. Thus, x = −4 and y = −1. Since z = x2 + y 2 = 16 + 1 = 17, the point on the
surface is (−4, −1, −17).
14. F (x, y, z) = x3 + y 2 + z; ∇F = 3x2 i + 2yj + k. We want ∇F = c(27i + 8j + k) or 3x2 = 27c, 2y = 8c, 1 = c.
From c = 1 we obtain x = ±3 and y = 4. Since z = 15 − x3 − y 2 = 15 − (±3)3 − 16 = −1 ∓ 27, the points on
the surface are (3, 4, −28) and (−3, 4, 26).
15. F (x, y, z) = x2 + y 2 + z 2 ; ∇F = 2xi + 2yj + 2zk. ∇F (−2, 2, 1) = −4i + 4j + 2k. The equation of the tangent
plane is −4(x + 2) + 4(y − 2) + 2(z − 1) = 0 or −2x + 2y + z = 9.
16. F (x, y, z) = 5x2 − y 2 + 4z 2 ; ∇F = 10xi − 2yj + 8zk; ∇F (2, 4, 1) = 20i − 8j + 8k. The equation of the tangent
plane is 20(x − 2) − 8(y − 4) + 8(z − 1) = 0 or 5x − 2y + 2z = 4.
17. F (x, y, z) = x2 − y 2 − 3z 2 ; ∇F = 2xi − 2yj − 6zk; ∇F (6, 2, 3) = 12i − 4j − 18k. The equation of the tangent
plane is 12(x − 6) − 4(y − 2) − 18(z − 3) = 0 or 6x − 2y − 9z = 5.
18. F (x, y, z) = xy + yz + zx; ∇F = (y + z)i + (x + z)j + (y + x)k; ∇F (1, −3, −5) = −8i − 4j − 2k. The equation
of the tangent plane is −8(x − 1) − 4(y + 3) − 2(z + 5) = 0 or 4x + 2y + z = −7.
19. F (x, y, z) = x2 + y 2 + z; ∇F = 2xi + 2yj + k; ∇F (3, −4, 0) = 6i − 8j + k. The equation of the tangent plane
is 6(x − 3) − 8(y + 4) + z = 0 or 6x − 8y + z = 50.
20. F (x, y, z) = xz; ∇F = zi+xk; ∇F (2, 0, 3) = 3i+2k. The equation of the tangent plane is 3(x−2)+2(z −3) = 0
or 3x + 2z = 12.
√
√
√
2
21. F (x, y, z) = cos(2x + y) − z; ∇F = −2 sin(2x + y)i − sin(2x + y)j − k; ∇F (π/2, π/4, −1/ 2 ) = 2 i +
j − k.
2
√ √
2
π
π
1
The equation of the tangent plane is 2 x −
= 0,
+
y−
− z+√
2
2
4
2
√
π 1
5π
π √
2 x−
= 0, or 2x + y − 2 z =
+ y−
− 2 z+√
+ 1.
2
4
4
2
22. F (x, y, z) = x2 y 3 + 6z; ∇F = 2xy 3 i + 3x2 y 2 j + 6k; ∇F (2, 1, 1) = 4i + 12j + 6k. The equation of the tangent
plane is 4(x − 2) + 12(y − 1) + 6(z − 1) = 0 or 2x + 6y + 3z = 13.
462
9.6 Tangent Planes and Normal Lines
√
√
√
√
2x
2y
i+ 2
j − k; ∇F (1/ 2 , 1/ 2 , 0) = 2 i + 2 j − k.
2
2
+y
x +y
√
√
1
1
The equation of the tangent plane is 2 x − √
+ 2 y− √
− (z − 0) = 0,
2
2
√
√
√
1
1
2 x− √
+2 y− √
− 2 z = 0, or 2x + 2y − 2 z = 2 2 .
2
2
√
24. F (x, y, z) = 8e−2y sin 4x − z; ∇F = 32e−2y cos 4xi − 16e−2y sin 4xj − k; ∇F (π/24, 0, 4) = 16 3 i − 8j − k.
The equation of the tangent plane is
√
√
√
2 3π
16 3(x − π/24) − 8(y − 0) − (z − 4) = 0 or 16 3 x − 8y − z =
− 4.
3
23. F (x, y, z) = ln(x2 + y 2 ) − z; ∇F =
x2
25. The gradient of F (x, y, z) = x2 + y 2 + z 2 is ∇F = 2xi + 2yj + 2zk, so the normal vector to the surface at
(x0 , y0 , z0 ) is 2x0 i + 2y0 j + 2z0 k. A normal vector to the plane 2x + 4y + 6z = 1 is 2i + 4j + 6k. Since we want
the tangent plane to be parallel to the given plane, we find c so that 2x0 = 2c, 2y0 = 4c, 2z0 = 6c or x0 = c,
√
y0 = 2c, z0 = 3c. Now, (x0 , y0 , z0 ) is on the surface, so c2 + (2c)2 + (3c)2 = 14c2 = 7 and c = ±1/ 2 . Thus,
√
√
√
√
√
√
the points on the surface are ( 2/2, 2 , 3 2/2) and − 2/2, − 2 , −3 2/2).
26. The gradient of F (x, y, z) = x2 − 2y 2 − 3z 2 is ∇F (x, y, z) = 2xi − 4yj − 6zk, so a normal vector to the surface at
(x0 , y0 , z0 ) is ∇F (x0 , y0 , z0 ) = 2x0 i−4y0 j−6z0 k. A normal vector to the plane 8x+4y+6z = 5 is 8i+4j+6k. Since
we want the tangent plane to be parallel to the given plane, we find c so that 2x0 = 8c, −4y0 = 4c, −6z0 = 6c
or x0 = 4c, y0 = −c, z0 = −c. Now, (x0 , y0 , z0 ) is on the surface, so (4c)2 − 2(−c)2 − 3(−c)2 = 11c2 = 33 and
√
√
√
√
√ √ √
c = ± 3 . Thus, the points on the surface are (4 3 , − 3 , − 3) and (−4 3 , 3 , 3 ).
27. The gradient of F (x, y, z) = x2 + 4x + y 2 + z 2 − 2z is ∇F = (2x + 4)i + 2yj + (2z − 2)k, so a normal to the
surface at (x0 , y0 , z0 ) is (2x0 + 4)i + 2y0 j + (2z0 − 2)k. A horizontal plane has normal ck for c = 0. Thus, we
want 2x0 + 4 = 0, 2y0 = 0, 2z0 − 2 = c or x0 = −2, y0 = 0, z0 = c + 1. since (x0 , y0 , z0 ) is on the surface,
(−2)2 + 4(−2) + (c + 1)2 − 2(c + 1) = c2 − 5 = 11 and c = ±4. The points on the surface are (−2, 0, 5) and
(−2, 0, −3).
28. The gradient of F (x, y, z) = x2 + 3y 2 + 4z 2 − 2xy is ∇F = (2x − 2y)i + (6y − 2x)j + 8zk, so a normal to the
surface at (x0 , y0 , z0 ) is 2(x0 − y0 )i + 2(3y0 − x0 )j + 8z0 k.
(a) A normal to the xz plane is cj for c = 0. Thus, we want 2(x0 − y0 ) = 0, 2(3y0 − x0 ) = c, 8z0 = 0 or x0 = y0 ,
3y0 − x0 = c/2, z0 = 0. Solving the first two equations, we obtain x0 = y0 = c/4. Since (x0 , y0 , z0 ) is on
√
the surface, (c/4)2 + 3(c/4)2 + 4(0)2 − 2(c/4)(c/4) = 2c2 /16 = 16 and c = ±16/ 2 . Thus, the points on
√
√
√
√
the surface are (4/ 2 , 4/ 2 , 0) and (−4/ 2 , −4/ 2 , 0).
(b) A normal to the yz-plane is ci for c = 0. Thus, we want 2(x0 − y0 ) = c, 2(3y0 − x0 ) = 0, 8z0 = 0 or x0 − y0 =
c/2, x0 = 3y0 , z0 = 0. Solving the first two equations, we obtain x0 = 3c/4 and y0 = c/4. Since (x0 , y0 , z0 )
√
is on the surface, (3c/4)2 + 3(c/4)2 + 4(0)2 − 2(3c/4)(c/4) = 6c2 /16 = 16 and c = ±16 6 . Thus, the points
√
√
√
√
√
√
on the surface are (12/ 6 , 4/ 6 , 0) on the surface are (12/ 6 , 4/ 6 , 0) and (−12/ 6 , −4/ 6 , 0).
(c) A normal to the xy-plane is ck for c = 0. Thus, we want 2(x0 − y0 ) = 0, 2(3y0 − x0 ) = 0, 8z0 = c or
x0 = y0 , 3y0 − x0 = 0, z0 = c/8. Solving the first two equations, we obtain x0 = y0 = 0. Since (x0 , y0 , z0 )
is on the surface, 02 + 3(0)2 + 4(c/8)2 − 2(0)(0) = c2 /16 = 16 and c = ±16. Thus, the points on the surface
are (0, 0, 2) and (0, 0, −2).
29. If (x0 , y0 , z0 ) is on x2 /a2 + y 2 /b2 + z 2 /c2 = 1, then x20 /a2 + y02 /b2 + z02 /c2 = 1 and (x0 , y0 , z0 ) is on the plane
xx0 /a2 + yy0 /b2 + zz0 /c2 = 1. A normal to the surface at (x0 , y0 , z0 ) is
∇F (x0 , y0 , z0 ) = (2x0 /a2 )i + (2y0 /b2 )j + (2z0 /c2 )k.
463
9.6
Tangent Planes and Normal Lines
A normal to the plane is (x0 /a2 )i + (y0 /b2 )j + (z0 /c2 )k. Since the normal to the surface is a multiple of the
normal to the plane, the normal vectors are parallel and the plane is tangent to the surface.
30. If (x0 , y0 , z0 ) is on x2 /a2 − y 2 /b2 + z 2 /c2 = 1, then x20 /a2 − y02 /b2 + z02 /c2 = 1 and (x0 , y0 , z0 ) is on the plane
xx0 /a2 − yy0 /b2 + zz0 /c2 = 1. A normal to the surface at (x0 , y0 , z0 ) is
∇F (x0 , y0 , z0 ) = (2x0 /a2 )i − (2y0 /b2 )j + (2z0 /c2 )k.
A normal to the plane is (x0 /a2 )i − (y0 /b2 )j + (z0 /c2 )k. Since the normal to the surface is a multiple of the
normal to the plane, the normal vectors are parallel, and the plane is tangent to the surface.
31. Let F (x, y, z) = x2 + y 2 − z 2 . Then ∇F = 2xi + 2yj − 2zk and a normal to the surface at (x0 , y0 , z0 ) is
x0 i + y0 j − z0 k. An equation of the tangent plane at (x0 , y0 , z0 ) is x0 (x − x0 ) + y0 (y − y0 ) − z0 (z − z0 ) = 0 or
x0 x + y0 y − z0 z = x20 + y02 − z02 . Since (x0 , y0 , z0 ) is on the surface, z02 = x20 + y02 and x20 + y02 − z02 = 0. Thus,
the equation of the tangent plane is x0 x + y0 y − z0 z = 0, which passes through the origin.
√
√
1
1
1
√
32. Let F (x, y, z) = x + y + z . Then ∇F = √ i + √ j + √ k and a normal to the surface at (x0 , y0 , z0 )
2 y
2 x
2 z
1
1
1
is √ i + √ j + √ k. An equation of the tangent plane at (x0 , y0 , z0 ) is
2 x0
2 y0
2 z0
1
1
1
√ (x − x0 ) + √ (y − y0 ) + √ (z − z0 ) = 0
2 x0
2 y0
2 z0
or
√
√
√
1
1
1
√
√ x + √ y + √ z = x0 + y0 + z0 = a .
x0
y0
z0
√ √
√ √
√ √
√
√ √
√ √
√
The sum of the intercepts is x0 a + y0 a + z0 a = ( x0 + y0 + z0 ) a = a · a = a.
33. F (x, y, z) = x2 + 2y 2 + z 2 ; ∇F = 2xi + 4yj + 2zk; ∇F (1, −1, 1) = 2i − 4j + 2k. Parametric equations of the
line are x = 1 + 2t, y = −1 − 4t, z = 1 + 2t.
34. F (x, y, z) = 2x2 − 4y 2 − z; ∇F = 4xi − 8yj − k; ∇F (3, −2, 2) = 12i + 16j − k. Parametric equations of the
line are x = 3 + 12t, y = −2 + 16t, z = 2 − t.
35. F (x, y, z) = 4x2 + 9y 2 − z; ∇F = 8xi + 18yj − k; ∇F (1/2, 1/3, 3) = 4i + 6j − k. Symmetric equations of the
x − 1/2
y − 1/3
z−3
line are
=
=
.
4
6
−1
36. F (x, y, z) = x2 + y 2 − z 2 ; ∇F = 2xi + 2yj − 2zk; ∇F (3, 4, 5) = 6i + 8j − 10k. Symmetric equations of the line
x−3
y−4
z−5
are
=
=
.
6
8
−10
37. A normal to the surface at (x0 , y0 , z0 ) is ∇F (x0 , y0 , z0 ) = 2x0 i+2y0 j+2z0 k. Parametric equations of the normal
line are x = x0 + 2x0 t, y = y0 + 2y0 t, z = z0 + 2z0 t. Letting t = −1/2, we see that the normal line passes
through the origin.
38. The normal lines to F (x, y, z) = 0 and G(x, y, z) = 0 are Fx i + Fy j + Fz k and Gx i + Gy j + Gz k, respectively.
These vectors are orthogonal if and only if their dot product is 0. Thus, the surfaces are orthogonal at P if and
only if Fx Gx + Fy Gy + Fz Gz = 0.
39. Let F (x, y, z) = x2 + y 2 + z 2 − 25 and G(x, y, z) = −x2 + y 2 + z 2 . Then
Fx Gx + Fy Gy + Fz gz = (2x)(−2x) + (2y)(2y) + (2z)(2z) = 4(−x2 + y 2 + z 2 ).
For (x, y, z) on both surfaces, F (x, y, z) = G(x, y, z) = 0. Thus, Fx Gx + Fy Gy + Fz Gz = 4(0) = 0 and the
surfaces are orthogonal at points of intersection.
464
9.7
Divergence and Curl
40. Let F (x, y, z) = x2 − y 2 + z 2 − 4 and G(x, y, z) = 1/xy 2 − z. Then
Fx Gx + Fy Gy + Fz Gz = (2x)(−1/x2 y 2 ) + (−2y)(−2/xy 3 ) + (2z)(−1)
= −2/xy 2 + 4/xy 2 − 2z = 2(1/xy 2 − z).
For (x, y, z) on both surfaces, F (x, y, z) = G(x, y, z) = 0. Thus, Fx Gx + Fy Gy + Fz Gz = 2(0) and the surfaces
are orthogonal at points of intersection.
EXERCISES 9.7
Divergence and Curl
y
1.
y
2.
3
3
x
-3
x
3
-3
-3
-3
y
3.
3
y
4.
3
3
x
-3
x
3
-3
-3
-3
y
5.
3
y
6.
3
3
x
-3
x
3
-3
-3
3
-3
465
9.7
Divergence and Curl
7. curl F = (x − y)i + (x − y)j; div F = 2z
8. curl F = −2x2 i + (10y − 18x2 )j + (4xz − 10z)k; div F = 0
9. curl F = 0; div F = 4y + 8z
10. curl F = (xe2y + ye−yz + 2xye2y )i − ye2y j + 3(x − y)2 k; div F = 3(x − y)2 − ze−yz
11. curl F = (4y 3 − 6xz 2 )i + (2z 3 − 3x2 )k; div F = 6xy
12. curl F = −x3 zi + (3x2 yz − z)j + 32 x2 y 2 − y − 15y 2 k; div F = (x3 y − x) − (x3 y − x) = 0
13. curl F = (3e−z − 8yz)i − xe−z j; div F = e−z + 4z 2 − 3ye−z
yz
− 3z + 3xy 2 z 2
x
15. curl F = (xy 2 ey + 2xyey + x3 yez + x3 yzez )i − y 2 ey j + (−3x2 yzez − xex )k; div F = xyex + yex − x3 zez
14. curl F = (2xyz 3 + 3y)i + (y ln x − y 2 z 3 )j + (2 − z ln x)k; div F =
16. curl F = (5xye5xy + e5xy + 3xz 3 sin xz 3 − cos xz 3 )i + (x2 y cos yz − 5y 2 e5xy )j
+ (−z 4 sin xz 3 − x2 z cos yz)k;
div F = 2x sin yz
17. div r = 1 + 1 + 1 = 3
i
j
k 18. curl r = ∂/∂x ∂/∂y ∂/∂z = 0i − 0j + 0k = 0
x
y
z i
j
k ∂
∂
19. a × ∇ = a1
− a3
a2
a3 = a2
∂z
∂y
∂/∂x ∂/∂y ∂/∂z i
j
∂
∂
∂
∂
(a × ∇) × r = a2
− a3 a3
− a1
a1
∂z
∂x
∂z
∂y
x
y
i + a3
∂
∂
− a1
∂x
∂z
j + a1
∂
∂
− a2
∂y
∂x
k
∂ − a2
∂x z
k
= (−a1 − a1 )i − (a2 + a2 )j + (−a3 − a3 )k = −2a
20. ∇ × (a × r) = (∇ · r)a − (∇ · a)r = (1 + 1 + 1)a − a1
∂
∂
∂
+ a2
+ a3
∂x
∂y
∂z
r = 3a − (a1 i + a2 j + a3 k) = 2a
∂/∂x ∂/∂y ∂/∂z ∂
∂
∂
21. ∇ · (a × r) = a1
(a2 z − a3 y) −
(a1 z − a3 x) +
(a1 y − a2 x) = 0
a2
a3 =
∂x
∂y
∂z
x
y
z
i
j
k 22. ∇ × r = ∂/∂x ∂/∂y ∂/∂z = 0; a × (∇ × r) = a × 0 = 0
x
y
z i
j k 23. r × a = x y z = (a3 y − a2 z)i − (a3 x − a1 z)j + (a2 x − a1 y)k; r · r = x2 + y 2 + z 2
a1 a2 a3 i
j
k
∂/∂y
∂/∂z ∇ × [(r · r)a] = ∂/∂x
(r · r)a1 (r · r)a2 (r · r)a3 = (2ya3 − 2za2 )i − (2xa3 − 2za1 )j + (2xa2 − 2ya1 )k = 2(r × a)
24. r · a = a1 x + a2 y + a3 z; r · r = x2 + y 2 + z 2 ; ∇ · [(r · r)a] = 2xa1 + 2ya2 + 2za3 = 2(r · a)
466
9.7
Divergence and Curl
25. Let F = P (x, y, z)i + Q(x, y, z)j + R(x, y, z)k and G = S(x, y, z)i + T (x, y, z)j + U (x, y, z)k.
∇ · (F + G) = ∇ · [(P + S)i + (Q + T )j + (R + U )k] = Px + Sx + Qy + Ty + Rz + Uz
= (Px + Qy + Rz ) + (Sx + Ty + Uz ) = ∇ · F + ∇ · G
26. Let F = P (x, y, z)i + Q(x, y, z)j + R(x, y, z)k and G = S(x, y, z)i + T (x, y, z)j + U (x, y, z)k.
i
j
k ∇ × (F + G) = ∂/∂x ∂/∂y
∂/∂z P + S Q + T R + U = (Ry + Uy − Qz − Tz )i − (Rx + Ux − Pz − Sz )j + (Qx + Tx − Py − Sy )k
= (Ry − Qz )i − (Rx − Pz )j + (Qx − Py )k + (Uy − Tz )i − (Ux − Sz )j + (Tx − Sy )k
=∇×F+∇×G
27. ∇ · (f F) = ∇ · (f P i + f Qj + f Rk) = f Px + P fx + f Qy + Qfy + f Rz + Rfz
= f (Px + Qy + Rz ) + (P fx + Qfy + Rfz ) = f (∇ · F) + F · (∇f )
i
j
k 28. ∇ × (f F) = ∂/∂x ∂/∂y ∂/∂z fP
fQ
fR = (f Ry + Rfy − f Qz − Qfz )i − (f Rx + Rfx − f Pz − P fz )j + (f Qx + Qfx − f Py − P fy )k
= (f Ry − f Qz )i − (f Rx − f Pz )j + (f Qx − f Py )k + (Rfy − Qfz )i
− (Rfx − P fz )j + (Qfx − P fy )k
i
= f [(Ry − Qz )i − (Rx − Pz )j + (Qx − Py )k + fx
P
j
fy
Q
k fz = f (∇ × F) + (∇f ) × F
R
29. Assuming continuous second partial derivatives,
i
curl (grad f ) = ∇ × (fx i + fy j + fz k) = ∂/∂x
fx
j
∂/∂y
fy
k ∂/∂z fz = (fzy − fyz )i − (fzx − fxz )j + (fyx − fxy )k = 0.
30. Assuming continuous second partial derivatives,
div (curl F) = ∇ · [(Ry − Qz )i − (Rx − Pz )j + (Qx − Py )k]
= (Ryx − Qzx − (Rxy − Pzy ) + (Qxz − Pyz ) = 0.
31. Let F = P (x, y, z)i + Q(x, y, z)j + R(x, y, z)k and G = S(x, y, z)i + T (x, y, z)j + U (x, y, z)k.
i j k
F × G = P Q R = (QU − RT )i − (P U − RS)j + (P T − QS)k
S T U
div (F × G) = (QUx + Qx U − RTx − Rx T ) − (P Uy + Py U − RSy − Ry S) + (P Tz + Pz T − QSz − Qz S)
= S(Ry − Qz ) + T (Pz − Rx ) + U (Qx − Py ) − P (Uy − Tz ) − Q(Sz − Ux ) − R(Tx − Sy )
= G · (curl F) − F · (curl G)
32. Using Problems 26 and 29,
curl (curl F + grad f ) = ∇ × (curl F + grad f ) = ∇ × (curl F) + ∇ × (grad f )
= curl (curl F) + curl (grad f ) = curl (curl F) + 0 = curl (curl F).
467
9.7
Divergence and Curl
33. ∇ · ∇f = ∇ · (fx i + fy j + fz k) = fxx + fyy + fzz
34. Using Problem 27, ∇ · (f ∇f ) = f (∇ · ∇f ) + ∇f · ∇f = f (∇2 f ) + |∇f |2 .
35. curl F = −8yzi − 2zj − xk; curl (curl F) = 2i − (8y − 1)j + 8zk
36. (a) For F = P i + Qj + Rk,
curl (curl F) = (Qxy − Pyy − Pzz + Rxz )i + (Ryz − Qzz − Qxx + Pyx )j
+ (Pzx − Rxx − Ryy + Qzy )k
and
−∇2 F + grad (div F) = −(Pxx + Pyy + Pzz )i − (Qxx + Qyy + Qzz )j − (Rxx + Ryy + Rzz )k
+ grad (Px + Qy + Rz )
= −Pxx i − Qyy j − Rzz k + (−Pyy − Pzz )i + (−Qxx − Qzz )j
+ (−Rxx − Ryy )k + (Pxx + Qyx + Rzx )i + (Pxy + Qyy + Rzy )j
+ (Pxz + Qyz + Rzz )k
= (−Pyy − Pzz + Qyx + Rzx )i + (−Qxx − Qzz + Pxy + Rzy )j
+ (−Rxx − Ryy + Pxz + Qyz )k.
Thus, curl (curl F) = −∇2 F + grad (div F).
(b) For F = xyi + 4yz 2 j + 2xzk, ∇2 F = 0i + 8yj + 0k, div F = y + 4z 2 + 2x, and grad (div F) = 2i + j + 8zk.
Then curl (curl F) = −8yj + 2i + j + 8zk = 2i + (1 − 8y)j + 8zk.
37.
∂f
= −x(x2 + y 2 + z 2 )−3/2
∂x
∂f
= −y(x2 + y 2 + z 2 )−3/2
∂y
∂f
= −z(x2 + y 2 + z 2 )−3/2
∂z
∂2f
= 3x2 (x2 + y 2 + z 2 )−5/2 − (x2 + y 2 + z 2 )−3/2
∂x2
∂2f
= 3y 2 (x2 + y 2 + z 2 )−5/2 − (x2 + y 2 + z 2 )−3/2
∂y 2
∂2f
= 3z 2 (x2 + y 2 + z 2 )−5/2 − (x2 + y 2 + z 2 )−3/2
∂z 2
Adding the second partial derivatives gives
∂2f
∂2f
∂2f
3(x2 + y 2 + z 2 )
+ 2 + 2 = 2
− 3(x2 + y 2 + z 2 )−3/2
2
∂x
∂y
∂z
(x + y 2 + z 2 )5/2
= 3(x2 + y 2 + z 2 )−3/2 − 3(x2 + y 2 + z 2 )−3/2 = 0
except when x = y = z = 0.
38. fx =
1
4y 2
1+ 2
(x + y 2 − 1)2
−
4xy
(x2 + y 2 − 1)2
=−
4xy
(x2 + y 2 − 1)2 + 4y 2
[(x2 + y 2 − 1)2 + 4y 2 ]4y − 4xy[4x(x2 + y 2 − 1)]
12x4 y − 4y 5 + 8x2 y 3 − 8x2 y − 8y 3 − 4y
=
[(x2 + y 2 − 1)2 + 4y 2 ]2
[(x2 + y 2 − 1)2 + 4y 2 ]2
2(x2 + y 2 − 1) − 4y 2
2(x2 − y 2 − 1)
1
=
fy =
(x2 + y 2 − 1)2
(x2 + y 2 − 1)2 + 4y 2
4y 2
1+ 2
2
2
(x + y − 1)
fxx = −
468
9.7
fyy =
=
Divergence and Curl
[(x2 + y 2 − 1)2 + 4y 2 ](−4y) − 2(x2 − y 2 − 1)[4y(x2 + y 2 − 1) + 8y]
[(x2 + y 2 − 1)2 + 4y 2 ]2
−12x4 y + 4y 5 − 8x2 y 3 + 8x2 y + 8y 3 + 4y
[(x2 + y 2 − 1)2 + 4y 2 ]2
∇2 f = fxx + fyy = 0
i
39. curl F = −Gm1 m2 ∂/∂x
x/|r|3
j
∂/∂y
y/|r|3
k ∂/∂z z/|r|3 = −Gm1 m2 [(−3yz/|r|5 + 3yz/|r|5 )i − (−3xz/|r|5 + 3xz/|r|5 )j + (−3xy/|r|5 + 3xy/|r|5 )k]
=0
x2 − 2y 2 + z 2
x2 + y 2 − 2z 2
−2x2 + y 2 + z 2
+
+
=0
|r|5/2
|r|5/2
|r|5/2
i
j
k
1
1
1
ω × r) = ∂/∂x
40. curl v = curl (ω
∂/∂y
∂/∂z 2
2
2
ω2 z − ω3 y ω 3 x − ω1 z ω1 y − ω2 x 1
= [(ω1 + ω1 )i − (−ω2 − ω2 )j + (ω3 + ω3 )k] = ω1 i + ω2 j + ω3 k = ω
2
div F = −Gm1 m2
41. Using Problems 31 and 29,
∇ · F = div (∇f × ∇g) = ∇g · (curl ∇f ) − ∇f · (curl ∇g) = ∇g · 0 − ∇f · 0 = 0.
42. Recall that a · (a × b) = 0. Then, using Problems 31, 29, and 28,
∇ · F = div (∇f × f ∇g) = f ∇g · (curl ∇f ) − ∇f · (curl f ∇g) = f ∇g · 0 − ∇f · (∇ × f ∇g)
= −∇f · [f (∇ × ∇g) + (∇f × ∇g)] = −∇f · [f curl ∇g + (∇f × ∇g]
= −∇f · [f 0 + (∇f × ∇g)] = −∇f · (∇f × ∇g) = 0.
43. (a) Expressing the vertical component of V in polar coordinates, we have
2xy
2r2 sin θ cos θ
sin 2θ
=
=
.
(x2 + y 2 )2
r4
r2
Similarly,
x2 − y 2
r2 (cos2 θ − sin2 θ)
cos 2θ
=
=
.
(x2 + y 2 )2
r4
r2
Since limr→∞ (sin 2θ)/r2 = limr→∞ (cos 2θ)/r2 = 0, V ≈ Ai for r large or (x, y) far from the origin.
2Axy
x2 − y 2
, Q(x, y) = − 2
(b) Identifying P (x, y) = A 1 − 2
, and R(x, y) = 0, we have
(x − y 2 )2
(x + y 2 )2
Py =
2Ay(3x2 − y 2 )
,
(x2 + y 2 )3
Qx =
2Ay(3x2 − y 2 )
,
(x2 + y 2 )3
and Pz = Qz = Rx = Ry = 0.
Thus, curl V = (Ry − Qz )i + (Pz − Rx )j + (Qx − Py )k = 0 and V is irrotational.
(c) Since Px =
2Ax(x2 − 3y 2 )
2Ax(3y 2 − x2 )
,
Q
=
, and Rz = 0, ∇ · F = Px + Qy + Rz = 0 and V is
y
(x2 + y 2 )3
(x2 + y 2 )3
incompressible.
469
9.7
Divergence and Curl
44. We first note that curl (∂H/∂t) = ∂(curl H)/∂t and curl (∂E/∂t) = ∂(curl E)/∂t. Then, from Problem 36,
−∇2 E = −∇2 E + 0 = −∇2 E + grad 0 = −∇2 E + grad (div E) = curl (curl E)
−
= curl
and ∇2 E =
1
c2
1 ∂H
c ∂t
=−
1 ∂
1 ∂
curl H = −
c ∂t
c ∂t
1 ∂E
c ∂t
=−
1 ∂2E
c2 ∂t
∂ 2 E/∂t2 . Similarly,
1 ∂E
c ∂t
−∇2 H = −∇2 H + grad (div H) = curl (curl H) = curl
=
1 ∂
c ∂t
−
1 ∂H
c ∂t
=−
=
1 ∂
curl E
c ∂t
1 ∂2H
c2 ∂t2
1 2
∂ H/∂t2 .
c2
45. We note that div F = 2xyz − 2xyz + 1 = 1 = 0. If F = curl G, then div(curl G) = div F = 1. But, by
Problem 30, for any vector field G, div(curl G) = 0. Thus, F cannot be the curl of G.
and ∇2 H =
EXERCISES 9.8
Line Integrals
√
π/4
1
125 2
sin3 t = −
3
6
C
0
0
0
π/4
π/4
π/4
1
2xy dy =
2(5 cos t)(5 sin t)(5 cos t) dt = 250
cos2 t sin t dt = 250 − cos3 t 3
C
0
0
0
√ √
2
125
250
1−
=
(4 − 2)
=
3
4
6
π/4
π/4
2xy ds =
2(5 cos t)(5 sin t) 25 sin2 t + 25 cos2 t dt = 250
sin t cos t dt
1.
π/4
0
C
= 250
1
sin2 t
2
sin2 t cos t dt = −250
0
π/4
= 125
2
0
1
(x3 + 2xy 2 + 2x) dx =
2.
π/4
2(5 cos t)(5 sin t)(−5 sin t) dt = −250
2xy dx =
1
[8t3 + 2(2t)(t4 ) + 2(2t)]2 dt = 2
0
C
(8t3 + 4t5 + 4t) dt
0
1
2 6
28
2
= 2 2t + t + 2t =
3
3
0
1
1
(x3 + 2xy 2 + 2x) dy =
[8t3 + 2(2t)(t4 ) + 2(2t)]2t dt = 2
(8t4 + 4t6 + 4t2 ) dt
4
0
C
8 5 4 7 4 3
t + t + t
5
7
3
=2
3
2
=8
4
[8t + 2(2t)(t ) + 2(2t)] 4 +
0
C
1
3
(x + 2xy + 2x) ds =
0
1
= 736
105
0
1
(1 + t2 )7/2
7
4t2
1
= 8 (27/2 − 1)
7
0
470
1
t(1 + t2 )5/2 dt
dt = 8
0
9.8
0
(3x2 + 6y 2 ) dx =
3.
0
[3x2 + 6(2x + 1)2 ] dx =
−1
C
−1
Line Integrals
0
(27x2 + 24x + 6) dx = (9x3 + 12x2 + 6x) −1
= −(−9 + 12 − 6) = 3
0
(3x2 + 6y 2 ) dy =
[3x2 + 6(2x + 1)2 ]2 dx = 6
C
−1
0
(3x2 + 6y 2 ) ds =
−1
C
C
x2
dx =
y3
C
x2
dy =
y3
4.
C
x2
ds =
y3
5.
8
x2
8
dx =
2
27x /8
27
1
8
1
8
x2
27x2 /8
1
8
dx =
1
1+
x−2/3
56
27
8
−1/3
x
1
8
dx =
27
8
x
t(− sin t) dt
8
8
8 3/2
2/3 3/2 dx =
=
(1 + x )
(5 − 23/2 )
27
27
1
Integration by parts
π/2
= (t cos t − sin t) = −1
0
π/2
t cos t dt
Integration by parts
0
0
1
1 3
8
t (t2 )(2t)t2 dt =
3
3
4
1 3
t
3
4
1 3
t
3
4
1 3
t
3
1
4xyz dy =
0
C
1
4xyz dz =
0
C
4xyz ds =
1
0
√
π2 2
t dt =
8
1
8 9 8
t =
27
27
0
0
1
1
16
2 2
(t2 )(2t)2t dt =
t7 dt = t8 =
3 0
3
3
0
1
1
16
16 7 16
(t2 )(2t)2 dt =
t6 dt =
=
t
3 0
21 0 21
8 1 6 2
8
(t2 )(2t) t4 + 4t2 + 4 dt =
t (t + 2) dt =
3 0
3
4
0
C
π/2
0
4xyz dx =
C
1+
x2/3
1
π/2 π
= (t sin t + cos t) = − 1
2
0
π/2
π/2
1 π2
z dz =
t dt = t2 =
2
8
C
0
0
π/2
√ 2
2
z ds =
t sin t + cos t + 1 dt = 2
6.
−1/3
π/2
z dy =
C
8
4 2/3 4
dx = x
=3
9
1
0
C
x2
8
x−1/3 dx =
27x2 /8
27
z dx =
C
√
√
[3x2 + 6(2x + 1)2 ] 1 + 4 dx = 3 5
1
t8 dt =
1 9 2 7
t + t
9
7
1
= 200
189
0
7. Using x as the parameter, dy = dx and
2
−1
C
=
2
(2x + x + 3 + x2 + 3x) dx =
(2x + y) dx + xy dy =
1 3
x + 3x2 + 3x
3
471
2
= 21.
−1
(x2 + 6x + 3) dx
−1
9.8
Line Integrals
8. Using x as the parameter, dy = 2x dx and
2
2
(2x + y) dx + xy dy =
(2x + x + 1) dx +
−1
C
2
2
(2x4 + 3x2 + 2x + 1) dx
x(x + 1) 2x dx =
−1
2
2 5
x + x3 + x2 + x
5
=
2
−1
=
−1
141
.
5
9. From (−1, 2) to (2, 2) we use x as a parameter with y = 2 and dy = 0. From (2, 2) to (2, 5) we use y as a
parameter with x = 2 and dx = 0.
2
5
5
2
(2x + y) dx + xy dy =
(2x + 2) dx +
2y dy = (x2 + 2x) + y 2 = 9 + 21 = 30
−1
C
−1
2
2
10. From (−1, 2) to (−1, 0) we use y as a parameter with x = −1 and dx = 0. From (−1, 0) to (2, 0) we use x as a
parameter with y = dy = 0. From (2, 0) to (2, 5) we use y as a parameter with x = 2 and dx = 0.
0
2
5
0
2
5
1
(2x + y) dx + xy dy =
(−1) y dy +
2x dx +
2y dy = − y 2 + x2 + y2 2
2
−1
0
C
2
−1
0
= 2 + 3 + 25 = 30
11. Using x as a the parameter, dy = 2x dx.
y dx + x dy =
1
0
C
1
x2 dx +
0
12. Using x as the parameter, dy = dx.
y dx + x dy =
1
1
3x2 dx = x3 = 1
0
0
1
x dx +
1
x dx =
0
C
1
x(2x) dx =
0
1
2x dx = x2 = 1
0
0
13. From (0, 0) to (0, 1) we use y as a parameter with x = dx = 0. From (0, 1) to (1, 1) we use x as a parameter
with y = 1 and dy = 0.
1
y dx + x dy = 0 +
1 dx = 1
0
C
14. From (0, 0) to (1, 0) we use x as a parameter with y = dy = 0. From (1, 0) to (1, 1) we use y as a parameter
with x = 1 and dx = 0.
1
y dx + x dy = 0 +
1 dy = 1
0
C
9
1
(6t + 2t2 ) t−1/2 dt +
2
4
9
= (2t3/2 + 2t5/2 ) = 460
9
√
4 t t dt =
(3t1/2 + 5t3/2 ) dt
9
(6x2 + 2y 2 ) dx + 4xy dy =
15.
C
4
4
4
(−y 2 ) dx + xy dy =
16.
C
2
(−t6 ) 2 dt +
0
(2t)(t3 )3t2 dt =
0
=
4 9
1
y + y3 + y2
9
2
4t6 dt =
1
2(y 6 )y 2y dy +
−1
C
2
0
1
2x3 y dx + (3x + y) dy =
17.
2
1
(3y 2 + y) dy =
−1
1
−1
(4y 8 + 3y 2 + y) dy
−1
=
26
9
472
2
4 7 512
t =
7
7
0
9.8
18.
2
−1
C
2
4(y 3 + 1)3y 2 dy +
4x dx + 2y dy =
Line Integrals
2
(12y 5 + 12y 2 + 2y) dy
2y dy =
−1
−1
2
= (2y 6 + 4y 3 + y 2 ) = 165
−1
19. From (−2, 0) to (2, 0) we use x as a parameter with y = dy = 0. From (2, 0) to (−2, 0) we parameterize the
semicircle as x = 2 cos θ and y = 2 sin θ for 0 ≤ θ ≤ π.
2
π
π
2
2
2
x dx +
4(−2 sin θ dθ) −
8 cos θ sin θ(2 cos θ dθ)
ˇ (x + y ) dx − 2xy dy =
−2
C
0
0
2
π
1 3 = x −8
(sin θ + 2 cos2 θ sin θ) dθ
3
0
−2
π
16
2
64
16 80
=
− 8 − cos θ − cos3 θ =
−
=−
3
3
3
3
3
0
20. We start at (0, 0) and use x as a parameter.
1
2
2
2
4
(x + x ) dx − 2
ˇ (x + y ) dx − 2xy dy =
C
0
1
(x2 + x) dx
xx (2x dx) +
0
0
0
2
1
√
x x
1 −1/2
dx
x
2
1
1
0
1
3
2
4
2
=
(x − 3x ) dx +
x dx =
(−3x4 ) dx = − x5
5
0
1
0
−2
1
= −3
5
0
21. From (1, 1) to (−1, 1) and (−1, −1) to (1, −1) we use x as a parameter with y = 1 and y = −1, respectively,
and dy = 0. From (−1, 1) to (−1, −1) and (1, −1) to (1, 1) we use y as a parameter with x = −1 and x = 1,
respectively, and dx = 0.
2 3
2
x
y
dx
−
xy
dy
=
ˇ
C
−1
x2 (1) dx +
1
−1
1 1
= x3 + y3
3
3
1
−1
−(−1)y 2 dy +
1
−1
1
− x3
3
1
1
−1
1
1
x2 (−1)3 dx +
−1
−1
1
1 8
− y 3 = −
3
3
−1
−(1)y 2 dy
22. From (2, 4) to (0, 4) we use x as a parameter with y = 4 and dy = 0. From (0, 4) to (0, 0) we use y as a parameter
with x = dx = 0. From (0, 0) to (2, 4) we use y = 2x and dy = 2 dx.
0
0
2
2
2 3
2
2
2
3
x
y
dx
−
xy
dy
=
x
(64)
dx
−
0
dy
+
x
(8x
)
dx
−
x(4x2 )2 dx
ˇ
C
2
0
64 3 4
=
x + x6
3
3
2
23. ˇ (x2 − y 2 ) ds =
C
0
4
2
− 2x4
0
0
0
2
= − 512 + 256 − 32 = − 352
3
3
3
0
2π
(25 cos2 θ − 25 sin2 θ) 25 sin2 θ + 25 cos2 θ dθ = 125
2π
(cos2 θ − sin2 θ) dθ
0
2π
125
= 125
cos 2θ dθ =
sin 2θ = 0
2
0
0
π
π
π
π
24. ˇ y dx − x dy =
3 sin t(−2 sin t) dt −
2 cos t(3 cos t) dt = −6
(sin2 t + cos2 t) dt = −6
dt = −6π
C
0
0
0
0
Thus,
y dx − x dy = 6π.
2π
−C
473
9.8
Line Integrals
25. We parameterize the line segment from (0, 0, 0) to (2, 3, 4) by x = 2t, y = 3t, z = 4t for 0 ≤ t ≤ 1. We
parameterize the line segment from (2, 3, 4) to (6, 8, 5) by x = 2 + 4t, y = 3 + 5t, z = 4 + t, 0 ≤ t ≤ 1.
1
1
1
1
3t(2 dt) +
4t(3 dt) +
2t(4 dt) +
(3 + 5t)(4 dt)
ˇ y dx + z dy + x dz =
C
0
+
0
1
0
0
1
(4 + t)(5 dt) +
(2 + 4t) dt
0
0
1
55 2
123
=
(55t + 34) dt =
t + 34t =
2
2
0
0
2
2
2
5 2
5
26.
y dx + z dy + x dz =
t3 (3 dt) +
(3t)
t (3t2 dt) +
t dt
4
2
C
0
0
0
2
2
3 4 3 5 5 3 15 4 15 2
3
3t + t +
=
t dt =
t + t + t = 56
4
2
4
4
2
0
0
1
27. From (0, 0, 0) to (6, 0, 0) we use x as a parameter with y = dy = 0 and z = dz = 0. From (6, 0, 0) to (6, 0, 5) we
use z as a parameter with x = 6 and dx = 0 and y = dy = 0. From (6, 0, 5) to (6, 8, 5) we use y as a parameter
with x = 6 and dx = 0 and z = 5 and dz = 0.
6
5
8
y dx + z dy + x dz =
0+
6 dz +
5 dy = 70
0
C
0
0
28. We parametrize the line segment from (0, 0, 0) to (6, 8, 0) by x = 6t, y = 8t, z = 0 for 0 ≤ t ≤ 1. From (6, 8, 0)
to (6, 8, 5) we use z as a parameter with x = 6, dx = 0, and y = 8, dy = 0.
1
5
1
y dx + z dy + x dz =
8t(6 dt) +
6 dz = 24t2 + 30 = 54
0
C
0
0
29. F = e3t i − (e−4t )et j = e3t i − e−3t j; dr = (−2e−2t i + et j) dt; F · dr = (−2et − e−2t ) dt;
ln 2
ln 2
3
19
1 −2t 31
t
−2t
t
=−
F · dr =
(−2e − e ) dt = −2e + e
=− − −
2
8
2
8
0
C
0
3
6
30. F = et i + tet j + t3 et k; dr = (i + 2tj + 3t2 k) dt;
1
3
6
2 3 1 6
F · dr =
(et + 2t2 et + 3t5 et ) dt = et + et + et
3
2
C
0
1
= 13 (e − 1)
6
0
1
31. Using x as a parameter, r(x) = xi + ln xj. Then F = ln xi + xj, dr = (i + j) dx, and
x
e
e
W =
F · dr =
(ln x + 1) dx = (x ln x) = e.
1
1
C
32. Let r1 = (−2 + 2t)i + (2 − 2t)j and r2 = 2ti + 3tj for 0 ≤ t ≤ 1. Then
dr1 = 2i − 2j,
dr2 = 2i + 3j,
F1 = 2(−2 + 2t)(2 − 2t)i + 4(2 − 2t)2 j = (−8t2 + 16t − 8)i + (16t2 − 32t + 16)j,
F2 = 2(2t)(3t)i + 4(3t)2 j = 12t2 i + 36t2 j,
and
F1 · dr1 +
W =
C1
=
0
1
F2 · dr2 =
C2
1
(−16t2 + 32t − 16 − 32t2 + 64t − 32) dt +
0
1
(84t2 + 96t − 48) dt = (28t3 + 48t2 − 48t) = 28.
0
474
1
(24t2 + 108t2 ) dt
0
9.8
Line Integrals
33. Let r1 = (1 + 2t)i + j, r2 = 3i + (1 + t)j, and r3 = (3 − 2t)i + (2 − t)j for 0 ≤ t ≤ 1.
Then
dr1 = 2i,
dr3 = −2i − j,
dr2 = j,
F1 = (1 + 2t + 2)i + (6 − 2 − 4t)j = (3 + 2t)i + (4 − 4t)j,
F2 = (3 + 2 + 2t)i + (6 + 6t − 6)j = (5 + 2t)i + 6tj,
F3 = (3 − 2t + 4 − 2t)i + (12 − 6t − 6 + 4t)j = (7 − 4t)i + (6 − 2t)j,
and
F2 · dr2 +
C1
C2
1
=
=
5
(−14 + 8t − 6 + 2t)dt
0
1
(−14 + 20t) dt = (−14t + 10t2 ) = −4.
0
0
4
1
6tdt +
0
1
F3 · dr3
C3
1
(6 + 4t)dt +
0
3
F1 · dr1 +
W =
F · dr =
2
34. F = t i + t j + t k; dr = 3t i + 2tj + k; W =
3
5
5
(3t + 2t + t ) dt =
1
C
3
5
3
6t5 dt = t6 = 728
1
1
35. r = 3 cos ti + 3 sin tj, 0 ≤ t ≤ 2π; dr = −3 sin ti + 3 cos tj; F = ai + bj;
2π
2π
W =
F · dr =
(−3a sin t + 3b cos t) dt = (3a cos t + 3b sin t) = 0
0
0
C
36. Let r = ti + tj + tk for 1 ≤ t ≤ 3. Then dr = i + j + k, and
c
ct
c
(ti + tj + tk) = √
(i + j + k) = √
(i + j + k),
3
|r|
3 3 t2
( 3t2 )3
3
3
c
1
c
c
c
1 1
√
√
(1 + 1 + 1) dt = √
dt
=
=√
−
− +1
t 1
3
3 3 t2
3 1 t2
3
3
F=
F · dr =
W =
1
C
y 2 dx + xy dy =
37.
3
1
0
C1
1
(4t + 2)2 2 dt +
1
(64t2 + 64t + 16) dt
(2t + 1)(4t + 2)4 dt =
0
2c
= √ .
3 3
0
1
64 3
64
208
=
t + 32t2 + 16t =
+ 32 + 16 =
3
3
3
0
√3
√3
√3
√3
8
8
208
y 2 dx + xy dy =
4t4 (2t) dt +
2t4 (4t) dt =
16t5 dt = t6 = 72 − =
3
3
3
C2
1
1
1
1
e3
e3
e3
e3
208
1
2
8
8
8
(ln t)2 dt = (ln t)3 = (27 − 1) =
y 2 dx + xy dy =
4(ln t)2 dt +
2(ln t)2 dt =
t
t
t
3
3
3
C3
e
e
e
e
√
2
2
2
√
√
√
1 3 16 5
38.
t =
xy ds =
t(2t) 1 + 4 dt = 2 5
t2 dt = 2 5
3
3
C1
0
0
0
2
2
xy ds =
t(t2 ) 1 + 4t2 dt =
t3 1 + 4t2 dt
u = 1 + 4t2 , du = 8t dt, t2 = 14 (u − 1)
0
C2
0
√
17
391 17 + 1
2 5/2 2 3/2 =
u − u
5
3
120
1
1
1
√
3
3
3
√
√ 1
√
16 5
xy ds =
(2t − 4)(4t − 8) 4 + 16 dt = 16 5
(t − 2)2 dt = 16 5 (t − 2)3 =
3
3
2
2
C3
2
=
17
1
1
1
(u − 1)u1/2 du =
4
8
32
17
(u3/2 − u1/2 ) du =
1
32
C1 and C3 are different parameterizations of the same curve, while C1 and C2 are different curves.
475
9.8
Line Integrals
39. Since v · v = v 2 ,
d 2
d
dv dv
dv
v = (v · v) = v ·
+
·v =2
· v. Then
dt
dt
dt
dt
dt
b
b
b
dv
1
dr
W =
F · dr =
ma ·
dt = m
· v dt = m
dt
dt
C
a
a
a 2
b
1
1
1
= m(v 2 ) = m[v(b)]2 − m[v(a)]2 .
2
2
2
a
d 2
v
dt
dt
40. We are given ρ = kx. Then
π
π
π
2
2
m=
ρ ds =
kx ds = k
(1 + cos t) sin t + cos t dt = k
(1 + cos t) dt
C
0
0
0
π
= k(t + sin t) = kπ.
0
41. From Problem 40, m = kπ and ds = dt.
π
π
1
Mx =
yρ ds =
kxy ds = k
(1 + cos t) sin t dt = k − cos t + sin2 t = 2k
2
0
C
C
0
π
π
My =
xρ ds =
kx2 ds = k
(1 + cos t)2 dt = k
(1 + 2 cos t + cos2 t) dt
0
0
C
C
π
1
1
3
= k t + 2 sin t + t + sin 2t = kπ
2
4
2
0
3kπ/2
2k
3
2
= ; ȳ = Mx /m =
= . The center of mass is (3/2, 2/π).
kπ
2
kπ
π
42. On C1 , T = i and F · T = compT F ≈ 1. On C2 , T = −j and F · T = compT F ≈ 2. On C3 , T = −i and
F · T = compT F ≈ 1.5. Using the fact that the lengths of C1 , C2 , and C3 are 4, 5, and 5, respectively, we have
W =
F · T ds =
F · T ds +
F · T ds +
F · T ds ≈ 1(4) + 2(5) + 1.5(5) = 21.5 ft-lb.
x̄ = My /m =
C
C1
C2
C3
EXERCISES 9.9
Independence of Path
1
1. (a) Py = 0 = Qx and the integral is independent of path. φx = x2 , φ = x3 + g(y),
3
(2,2)
(2,2)
1
1
1
1
16
φy = g (y) = y 2 , g(y) = y 3 , φ = x3 + y 3 ,
x2 dx + y 2 dy = (x3 + y 3 ) =
3
3
3
3
3
(0,0)
(0,0)
(2,2)
2
2 2
16
(b) Use y = x for 0 ≤ x ≤ 2.
x2 dx + y 2 dy =
(x2 + x2 ) dx = x3 =
3
3
0
(0,0)
0
2. (a) Py = 2x = Qx and the integral is independent of path. φx = 2xy, φ = x2 y + g(y),
(2,4)
(2,4)
φy = x2 + g (y) = x2 , g(y) = 0, φ = x2 y,
2xy dx + x2 dy = x2 y = 16 − 1 = 15
(1,1)
(1,1)
(b) Use y = 3x − 2 for 1 ≤ x ≤ 2.
(2,4)
2
2
2
2xy dx + x2 dy =
[2x(3x − 2) + x2 (3)] dx =
(9x2 − 4x) dx = (3x3 − 2x2 ) = 15
(1,1)
1
1
476
1
9.9
3. (a) Py = 2 = Qx and the integral is independent of path. φx = x + 2y, φ =
Independence of Path
1 2
x + 2xy + g(y),
2
1
1
1
φy = 2x + g (y) = 2x − y, g(y) = − y 2 , φ = x2 + 2xy − y 2 ,
2
2
2
(3,2)
(3,2)
1 2
1
x + 2xy − y 2 (x + 2y) dx + (2x − y) dy =
= 14
2
2
(1,0)
(1,0)
(b) Use y = x − 1 for 1 ≤ x ≤ 3.
(3,2)
3
(x + 2y) dx + (2x − y) dy =
(1,0)
[x + 2(x − 1) + 2x − (x − 1) dx
1
3
=
3
(4x − 1) dx = (2x2 − x) = 14
1
1
4. (a) Py = − cos x sin y = Qx and the integral is independent of path. φx = cos x cos y,
φ = sin x cos y + g(y), φy = − sin x sin y + g (y) = 1 − sin x sin y, g(y) = y, φ = sin x cos y + y,
(π/2,0)
(π/2,0)
cos x cos y dx + (1 − sin x sin y) dy = (sin x cos y + y) =1
(0,0)
(0,0)
(b) Use y = 0 for 0 ≤ x ≤ π/2.
(π/2,0)
π/2
cos x cos y dx + (1 − sin x sin y) dy =
(0,0)
π/2
cos x dx = sin x =1
0
0
1
x
x
x
5. (a) Py = 1/y 2 = Qx and the integral is independent of path. φx = − , φ = − +g(y), φy = 2 +g (x) = 2 ,
y
y
y
y
(4,4)
(4,4)
x
1
x x
g(y) = 0, φ = − ,
− dx + 2 dy = (− ) =3
y
y
y
y (4,1)
(4,1)
(b) Use x = 4 for 1 ≤ y ≤ 4.
(4,4)
(4,1)
1
x
− dx + 2 dy =
y
y
4
4 4
dy = − = 3
y2
y 1
4
1
6. (a) Py = −xy(x2 + y 2 )−3/2 = Qx and the integral is independent of path. φx =
y
x2 + y 2 + g(y), φy =
φ=
(3,4)
x dx + y dy
x2 + y 2
(1,0)
=
x2
+
y2
+ g (y) =
y
x2
+ y2
, g(y) = 0, φ =
x
x2
+ y2
,
x2 + y 2 ,
(3,4)
x2 + y 2 =4
(1,0)
(b) Use y = 2x − 2 for 1 ≤ x ≤ 3.
(3,4)
(1,0)
x dx + y dy
x2 + y 2
=
1
3
x + (2x − 2)2
x2 + (2x − 2)2
dx =
1
3
√
5x − 4
=
5x2 − 8x + 4
3
5x2 − 8x + 4 = 4
1
7. (a) Py = 4xy = Qx and the integral is independent of path. φx = 2y 2 x − 3, φ = x2 y 2 − 3x + g(y),
φy = 2x2 y + g (y) = 2x2 y + 4, g(y) = 4y, φ = x2 y 2 − 3x + 4y,
(3,6)
(3,6)
(2y 2 x − 3) dx + (2yx2 + 4) dy = (x2 y 2 − 3x + 4y) = 330
(1,2)
(1,2)
477
9.9
Independence of Path
(b) Use y = 2x for 1 ≤ x ≤ 3.
(3,6)
2
2
(2y x − 3) dx + (2yx + 4) dy =
(1,2)
3
1
3
=
[2(2x)2 x − 3] + [2(2x)x2 + 4]2 dx
3
(16x3 + 5) dx = (4x4 + 5x) = 330
1
1
8. (a) Py = 4 = Qx and the integral is independent of path. φx = 5x + 4y, φ =
5 2
x + 4xy − 2y 4 ,
2
(0,0)
5 2
7
4
=
x + 4xy − 2y 2
2
(−1,1)
5 2
x + 4xy + g(y),
2
φy = 4x + g (y) = 4x − 8y 3 , g(y) = −2y 4 , φ =
(0,0)
(5x + 4y) dx + (4x − 8y 3 ) dy =
(−1,1)
(b) Use y = −x for −1 ≤ x ≤ 0.
(0,0)
(5x + 4y) dx + (4x − 8y 3 ) dy =
0
−1
(−1,1)
[(5x − 4x) + (4x + 8x3 )(−1)] dx
0
=
−1
3
− x2 − 2x4
2
(−3x − 8x3 ) dx =
0
=
−1
7
2
9. (a) Py = 3y 2 + 3x2 = Qx and the integral is independent of path. φx = y 3 + 3x2 y,
φ = xy 3 + x3 y + g(y), φy = 3xy 2 + x3 + g (y) = x3 + 3y 2 x + 1, g(y) = y, φ = xy 3 + x3 y + y,
(2,8)
(2,8)
(y 3 + 3x2 y) dx + (x3 + 3y 2 x + 1) dy = (xy 3 + x3 y + y) = 1096
(0,0)
(0,0)
(b) Use y = 4x for 0 ≤ x ≤ 2.
(2,8)
2
(y 3 + 3x2 y) dx + (x3 + 3y 2 x + 1) dy =
[(64x3 + 12x3 ) + (x3 + 48x3 + 1)(4)] dx
(0,0)
0
=
2
2
(272x3 + 4) dx = (68x4 + 4x) = 1096
0
0
10. (a) Py = −xy cos xy − sin xy − 20y 3 = Qx and the integral is independent of path.
φx = 2x − y sin xy − 5y 4 , φ = x2 + cos xy − 5xy 4 + g(y),
φy = −x sin xy − 20xy 3 + g (y) = −20xy 3 − x sin xy, g(y) = 0, φ = x2 + cos xy − 5xy 4 ,
(1,0)
(1,0)
(2x − y sin xy − 5y 4 ) dx − (20xy 3 + x sin xy) dy = (x2 + cos xy − 5xy 4 ) = −3
(−2,0)
(−2,0)
(b) Use y = 0 for −2 ≤ x ≤ 1.
(1,0)
(2x − y sin xy − 5y 4 ) dx − (20xy 3 + x sin xy) dy =
1
−2
(−2,0)
1
2x dx = x2 = −3
−2
11. Py = 12x3 y 2 = Qx and the vector field is a gradient field. φx = 4x3 y 3 + 3, φ = x4 y 3 + 3x + g(y),
φy = 3x4 y 2 + g (y) = 3x4 y 2 + 1, g(y) = y, φ = x4 y 3 + 3x + y
12. Py = 6xy 2 = Qx and the vector field is a gradient field. φx = 2xy 3 , φ = x2 y 3 + g(y),
φy = 3x2 y 2 + g (y) = 3x2 y 2 + 3y 2 , g(y) = y 3 , φ = x2 y 3 + y 3
13. Py = −2xy 3 sin xy 2 + 2y cos xy 2 , Qx = −2xy 3 cos xy 2 − 2y sin xy 2 and the vector field is not a gradient field.
14. Py = −4xy(x2 + y 2 + 1)−3 = Qx and the vector field is a gradient field.
478
9.9
Independence of Path
1
φx = x(x2 + y 2 + 1)−2 , φ = − (x2 + y 2 + 1)−1 + g(y), φy = y(x2 + y 2 + 1)−2 + g (y) = y(x2 + y 2 + 1)−2 ,
2
1
g(y) = 0, φ = − (x2 + y 2 + 1)−1
2
1
15. Py = 1 = Qx and the vector field is a gradient field. φx = x3 + y, φ = x4 + xy + g(y), φy = x + g (y) = x + y 3 ,
4
1 4
1 4
1 4
g(y) − y , φ = x + xy + y
4
4
4
16. Py = 4e2y , Qx = e2y and the vector field is not a gradient field.
−y
17. Since Py = −e = Qx , F is conservative and
F · dr is independent of the path. Thus, instead of the given
C
curve we may use the simpler curve C1 : y = x, 0 ≤ x ≤ 1. Then
W =
(2x + e−y ) dx + (4y − xe−y ) dy
C1
=
1
(2x + e−x ) dx +
0
1
(4x − xe−x ) dx
Integration by parts
0
1
1
= (x2 − e−x ) + (2x2 + xe−x + e−x ) 0
0
= [(1 − e−1 ) − (−1)] + [(2 + e−1 + e−1 ) − (1)] = 3 + e−1 .
18. Since Py = −e−y = Qx , F is conservative and
F · dr is independent of the path. Thus, instead of the given
C
curve we may use the simpler curve C1 : y = 0, −2 ≤ −x ≤ 2. Then dy = 0 and
−2
−2
−y
−y
W =
(2x + e ) dx + (4y − xe ) dy =
(2x + 1) dx = (x2 + x) = (4 − 2) − (4 + 2) = −4.
2
2
C1
19. Py = z = Qx , Qz = x = Ry , Rx = y = Pz , and the integral is independent of path. Parameterize the line
segment between the points by x = 1 + t, y = 1 + 3t, z = 1 + 7t, for 0 ≤ t ≤ 1. Then dx = dt, dy = 3 dt,
dz = 7 dt and
(2,4,8)
1
yz dx + xz dy + xy dz =
(1,1,1)
[(1 + 3t)(1 + 7t) + (1 + t)(1 + 7t)(3) + (1 + t)(1 + 3t)(7)] dt
0
=
1
1
(11 + 62t + 63t2 ) dt = (11t + 31t2 + 21t3 ) = 63.
0
0
20. Py = 0 = Qx , Qz = 0 = Ry , Rx = 0 = Pz and the integral is independent of path. Parameterize the line
segment between the points by x = t, y = t, z = t, for 0 ≤ t ≤ 1. Then dx = dy = dz = dt and
(1,1,1)
1
1
2x dx + 3y 2 dy + 4z 3 dz =
(2t + 3t2 + 4t3 ) dt = (t2 + t3 + t4 ) = 3.
(0,0,0)
0
0
21. Py = 2x cos y = Qx , Qz = 0 = Ry , Rx = 3e3z = Pz , and the integral is independent of path. Integrating
φx = 2x sin y + e3z we find φ = x2 sin y + xe3z + g(y, z). Then φy = x2 cos y + gy = Q = x2 cos y, so gy = 0,
g(y, z) = h(z), and φ = x2 sin y + xe3z + h(z). Now φz = 3xe3z + h (z) = R = 3xe3z + 5, so h (z) = 5 and
h(z) = 5z. Thus φ = x2 sin y + xe3z + 5z and
(2,π/2,1)
(2x sin y + e3z ) dx + x2 cos y dy + (3xe3z + 5) dz
(1,0,0)
(2,π/2,1)
= (x2 sin y + xe3z + 5z) = [4(1) + 2e3 + 5] − [0 + 1 + 0] = 8 + 2e3 .
(1,0,0)
479
9.9
Independence of Path
22. Py = 0 = Qx , Qz = 0 = Ry , Rx = 0 = Pz , and the integral is independent of path. Parameterize the line
segment between the points by x = 1 + 2t, y = 2 + 2t, z = 1, for 0 ≤ t ≤ 1. Then dx = 2 dt, dz = 0 and
(3,4,1)
1
1
(2x + 1) dx + 3y 2 dy + dz =
[(2 + 4t + 1)2 + 3(2 + 2t)2 2] dt
z
(1,2,1)
0
1
1
=
(24t2 + 56t + 30) dt = (8t3 + 28t2 + 30t) = 66.
0
0
23. Py = 0 = Qx , Qz = 0 = Ry , Rx = 2e2z = Pz and the integral is independent of path. Parameterize the line
segment between the points by x = 1 + t, y = 1 + t, z = ln 3, for 0 ≤ t ≤ 1. Then dx = dy = dt, dz = 0 and
(2,2 ln 3)
1
1
e2z dx + 3y 2 dy + 2xe2z dz =
[e2 ln 3 + 3(1 + t)2 ] dt = [9t + (1 + t)3 ] = 16.
(1,1,ln 3)
0
0
24. Py = 0 = Qx , Qz = 2y = Ry , Rx = 2x = Pz and the integral is independent of path. Parameterize the line
segment between the points by x = −2(1 − t), y = 3(1 − t), z = 1 − t, for 0 ≤ t ≤ 1. Then dx = 2 dt, dy = −3 dt,
dz = −dt, and
(0,0,0)
2xz dx + 2yz dy + (x2 + y 2 ) dz
(−2,3,1)
1
[−4(1 − t)2 (2) + 6(1 − t)2 (−3) + 4(1 − t)2 (−1) + 9(1 − t)2 (−1)] dt
=
0
=
1
1
−39(1 − t)2 dt = 13(1 − t)3 = −13.
0
0
25. Py = 1−z sin x = Qx , Qz = cos x = Ry , Rx = −y sin x = Pz and the integral is independent of path. Integrating
φx = y − yz sin x we find φ = xy + yz cos x + g(y, z). Then φy = x + z cos x + gy (y, z) = Q = x + z cos x, so
gy = 0, g(y, z) = h(z), and φ = xy + yz cos x + h(z). Now φz = y cos x + h(z) = R = y cos x, so h(z) = 0 and
φ = xy + yz cos x. Since r(0) = 4j and r(π/2) = πi + j + 4k,
(π,1,4)
F · dr = (xy + yz cos x) = (π − 4) − (0 + 0) = π − 4.
C
(0,4,0)
26. Py = 0 = Qx , Qz = 0 = Ry Rx = −ez = Pz and the integral is independent of path. Integrating φx = 2 − ez we
find φ = 2x − xez + g(y, z). Then φy = gy = 2y − 1, so g(y, z) = y 2 − y + h(z) and φ = 2x − xez + y 2 − y + h(z).
Now φz = −xez + h (z) = R = 2 − xez , so h (z) = 2, h(z) = 2z, and φ = 2x − xez + y 2 − y + 2z. Thus
(2,4,8)
z
2
F · dr = (2x − xe + y − y + 2z) = (4 − 2e4 + 16 − 4 + 16) − (−2 + e−1 + 1 − 1 − 2) = 36 − 2e4 − e−1 .
C
(−1,1,−1)
27. Since Py = Gm1 m2 (2xy/|r|5 ) = Qx , Qz = Gm1 m2 (2yz/|r|5 ) = Ry , and Rx = Gm1 m2 (2xz/|r|5 ) = Pz , the
force field is conservative.
x
φx = −Gm1 m2 2
, φ = Gm1 m2 (x2 + y 2 + z 2 )−1/2 + g(y, z),
2
(x + y + z 2 )3/2
y
y
φy = −Gm1 m2 2
+ gy (y, z) = −Gm1 m2 2
, g(y, z) = h(z),
(x + y 2 + z 2 )3/2
(x + y 2 + z 2 )3/2
φ = Gm1 m2 (x2 + y 2 + z 2 )−1/2 + h(z),
z
z
+ h (z) = −Gm1 m2 2
,
φz = −Gm1 m2 2
2
2
3/2
2
(x + y + z )
(x + y + z 2 )3/2
Gm1 m2
Gm1 m2
h(z) = 0, φ =
=
2
2
2
|r|
x +y +z
480
9.10
Double Integrals
28. Since Py = 24xy 2 z = Qx , Qz = 12x2 y 2 = Ry , and Rx = 8xy 3 = Pz , F is conservative. Thus, the work done
between two points is independent of the path. From φx = 8xy 3 z we obtain φ = 4x2 y 3 z which is a potential
function for F. Then
(1,√3 ,π/3)
(0,2,π/2)
(1,√3 ,π/3)
√
2 3 W =
F · dr = 4x y z = 4 3 π and W =
F · dr = 0.
(2,0,0)
(2,0,0)
F · dr =
29. Since F is conservative,
−C2
C1
C is composed of C1 and C2 ,
ˇ F · dr =
C
(2,0,0)
F · dr. Then, since the simply closed curve
F · dr +
C1
F · dr =
C2
F · dr −
C1
−C2
F · dr = 0.
30. From F = (x2 + y 2 )n/2 (xi + yj) we obtain Py = nxy(x2 + y 2 )n/2−1 = Qx , so that F is conservative. From
φx = x(x2 + y 2 )n/2 we obtain the potential function φ = (x2 + y 2 )(n+2)/2 /(n + 2). Then
(x ,y )
(x2 ,y2 )
(x2 + y 2 )(n+2)/2 2 2
1 2
W =
F · dr =
=
(x2 + y22 )(n+2)/2 − (x21 + y12 )(n+2)/2 .
n+2
n+2
(x1 ,y1 )
(x1 ,y1 )
31. From the solution to Problem 39 in Exercises 9.8,
dv dr
dv
1 d 2
·
=
· v =
v .
dt
dt
dt
2 dt
Then, using
dp
∂p dx ∂p dy
dr
=
+
= ∇p ·
, we have
dt
∂x dt
∂y dt
dt
dv dr
dr
m
·
dt + ∇p ·
= 0 dt
dt dt
dt
1
dp
d 2
m
v dt +
dt = constant
2
dt
dt
1
mv 2 + p = constant.
2
32. By Problem 31, the sum of kinetic and potential energies in a conservative force field is constant. That is, it is
independent of points A and B, so p(B) + K(B) = p(A) + K(A).
EXERCISES 9.10
Double Integrals
3
(6xy − 5ey ) dx = (3x2 y − 5xey ) 3
1.
−1
−1
= (27y − 15ey ) − (3y + 5ey ) = 24y − 20ey
2
1
1
ln | sec xy| = ln | sec 2x − sec x|
x
x
1
1
3x
3x
2
3.
x3 exy dy = x2 exy = x2 (e3x − ex )
2
2.
tan xy dy =
1
1
4.
y
√
3
y
y 3
(8x3 y − 4xy 2 ) dx = (2x4 y − 2x2 y 2 ) √ = (2y 13 − 2y 8 ) − (2y 3 − 2y 3 ) = 2y 13 − 2y 8
y
481
9.10 Double Integrals
2x
x
x
x
xy
dy = ln(x2 + y 2 ) = [ln(x2 + 4x2 ) − ln x2 ] = ln 5
2
+
y
2
2
2
0
0
x
x
3
2
x
x
x
6.
e2y/x dy = e2y/x = (e2x/x − e2x /x ) = (e2 − e2x )
3
2
2
2
x
3
x
sec y
sec y
7.
(2x + cos y) dx = (x2 + x cos y) = sec2 y + sec y cos y − tan2 y − tan y cos y
2x
5.
x2
tan y
tan y
= sec y + 1 − tan2 y − sin y = 2 − sin y
2
8.
1
√
y ln x dx
y
Integration by parts
1
= y(x ln x − x) √
9.
y
√
√
√
√
= y(0 − 1) − y( y ln y − y ) = −y − y y
10.
1
0
0
0
2
(x + 1) dy dx =
0
R
12.
x
1
1 3 3 1 1 6
1 7 1
x dx =
x y dx =
x =
3
3
21
21
0
0
0
4−x
(x + 1) dA =
2
4−x
(xy + y) dx
x
0
x
2
[(4x − x2 + 4 − x) − (x2 + x)] dx =
=
0
1
(2x − 2x2 + 4) dx
1
(2x + 4y + 1) dy dx =
x3
0
R
x2
(2x + 4y + 1) dA =
2
0
2
= 20
3
0
2
x − x3 + 4x
3
2
=
15.
1
x3 y 2 dy dx =
R
14.
11.
x
x3 y 2 dA =
13.
1
ln y − 1
2
x2
(2xy + 2y 2 + y) dx
x3
0
1
[(2x3 + 2x4 + x2 ) − (2x4 + 2x6 + x3 )] dx
=
0
1
(x + x − 2x ) dx =
3
=
2
6
0
1 4 1 3 2 7
x + x − x
4
3
7
1
0
1 1 2
25
= + − =
4 3 7
84
1 x
1
1
x
y
y
y 16.
xe dA =
xe dy dx =
xe dx =
(xex − x) dx
0
R
2
8
2xy dA =
R
0
Integration by parts
1
1
xex − ex − x2 =
2
0
=
17.
0
2xy dy dx =
0
x3
0
2
0
e−e−
0
1
2
− (−1) =
1
2
2
8
xy 2 3 dx =
(64x − x7 )dx =
x
0
482
1
32x2 − x8
8
2
= 96
0
9.10
18.
R
x
√ dA =
y
1
−1
3−x2
xy −1/2 dy dx =
1
−1
x2 +1
2
√ 3−x
2x y 2 dx
x +1
1
=2
−1
(x 3 − x2 − x x2 + 1 ) dx
1
1
1 2
2 3/2
3/2 = 2 − (3 − x ) − (x + 1)
3
3
−1
2
= − [(23/2 + 23/2 ) − (23/2 + 23/2 )] = 0
3
1 1
1
1
1
y
y
19.
dA =
dx dy =
ln(1 + xy) dy =
ln(1 + y) dy
0
R 1 + xy
0
0 1 + xy
0
0
1
= [(1 + y) ln(1 + y) − (1 + y)] = (2 ln 2 − 2) − (−1) = 2 ln 2 − 1
0
20.
πx
sin
dA =
y
R
21.
y2
πx
sin
dx dy =
y
1
0
2
y
y
=
− cos πy +
dy
π
π
1
−
=
2
1
0
√
2
=
−
1
√
3
x2 + 1 dy dx =
−x
y
0
3
0
1
2
+
π3
π
1
1
+
π3
2π
−
x
x2 + 1 dx
(x x2 + 1 + x x2 + 1 ) dx =
=
0
Integration by parts
x
y 2
dy
y
πx
− cos
π
y
y
1
y2
sin πy − 3 cos πy +
2
π
π
2π
3π 2 − 4
=
2π 3
√3 x2 + 1 dA =
R
2
√
−x
3
2x x2 + 1 dx
0
√3
2 2
2
14
= (x + 1)3/2 = (43/2 − 13/2 ) =
3
3
3
0
1
π/4 1
π/4
π/4
1 2 1
22.
x dA =
x dx dy =
dy =
(1 − tan2 y) dy
x 2
2 0
R
0
tan y
0
tan y
π/4 1 π
π 1
1 π/4
1
=
(2 − sec2 y) dy = (2y − tan y) =
−1 = −
2 0
2
2 2
4
2
0
23. The correct integral is (c).
√ 2
2
(4 − y) dx dy = 2
V =2
−2
= 2 2y
4−y
0
4 − y 2 + 8 sin−1
2
−2
√4−y2
(4 − y)x dy = 2
0
2
−2
(4 − y) 4 − y 2 dy
2
y 1
+ (4 − y 2 )3/2 = 2(4π − (−4π)] = 16π
2 3
−2
24. The correct integral is (b).
2 √4−y2
2
2
√4−y2
2 1/2
2 1/2 V =8
(4 − y ) dx dy = 8
(4 − y ) x dy = 8
(4 − y 2 ) dy
0
0
1
= 8 4y − y 3
3
2
= 128
3
0
0
0
483
0
Double Integrals
9.10 Double Integrals
25. Setting z = 0 we have y = 6 − 2x.
3 6−2x
V =
(6 − 2x − y) dy dx =
6−2x
1 2 6y − 2xy − y dx
2
0
0
0
0
3
3
1
2
=
[6(6 − 2x) − 2x(6 − 2x) − (6 − 2x) ] dx =
(18 − 12x + 2x2 ) dx
2
0
0
3
2
= 18x − 6x2 + x3 = 18
3
0
3
26. Setting z = 0 we have y = ±2.
3 2
3
1
V =
4y − y 3
(4 − y 2 ) dy dx =
3
0
0
0
2
dx =
0
0
3
16
dx = 16
3
27. Solving for z, we have x = 2 − 12 x + 12 y. Setting z = 0, we see that this surface (plane) intersects
the xy-plane in the line y = x − 4. since z(0, 0) = 2 > 0, the surface lies above the xy-plane over
the quarter-circular region.
√4−x2
2 √4−x2
2
1
1
1
1 2 V =
2 − x + y dy dx =
2y − xy + y dx
2
2
2
4
0
0
0
0
2
2
1
1
1 2
1 3 −1 x
2 3/2
2
2
2
=
2 4 − x − x 4 − x + 1 − x dx = x 4 − x + 4 sin
+ (4 − x ) + x − x 2
4
2 6
12
0
0
=
2π + 2 −
2
3
−
4
= 2π
3
28. Setting z = 0 we have y = 3. Using symmetry,
3
√3 3
√3
√3
1 2 9
1
V =2
− 3x2 + x4 dx
3y − y dx = 2
(3 − y)dy dx = 2
2
2
2
2
0
x
0
0
x2
√
√ 3
√
9
1 5 9√
9√
24 3
3
=2
3−3 3+
3 =
x−x + x =2
.
2
10
2
10
5
0
29. Note that z = 1 + x2 + y 2 is always positive. Then
3−3x
1 3−3x
1
1 3 2
2
2
y+x y+ y V =
(1 + x + y )dy dx =
dx
3
0
0
0
0
1
1
=
[(3 − 3x) + x2 (3 − 3x) + 9(1 − x)3 ] dx =
(12 − 30x + 30x2 − 12x3 ) dx
0
1
= (12x − 15x2 + 10x3 − 3x4 ) = 4.
0
0
30. In the first octant, z = x + y is nonnegative. Then
√9−x2
3 √9−x2
3
1 2 V =
xy + y (x + y) dy dx =
dx
2
0
0
0
0
3
3
1
9 1 2
9
1 3 2 3/2
2
=
x 9 − x + − x dx = − (9 − x ) + x − x =
2 2
3
2
6
0
0
484
27 9
−
2
2
− (−9) = 18.
9.10
Double Integrals
31. In the first octant z = 6/y is positive. Then
5
6 5
6
6
6
6
6x dy
V =
dy
=
30
dx dy =
=
30
ln
y
= 30 ln 6.
y 0
1
1
0 y
1
1 y
32. Setting z = 0, we have x2 /4 + y 2 /16 = 1. Using symmetry,
2
√
2 4−x2
2√4−x2
1 3 4y − x y − y dx
12
0
0
2
2 3/2
2
dx
4 − x − (4 − x )
Trig substitution
3
1
4 − x − y 2 dy dx = 4
4
2
V =4
0
0
2
=4
8
4 − x2 − 2x2
0
2
2
2
1
1
x −1 x
−1 x
2
2
2
2
2
− x(2x − 4) 4 − x − 4 sin
+ x(2x − 20) 4 − x − 4 sin
= 4 4x 4 − x + 16 sin
2 4
2 12
2 0
=4
16π 4π 4π
−
−
2
2
2
− (0) = 16π.
33. Note that z = 4 − y 2 is positive for |y| ≤ 1. Using symmetry,
√2x−x2
2 √2x−x2
2
1 3 2
V =2
(4 − y ) dy dx = 2
dx
4y − y 3
0
0
0
0
2
1
2
2
2
=2
4 2x − x − (2x − x ) 2x − x dx
3
0
2
1
=2
4 1 − (x − 1)2 − [1 − (x − 1)2 ] 1 − (x − 1)2 dx
u = x − 1, du = dx
3
0
1
1
11
1
1
2
2
2
4 1 − u − (1 − u ) 1 − u du = 2
=2
1 − u2 + u2 1 − u2 du
3
3
3
−1
−1
Trig substitution
11
11
=2
u 1 − u2 +
sin u +
6
6
11 π
1 π
11
=2
+
− −
6 2
24 2
6
1
1
1
x(2x2 − 1) 1 − u2 +
sin−1 u 24
24
−1
π
1 π
15π
−
=
.
2
24 2
4
34. From z = 1 − x2 and z = 1 − y 2 we have 1 − x2 = 1 − y 2 or y = x (in the first octant).
Thus, the surfaces intersect in the plane y = x. Using symmetry,
1
1 1
1
1
1
1 3 2
2
V =2
− x + x3 dx
y − y dx = 2
(1 − y ) dy dx = 2
3
3
3
0
x
0
0
x
1
2
1
1
1
=2
x − x2 + x4 = .
3
2
12
2
0
1
1
2
35.
x
0
x
1+
y4
1
y
2
dy dx =
x
0
1
=
3
0
1
y3
0
1+
y4
1
dx dy =
0
1 3
x
3
1+
y4
y
dy
0
1
1 1
1 √
4 3/2 4
1 + y dy =
=
(1 + y )
(2 2 − 1)
3 6
18
0
485
9.10 Double Integrals
1
2
36.
0
e−y/x dx dy =
2y
2
0
0
2
e−y/x dy dx =
y2
0
√
x
0
0
4
38.
√
1 − x2 − y 2 dy dx =
1
1
40.
0
−
√
x 1 − x2 − y 2 dx dy
1−y 2
2
√
1
y
1
dx dy =
4
0
0 1+y
1
1
π
−1 2 = tan y =
2
8
0
x
4
−1
√1−y2
1
1
dy dx =
1 + y4
39.
0
1
√1−y2
1
1 1
2
2 3/2 =
− (1 − x − y )
(0 − 0)dy = 0
√ 2 dy = − 3
3
−1
−1
− 1−y
0
4
= 2 sin 8
3
0
2
sin x3/2
3
x cos x3/2 dx =
1−x2
x
√
− 1−x2
−1
0
√x
y cos x3/2 dx
4
0
0
1
0
cos x3/2 dy dx =
=
x/2 2
−xe−y/x =
(−xe−1/2 + x) dx
0
4
√
2
0
cos x3/2 dx dy =
37.
2
1
−1/2
−1/2 2 (1 − e
)x dx = (1 − e
)x = 2(1 − e−1/2 )
2
4
x/2
0
2
=
2
x3 + 1 dx dy =
1
0
y
1
x y
dy
=
dy
4
1 + y 4 0
0 1+y
x2
2
x3 + 1 dy dx =
0
0
y
y
0
x2
x3 + 1 dx
0
2
= 2 (93/2 − 13/2 ) = 52
9
9
0
2
2 3
(x + 1)3/2
9
0
4
3 4
3
3
3
1 2 41. m =
xy dx dy =
8y dy = 4y 2 = 36
x y dy =
0
0
0
0 2
0
0
3
4
3 4
3
3
32 2 1 3 64
2
My =
x y dy =
y dy =
y = 96
x y dx dy =
3
0
0
0 3
0 3
0
0
4 3
3
3 4
3
1 2 2 8 Mx =
x y =
xy 2 dx dy =
8y 2 dy = y 3 = 72
2
3
0
0
0
0
0
0
=
x2
x3 + 1 dx =
x̄ = My /m = 96/36 = 8/3; ȳ = Mx /m = 72/36 = 2. The center of mass is (8/3, 2).
2 4−2x
2
2
4−2x
42. m =
x2 dy dx =
x2 y dx =
x2 (4 − 2x) dx
0
0
0
0
0
2
8
4 3 1 4 32
2
3
=
x − x =
−8=
(4x − 2x ) dx =
3
2
3
3
0
0
2 4−2x
2
2
2
4−2x
3
3 3
My =
x dy dx =
x y
dx =
x (4 − 2x) dx =
(4x3 − 2x4 ) dx
2
0
0
0
0
0
2
2
64
16
= x4 − x5 = 16 −
=
5
5
5
0
2 4−2x
2
1 2 2 4−2x
2
x y dy dx =
dx =
Mx =
x y 0
0
0 2
0
2
1
4 3
=2
(4x2 − 4x3 + x4 ) dx = 2
x − x4 + x5
3
5
0
0
1 2 2
1 2
2
x (4 − 2x) dx =
(16x2 − 16x3 + 4x4 ) dx
2 0
2 0
2
= 2 32 − 16 + 32 = 32
3
5
15
0
486
9.10
x̄ = My /m =
Double Integrals
16/5
32/15
= 6/5; ȳ = Mx /m =
= 4/5. The center of mass is (6/5, 4/5).
8/3
8/3
43. Since both the region and ρ are symmetric with respect to the line x = 3, x̄ = 3.
3 6−y
3
3
3
6−y
m=
2y dx dy =
2xy dy =
2y(6 − y − y) dy =
(12y − 4y 2 ) dy
0
y
3
= 18
y
0
4
6y 2 − y 3
3
0
3 6−y
2
Mx =
2y dx dy =
=
0
y
3
= (4y 3 − y 4 ) = 27
3
0
0
6−y
2xy dx dy =
2
y
0
3
2y (6 − y − y) dy =
0
3
(12y 2 − 4y 3 ) dy
2
0
0
ȳ = Mx /m = 27/18 = 3/2. The center of mass is (3, 3/2).
44. Since both the region and ρ are symmetric with respect to the y-axis, x̄ = 0. Using
symmetry,
y
3 y
3
3
1 3
1 3
m=
(x2 + y 2 ) dx dy =
x + xy 2 dy =
y + y 3 dy
3
3
0
0
0
0
0
3
4 3 3
1 4 =
y dy = y = 27.
3 0
3
0
y
3 y
3
3
4 3 4
1 3
1 4
2
3
3
4
Mx =
(x y + y ) dx dy =
y dy
x y + xy dy =
y + y dy =
3
3
3 0
0
0
0
0
0
3
4 5 324
=
y
=
15 0
5
324/5
= 12/5. The center of mass is (0, 12/5).
27
x2
1 x2
1
1
1 2 1
45. m =
xy + y dx =
x3 + x4 dx
(x + y) dy dx =
2
2
0
0
0
0
0
1
1 4
1
7
=
x + x5 =
4
10
20
0
x2
1 x2
1
1
1 2 1
2
2
My =
x y + xy dx =
x4 + x5 dx
(x + xy) dy dx =
2
2
0
0
0
0
0
1
1 5
1
17
=
x + x6 =
5
12
60
0
x2 1
1 x2
1
1 2 1 3 1 5 1 6
2
Mx =
xy + y =
x + x dx =
(xy + y ) dy dx =
2
3
2
3
0
0
0
0
0
ȳ = Mx /m =
x̄ = My /m =
4
46. m =
1
= 11
84
0
17/60
11/84
= 17/21; ȳ = Mx /m =
= 55/147. The center of mass is (17/21, 55/147).
7/20
7/20
x
(y + 5) dy dx =
0
=
√
1 6
1
x + x7
12
21
0
1 2 10 3/2
x + x
4
3
0
4
= 92
3
0
4
1 2
y + 5y
2
√x
dx =
0
0
487
4
√
1
x+5 x
2
dx
9.10 Double Integrals
4
√
My =
x
4
(xy + 5x) dy dx =
0
0
4
= 224
3
0
0
1 3
x + 2x5/2
6
4 √x
Mx =
(y 2 + 5y) dy dx =
=
0
0
2 5/2 5 2
x + x
15
4
=
x̄ = My /m =
4
= 364
15
0
4
0
1 2
xy + 5xy
2
1 3 5 2
y + y
3
2
√ x
dx =
0
0
√ x
dx =
0
0
4
4
1 2
x + 5x3/2 dx
2
1 3/2 5
x + x dx
3
2
224/3
364/15
= 56/23; ȳ = Mx /m =
= 91/115. The center of mass is (56/23, 91/115).
92/3
92/3
47. The density is ρ = ky. Since both the region and ρ are symmetric with respect to the
y-axis, x̄ = 0. Using symmetry,
1−x2
1 1−x2
1
1
1 2 m=2
ky dy dx = 2k
dx = k
(1 − x2 )2 dx
y 0
0
0 2
0
0
1
1
2
1
2 1
8
=k
(1 − 2x2 + x4 ) dx = k x − x3 + x5 = k 1 − +
=
k
3
5
3
5
15
0
0
1−x2
1 1−x2
1
1
1
1 3 2
2
2
2 3
Mx = 2
ky dy dx = 2k
dx = k
(1 − x ) dx = k
(1 − 3x2 + 3x4 − x6 ) dx
y 3 0
3 0
0
0
0 3
0
1
2
3 5 1 7 2
3 1
32
3
= k x−x + x − x = k 1−1+ −
=
k
3
5
7
3
5
7
105
0
ȳ = Mx /m =
32k/105
= 4/7. The center of mass is (0, 4/7).
8k/15
48. The density is ρ = kx.
π sin x
m=
kx dy dx =
0
0
0
π
sin x
kxy dx =
0
π
kx sin x dx
0
Integration by parts
π
= k(sin x − x cos x) = kπ
0
π sin x
π
π
sin x
My =
kx2 dy dx =
kx2 y dx =
kx2 sin x dx
Integration by parts
0
0
0
0
0
π
= k(−x2 cos x + 2 cos x + 2x sin x) = k[(π 2 − 2) − 2] = k(π 2 − 4)
0
sin x
π sin x
π
π
π
1
1
1
2
2 Mx =
kxy dy dx =
dx =
kxy kx sin x dx =
kx(1 − cos 2x) dx
0
0
0 2
0 2
0 4
0
π
π
1
= k
x dx −
x cos 2x dx
Integration by parts
4
0
0
π 1
1
1 2
1 1 π 1
= k
π = kπ 2
= k x2 − (cos 2x + 2x sin 2x) 4 2
4
4
2
8
0
0
x̄ = My /m =
k(π 2 − 4)
kπ 2 /8
= π − 4/π; ȳ = Mx /m =
= π/8. The center of mass is (π − 4/π, π/8).
kπ
kπ
488
9.10
Double Integrals
ex
1
1
1 4 1 4x
1 4x 1 4
49. m =
y dy dx =
y dx =
e dx =
e =
(e − 1)
4
4
16
16
0
0
0
0
0
0
ex
1 ex
1
1
1 4 1 4x
My =
xy 3 dy dx =
Integration by parts
xy dx =
xe dx
4
0
0
0
0 4
0
1
1 1 4x
1
1 3 4
1
1
=
xe − e4x =
e +
=
(3e4 + 1)
4 4
16
4
16
16
64
0
ex
1
1 ex
1
1
1 5 1 5x
1 5x 1 5
4
Mx =
y dy dx =
y dx =
e dx =
e =
(e − 1)
25
25
0
0
0 5
0 5
0
0
1
ex
1
3
x̄ = My /m =
(3e4 + 1)/64
(e5 − 1)/25
3e4 + 1
16(e5 − 1)
=
; ȳ = Mx /m = 4
=
4
4
(e − 1)/16
4(e − 1)
(e − 1)/16
25(e4 − 1)
The center of mass is
3e4 + 1 16(e5 − 1)
,
4(e4 − 1) 25(e4 − 1)
≈ (0.77, 1.76).
50. Since both the region and ρ are symmetric with respect to the y-axis, x̄ = 0. Using
symmetry,
3
m=2
√
9−x2
2
√9−x2
x y
dx = 2
3
2
x dy dx = 2
0
0
3
0
0
x2
9 − x2 dx
0
Trig substitution
3
81π
x
81
81 π
−1 x 2
2
(2x − 9) 9 − x +
sin
=
.
=2
=
8
8
3 0
4 2
2
√9−x2
3 √9−x2
3
3
1 2 2 2
Mx = 2
x y dy dx = 2
dy dx =
x2 (9 − x2 ) dx =
x y 2
0
0
0
0
0
162/5
ȳ = Mx /m =
= 16/5π. The center of mass is (0, 16/5π).
81π/8
1 y−y2
1
1
y−y2
51. Ix =
2xy 2 dx dy =
x2 y 2 dy =
(y − y 2 )2 y 2 dy
0
0
0
1
(y 4 − 2y 5 + y 6 ) dy =
=
0
1
√
0
=
x
x2 y 2 dy dx =
52. Ix =
1
3
x2
2 9/2 1 9
x − x
9
9
0
1
0
0
1
1 5 1 6 1 7 1
y − y + y =
5
3
7
105
0
√ x
1
1 2 3 1
(x7/2 − x8 ) dx
x y dx =
3
3
2
0
x
1
= 1
27
0
53. Using symmetry,
π/2 cos x
2
Ix = 2
ky dy dx = 2k
cos x
π/2
1 3 2
dx = k
cos3 x dx
y
3 0
3 0
0
0
0
π/2
π/2
2
2
4
1
2
3
= k
cos x(1 − sin x) dx = k sin x − sin x = k.
3 0
3
3
9
0
√
√
4−x2
2 4−x2
2
1 4 1 2
y 3 dy dx =
dx =
(4 − x2 )2 dx
54. Ix =
y 4
4
0
0
0
0
0
2
1 2
1
1
1
8
64 32
=
(16 − 8x2 + x4 ) dx =
16x − x3 + x5 =
32 −
+
4 0
4
3
5
4
3
5
0
π/2
489
1
3x2 − x5
5
3
= 162
5
0
9.10 Double Integrals
=8 1−
4
√
=
y
x y dx dy =
0
0
1
3
0
1
√
x
1
x4 dy dx =
56. Iy =
x2
0
1
√y
1 3 1 4 3/2
1 4 5/2
x y dy =
y y dy =
y dy
3
3 0
3 0
0
4
= 2 (47/2 ) = 256
21
21
0
2 7/2
y
7
64
15
4
2
55. Iy =
=
2 1
+
3 5
1
√x
x4 y dx =
(x9/2 − x6 ) dx =
2
x
0
3
3
57. Iy =
0
1
2
=
0
y
81y +
27 2 2 5
y − y
2
5
1
= 941
10
0
1
= 3
77
0
1
3
(x + x y) dy =
(81 + 27y − 2y 4 ) dy
4
(4x + 3x y) dx dy =
0
2 11/2 1 7
− x
x
11
7
3
y
0
58. The density is ρ = ky. Using symmetry,
1−x2
1 1−x2
1
1
1 2 2 2
Iy = 2
kx y dy dx = 2
dx = k
x2 (1 − x2 )2 dx
kx y 0
0 2
0
0
0
1
1
1
2
1 7 8k
2
4
6
3
5
=k
(x − 2x + x ) dx = k
x − x + x =
.
3
5
7
105
0
0
59. Using symmetry,
a √a2 −y2
m=2
x dx dy = 2
0
0
a
= 2 a3 .
3
0
1
a2 y − y 3
3
√
a a2 −y2
=
a
0
√a2 −y2
a
1 2 dy =
(a2 − y 2 ) dy
x 2
0
0
√a2 −y2
1 4 1 a 2
Iy = 2
x dx dy = 2
dy =
(a − y 2 )2 dy
x 4
2
0
0
0
0
0
a
1 a 4
1
2
1
4 5
2 2
4
4
2 3
5
=
(a − 2a y + y ) dy =
a y − a y + y =
a
2 0
2
3
5
15
0
Iy
4a5 /15
2
Rg =
=
=
a
3
m
2a /3
5
a
3
a
a−x
1
60. m =
k dy dx =
ky dx = k
(a − x) dx = k ax − x2
2
0
0
0
0
0
a−x
a a−x
a
a
1
1 3 Ix =
ky 2 dy dx =
dx = k
(a − x)3 dx
ky 3 0
0
0
0 3
0
a
1
1
3
1
= k
(a3 − 3a2 x + 3ax2 − x3 ) dx = k a3 x − a2 x2 + ax3 − x4
3 0
3
2
4
Ix
ka4 /12
1
=
=
a
Rg =
m
ka2 /2
6
a
a−x
a
490
a
= 1 ka2
2
0
a
= 1 ka4
12
0
9.10
Double Integrals
61. (a) Using symmetry,
a b√a2 −x2 /a
4b3 a 2
2
Ix = 4
y dy dx = 3
(a − x2 )3/2 dx
x = a sin θ, dx = a cos θ dθ
3a 0
0
0
π/2
π/2
π/2
1 1
1
4
4
1
1 + cos 2θ + + cos 4θ dθ
= ab3
cos4 θ dθ = ab3
(1 + cos 2θ)2 dθ = ab3
3
3
4
3
2 2
0
0
0
π/2
1
3
ab3 π
1
1
= ab3
θ + sin 2θ + sin 4θ =
.
3
2
2
8
4
0
(b) Using symmetry,
a b√a2 −x2 /a
4b a 2
2
Iy = 4
x dy dx =
x a2 − x2 dx
x = a sin θ, dx = a cos θ dθ
a 0
0
0
π/2
π/2
1
(1 − cos2 2θ) dθ
= 4a3 b
sin2 θ cos2 θ dθ = 4a3 b
4
0
0
π/2
π/2
1 1
1
a3 bπ
1
3
3
1 − − cos 4θ dθ = a b
=a b
θ − sin 4θ =
.
2 2
2
8
4
0
0
(c) Using m = πab, Rg =
(d) Rg =
1
2
Iy /m =
Ix /m =
a3 bπ/πab =
1
2
ab3 π/πab =
1
b.
2
1
a
2
62. The equation of the ellipse is 9x2 /a2 + 4y 2 /b2 = 1 and the equation of the parabola is y = ±(9bx2 /8a2 − b/2).
Letting Ie and Ip represent the moments of inertia of the ellipse and parabola, respectively, about the x-axis,
we have
Ie = 2
−a/3
3
=
4
b a
12a3 3
and
√
b a2 −9x2 /2a
0
2a/3
0
0
cos4 θ dθ =
−π/3
b/2−9bx2 /8a2
b3
12a3
3
0
0
0
2a/3
0
−a/3
(a2 − 9x2 )3/2 dx
a
a
sin θ, dx = cos θ dθ
3
3
b a 3π
ab π
=
36 16
192
2
y dy dx =
3
0
2a/3
9b 2
b
−
x
2 8a2
27
b3
243 4
729 6
1 − 2 x2 +
dx
=
x
−
x
4a
16a4
64a6
12
3
2 b3
dx =
3 8
ab3 π 8ab3
+
.
192
315
1
1 4
63. From the solution to Problem 60, m = ka2 and Ix =
ka .
2
12
a a−x
a
a−x
a
Iy =
kx2 dy dx =
kx2 y dx = k
x2 (a − x) dx
0
0
0
0
0
a
1 3 1 4 1 4
=k
ax − x =
ka
3
4
12
0
1 4
1
1
I0 = Ix + Iy =
ka + ka4 = ka4
12
12
6
491
2a/3
1−
0
9 2
x
4a2
243 5
729 7
9
x −
x
x − 2 x3 +
4a
80a4
64a6
b3 32a
8ab3
=
=
.
12 105
315
Then Ix = Ie + Ip =
x=
3
2
Ip = 2
b3
=
12
y 2 dy dx =
3
dx
2a/3
0
9.10 Double Integrals
64. From the solution to Problem 52, Ix =
3
158
1
+
=
.
27 77
2079
Ix + Iy =
1
3
, and from the solution to Problem 56, Iy =
. Thus, I0 =
27
77
65. The density is ρ = k/(x2 + y 2 ). Using symmetry,
√2 6−y2
√2
6−y2
k
2
2
I0 = 2
(x + y ) 2
dx
dy
=
2
kx
2 dy
2
x
+
y
y +2
2
0
0
y +2
√
2
= 2k
0
3
2
(6 − y 2 − y 2 − 2) dy = 2k 4y − y 3
3
4
3
k(x2 + y 2 ) dx dy = k
66. I0 =
0
y
=k
0
3
0
1 3
x + xy 2
3
64
1
+ 4y 2 − y 3 − y 3 dy = k
3
3
√ 2
= 2k
0
8√
2
3
√
16 2
=
k.
3
4
dy
y
64
1
4
y + y3 − y4
3
3
3
3
= 73k
0
1
1
67. From the solution to Problem 60, m = ka2 , and from the solution to Problem 63, I0 = ka4 . Then
2
6
ka4 /6
1
Rg = I0 /m =
=
a.
ka2 /2
3
68. Since the plate is homogeneous, the density is ρ = m/ ω. Using symmetry,
ω/2
/2 ω/2
m 2
4m /2
1
I0 = 4
x2 y + y 3 dx
(x + y 2 ) dy dx =
ω
ω 0
3
0
0
0
/2
4m /2 ω 2 ω 3
4m ω 3 ω 3
ω3
4m ω 3
=
x +
dx =
x +
x +
=
ω 0
2
24
ω 6
24
ω
48
48
0
2
=m
+ ω2
.
12
EXERCISES 9.11
Double Integrals in Polar Coordinates
1. Using symmetry,
π/2 3+3 sin θ
A=2
r dr dθ = 2
3+3 sin θ
π/2
1 2 dθ =
9(1 + sin θ)2 dθ
r 2
−π/2 0
−π/2
−π/2
0
π/2
π/2
1
1
2
=9
(1 + 2 sin θ + sin θ) dθ = 9 θ − 2 cos θ + θ − sin 2θ 2
4
−π/2
−π/2
π
27π
3 π 3
−
−
=
=9
2 2
2
2
2
π/2
492
9.11 Double Integrals in Polar Coordinates
2. Using symmetry,
π 2+cos θ
A=2
r dr dθ = 2
2+cos θ
π
1 2 dθ =
(2 + cos θ)2 dθ
r 0
0
0 2
0
0
π
π
1
1
2
=
(4 + 4 cos θ + cos θ) dθ = 4θ + 4 sin θ + θ + cos 2θ 2
4
π
0
0
π 1
4π + +
2
4
=
1
4
−
9π
=
.
2
3. Solving r = 2 sin θ and r = 1, we obtain sin θ = 1/2 or θ = π/6. Using symmetry,
π/6 2 sin θ
π/2 1
A=2
r dr dθ + 2
r dr dθ
0
0
0
π/6
2 sin θ
1
π/6
π/2
π/6
π/2
1 2 1 2 =2
dθ + 2
4 sin2 θ dθ +
dθ
r r dθ =
2
0
π/6 2
0
π/6
0
0
√
π/6 π π π √3 π
4π − 3 3
= (2θ − sin 2θ) +
−
= −
+ =
2
6
3
2
3
6
0
π/4
8 sin 4θ
4. A =
8 sin 4θ
1 2 1 π/4
dθ =
64 sin2 4θ dθ
r
2 0
2 0
π/4
r dr dθ =
0
0
0
1
1
θ−
sin 8θ
2
16
= 32
π/4
= 4π
0
5. Using symmetry,
π/6 5 cos 3θ
V =2
4r dr dθ = 4
0
0
2π
2π
9 − r2 r dr dθ =
0
0
1
=−
3
0
2π
3/2
(5
0
2π
16 − r2 r dr dθ =
0
1
=−
3
1
2π
(7
0
8. V =
0
0
2π
3/2
0
5
√
25 cos2 3θ dθ
0
2
1
− (9 − r2 )3/2 dθ
3
0
√
1
2π(27 − 5 5 )
3/2
− 27) dθ = (27 − 5 )2π =
3
3
3
7. V =
π/6
π/6
= 25π
3
0
2
6. V =
5 cos 3θ
r dθ = 4
2
0
0
1
1
θ+
sin 6θ
2
12
= 100
π/6
− 15
3/2
2π
3
1
− (16 − r2 )3/2 dθ
3
1
√
√
1
2π(15 15 − 7 7 )
3/2
3/2
) dθ = (15 − 7 )2π =
3
3
r2 r dr dθ =
0
2π
5
2π
1 3 125
250π
r dθ =
dθ =
3
3
3
0
0
493
9.11 Double Integrals in Polar Coordinates
1+cos θ
1 3
dθ
r sin θ 3
0
0
0
0
π/2
1 π/2
1
1
5
1
=
(1 + cos θ)3 sin θ dθ =
− (1 + cos θ)4 = − (1 − 24 ) =
3 0
3
4
12
4
0
π/2
1+cos θ
9. V =
π/2
(r sin θ)r dr dθ =
10. Using symmetry,
π/2 cos θ
2
V =2
(2 + r ) r dr dθ =
0
0
π/2
=2
0
2 cos2 θ +
0
0
1
cos2 θ + cos4 θ dθ = 2
4
π/2
=
π/2
1
r + r4
4
2
0
π/2
cos θ
dθ
0
1
cos2 θ +
4
1 + cos 2θ
2
2
dθ
1 1
1
+ cos 2θ + cos2 2θ dθ
8 4
8
π/2
1
19π
1
1
1
1
sin 2θ + θ + sin 2θ + θ +
sin 4θ =
.
2
8
8
16
64
32
0
3
π/2 3
π/2
π/2
1 2 1
11. m =
kr dr dθ = k
8 dθ = 2kπ
r dθ = k
2
2 0
0
1
0
1
3
π/2 3
π/2 3
π/2
1 3
My =
kxr dr dθ = k
r2 cos θ dr dθ = k
r cos θ dθ
3
0
1
0
1
0
1
π/2
π/2
1
26
26
= k
26 cos θ dθ =
k sin θ =
k
3 0
3
3
0
13
26k/3
x̄ = My /m =
=
. Since the region and density function are symmetric about the ray θ = π/4,
2kπ
3π
=
θ+
ȳ = x̄ = 13/3π and the center of mass is (13/3π, 13/3π).
12. The interior of the upper-half circle is traced from θ = 0 to π/2. The density is kr.
Since both the region and the density are symmetric about the polar axis, ȳ = 0.
cos θ
π/2 cos θ
π/2
1 3 k π/2
2
m=
r kr dr dθ = k
dθ =
cos3 θ dθ
3
3 0
0
0
0
0
π/2
k 2 1
2k
2
=
+ cos θ sin θ =
3 3 3
9
0
π/2 cos θ
π/2 cos θ
π/2
1 4
My = k
(r cos θ)(r)(r dr dθ) = k
r3 cos θ dr dθ = k
r cos θ
4
0
0
0
0
0
π/2
k π/2
k
1
2k
2
3
5
5
=
cos θ dθ =
sin θ − sin θ + sin θ =
4 0
4
3
5
15
0
Thus, x̄ =
cos θ
dθ
2k/15
= 3/5 and the center of mass is (3/5, 0).
2k/9
13. In polar coordinates the line x = 3 becomes r cos θ = 3 or r = 3 sec θ. The angle of
√
inclination of the line y = 3 x is π/3.
3 sec θ
π/3 3 sec θ
π/3
1 4 81 π/3
m=
r2 r dr dθ =
dθ =
sec4 θ dθ
r 4
4
0
0
0
0
0
π/3
π/3
√
81
81
81 √
81 √
1
=
(1 + tan2 θ) sec2 θ dθ =
3
tan θ + tan3 θ =
( 3 + 3) =
4 0
4
3
4
2
0
494
0
9.11 Double Integrals in Polar Coordinates
π/3
3 sec θ
π/3
My =
4
xr r dr dθ =
0
0
3 sec θ
2
r cos θ dr dθ =
0
0
0
π/3
3 sec θ
1 5
r cos θ dθ
5
0
243
243
243 √
486 √
=
(2 3 ) =
sec5 θ cos θ dθ =
sec4 θ dθ =
3
5 0
5 0
5
5
3 sec θ
π/3 3 sec θ
π/3 3 sec θ
π/3
1 5
Mx =
yr2 r dr dθ =
r4 sin θ dθ =
r sin θ 5
0
0
0
0
0
0
π/3
π/3
π/3
243
243
243
=
sec5 θ sin θ dθ =
tan θ sec4 θ dθ =
tan θ(1 + tan2 θ) sec2 θ dθ
5 0
5 0
5 0
π/3
243 π/3
243 1
1
243 3 9
729
=
(tan θ + tan3 θ) sec2 θ dθ =
tan2 θ + tan4 θ =
+
=
5 0
5
2
4
5
2
4
4
0
√
√
√
My
Mx
486 3/5
729/4
√
x̄ =
= 12/5; ȳ =
= 3 3/2. The center of mass is (12/5, 3 3/2).
=
= √
m
m
81 3/2
81 3/2
π/3
π/3
14. Since both the region and the density are symmetric about the x-axis, ȳ = 0. Using
symmetry,
4 cos 2θ
π/4 4 cos 2θ
π/4
π/4
1 2 m=2
kr dr dθ = 2k
dθ = 16k
cos2 2θ dθ
r
2 0
0
0
0
0
π/4
1
1
= 16k
θ + sin 4θ = 2kπ
2
8
0
4 cos 2θ
π/4 4 cos 2θ
π/4 4 cos 2θ
π/4
1 3
2
My = 2
kxr dr dθ = 2k
r cos θ dr dθ = 2k
dθ
r cos θ 3
0
0
0
0
0
0
π/4
π/4
128
128
3
=
cos 2θ cos θ dθ =
(1 − 2 sin2 θ)3 cos θ dθ
k
k
3
3
0
0
π/4
π/4
128
128
12
8
=
(1 − 6 sin2 θ + 12 sin4 θ − 8 sin6 θ) cos θ dθ =
k
k sin θ − 2 sin3 θ +
sin5 θ − sin7 θ 3
3
5
7
0
0
√
√
√
√ √
2
2 3 2
2
1024
128
k
−
+
−
=
=
2k
3
2
2
10
14
105
√
√
√
512 2
1024 2 k/105
x̄ = My /m =
=
. The center of mass is (512 2/105π, 0) or approximately (2.20, 0).
2kπ
105π
15. The density is ρ = k/r.
π/2 2+2 cos θ
π/2 2+2 cos θ
k
m=
dr dθ
r dr dθ = k
r
0
2
0
2
π/2
π/2
=k
2 cos θ dθ = 2k(sin θ) = 2k
0
π/2
0
2+2 cos θ
My =
x
0
=
1
k
2
2
k
r dr dθ = k
r
π/2
2+2 cos θ
π/2
r cos θ dr dθ = k
0
2
π/2
0
π/2
(2 cos2 θ + cos θ − sin2 θ cos θ) dθ
(8 cos θ + 4 cos2 θ) cos θ dθ = 2k
0
1
1
= 2k θ + sin 2θ + sin θ − sin3 θ
2
3
0
π/2
= 2k
0
2+2 cos θ
1 2 cos θ dθ
r 2
2
π 2
+
2
3
495
=
3π + 4
k
3
9.11 Double Integrals in Polar Coordinates
2+2 cos θ
1 2 Mx =
r
r sin θ dr dθ = k
sin θ dθ
2 2
0
2
0
2
0
π/2
π/2
1
1
4
1
4
8
2
2
3
= k
(8 cos θ + 4 cos θ) sin θ dθ = k −4 cos θ − cos θ = k − −4 −
= k
2 0
2
3
2
3
3
0
π/2
2+2 cos θ
k
y r dr dθ = k
r
π/2
2+2 cos θ
π/2
(3π + 4)k/3
8k/3
3π + 4
4
=
; ȳ = Mx /m =
= . The center of mass is ((3π + 4)/6, 4/3).
2k
6
2k
3
2+2 cos θ
π 2+2 cos θ
π
π
1 2 16. m =
r kr dr dθ = k
dθ = 2k
(1 + cos θ)2 dθ
0
0
0 2
0
0
π
π
1
1
2
= 2k
(1 + 2 cos θ + cos θ) dθ = 2k θ + 2 sin θ + θ + sin 2θ = 3πk
2
4
0
0
2+2 cos θ
π 2+2 cos θ
π 2+2 cos θ
π
1 3 2
My =
r kxr dr dθ = k
r cos θ dr dθ = k
cos θ dθ
0
0
0
0
0 3
0
π
π
8
8
3
= k
(1 + cos θ) cos θ dθ = k
(cos θ + 3 cos2 θ + 3 cos3 θ + cos4 θ) dθ
3 0
3 0
π
8
3
3
1
1
3
3
= 8 k 15 π = 5πk
= k sin θ +
θ + sin 2θ + (3 sin θ − sin θ) +
θ + sin 2θ +
sin 4θ
3
2
4
8
4
32
3
8
0
2+2 cos θ
π 2+2 cos θ
π 2+2 cos θ
π
1 3 Mx =
kyr dr dθ = k
r2 sin θ dr dθ = k
sin θ dθ
r 0
0
0
0
0 3
0
π
π
8
8
= k
(1 + cos θ)3 sin θ dθ = k
(1 + 3 cos θ + 3 cos2 θ + cos3 θ) sin θ dθ
3 0
3 0
π
8
1
8 1
3
15
32
= k − cos θ − cos2 θ − cos3 θ − cos4 θ = k
− −
=
k
3
2
4
3
4
4
3
0
x̄ = My /m =
5πk
32k/3
= 5/3; ȳ = Mx /m =
= 32/9π. The center of mass is (5/3, 32/9π).
3πk
3πk
a
2π a
2π a
2π
1 4 2 2
2
3
17. Ix =
y kr dr dθ = k
r sin θ dr dθ = k
r sin θ dθ
4
0
0
0
0
0
0
2π
4 2π
4
4
ka
ka
kπa
1
1
=
sin2 θ dθ =
θ − sin 2θ =
4 0
4
2
4
4
0
2π a
2π a
1
r3
18. Ix =
y2
r
dr
dθ
=
sin2 θ dr dθ
4
1 + r4
0
0
0
0 1+r
2π
a
2π
1
1
1
1
π
2
4 4
=
ln(1 + r ) sin θ dθ = ln(1 + a )
θ − sin 2θ = ln(1 + a4 )
4
4
2
4
4
0
0
0
x̄ = My /m =
19. Solving a = 2a cos θ, cos θ = 1/2 or θ = π/3. The density is k/r3 . Using symmetry,
π/3 2a cos θ
π/3 2a cos θ
k
x2 3 r dr dθ = 2k
cos2 θ dr dθ
Iy = 2
r
0
a
0
a
π/3
π/3
2
1
1
3
3
2
= 2k
(2a cos θ − a cos θ) dθ = 2ak 2 sin θ − sin θ − θ − sin 2θ 3
2
4
0
0
√
√ √
√
3 π
3
5ak 3 akπ
= 2ak
3−
− −
=
−
4
6
8
4
3
496
9.11 Double Integrals in Polar Coordinates
20. Solving 1 = 2 sin 2θ, we obtain sin 2θ = 1/2 or θ = π/12 and θ = 5π/12.
5π/12 2 sin 2θ
5π/12 2 sin 2θ
Iy =
x2 sec2 θ r dr dθ =
r3 dr dθ
1
π/12
1
π/12
2 sin 2θ
5π/12
5π/12
1 4 1
1
3
=
r θ − sin 4θ +
sin 8θ
dθ = 4
sin4 2θ dθ = 2
4
4
4
32
π/12
π/12
1
√
√ √
√ √
3
3
3
3
5π
π
8π + 7 3
=2
+
−
−
−
+
=
16
8
64
16
8
64
16
5π/12
π/12
21. From the solution to Problem 17, Ix = kπa4 /4. By symmetry, Iy = Ix . Thus I0 = kπa4 /2.
22. The density is ρ = kr.
π θ
2
I0 =
r (kr)r dr dθ = k
0
1
k
5
=
0
π
1
k
5
θ5 dθ =
r dr dθ = k
0
0
0
π
6
= kπ
30
0
1 6
θ
6
π
4
0
π
θ
θ
1 5 dθ
r
5 0
23. The density is ρ = k/r.
3 1/r
3 1/r
3
k
I0 =
r2 r dθ dr = k
r2 dθ dr = k
r2
r
1
0
1
0
1
1
r
dr = k
1 2
r
2
3
= 4k
1
2a cos θ
π
π 2a cos θ
π
1 4 2
24. I0 =
r kr dr dθ = k
dθ = 4ka4
cos4 θ dθ
r 0
0
0 4
0
0
π
3
3π
1
1
3kπa4
= 4ka4
θ + sin 2θ +
sin 4θ = 4ka4
=
8
4
32
8
2
0
3
√
9−x2
x2
25.
−3
+
y2
2/2
√1−y2
x2 + y 2
π/4 y
1
√1−y2
27.
x2 +y 2
e
0
π/4
0
1
0
π/4
2
r sin θ dr dθ =
1
3
0
0
1
1
θ − sin 2θ
2
4
π/2
1
dx dy =
0
=
π/4
= π−2
24
0
r2
1
1 3 2 1 π/4 2
sin θ dθ
r sin θ dθ =
3
3 0
0
e r dr dθ =
0
1
2
0
π/2
(e − 1) dθ =
0
3
π
1 3 r
dθ = 9
dθ = 9π
3 0
0
r2 sin2 θ
r dr dθ
|r|
1
2
0
π
0
dx dy =
=
=
0
y2
3
|r|r dr dθ =
0
26.
0
dy dx =
0
√
π
0
π/2
1
1 r2 e dθ
2
0
π(e − 1)
4
497
9.11 Double Integrals in Polar Coordinates
28.
√
√
π
π−x2
2
π
29.
√
0
√
1
4−x2
1−x2
π
(sin r )r dr dθ =
0
π
2
0
=−
1
√
sin(x + y ) dy dx =
√
− π
2
1
2
0
0
√π
1
2 − cos r dθ
2
0
π
(−1 − 1) dθ = π
0
2 √4−x2
x2
x2
dy dx +
dy dx
2
2
2
x +y
x + y2
1
0
π/2 2 2
π/2 2
r cos2 θ
=
r
dr
dθ
=
r cos2 θ dr dθ
r2
0
1
0
1
2
π/2
1 2 3 π/2
3 1
1
=
cos2 θ dθ =
r cos2 θ dθ =
θ + sin 2θ
2
2 0
2 2
4
0
1
√2y−y2
π/2
= 3π
8
0
(1 − x2 − y 2 ) dx dy
30.
0
0
π/4
2 sin θ
π/2
csc θ
(1 − r )r dr dθ +
(1 − r2 )r dr dθ
2
=
0
0
π/4
0
2 sin θ
dθ
+
1 2 1 4
r − r
2
4
=
0
π/4
π/2
π/4
0
π/4
csc θ
dθ
1 2 1 4
r − r
2
4
0
π/2
1
1
csc2 θ − csc4 θ dθ
2
4
0
π/4
π/2
1
3
1
1
1
1
3
= θ − sin 2θ −
θ − sin 2θ + sin 4θ
+ − cot θ −
− cot θ − cot θ
2
2
8
2
4
3
π/4
1
1
16 − 3π
π 1
+ 0− − +
=
= − +
8
2
4 12
24
(2 sin2 θ − 4 sin4 θ) dθ +
=
5
√
25−x2
31.
π
5
(4x + 3y) dy dx =
−5
0
π
0
π
=
0
√1−y2
32.
0
0
dx dy =
1 + x2 + y 2
π/2 1
=
1−
0
0
1
1+r
π/2
(1 − ln 2)dθ =
=
0
π
2
1
5
2
1
0
5
4 3
3
(4r cos θ + 3r sin θ) dr dθ =
r cos θ + r sin θ dθ
3
0
0
0
π
500
500
cos θ + 125 sin θ dθ =
sin θ − 125 cos θ = 250
3
3
0
=
(4r cos θ + 3r sin θ)r dr dθ
0
π/2
1
1
r dr dθ
1
+
r
0
0
π/2
1
dr dθ =
[r − ln(1 + r)] dθ
0
0
π
(1 − ln 2)
2
33. The volume of the cylindrical portion of the tank is Vc = π(4.2)2 19.3 ≈ 1069.56 m3 . We take the equation of
the ellipsoid to be
x2
z2
+
=1
2
(4.2)
(5.15)2
or z = ±
498
5.15
4.2
(4.2)2 − x2 − y 2 .
9.12
Green’s Theorem
The volume of the ellipsoid is
5.15
10.3 2π 4.2
(4.2)2 − x2 − y 2 dx dy =
[(4.2)2 − r2 ]1/2 r dr dθ
4.2
4.2
R
0
0
4.2 2π 10.3
1 2
10.3 1 2π
=
−
dθ =
[(4.2)2 − r2 ]3/2 (4.2)3 dθ
4.2 0
2 3
4.2
3
0
0
Ve = 2
=
2π 10.3
(4.2)3 ≈ 380.53.
3 4.2
The volume of the tank is approximately 1069.56 + 380.53 = 1450.09 m3 .
34.
π/2
π/2
2
r2 (cos θ + sin θ) dr dθ
(r cos θ + r sin θ) r dr dθ =
0
R
2
(x + y) dA =
2 sin θ
π/2
=
0
0
π/2
2
1 3
8
r (cos θ + sin θ) dθ =
3
3 0
2 sin θ
2 sin θ
(cos θ + sin θ − sin3 θ cos θ − sin4 θ) dθ
8
=
3
1
3
1
3
sin θ − cos θ − sin4 θ + sin3 θ cos θ − θ +
sin 2θ
4
4
8
16
8
1 3π
28 − 3π
=
1− −
− (−1) =
3
4
16
6
35. I 2 =
∞
0
=
∞
e−(x
+y 2 )
t→∞
π/2
∞
e−r r dr dθ =
2
dx dy =
0
π/2
lim
0
2
0
2
1
1
− e−t +
2
2
0
dθ =
0
π/2
1
π
dθ = ;
2
4
π/2
0
t
2 1
lim − e−r dθ
t→∞
2
0
0
√
π
I=
2
π/2
EXERCISES 9.12
Green’s Theorem
1. The sides of the triangle are C1 : y = 0, 0 ≤ x ≤ 1; C2 : x = 1, 0 ≤ y ≤ 3; C3 : y = 3x,
0 ≤ −x ≤ 1.
1
3
0
0
(x
−
y)
dx
+
xy
dy
=
x
dx
+
y
dy
+
(x
−
3x)
dx
+
x(3x) 3 dx
ˇ
C
0
0
1
+
1
1
3
1
0 1 9
1 2 =
y + (−x2 ) + (3x2 ) = + + 1 − 3 = 3
2
2 2
0
1
0
0
3x
1 3x
1
1
1 2
9 2
(y + 1) dA =
(y + 1) dy dx =
y + y dx =
x + 3x dx
2
2
R
0
0
0
0
0
1
3 3 3 2 =
x + x =3
2
2
0
1 2
x
2
2. The sides of the rectangle are C1 : y = 0, −1 ≤ x ≤ 1; C2 : x = 1, 0 ≤ y ≤ 1; C3 : y = 1, 1 ≥ x ≥ −1;
C4 : x = −1, 1 ≥ y ≥ 0.
499
9.12 Green’s Theorem
1
1
1
1
2
2
3x
y
dx
+
(x
−
5y)
dy
=
0
dx
+
(1
−
5y)
dy
=
0
dx
+
(1 − 5y) dy
ˇ
C
−1
0
−1
0
1
0
−1
0
−1
5
5
=
3x2 dx +
(1 − 5y) dy = y − y 2 + x3 + y − y 2 = −2
2
2
1
1
1
0
1
1 1
1
1
1
(2x − 3x2 ) dA =
(2x − 3x2 ) dx dy =
(x2 − x3 ) dy =
(−2) dy = −2
−1
0
R
3. ˇ
2π
− y dx + x dy =
2
−1
0
2π
2
2
9 cos2 t(3 cos t) dt
(−9 sin t)(−3 sin t) dt +
C
0
0
0
2π
[(1 − cos2 t) sin t + (1 − sin2 t) cos t] dt
= 27
0
2π
1
1
3
3
= 27 − cos t + cos t + sin t − sin t = 27(0) = 0
3
3
0
2π 3
2π 3
(2x + 2y) dA = 2
(r cos θ + r sin θ)r dr dθ = 2
r2 (cos θ + sin θ) dr dθ
0
R
0
0
0
3
2π
1 3
=2
(cos θ + sin θ) dθ
r (cos θ + sin θ) dθ = 18
3
0
0
0
2π
= 18(sin θ − cos θ) = 18(0) = 0
2π
0
4. The sides of the region are C1 : y = 0, 0 ≤ x ≤ 2; C2 : y = −x + 2, 2 ≥ x ≥ 1;
√
C3 : y = x , 1 ≥ x ≥ 0.
2
1
1
2
2
0 dx +
−2(−x + 2) dx +
4x(−x + 2)(−dx)
ˇ − 2y dx + 4xy dy =
C
0
2
0
−2x dx +
+
1
2
0
√
4x x
1
√
2 x
1
2 8
10
+ +1−1=
3 3
3
1 2−y
1
8y dA =
8y dx dy =
8y(2 − y − y 2 ) dy =
dx
=0+
R
0
y2
8
8y 2 − y 3 − 2y 4
3
0
1
= 10
3
0
5. P = 2y, Py = 2, Q = 5x, Qx = 5
(5 − 2) dA = 3
dA = 3(25π) = 75π
ˇ 2y dx + 5x dy =
C
R
R
6. P = x + y 2 , Py = 2y, Q = 2x2 − y, Qx = 4x
2
2
(x
+
y
)
dx
+
(2x
−
y)
dy
=
(4x
−
2y)
dA
=
ˇ
C
−2
=
−2
R
2
=
2
4
(4xy − y 2 ) 2 dx =
4
(4x − 2y) dy dx
x2
2
x
1
8x2 − 16x − x4 + x5
5
500
−2
2
(16x − 16 − 4x3 + x4 ) dx
−2
=−
96
5
9.12
7. P = x4 − 2y 3 , Py = −6y 2 , Q = 2x3 − y 4 , Qx = 6x2 . Using polar coordinates,
2π 2
4
3
3
4
2
2
(x
−
2y
)
dx
+
(2x
−
y
)
dy
=
(6x
+
6y
)
dA
=
6r2 r dr dθ
ˇ
C
R
2π
2
dθ =
3 4
r
2
=
0
0
2π
0
24 dθ = 48π.
0
0
8. P = x − 3y, Py = −3, Q = 4x + y, Qx = 4
(x
−
3y)
dx
+
4(x
+
y)
dy
=
(4 + 3) dA = 7(10) = 70
ˇ
C
R
9. P = 2xy, Py = 2x, Q = 3xy 2 , Qx = 3y 2
2
2
2xy
dx
+
3xy
dy
=
(3y
−
2x)
dA
=
ˇ
C
2x
(3y 2 − 2x) dy dx
2
2
2x
(y 3 − 2xy) dx =
(8x3 − 4x2 − 8 + 4x) dx
=
2
1
=
1
R
2
2
1
2
= 40 − − 16
3
3
1
4
2x − x3 − 8x + 2x2
3
4
=
56
3
10. P = e2x sin 2y, Py = 2e2x cos 2y, Q = e2x cos 2y, Qx = 2e2x cos 2y
2x
2x
=
e
sin
2y
dx
+
e
cos
2y
dy
=
0 dA = 0
ˇ
C
R
11. P = xy, Py = x, Q = x2 , Qx = 2x. Using polar coordinates,
π/2 1
2
(2x − x) dA =
r cos θ r dr dθ
ˇ xy dx + x dy =
C
−π/2
R
π/2
=
−π/2
1 3
r cos θ
3
0
1
dθ =
π/2
−π/2
0
π/2
1
2
1
=
cos θ dθ = sin θ 3
3
3
−π/2
2
2
12. P = ex , Py = 0, Q = 2 tan−1 x, Qx =
1 + x2
0 1
2
2
x2
−1
e
dx
+
2
tan
x
dy
=
dA
=
dy dx
2
ˇ
1
+
x
1
+
x2
C
−1 −x
R
1
0
0
2y
2
2x
=
dx
dx
=
+
1 + x2 −x
1 + x2
1 + x2
−1
−1
0
π
π
= [2 tan−1 x + ln(1 + x2 )] = 0 − − + ln 2 = − ln 2
2
2
−1
1
13. P = y 3 , Py = y 2 , Q = xy + xy 2 , Qx = y + y 2
3
1/√2 1−y2
1 3
2
y dA =
y dx dy
ˇ 3 y dx + (xy + xy ) dy =
C
0
R
y2
1/√2
1/√2
1−y2
=
(xy) 2
dy =
(y − y 3 − y 3 ) dy
0
=
1 2 1 4
y − y
2
2
y
0
1/√2
1 1
1
= − =
4 8
8
0
501
Green’s Theorem
9.12 Green’s Theorem
14. P = xy 2 , Py = 2xy, Q = 3 cos y, Qx = 0
2
xy
dx
+
3
cos
y
dy
=
(−2xy)
dA
=
−
ˇ
C
=−
0
0
R
1
1
x2
2xy dy dx
x3
1
x2
(xy) dx = −
(x3 − x4 ) dx =
x3
0
1 4 1 5
x − x
4
5
1
=− 1
20
0
15. P = ay, Py = a, Q = bx, Qx = b. ˇ ay dx + bx dy =
(b − a) dA = (b − a) × (area bounded by C)
C
R
16. P = P (x), Py = 0, Q = Q(y), Qx = 0. ˇ P (x) dx + Q(y) dy =
0 dA = 0
C
R
17. For the first integral: P = 0, Py = 0, Q = x, Qx = 1; ˇ x dy =
1 dA = area of R.
C R
For the second integral: P = y, Py = 1, Q = 0, Qx = 0; − ˇ y dx = −
−1 dA = area of R.
C
R
Thus, ˇ x dy = − ˇ y dx.
C
C
1
1
18. P = −y, Py = −1, Q = x, Qx = 1.
−
y
dx
+
x
dy
=
2
dA
=
dA = area of R
2 ˇC
2
R
R
2π
2π
2
3
2
19. A =
dA = ˇ x dy =
a cos t(3a sin t cos t dt) = 3a
sin2 t cos4 t dt
C
R
0
2π
1
1
1
3
= 3a2
t−
sin 4t +
sin3 2t = πa2
16
64
48
8
0
2π
20. A =
dA = ˇ x dy =
a cos t(b cos t dt) = ab
C
R
0
0
2π
2
cos t dt = ab
0
1
1
t + sin 2t
2
4
2π
= πab
0
21. (a) Parameterize C by x = x1 + (x2 − x1 )t and y = y1 + (y2 − y1 )t for 0 ≤ t ≤ 1. Then
1
1
−y dx + x dy =
−[y1 + (y2 − y1 )t](x2 − x1 ) dt +
[x1 + (x2 − x1 )t](y2 − y1 ) dt
0
C
0
1
1
1
1
2 2 = −(x2 − x1 ) y1 t + (y2 − y1 )t + (y2 − y1 ) x1 t + (x2 − x1 )t 2
2
0
0
1
1
= −(x2 − x1 ) y1 + (y2 − y1 ) + (y2 − y1 ) x1 + (x2 − x1 ) = x1 y2 − x2 y1 .
2
2
(b) Let Ci be the line segment from (xi , yi ) to (xi+1 , yi+1 ) for i = 1, 2, . . . , n − 1, and C2 the line segment
from (xn , yn ) to (x1 , y1 ). Then
1
A = ˇ − y dx + x dy
Using Problem 18
2 C
1
=
−y dx + x dy +
−y dx + x dy + · · · +
−y dx + x dy +
−y dx + x dy
2 C1
C2
Cn−1
Cn
=
1
1
1
1
(x1 y2 − x2 y1 ) + (x2 y3 − x3 y2 ) + (xn−1 yn − xn yn−1 ) + (xn y1 − x1 yn ).
2
2
2
2
22. From part (b) of Problem 21
1
1
1
1
[(−1)(1) − (1)(3)] + [(1)(2) − (4)(1)] + [(4)(5) − (3)(2)] + [(3)(3) − (−1)(5)]
2
2
2
2
1
= (−4 − 2 + 14 + 14) = 11.
2
A=
502
9.12
23. P = 4x2 − y 3 , Py = −3y 2 ; Q = x3 + y 2 , Qx = 3x2 .
2
3
3
2
2
2
(3x + 3y ) dA =
ˇ (4x − y ) dx + (x + y ) dy =
C
0
R
2π
2π
2
2π
2
3r (r dr dθ) =
1
0
3 4
r
4
Green’s Theorem
2
dθ
1
45π
45
dθ =
4
2
=
0
24. P = cos x2 − y, Py = −1; Q = y 3 + 1 , Qx = 0
√
2
3 + 1 dy =
(cos
x
−
y)
dx
+
y
(0
+
1)
dA
dA = (6 2)2 − π(2)(4) = 72 − 8π
ˇ
C
R
R
25. We first observe that Py = (y 4 − 3x2 y 2 )/(x2 = y 2 )3 = Qx . Letting C be the circle x2 + y 2 =
ˇ
C
1
4
we have
−y 3 dx + xy 2 dy
−y 3 dx + xy 2 dy
=
ˇ (x2 + y 2 )2
(x2 + y 2 )2
C
x=
1
4
cos t, dx = − 14 sin t dt, y =
1
4
sin t, dy =
1
4
cos t dt
1
1
− 64
sin3 t(− 14 sin t dt) + 14 cos t( 16
sin2 t)( 14 cos t dt)
1/256
0
2π
2π
4
2
2
=
(sin t + sin t cos t) dt =
(sin4 t + (sin2 t − sin4 t) dt
2π
=
0
0
2π
2
=
sin t dt =
0
1
1
t − sin 2t
2
4
2π
=π
0
26. We first observe that Py = [4y 2 − (x + 1)2 ]/[(x + 1)2 + 4y 2 ]2 = Qx . Letting C be the ellipse (x + 1)2 + 4y 2 = 4
we have
−y
−y
x+1
x+1
ˇ (x + 1)2 + 4y 2 dx + (x + 1)2 + 4y 2 dy = ˇ (x + 1)2 + 4y 2 dx + (x + 1)2 + 4y 2 dy
C
C
x + 1 = 2 cos t, dx = −2 sin t dt, y = sin t, dy = cos t dt
2π − sin t
2 cos t
1 2π
=
(sin2 t + cos2 t) dt = π.
(−2 sin t) +
cos t dt =
4
4
2 0
0
27. Writing
x2 dA =
(Qx − Py )dA we identify Q = 0 and P = −x2 y. Then, with C: x = 3 cos t,
R
R
y = 2 sin t, 0 ≤ t ≤ 2π, we have
2π
x2 dA = ˇ P dx + Q dy = ˇ − x2 y dx = −
9 cos2 t(2 sin t)(−3 sin t) dt
C
C
0
R
2π
54 2π
27
27 2π
2
2
2
=
4 sin t cos t dt =
sin 2t dt =
(1 − cos 4t) dt
4 0
2 0
4 0
2π
27
27π
1
=
t − sin 4t =
.
4
4
2
0
[1 − 2(y − 1)] dA =
28. Writing
R
(Qx − Py ) dA we identify Q = x and P = (y − 1)2 .
R
C1 : x = cos t, y − 1 = sin t, −π/2 ≤ t ≤ π/2, and C2 : x = 0, 2 ≥ y ≥ 0,
503
Then, with
9.12 Green’s Theorem
[1 − 2(y − 1)] dA =
R
P dx + Q dy +
C1
(y − 1)2 dx + x dy +
P dx + Q dy =
C2
C1
π/2
π/2
2
=
[sin t(− sin t) + cos t cos t] dt =
−π/2
−π/2
0 dy
C2
[cos2 t − (1 − cos2 t) sin t] dt
1
2
=
(1 + cos 2t) − sin t + cos t sin t dt
−π/2 2
π/2
1
π π π
1
1
3
=
= − −
t + sin 2t + cos t − cos t = .
2
4
3
4
4
2
−π/2
3π
3
29. P = x − y, Py = −1, Q = x + y, Qx = 1; W = ˇ F · d r =
2 dA = 2 × area = 2
= π
4
2
C
R
π/2
30. P = −xy 2 , Py = −2xy, Q = x2 y, Qx = 2xy. Using polar coordinates,
π/2 2
W = ˇ F · dr =
4xy dA =
4(r cos θ)(r sin θ)r dr dθ =
C
π/2
= 15
0
0
R
1
π/2
2
(r4 cos θ sin θ) dθ
1
0
π/2
15
15
2 sin θ =
.
sin θ cos θ dθ =
2
2
0
B
31. Since
P dx + Q dy is independent of path, Py = Qx by Theorem 9.9. Then, by Green’s Theorem
A
ˇ P dx + Q dy =
C
(Qx − Py ) dA =
R
0 dA = 0.
R
32. Let P = 0 and Q = x2 . Then Qx − Py = 2x and
x dA
1
1
2
x dy =
2x dA = R
= x̄.
ˇ
2A C
2A
A
R
Let P = y 2 and Q = 0. Then Qx − Py = −2y and
y dA
1
1
2
−
y dx = −
−2y dA = R
= ȳ.
2A ˇC
2A
A
R
33. Using Green’s Theorem,
2π
1+cos θ
W = ˇ F · dr = ˇ − y dx + x dy =
2 dA = 2
C
C
0
0
R
1+cos θ
2π
2π
1 2 =2
dθ =
(1 + 2 cos θ + cos2 θ) dθ
r 2
0
0
0
2π
1
1
= θ + 2 sin θ + θ + sin 2θ = 3π.
2
4
0
504
r dr dθ
9
Vector Calculus
DO NOT USE THIS PAGE
504
9.13
Surface Integrals
EXERCISES 9.13
Surface Integrals
1
1
3
1. Letting z = 0, we have 2x + 3y = 12. Using f (x, y) = z = 3 − x − y we have fx = − ,
2
4
2
3
29
2
2
fy = − , 1 + fx + fy =
. Then
4
16
√ 6
√ 6
6 4−2x/3 2
29
29
1 2 A=
4 − x dx =
29/16 dy dx =
4x − x 4
3
4
3
0
0
0
0
√
√
29
=
(24 − 12) = 3 29 .
4
2. We see from the graph in Problem 1 that the plane is entirely above the region bounded by
1
3
1
3
r = sin 2θ in the first octant. Using f (x, y) = z = 3 − x − y we have fx = − , fy = − ,
2
4
2
4
29
2
2
1 + fx + fy =
. Then
16
√ π/2
√ π/2
sin 2θ
π/2 sin 2θ 29
1 2 29
A=
29/16 r dr dθ =
dθ =
sin2 2θ dθ
r 4
2
8
0
0
0
0
0
√ π/2 √
29 1
29 π
1
=
θ − sin 4θ =
.
8
2
8
32
0
√
16 − x2 we see that for 0 ≤ x ≤ 2 and 0 ≤ y ≤ 5, z > 0.
x
Thus, the surface is entirely above the region. Now fx = − √
, fy = 0,
16 − x2
2
x
16
1 + fx2 + fy2 = 1 +
=
and
2
16 − x
16 − x2
2
5 2
5
5
4
π
10π
−1 x √
A=
dx dy = 4
sin
dy
=
4
dy =
.
2
4 0
3
16 − x
0
0
0
0 6
3. Using f (x, y) = z =
4. The region in the xy-plane beneath the surface is bounded by the graph of x2 + y 2 = 2.
Using f (x, y) = z = x2 + y 2 we have fx = 2x, fy = 2y, 1 + fx2 + fy2 = 1 + 4(x2 + y 2 ).
Then,
2π
A=
0
0
√
2
1 + 4r2 r dr dθ =
0
2π
√ 2
2π
1
1
13π
2 3/2 (27 − 1)dθ =
(1 + 4r )
.
dθ = 12
12
3
0
0
5. Letting z = 0 we have x2 + y 2 = 4. Using f (x, y) = z = 4 − (x2 + y 2 ) we have fx = −2x,
fy = −2y, 1 + fx2 + fy2 = 1 + 4(x2 + y 2 ). Then
2
2π 2 2π
1
2 3/2 2
A=
1 + 4r r dr dθ =
(1 + 4r )
dθ
3
0
0
0
0
2π
1
π
=
(173/2 − 1)dθ = (173/2 − 1).
12 0
6
505
9.13 Surface Integrals
6. The surfaces x2 + y 2 + z 2 = 2 and z 2 = x2 + y 2 intersect on the cylinder 2x2 + 2y 2 = 2
or x2 + y 2 = 1. There are portions of the sphere within the cone both above and
x
below the xy-plane. Using f (x, y) = 2 − x2 − y 2 we have fx = − ,
2 − x2 − y 2
y
2
fy = − , 1 + fx2 + fy2 =
. Then
2 − x2 − y 2
2 − x2 − y 2
√
1
2π 1
√ 2π 2
√
A=2
r dr dθ = 2 2
− 2 − r2 dθ
2
0
2
−
r
0
0
0
2π √
√
√ √
=2 2
( 2 − 1)dθ = 4π 2( 2 − 1).
0
7. Using f (x, y) = z = 25 − x2 − y 2 we have fx = − y
fy = − 25 − x2 − y 2
5
, 1 + fx2 + fy2 =
√25−y2 /2
A=
0
=5
0
5
5
25 − x2 − y 2
,
25
. Then
25 − x2 − y 2
25 − x2 − y 2
0
x
dx dy = 5
5
sin−1
0
√25−y2 /2
dy
2
25 − y 0
x
π
25π
dy =
.
6
6
8. In the first octant, the graph of z = x2 − y 2 intersects the xy-plane in the line y = x. The
surface is in the firt octant for x > y. Using f (x, y) = z = x2 − y 2 we have fx = 2x,
fy = −2y, 1 + fx2 + fy2 = 1 + 4x2 + 4y 2 . Then
2
π/4 2 π/4
1
A=
1 + 4r2 r dr dθ =
(1 + 4r2 )3/2 dθ
12
0
0
0
0
π/4
1
π
=
(173/2 − 1)dθ =
(173/2 − 1).
12 0
48
9. There are portions of the sphere within the cylinder both above and below the xy-plane.
x
y
Using f (x, y) = z = a2 − x2 − y 2 we have fx = − , fy = − ,
12 − x2 − y 2
a2 − x2 − y 2
a2
1 + fx2 + fy2 = 2
. Then, using symmetry,
a − x2 − y 2
π/2 a sin θ
π/2 a sin θ
a
√
A=2 2
r dr dθ = 4a
− a2 − r2 dθ
0
a2 − r2
0
0
0
π/2
π/2
2
2
= 4a
(a − a 1 − sin θ )dθ = 4a
(1 − cos θ)dθ
0
0
π/2
π
= 4a2 (θ − sin θ) = 4a2
− 1 = 2a2 (π − 2).
2
0
10. There are portions of the cone within the cylinder both above and below the xy-plane. Using
x
y
f (x, y) = 12 x2 + y 2 , we have fx = , fy = , 1 + fx2 + fy2 = 54 .
2 x2 + y 2
2 x2 + y 2
Then, using symmetry,
506
9.13
A=2 2
Surface Integrals
√ π/2 1 2 2 cos θ
5
dθ
r dr dθ = 2 5
r 4
2 0
0
0
0
π/2
√ π/2
√
√
1
1
2
=4 5
θ + sin 2θ = 5 π.
cos θ dθ = 4 5
2
4
0
0
π/2
2 cos θ
11. There are portions of the surface in each octant with areas equal to the area of the portion
y
in the first octant. Using f (x, y) = z = a2 − y 2 we have fx = 0, fy = ,
2
a − y2
a2
1 + fx2 + fy2 = 2
. Then
a − y2
√a2 −y2
a √a2 −y2
a
a
a
x
A=8
dx dy = 8a
dy = 8a
dy = 8a2 .
a2 − y 2
a2 − y 2 0
0
0
0
0
12. From Example 1, the area of the portion of the hemisphere within x2 + y 2 = b2 is 2πa(a −
the area of the sphere is A = 2 lim 2πa(a − a2 − b2 ) = 2(2πa2 ) = 4πa2 .
√
a2 − b2 ). Thus,
b→a
13. The projection of the surface onto the xz-plane is shown in the graph. Using f (x, z) =
√
x
y = a2 − x2 − z 2 we have fx = − √
,
2
a − x2 − z 2
z
a2
fz = − √
, 1 + fx2 + fz2 = 2
. Then
2
2
2
a − x2 − z 2
a −x −z
√a2 −c21
2π √a2 −c21
2π a
A=
r dr dθ = a
− a2 − r2 √
dθ
√ 2 2 √ 2
a − r2
a −c2
0
0
a2 −c22
2π
=a
(c2 − c1 ) dθ = 2πa(c2 − c1 ).
0
14. The surface area of the cylinder x2 + z 2 = a2 from y = c1 to y = c2 is the area of a cylinder of radius a and
height c2 − c1 . This is 2πa(c2 − c1 ).
√
15. zx = −2x, zy = 0; dS = 1 + 4x2 dA
4 √2 x dS =
x 1 + 4x2 dx dy =
0
S
=
0
0
4
√ 2
1
(1 + 4x2 )3/2 dy
12
0
4
0
13
26
dy =
6
3
16. See Problem 15.
xy(9 − 4z) dS =
xy(1 + 4x2 ) dS =
S
=
0
S
4
0
4
√
2
xy(1 + 4x2 )3/2 dx dy
0
√2
4
4
y
242
121 1 2 484
121 4
2 5/2 dy =
y dy =
(1 + 4x )
y dy =
y =
20
10 0
10 2
5
0 20
0
0
507
9.13 Surface Integrals
17. zx = x
y
, zy = √
2 dA.
x2 + y 2
x2 + y 2
Using polar coordinates,
√
√ 2π 1
3
2
2 3/2
xz dS =
x(x + y )
2 dA = 2
(r cos θ)r3/2 r dr dθ
S
R
√ = 2
2π
0
√ = 2
x
0
2π
0
√ r7/2 cos θ dr dθ = 2
1
0
0
18. zx = ; dS =
2π
0
√
1
2 9/2
r cos θ dθ
9
0
2π
2
2 2
cos θ dθ =
sin θ = 0.
9
9
0
y
, zy = √
2 dA.
x2 + y 2
x2 + y 2
Using polar coordinates,
√
(x + y + z) dS =
(x + y + x2 + y 2 ) 2 dA
S
R
=
√ 2
2π
; dS =
4
(r cos θ + r sin θ + r)r dr dθ
0
√ = 2
1
2π
4
√ r (1 + cos θ + sin θ) dr dθ = 2
2π
2
0
1
0
4
1 3
r (1 + cos θ + sin θ) dθ
3
1
√ 2π
√
√
63 2 2π
=
(1 + cos θ + sin θ) dθ = 21 2(θ + sin θ − cos θ) = 42 2 π.
3
0
0
19. z =
x
y
36 − x2 − y 2 , zx = − , zy = − ;
2
2
36 − x − y
36 − x2 − y 2
dS =
1+
x2
y2
6
+
dA = dA.
2
2
36 − x − y
36 − x2 − y 2
36 − x2 − y 2
Using polar coordinates,
2
2
(x + y )z dS =
(x2 + y 2 ) 36 − x2 − y 2 S
dA
36 − x2 − y 2
6
2π
1 4 dθ = 6
324 dθ = 972π.
r
4 0
0
R
2π
6
2π
2
=6
r r dr dθ = 6
0
0
0
√
20. zx = 1, zy = 0; dS = 2 dA
1 1−x2
√
√ z 2 dS =
(x + 1)2 2 dy dx = 2
S
−1
√ = 2
0
1
√ (1 − x2 )(x + 1)2 dx = 2
−1
6
1
−1
1−x2
y(x + 1)2 dx
0
1
−1
(1 + 2x − 2x3 − x4 ) dx
√
1
√
1 4 1 5 8 2
2
= 2 x+x − x − x =
2
5
5
−1
508
9.13
21. zx = −x, zy = −y; dS = 1 + x2 + y 2 dA
1 1 xy dS =
xy 1 + x2 + y 2 dx dy =
0
S
0
0
1
Surface Integrals
1
1
y(1 + x2 + y 2 )3/2 dy
3
0
1
1
1
=
y(2 + y 2 )3/2 − y(1 + y 2 )3/2 dy
3
3
0
1
1
1
1 5/2
2 5/2
2 5/2 7/2
=
(2 + y ) − (1 + y )
= 15 (3 − 2 + 1)
15
15
0
1 1 2 1 2
+ x + y , zx = x, zy = y; dS = 1 + x2 + y 2 dA
2 2
2
Using polar coordinates,
2z dS =
(1 + x2 + y 2 ) 1 + x2 + y 2 dA
22. z =
S
R
π/2
1
=
π/3
π/2
(1 + r2 ) 1 + r2 r dr dθ
0
1
2 3/2
=
(1 + r )
0
π/3
π/2
r dr dθ =
π/3
√
4 2 − 1 π π
=
−
5
2
3
√
(4 2 − 1)π
=
.
30
1
1
1 π/2 5/2
2 5/2 (1 + r )
(2 − 1) dθ
dθ = 5
5
π/3
0
√
23. yx = 2x, yz = 0; dS = 1 + 4x2 dA
3 2
√
24 y z dS =
24xz 1 + 4x2 dx dz =
S
0
0
3
2
2z(1 + 4x2 )3/2 dz
0
3
= 2(173/2 − 1)
z dz = 2(173/2 − 1)
0
24. xy = −2y, xz = −2z; dS =
0
1 + 4y 2 + 4z 2 dA
Using polar coordinates,
2
2 1/2
(1 + 4y + 4z ) dS =
π/2
=
0
2
(1 + 4r2 )r dr dθ
0
S
0
3
1 2 3/2
z
= 9(17 − 1)
2
1
π/2
π/2
2
1
1
3π
(1 + 4r2 )2 dθ =
.
12 dθ =
16
16 0
8
1
1
1
3
25. Write the equation of the surface as y = (6−x−3z). Then yx = − , yz = − ; dS = 1 + 1/4 + 9/4 =
2
2
2
√
2 6−3z
1
14
3z 2 + 4z (6 − x − 3z)
(3z 2 + 4yz) dS =
dx dz
2
2
0
0
S
√ 2
6−3z
14
=
[3z 2 x − z(6 − x − 3z)2 ] dz
2
0
0
√ 2
2
14
=
[3z (6 − 3z) − 0] − [0 − z(6 − 3z)2 ] dz
2
0
√ 2
√
2 √14
√
14
14
=
(36z − 18z 2 ) dz =
(18z 2 − 6z 3 ) =
(72 − 48) = 12 14
2
2
2
0
0
509
√
14
.
2
9.13 Surface Integrals
26. Write the equation of the surface as x = 6 − 2y − 3z. Then xy = −2, xz = −3; dS =
2 3−3z/2
3−3z/2
√
√ 2
(3z 2 + 4yz) dS =
(3z 2 + 4yz) 14 dy dz = 14
(3yz + 2y 2 z) dz
0
S
√
0
0
√
1+4+9 =
0
√
z
45
9
z 2
dz = 14
27z − z 2 + z 3
+ 18z 1 −
2
2
2
2
0
0
2
√
√
√
27 2 15 3 9 4 = 14
= 14(54 − 60 + 18) = 2 14
z − z + z
2
2
8
0
=
14
2
9z 1 −
2
27. The density is ρ = kx2 . The surface is z = 1−x−y. Then zx = −1, zy = −1; dS =
1 1−x √
√ 1 1 3 1−x
2
2
m=
x kx dS = k
x 3 dy dx = 3 k
dx
S
0
0
0 3
0
√
√
1 √
1
1
3
3
3
=
k
k − (1 − x)4 =
k
(1 − x)3 dx =
3
3
4
12
0
0
√
dz
3 dA.
x
y
, zy = − ;
4 − x2 − y 2
4 − x2 − y 2
28. zx = − dS =
1+
x2
y2
2
+
dA = dA.
4 − x2 − y 2
4 − x2 − y 2
4 − x2 − y 2
Using symmetry and polar coordinates,
π/2 2
2
m=4
|xy| dS = 4
(r2 cos θ sin θ) √
r dr dθ
4
− r2
S
0
0
π/2 2
=4
r2 (4 − r2 )−1/2 sin 2θ(r dr) dθ
u = 4 − r2 , du = −2r dr, r2 = 4 − u
0
0
π/2 0
1
=4
(4 − u)u
sin 2θ − du dθ = −2
(4u−1/2 − u1/2 ) sin 2θ du dθ
2
0
4
0
4
0
π/2
π/2 π/2 32
2 3/2 64
64
1
1/2
= −2
8u − u
− sin 2θ dθ =
sin 2θ dθ = −2
− cos 2θ =
.
3
3
3
2
3
0
0
4
0
π/2
0
−1/2
29. The surface is g(x, y, z) = y 2 + z 2 − 4 = 0. ∇g = 2yj + 2zk,
yj + zk
|∇g| = 2 y 2 + z 2 ; n = ;
y2 + z2
yz
3yz
2yz
+
=
; z = 4 − y 2 , zx = 0,
F·n= 2
2
2
2
2
2
y +z
y +z
y +z
y2
2
dA = dA
4 − y2
4 − y2
4 − y2
3yz
2
3y 4 − y 2
2
Flux =
F · n dS =
dA =
dA
y2 + z2 4 − y2
y2 + 4 − y2 4 − y2
S
R
R
2
3 2
3
3
3 2 =
3y dy dx =
6 dx = 18
y dx =
0
0
0 2
0
0
zy = − y
; dS =
1+
510
√
14 .
9.13
Surface Integrals
30. The surface is g(x, y, z) = x2 + y 2 + z − 5 = 0. ∇g = 2xi + 2yj + k,
2xi + 2yj + k
z
1 + 4x2 + 4y 2 ; n = ; F·n= ;
1 + 4x2 + 4y 2
1 + 4x2 + 4y 2
zx = −2x, zy = −2y, dS = 1 + 4x2 + 4y 2 dA. Using polar coordinates,
z
Flux =
F · n dS =
1 + 4x2 + 4y 2 dA =
(5 − x2 − y 2 ) dA
2 + 4y 2
1
+
4x
S
R
R
2
2π 2
2π 2π
5 2 1 4 =
r − r dθ =
(5 − r2 )r dr dθ =
6 dθ = 12π.
2
4
0
0
0
0
0
|∇g| =
2xi + 2yj + k
2x2 + 2y 2 + z
31. From Problem 30, n = . Then F · n = . Also, from Problem 30,
1 + 4x2 + 4y 2
1 + 4x2 + 4y 2
1 + 4x2 + 4y 2 dA. Using polar coordinates,
2x2 + 2y 2 + z Flux =
F · n dS =
1 + 4x2 + 4y 2 dA =
(2x2 + 2y 2 + 5 − x2 − y 2 ) dA
1 + 4x2 + 4y 2
S
R
R
2
2π 2
2π 2π
1 4 5 2 2
=
(r + 5)r dr dθ =
14 dθ = 28π.
r + r dθ =
4
2
0
0
0
0
0
√
−i + k
32. The surface is g(x, y, z) = z − x − 3 = 0. ∇g = −i + k, |∇g| = 2 ; n = √
;
2
√
1
1
F · n = √ x3 y + √ xy 3 ; zx = 1, zy = 0, dS = 2 dA. Using polar coordinates,
2
2
√
1
√ (x3 y + xy 3 ) 2 dA =
Flux =
F · n dS =
xy(x2 + y 2 ) dA
2
S
R
R
π/2 2 cos θ
π/2 2 cos θ
=
(r2 cos θ sin θ)r2 r dr dθ =
r5 cos θ sin θ dr dθ
dS =
0
=
0
0
π/2
0
0
2 cos θ
π/2
1 π/2
32
4
1 6
1
7
8
dθ =
64 cos θ sin θ dθ =
r cos θ sin θ − cos θ = .
6
6 0
3
8
3
0
0
33. The surface is g(x, y, z) = x2 + y 2 + z − 4. ∇g = 2xi + 2yj + k,
2xi + 2yj + k
x3 + y 3 + z
4x2 + 4y 2 + 1 ; n = ; F·n= ;
4x2 + 4y 2 + 1
4x2 + 4y 2 + 1
zx = −2x, zy = −2y, dS = 1 + 4x2 + 4y 2 dA. Using polar coordinates,
Flux =
F · n dS =
(x3 + y 3 + z) dA =
(4 − x2 − y 2 + x3 + y 3 ) dA
|∇g| =
S
2π R
R
2
(4 − r2 + r3 cos3 θ + r3 sin3 θ) r dr dθ
=
0
0
2
1 4 1 5
1 5 3
3
=
2r − r + r cos θ + r sin θ dθ
4
5
5
0
0
2π 32
32
2π
=
4+
cos3 θ +
sin3 θ dθ = 4θ + 0 + 0 = 8π.
5
5
0
0
2π
2
511
9.13 Surface Integrals
√
√
34. The surface is g(x, y, z) = x + y + z − 6. ∇g = i + j + k, |∇g| = 3 ; n = (i + j + k)/ 3 ;
√
√
√
F · n = (ey + ex + 18y)/ 3 ; zx = −1, zy = −1, dS = 1 + 1 + 1 dA = 3 dA.
F · n dS =
Flux =
=
S
6
6
6−x
(ey + ex + 18y) dA =
(ey + ex + 18y) dy dx
0
r
0
6
6−x
(ey + yex + 9y 2 ) dx =
[e6−x + (6 − x)ex + 9(6 − x)2 − 1] dx
0
0
0
6
= [−e6−x + 6ex − xex + ex − 3(6 − x)3 − x] 0
= (−1 + 6e − 6e + e − 6) − (−e + 6 + 1 − 648) = 2e6 + 634 ≈ 1440.86
6
6
6
6
35. For S1 : g(x, y, z) = x2 + y 2 − z, ∇g = 2xi + 2yj − k, |∇g| =
2xi + 2yj − k
4x2 + 4y 2 + 1 ; n1 = ;
4x2 + 4y 2 + 1
2xy 2 + 2x2 y − 5z
F · n1 = ; zx = 2x, zy = 2y, dS1 = 1 + 4x2 + 4y 2 dA. For S2 : g(x, y, z) = z − 1,
2
2
4x + 4y + 1
∇g = k, |∇g| = 1; n2 = k; F · n2 = 5z; zx = 0, zy = 0, dS2 = dA. Using polar coordinates and R: x2 + y 2 ≤ 1
we have
F · n1 dS1 +
Flux =
F · n2 dS2 =
S1
(2xy + 2x y − 5z) dA +
2
S2
2
R
5z dA
R
[2xy 2 + 2x2 y − 5(x2 + y 2 ) + 5(1)] dA
=
R
2π
1
(2r3 cos θ sin2 θ + 2r3 cos2 θ sin θ − 5r2 + 5)r dr dθ
=
0
0
1
2 5
2 5
5 4 5 2 2
2
=
r cos θ sin θ + r cos θ sin θ − r + r dθ
5
5
4
2
0
0
2π
2π
2π
2
5
1
5 2 1
2
3
2
3
=
(cos θ sin θ + cos θ sin θ) +
dθ =
sin θ − cos θ + θ 5
4
5 3
3
4 0
0
0
2
1
1
5
5
=
− − −
+ π = π.
5
3
3
2
2
36. For S1 : g(x, y, z) = x2 + y 2 + z − 4, ∇g = 2xi + 2yj + k, |∇g| = 4x2 + 4y 2 + 1 ;
2xi + 2yj + k
n1 = ; F · n1 = 6z 2 / 4x2 + 4y 2 + 1 ; zx = −2x, zy = −2y,
2
2
4x + 4y + 1
dS1 = 1 + 4x2 + 4y 2 dA. For S2 : g(x, y, z) = x2 + y 2 − z, ∇g = 2xi + 2yj − k,
2xi + 2yj − k
|∇g| = 4x2 + 4y 2 + 1 ; n2 = ; F · n2 = −6z 2 / 4x2 + 4y 2 + 1 ; zx = 2x, zy = 2y,
4x2 + y 2 + 1
dS2 = 1 + 4x2 + 4y 2 dA. Using polar coordinates and R: x2 + y 2 ≤ 2 we have
2
Flux =
F · n1 dS1 +
F · n2 dS2 =
6z dA +
−6z 2 dA
2π
S1
S1
R
2π
[6(4 − x − y ) − 6(x + y ) ] dA = 6
2
=
R
=6
0
2π
2 2
2
√
2
[(4 − r2 )2 − r4 ] r dr dθ
2 2
0
0
√2
2π
2π
√
1
1 6 2 3
− (4 − r ) − r dθ = −
[(23 − 43 ) + ( 2 )6 ] dθ =
48 dθ = 96π.
6
6
0
0
0
512
9.13
Surface Integrals
37. The surface is g(x, y, z) = x2 + y 2 + z 2 − a2 = 0. ∇g = 2xi + 2yj + 2zk,
xi + yj + zk
|∇g| = 2 x2 + y 2 + z 2 ; n = ;
x2 + y 2 + z 2
xi + yj + zk
2x2 + 2y 2 + 2z 2
F · n = −(2xi + 2yj + 2zk) · =− = −2 x2 + y 2 + z 2 = −2a.
x2 + y 2 + z 2
x2 + y 2 + z 2
Flux =
−2a dS = −2a × area = −2a(4πa2 ) = −8πa3
S
38. n1 = k, n2 = −i, n3 = j, n4 = −k, n5 = i, n6 = −j; F · n1 = z = 1, F · n2 = −x = 0, F · n3 = y = 1,
F · n4 = −z = 0, F · n5 = x = 1, F · n6 = −y = 0; Flux =
1 dS +
1 dS +
1 dS = 3
S1
S3
S5
xi + yj + zk
a
39. Refering to the solution to Problem 37, we find n = and dS = dA.
2
2
2
2
x +y +z
a − x2 − y 2
Now
r
r
kq
kq
kq
kq
F · n = kq 3 ·
= 2
= 4 |r|2 = 2 = 2
|r| |r|
|r|
|r|
x + y2 + z2
a
and
F · n dS =
Flux =
S
40. We are given σ = kz. Now zx − S
x
16 −
x2
−
y2
kq
kq
kq
dS = 2 × area = 2 (4πa2 ) = 4πkq.
a2
a
a
, zy = − y
16 − x2 − y 2
;
x2
y2
4
+
dA = dA
2
2
2
2
16 − x − y
16 − x − y
16 − x2 − y 2
Using polar coordinates,
2π 3
4
Q=
kz dS = k
16 − x2 − y 2 dA = 4k
r dr dθ
16 − x2 − y 2
S
R
0
0
3
2π
2π
1 2 9
= 4k
r dθ = 4k
dθ = 36πk.
2
2
0
0
0
√
√
41. The surface is z = 6 − 2x − 3y. Then zx = −2, zy = −3, dS = 1 + 4 + 9 = 14 dA.
The area of the surface is
3 2−2x/3 √
√ 3
2
A(s) =
2 − x dx
dS =
14 dy dx = 14
3
0
0
0
S
3
√
√
1
= 14 2x − x2 = 3 14 .
3
0
2−2x/3
3 2−2x/3 √
1
1
1 3
x̄ = √
x dS = √
14 x dy dx =
xy dx
3 0
3 14
3 14 0 0
S
0
3
1 3
2
1
2
=
2x − x2 dx =
x2 − x3 = 1
3 0
3
3
9
0
2−2x/3
3 2−2x/3 √
1
1
1 3 1 2 ȳ = √
y dS = √
14 y dy dx =
dx
y 3 0 2
3 14
3 14 0 0
S
0
2
3 3
1 3
2
1
1
2
=2
=
2− x
dx =
−
2− x
6 0
3
6
2
3
3
0
dS =
1+
513
9.13 Surface Integrals
3 2−2x/3
√
1
1
z̄ = √
z dS = √
(6 − 2x − 3y) 14 dy dx
3 14
3 14 0 0
S
2−2x/3
3
3
3 2 2 2
1
1
1 3
2 3 2
6y − 2xy − y 6 − 4x + x
dx =
=
6x − 2x + x = 2
dx =
3 0
2
3 0
3
3
9
0
0
The centroid is (1, 2/3, 2).
42. The area of the hemisphere is A(s) = 2πa2 . By symmetry, x̄ = ȳ = 0.
x
y
zx = − , zy = − ;
a2 − x2 − y 2
a2 − x2 − y 2
x2
y2
a
+ 2
dA = dA
2
2
2
2
2
−x −y
a −x −y
a − x2 − y 2
Using polar coordinates,
2π a
z dS
a
1
1
2
2
2
z=
=
a −x −y dA =
r dr dθ
2
2πa2
2πa 0
a2 − x2 − y 2
S 2πa
R
0
a
2π
2π
1
1 2 1 2
1
a
=
r
s dθ = .
dθ =
2πa 0 2 0
2πa 0 2
2
dS =
1+
a2
The centroid is (0, 0, a/2).
43. The surface is g(x, y, z) = z − f (x, y) = 0. ∇g = −fx i − fy j + k, |∇g| = fx2 + fy2 + 1 ;
−fx i − fy j + k
−P fx − Qfy + R
n= ; F·n= ; dS = 1 + fx2 + fy2 dA
1 + fx2 + fy2
1 + fx2 + fy2
−P fx − Qfy + R F · n dS =
1 + fx2 + fy2 dA =
(−P fx − Qfy + R) dA
S
R
R
1 + fx2 + fy2
EXERCISES 9.14
Stokes’ Theorem
1. Surface Integral: curl F = −10k. Letting g(x, y, z) = z − 1, we have ∇g = k and n = k. Then
(curl F) · n dS =
(−10) dS = −10 × (area of S) = −10(4π) = −40π.
S
S
Line Integral: Parameterize the curve C by x = 2 cos t, y = 2 sin t, z = 1, for 0 ≤ t ≤ 2π. Then
2π
F
·
dr
=
5y
dx
−
5x
dy
+
3
dz
=
[10 sin t(−2 sin t) − 10 cos t(2 cos t)] dt
ˇ
ˇ
C
C
2π
0
(−20 sin t − 20 cos t) dt =
2
=
0
0
514
2π
−20 dt = −40π.
2
9.14 Stokes’ Theorem
2. Surface Integral: curl F = 4i − 2j − 3k. Letting g(x, y, z) = x2 + y 2 + z − 16,
∇g = 2xi + 2yj + k, and n = (2xi + 2yj + k)/ 4x2 + 4y 2 + 1 . Thus,
(curl F) · n dS =
S
S
8x − 4y − 3
dS.
4x2 + 4y 2 + 1
Letting the surface be z = 16 − x − y , we have zx = −2x, zy = −2y, and
dS = 1 + 4x2 + 4y 2 dA. Then, using polar coordinates,
2π 4
(curl F) · n dS =
(8x − 4y − 3) dA =
(8r cos θ − 4r sin θ − 3) r dr dθ
2
S
2
0
R
0
4
2π 8 3
512
256
4 3
3 2 =
r cos θ − r sin θ − r dθ =
cos θ −
sin θ − 24 dθ
3
3
2
3
3
0
0
0
2π
512
256
=
sin θ +
cos θ − 24θ = −48π.
3
3
0
2π
Line Integral: Parameterize the curve C by x = 4 cos t, y = 4 sin t, z = 0, for 0 ≤ t ≤ 2π. Then,
2π
[−12 cos t(4 cos t)] dt
ˇ F · dr = ˇ 2z dx − 3x dy + 4y dz =
C
C
2π
=
0
2π
−48 cos2 t dt = (−24t − 12 sin 2t) = −48π.
0
0
3. Surface Integral: curl F = i + j + k. Letting g(x, y, z) = 2x + y +
2z − 6, we have
5
∇g = 2i + j + 2k and n = (2i + j + 2k)/3. Then
(curl F) · n dS =
dS. Letting
S
S 3
the surface be z = 3 − 12 y − x we have zx = −1, zy = − 12 , and
dS = 1 + (−1)2 + (− 12 )2 dA = 32 dA. Then
(curl F) · n dS =
S
R
5
3
3
5
5
45
dA = × (area of R) = (9) =
.
2
2
2
2
Line Integral: C1 : z = 3 − x, 0 ≤ x ≤ 3, y = 0; C2 : y = 6 − 2x, 3 ≥ x ≥ 0, z = 0; C3 : z = 3 − y/2, 6 ≥ y ≥ 0,
x = 0.
z
dx
+
x
dy
+
y
dz
=
z
dx
+
x
dy
+
y dz
ˇ
C
C1
C2
3
(3 − x) dx +
=
0
C3
0
x(−2 dx) +
3
0
y(−dy/2)
6
0
0
3
−x2 − 1 y 2 = 9 − (0 − 9) − 1 (0 − 36) = 45
4 6 2
4
2
0
3
4. Surface Integral: curl F = 0 and
(curl F) · n dS = 0.
=
1
3x − x2
2
S
Line Integral: the curve is x = cos t, y = sin t, z = 0, 0 ≤ t ≤ 2π.
2π
x
dx
+
y
dy
+
z
dz
=
[cos t(− sin t) + sin t(cos t)]dt = 0.
ˇ
C
0
515
9.14 Stokes’ Theorem
√
5. curl F = 2i + j. A unit vector normal to the plane is n = (i + j + k)/ 3 . Taking the
equation of the plane to be z = 1 − x − y, we have zx = zy = −1. Thus,
√
√
dS = 1 + 1 + 1 dA = 3 dA and
√
√ √
F
·
dr
=
(curl
F)
·
n
dS
=
3
dS
=
3
3 dA
ˇ
C
S
S
R
= 3 × (area of R) = 3(1/2) = 3/2.
√
6. curl F = −2xzi + z 2 k. A unit vector normal to the plane is n = (j + k)/ 2 . From z = 1 − y, we have zx = 0
√
√
and zy = −1. Thus, dS = 1 + 1 dA = 2 dA and
√
1
√ z 2 2 dA =
(1 − y)2 dA
2
S
R
R
1
2 1
2
2
1
1
2
=
(1 − y)2 dy dx =
− (1 − y)3 dx =
dx = .
3
3
0
0
0
0 3
0
ˇ F · dr =
C
(curl F) · n dS =
√
7. curl F = −2yi − zj − xk. A unit vector normal to the plane is n = (j + k)/ 2 . From z = 1 − y we have zx = 0
√
√
and zy = −1. Then dS = 1 + 1 dA = 2 dA and
√
1
− √ (z + x)
2 dA =
(y − x − 1) dA
2
S
R
R
1
2 1
2
2
1 2
1
=
−x −
(y − x − 1) dy dx =
y − xy − y dx =
dx
2
2
0
0
0
0
0
2
1 2 1
= − x − x = −3.
2
2
0
ˇ F · dr =
C
(curl F) · n dS =
8. curl F = 2i + 2j + 3k. Letting g(x, y, z) = x + 2y + z − 4, we have ∇g = i + 2j + k
√
and n = (i + 2j + k)/ 6 . From z = 4 − x − 2y we have zx = −1 and zy = −2. Then
√
dS = 6 dA and
√
1
√
F
·
dr
=
(curl
F)
·
n
dS
=
6
dA
=
9 dA = 9(4) = 36.
(9)
ˇ
6
C
S
R
R
√
9. curl F = (−3x2 − 3y 2 )k. A unit vector normal to the plane is n = (i + j + k)/ 3 . From
√
z = 1 − x − y, we have zx = zy = −1 and dS = 3 dA. Then, using polar coordinates,
√
√
√
(curl F) · n dS =
(− 3 x2 − 3 y 2 ) 3 dA
ˇ F · dr =
C
S
R
2π
(−x2 − y 2 ) dA = 3
=3
=3
0
R
2π
1
(−r2 )r dr dθ
0
0
1
2π
1 4 1
3π
− r dθ = 3
− dθ = −
.
4
4
2
0
0
2yj + k
10. curl F = 2xyzi − y 2 zj + (1 − x2 )k. A unit vector normal to the surface is n = . From z = 9 − y 2 we
4y 2 + 1
have zx = 0, zy = −2y and dS = 1 + 4y 2 dA. Then
516
9.14 Stokes’ Theorem
ˇ F · dr =
C
(curl F) · n dS =
3
y/2
(−2y 3 z + 1 − x2 ) dA =
S
[−2y 3 (9 − y 2 ) + 1 − x2 ] dx dy
0
R
0
y/2
3
1
1
1
=
−18y 3 x + 2y 5 x + x − x3 −9y 4 + y 6 + y − y 3 dy
dy =
3
2
24
0
0
0
3
9
1
1
1
= − y 5 + y 7 + y 2 − y 4 ≈ 123.57.
5
7
4
96
0
3
11. curl F = 3x2 y 2 k. A unit vector normal to the surface is
8xi + 2yj + 2zk
4xi + yj + zk
n= =
.
2
2
2
64x + 4y + 4z
16x2 + y 2 + z 2
4x
From zx = − 4 − 4x2 − y 2
y
, zy = − 4 − 4x2 − y 2
ˇ F · dr =
C
3x2 y 2 dA
=
16x2 + y 2 + z 2
R
3x2 y 2 z
(curl F) · n dS =
S
1 + 3x2
dA. Then
4 − 4x2 − y 2
we obtain dS = 2
2
1 + 3x2
4 − 4x2 − y 2
dA
Using symmetry
R
1
√
2 1−x2
1
2 2
= 12
x y dy dx = 12
0
0
0
1 2 3
x y
3
2√1−x2
dx
0
1
x2 (1 − x2 )3/2 dx
= 32
x = sin t, dx = cos t dt
0
π/2
sin2 t cos4 t dt = π.
= 32
0
12. curl F = i + j + k. Taking the surface S bounded by C to be the portion of the plane
√
√
x + y + z = 0 inside C, we have n = (i + j + k)/ 3 and dS = 3 dA.
√
√ √
(curl F) · n dS =
3 dS = 3
3 dA = 3 × (area of R)
ˇ F · dr =
C
S
S
R
The region R is obtained by eliminating z from the equations of the plane and the sphere.
This gives x2 + xy + y 2 = 12 . Rotating axes, we see that R is enclosed by the ellipse
X 2 /(1/3) + Y 2 /1 = 1 in a rotated coordinate system. Thus,
√
1
√
F
·
dr
=
3
×
(area
of
R)
=
3
π
1
= 3 π.
ˇ
3
C
13. Parameterize C by x = 4 cos t, y = 2 sin t, z = 4, for 0 ≤ t ≤ 2π. Then
2
(curl F) · n dS = ˇ F · dr = ˇ 6yz dx + 5x dy + yzex dz
S
C
2π
=
C
[6(2 sin t)(4)(−4 sin t) + 5(4 cos t)(2 cos t) + 0] dt
0
2π
0
0
517
2π
(5 − 29 sin2 t) dt = −152π.
(−24 sin2 t + 5 cos2 t) dt = 8
=8
9.14 Stokes’ Theorem
14. Parameterize C by x = 5 cos t, y = 5 sin t, z = 4, for 0 ≤ t ≤ 2π. Then,
(curl F) · n dS = ˇ F · r = ˇ y dx + (y − x) dy + z 2 dz
S
C
2π
C
[(5 sin t)(−5 sin t) + (5 sin t − 5 cos t)(5 cos t)] dt
=
0
2π
(25 sin t cos t − 25) dt =
=
0
25
sin2 t − 25t
2
2π
= −50π.
0
15. Parameterize C by C1 : x = 0, z = 0, 2 ≥ y ≥ 0; C2 : z = x, y = 0, 0 ≤ x ≤ 2;
C3 : x = 2, z = 2, 0 ≤ y ≤ 2; C4 : z = x, y = 2, 2 ≥ x ≥ 0. Then
(curl F) · n dS = ˇ F · r = ˇ 3x2 dx + 8x3 y dy + 3x2 y dz
C
C
S
2
=
0 dx + 0 dy + 0 dz +
3x dx +
64 dy +
3x2 dx + 6x2 dx
C1
2
C2
3x2 dx +
=
0
2
0
64 dy +
0
C3
C4
2
0
2
9x2 dx = x3 + 64y + 3x3 = 112.
0
2
0
2
16. Parameterize C by x = cos t, y = sin t, z = sin t, 0 ≤ t ≤ 2π. Then
(curl F) · n dS = ˇ F · r = ˇ 2xy 2 z dx + 2x2 yz dy + (x2 y 2 − 6x) dz
S
C
2π
C
[2 cos t sin2 t sin t(− sin t) + 2 cos2 t sin t sin t cos t
=
0
+ (cos2 t sin2 t − 6 cos t) cos t] dt
2π
(−2 cos t sin4 t + 3 cos3 t sin2 t − 6 cos2 t) dt = −6π.
=
0
2
1
17. We take the surface to be z = 0. Then n = k and dS = dA. Since curl F =
i + 2zex j + y 2 k,
2
1+y
2
2 x
−1
2
z
e
dx
+
xy
dy
+
tan
y
dz
=
(curl
F)
·
n
dS
=
y
dS
=
y 2 dA
ˇ
C
S
2π
S
3
2
=
2
r sin θ r dr dθ =
0
=
81
4
0
2π
0
sin2 θ dθ =
0
R
2π
3
1 4 2 r sin θ dθ
4
0
81π
.
4
2xi + 2yj + k
18. (a) curl F = xzi − yzj. A unit vector normal to the surface is n = and
4x2 + 4y 2 + 1
dS = 1 + 4x2 + 4y 2 dA. Then, using x = cos t, y = sin t, 0 ≤ t ≤ 2π, we have
2
2
(curl F) · n dS =
(2x z − 2y z) dA =
(2x2 − 2y 2 )(1 − x2 − y 2 ) dA
S
R
R
=
(2x2 − 2y 2 − 2x4 + 2y 4 ) dA
R
2π
1
(2r2 cos2 θ − 2r2 sin2 θ − 2r4 cos4 θ + 2r4 cos4 θ) r dr dθ
=
0
0
2π
1
[r3 cos 2θ − r5 (cos2 θ − sin2 θ)(cos2 θ + sin2 θ)] dr dθ
=2
0
0
518
9.15 Triple Integrals
2π
1
=2
0
1
=
6
5
0
2π
(r cos 2θ − r cos 2θ) dr dθ = 2
3
cos 2θ
0
1 4 1 6
r − r
4
6
1
dθ
0
2π
cos 2θ dθ = 0.
0
(b) We take the surface to be z = 0. Then n = k, curl F · n = curl F · k = 0 and
(curl F) · n dS = 0.
S
(c) By Stokes’ Theorem, using z = 0, we have
(curl F) · n dS = ˇ F · dr = ˇ xyz dz = ˇ xy(0) dz = 0.
C
S
C
C
EXERCISES 9.15
Triple Integrals
4
2
1
1 2
(x + y + z)dx dy dz =
x + xy + xz dy dz
2
−1
2
−2
−1
4 2
4
4
4
2
2
=
(2y + 2z) dy dz =
(y + 2yz) dz =
8z dz = 4z 2 = 48
1.
−2
2
3
x
1
−2
2
xy
2.
3
1
2
=
1
3
2
−2
2
x
24xy dz dy dx =
1
4
xy
24xyz dy dx =
2
1
3
2
2
x
(24x2 y 2 − 48xy)dy dx
1
1
3
x
(8x2 y 3 − 24xy 2 ) dx =
(8x5 − 24x3 − 8x2 + 24x) dx
1
1
1
3
4 6
8 3
14
1552
4
2
=
x − 6x − x + 12x = 522 −
=
3
3
3
3
1
6−x
6 6−x 6−x−z
6 6−x
6
1
6z − xz − z 2 3.
dy dz dx =
(6 − x − z)dz dx =
dx
2
0
0
0
0
0
0
0
6
6
1
1 2
2
=
6(6 − x) − x(6 − x) − (6 − x) dx =
18 − 6x + x dx
2
2
0
0
6
1
= 18x − 3x2 + x3 = 36
6
0
1
1−x
√
y
1
2 3
4.
0
0
√ y
x z dy dx =
1
2 4
4x z dz dy dx =
0
1−x
0
0
0
0
1−x
x2 y 2 dy dx
0
1−x
1 2 3 1 1 2
1 1 2
dx =
x (1 − x)3 dx =
(x − 3x3 + 3x4 − x5 )dx
x y 3
3
3
0
0
0
0
1
1 1 3 3 4 3 5 1 6 1
=
x − x + x − x =
3 3
4
5
6
180
0
1
=
519
9.15 Triple Integrals
π/2
y2
y
5.
0
0
0
π/2 y2
π/2
x
x
x
cos dz dx dy =
y cos dx dy =
y 2 sin
y
y
y
0
0
0
π/2
=
y 2 sin y dy
Integration by parts
y 2
dy
0
0
π/2
= (−y 2 cos y + 2 cos y + 2y sin y) = π − 2
0
√
2
6.
√
0
2
√
2
ex
2
x dz dx dy =
0
y
2
√
0
√
2
2
xex dx dy =
0
y
2
√2
1 x2 1
(e4 − ey )dy
e dy =
√
2
2 0
y
√2 1
√
√
√
√
1
1
4
y = (ye − e ) = [(e4 2 − e 2 ) − (−1)] = (1 + e4 2 − e 2 )
2
2
2
0
1
1
2−x2 −y 2
7.
1
1
z
xye dz dx dy =
0
0
0
2−x2 −y2
xye dx dy =
1
1
0
0
0
2
(xye2−x
z
0
−y 2
− xy)dx dy
0
1
1
2
2
2
2
1
1
1
1
1
− ye2−x −y − x2 y dy =
− ye1−y − y + ye2−y dy
2
2
2
2
2
0
0
0
1 1 1−y2 1 2 1 2−y2 1 2 1
1 1 1
1
1 2
=
− y − e
e
= 4 − 4 − 4e − 4e − 4e = 4e − 2e
4
4
4
0
1
=
4
1/2
x2
8.
0
0
0
1
x2 − y 2
4
1/2
dy dx dz =
0
sin−1
0
x2
4 1/2
y dx
dz
=
sin−1 x dx dz
x 0
0
0
Integration by parts
4
=
9.
−1
(x sin
0
5
3
x+
1/2
1 − x2 ) dz =
0
y+2
z dV =
5
0
3
z dx dy dz =
0
D
=
0
1
5
0
y
3
2yz dz =
1
2
4
0
2
x2
4
1
0
4−y
2
4
(x2 + y 2 ) dz dy dx = 2
0
y
1
5
4z dz = 2z 2 = 50
0
10. Using symmetry,
(x2 + y 2 ) dV = 2
D
5
y+2
xz √
√
3
1 π
π
+
− 1 dz = + 2 3 − 4
2 6
2
3
5 3
dy dz =
2z dy dz
4
0
0
x2
4−y
(x2 + y 2 )z dy dx
(4x2 − x2 y + 4y 2 − y 3 ) dy dx
=2
x2
0
4
1
4
1
4x2 y − x2 y 2 + y 3 − y 4 dx
2
3
4
0
x2
2 64
5 6 1 8
2
4
=2
8x +
dx
− 4x + x − x
3
6
4
0
2
8 3 64
5 7
1 9 23,552
4 5
=2
x + x− x − x + x =
.
3
3
5
42
36
315
0
2
=2
520
0
9.15 Triple Integrals
4
2−x/2
4
11. The other five integrals are
4
z
F (x, y, z) dz dy dx,
0
0
(z−x)/2
x+2y
4
4
(z−x)/2
F (x, y, z) dy dx dz,
0
4
0
0
z/2 z−2y
0
0
F (x, y, z) dy dz dx,
0
2
x
4
0
z−2y
F (x, y, z) dx dy dz,
0
F (x, y, z) dx dz dy.
0
3
2y
0
√36−4y2 /3 12. The other five integrals are
3
2
√
0
0
36−9x2 /2
3
3
F (x, y, z) dz dx dy,
√
2
1
3
36−4y /3
F (x, y, z) dy dx dz,
1
0
0
F (x, y, z) dx dy dz,
1
3 3 √36−4y2 /3
0
2
0
√
3
36−9x2 /2
F (x, y, z) dx dz dy,
0
1
0
F (x, y, z) dy dz dx.
0
2
8
13. (a) V =
4
dz dy dx
x3
0
14. Solving z =
√
1
0
0
8
4
y 1/3
(b) V =
dx dz dy
0
1
3
2−z
(b) V =
3
dy dx dz
0
1
z2
0
3
1
√
0
0
15.
16.
17.
18.
19.
20.
521
0
x2
z2
0
3
2
dz dx dy +
0
8
dx dz dy
x
(c) V =
2−z
x and x + z = 2, we obtain x = 1, z = 1. (a) V =
2
dy dx dz
0
0
(c) V =
0
0
4
2−x
dz dx dy
0
1
0
The region in the first
octant is shown.
9.15 Triple Integrals
√
21. Solving x = y 2 and 4 − x = y 2 , we obtain x = 2, y = ± 2 . Using symmetry,
3 √2 4−y2
3 √2
V =2
dx dy dz = 2
(4 − 2y 2 )dy dz
0
=2
0
√
2
4−x2 0
3
y2
√
4−x2
dz dy dx =
0
0
2
√
0
4−x2
=
0
2
(x + y) dy dx =
0
0
0
3
0
0
2
x+y
22. V =
0
√ 2
2 3 4y − y dz = 2
3
0
√
√
8 2
dz = 16 2 .
3
x+y
z
dy dx
0
0
√4−x2
1 2 xy + y dx
2
0
2
1
1
1
=
x 4 − x2 + (4 − x2 ) dx = − (4 − x2 )3/2 + 2x − x3
2
3
6
0
4
8
16
= 4−
− −
=
3
3
3
2
0
23. Adding the two equations, we obtain 2y = 8. Thus, the paraboloids intersect
in the plane y = 4. Their intersection is a circle of radius 2. Using symmetry,
2 √4−x2 8−x2 −z2
2 √4−x2
V =4
dy dz dx = 4
(8 − 2x2 − 2z 2 ) dz dx
x2 +z 2
0
0
0
0
√4−x2
2
2
4
2
2
3 =4
(4 − x2 )3/2 dx
2(4 − x )z − z dx = 4
3
0
0 3
0
2
x 16 x
=
= 16π.
− (2x2 − 20) 4 − x2 + 6 sin−1
3
8
2 0
Trig substitution
24. Solving x = 2, y = x, and z = x2 + y 2 , we obtain the point (2, 2, 8).
x
2 x x2 +y2
2 x
2
1 3 2
2
2
V =
x y + y dx
dz dy dx =
(x + y ) dy dx =
3
0
0
0
0
0
0
0
2
2
4 3
1 16
=
x dx = x4 =
.
3
3
0 3
0
25. We are given ρ(x, y, z) = kz.
8 4 y1/3
m=
kz dx dz dy = k
0
0
0
0
4
y1/3
xz dz dy = k
0
0
0
8
4
8
8
1 1/3 2 3 4/3 1/3
y z dy = 8k
y
y dy = 8k
= 96k
2
4
0
0
0
8
=k
0
8
8
4
y 1/3
8
2
Mxy =
=k
0
0
8
0
4
y 1/3 z dz dy
0
y1/3
xz dz dy = k
8
4
2
kz dx dz dy = k
0
4
0
0
0
0
y 1/3 z 2 dz dy
0
4
8
8
1 1/3 3 64
64
3 4/3 y 1/3 dy =
y z dy =
k
k
y
= 256k
3
3
3
4
0
0
0
522
9.15 Triple Integrals
8
4
Mxz =
8
4
kyz dx dz dy = k
0
0
0
0
y1/3
xyz dz dy = k
0
0
0
8
4
y 4/3 z dz dy
0
4
8
8
1 4/3 2 3 7/3 3072
4/3
=k
y z dy = 8k
y
y dy = 8k
= 7 k
2
7
0
0
0
0
y1/3
8 4 y1/3
8 4
8 4
1 2 1
=
kxz dx dz dy = k
dz dy = k
y 2/3 z dz dy
x z
2
0
0
0
0
0 2
0
0
0
4
8
8
8
1
1 2/3 2 3
2/3
5/3
= 384 k
= k
y dy = 4k
y z dy = 4k
y
2 0 2
5
5
0
0
0
Myz
y 1/3
8
384k/5
3072k/7
256k
= 4/5; ȳ = Mxz /m =
= 32/7; z̄ = Mxy /m =
= 8/3
96k
96k
96k
The center of mass is (4/5, 32/7, 8/3).
x̄ = Myz /m =
26. We use the form of the integral in Problem 14(b) of this section. Without loss of generality, we take ρ = 1.
1
1 2−z 3
1 2−z
1
1 2 1 3 7
2
m=
dy dx dz =
3 dx dz = 3
(2 − z − z ) dz = 3 2z − z − z
=
2
3
2
0
z2
0
0
z2
0
0
1 2−z 3
1 2−z
1 2−z
3
Mxy =
z dy dx dz =
yz dx dz =
3z dx dz
z2
0
0
z2
0
z2
0
0
z2
0
0
Myz
z2
0
z2
1
1
2−z
1 3 1 4 5
2
3
2
=3
xz dz = 3
(2z − z − z ) dz = 3 z − z − z
=
2
3
4
4
z
0
0
0
3
1 2−z 3
1 2−z
1 2−z
1 2 9
=
y dy dx dz =
dx dz
y dx dz =
2
2 0 z2
0
z2
0
0
z2
0
1
9 1
9
21
1 2 1 3 2
=
(2 − z − z ) dz =
2z − z − z
= 4
2 0
2
2
3
0
1 2−z 3
1 2−z
1 2−x
3
=
x dy dx dz =
xy dx dz =
3x dx dz
Mxz
0
1
0
2−z
1
1 2 3 1
3
1 3 1 5 16
2
4
2
x
4z − 2z + z − z
dz =
(4 − 4z + z − z ) dz =
= 5
2 z 2
2 0
2
3
5
0
1
=3
0
16/5
21/4
5/4
= 32/35, ȳ = Mxz /m =
= 3/2, z̄ = Mxy /m =
= 5/14.
7/2
7/2
7/2
The centroid is (32/35, 3/2, 5/14).
x̄ = Myz /m =
27. The density is ρ(x, y, z) = ky. Since both the region and the density function are symmetric
with respect to the xy-and yz-planes, x̄ = z̄ = 0. Using symmetry,
3 2 √4−x2
3 2
3 2 √4−x2
m=4
ky dz dx dy = 4k
yz dx dy = 4k
y 4 − x2 dx dy
0
0
0
0
0
0
0
0
2
3
3 3
x
1 2 −1 x 2
= 4k
y
4 − x + 2 sin
dy = 4k
πy dy = 4πk
y = 18πk
2
2 0
2
0
0
0
√
3 2 4−x2
3 2
3 2 √4−x2
2
2 Mxz = 4
ky dz dx dy = 4k
y z
dx dy = 4k
y 2 4 − x2 dx dy
0
0
3
y2
= 4k
0
ȳ = Mxz /m =
0
x
2
0
4 − x2 + 2 sin−1
0
0
0
0
2
3
3
x 1 3 2
dy = 4k
πy dy = 4πk
y = 36πk.
2 0
3
0
0
36πk
= 2. The center of mass is (0, 2, 0).
18πk
523
9.15 Triple Integrals
28. The density is ρ(x, y, z) = kz.
y+2
1 2 kz dz dy dx = k
dy dx
z 0
x2 0
0
x2 2
0
x
1 x
1
1
1
1
2
3 (y + 2) dy dx = k
k
(y + 2) dx
2 0 x2
2 0 3
x2
1
1
1
1
k
[(x + 2)3 − (x2 + 2)3 ] dx = k
[(x + 2)3 − (x6 + 6x4 + 12x2 + 8)]dx
6 0
6 0
1
1 1
1 7 6 5
407
4
3
k (x + 2) − x − x − 4x − 8x =
k
6 4
7
5
840
0
m=
=
=
=
1
1
x
y+2
x
x
1
x
kz 2 dz dy dx = k
0
=
1
y+2
Mxy =
=
1
k
3
x2
1
0
1
k
12
0
x2
0
y+2
1 x
1 3 1
dy dx = k
(y + 2)3 dy dx
z 3
3
2
0
x
0
x
1
1
1
4 [(x + 2)4 − (x2 + 2)4 ] dx
(y + 2) dx =
k
4
12 0
x2
1
[(x + 2)4 − (x8 + 8x6 + 24x4 + 32x2 + 16)] dx
0
1
1
1 9 8 7 24 32 3
1493
1
5
=
k (x + 2) − x − x −
− x − 16x =
k
12 5
9
7
5
3
1890
0
y+2
1 x
1 2 1
dy dx = k
y(y + 2)2 dy dx
yz 2
2
2
2
2
0
x
0
0
x
0
x
0
x
1 x
1
1
1
1
4
= k
(y 3 + 4y 2 + 4y) dy dx = k
y 4 + y 3 + 2y 2 dx
2 0 x2
2 0
4
3
x2
1
x
y+2
Mxz =
1
x
kyz dz dy dx = k
1
4
4
− x8 − x6 − 74x4 + x3 + 2x2 dx
4
3
3
0
1
1
4 7
7 5 1 4 2 3 68
1 9
= k − x − x − x + x + x =
k
2
36
21
20
3
3
315
0
=
1
k
2
1
1
x
Myz =
=
=
1
x
kxz dz dy dx = k
0
=
y+2
1
k
2
1
k
6
x2
1
0
0
0
x2
y+2
1 x
1 2 1
dy dx = k
x(y + 2)2 dy dx
xz 2
2
2
0
x
0
x
1
1
1
3 [x(x + 2)3 − x(x2 + 2)3 ] dx
x(y + 2) dx = k
3
6 0
x2
1
[x4 + 6x3 + 12x2 + 8x − x(x2 + 2)3 ] dx
0
1
1 1 5 3 4
1
21
k x + x + 4x3 + 4x2 − (x2 + 2)4 =
k
6 5
2
8
80
0
x̄ = Myz /m =
21k/80
68k/315
= 441/814, ȳ = Mxz /m =
= 544/1221,
407k/840
407k/840
z̄ = Mxy /m =
1493k/1890
= 5972/3663. The center of mass is (441/814, 544/1221, 5972/3663).
407k/840
524
9.15 Triple Integrals
1
29. m =
√
1−x2
−1
8−y
(x + y + 4) dz dy dx
√
− 1−x2
2+2y
30. Both the region and the density function
are symmetric with respect to the xz- and
2 √1+z2 √1+z2 −y2
yz-planes. Thus, m = 4
z 2 dx dy dz.
−1
0
0
31. We are given ρ(x, y, z) = kz.
8 4 y1/3
Iy =
kz(x2 + z 2 )dx dz dy = k
y1/3
1 3
dz dy
x z + xz 3 3
0
0
0
0
0
0
4
8 4
8
1
1 2 1 1/3 4 =k
yz + y 1/3 z 3 dz dy = k
yz + y z
dy
3
6
4
0
0
0
0
8
8
8
4
2560
=k
y + 64y 1/3 dy = k
y 2 + 48y 4/3 =
k
3
3
3
0
0
√
2560k/3
4 5
From Problem 25, m = 96k. Thus, Rg = Iy /m =
=
.
96k
3
32. We are given ρ(x, y, z) = k.
3
1 2−z 3
1 2−z 1 2−z
1 3
2
2
2
dx dz = k
Ix =
k(y + z )dy dx dz = k
(9 + 3z 2 ) dx dz
y + yz
3
0
z2
0
0
z2
0
z2
0
1
1
2−z
2 =k
(9x + 3xz ) dz = k
(18 − 9z − 3z 2 − 3z 3 − 3z 4 ) dz
z2
0
8
4
0
1
9
3
3
223
= k 18z − z 2 − z 3 − z 4 − z 5 =
k
2
4
5
20
0
1
1 2−z 3
1 2−z
1
1
1
7
m=
k dy dx dz = k
3 dx dz = 3k
(2 − z − z 2 ) dz = 3k 2z − z 2 − z 3 = k
2
3
2
2
2
0
z
0
0
z
0
0
Ix
=
m
Rg =
1
223k/20
=
7k/2
1−x
223
70
1−x−y
1
1−x
(x2 + y 2 )(1 − x − y) dy dx
(x2 + y 2 ) dz dy dx = k
33. Iz = k
0
0
1
0
0
0
1−x
(x2 − x3 − x2 y + y 2 − xy 2 − y 3 ) dy dx
=k
0
0
1
1−x
1 2 2 1
1 4 3
(x − x )y − x y + (1 − x)y − y dx
2
3
4
0
2
=k
0
=k
0
1
3
1
1 2
1
1
1
1
1
1
k
x − x3 + x4 + (1 − x)4 dx = k x6 − x4 + x5 − (1 − x)5 =
2
2
12
6
4
10
60
30
0
525
9.15 Triple Integrals
34. We are given ρ(x, y, z) = kx.
1 2 4−z
Iy =
kx(x2 + z 2 ) dy dx dz = k
0
0
z
1
2
3
36.
37.
38.
39.
40.
41.
42.
0
1
2
4−z
(x3 + xz 2 )y dx dz
0
z
2
1 4 1 2 2
=k
(x + xz )(4 − 2z) dx dz = k
x + x z (4 − 2z) dz
4
2
0
0
0
0
1
1
1
2
1
41
=k
(4 + 2z 2 )(4 − 2z) dz = 4k
(4 − 2z + 2z 2 − z 3 ) dz = 4k 4z − z 2 + z 3 − z 4 =
k
3
4
3
0
0
0
√
√
√
√
x = 10 cos 3π/4 = −5 2 ; y = 10 sin 3π/4 = 5 2 ; (−5 2 , 5 2 , 5)
√
√
x = 2 cos 5π/6 = − 3 ; y = 2 sin 5π/6 = 1; (− 3 , 1, −3)
√
√
√
√
x = 3 cos π/3 = 3/2; y = 3 sin π/3 = 3/2; ( 3/2, 3/2, −4)
√
√
√
√
x = 4 cos 7π/4 = 2 2 ; y = 4 sin 7π/4 = −2 2 ; (2 2 , −2 2 , 0)
√
With x = 1 and y = −1 we have r2 = 2 and tan θ = −1. The point is ( 2 , −π/4, −9).
√
√
With x = 2 3 and y = 2 we have r2 = 16 and tan θ = 1/ 3 . The point is (4, π/6, 17).
√
√
√
√
With x = − 2 and y = 6 we have r2 = 8 and tan θ = − 3 . The point is (2 2 , 2π/3, 2).
√
With x = 1 and y = 2 we have r2 = 5 and tan θ = 2. The point is ( 5 , tan−1 2, 7).
35.
1
2
43. r2 + z 2 = 25
44. r cos θ + r sin θ − z = 1
45. r2 − z 2 = 1
46. r2 cos2 θ + z 2 = 16
47. z = x2 + y 2
48. z = 2y
49. r cos θ = 5, x = 5
√
√
√
50. tan θ = 1/ 3 , y/x = 1/ 3 , x = 3 y, x > 0
51. The equations are r2 = 4, r2 + z 2 = 16, and z = 0.
2π 2 √16−r2
2π 2 V =
r dz dr dθ =
r 16 − r2 dr dθ
0
0
52. The equation is z = 10 − r2 .
2π 3 10−r2
V =
r dz dr dθ =
2π
0
0
2π
=
0
0
0
2π
=
0
0
2
2π
√
√
1
1
2π
2 3/2 − (16 − r )
dθ
=
(64 − 24 3 ) dθ =
(64 − 24 3 )
3
3
3
0
0
1
3
0
0
0
81
81π
dθ =
.
4
2
53. The equations are z = r2 , r = 5, and z = 0.
2π 5 r2
2π 5
V =
r dz dr dθ =
r3 dr dθ =
0
=
0
0
2π
2π
r(9 − r2 ) dr dθ =
0
0
0
0
2π
5
1 4 r dθ
4
0
625
625π
dθ =
4
2
526
9 2 1 4
r − r
2
4
3
dθ
0
9.15 Triple Integrals
54. Substituting the first equation into the second, we see that the surfaces intersect
in the plane y = 4. Using polar coordinates in the xz-plane, the equations of the
surfaces become y = r2 and y = 12 r2 + 2.
2π 2 r2 /2+2
2π 2 2
r
V =
+ 2 − r2 dr dθ
r dy dr dθ =
r
2
0
0
r2
0
0
2
2π 2 2π 2π
1 3
1 4 2
=
2r − r dr dθ =
r − r dθ =
2 dθ = 4π
2
8
0
0
0
0
0
√
55. The equation is z = a2 − r2 . By symmetry, x̄ = ȳ = 0.
2π a √a2 −r2
2π a m=
r dz dr θ =
r a2 − r2 dr dθ
0
=
0
0
a
dθ =
2π
0
1 3
1
2
− (a2 − r2 )3/2
a dθ = πa3
3
3
3
0
0
√a2 −r2
2π a √a2 −r2
2π a
1 2 1 2π a
rz =
zr dz dr dθ =
dr dθ =
r(a2 − r2 ) dr dθ
2 0
0
0
0
0
0 2
0
0
a
1 2π 1 2 2 1 4 1 2π 1 4
1
=
a r − r dθ =
a dθ = πa4
2 0
2
4
2
4
4
0
0
0
Mxy
0
2π
z̄ = Mxy /m =
πa4 /4
= 3a/8. The centroid is (0, 0, 3a/8).
2πa3 /3
56. We use polar coordinates in the yz-plane. The density is ρ(x, y, z) = kz. By symmetry,
ȳ = z̄ = 0.
5
2π 4 5
2π 4
1 2 k 2π 4
m=
rz dr dθ =
kxr dx dr dθ = k
25r dr dθ
2 0
0
0
0
0
0 2
0
0
4
25k 2π 1 2 25k 2π
=
dθ =
8 dθ = 200kπ
r
2 0 2 0
2 0
5
2π 4 5
2π 4
2π 4
1 3 1
2
Myz =
kx r dx dr dθ = k
125r dr dθ
rx dr dθ = k
3 0
0
0
0
0
0 3
0
0
4
2π
2π
1
125 2 1
2000
= k
1000 dθ =
r dθ = k
kπ
3 0
2
3
3
0
0
x̄ = Myz /m =
2000kπ/3
= 10/3. The center of mass of the given solid is (10/3, 0, 0).
200kπ
√
√
57. The equation is z = 9 − r2 and the density is ρ = k/r2 . When z = 2, r = 5 .
2π √5 √9−r2
2π √5 √9−r2
2
2
Iz =
r (k/r )r dz dr dθ = k
rz dr dθ
0
0
2π =k
√
2
5
(r
0
=k
0
9 − r2 − 2r) dr dθ = k
0
2π
0
0
2π
0
2
√5
1
2 3/2
2 − (9 − r ) − r dθ
3
0
8
4
dθ = πk
3
3
527
9.15 Triple Integrals
58. The equation is z = r and the density is ρ = kr.
2π 1 1
2π 2
2
Ix =
(y + z )(kr)r dz dr dθ = k
0
0
r
0
0
1
1
(r4 sin2 θ + r2 z 2 ) dz dr dθ
r
1
1
(r4 sin2 θ)z + r2 z 3 dr dθ
3
r
0 2π 0 1 1 2
1
2
4
5
r sin θ + r − r sin2 θ − r5 dr dθ
=k
3
3
0
0
1
2π 2π 1 5 2
1
1 3 1 6 2
1 6 1
=k
r sin θ + r − r sin θ − r dθ = k
sin2 θ +
dθ
5
9
6
18
30
18
0
0
0
2π
1
13
1
1
=k
θ−
sin 2θ + θ =
πk
60
120
18
90
0
√
(a) x = (2/3) sin(π/2) cos(π/6) = 3/3; y = (2/3) sin(π/2) sin(π/6) = 1/3;
√
z = (2/3) cos(π/2) = 0; ( 3/3, 1/3, 0)
√
√
(b) With x = 3/3 and y = 1/3 we have r2 = 4/9 and tan θ = 3/3. The point is (2/3, π/6, 0).
√
√
(a) x = 5 sin(5π/4) cos(2π/3) = 5 2/4; y = 5 sin(5π/4) sin(2π/3) = −5 6/4;
√
√
√
√
z = 5 cos(5π/4) = −5 2/2; (5 2/4, −5 6/4, −5 2/2)
√
√
√
(b) With x = 5 2/4 and y = −5 6/4 we have r2 = 25/2 and tan θ = − 3 .
√
√
The point is (5/ 2 , 2π/3, −5 2/2).
√
(a) x = 8 sin(π/4) cos(3π/4) = −4; y = 8 sin(π/4) sin(3π/4) = 4; z = 8 cos(π/4) = 4 2 ;
√
(−4, 4, 4 2)
√
√
(b) With x = −4 and y = 4 we have r2 = 32 and tan θ = −1. The point is (4 2 , 3π/4, 4 2 ).
√
(a) x = (1/3) sin(5π/3) cos(π/6) = −1/4; y = (1/3) sin(5π/3) sin(π/6) = − 3/12;
√
z = (1/3) cos(5π/3) = 1/6; (−1/4, − 3/12, 1/6)
√
√
√
(b) With x = −1/4 and y = − 3/12 we have r2 = 1/12 and tan θ = 3/3. The point is (1/2 3 , π/6, 1/6).
√
With x = −5, y = −5, and z = 0, we have ρ2 = 50, tan θ = 1, and cos φ = 0. The point is (5 2 , π/2, 5π/4).
√
√
√
With x = 1, y = − 3 , and z = 1, we have ρ2 = 5, tan θ = − 3 , and cos φ = 1/ 5 . The point is
√
√
( 5 , cos−1 1/ 5 , −π/3).
√
√
√
With x = 3/2, y = 1/2, and z = 1, we have ρ2 = 2, tan θ = 1/ 3 , and cos φ = 1/ 2 . The point is
√
( 2 , π/4, π/6).
√
With x = − 3/2, y = 0, and z = −1/2, we have ρ2 = 1, tan θ = 0, and cos φ = −1/2. The point is (1, 2π/3, 0).
2π
1
=k
59.
60.
61.
62.
63.
64.
65.
66.
67. ρ = 8
68. ρ2 = 4ρ cos φ; ρ = 4 cos φ
√
69. 4z 2 = 3x2 + 3y 2 + 3z 2 ; 4ρ2 cos2 φ = 3ρ2 ; cos φ = ± 3/2; φ = π/6 or equivalently, φ = 5π/6
70. −x2 − y 2 − z 2 = 1 − 2z 2 ; −ρ2 = 1 − 2ρ2 cos2 φ; ρ2 (2 cos2 φ − 1) = 1
71. x2 + y 2 + z 2 = 100
72. cos φ = 1/2; ρ2 cos2 φ = ρ2 /4; 4z 2 = x2 + y 2 + z 2 ; x2 + y 2 = 3z 2
73. ρ cos φ = 2; z = 2
74. ρ(1 − cos2 φ) = cos φ; ρ2 − ρ2 cos2 φ = ρ cos φ; x2 + y 2 + z 2 − z 2 = z; z = x2 + y 2
528
9.15 Triple Integrals
75. The equations are φ = π/4 and ρ = 3.
2π π/4 3
V =
ρ2 sin φ dρ dφ dθ =
0
=
0
0
π/4
2π
−9 cos φ dθ = −9
0
0
0
2π
0
2π
√
π/4
0
3
2π π/4
1 3
9 sin φ dφ dθ
ρ sin φ dφ dθ =
3
0
0
0
√
2
− 1 dθ = 9π(2 − 2 )
2
76. The equations are ρ = 2, θ = π/4, and θ = π/3.
2
π/3 π/2 2
π/3 π/2
1 3
2
ρ sin φ dρ dφ dθ =
ρ sin φ dφ dθ
3
π/4
0
0
π/4
0
0
π/3 π/2
π/2
8
8 π/3
=
− cos φ dθ
sin φ dφ dθ =
3
3
0
π/4
0
π/4
8 π/3
2π
=
(0 + 1) dθ =
3 π/4
9
77. From Problem 69, we have φ = π/6. Since the figure is in the first octant and z = 2 we also
have θ = 0, θ = π/2, and ρ cos φ = 2.
2 sec φ
π/2 π/6 2 sec φ
π/2 π/6
1 3
V =
ρ sin φ ρ2 sin φ dρ dφ dθ =
dφ dθ
3
0
0
0
0
0
0
8 π/2 π/6
8 π/2 π/6
=
sec3 φ sin φ dφ dθ =
sec2 φ tan φ dφ dθ
3 0
3 0
0
0
π/6
8 π/2 1
2
4 π/2 1
2
=
tan φ dθ = π
dθ =
3 0
2
3 0
3
9
0
78. The equations are ρ = 1 and φ = π/4. We find the volume above the xy-plane and double.
1
2π π/2 1
2π π/2
1 3
V =2
ρ sin φ dφ dθ
ρ2 sin φ dρ dφ dθ = 2
0
π/4
0
0
π/4 3
0
√
2π π/2
2π
2π √
π/2
2
2
2π 2
2
2
=
dθ =
sin φ dφ dθ =
− cos φ dθ =
3 0
3 0
3 0
2
3
π/4
π/4
79. By symmetry, x̄ = ȳ = 0. The equations are φ = π/4 and ρ = 2 cos φ.
2 cos φ
2π π/4 2 cos φ
2π π/4
1 3
m=
ρ sin φ ρ2 sin φ dρ dφ dθ =
dφ dθ
3
0
0
0
0
0
0
π/4
8 2π π/4
8 2π 1
=
sin φ cos3 φ dφ dθ =
− cos4 φ dθ
3 0
3 0
4
0
0
2 2π 1
=−
− 1 dθ = π
3 0
4
2π π/4 2 cos φ
2π π/4 2 cos φ
zρ2 sin φ dρ dφ dθ =
ρ3 sin φ cos φ dρ dφ dθ
Mxy =
0
0
0
0
0
0
2 cos φ
2π π/4
1 4
=
dφ dθ = 4
cos5 φ sin φ dφ dθ
ρ sin φ cos φ 4
0
0
0
0
0
π/4
2π
7
1
2 2π 1
=4
− 1 dθ = π
− cos6 φ dθ = −
6
3 0
8
6
0
0
2π
z̄ = Mxy /m =
π/4
7π/6
= 7/6. The centroid is (0, 0, 7/6).
π
529
9.15 Triple Integrals
80. We are given density = kz. By symmetry, x̄ = ȳ = 0. The equation is ρ = 1.
2π π/2 1
2π π/2 1
m=
kzρ2 sin φ dρ dφ dθ = k
ρ3 sin φ cos φ dρ dφ dθ
0
0
0
0
0
0
1
2π π/2
1 4
1
=k
ρ sin φ cos φ dφ dθ = k
sin φ cos φ dφ dθ
4
4 0
0
0
0
0
π/2
2π
2π
1
1
1
kπ
2
= k
dθ = k
dθ =
sin φ 4 0 2
8
4
0
0
2π π/2 1
2π π/2 1
Mxy =
kz 2 ρ2 sin φ dρ dφ dθ = k
ρ4 cos2 φ sin φ dρ dφ dθ
2π
π/2
0
0
0
0
0
0
1
2π π/2
1 5
1
=k
ρ cos2 φ sin φ dφ dθ = k
cos2 φ sin φ dφ dθ
5
5
0
0
0
0
0
π/2
2π
2π
1
1
1
2
= k
− cos3 φ dθ = − k
(0 − 1) dθ =
kπ
5 0
3
15 0
15
0
2π
z̄ = Mxy /m =
π/2
2kπ/15
= 8/15. The center of mass is (0, 0, 8/15).
kπ/4
81. We are given density = k/ρ.
5
2π cos−1 4/5 5
2π cos−1 4/5
k 2
1 2
m=
dφ dθ
ρ sin φ dρ dφ dθ = k
ρ sin φ 2
0
0
4 sec φ ρ
0
0
4 sec φ
2π cos−1 4/5
1
= k
(25 sin φ − 16 tan φ sec φ) dφ dθ
2 0
0
2π
2π
cos−1 4/5
1
1
= k
(−25 cos φ − 16 sec φ) dθ = k
[−25(4/5) − 16(5/4) − (−25 − 16)] dθ
2 0
2 0
0
2π
1
= k
dθ = kπ
2 0
82. We are given density = kρ.
2π π a
Iz =
(x2 + y 2 )(kρ)ρ2 sin φ dρ dφ dθ
0
0
2π
0
π
a
(ρ2 sin2 φ cos2 θ + ρ2 sin2 φ sin2 θ)ρ3 sin φ dρ dφ dθ
=k
0
0
0
a
1 6 3 1 6 2π π 3
=k
ρ sin φ dρ dφ dθ = k
sin φ dφ dθ
ρ sin φ dφ dθ = ka
6
0
0
0
0
0 6
0
0
0
π
2π π
2π 2π
1
4
1
1
1
4π 6
= ka3
− cos φ + cos3 φ dθ = ka3
(1 − cos2 φ) sin φ dφ dθ = ka3
dθ =
ka
6
6
3
6
3
9
0
0
0
0
0
2π
π
a
5
2π
π
3
530
9.16 Divergence Theorem
EXERCISES 9.16
Divergence Theorem
1. div F = y + z + x
The Triple Integral:
1
div F dV =
0
D
1
1
(x + y + z) dx dy dz
0
0
1
1 2
=
x + xy + xz dy dz
2
0
0
0
1
1 1
1
1
1
1 2
=
+ y + z dy dz =
y + y + yz dz
2
2
2
0
0
0
0
1
1
3
1
1
=
(1 + z) dz = (1 + z 2 ) = 2 − =
2
2
2
0
0
1
1
The Surface Integral: Let the surfaces be S1 in z = 0, S2 in z = 1, S3 in y = 0, S4 in y = 1, S5 in x = 0,
and S6 in x = 1. The unit outward normal vectors are −k, k, −j, j, −i and i, respectively. Then
F · n dS =
F · (−k) dS1 +
F · k dS2 +
F · (−j) dS3 +
F · j dS4
S
S1
S2
F · (−i) dS5 +
+
S5
=
z dS4 +
S2
=
xy dS6
S4
1
1
1
x dx dy +
0
0
1
=
0
1
dy +
2
1
0
1
z dz dx +
0
y dS6
S6
0
1
dx +
2
1
0
0
1
3
dz = .
2
2
2. div F = 6y + 4z
The Triple Integral:
1
div F dV =
0
D
1−x
1
0
1−x
=
0
1−x−y
(6y + 4z) dz dy dx
0
1
1−x−y
(6yz + 2z 2 ) dy dx
0
0
1−x
(−4y 2 + 2y − 2xy + 2x2 − 4x + 2) dy dx
=
0
0
531
1
y dy dz
0
yz dS4
S4
S6
x dS2 +
1
S3
(−xy) dS5 +
S5
(−yz) dS3 +
S2
+
=
xz dS2 +
S1
S4
F · i dS6
S6
(−xz) dS1 +
S3
9.16 Divergence Theorem
1−x
4 3
2
2
2
=
− y + y − xy + 2x y − 4xy + 2y dx
3
0
0
1
1
5 3
5
5
5 4 5 3 5 2 5
2
=
− x + 5x − 5x +
dx = − x + x − x + x =
3
3
12
3
2
3
12
0
0
1
The Surface Integral: Let the surfaces be S1 in the plane x + y + z = 1, S2 in z = 0, S3 in x = 0, and S4 in
√
y = 0. The unit outward normal vectors are n1 = (i + j + k)/ 3 , n2 = −k, n3 = −i, and n4 = −j, respectively.
√
Now on S1 , dS1 = 3 dA1 , on S3 , x = 0, and on S4 , y = 0, so
F · n dS =
F · n1 dS1 +
F · (−k) dS2 +
F · (−j) dS3 +
F · (−i) dS4
S
=
S1
1
S2
(6xy + 4y(1 − x − y) + xe−y ) dy dx +
+
(−6xy) dS3 +
(−4yz) dS4
0
0
1
S3
S4
4
xy + 2y − y 3 − xe−y
3
2
=
0
S3
1−x
2
1−x
dx +
1
1
0
1−x
(−xe−y ) dy dx
0
1−x
xe−y dx + 0 + 0
0
0
0
S4
1
4
[x(1 − x)2 + 2(1 − x)2 − (1 − x)3 − xex−1 + x] dx +
3
0
1
1 2 2 3 1 4 2
1
5
3
4 =
x − x + x − (1 − x) + (1 − x) =
.
2
3
4
3
3
12
0
1
(xex−1 − x) dx
=
3. div F = 3x2 + 3y 2 + 3z 2 . Using spherical coordinates,
F · n dS =
3(x2 + y 2 + z 2 ) dV =
S
2π
a
3ρ2 ρ2 sin φ dρ dφ dθ
0
D
π
0
0
0
a
3 5
3a5 2π π
=
sin φ dφ dθ
ρ sin φ dφ dθ =
5 0
0
0 5
0
0
π
3a5 2π
6a5 2π
12πa5
=
− cos φ dθ =
dθ =
.
5 0
5 0
5
0
2π
π
4. div F = 4 + 1 + 4 = 9. Using the formula for the volume of a sphere,
4 3
F · n dS =
9 dV = 9
π2 = 96π.
3
S
D
5. div F = 2(z − 1). Using cylindrical coordinates,
2π F · n dS =
2(z − 1) V =
S
0
D
2π
=
4
16r dr dθ =
0
0
0
6. div F = 2x + 2z + 12z 2 .
F · n dS =
div F dV =
S
3
=
0
=
0
D
0
3
2
3
2
0
2π
4
5
2π
2(z − 1) dz r dr dθ =
1
4
8r2 dθ = 128
0
0
2π
dθ = 256π.
0
1
(2x + 2z + 12z 2 ) dx dy dz
0
0
1
(x2 + 2xz + 12xz 2 ) dy dz =
0
3
0
2
(1 + 2z + 12z 2 ) dy dz
0
3
2(1 + 2z + 12z 2 ) dz = (2z + 2z 2 + 8z 3 ) = 240
0
0
532
0
4
5
(z − 1)2 r dr dθ
1
9.16 Divergence Theorem
7. div F = 3z 2 . Using cylindrical coordinates,
2π √3 F · n dS =
div F dV =
S
D
√
2π
0
3
=
0
0
0
3
3
9
9
0
3
(x3 − 18x2 + 81x) dx =
0
9. div F =
r(4 − r2 )3/2 dr dθ
0
0
2π
1
− (1 − 32) dθ =
5
2π
0
9
−x(9 − y) dx =
2
0
3
=
3
31
62π
dθ =
.
5
5
3
x(9 − x)2 dx
2
x2
0
√
9−y
2x(9 − y) dy dx =
=
2x dz dy dx
x2
0
D
2π
0
0
8. div F = 2x.
F · n dS =
div F dV =
S
0
√3
dθ
=
1
− (4 − r2 )5/2
5
=
4−r 2
3z 2 r dz dr dθ
√4−r2
rz 3 dr dθ =
0
2π
0
√
x
1 4
81 2
x − 6x3 +
x
4
2
1
. Using spherical coordinates,
x2 + y 2 + z 2
2π π F · n dS =
div F dV =
0
3
= 891
4
0
b
1 2
ρ sin φ dρ dφ dθ
2
ρ
D
0
0
a
2π π
2π
π
=
(b − a) sin φ dφ dθ = (b − a)
− cos φ dθ
S
0
0
0
0
2π
= (b − a)
2 dθ = 4π(b − a).
0
F · n dS =
10. Since div F = 0,
0 dV = 0.
S
D
11. div F = 2z + 10y − 2z = 10y.
F·n dS =
10y dV =
S
2
2−x2 /2
0
2
(80 − 40z) dz dx =
0
0
D
2
2−x
=
0
2
3
30xyz 2
0
x+y
0
2
2−x2 /2
2−x
0
4−z
5y 2 dz dx
z
0
2
2−x2 /2
2
(80z − 20z 2 ) dx =
(80 − 5x4 ) dx = (80x − x5 ) = 128
0
12. div F = 30xy.
F · n dS =
30xy dV =
S
0
z
2−x2 /2
=
4−z
10y dy dz dx =
0
D
2
0
0
3
30xy dz dy dx
0
x+y
dy dx
2−x
(90xy − 30x2 y − 30xy 2 ) dy dx
=
0
0
2
=
0
0
2−x
(45xy 2 − 15x2 y 2 − 10xy 3 ) dx
2
(−5x4 + 45x3 − 120x2 + 100x) dx =
=
−x5 +
0
533
45 4
x − 40x3 + 50x2
4
2
= 28
0
9.16 Divergence Theorem
13. div F = 6xy 2 + 1 − 6xy 2 = 1. Using cylindrical coordinates,
π 2 sin θ 2r sin θ
F · n dS =
dV =
dz r dr dθ =
S
0
D
2 sin θ
(2r sin θ − r2 )r dr dθ
0
0
π
2 3
16
1 4 4
4
=
dθ =
r sin θ − r sin θ − 4 sin θ dθ
3
4
3
0
0
0
π
4 π 4
1
1
4 3
π
=
θ − sin 2θ +
sin 4θ =
sin θ dθ =
3 0
3 8
4
32
2
0
π
r2
2 sin θ
0
π
14. div F = y 2 + x2 . Using spherical coordinates, we have x2 + y 2 = ρ2 sin2 φ and z = ρ cos φ or ρ = z sec φ. Then
2π π/4 4 sec φ
2
2
F · n dS =
(x + y ) dS =
ρ2 sin2 φρ2 sin φ dρ dφ dθ
S
0
D
2π
π/4
=
0
0
2π 2 sec φ
π/4
0
2π
0
2 sec φ
=
0
0
4 sec φ
1 5 3 dφ dθ =
ρ sin φ 5
992
992
tan3 φ sec2 φ dφ dθ =
5
5
π/4
0
0
2π
992
sec5 φ sin3 φ dφ dθ
5
π/4
496π
1
992 2π 1
tan4 φ dθ =
.
dθ =
4
5
4
5
0
0
−2x2 + y 2 + z 2
x2 − 2y 2 + z 2
x2 + y 2 − 2z 2
15. (a) div E = q
=0
+
+
(x2 + y 2 + z 2 )5/2
(x2 + y 2 + z 2 )5/2
(x2 + y 2 + z 2 )5/2
(E · n) dS =
div E dV =
0 dV = 0
S∪Sa
D
(E · n) dS +
(b) From (a),
D
(E · n) dS = 0 and
S
Sa
(E · n) dS = −
(E · n) dS. On Sa ,
S
Sa
|r| = a, n = −(xi + yj + zk)/a = −r/a and E · n = (qr/a3 ) · (−r/a) = −qa2 /a4 = −q/a2 . Thus
q
q
q
q
− 2 dS = 2
(E · n) dS = −
dS = 2 × (area of Sa ) = 2 (4πa2 ) = 4πq.
a
a
a
a
S
Sa
Sa
16. (a) By Gauss’ Law
(E · n) dS =
4πρ dV , and by the Divergence Theorem
D (E · n) dS =
div E dV . Thus
4πρ dV =
div E dV and
(4πρ − div E) dV = 0.
S
D
D
D
D
Since this holds for all regions D, 4πρ − div E = 0 and div E = 4πρ.
(b) Since E is irrotational, E = ∇φ and ∇2 φ = ∇ · ∇φ = ∇E = div E = 4πρ.
17. Since div a = 0, by the Divergence Theorem
(a · n) dS =
div a dV =
0 dV = 0.
S
D
D
18. By the Divergence Theorem and Problem 30 in Section 9.7,
(curl F · n) dS =
div (curl F) dV =
0 dV = 0.
S
D
D
19. By the Divergence Theorem and Problem 27 in Section 9.7,
(f ∇g) · n dS =
div (f ∇g) dV =
∇ · (f ∇g) dV =
[f (∇ · ∇g) + ∇g · ∇f ] dV
S
D
D
D
=
(f ∇2 g + ∇g · ∇f ) dV.
D
534
9.17 Change of Variables in Multiple Integrals
20. By the Divergence Theorem and Problems 25 and 27 in Section 9.7,
(f ∇g − g∇f ) · n dS =
div (f ∇g − g∇f ) dV =
∇ · (f ∇g − g∇f ) dV
S
D
D
=
[f (∇ · ∇g) + ∇g · ∇f − g(∇ · ∇f ) − ∇f · ∇g] dV
D
=
(f ∇2 g − g∇2 f ) dV.
D
21. If G(x, y, z) is a vector valued function then we define surface integrals and triple integrals of G component-wise.
In this case, if a is a constant vector it is easily shown that
a · G dS = a ·
G dS and
a · G dV = a ·
G dV.
S
S
Now let F = f a. Then
D
D
F · n dS =
S
(f a) · n dS =
a · (f n) dS
S
S
and, using Problem 27 in Section 9.7 and the fact that ∇ · a = 0, we have
div F dV =
∇ · (f a) dV =
[f (∇ · a) + a · ∇f ] dV =
a · ∇f dV.
D
D
D
D
By the Divergence Theorem,
a · (f n) dS =
F · n dS =
div F dV =
a · ∇f dV
S
and
S
a·
f n dS
D
=a·
∇f dV
S
D
or a ·
f n dS −
D
∇f dV
S
= 0.
D
Since a is arbitrary,
f n dS −
S
22. B + W = −
∇f dV = 0
and
D
∇f dV.
f n dS =
S
pn dS + mg = mg −
S
D
∇p dV = mg −
D
ρg dV = mg −
D
ρ dV
g
D
= mg − mg = 0
EXERCISES 9.17
Change of Variables in Multiple Integrals
1. T : (0, 0) → (0, 0); (0, 2) → (−2, 8); (4, 0) → (16, 20); (4, 2) → (14, 28)
2. Writing x2 = v − u and y = v + u and solving for u and v, we obtain u = (y − x2 )/2 and v = (x2 + y)/2.
√
Then the images under T −1 are (1, 1) → (0, 1); (1, 3) → (1, 2); ( 2 , 2) → (0, 2).
535
9.17 Change of Variables in Multiple Integrals
3. The uv-corner points (0, 0), (2, 0), (2, 2) correspond to xy-points (0, 0), (4, 2), (6, −4).
v = 0: x = 2u, y = u =⇒ y = x/2
u = 2: x = 4 + v, y = 2 − 3v =⇒
y = 2 − 3(x − 4) = −3x + 14
v = u: x = 3u, y = −2u =⇒ y = −2x/3
v
y
2
2
y=x/2
y=14-3x
u=2
-2
S
y=-2x/3
v=0
1
-4
2 u
v
4. Solving for x and y we see that the transformation
is x = 2u/3+v/3, y = −u/3+v/3. The uv-corner
6 x
3
v=u
1
y
6
6
v=5
points (−1, 1), (4, 1), (4, 5), (−1, 5) correspond
to the xy-points (−1/3, 2/3), (3, −1), (13/3, 1/3),
(1, 2).
v = 1: x + 2y = 1;
v = 5: x + 2y = 5;
u = −1: x − y = −1;
u = 4: x − y = 4
u=-1
S
3
3
u=4
-2
x+2y=5
x-y=-1
v=1
4
2
-2
u
x+2y=1
v
5. The uv-corner points (0, 0), (1, 0), (1, 2),
(0, 2) correspond to the xy-points (0, 0),
(1, 0), (−3, 2), (−4, 0).
y
x=y2 /4-4
v=2
2
u=0
-4
S
2
x=1-y2
u=1
2u
v=0
-2
3 x-y=4 x
-4
-2 y=0
2x
v = 0: x = u2 , y = 0 =⇒ y = 0 and 0 ≤ x ≤ 1
u = 1: x = 1 − v 2 , y = v =⇒ x = 1 − y 2
v = 2: x = u2 − 4, y = 2u =⇒ x = y 2 /4 − 4
u = 0: x = −v 2 , y = 0 =⇒ y = 0 and −4 ≤ x ≤ 0
6. The uv-corner points (1, 1), (2, 1), (2, 2), (1, 2)
correspond to the xy-points (1, 1), (2, 1), (4, 4), (2, 4).
v = 1: x = u, y = 1 =⇒ y = 1, 1 ≤ x ≤ 2
u = 2: x = 2v, y = v 2 =⇒ y = x2 /4
v = 2: x = 2u, y = 4 =⇒ y = 4, 2 ≤ x ≤ 4
u = 1: x = v, y = v 2 =⇒ y = x2
∂(x, y) −ve−u
=
7.
∂(u, v) veu
v
y
4
4
y=4
y=x2
v=2
2
u=1
S
u=2
v=1
2
y=1
4
u
e−u = −2v
eu ∂(x, y) 3e3u sin v
8.
=
∂(u, v) 3e3u cos v
e3u cos v = −3e6u
−e3u sin v y 2
3y 2
∂(u, v) −2y/x3 1/x2 =
−
=
−3
= −3u2 ;
= 2 2
∂(x, y)
x4
x2
−y /x 2y/x −4xy 2(y 2 − x2 )
2
4
∂(u, v) (x + y 2 )2 (x2 + y 2 )2 =
10.
=
∂(x, y) 4xy
2(y 2 − x2 ) (x2 + y 2 )2
2
(x + y 2 )2 (x2 + y 2 )2
9.
536
2
y=x /4
2
∂(x, y)
1
1
=− 2
=
2
∂(u, v)
−3u
3u
2
4
x
9.17 Change of Variables in Multiple Integrals
From u = 2x/(x2 + y 2 ) and v = −2y(x2 + y 2 ) we obtain u2 + v 2 = 4/(x2 + y 2 ). Then x2 + y 2 = 4/(u2 + v 2 ) and
∂(x, y)/∂(u, v) = (x2 + y 2 )2 /4 = 4/(u2 + v 2 )2 .
11. (a) The uv-corner points (0, 0), (1, 0), (1, 1), (0, 1) correspond to the xy-points (0, 0), (1, 0), (0, 1), (0, 0).
v = 0:
u = 1:
v = 1:
u = 0:
x = u, y = 0 =⇒ y = 0, 0 ≤ x ≤ 1
x = 1 − v, y = v =⇒ y = 1 − x
x = 0, y = u =⇒ x = 0, 0 ≤ y ≤ 1
x = 0, y = 0
v
y
v=1
S
u=0
u
v=0
y=1-x
x=0
u=1
x
y=0
(b) Since the segment u = 0, 0 ≤ v ≤ 1 in the uv-plane maps to the origin in the xy-plane, the transformation
is not one-to-one.
∂(x, y) 1 − v v 12.
=
= u. The transformation is 0 when u is 0, for 0 ≤ v ≤ 1.
∂(u, v)
−u u y
x + y = −1 =⇒ v = −1
R43
x − 2y = 6 =⇒ u = 6
R R3
-6
-3 R1
3
6 x
x + y = 3 =⇒ v = 3
R2
-3
x − 2y = −6 =⇒ u = −6
∂(u, v) 1 −2 ∂(x, y)
1
=
= 3 =⇒
=
∂(x, y)
∂(u, v)
3
1 1
3 6
3
1
1
1
1
(x + y) dA =
v
v du dv = (12)
v dv = 4
dA =
v2
3
3
3
2
−1 −6
−1
R
S
v
13. R1:
R2:
R3:
R4:
2
-3
3
= 16
−1
v
6
2
R4
R3: y = −3x + 6 =⇒ v = 6
R4: y = x =⇒ u = 0
∂(u, v) 1 −1 ∂(x, y)
1
=
= 4 =⇒
=
∂(x, y)
∂(u, v)
4
3 1 u
3
-2
y
14. R1: y = −3x + 3 =⇒ v = 3
R2: y = x − π =⇒ u = π
S
S
R3
R1
3
3x
R
R2
-2
4u
2
R
cos 12 (x − y)
dA =
3x + y
1
=
2
S
3
1
1 6 2 sin u/2
1 6 π cos u/2
dA =
du dv =
4
4 3 0
v
4 3
v
6
1
dv
1
= ln v = ln 2
v
2
2
3
cos u/2
v
6
15. R1: y = x2 =⇒ u = 1
R2: x = y 2 =⇒ v = 1
R3: y = 12 x2 =⇒ u = 2
y
0
v
2
2
R4
R1
R
R3
S
R2
R4: x = 12 y 2 =⇒ v = 2
∂(u, v) 2x/y
=
∂(x, y) −y 2 /x2
π
dv
x
−x2 /y 2 ∂(x, y)
1
=
= 3 =⇒
∂(u,
v)
3
2y/x
537
u
9.17 Change of Variables in Multiple Integrals
y2
dA =
x
R
2
=1
2
1
1
1 2 2
1 2
1
v
v du dv =
v dv = v 2
dA =
3
3
3
6
S
1
1
1
16. R1: x2 + y 2 = 2y =⇒ v = 1
R2: x2 + y 2 = 2x =⇒ u = 1
R3: x2 + y 2 = 6y =⇒ v = 1/3
R4: x2 + y 2 = 4x =⇒ u = 1/2
2(y 2 − x2 )
−4xy 2
∂(u, v) (x + y 2 )2 (x2 + y 2 )2 −4
=
= 2
2
2 ∂(x, y) −4xy
(x
+ y 2 )2
2(x − y ) 2
(x + y 2 )2 (x2 + y 2 )2
y
v
3
2
R4
R
R3
R1
S
R2
3x
2u
Using u2 + v 2 = 4/(x2 + y 2 ) we see that ∂(x, y)/∂(u, v) = −4/(u2 + v 2 )2 .
2
2 −3
(x + y )
dA =
R
S
4
2
u + v2
−3 1 1
−4
115
dA = 1
(u2 + v 2 ) du dv =
(u2 + v 2 )2 16 1/3 1/2
5184
17. R1: 2xy = c =⇒ v = c
R2: x2 − y 2 = b =⇒ u = b
R3: 2xy = d =⇒ v = d
v
v
d
d
S
R4: x2 − y 2 = a =⇒ u = a
c
c
∂(u, v) 2x −2y =
= 4(x2 + y 2 )
∂(x, y)
2y 2x R3
R4
R
R2
R1
a
a
b u
b u
∂(x, y)
1
=⇒
=
∂(u, v)
4(x2 + y 2 )
(x2 + y 2 ) dA =
R
(x2 + y 2 )
S
1
1
dA =
2
2
4(x + y )
4
d
b
du dv =
c
a
y
18. R1: xy = −2 =⇒ v = −2
R2: x2 − y 2 = 9 =⇒ u = 9
R3: xy = 2 =⇒ v = 2
R4: x2 − y 2 = 1 =⇒ u = 1
2
R3
R
R4
2
(x + y ) sin xy dA =
R
2
2
(x + y ) sin v
S
4 x
S
5
-2
10 u
-2
2
2
R1
∂(x, y)
1
=
2
∂(u, v)
2(x + y 2 )
v
R2
2
∂(u, v) 2x −2y = 2(x2 + y 2 )
=
∂(x, y) y
x =⇒
1
(b − a)(d − c)
4
1
2(x2 + y 2 )
538
1
dA =
2
2
−2
1
9
1
sin v du dv =
2
2
8 sin v dv = 0
−2
9.17 Change of Variables in Multiple Integrals
19. R1: y = x2 =⇒ v + u = v − u =⇒ u = 0
y
v
R2
R2 : y = 4 − x2 =⇒ v + u = 4 − (v − u)
=⇒ v + u = 4 − v + u =⇒ v = 2
R3
2
R
2
S
R3: x = 1 =⇒ v − u = 1 =⇒ v = 1 + u
1
1
∂(x, y) − 2√v − u 2√v − u 1
=
= − √v − u
∂(u, v) 1
1
R1
x
1
1
u
1 1 1
v − u 1 1 2 1
√
−
dv
du
=
=
[ln 2 − ln(1 + u)] du
dA
2v 2 0 1+u v
2 0
v − u
S
1 1
1
1
1
1 1
= ln 2 − [(1 + u) ln(1 + u) − (1 + u)] = ln 2 − [2 ln 2 − 2 − (0 − 1)] = − ln 2
2
2
2
2
2 2
0
√
x
dA =
y + x2
R
20. Solving x = 2u − 4v, y = 3u + v for u and v we
1
3
obtain u = 14
x + 27 y, v = − 14
x + 17 y. The xy-4
R
S
1
= 14
0
0
5
1
+ 2u − u2 du =
2
2
-2
1−u
0
1
35 3 28
2
7u + 14u − u =
3
3
0
R2: y = x =⇒ v = 1
R3: y = 4/x =⇒ u = 4
R4: y = 4x =⇒ v = 4
x 2y
∂(u, v) y
∂(x, y)
x
=
=
=⇒
=
∂(x, y)
x
∂(u, v)
2y
−y/x2 1/x R
y 4 dA =
8
u2 v 2
S
1
2v
du dv =
1
2
1
(3u + v) dv du = 14
21. R1: y = 1/x =⇒ u = 1
v
4
R3
1
1
4
S
R
2
R2
2
0
4
R1
1
2
1−u
du
y
2
u2 v du dv =
2u
1
3uv + v 2
2
R4
4
S
2 x
0
R2
R
R1
the uv-points (0, 1), (0, 0), (1, 0).
1
3
R3
corner points (−4, 1), (0, 0), (2, 3) correspond to
∂(x, y) 2 −4 =
= 14
∂(u, v) 3 1 y dA =
(3u + v)(14) dA = 14
v2
y
4x
4
1 3 1 4
21 2
u v dv =
v
63v dv =
3
6
4
1
1
2
4u
4
= 315
4
1
22. Under the transformation u = y + z, v = −y + z, w = x − y the parallelepiped D is mapped to the parallelepiped
E: 1 ≤ u ≤ 3, −1 ≤ v ≤ 1, 0 ≤ w ≤ 3.
0 1 1
∂(u, v, w) ∂(x, y, z)
1
= 0 −1 1 = 2 =⇒
=
∂(x, y, z)
∂(u, v, w)
2
1 −1 0 539
9.17 Change of Variables in Multiple Integrals
1
1 3 1 3
(2u + 2v + 2w) dV =
(2u + 2v + 2w) du dv dw
2
2 0 −1 1
E
3
1 3 1 2
1 3 1
=
(u + 2uv + 2uw) dv dw =
(8 + 4v + 4w) dv dw
2 0 −1
2 0 −1
1
3
3
1
3
2
=
(4v + v + 2vw) dw =
(8 + 4w) dw = (8w + 2w2 ) = 42
(4z + 2x − 2y) dV =
D
−1
0
0
0
23. We let u = y − x and v = y + x.
R1: y = 0 =⇒ u = −x, v = x =⇒ v = −u
R2: x + y = 1 =⇒ v = 1
R3: x = 0 =⇒ u = y, v = y =⇒ v = u
∂(u, v) −1 1 ∂(x, y)
1
=
= −2 =⇒
=−
∂(x, y) 1 1 ∂(u, v)
2
y
=
1
2
1
R
2
−2xy+x2
dA =
S
eu | − 1| dA =
2
2
−v
y
v
2
2
R3
-2
u
2
eu dv du =
0
v
veu/v dv
R
0
1 u
-1
1
(e − e−1 )
4
R3: y = x + 2 =⇒ u = 2
∂(u, v) −1 1 ∂(x, y)
=
= −1 =⇒
= −1
∂(x, y)
∂(u, v)
0 1
ey
1 x
0
24. We let u = y − x and v = y.
R1: y = 0 =⇒ v = 0, u = −x =⇒ v = 0, 0 ≤ u ≤ 2
R2: x = 0 =⇒ v = u
S
R
R1
e(y−x)/(y+x) dA =
R
1
R2
-1
1
1 1 v u/v
eu/v − dA =
e
du dv
2
2 0 −v
S
1
1 1
1
−1
−1 1 2 =
v(e − e ) dv = (e − e ) v =
2 0
2
2
0
v
1
R3
2
2
0
25. Noting that R2, R3, and R4 have equations y+2x = 8, y−2x = 0,
and y + 2x = 2, we let u = y/x and v = y + 2x.
R1: y = 0 =⇒ u = 0, v = 2x =⇒ u = 0, 2 ≤ v ≤ 8
R2: y + 2x = 8 =⇒ v = 8
R3: y − 2x = 0 =⇒ u = 2
R4: y + 2x = 2 =⇒ v = 2
∂(u, v) −y/x2 1/x ∂(x, y)
y + 2x
x2
=⇒
=−
=
=−
2
∂(x, y)
x
∂(u, v)
y + 2x
2
1
2
x (6x + 3y) dA = 3
(y + 2x) −
x2 dA
dA = 3
y
+
2x R
S
S
x
R1
ueu du =
S
R2
2u
2
1 u2 1
e = (e4 − 1)
2
2
0
v8
y
4
R3
R2
R
R4
R1
4x
From y = ux we see that v = ux + 2x and x = v/(u + 2). Then
8
2 8
2
2
v3
du
504
2
2
2
3
x dA = 3
v (u + 2) dv du =
du = 504
=−
2
2
u+2
S
0
2
0 (u + 2)
0 (u + 2)
2
540
S
4
2
2
= 126.
0
4u
9.17 Change of Variables in Multiple Integrals
26. We let u = x + y and v = x − y.
y
v
2
2
R3
R4
= 1 =⇒ u = 1
R
= 1 =⇒ v = 1
R1
R2
= 3 =⇒ u = 3
4 x
= −1 =⇒ v = −1
-2
-2
∂(u, v) 1 1 ∂(x, y)
1
=
= −2 =⇒
=−
∂(x, y)
∂(u, v)
2
1 −1
1 3 1 4 v
1 3 4 v 1
4 x−y
4 v 1
(x + y) e
dA =
u e − dA =
u e dv du =
u e du
2
2 1 −1
2 1
−1
R
S
3
e − e−1 3 4
121
e − e−1 5 242(e − e−1 )
=
u =
=
(e − e−1 )
u du =
2
10
10
5
1
1
R1:
R2:
R3:
R4:
x+y
x−y
x+y
x−y
S
4 u
27. Let u = xy and v = xy 1.4 . Then xy 1.4 = c =⇒ v = c; xy = b =⇒ u = b; xy 1.4 = d =⇒ v = d;
xy = a =⇒ u = a.
x
∂(u, v) y
= 0.4xy 1.4 = 0.4v =⇒ ∂(x, y) = 5
=
∂(x, y) y 1.4 1.4xy 0.4 ∂(u, v)
2v
d b
d
5
5
dv
5
5
dA =
dA =
du dv = (b − a)
= (b − a)(ln d − ln c)
2v
2v
2
v
2
R
S
c
a
c
28. The image of the ellipsoid x2 /a2 + y 2 /b2 + z 2 /c2 = 1 under the transformation u = x/a, v = y/b, w = z/c, is
the unit sphere u2 + v 2 + w2 = 1. The volume of this sphere is 43 π. Now
a 0 0
∂(x, y, z)
= 0 b 0 = abc
∂(u, v, w) 0 0 c
and
dV =
D
abc dV = abc
E
dV = abc
E
4
π
3
=
4
πabc.
3
∂(x, y) 5 0 29. The image of the ellipse is the unit circle x + y = 1. From
=
= 15 we obtain
∂(u, v) 0 3 2
2π 1
x
15 2π 4 1
y2
2
2
2
(u + v )15 dA = 15
r r dr dθ =
r dθ
+
dA =
25
9
4 0
0
R
S
0
0
2π
15
15π
=
dθ =
.
4 0
2
2
sin φ cos θ
∂(x, y, z) 30.
=
∂(ρ, φ, θ) sin φ sin θ
cos φ
ρ cos φ cos θ
ρ cos φ sin θ
−ρ sin φ
2
−ρ sin φ sin θ ρ sin φ cos θ 0
= cos φ(ρ2 sin φ cos φ cos2 θ + ρ2 sin φ cos φ sin2 θ) + ρ sin φ(ρ sin2 φ cos2 θ + ρ sin2 φ sin2 θ)
= ρ2 sin φ cos2 φ(cos2 θ + sin2 θ) + ρ2 sin3 φ(cos2 θ + sin2 θ) = ρ2 sin φ(cos2 φ + sin2 φ) = ρ2 sin φ
541
CHAPTER
9 REVIEW
EXERCISES
9.17 Change
of Variables
in Multiple Integrals
CHAPTER 9 REVIEW EXERCISES
1. True; |v(t)| =
√
2
2. True; for all t, y = 4.
3. True
4. False; consider r(t) = t2 i. In this case, v(t) = 2ti and a(t) = 2i. Since v · a = 4t, the velocity and acceleration
vectors are not orthogonal for t = 0.
5. False; ∇f is perpendicular to the level curve f (x, y) = c.
6. False; consider f (x, y) = xy at (0, 0).
7. True; the value is 4/3.
8. True; since 2xy dx − x2 dy is not exact.
√
9. False;
x dx + x2 dy = 0 from (−1, 0) to (1, 0) along the x-axis and along the semicircle y = 1 − x2 , but
C
since x dx + x2 dy is not exact, the integral is not independent of path.
10. True
11. False; unless the first partial derivatives are continuous.
12. True
13. True
14. True; since curl F = 0 when F is a conservative vector field.
15. True
16. True
17. True
18. True
19. F = ∇φ = −x(x2 + y 2 )−3/2 i − y(x2 + y 2 )−3/2 j
i
j
k 20. curl F = ∂/∂x ∂/∂y ∂/∂z = 0
f (x) g(y) h(z) 21. v(t) = 6i + j + 2tk; a(t) = 2k. To find when the particle passes through the plane, we solve −6t + t + t2 = −4
or t2 − 5t + 4 = 0. This gives t = 1 and t = 4. v(1) = 6i + j + 2k, a(1) = 2k; v(4) = 6i + j + 8k, a(4) = 2k
22. We are given r(0) = i + 2j + 3k.
r(t) = v(t) dt = (−10ti + (3t2 − 4t)j + k) dt = −5t2 i + (t3 − 2t2 )j + tk + c
i + 2j + 3k = r(0) = c
r(t) = (1 − 5t2 )i + (t3 − 2t2 + 2)j + (t + 3)k
r(2) = −19i + 2j + 5k
√
√
√
√
23. v(t) = a(t) dt = ( 2 sin ti + 2 cos tj) dt = − 2 cos ti + 2 sin tj + c;
√
√
−i + j + k = v(π/4) = −i + j + c, c = k; v(t) = − 2 cos ti + 2 sin tj + k;
542
CHAPTER 9 REVIEW EXERCISES
√
√
r(t) = − 2 sin ti − 2 cos tj + tk + b; i + 2j + (π/4)k = r(π/4) = −i − j + (π/4)k + b, b = 2i + 3j;
√
√
r(t) = (2 − 2 sin t)i + (3 − 2 cos t)j + tk; r(3π/4) = i + 4j + (3π/4)k
√
24. v(t) = ti + t2 j − tk; |v| = t t2 + 2 , t > 0; a(t) = i + 2tj − k; v · a = t + 2t3 + t = 2t + 2t3 ;
√
√
√
2t
2t + 2t3
t2 2
2 + 2t2
2
2
2
v × a = t i + t k, |v × a| = t 2 ; aT = √
=√
, aN = √
=√
;
2
2
2
2
t t +2
t +2
t t +2
t +2
√
√
t2 2
2
κ= 3 2
=
3/2
2
t (t + 2)
t(t + 2)3/2
25.
26. r (t) = sinh ti + cosh tj + k, r (1) = sinh 1i + cosh 1j + k;
√
√
|r (t)| = sinh2 t + cosh2 t + 1 =
2 cosh2 t = 2 cosh t; |r (1)| = 2 cosh 1;
1
1
1
1
T(t) = √ tanh ti + √ j + √ sech tk, T(1) = √ (tanh 1i + j + sech 1k);
2
2
2
2
dT
1
1
d
1
1
= √ sech2 ti − √ sech t tanh tk;
T(1) = √ sech2 1i − √ sech 1 tanh 1k,
dt
dt
2
2
2
2
d
sech 1 1
T(1) = √
sech2 1 + tanh2 1 = √ sech 1; N(1) = sech 1i − tanh 1k;
dt
2
2
1
1
1
B(1) = T(1) × N(1) = − √ tanh 1i + √ (tanh2 1 + sech2 1)j − √ sech 1k
2
2
2
1
= √ (− tanh 1i + j − sech 1k)
2
√
d
(sech 1)/ 2
1
κ = T(1) /|r (1)| = √
= sech2 1
dt
2
2 cosh 1
2
6
1
27. ∇f = (2xy − y 2 )i + (x2 − 2xy)j; u = √ i + √ j = √ (i + 3j);
40
40
10
1
1
Du f = √ (2xy − y 2 + 3x2 − 6xy) = √ (3x2 − 4xy − y 2 )
10
10
2x
2y
2z
2
−4x + 2y + 4z
1
2
28. ∇F = 2
i+ 2
j+ 2
k; u = − i + j + k; Du F =
x + y2 + z2
x + y2 + z2
x + y2 + z2
3
3
3
3(x2 + y 2 + z 2 )
29. fx = 2xy 4 , fy = 4x2 y 3 .
(a) u = i, Du (1, 1) = fx (1, 1) = 2
√
√
√
(b) u = (i − j)/ 2 , Du (1, 1) = (2 − 4)/ 2 = −2/ 2
(c) u = j, Du (1, 1) = fy (1, 1) = 4
30. (a)
dw
∂w dx ∂w dy ∂w dz
=
+
+
dt
∂x dt
∂y dt
∂z dt
=
=
x
x2
+
y2
+
z2
6 cos 2t + y
x2
+
y2
+
z2
(−8 sin 2t) + (6x cos 2t − 8y sin 2t + 15zt2 )
x2 + y 2 + z 2
543
z
x2
+ y2 + z2
15t2
CHAPTER 9 REVIEW EXERCISES
∂w
∂w ∂x ∂w ∂y ∂w ∂z
=
+
+
∂t
∂x ∂t
∂y ∂t
∂z ∂t
(b)
x
6
2t
y
=
cos + 2
2
2
2
r
x +y +z r
x + y2 + z2
2r
6x
2t 8yr
2 3
cos + 2 sin
+ 15zt r
r
r
t
t
=
x2 + y 2 + z 2
8r
2r
sin
2
t
t
z
+
x2
+ y2 + z2
15t2 r3
√
π
1
31. F (x, y, z) = sin xy − z; ∇F = y cos xyi + x cos xyj − k; ∇F (1/2, 2π/3, 3/2) = i + j − k. The equation of
3
4
the tangent plane is
√ π
3
1
1
2π
x−
+
y−
− z−
=0
3
2
4
3
2
√
or 4πx + 3y − 12z = 4π − 6 3 .
32. We want to find a normal to the surface that is parallel to k. ∇F = (y − 2)i + (x − 2y)j + 2zk. We need y − 2 = 0
and x − 2y = 0. The tangent plane is parallel to z = 2 when y = 2 and x = 4. In this case z 2 = 5. The points
√
√
are (4, 2, 5 ) and (4, 2, − 5 ).
1
1 2x 1 1 2x
1
1
33. (a) V =
1 − x2 dy dx =
y 1 − x2 dx =
x 1 − x2 dx = − (1 − x2 )3/2 =
3
3
x
0
x
0
0
0
1 y 2 1 (b) V =
1 − x2 dx dy +
1 − x2 dx dy
0
y/2
1
34. We are given ρ = k(x2 + y 2 ).
1 x2
m=
k(x2 + y 2 ) dy dx = k
y/2
x2
1 3 x y + y dx
3
0
0
x3
x3
1
1
1 5
1 6
1 9
1 7 1 6
1 10 k
4
5
=k
x + x − x − x dx = k
x + x − x − x
= 21
3
3
5
21
6
30
0
0
x2
1 x2
1
1
1
1
1
My =
x3 y + xy 3 dx = k
x5 + x7 − x6 − x10 dx
k(x3 + xy 2 ) dy dx = k
3
3
3
x3
0
0
0
x3
1
1 6
1
1
1
65k
=k
x + x8 − x7 − x11 =
6
24
7
33
1848
0
x2
1 x2
1
1
1 2 2 1 4 1 6 1 8 1 8 1 12
2
3
Mx =
dx
k(x y + y ) dy dx = k
x y + y dx = k
x + x − x − x
2
4
2
4
2
4
x3
0
0
0
x3
1
1 7
1 9
1 13 20k
=k
=
x − x − x
14
36
52
819
0
x̄ = My /m =
1
2
65k/1848
20k/819
= 65/88; ȳ = Mx /m =
= 20/39 The center of mass is (65/88, 20/39).
k/21
k/21
x2
k(x4 + x2 y 2 ) dy dx = k
35. Iy =
0
=k
1
x3
1 7
1
1
1
x + x9 − x8 − x12
7
27
8
36
0
1
1
x4 y + x2 y 3
3
1
= 41 k
1512
0
x2
dx = k
3
544
x
0
1
1
1
x6 + x8 − x7 − x11 dx
3
3
CHAPTER 9 REVIEW EXERCISES
36. (a) Using symmetry,
a √a2 −x2 √a2 −x2 −y2
V =8
dz dy dx = 8
0
0
a
0
0
√
a2 −x2
a2 − x2 − y 2 dy dx
0
Trig substitution
a y
y
a2 − x2
=8
a2 − x2 − y 2 +
sin−1 √
2 − x2
2
2
a
0
a
1 3 4
2
= 2π a x − x = πa3
3
3
√a2 −x2
dx
=
8
0
0
a
π a2 − x2
dx
2
2
0
(b) Using symmetry,
2π a V =2
0
√
a2 −r 2
r dz dr dθ = 2
0
0
2π
a
r
0
2π
a2 − r2 dr dθ
0
a
1 2
2
4
2 3/2 =2
− (a − r )
dθ
=
a3 dθ = πa3
3
3 0
3
0
0
a
2π π a
2π π
1 3
(c) V =
ρ2 sin φ dρ dφ dθ =
ρ sin φ dφ dθ
0
0
0
0
0 3
0
2π π
2π
π
1
1
1 2π 3
4
=
a3 sin φ dφ dθ =
−a3 cos φ dθ =
2a dθ = πa3
3 0
3 0
3 0
3
0
0
37. We use spherical coordinates.
3 sec φ
2π π/4
3 sec φ
2π π/4
1 3
V =
ρ2 sin φ dρ dφ dθ =
dφ dθ
ρ sin φ 0
tan−1 1/3 0
0
tan−1 1/3 3
0
2π π/4
1 2π π/4
=
27 sec3 φ sin φ dφ dθ = 9
tan φ sec2 φ dφ dθ
3 0
tan−1 1/3
0
tan−1 1/3
π/4
2π
1
1
9 2π
=9
tan2 φ dθ = 8π
1−
dθ =
2
2 0
9
0
tan−1 1/3
2
2π π/6 2
2π π/6
1 3
2
38. V =
ρ sin φ dρ dφ dθ =
ρ sin φ dφ dθ
3
0
0
1
0
0
1
2π π/6 2π π/6
π/6
1
7
8
7 2π
=
sin φ − sin φ dφ dθ =
sin φ dφ dθ =
− cos φ dθ
3
3
3
3
0
0
0
0
0
0
√
√ √
7 2π
7
7π
3
3
=
−
− (−1) dθ =
1−
2π =
(2 − 3 )
3 0
2
3
2
3
2π
39. 2xy + 2xy + 2xy = 6xy
i
j
k 40. ∂/∂x ∂/∂y ∂/∂z = 2xzi − 2yzj + (y 2 − x2 )k
2
x y
xy 2 2xyz 41.
∂
∂
∂ 2
(2xz) −
(2yz) +
(y − x2 ) = 0
∂x
∂y
∂z
42. ∇(6xy) = 6yi + 6xj
2π
2π √
z2
4t2
2
2
43.
ds =
4 sin 2t + 4 cos 2t + 4 dt =
8 2 t2 dt =
2
2
cos2 2t + sin2 2t
C x +y
π
π
0
√ 0
√
√
44.
(xy + 4x) ds =
[x(2 − 2x) + 4x] 1 + 4 dx = 5
(6x − 2x2 ) dx = 5 3x2 −
C
1
1
545
√
√
2π
8 2 3 56 2 π 3
t =
3
3
π
√
0
2 3 7 5
x =−
3
3
1
CHAPTER 9 REVIEW EXERCISES
45. Since Py = 6x2 y = Qx , the integral is independent of path.
φx = 3x2 y 2 , φ = x3 y 2 + g(y), φy = 2x3 y + g (y) = 2x3 y − 3y 2 ; g(y) = −y 3 ; φ = x3 y 2 − y 3 ;
(1,−2)
(1,−2)
3x2 y 2 dx + (2x3 y − 3y 2 ) dy = (x3 y 2 − y 3 ) = 12
(0,0)
(0,0)
46. Let x = a cos t, y = a sin t, 0 ≤ t ≤ 2π. Then using dx = −a sin t dt, dy = a cos t dt, x2 + y 2 = a2 we have
2π
2π
2π
−y dx + x dy
1
2
2
=
[−a
sin
t(−a
sin
t)
+
a
cos
t(a
cos
t)]
dt
=
(sin
t
+
cos
t)
dt
=
dt = 2π.
ˇ
x2 + y 2
a2
C
0
0
0
y sin πz dx+x2 ey dy + 3xyz dz
47.
C
1
2
=
2 t2
3
2 3
2
0
1
2
(t2 sin πt3 + 2t3 et + 9t8 ) dt
[t sin πt + t e (2t) + 3tt t (3t )] dt =
0
1
1
2
1
= −
t3 et dt
cos πt3 + t9 + 2
3π
0
0
1
2
2
2 t2
t2 =
+ 1 + (t e − e ) =
+2
3π
3π
0
Integration by parts
48. Parameterize C by x = cos t, y = sin t; 0 ≤ t ≤ 2π. Then
2π
F
·
dr
=
[4
sin
t(−
sin
t
dt)
+
6
cos
t(cos
t)
dt]
=
ˇ
C
0
2π
(6 cos2 t − 4 sin2 t) dt
0
2π
5
=
(10 cos t − 4) dt = 5t + sin 2t − 4t = 2π.
2
0
0
Using Green’s Theorem, Qx − Py = 6 − 4 = 2 and ˇ F · dr =
2 dA = 2(π · 12 ) = 2π.
C
2π
2
R
π
π
π
49. Let r1 = ti and r2 = i + πtj for 0 ≤ t ≤ 1. Then dr1 = i, dr2 = πj, F1 = 0,
2
2
2
π
π
π
F2 = sin πti + πt sin j = sin πti + πtj,
2
2
2
and
1
1
1 2 2 π2
2
W =
F1 · dr1 +
F2 · dr2 =
π t dt = π t =
.
2
2
C1
C2
0
0
50. Parameterize the line segment from (−1/2, 1/2) to (−1, 1) using y = −x as x goes from −1/2 to −1. Parameterize
the line segment from (−1, 1) to (1, 1) using y = 1 as x goes from −1 to 1. Parameterize the line segment from
√
√
(1, 1) to (1, 3 ) using x = 1 as y goes from 1 to 3 . Then
F · dr =
W =
C
−1
−1/2
F · (dxi − dxj) +
1
−1
F · (dxi) +
√
3
F · (dyj)
1
1
√3
2
2
1
1
dx +
=
− 2
dy
dx +
2 + (−x)2
2
2+1
x
x
+
(−x)
x
1
+
y2
−1/2
−1
1
−1
1
√3
1
2
1
=
dx +
dx +
dy
2
2
2x
1
+
x
1
+
y2
−1/2
−1
1
−1
1
√ 3
π
1 1
π
13π − 6
−1
−1 =−
+
2
tan
x
+
tan
y
=
−
+
2
+
=
.
2x −1/2
2
2
12
12
−1
1
−1
546
CHAPTER 9 REVIEW EXERCISES
√
51. zx = 2x, zy = 0; dS = 1 + 4x2 dA
3 2 2
3
z
x
1 1
dS =
(1 + 4x2 )3/2
1 + 4x2 dx dy =
xy
xy
y
12
S
1
1
1
√
√
3
3 3/2
1
17 − 53/2
17 17 − 5 5
=
dy =
ln y 12 1
y
12
1
√
√
17 17 − 5 5
=
ln 3
12
2
dy
1
52. n = k, F · n = 3;
flux = S F · n dS = 3 S dS = 3 × (area of S) = 3(1) = 3
53. The surface is g(x, y, z) = x2 + y 2 + z 2 − a2 = 0. ∇g = 2(xi + yj + zk) = 2r, n = r/|r|,
−xi − yj − zk
F = c∇(1/|r|) + c∇(x2 + y 2 + z 2 )−1/2 = c 2
= −cr/|r|3
(x + y 2 + z 2 )3/2
r
r
|r|2
c
c
r·r
F·n=− 3 ·
= −c 4 = −c 4 = − 2 = − 2
|r| |r|
|r|
|r|
|r|
a
c
c
c
flux =
F · n dS = − 2
dS = − 2 × (area of S) = − 2 (4πa2 ) = −4πc
a
a
a
S
S
54. In Problem 53, F is not continuous at (0, 0, 0) which is in any acceptable region containing the sphere.
55. Since F = c∇(1/r), div F = ∇ · (c∇(1/r)) = c∇2 (1/r) = c∇2 [(x2 + y 2 + z 2 )−1/2 ] = 0 by Problem 37 in
Section 9.7. Then, by the Divergence Theorem,
flux F =
F · n dS =
div F dV =
0 dV = 0.
S
D
D
56. Parameterize C by x = 2 cos t, y = 2 sin t, z = 5, for 0 ≤ t ≤ 2π. Then
(curl F · n) dS = ˇ F · dr = ˇ 6x dx + 7z dy + 8y dz
S
C
2π
=
C
[12 cos t(−2 sin t) + 35(2 cos t)] dt
0
=
2π
2π
(70 cos t − 24 sin t cos t) dt = (70 sin t − 12 sin2 t) = 0.
0
0
57. Identify F = −2yi + 3xj + 10zk. Then curl F = 5k. The curve C lies in the plane z = 3,
so n = k and dS = dA. Thus,
F
·
dr
=
(curl
F)
·
n
dS
=
5 dA = 5 × (area of R) = 5(25π) = 125π.
ˇ
C
S
R
58. Since curl F = 0, ˇ F · dr =
C
(curl F · n) dS =
S
0 dS = 0.
S
59. div f = 1 + 1 = 1 = 3;
F · n dS =
div F dV =
3 dV = 3 × (volume of D) = 3π
S
D
D
547
CHAPTER 9 REVIEW EXERCISES
60. div F = x2 + y 2 + z 2 . Using cylindrical coordinates,
F · n dS =
S
D
1
(r2 + z 2 )r dz dr dθ
0
D
1
0
0
1
2π 1 1 3 1
3
3
=
r z + rz
r
dr
dθ
=
+
r
dr dθ
3
3
0
0
0
0
0
1
2π 2π
5π
1 4 1 2 5
=
r + r dθ =
dθ =
.
4
6
12
6
2π
1
0
0
0
61. div F = 2x + 2(x + y) − 2y = 4x
F · n dS =
div F dV =
4x dV =
S
2π
(x2 + y 2 + z 2 ) dV =
div F dV =
D
1
D
1−x2
1
1
0
0
=
1
0
1−x2
(8xz − 2xz 2 ) dx =
0
0
1−x2
2−z
4x dy dz dx
0
0
1−x2
4x(2 − z) dz dx =
=
(8x − 4xz) dz dx
01
0
[8x(1 − x2 ) − 2x(1 − x2 )2 ] dx
0
1
1
5
= −2(1 − x2 )2 + (1 − x2 )3 =
3
3
0
62. For S1 , n = (xi + yj)/ x2 + y 2 ; for S2 , n2 = −k and z = 0; and for S3 , n3 = k and z = c.
Then
F · n dS =
F · n1 dS1 +
F · n2 dS2 +
F · n3 dS3
S
S1
S2
S3
x +y
dS1 +
(−z 2 − 1) dS2 +
(z 2 + 1) dS3
2
2
x +y
S1
S2
S3
=
x2 + y 2 dS1 +
(−1) dS2 +
(c2 + 1) dS3
2
2
=
S1
dS1 −
=a
S1
S2
dS2 + (c2 + 1)
S2
S3
dS3
S3
= a(2πac) − πa2 + (c2 + 1)πa2 = 2πa2 c + πa2 c2 .
63. x = 0 =⇒ u = 0, v = −y 2 =⇒ u = 0, −1 ≤ v ≤ 0
x = 1 =⇒ u = 2y, v = 1 − y 2 = 1 − u2 /4
y = 0 =⇒ u = 0, v = x2 =⇒ u = 0, 0 ≤ v ≤ 1
y = 1 =⇒ u = 2x, v = x2 − 1 = u2 /4 − 1
∂(u, v) 2y 2x ∂(x, y)
1
=
= −4(x2 + y 2 ) =⇒
=−
2
∂(x, y)
∂(u, v)
4(x + y 2 )
2x −2y
2 1−u2 /4
√ 1
2
2 3 2
2
2 3
2
dA = 1
(x + y ) x − y dA =
(x + y ) v −
v 1/3 dv du
4(x2 + y 2 ) 4 0 u2 /4−1
R
S
1−u2 /4
2
1 2 3 4/3 3
=
v
(1 − u2 /4)4/3 − (u2 /4 − 1)4/3 du
du =
4 0 4
16 0
u2 /4−1
2
3
=
(1 − u2 /4)4/3 − (1 − u2 /4)4/3 du = 0
16 0
548
CHAPTER 9 REVIEW EXERCISES
64. y = x =⇒ u + uv = v + uv =⇒ v = u
x = 2 =⇒ u + uv = 2 =⇒ v = (2 − u)/u
y = 0 =⇒ v + uv = 0 =⇒ v = 0 or u = −1
∂(x, y) 1 + w
=
∂(u, v) v
(we take v = 0)
u
=1+u+v
1 + u
Using x = u + uv and y = v + uv we find
(x − y)2 = (u + uv − v − uv)2 = (u − v)2 = u2 − 2uv + v 2
x + y = u + uv + v + uv = u + v + 2uv
(x − y) + 2(x + y) + 1 = u2 + 2uv + v 2 + 2(u + v) + 1 = (u + v)2 + 2(u + v) + 1 = (u + v + 1)2 .
2
Then
1 2/(1+v)
1
dA =
(u + v + 1) dA =
du dv
(x − y)2 + 2(x + y) + 1
S u+v+1
0
v
1
1
1 2 2
1
=
− v dv = 2 ln(1 + v) − v = 2 ln 2 − .
1+v
2
2
0
0
1
R
65. The equations of the spheres are x2 + y 2 + z 2 = a2 and x2 + y 2 + (z − a)2 = 1. Subtracting these equations, we
obtain (z − a)2 − z 2 = 1 − a2 or −2az + a2 = 1 − a2 . Thus, the spheres intersect on the plane z = a − 1/2a.
The region of integration is x2 + y 2 + (a − 1/2a)2 = a2 or r2 = 1 − 1/4a2 . The area is
2π √1−1/4a2
√1−1/4a2
2
2 −1/r
2
2 1/2 A=a
(a − r )
r dr dθ = 2πa[−(a − r ) ] 0
0
= 2πa a − a2 − 1 −
1
4a2
1/2
0
2 1/2 1
= 2πa a −
a−
= π.
2a
66. (a) Both states span 7 degrees of longitude and 4 degrees of latitude, but Colorado is larger because it lies
to the south of Wyoming. Lines of longitude converge as they go north, so the east-west dimensions of
Wyoming are shorter than those of Colorado.
(b) We use the function f (x, y) =
R2 − x2 − y 2 to describe the northern
hemisphere, where R ≈ 3960 miles is the radius of the Earth. We need to
compute the surface area over a polar rectangle P of the form θ1 ≤ θ ≤ θ2 ,
R cos φ2 ≤ r ≤ R cos φ1 . We have
−x
fx = 2
R − x2 − y 2
and
fy = −y
R2
− x2 − y 2
so that
1 + fx2 + fy2 =
1+
R2
x2 + y 2
R
=√
.
2
2
2
−x −y
R − r2
549
CHAPTER 9 REVIEW EXERCISES
Thus
A=
1+
fx2
+
fy2
θ2
R cos φ1
dA =
P
θ1
R cos φ2
√
R
r dr dθ
− r2
R2
R cos φ2
= (θ2 − θ1 )R R2 − r2 = (θ2 − θ1 )R2 (sin φ2 − sin φ1 ).
R cos φ1
◦
The ratio of Wyoming to Colorado is then
Colorado.
sin 45 − sin 41◦
≈ 0.941. Thus Wyoming is about 6% smaller than
sin 41◦ − sin 37◦
(c) 97,914/104,247 ≈ 0.939, which is close to the theoretical value of 0.941. (Our formula for the area says that
the area of Colorado is approximately 103,924 square miles, while the area of Wyoming is approximately
97,801 square miles.)
550
Part III
Systems of Differential Equations
10
Systems of Linear
Differential Equations
EXERCISES 10.1
Preliminary Theory
x
3 −5
. Then X =
X.
y
4
8
x
4 −7
X=
. Then X =
X.
y
5
0
 


x
−3
4 −9
 


X =  y . Then X =  6 −1
0  X.
z
10
4
3
 


x
1 −1 0
 


X =  y . Then X =  1
0 2  X.
z
−1
0 1
 
   




x
t
−1
1 −1
1
0
 
   




X =  y . Then X =  2
1 −1  X +  −3t2  +  0  +  0 .
z
−t
2
1
1
1
t2
 



 −t
x
−3 4
0
e sin 2t
 



 −t
X =  y . Then X =  5 9 0  X +  4e cos 2t .
z
0 1 6
−e−t
1. Let X =
2. Let
3. Let
4. Let
5. Let
6. Let
7.
8.
9.
10.
11.
dx
dy
= 4x + 2y + et ;
= −x + 3y − et
dt
dt
dx
dy
dz
= 7x + 5y − 9z − 8e−2t ;
= 4x + y + z + 2e5t ;
= −2y + 3z + e5t − 3e−2t
dt
dt
dt
dx
dy
dz
= x − y + 2z + e−t − 3t;
= 3x − 4y + z + 2e−t + t;
= −2x + 5y + 6z + 2e−t − t
dt
dt
dt
dx
dy
= 3x − 7y + 4 sin t + (t − 4)e4t ;
= x + y + 8 sin t + (2t + 1)e4t
dt
dt
Since
−5
3 −4
−5
X =
e−5t and
X=
e−5t
−10
4 −7
−10
551
10.1 Preliminary Theory
we see that
X =
12. Since
X =
5 cos t − 5 sin t
2 cos t − 4 sin t
et
and
we see that
X =
13. Since
X =
3
2
−3
e
and
X =
X =
5
−1
t
e +
4
−4
X =
15. Since
 
0
 
X =  0 
0
1

 6
−1
and

1

X = 6
−1


cos t


X =  12 sin t − 12 cos t 
− cos t − sin t
we see that

1

X =  1
−2
0
1
0
et
X=
3
2
−3
e−3t/2
X.
1
0
X=
5
−1
t
e +
4
−4
tet
X.
−2

 1
−2
and

 
1
0

 
0 X = 0
2
−1
1
5 cos t − 5 sin t
2 cos t − 4 sin t
−1
1
0
2
−1
−2

1
4
2
−1
2
−1
X=
X.
−1
1

we see that
16. Since
5
4
1/4
and
we see that
5
4
−1
1
te
X.
−1
t
−2
−2
−2
−2
−3t/2
we see that
14. Since
−4
−7
3
4
−1
0

1

0  X.
−1

cos t



0  X =  12 sin t − 12 cos t 
0
1
1
0
−1


− cos t − sin t

1

0  X.
−1
17. Yes, since W (X1 , X2 ) = −2e−8t = 0 the set X1 , X2 is linearly independent on −∞ < t < ∞.
18. Yes, since W (X1 , X2 ) = 8e2t = 0 the set X1 , X2 is linearly independent on −∞ < t < ∞.
19. No, since W (X1 , X2 , X3 ) = 0 the set X1 , X2 , X3 is linearly dependent on −∞ < t < ∞.
20. Yes, since W (X1 , X2 , X3 ) = −84e−t = 0 the set X1 , X2 , X3 is linearly independent on −∞ < t < ∞.
21. Since
Xp =
2
−1
and
1
3
4
2
Xp +
552
2
−4
t+
−7
−18
=
2
−1
10.1
we see that
Xp =
22. Since
Xp
1
3
0
=
0
4
2
Xp
Xp
2
1
t
=
e +
tet
0
−1
=

we see that
Then

6


X1 =  −1  e−t ,
−5
2
1
1
−1
2
2
−6
1
2
2
1

t+
Xp +
Xp +
2
3
1

−4


and
1
−1
1
4


2
1
2
3
1

Xp =  −4
−6
25. Let
and
Xp =

3 cos 3t


Xp = 
0

−3 sin 3t
2
−4
we see that
24. Since
Xp +
and
we see that
23. Since
1
4
−5
2
−7
−18
−5
2
Preliminary Theory
.
0
=
0
.
1
2
1
t
t
Xp −
e =
e +
tet
7
0
−1
Xp −
1
et .
7





3
−1
3 cos 3t





0  Xp +  4  sin 3t = 
0

3
0
−3 sin 3t



3
−1



0  Xp +  4  sin 3t.
0
3

−3


X2 =  1  e−2t ,
1
 
2
  3t
X3 =  1  e ,
1

0

and A =  1
1
6
0
1

0

1.
0


−6


X1 =  1  e−t = AX1 ,
5


6


X2 =  −2  e−2t = AX2 ,
−2
 
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 −∞ < t < ∞.
26. Let
X1 =
X2 =
1
√
−1 − 2
1
√
−1 + 2
√
e
553
2t
√
e−
,
2t
,
10.1 Preliminary Theory
Xp =
1 2
−2
1
t +
t+
,
0
4
0
and
A=
Then
−1
−1
−1
1
.
√
√
2
√
e 2 t = AX1 ,
−2 − 2
√
√
− 2
√
X2 =
e− 2 t = AX2 ,
−2 + 2
2
−2
1 2
4
−1
Xp =
t+
= AXp +
t +
t+
,
0
4
1
−6
5
X1 =
√
and W (X1 , X2 ) = 2 2 = 0 so that Xp is a particular solution and X1 and X2 form a fundamental set on
−∞ < t < ∞.
EXERCISES 10.2
Homogeneous Linear Systems
1. The system is
X =
1
4
2
3
X
and det(A − λI) = (λ − 5)(λ + 1) = 0. For λ1 = 5 we obtain
−4
2 0
1 − 12 0
=⇒
4 −2 0
0
0 0
For λ2 = −1 we obtain
2
2
0
4
4
0
Then
=⇒
1 1
0 0
0
so that K1 =
so that K2 =
0
1
.
2
−1
1
.
1
−1
5t
X = c1
e + c2
e−t .
2
1
2. The system is
X =
2
1
2
3
X
and det(A − λI) = (λ − 1)(λ − 4) = 0. For λ1 = 1 we obtain
1 2 0
−2
1 2 0
.
=⇒
so that K1 =
1 2 0
0 0 0
1
For λ2 = 4 we obtain
−2
2
0
1
−1
0
=⇒
−1
0
1
0
0
0
554
so that K2 =
1
.
1
10.2 Homogeneous Linear Systems
Then
X = c1
−2
1
3. The system is
et + c2
X =
−4
− 52
1
e4t .
1
2
2
X
and det(A − λI) = (λ − 1)(λ + 3) = 0. For λ1 = 1 we obtain
−5 2 0
2
−5 2 0
=
.
=⇒
so
that
K
1
5
−2 1 0
0 0 0
5
For λ2 = −3 we obtain
−1
2
0
− 52
5
0
=⇒
Then
−1
0
2
X = c1
5
4. The system is
X =
0
2
0
2
.
so that K2 =
1
0
et + c2
2
1
− 52
2
3
4
−2
e−3t .
X
and det(A − λI) = 12 (λ + 1)(2λ + 7) = 0. For λ1 = −7/2 we obtain
1 2 0
−2
1 2 0
=
.
=⇒
so
that
K
1
3
3
0 0 0
1
0
4
2
For λ2 = −1 we obtain
− 32
2
0
3
4
−1
0
=⇒
Then
X = c1
−3
0
−2
4
0
1
5. The system is
X =
0
4
.
so that K2 =
3
0
e−7t/2 + c2
10
8
−5
−12
4
3
e−t .
X
and det(A − λI) = (λ − 8)(λ + 10) = 0. For λ1 = 8 we obtain
2 −5 0
5
1 − 52 0
.
=⇒
so that K1 =
8 −20 0
0
0 0
2
For λ2 = −10 we obtain
−5
0
8 −2
0
20
=⇒
Then
X = c1
1 − 14
0
0
0
so that K2 =
0
5
1
e8t + c2
e−10t .
2
4
6. The system is
X =
−6
−3
555
2
1
X
1
.
4
10.2 Homogeneous Linear Systems
and det(A − λI) = λ(λ + 5) = 0. For λ1 = 0 we obtain
−6 2 0
1 − 13 0
=⇒
−3 1 0
0
0 0
For λ2 = −5 we obtain
−1
2
0
−3
6
0
=⇒
1 −2
0
0
Then
X = c1
7. The system is
0
0
so that K1 =
1
.
3
2
so that K2 =
.
1
1
2
+ c2
e−5t .
3
1

1

X =  0
0
1
2
1

−1

0 X
−1
and det(A − λI) = (λ − 1)(2 − λ)(λ + 1) = 0. For λ1 = 1, λ2 = 2, and λ3 = −1 we obtain
 
 
 
1
2
1
 
 
 
K1 =  0  , K2 =  3  , and K3 =  0  ,
0
1
2
so that
 
 
 
1
2
1
  t
  2t
  −t
X = c1  0  e + c2  3  e + c3  0  e .
0
1
2
8. The system is

2

X = 5
0
−7
10
5

0

4X
2
and det(A − λI) = (2 − λ)(λ − 5)(λ − 7) = 0. For λ1 = 2, λ2 = 5, and λ3 = 7 we obtain






4
−7
−7






K1 =  0  , K2 =  3  , and K3 =  5  ,
−5
5
5
so that






4
−7
−7






X = c1  0  e2t + c2  3  e5t + c3  5  e7t .
−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

 



X = c1  0  e−t + c2  4  e3t + c3  −1  e−2t .
1
3
3
556
10.2 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





X = c1  0  e−t + c2 
−1
−12


4




6  e−t/2 + c3  2  e−3t/2 .
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
 




X = c1  1  e3t + c2  −1  e−5t + c3  −2  e6t .
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
X = c1
0
1
e−t/2 + c2
If
X(0) =
1
1
et/2 .
3
5
then c1 = 2 and c2 = 3.
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
so that


 


5
2
−2

 



X = c1  −3  e2t + c2  0  e3t + c3  0  e−t .
2
1
1
557
10.2 Homogeneous Linear Systems
If
 
1
 
X(0) =  3 
0
then c1 = −1, c2 = 5/2, and c3 = −1/2.






0.382175
0.405188
−0.923562






15. 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


16. X = c1  0.239535  e
+ c2  −0.275304  e
+ c3  −0.826218  e−2.92362t






 0.195825 
 0.176045 
 −0.346439 
0.0508861
0.338775
0.31957




0.313235
−0.301294
 0.466599 
0.64181 



 2.02882t


0.31754  e
+ c5  0.222136  e−0.155338t



 0.0534311 
0.173787 



+c4 


−0.599108
−0.799567
17. (a)
(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).
18. 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.
558
10.2 Homogeneous Linear Systems
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.
19. We have det(A − λI) = λ2 = 0. For λ1 = 0 we obtain
1
K=
.
3
A solution of (A − λ1 I)P = K is
1
2
P=
so that
X = c1
1
3
1
+ c2
3
t+
1
2
.
20. 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
X = c1
1
1
−t
e
0
1
5
1
+ c2
1
−t
te
+
0
1
5
e−t .
21. 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
− 13
0
1
1
1
−3
e2t + c2
te2t +
e2t .
1
1
0
22. 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
1
P=
2
0
559
10.2 Homogeneous Linear Systems
so that
1
3
3
6t
6t
X = c1
e + c2
te + 2 e6t .
2
2
0
23. 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
 
1
 
K2 =  0 
1
Then
 
1
 
and K3 =  1  .
0
 
 
 
1
1
1
  t
  2t
  2t
X = c1  1  e + c2  0  e + c3  1  e .
1
1
0
24. 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
Then


0


K2 =  −2 
1


1


and K3 =  −2  .
0
 




2
0
1
 




X = c1  1  e8t + c2  −2  e−t + c3  −2  e−t .
2
1
0
25. 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
 
P =  12 
0
560
10.2 Homogeneous Linear Systems
so that
5 





−4
−2
−2
2
  






X = c1  −5  + c2  0  e5t + c3  0  te5t +  12  e5t  .
2
1
1
0

26. 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
A solution of (A − λ2 I)P = K is
so that


0


P =  −1 
0
 





 
1
0
0
0
 





 
X = c1  0  et + c2  −1  e2t + c3  −1  te2t +  −1  e2t  .
0
1
1
0
27. 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
 
1
0
2
 
 
P =  1  and Q =  0 
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
0
1
1
1
28. 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
0
 
 
P =  1  and Q =  0 
0
1
561
10.2 Homogeneous Linear Systems
so that
 
 
 
  
 
  
1
1
1
0
0
0
2
  4t
  4t   4t 
  t 4t   4t   4t 
X = c1  0  e + c2  0  te +  1  e  + c3  0  e +  1  te +  0  e  .
2
0
0
0
0
0
1
29. 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
P=
so that
1
1
2
2
1
4t
4t
X = c1
e + c2
te +
e4t .
1
1
1
If
X(0) =
−1
6
then c1 = −7 and c2 = 13.
30. 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
 
1
 
K2 =  0 
1

 
0
 
and K3 =  1 
0
 
 
1
0

 −t
  t
  t
X = c1  0  e + c2  0  e + c3  1  e .
1
1
0
If
−1

 
1
 
X(0) =  2 
5
then c1 = 2, c2 = 3, and c3 = 2.
31. 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
 
 
 
 
 
 
K1 =  0  ,
K2 =  1  ,
and
K3 =  0  .
 
 
 
0
0
1
0
0
0
32. In Problem 20 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.
562
10.2 Homogeneous Linear Systems
In Problem 21 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.
In Problems 33-46 the form of the
answer
will vary according to the choice of eigenvector.
1
Problem 33, if K1 is chosen to be
the solution has the form
2−i
cos t
sin t
X = c1
e4t + c2
e4t .
2 cos t + sin t
2 sin t − cos t
33. We have det(A − λI) = λ2 − 8λ + 17 = 0. For λ1 = 4 + i we obtain
2+i
K1 =
5
so that
X1 =
2+i
5
(4+i)t
e
Then
X = c1
=
2 cos t − sin t
5 cos t
2 cos t − sin t
e +i
5 cos t
4t
4t
e + c2
cos t + 2 sin t
5 sin t
cos t + 2 sin t
e4t .
5 sin t
e4t .
34. We have det(A − λI) = λ2 + 1 = 0. For λ1 = i we obtain
−1 − i
K1 =
2
so that
X1 =
−1 − i
2
Then
X = c1
eit =
sin t − cos t
2 cos t
sin t − cos t
2 cos t
+ c2
+i
− cos t − sin t
2 sin t
− cos t − sin t
2 sin t
.
.
35. We have det(A − λI) = λ2 − 8λ + 17 = 0. For λ1 = 4 + i we obtain
−1 − i
K1 =
2
so that
X1 =
−1 − i
2
e(4+i)t =
sin t − cos t
2 cos t
563
e4t + i
− sin t − cos t
2 sin t
e4t .
For example, in
10.2 Homogeneous Linear Systems
Then
X = c1
sin t − cos t
2 cos t
e4t + c2
− sin t − cos t
2 sin t
e4t .
36. We have det(A − λI) = λ2 − 10λ + 34 = 0. For λ1 = 5 + 3i we obtain
1 − 3i
K1 =
2
so that
X1 =
1 − 3i
2
e(5+3i)t =
Then
X = c1
cos 3t + 3 sin 3t
2 cos 3t
cos 3t + 3 sin 3t
2 cos 3t
e5t + i
e5t + c2
sin 3t − 3 cos 3t
2 sin 3t
sin 3t − 3 cos 3t
2 sin 3t
e5t .
e5t .
37. We have det(A − λI) = λ2 + 9 = 0. For λ1 = 3i we obtain
4 + 3i
K1 =
5
so that
X1 =
4 + 3i
5
Then
e
X = c1
3it
=
4 cos 3t − 3 sin 3t
5 cos 3t
4 cos 3t − 3 sin 3t
+ c2
5 cos 3t
+i
4 sin 3t + 3 cos 3t
5 sin 3t
4 sin 3t + 3 cos 3t
.
.
5 sin 3t
38. We have det(A − λI) = λ2 + 2λ + 5 = 0. For λ1 = −1 + 2i we obtain
2 + 2i
K1 =
1
so that
X1 =
2 + 2i
1
(−1+2i)t
e
Then
X = c1
=
2 cos 2t − 2 sin 2t
cos 2t
2 cos 2t − 2 sin 2t
cos 2t
−t
e
+ c2
−t
e
+i
2 cos 2t + 2 sin 2t
sin 2t
2 cos 2t + 2 sin 2t
sin 2t
39. We have det(A − λI) = −λ λ2 + 1 = 0. For λ1 = 0 we obtain
 
1
 
K1 =  0  .
0
For λ2 = i we obtain
so that


−i
 
K2 =  i 
1






−i
sin t
− cos t
 




X2 =  i  eit =  − sin t  + i  cos t  .
1
cos t
sin t
564
e−t .
e−t .
10.2 Homogeneous Linear Systems
Then
 




1
sin t
− cos t
 




X = c1  0  + c2  − sin t  + c3  cos t  .
0
cos t
sin t
40. 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
Then


−2 − i


K2 =  −3i 
2




−2 cos 2t + sin 2t
− cos 2t − 2 sin 2t

 t

 t
X2 = 
3 sin 2t
−3 cos 2t
e + i
e .
2 cos 2t
2 sin 2t






0
−2 cos 2t + sin 2t
− cos 2t − 2 sin 2t



 t

 t
X = c1  −2  e−3t + c2 
3 sin 2t
−3 cos 2t
 e + c3 
e .
1
2 cos 2t
2 sin 2t
41. 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
42. We have det(A − λI) = −(λ − 6)(λ2 − 8λ + 20) = 0. For λ1 = 6 we obtain
 
0
 
K1 =  1  .
0
565
10.2 Homogeneous Linear Systems
For λ2 = 4 + 2i we obtain


−i
 
K2 =  0 
2
so that





−i
sin 2t
− cos 2t
 



 4t
X2 =  0  e(4+2i)t =  0  e4t + i 
0
e .
2
2 cos 2t
2 sin 2t
Then
 




0
sin 2t
− cos 2t
 



 4t
X = c1  1  e6t + c2  0  e4t + c3 
0
e .
0
2 cos 2t
2 sin 2t

43. 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
so that


4 + 3i


K2 =  −5 
0






4 + 3i
4 cos 3t − 3 sin 3t
4 sin 3t + 3 cos 3t



 −2t

 −2t
X2 =  −5  e(−2+3i)t = 
−5 cos 3t
+ i
−5 sin 3t
e
e .
0
0
0
Then

28





X = c1  −5  e2t + c2 
25
4 cos 3t − 3 sin 3t
−5 cos 3t
0


 −2t

+ c3 
e
4 sin 3t + 3 cos 3t
−5 sin 3t
0

 −2t
e .
44. We have det(A − λI) = −(λ + 2)(λ2 + 4) = 0. For λ1 = −2 we obtain


0


K1 =  −1  .
1
For λ2 = 2i we obtain
so that


−2 − 2i


K2 = 
1

1






−2 − 2i
−2 cos 2t + 2 sin 2t
−2 cos 2t − 2 sin 2t

 2it 



X2 = 
1
cos 2t
sin 2t
e = 
 + i
.
1
cos 2t
sin 2t
566
10.2 Homogeneous Linear Systems
Then

0





X = c1  −1  e−2t + c2 
−2 cos 2t + 2 sin 2t
cos 2t
cos 2t
1




 + c3 
−2 cos 2t − 2 sin 2t
sin 2t
sin 2t


.
45. We have det(A − λI) = (1 − λ)(λ2 + 25) = 0. For λ1 = 1 we obtain


25


K1 =  −7  .
6
For λ2 = 5i we obtain
so that


1 + 5i


K2 =  1 
1






1 + 5i
cos 5t − 5 sin 5t
sin 5t + 5 cos 5t






X2 =  1  e5it = 
cos 5t
sin 5t
 + i
.
1
cos 5t
sin 5t
Then

25





X = c1  −7  et + c2 



 + c3 
cos 5t
cos 5t
6
If

cos 5t − 5 sin 5t
sin 5t + 5 cos 5t
sin 5t
sin 5t


.


4


X(0) =  6 
−7
then c1 = c2 = −1 and c3 = 6.
46. We have det(A − λI) = λ2 − 10λ + 29 = 0. For λ1 = 5 + 2i we obtain
K1 =
so that
X1 =
1
1 − 2i
and
X = c1
If X(0) =
e(5+2i)t =
1
1 − 2i
cos 2t
cos 2t + 2 sin 2t
cos 2t
cos 2t + 2 sin 2t
e5t + c3
−2
, then c1 = −2 and c2 = 5.
8
567
e5t + i
sin 2t
sin 2t − 2 cos 2t
sin 2t
sin 2t − 2 cos 2t
e5t .
e5t .
10.2 Homogeneous Linear Systems
47.
48. (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 1 0
.
=⇒
so that K1 =
1
1 1 0
0 0 0
For λ2 = 2 we obtain
−1
1
0
1
−1
0
=⇒
Then
−1
0
1
0
X = c1
−1
1
0
1
.
so that K2 =
1
0
+ c2
1
e2t .
1
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
1
K=
A solution of (A − λ1 I)P = K is
P=
so that
X = c1
−1
1
−1
0
+ c2
568
.
−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.
t+
−1
0
.
10.2 Homogeneous Linear Systems
49. 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
The general solution of the system is

c2 te2t + c1 e2t



c2 e2t




2t
X=
c3 e



 c5 te2t + c4 e2t 
c5 e2t
 
 
  
1
1
0
0
 0 
1 
 
 
  
 
 
  
= c1  0  e2t + c2  0  te2t +  0  e2t 
 
 
  
0
 0 
0 
0
0
 
 
 
  
0
0
0
0
0
0
 0 
0 
 
 
 
  
 
 
 
  
+ c3  1  e2t + c4  0  e2t + c5  0  te2t +  0  e2t 
 
 
 
  
0
1
 1 
0 
0
0
0
1

  
0

1 

  

  
= c1 K1 e2t + c2 K1 te2t +  0  e2t 

  

0 
0
0
  
0

0 

  

  
+ c3 K2 e2t + c4 K3 e2t + c5 K3 te2t +  0  e2t  .

  

0 

1
2t
There are three solutions of the form X = Ke , 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
569
10.2 Homogeneous Linear Systems
(A − 2I)K2 = 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
0
p5
so p2 = 1 and p5 = 0, while p1 , p3 , and p4 are arbitrary. Choosing p1 = p3 = p4 = 0 we have
 
1
0
 
 
P = 0.
 
0
0
Therefore a solution is
 
 
1
1
0
0
 
 
 
 
X =  0  te2t +  0  e2t .
 
 
0
0
0
Repeating for K3 we find
0
 
0
0
 
 
P = 0,
 
0
1
so another solution is
 
 
0
0
0
0
 
 
 
 
X =  0  te2t +  0  e2t .
 
 
1
0
0
1
50. 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.
51. Suppose the eigenvalues are α ± iβ, β > 0. In Problem 36 the eigenvalues are 5 ± 3i, in Problem 37 they are
±3i, and in Problem 38 they are −1 ± 2i. From Problem 47 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.
570
10.3
Solution by Diagonalization
52. (a) The given system can be written as
x1 = −
k1 + k2
k2
x1 +
x2 ,
m1
m1
In terms of matrices this is X = AX where
x1
X=
x2
x2 =


and A = 
−
k2
k2
x1 −
x2 .
m2
m2
k1 + k2
m1
k2
m2
k2 
m1 
.
k2
−
m2
(b) If X = Keωt then X = ω 2 Keωt and AX = AKeωt so that X = AX becomes ω 2 Keωt = AKeωt or
(A − ω 2 I)K = 0. Now let ω 2 = λ.
−5
2
(c) When m1 = 1, m2 = 1, k1 = 3, and k2 = 2 we obtain A =
. The eigenvalues and corresponding
2 −2
√
1
−2
eigenvectors of A are λ1 = −1, λ2 = −6, K1 =
, K2 =
. Since ω1 = i, ω2 = −i, ω3 = 6 i, and
2
1
√
ω4 = − 6 i a solution is
√
√
1
1
−2
−2
X = c1
eit + c2
e−it + c3
e 6 it + c4
e− 6 it .
2
2
1
1
√
√
√
(d) Using eit = cos t + i sin t and e 6 it = cos 6 t + i sin 6 t the preceding solution can be rewritten as
1
1
(cos t + i sin t) + c2
(cos t − i sin t)
X = c1
2
2
√
√
√
√
−2
−2
+ c3
(cos 6 t + i sin 6 t) + c4
(cos 6 t + i sin 6 t)
1
1
1
1
= (c1 + c2 )
cos t + i(c1 − c2 )
sin t
2
2
√
√
−2
−2
cos 6 t + i(c3 − c4 )
sin 6 t
+ (c3 + c4 )
1
1
√
√
1
1
−2
−2
= b1
cos t + b2
sin t + b3
cos 6 t + b4
sin 6 t
2
2
1
1
where b1 = c1 + c2 , b2 = i(c1 − c2 ), b3 = c3 + c4 , and b4 = i(c3 − c4 ).
EXERCISES 10.3
Solution by Diagonalization
1. λ1 = 7, λ2 = −4, K1 =
X = PY =
3
−2
1
3
3
1
, K2 =
c1 e7t
c2 e−4t
3
=
−2
,P=
3
−2
1
3
3c1 e7t − 2c2 e−4t
c1 e7t + 3c2 e−4t
;
= c1
571
3
−2
e7t + c2
e−4t
1
3
10.3 Solution by Diagonalization
1
1
1 1
, K2 =
,P=
;
−1
1
−1 1
1
c1
c1 + c2 et
1
1
=
=
c
+
c
et
1
2
t
t
1
−1
1
c2 e
−c1 + c2 e
2. λ1 = 0, λ2 = 1, K1 =
X = PY =
1
−1
1
1
1 1
3. λ1 = , λ2 = , K1 =
, K2 =
,P=
;
−2
2
−2 2
1 1
c1 et/2
c1 et/2 + c2 e3t/2
1
1
t/2
X = PY =
=
= c1
e + c2
e3t/2
3t/2
t/2
3t/2
−2 2
−2
2
c2 e
−2c1 e + 2c2 e
1
2
3
2
√
√
4. λ1 = − 2 , λ2 = 2 , K1 =
−1
−1
−1
√ , K2 =
√ ,P=
√
1+ 2
1− 2
1+ 2
−1
√ ;
1− 2
√
√
√
−c1 e− 2 t − c2 e 2 t
−1
−1
c1 e− 2 t
√
√
√
√
√
√
√
=
X = PY =
1+ 2 1− 2
c2 e 2 t
(1 + 2 )c1 e− 2 t + (1 − 2 )c2 e 2 t
√
√
−1
−1
√
√
= c1
e− 2 t + c2
e 2t
1+ 2
1− 2


 
 

−1
1
0
−1


 
 

5. λ1 = −4, λ2 = 2, λ3 = 6, K1 =  1 , K2 =  1 , K3 =  0 , P =  1
0
1
1
0

−1

X = PY =  1
0
1
1
1



 

−1
1
1
−1




 

6. λ1 = −1, λ2 = 1, λ3 = 4, K1 =  0 , K2 =  −2 , K3 =  1 , P =  0
1
1
1
1
−1

X = PY =  0
1
1
−2
1

0

0 ;
1
1

 



 
 
0
c1 e−4t
−c1 e−4t + c2 e2t
−1
1
0

 



 
 
0   c2 e2t  =  c1 e−4t + c2 e2t  = c1  1  e−4t + c2  1  e2t + c3  0  e6t
1
0
0
1
c3 e6t
c2 e2t + c3 e6t


1
1

1 1

−2 1 ;
1 1

 





 
1
c1 e−t
−c1 + c2 et + c3 e4t
−1
1
1

 





 
1   c2 et  = 
−2c2 et + c3 e4t  = c1  0  e−t +c2  −2  et +c3  1  e4t
1
1
1
1
c3 e4t
c1 e−t + c2 et + c3 e4t
 





1
−1
−1
1
 





7. λ1 = −1, λ2 = 2, λ3 = 2, K1 =  1 , K2 =  1 , K3 =  0 , P =  1
1
0
1
1

1

X = PY =  1
−1
1
1
0
−1
1
0

−1

0 ;
1

 

−1
c1 e−t
c1 e−t − c2 e2t − c3 e2t

 

0   c2 e2t  =  c1 e−t + c2 e2t

1
c3 e2t
c1 e−t
+ c3 e2t
 




1
−1
−1
 




= c1  1  e−t + c2  1  e2t + c3  0  e2t
1
0
1
572
10.3
Solution by Diagonalization






 
−1
−1
−1
1
 1
 0
 0
1






 
8. λ1 = 0, λ2 = 0, λ3 = 0, λ4 = 4, K1 = 
, K2 = 
, K3 = 
, K4 =  ,
 0
 1
 0
1
0


−1 −1 −1 1
 1
0
0 1


P=
;
 0
1
0 1
0

0
1
1
−1
−1
−1
1


c1
0

−c1 − c2 − c3 + c4 e4t
1
1

 1

 
0
0 1
+ c4 e4t 

  c2   c1

X = PY = 

=
4t 
 0




1
0 1
c3
c2
+ c4 e 
4t
0
0
1 1
c4 e
c3 + c4 e4t






 
−1
−1
−1
1
 1
 0
 0
1



 



= c1 
 + c2 
 + c3 
 + c4   e4t
 0
 1
 0
1
0
0
1
1
 
 
 


1
2
3
1 2 3
 
 
 


9. λ1 = 1, λ2 = 2, λ3 = 3, K1 =  1 , K2 =  2 , K3 =  4 , P =  1 2 4 ;
1
3
5
1 3 5







 
 
1 2 3
c1 et
c1 et + 2c2 e2t + 3c3 e3t
1
2
3



  t
  2t
  3t
2t 
t
2t
3t 
X = PY =  1 2 4   c2 e  =  c1 e + 2c2 e + 4c3 e  = c1  1  e + c2  2  e + c3  4  e
1
3
c3 e3t
5
c1 et + 3c2 e2t + 5c3 e3t
1
3
5








1
1
1
1
1
1
√
√
√
√ 


 √ 
√ 

10. λ1 = 0, λ2 = −2 2 , λ3 = 2 2 , K1 =  0 , K2 =  − 2 , K3 =  2 , P =  0 − 2
2 ;
−1
1
1
−1
1
1
√
√


 

1
1
1
c1
c1 + c2 e−2 2 t + c3 e2 2 t
√
√
√
√
√
√
√


 

X = PY =  0 − 2
2   c2 e−2 2 t  =  − 2 c2 e−2 2 t + 2 c3 e2 2 t 
√
√
√
−1
1
1
c3 e2 2 t
−c1 + c2 e−2 2 t + c3 e2 2 t






1
1
1
√


 √ 
√  √
= c1  0  + c2  − 2  e−2 2 t + c3  2  e2 2 t
−1
1
1
m1 0
11. (a) Since M =
0 m2
system in the form
is a diagonal matrix with nonzero diagonal entries, it has an inverse. Writing the
m1 x1 + (k1 + k2 )x1 − k2 x2 = 0
m2 x2 − k2 x1 + k2 x2 = 0
we see that K =
k1 + k2
−k2
−k2
k2
.
(b) Since M has an inverse, MX + KX = 0 can be written as X + M−1 KX = 0 or X + BX = 0 where
573
10.3 Solution by Diagonalization
B = M−1 K =
 1
 m1
0

0

1
m2
−k2
k2
k1 + k2
−k2
 k1 + k2

=
−
m1
k2
−
m2
k2 
m1 
.
k2
m2
5 −2
(c) With m1 = 1, m2 = 1, k1 = 3, and k2 = 2 we have B =
. The eigenvalues of B are λ1 = 1 and
−2
2
1
−2
λ2 = 6 with corresponding eigenvectors
and
. Letting X = PY the system can be written
2
1
1 −2
1 0
−1
−1
PY + BPY = 0 or Y + P BPY = 0 where
and P BP =
. The system is then
2
1
0 6
1 0
Y +
Y = 0, which is uncoupled and equivalent to y1 + y1 = 0 and y2 + 6y2 = 0. The solutions
0 6
√
√
are y1 = c1 cos t + c2 sin t and y2 = c3 cos 6 t + c4 sin 6 t.
(d) From
X = PY =
we have
1
2
−2
1
y1
=
y2
y1 − 2y2
2y1 + y2
√
√
6 t − 2c4 sin 6 t
√
√
x1 = 2c1 cos t + 2c2 sin t + c3 cos 6 t + c4 sin 6 t
x1 = c1 cos t + c2 sin t − 2c3 cos
which is the same as
X = c1
1
2
cos t + c2
1
2
sin t + c3
−2
1
cos
√
6 t + c4
−2
1
√
sin 6 t.
EXERCISES 10.4
Nonhomogeneous Linear Systems
1. Solving
2 − λ
det(A − λI) = −1
3 = λ2 − 1 = (λ − 1)(λ + 1) = 0
−2 − λ we obtain eigenvalues λ1 = −1 and λ2 = 1. Corresponding eigenvectors are
−1
−3
K1 =
and K2 =
.
1
1
Thus
Xc = c1
−1
1
e−t + c2
Substituting
Xp =
a1
b1
574
−3
1
et .
10.4 Nonhomogeneous Linear Systems
into the system yields
2a1 + 3b1 = 7
−a1 − 2b1 = −5,
from which we obtain a1 = −1 and b1 = 3. Then
−1
−1
−3
−t
t
X(t) = c1
.
e + c2
e +
3
1
1
2. Solving
5 − λ
det(A − λI) = −1
9 = λ2 − 16λ + 64 = (λ − 8)2 = 0
11 − λ we obtain the eigenvalue λ = 8. A corresponding eigenvector is
3
K=
.
1
Solving (A − 8I)P = K we obtain
2
P=
.
1
Thus
Xc = c1
3
3
8t
e + c2
1
1
Substituting
Xp =
a1
2
te +
e8t .
1
8t
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
3
2
2
X(t) = c1
e8t + c2
te8t +
.
e8t +
1
1
1
− 12
3. Solving
1 − λ
det(A − λI) = 3
= λ2 − 2λ − 8 = (λ − 4)(λ + 2) = 0
1 − λ
3
we obtain eigenvalues λ1 = −2 and λ2 = 4. Corresponding eigenvectors are
1
1
K1 =
and K2 =
.
−1
1
Thus
Xc = c1
Substituting
Xp =
a3
b3
1
−1
−2t
e
t2 +
1
+ c2
e4t .
1
a2
b2
t+
a1
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
575
10.4 Nonhomogeneous Linear Systems
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
1
1
−2
−4
2
4
e−2t + c2
e4t +
t
t
+
.
X(t) = c1
+
3
3
−1
1
− 14
4
4
4. Solving
1 − λ
det(A − λI) = 4
−4 = λ2 − 2λ + 17 = 0
1 − λ
we obtain eigenvalues λ1 = 1 + 4i and λ2 = 1 − 4i. Corresponding eigenvectors are
i
−i
K1 =
and K2 =
.
1
1
Thus
0
−1
−1
0
t
Xc = c1
cos 4t +
sin 4t e + c2
cos 4t −
sin 4t et
1
0
0
1
− sin 4t
− cos 4t
et + c2
et .
= c1
cos 4t
− sin 4t
Substituting
Xp =
a3
b3
t+
a2
b2
+
a1
b1
e6t
into the system yields
a3 − 4b3 = −4
4a3 + b3 = 1
a2 − 4b2 = a3
−5a1 − 4b1 = −9
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
4 − sin 4t
− cos 4t
1
0
et + c2
et +
+
e6t .
X(t) = c1
t + 17
1
cos 4t
− sin 4t
1
1
17
5. Solving
4 − λ
det(A − λI) = 9
= λ2 − 10λ + 21 = (λ − 3)(λ − 7) = 0
6 − λ
1
3
we obtain the eigenvalues λ1 = 3 and λ2 = 7. Corresponding eigenvectors are
1
1
K1 =
and K2 =
.
−3
9
Thus
Xc = c1
1
−3
e3t + c2
Substituting
Xp =
a1
b1
1
e7t .
9
et
into the system yields
1
3a1 + b1 = 3
3
9a1 + 5b1 = −10
from which we obtain a1 = 55/36 and b1 = −19/4. Then
55 1
1
36
e3t + c2
e7t +
et .
X(t) = c1
−3
9
− 19
4
576
10.4 Nonhomogeneous Linear Systems
6. Solving
−1 − λ
det(A − λI) = −1
5 = λ2 + 4 = 0
1 − λ
we obtain the eigenvalues λ1 = 2i and λ2 = −2i. Corresponding eigenvectors are
5
5
K1 =
and K2 =
.
1 + 2i
1 − 2i
Thus
Xc = c1
5 cos 2t
cos 2t − 2 sin 2t
Substituting
Xp =
a2
b2
+ c2
cos t +
5 sin 2t
2 cos 2t + sin 2t
a1
b1
.
sin t
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
1
5 cos 2t
5 sin 2t
−3
−3
X(t) = c1
+ c2
+
cos t +
sin t.
1
2
cos 2t − 2 sin 2t
2 cos 2t + sin 2t
−3
3
7. Solving
1 − λ
det(A − λI) = 0
0
1
2−λ
0
1 3 = (1 − λ)(2 − λ)(5 − λ) = 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
 
Xp =  b1  e4t
c1
into the system yields
−3a1 + b1 + c1 = −1
−2b1 + 3c1 = 1
c1 = −2
577
10.4 Nonhomogeneous Linear Systems
from which we obtain c1 = −2, b1 = −7/2, and a1 = −3/2. Then
 
 
 
 3
1
1
1
−2
  t
  2t
  5t  7  4t
X(t) = C1  0  e + C2  1  e + C3  2  e +  − 2  e .
0
0
2
−2
8. Solving
−λ
det(A − λI) = 0
5
= −(λ − 5)2 (λ + 5) = 0
−λ 0
5−λ
5
0
0
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
 
 


Xc = C1  0  e5t + C2  1  e5t + C3  0  e−5t .
1
1
−1
Substituting

a1
 
X p =  b1 
c1

into the system yields
5c1 = −5
5b1 = 10
5a1 = −40
from which we obtain c1 = −1, b1 = 2, and a1 = −8. Then
 
 




1
1
1
−8
 
 




X(t) = C1  0  e5t + C2  1  e5t + C3  0  e−5t +  2  .
1
1
−1
−1
9. Solving
−1 − λ
det(A − λI) = 3
−2 = λ2 − 3λ + 2 = (λ − 1)(λ − 2) = 0
4 − λ
we obtain the eigenvalues λ1 = 1 and λ2 = 2. Corresponding eigenvectors are
1
−4
and K2 =
.
K1 =
−1
6
Thus
Xc = c1
1
−1
et + c2
Substituting
Xp =
a1
b1
578
−4
6
e2t .
10.4 Nonhomogeneous Linear Systems
into the system yields
−a1 − 2b1 = −3
3a1 + 4b1 = −3
from which we obtain a1 = −9 and b1 = 6. Then
1
−4
−9
t
2t
e + c2
e +
X(t) = c1
.
−1
6
6
Setting
X(0) =
we obtain
−4
5
c1 − 4c2 − 9 = −4
−c1 + 6c2 + 6 = 5.
Then c1 = 13 and c2 = 2 so
X(t) = 13
10. (a) Let I =
i2
1
−1
and
Ic = c1
a1
then Ip =
b1
30
0
e2t +
−9
6
.
I = c1
0
0
−2
−2
−2
−5
2
−1
−t
e
I+
60
60
1
+ c2
e−6t .
2
so that
For I(0) =
−4
6
so that
i3
I =
et + 2
If Ip =
2
−t
e
−1
+ c2
1
2
−6t
e
+
30
0
.
we find c1 = −12 and c2 = −6.
(b) i1 (t) = i2 (t) + i3 (t) = −12e−t − 18e−6t + 30.
11. From
X =
we obtain
Xc = c1
Then
Φ=
so that
U=
1
3et
1
2et
−3
−2
3
2
X+
4
−1
1
3
+ c2
et .
1
2
and Φ−1 =
Φ−1 F dt =
and
Xp = ΦU =
−11
5e−t
−11
−11
579
−2
e−t
t+
3
−e−t
−11t
dt =
−5e−t
−15
−10
.
10.4 Nonhomogeneous Linear Systems
12. From
X =
we obtain
Xc = c1
Then
Φ=
so that
U=
et
e−t
et
3e−t
0
X+
t
4
1
1
et + c2
e−t .
1
3
−1
and Φ
Φ−1 F dt =
and
−1
−2
2
3
−2te−t
=
3 −t
2e
− 12 et
dt =
2tet
− 12 e−t
1 t
2e
2te−t + 2e−t
2tet − 2et
4
0
Xp = ΦU =
t+
.
8
−4
13. From
X =
we obtain
3
−5
3
4
−1
Xc = c1
Then
Φ=
10e3t/2
3e3t/2
so that
U=
2et/2
et/2
10
3
Xp = ΦU =
we obtain
Xc = c1
Then
Φ=
e2t cos 2t
2e2t cos 2t
2e2t sin 2t
U=
2
4
Xp = ΦU =
−1
2
=
t/2
te
− 15
2
sin 2t
2 cos 2t
e + c2
=
1
2
1
2
cos 4t
sin 4t
cos 2t sin 4t −
580
1 −2t
2e
dt =
1
8
1
4
e2t .
− 12 e−2t sin 2t
− 18 sin 2t cos 4t −
et/2 .
cos 2t
2 sin 2t
− 13
4 t
− 94
−1
− 12 e−3t/2
5 −t/2
2e
− 34 e−t
2t
and Φ
1
4
1 −3t/2
4e
− 34 e−t/2
dt =
+
X+
et/2
3 −t
4e
− 13
4
− 13
4
Φ−1 F dt =
and
− 13
2
− sin 2t
2 cos 2t
−e2t sin 2t
so that
2
et/2 .
1
e3t/2 + c2
and Φ
1
−1
−1
and
X =
X+
Φ−1 F dt =
14. From
1
8
− 18
cos 2t
sin 4t
cos 4t
cos 2t cos 4t
sin 2t cos 4t
e2t .
1 −2t
4e
cos 2t
1 −2t
4e
sin 2t
10.4 Nonhomogeneous Linear Systems
15. From
X =
we obtain
0
−1
2
3
1
−1
X+
et
2
1
t
e + c2
e2t .
Xc = c1
1
1
Then
Φ=
so that
2et
e2t
et
e2t
−1
U=
Φ
and Φ−1 =
F dt =
and
Xp = ΦU =
16. From
X =
we obtain
Xc = c1
Then
Φ=
so that
U=
e2t
e2t
2et
et
Φ−1 F dt =
2
−3e−t
0
−1
2
3
2
1
et + c2
and Φ
Xp = ΦU =
1
1
2t
3e−t
2
e−3t
1
1
=
e2t .
e−t
−e−2t
2e−t − e−4t
2e−2t
dt =
X+
X =
−e−t
−1
and
we obtain
dt =
−2e−2t + 2e−5t
17. From
e−t
−e−2t
4
3
tet +
et .
2
3
1 −3t
10 e
−3
X+
12
12
−2e−t + 14 e−4t
e−2t − 25 e−5t
.
3 −3t
− 20
e
−1
8
−1
−e−t
2e−2t
t
4
−2
3t
Xc = c1
e + c2
e−3t .
1
1
Then
Φ=
so that
e3t
U=
4e3t
−1
Φ
−2e−3t
e−3t
F dt =
and Φ−1 =
6te−3t
6te3t
and
Xp = ΦU =
− 16 e3t
−12
0
581
dt =
t+
1 −3t
6e
1 −3t
3e
2 3t
3e
−2te−3t − 23 e−3t
2te3t − 23 e3t
− 43
− 43
.
10.4 Nonhomogeneous Linear Systems
18. From
X =
we obtain
1
1
8
−1
X+
e−t
tet
4
−2
3t
Xc = c1
e + c2
e−3t .
1
1
Then
Φ=
so that
U=
−2e3t
4e3t
−1
and Φ
e−3t
e3t
1 e−4t + 1 te−2t 6
3
Φ−1 F dt =
Xp = ΦU =
19. From
X =
we obtain
Xc = c1
Then
Φ=
so that
et
1 t
2e
2
−1
1
and Φ
Xp = ΦU =
20. From
X =
Xc = c1
Then
Φ=
et
so that
U=
1
−1
tet
−et
1 t
2e
−1
Φ
=
3
−2
e + c2
F dt =
t
− tet
1
2
1
−1
e−t − 4te−t
582
et .
−2te−t
2e−t
2e−t
1
−2t
2e
+ 3te−2t
−3e−2t
e−t .
t
te +
3
−5
0
1
2
et .
e−t − 2te−t
−2te−t
2e−t
2e−t
2e−t
Xp = ΦU =
1
2
1
X+
1
and Φ−1 =
and
0
dt =
−2
2
−1
.
e−t − 2te−t
6e−2t
we obtain
tet +
2e−2t − 6te−2t
and
1 4t
24 e
−1
−1
− tet
1 −2t
12 e
2
X+
e−t
1
2 3t
3e
1 2t
− 12
e + 16 te4t −
− 18 e−t − 18 et
et + c2
−1
Φ−1 F dt =
U=
dt =
1 −4t
− 24
e
− 16 te−2t −
−tet − 14 et
1
tet
−et
3
−2
1 −3t
3e
− 16 e3t
− 16 e2t + 23 te4t
and
=
1 −3t
6e
dt =
.
3e−t + 4te−t
−2e−t
10.4 Nonhomogeneous Linear Systems
21. From
X =
we obtain
cos t
sin t
Xc = c1
Then
Φ=
so that
cos t
sin t
U=
sin t
− cos t
−1
Φ
−1
0
0
1
1
1
X =
Then
Φ=
so that
U=
− sin t
cos t
cos t
sin t
−1
Φ
−1
1
− sin t
t
e
X =
Xc = c1
Then
Φ=
− sin t
cos t
cos t
sin t
so that
U=
− sin t
cos t
e
Φ
X+
.
et .
=
dt =
583
cos t
sin t
e−t
3 cos t + 3 sin t
3 sin t − 3 cos t
cos t
sin t
cos t
sin t
=
et
et .
− sin t
cos t
0
0
F dt =
dt =
1
t
Xp = ΦU =
cos t
sin t
et .
and Φ
− sin t
cos t
et + c2
−1
and
t
−3
3
cos t
and Φ
−1
1
1
1
−1
sin t
− cos t
t
− ln | cos t|
sin t
−1
we obtain
dt =
−3 sin t + 3 cos t
3 cos t + 3 sin t
cos t
sin t
et + c2
Xp = ΦU =
and
23. From
=
.
3
X+
et
3
F dt =
t cos t − sin t ln | cos t|
cos t
t sin t + cos t ln | cos t|
1
tan t
sec t
0
sin t
− cos t
−1
Xp = ΦU =
Xc = c1
+ c2
and Φ
and
we obtain
X+
F dt =
22. From
tet .
cos t
sin t
e−t
10.4 Nonhomogeneous Linear Systems
24. From
X =
we obtain
Xc = c1
Then
1
2
2t +
1
2
Φ=
2
so that
U=
−1
Φ
e−2t
F dt =
25. From
X =
Xc = c1
1
0
Φ=
so that
U=
Φ−1 F dt =
and
Xp = ΦU =
cos t
− sin t
26. From
so that
U=
cos t
− sin t
−1
Φ
+ c2
− tan2 t
t+
sin t
cos t
0
−1
F dt =
and
Xp = ΦU =
cos t
− sin t
X+
1
cot t
+ c2
dt =
t − tan t
− ln | cos t|
sin t
cos t
ln | cos t|.
.
− sin t
cos t
cos t
sin t
0
ln | csc t − cot t|
sin t ln | csc t − cot t|
cos t ln | csc t − cot t|
584
− sin t
cos t
sin t
cos t
and Φ−1 =
0
csc t
.
−
dt =
e2t
cos t
sin t
=
− sin t
sin t tan t
1
0
sin t
e−2t .
cos t
and Φ
tan t
Φ=
2t + 1
−2
2t + 2 ln t
−2 ln t
0
sec t tan t
−1
Xc = c1
Then
cos t
t
X =
dt =
X+
we obtain
− sin t
sin t
cos t
4t + 3 ln t − 4t ln t
cos t
− sin t
2 + 2/t
−2/t
−4t − 1
4
2t + ln t − 2t ln t
0
−1
we obtain
and Φ−1 =
Xp = ΦU =
1 1 −2t
e
X+
3 t
and
Then
1
1
1
e−2t + c2
te−2t + 21 e−2t .
2
2
2
t+
1
−2
−6
2
8
.
10.4 Nonhomogeneous Linear Systems
27. From
X =
we obtain
2 sin t
cos t
Φ=
so that
U=
−1
Φ
and
Xp = ΦU =
2 cos t
− sin t
F dt =
3 sin t
3
2
so that
U=
cos t − sin t
cos t
Φ−1 F dt =
and
Xp = ΦU =
and Φ
te +
cos t − sin t
cos t + sin t
sin t
=
et .
1
2
sin t
cos t
1
2
cos t
− sin t
1
2
e ln | sin t| +
+ c2
tan t
1
2 cos t
− sin t
cos t + sin t
and Φ−1 =
dt =
et ln | cos t|.
.
sin t
e−t
ln | sin t| + ln | cos t|
t
3
2t
dt =
X+
2 cos t + sin t − sec t
2 sin t − cos t
cos t
− sin t
cos t
cos t + sin t
sin t − cos t
2 sin t − cos t − ln | sec t + tan t|
−2 cos t − sin t
sin2 t − cos2 t − cos t(ln | sec t + tan t|)

1

X = 1
0
we obtain

 t 
0
e

 2t 
0X +  e 
3
te3t
1
1
0


 
 
1
1
0


  2t
  3t
Xc = c1  −1  + c2  1  e + c3  0  e .
0
0
1

1
e2t
0
e2t
0

Φ =  −1

0

0 
e3t
so that
U=
3 sin t cos t − cos2 t − 2 sin2 t + (sin t − cos t) ln | sec t + tan t|
29. From
Then
−2
−1
et
cos t
− 12 sin t
1
1
2 cos t
− sin t
cot t − tan t
t
et + c2
3
2
Xc = c1
csc t
sec t
e
1
2
cos t
X+
−1
we obtain
1
t
X =
Φ=
− 12
28. From
Then
2
2 sin t
cos t
Xc = c1
Then
1
Φ−1 F dt =


and Φ−1 = 
1
2
1 −2t
e
2
0
1


t
2e
− 12 e2t
1 −t
2e
+
t
585
1
2


− 12
1 −2t
2e
0
1 t
2e

0

0 
e−3t
− 14 e2t




 dt =  − 12 e−t + 12 t 
1 2
2t
.
10.4 Nonhomogeneous Linear Systems
and


− 14 e2t + 12 te2t


Xp = ΦU =  −et + 14 e2t + 12 te2t  .
1 2 3t
2t e
30. From

we obtain
Then



−1
0



−1  X +  t 
1
2et
−1
1
−1
3

X =  1
1
 
 
 
1
1
1
  t
  2t
  2t
Xc = c1  1  e + c2  1  e + c3  0  e .
1
0
1

et

Φ =  et
et
e2t
e2t

e2t

0 
0
e2t

and Φ−1
U=
Φ−1 F dt =
−e−2t
0



−te−t − e−t + 2t
te−t + 2




2e−t

 −2e−t  dt = 

−te−2t
and

e−t

−e−2t 
e−t
0
e−2t
so that
−e−t
 −2t
=e
1 −2t
2 te
+ 14 e−2t

 3  
 
−4
2
− 12
2


   t   t

Xp = ΦU =  −1  t +  −1  +  2  e +  2  te .
0
− 12
− 34
2

31. From
X =
we obtain
Φ=
and
−e4t
e2t
e2t
e4t
−1
3
3
−1
X+
Φ−1 =
,
4e2t
4e4t
− 12 e−4t
1 −4t
2e
1 −2t
2e
1 −2t
2e
,
−2t
0
e
+ 2t − 1
X = ΦΦ (0)X(0) + Φ
Φ F ds = Φ ·
+Φ·
1
e2t + 2t − 1
0
2
−1
−2
2
=
te2t +
e2t +
te4t +
e4t .
2
1
2
0
−1
t
32. From
X =
we obtain
Φ=
and
X = ΦΦ−1 (1)X(1) + Φ
1
t
1
1
−1
−1
−1
1
1
1+t
t
Φ−1 F ds = Φ ·
X+
−1
,
Φ
−4
3
=
1/t
1/t
−t
1
586
+Φ·
1+t
−1
ln t
0
=
,
3
1
1
t−
+
ln t.
3
4
1
10.4 Nonhomogeneous Linear Systems
33. Let I =
i1
so that
i2
I =
and
−11
3
3
−3
100 sin t
0
I+
1
3
−2t
Ic = c1
e
e−12t .
+ c2
3
−1
Then
Φ=
U=
−1
Φ
e−2t
3e−12t
3e−2t
−e−12t
F dt =
10e2t sin t
30e12t sin t
Ip = ΦU =
so that
I = c1
If I(0) =
0
0
then c1 = 2 and c2 =
Φ−1 =
,
and
1
3
−2t
e
1 2t
10 e
dt =
sin t −
76
29
276
29
sin t −
168
29
+ c2
−1
−6
 −4

(b) Φ = 
 1
3e3t
e3t
2et
et
2
3t
0
0
2e
e3t



(c) Φ−1 (t)F(t) = 


Φ−1
cos t
e−12t + Ip .
− 13
0
1 −t
3 e
0
−4t
2
3
e
−4t
e


−1
 1



.
 0
0
− 13
−2e−t
2
3
−3t
e
0
2
3
8 −t
3 e
− 13 e−3t
1 −4t
3 e






2
1 2t
3 − 3 e
1 −2t
+ 83 e−t − 2et + 13 t 

3 e
,
1 −3t
2 −t

+ 3e
−3 e
1 −4t
1
2 −5t
−3t
e
+
e
−
te
3
3
3
− 16 e2t + 23 t
− 16 e−2t − 83 e−t − 2et +
2
− 15
−5t
e
−
1 −3t
− 23 e−t
9 e
1 −4t
1 −3t
+ 27
e
12 e

0
 1 e−t

= 3

0
0


Φ−1 (t)F(t)dt = 

Xp (t) = Φ(t)
,
6
29 .

−e4t
e4t 

,
0 

cos t
34. (a) The eigenvalues are 0, 1, 3, and 4, with corresponding eigenvectors


 
 
−6
2
3
 −4 
1
1


 
 
and

,
 ,
 ,
 1
0
2
2
0
1

,
2e2t (2 sin t − cos t)
6 12t
(12 sin t − cos t)
29 e
332
29
3
1 12t
− 10
e
3 12t
10 e
3 2t
10 e
−5e2t −
 −2e2t −

Φ−1 (t)F(t)dt = 


1 2
6 t
+
1
9


,

te−3t
1 −t
1 t
− 27
e − 19 tet + 13 t2 et
5 e
3 −t
1 t
+ 27
e + 19 tet + 16 t2 et
10 e
3 2t
− 2 e + 23 t + 29
−e2t + 43 t − 19
587
− 4t −
− 83 t
59
12
− 95
36



,

10.4 Nonhomogeneous Linear Systems

−6c1 + 2c2 et + 3c3 e3t − c4 e4t
 −4c + c et + c e3t + c e4t 
1
2
3
4


Xc (t) = Φ(t)C = 
,


c1 + 2c3 e3t

2c1 + c3 e3t
X(t) = Φ(t)C + Φ(t)

Φ−1 (t)F(t)dt
−6c1 + 2c2 et + 3c3 e3t − c4 e4t
 −4c + c et + c e3t + c e4t
1
2
3
4

=

c1 + 2c3 e3t


−5e2t −
  −2e2t −
 
+
 
2c1 + c3 e3t
1 −t
1 t
− 27
e − 19 tet + 13 t2 et
5 e
3 −t
1 t
+ 27
e + 19 tet + 16 t2 et
10 e
3 2t
− 2 e + 23 t + 29
−e2t + 43 t − 19
− 4t −
− 83 t
59
12
− 95
36







 
 


−6
2
3
−1
 −4 
1
1
 1

 
 


 4t
(d) X(t) = c1 
 + c2   et + c3   e3t + c4 
e
 1
0
2
 0
2
0
1
0

−5e2t −
 −2e2t −

+

1
3
3
−2
− 83 t
59
12
− 95
36





−2
34
, P−1 F =
;
1
−14
y1 = 34 + c1 e−t ,
y2 = −7 + c2 e−2t
20 + c1 e−t + 2c2 e−2t
1
2
2
34 + c1 e−t
20
−t
−2t
=
=
c
e
e
+
c
+
1
2
−7 + c2 e−2t
53 + 3c1 e−t + 7c2 e−2t
3
7
7
53
36. λ1 = −1, λ2 = 4, K1 =
X = PY =
− 4t −
1
2
1 2
7
, K2 =
,P=
, P−1 =
3
7
3 7
−3
−1
0
34
Y =
Y+
0 −2
−14
35. λ1 = −1, λ2 = −2, K1 =
X = PY =
1 −t
1 t
− 27
e − 19 tet + 13 t2 et
5 e
3 −t
1 t
+ 27
e + 19 tet + 16 t2 et
10 e
− 32 e2t + 23 t + 29
−e2t + 43 t − 19
3
1
3 1
1 1
, K2 =
,P=
, P−1 =
5 2
−2
1
−2 1
−1 0
0
Y =
Y+
0 4
et
0
−1
;
, P−1 F =
et
3
1
y1 = c1 e−t ,
y2 = − et + c2 e4t
3
1 t
− 3 e + 3c1 e−t + c2 e4t
3
1
1
c1 e−t
1 1
−t
4t
=
=
c
e
e
+
c
−
et
1
2
3 1
−2
1
1
− 13 et + c2 e4t
− 13 et − 2c1 e−t + c2 e4t
37. λ1 = 0, λ2 = 10, K1 =
1
1
1 1
1 1
, K2 =
,P=
, P−1 =
2 1
−1
1
−1 1
0
0
t−4
Y =
Y+
0 10
t+4
y1 =
1 2
t − 4t + c1 ,
2
y2 = −
588
t−4
−1
;
, P−1 F =
t+4
1
1
41
t−
+ c2 e10t
10
100
10.5
X = PY =
1 2
1 2
41
41
10t 1
2 t − 4t + c1
2 t − 10 t − 100 + c1 + c2 e
=
41
10t
1
41
1
− 12 t2 + 39
− 10
t − 100
+ c2 e10t
10 t − 100 − c1 + c2 e
1
1
1 2
1
1 −41
41 1
10t
+ c2
e +
t +
t−
2 −1
10
100 1
−1
1
39
1
−1
= c1
38. λ1 = −1, λ2 = 1, K1 =
Matrix Exponential
1
1
1 1
1 1
, K2 =
,P=
, P−1 =
2 1
−1
1
−1 1
−1 0
2 − 4e−2t
Y =
Y+
0 1
2 + 4e−2t
−1
2 − 4e−2t
, P−1 F =
;
2 + 4e−2t
1
4
y2 = −2 − e−2t + c2 et
3
8 −2t
1 1
2 + 4e−2t + c1 e−t
+ c1 e−t + c2 et
3e
=
X = PY =
−2t
−1 1
−2 − 43 e−2t + c2 et
−4 − 16
− c1 e−t + c2 et
3 e
1
1
0
1
8
−t
t
−2t
= c1
e + c2
e +
+
e
3 −2
−1
1
−4
y1 = 2 + 4e−2t + c1 e−t ,
EXERCISES 10.5
Matrix Exponential
1. For A =
1
0
0
2
we have
A2 =
1
0
0
2
3
2
A = AA =
A4 = AA3 =
and so on. In general
Ak =
1
0
1
0
1
0
1
0
0
2k
0
1 0
=
,
2
0 4
0
1 0
1
=
2
0 4
0
0
1 0
1
=
2
0 8
0
0
8
0
16
,
,
for k = 1, 2, 3, . . . .
Thus
A2 2 A3 3
A
t+
t +
t + ···
1!
2!
3!
1 0
1 1 0
1 1 0 3
1 1 0 2
=
+
t+
t +
t + ···
1! 0 2
2! 0 4
3! 0 8
0 1


t2
t3
t
1 + t + + + ···
0
0
e


2! 3!
=
=

0 e2t
(2t)2
(2t)3
0
1+t+
+
+ ···
2!
3!
eAt = I +
589
10.5 Matrix Exponential
and
−At
e
2. For A =
0
1
1
0
=
e−t
0
0
e−2t
.
we have
2
A =
0
1
1
0
A3 = AA2 =
1
1 0
=
=I
0
0 1
1
0 1
I=
=A
0
1 0
0
1
0
1
A4 = (A2 )2 = I
A5 = AA4 = AI = A,
and so on. In general,
Ak =
A,
I,
k = 1, 3, 5, . . .
k = 2, 4, 6, . . . .
Thus
A
A2 2 A3 3
t+
t +
t + ···
1!
2!
3!
1
1
= I + At + It2 + At3 + · · ·
2!
3!
1 2
1 4
1 3
1 5
= 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
e−At =
3. For
cosh(−t)
sinh(−t)

1

A= 1
−2
1
1
−2
sinh(−t)
cosh(−t)
=
cosh t
− sinh t
− sinh t
cosh t
.

1

1
−2
we have

1
1

2
A = 1
1
−2 −2
Thus, A3 = A4 = A5 = · · · = 0 and

1 0

At
e = I + At =  0 1
0 0

1
1

1 1
−2
−2
 
0
t
 
0 +  t
1
−2t
 
1
1
0
 
1
1 = 0
−2 −2
0
t
t
−2t
590
0
0
0

0

0.
0
 

t
t+1
t
t
 

t =  t
t+1
t
.
−2t
−2t −2t −2t + 1
10.5
Matrix Exponential
4. For

0

A = 3
0
0

0

0
5
1
0
we have


 

0
0 0 0
0 0 0

 

03 0 0 = 0 0 0
0
5 1 0
3 0 0


 
0 0 0
0 0 0
0 0


 
2
3
A = AA =  3 0 0   0 0 0  =  0 0
5 1 0
3 0 0
0 0
0

2
A = 3
5
0
0
1

0

0.
0
Thus, A4 = A5 = A6 = · · · = 0 and
1
eAt = I + At + A2 t2
2

 
1 0 0
0

 
=  0 1 0  +  3t
0 0 1
5t
 
0 0
0
 
0 0 +  0
3 2
t 0
2t
0
0


0 0

1 0.
t 1

1
 
0 0 = 
3t
3 2
0 0
2 t + 5t
5. Using the result of Problem 1,
X=
et
0
0
e2t
c1
= c1
c2
et
0
+ c2
0
et
.
6. Using the result of Problem 2,
X=
7. Using the result of Problem 3,

t+1
t

X= t
t+1
−2t
−2t
cosh t
sinh t
sinh t
cosh t
c1
= c1
c2
+ 5t
t
t
c1

cosh t
sinh t
+ c2
sinh t
cosh t
.





t+1
t
t
 






t
  c2  = c1  t  + c2  t + 1  + c3 
.
−2t + 1
−2t
−2t
−2t + 1
c3
8. Using the result of Problem 4,

1

X=
3t
3 2
2t



 
 
 

0 0
1
0
0
c1

 
 
 

1 0   c2  = c1 
3t
 + c2  1  + c3  0  .
3 2
t
1
t 1
c3
2 t + 5t
9. To solve
X =
1
0
0
2
X+
591
3
−1
10.5 Matrix Exponential
we identify t0 = 0, F(t) =
3
, and use the results of Problem 1 and equation (6) in the text.
−1
At
X(t) = e
t
e−As F(s) ds
At
C+e
t0
t
t −s
0
0
c1
e
e
=
+
2t
2t
0 e
0 e
0
c2
0
t
t −s c1 et
0
e
3e
=
+
ds
2t
2t
0 e
c2 e
−e−2s
0
=
et
c1 et
c2 e2t
=
c1 et
+
c2 e2t
=
+
c1 et
c2 e2t
et
0
0
e2t
et
0
e2t
0
+
−3 + 3et
1
2
−3e−s
1 −2s
2e
− 12 e2t
e−2s
3
−1
ds
t
0
−3e−t + 3
1 −2t
2e
= c3
0
− 12
1
0
et + c4
0
1
e2t +
−3
1
2
.
10. To solve
X =
we identify t0 = 0, F(t) =
1
0
0
2
X+
t
e4t
t
, and use the results of Problem 1 and equation (6) in the text.
e4t
t
X(t) = eAt C + eAt
e−As F(s) ds
t0
t −s
e
s
0
0
ds
e4s
0 e2t
0
e−2s
c2
0
t
t −s e
se
0
=
+
ds
2t
2t
c2 e
0 e
e2s
0
t
c1 et
e
−se−s − e−s t
0
=
+
1 2s
c2 e2t
0 e2t
0
2e
t
t −t
−t
e
−te − e + 1
0
c1 e
+
=
1 2t
1
0 e2t
c2 e2t
2e − 2
c1 et
−t − 1
−t − 1 + et
1
0
t
2t
=
+ 1 4t 1 2t = c3
e + c4
e +
.
1 4t
c2 e2t
0
1
2e − 2e
2e
=
et
0
0 e2t
c1 et
c1
+
et
11. To solve
X =
0
1
1
0
592
X+
1
1
10.5
Matrix Exponential
1
, and use the results of Problem 2 and equation (6) in the text.
1
t
At
At
X(t) = e C + e
e−As F(s) ds
we identify t0 = 0, F(t) =
t0
t
sinh t
cosh s − sinh s
1
=
+
ds
cosh t
−
sinh
s
cosh
s
1
c2
0
t
c1 cosh t + c2 sinh t
cosh s − sinh s
cosh t sinh t
=
ds
+
c1 sinh t + c2 cosh t
− sinh s + cosh s
sinh t cosh t
0
c1 cosh t + c2 sinh t
cosh t sinh t
sinh s − cosh s t
=
+
c1 sinh t + c2 cosh t
sinh t cosh t
− cosh s + sinh s 0
c1 cosh t + c2 sinh t
cosh t sinh t
sinh t − cosh t + 1
=
+
c1 sinh t + c2 cosh t
sinh t cosh t
− cosh t + sinh t + 1
c1 cosh t + c2 sinh t
sinh2 t − cosh2 t + cosh t + 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
= c1
+ c2
+
+
−
sinh t
cosh t
sinh t
cosh t
1
cosh t
sinh t
1
= c3
+ c4
−
.
sinh t
cosh t
1
cosh t
sinh t
sinh t
cosh t
c1
12. To solve
X =
cosh t
sinh t
0
1
1
0
X+
cosh t
sinh t
cosh t
, and use the results of Problem 2 and equation (6) in the text.
sinh t
t
At
At
X(t) = e C + e
e−As F(s) ds
we identify t0 = 0, F(t) =
=
t0
cosh t
sinh t
sinh t
cosh t
c1
+
cosh t
sinh t
t
cosh s
− sinh s
cosh s
sinh t cosh t
− sinh s
cosh s
sinh s
c2
0
t c1 cosh t + c2 sinh t
1
cosh t sinh t
=
ds
+
0
c1 sinh t + c2 cosh t
sinh t cosh t
0
t
c1 cosh t + c2 sinh t
cosh t sinh t
s =
+
c1 sinh t + c2 cosh t
sinh t cosh t
0 0
c1 cosh t + c2 sinh t
cosh t sinh t
t
=
+
c1 sinh t + c2 cosh t
sinh t cosh t
0
c1 cosh t + c2 sinh t
cosh t
sinh t
cosh t
t cosh t
=
+ c2
+t
.
+
= c1
c1 sinh t + c2 cosh t
sinh t
cosh t
sinh t
t sinh t
13. We have
 
 
    

1
0
0
c1
1
 
 
    

X(0) = c1  0  + c2  1  + c3  0  =  c2  =  −4  .
0
0
1
6
c3
593
ds
10.5 Matrix Exponential
Thus, the solution of the initial-value problem is






t+1
t
t






X =  t  − 4t + 1 + 6
t
.
−2t
−2t
−2t + 1
14. We have
X(0) = c3
Thus, c3 = 7 and c4 =
5
2
1
0
−3
c3 − 3
4
+ c4
+
=
=
.
1
1
0
1
3
c
+
4
2
2
, so
X=7
15. From sI − A =
s−4
4
−3
s+4
1
−3
5 0
et +
.
e2t +
1
2 1
0
2
we find
(sI − A)−1
and
 3/2
1/2
−
s−2 s+2
=
−1
1
+
s−2 s+2
3
At
e
2t
2e
=
− 12 e−2t
−e2t + e−2t
3/4
3/4 
−
s−2 s+2 

−1/2
3/2
+
s−2
s+2
3 2t
4e
− 34 e−2t
− 12 e2t + 32 e−2t
.
The general solution of the system is then
3
At
X=e
16. From sI − A =
2t
2e
C=
− 12 e−2t
3 2t
4e
− 34 e−2t
c1
c2
−e2t + e−2t − 12 e2t + 32 e−2t
3
1
3
3
−2
−4
2t
−2t
2t
2
4
= c1
e + c2
e−2t
+ c2
e + c1
e
3
1
−1
1
−2
2
1
1
3
1
1 3 e2t + − c1 − c2
e−2t
=
c1 + c2
2
4
2
4
−2
−2
3
1
2t
e + c4
e−2t .
= c3
−2
−2
s−4
2
−1 s − 1
we find

(sI − A)−1
2
1
−
s−3 s−2
=
1
1
−
s−3 s−2
and
eAt =
2e3t − e2t
e3t − e2t
The general solution of the system is then
594
−
2
2 
+
s−3 s−2

−1
2
+
s−3 s−2
−2e3t + 2e2t
−e3t + 2e2t
.
10.5
X = eAt C =
2e3t − e2t
−2e3t + 2e2t
c1
Matrix Exponential
e3t − e2t
−e3t + 2e2t
c2
2
−1
−2
2
3t
2t
3t
= c1
e + c1
e + c2
e + c2
e2t
1
−1
−1
2
2
1
= (c1 − c2 )
e3t + (−c1 + 2c2 )
e2t
1
1
2
1
3t
e + c4
e2t .
= c3
1
1
17. From sI − A =
s−5
9
−1 s + 1
we find

(sI − A)−1
3
1
+
s
−
2
(s
−
2)2

=
1
(s − 2)2
and
eAt =
e2t + 3te2t
te2t

9
(s − 2)2


1
3
−
s − 2 (s − 2)2
−
−9te2t
.
e2t − 3te2t
The general solution of the system is then
c1
−9te2t
e2t + 3te2t
te2t
e2t − 3te2t
c2
1
3
0
−9
2t
2t
2t
= c1
e + c1
te + c2
e + c2
te2t
0
1
1
−3
1 + 3t
−9t
e2t + c2
e2t .
= c1
t
1 − 3t
X = eAt C =
18. From sI − A =
s −1
2 s+2
we find
(sI − A)−1
 s+1+1
 (s + 1)2 + 1
=

−2
(s + 1)2 + 1
and
eAt =
e−t cos t + e−t sin t
−2e−t sin t

1
2
(s + 1) + 1 

s+1−1 
(s + 1)2 + 1
e−t sin t
−t
e cos t − e−t sin t
.
The general solution of the system is then
X = eAt C =
e−t cos t + e−t sin t
−2e−t sin t
e−t sin t
e−t cos t − e−t sin t
c1
c2
1
1
0
1
−t
−t
−t
= c1
e cos t + c1
e sin t + c2
e cos t + c2
e−t sin t
0
−2
1
−1
595
10.5 Matrix Exponential
= c1
cos t + sin t
−2 sin t
−t
e
+ c2
sin t
cos t − sin t
e−t .
19. The eigenvalues are λ1 = 1 and λ2 = 6. This leads to the system
et = b0 + b1
e6t = b0 + 6b1 ,
which has the solution b0 = 65 et − 15 e6t and b1 = − 15 et + 15 e6t . Then
4
At
e
= b0 I + b1 A =
t
5e
+ 15 e6t
2 t
5e
− 25 e6t
2 t
5e
− 25 e6t
1 t
5e
+ 45 e6t
.
The general solution of the system is then
4
At
X=e
t
5e
C=
2 t
5e
+ 15 e6t
− 25 e6t
c1
c2
− 25 e6t 15 et + 45 e6t
4
1
2
2
−5
6t
t
5
5
= c1 52 et + c1
e
e
e6t
+
c
+
c
2
2
1
4
2
−
5
5
5
5
2
2
1
1
1
2
et +
e6t
=
c1 + c2
c1 − c2
5
5
5
5
1
−2
2
1
et + c4
e6t .
= c3
1
−2
2 t
5e
20. The eigenvalues are λ1 = 2 and λ2 = 3. This leads to the system
e2t = b0 + 2b1
e3t = b0 + 3b1 ,
which has the solution b0 = 3e2t − 2e3t and b1 = −e2t + e3t . Then
At
e
= b0 I + b1 A =
2e2t − e3t
e2t − e3t
−2e2t + 2e3t
−e2t + 2e3t
.
The general solution of the system is then
At
X=e
−2e2t + 2e3t
c1
C=
2t
3t
2t
3t
e −e
−e + 2e
c2
2
−1
−2
2
= c1
e2t + c1
e3t + c2
e2t + c2
e3t
1
−1
−1
2
2
1
2t
= (c1 − c2 )
e + (−c1 + 2c2 )
e3t
1
1
2
1
e2t + c4
e3t .
= c3
1
1
2e2t − e3t
596
10.5
Matrix Exponential
21. The eigenvalues are λ1 = −1 and λ2 = 3. This leads to the system
e−t = b0 − b1
e3t = b0 + 3b1 ,
which has the solution b0 = 34 e−t + 14 e3t and b1 = − 14 e−t + 14 e3t . Then
eAt = b0 I + b1 A =
e3t
−2e−t + 2e3t
0
e−t
.
The general solution of the system is then
At
X=e
1
4
22. The eigenvalues are λ1 =
−2e−t + 2e3t
e3t
C=
c1
c2
0
e−t
1
−2
2
= c1
e3t + c2
e−t + c2
e3t
0
1
0
−2
1
−t
e + (c1 + 2c2 )
= c2
e3t
1
0
−2
1
e−t + c4
e3t .
= c3
1
0
and λ2 =
1
2
. This leads to the system
1
et/4 = b0 + b1
4
1
et/2 = b0 + b1 ,
2
which has the solution b0 = 2et/4 + et/2 and b1 = −4et/4 + 4et/2 . Then
eAt = b0 I + b1 A =
−2et/4 + 3et/2
6et/4 − 6et/2
−et/4 + et/2
3et/4 − 2et/2
.
The general solution of the system is then
At
X=e
C=
−2et/4 + 3et/2
6et/4 − 6et/2
c1
−et/4 + et/2
3et/4 − 2et/2
c2
−2
3
6
−6
= c1
et/4 + c1
et/2 + c2
et/4 + c2
et/2
−1
1
3
−2
2
3
t/4
= (−c1 + 3c2 )
e + (c1 − 2c2 )
et/2
1
1
2
3
et/4 + c4
et/2 .
= c3
1
1
23. From equation (3) in the text we have eDt = I + tD +
PeDt P−1 = PP−1 + t(PDP−1 ) +
t2 2 t 3 3
D + D + · · · so that
2!
3!
t2
t3
(PD2 P−1 ) + (PD3 P−1 ) + · · · .
2!
3!
597
10.5 Matrix Exponential
But PP−1 + I, PDP−1 = A and PDn P−1 = An (see Problem 37, Exercises 8.12). Thus,
PeDt P−1 = I + tA +
24. From equation

1
0

eDt =  ..
.
0
(3) in the text
 λ
0 ··· 0
1


0
1 ··· 0 
.. . .
.+ .
. ..   ..
.
0
··· 1
0
0
λ2
..
.
0
···
···
..
.
0
0
..
.
t2 2 t3 3
A + A + · · · = eAt .
2!
3!


λ21
 1 
 0

 + t2 
.
 2! 
 ..
· · · λn
0
0
λ22
..
.
0

0
0 

.. 

. 
· · · λ2n

···
···
..
.

1 3
+ t 
3! 




=





=

1 + λ1 t +
1
2
2! (λ1 t)
+ ···
0
..
.
0
eλ1 t
0
0
..
.
eλ2 t
..
.
0
0
0
1 + λ2 t +
1
2
2! (λ2 t)
+ ···
..
.
0
···
···
..
.
0
0
..
.
···
0
···
..
.
0
..
.
· · · 1 + λn t +
λ31
0
..
.
0
1
2
2! (λn t)
0
λ32
..
.
0
···





+ ···




· · · eλn t
25. From Problems 23 and 24 and equation (1) in the text
3t
3 −3t
3t
− 12 e−3t
c1
e5t
0
e
e
2e
At
Dt −1
X = e C = Pe P C =
3t
5t
5t
1 −5t
0 e
e
3e
c2
− 12 e−5t
2e
3 3t 1 5t
− 12 e3t + 12 e5t
c1
2e − 2e
=
.
3 3t
3 5t
1 3t
3 5t
c2
−2e + 2e
2e − 2e
26. From Problems 23 and 24 and equation (1) in the text
t
1 −t 1 −t t
−2e
c1
e
−e e3t
0
2e
At
Dt −1
X = e C = Pe P C =
t
3t
3t
1
1
3t
−3t
e e
0 e
c2
2e
2e
1 t 1 9t
− 12 et + 12 e3t
c1
2e + 2e
=
.
1 t
1 t
1 9t
1 3t
c2
−2e + 2e
2e + 2e
A={{4, 2},{3, 3}};
c={c1, c2};
m=MatrixExp[A t];
sol=Expand[m.c]
Collect[sol, {c1, c2}]//MatrixForm
598

0 

.. 
 + ···
. 
· · · λ3n

···
..
.

27. (a) The following commands can be used in Mathematica:
0
10.5
The output gives
2 t 3 6t
e + e
5
5
2
x(t) = c1
+ c2 − et +
5
3
3 t
3
y(t) = c1 − et + e6t + c2
e +
5
5
5
2 6t
e
5
Matrix Exponential
2 6t
.
e
5
The eigenvalues are 1 and 6 with corresponding eigenvectors
−2
1
and
,
3
1
so the solution of the system is
X(t) = b1
or
−2
3
1
e + b2
e6t
1
t
x(t) = −2b1 et + b2 e6t
y(t) = 3b1 et + b2 e6t .
If we replace b1 with − 15 c1 + 15 c2 and b2 with
exponential.
3
5 c1
+ 25 c2 , we obtain the solution found using the matrix
(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 )
29. 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.
30. Since A3 = 0, A is nilpotent. Since
t2
tk
+ · · · + Ak
+ ···,
2!
k!
= 0, then Ak = 0 for k ≥ m and
eAt = I + At + A2
if A is nilpotent and Am
eAt = I + At + A2
In this problem A3 = 0, so
eAt
t2
tm−1
+ · · · + Am−1
.
2!
(m − 1)!

 


1 0 0
−1 1 1
−1
t

 


= I + At + A2
=  0 1 0  +  −1 0 1  t +  0
2
0 0 1
−1 1 1
−1


1 − t − t2 /2 t
t + t2 /2


=
−t
1
t

2
−t − t2 /2
and the solution of X = AX is

c1


  
X(t) = eAt C = eAt  c2  = 
c3
t
0
0
0

1
 t2
0
2
1
1 + t + t2 /2
c1 (1 − t − t2 /2) + c2 t + c3 (t + t2 /2)
−c1 t + c2 + c3 t
c1 (−t − t2 /2) + c2 t + c3 (1 + t + t2 /2)
599


.
CHAPTER
10 REVIEW
EXERCISES
10.5 Matrix
Exponential
CHAPTER 10 REVIEW EXERCISES
1. If X = k
4
, then X = 0 and
5
1
4
4
8
24
8
0
k
−
=k
−
=
.
2 −1
5
1
3
1
0
We see that k =
1
3
.
2. Solving for c1 and c2 we find c1 = − 34 and c2 =
3. Since

4

 1
−1
6
3
−4
1
4
.
 



3
12
3

 



2 1 =  4 = 4 1,
−3
−1
−4
−1
6

we see that λ = 4 is an eigenvalue with eigenvector K3 . The corresponding solution is X3 = K3 e4t .
1
4. The other eigenvalue is λ2 = 1 − 2i with corresponding eigenvector K2 =
. The general solution is
−i
cos 2t
sin 2t
et + c2
et .
X(t) = c1
− sin 2t
cos 2t
1
0
5. We have det(A − λI) = (λ − 1)2 = 0 and K =
. A solution to (A − λI)P = K is P =
so that
−1
1
1
1
0
X = c1
et + c2
tet +
et .
−1
−1
1
6. We have det(A − λI) = (λ + 6)(λ + 2) = 0 so that
1
1
X = c1
e−6t + c2
e−2t .
−1
1
1
and
i
cos 2t
sin 2t
=
et + i
et .
− sin 2t
cos 2t
7. We have det(A − λI) = λ2 − 2λ + 5 = 0. For λ = 1 + 2i we obtain K1 =
X1 =
1
e(1+2i)t
i
Then
X = c1
cos 2t
− sin 2t
t
e + c2
sin 2t
cos 2t
et .
8. We have det(A − λI) = λ − 2λ + 2 = 0. For λ = 1 + i we obtain K1 =
2
X1 =
3−i
2
e(1+i)t =
3 cos t + sin t
2 cos t
600
et + i
3−i
2
and
− cos t + 3 sin t
2 sin t
et .
CHAPTER 10 REVIEW EXERCISES
Then
X = c1
3 cos t + sin t
2 cos t
− cos t + 3 sin t
2 sin t
et + c2
et .
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
√ 


 √ 

K1 =  5  , K2 =  12 2 i  , and K3 =  − 12 2 i  .
4
1
1
Thus

 



1
0
√
√ 


 
 √ 
X = c1  5  e−2t + c2  0  cos 2t −  12 2  sin 2t et
4
1
0

 


1
0
√
√
 

 √ 
+ c3  12 2  cos 2t +  0  sin 2t et
0
1
√
√






cos 2t
sin 2t
−7
√
√
√
√






= c1  5  e−2t + c2  − 12 2 sin 2t  et + c3  12 2 cos 2t  et .
√
√
4
cos 2t
sin 2t
−7

11. We have
Xc = c1
Then
Φ=
and
U=
e2t
0
Φ−1 F dt =
4e4t
e4t
1
0
2t
e + c2
,
−1
Φ
12. We have
Xc = c1
Φ=
and
U=
2 cos t
− sin t
Φ−1 F dt =
=
2 cos t
− sin t
2 sin t
cos t
e−2t
0
t
e + c2
e,
−1
Φ
cos t − sec t
sin t
=
601
2
1
2
dt =
,
15e−2t + 32te−2t
−e−4t − 4te−4t
,
.
2 sin t
cos t
1
t
−4e−2t
e−4t
dt =
11 + 16t
−1 − 4t
e4t .
16te−4t
Xp = ΦU =
1
2e−2t − 64te−2t
so that
Then
4
et .
cos t
− sin t
sin t
cos t
e−t ,
sin t − ln | sec t + tan t|
− cos t
,
CHAPTER 10 REVIEW EXERCISES
so that
Xp = ΦU =
13. We have
Xc = c1
Then
Φ=
−2 cos t ln | sec t + tan t|
−1 + sin t ln | sec t + tan t|
cos t + sin t
sin t − cos t
2 cos t
2 sin t
and
−1
U=
Φ
=
so that
Xp = ΦU =
14. We have
1
Xc = c1
Then
Φ=
−1
−1
−1
te2t + e2t
−e2t
−te2t
U=
−1
Φ
Φ
1
2 sin t +
1
2 sin t +
+
sin t
sin t + cos t
F dt =
Xp = ΦU =
− 12
1
2
15. (a) Letting
−te−2t
dt =

k1
sin t
sin t
,
dt
,
1
e2t .
0
−te−2t − e−2t
e−2t
1
t2 e2t +
cos t +
1
2
1
2
ln | csc t − cot t|.
te2t +
cos t −
1
2 csc t
1
2 csc t
e−2t
t−1
−1
.
−1
so that
1
Φ−1 =
,
− cos t
1
2 cos t +
1
2 cos t +
1
2
1
2
sin t
1
2 ln | csc t − cot t|
1
2 ln | csc t − cot t|
e2t + c2
=
1
2 sin t −
− 12 sin t −
et .
sin t − cos t
2 sin t
−1
e2t
and
+ c2
,
F dt =
cos t −
− 12 cos t −
1
2
cos t + sin t
2 cos t
−2
1
−t
−t
2t
2
,
,
te2t .

 
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
respectively. Thus,


−1


X1 =  0  e2t
1
0


−1


and X2 =  1  e2t
0
602
CHAPTER 10 REVIEW EXERCISES
are two solutions.
(b) From det(A − λI) = λ2 (3 − λ) = 0 we see that λ1 = 3, and 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 the eigenvector corresponding to λ1 = 3 is
 
1
 
K1 =  1  ,
1
the general solution of the system is
 




1
−1
−1
 




X = c1  1  e3t + c2  0  + c3  1  .
1
1
0
16. For X =
c1
c2
et we have X = X = IX.
603
11
Systems of Nonlinear
Differential Equations
EXERCISES 11.1
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.
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,
critical points are (0, 0), ( 1/ , 0) and (− 1/ , 0).
1/ , − 1/ . The
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.
604
11.1 Autonomous Systems
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, . . . .
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 10.2, Problem 1, x = c1 e5t − c2 e−t and y = 2c1 e5t + c2 e−t .
(b) From X(0) = (2, −1) it follows that c1 = 0 and c2 = 2. Therefore x = −2e−t and y = 2e−t .
(c)
18. (a) From Exercises 10.2, 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.
605
11.1 Autonomous Systems
(c)
19. (a) From Exercises 10.2, Problem 37, 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 one periodic with p = 2π/3.
(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.
(c)
20. (a) From Exercises 10.2, Problem 34, 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. Therefore x = −2 cos t and y = 2 cos t + 2 sin t.
(c)
21. (a) From Exercises 10.2, Problem 35, 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 .
606
11.1 Autonomous Systems
(c)
22. (a) From Exercises 10.2, Problem 38, 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.
(c)
23. Switching to polar coordinates,
dx
dy
1
x
+y
= (−xy − x2 r4 + xy − y 2 r4 ) = −r5
dt
dt
r
dθ
1
dy
1
dx
= 2 −y
+x
= 2 (y 2 + xyr4 + x2 − xyr4 ) = 1 .
dt
r
dt
dt
r
dr
1
=
dt
r
If we use separation of variables on
dr
= −r5 we obtain
dt
1/4
1
r=
and θ = t + c2 .
4t + c1
Since X(0) = (4, 0), r = 4 and θ = 0 when t = 0. It follows that c2 = 0 and c1 =
1
256
. The final solution can be
written as
4
,
1024t + 1
and so the solution spirals toward the origin as t increases.
r= √
4
θ=t
24. Switching to polar coordinates,
dr
1
dx
dy
1
=
x
+y
= (xy − x2 r2 − xy + y 2 r2 ) = r3
dt
r
dt
dt
r
dθ
1
dy
1
dx
= 2 −y
+x
= 2 (−y 2 − xyr2 − x2 + xyr2 ) = −1 .
dt
r
dt
dt
r
607
11.1 Autonomous Systems
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 =
written as
4
r= √
,
θ = −t.
1 − 32t
1
1
Note that r → ∞ as t → 32
, the curve is not a spiral.
. Because 0 ≤ t ≤ 32
1
16 .
The final solution can be
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
dθ
1
dy
1
dx
= 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
r= 1
1−
3
4
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
dθ
dx
xy
xy
1
dy
1
= 2 −y
+x
= 2 −y 2 +
(4 − r2 ) − x2 −
(4 − r2 ) = −1.
dt
r
dt
dt
r
r
r
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.
608
11.2 Stability of Linear Systems
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.
31. If X(t) = (x(t), y(t)) is a solution,
d
∂f dx ∂f dy
f (x(t), y(t)) =
+
= QP − P Q = 0,
dt
∂x dt
∂y dt
using the Chain Rule. Therefore f (x(t), y(t)) = c for some constant c, and the solution lies on a level curve of
the function f .
EXERCISES 11.2
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.
(b)
609
11.2 Stability of Linear Systems
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.
(b)
3. (a) All solutions are unstable spirals which become unbounded as t increases.
(b)
4. (a) All solutions are spirals which approach the origin.
(b)
610
11.2 Stability of Linear Systems
5. (a) All solutions approach (0, 0) from the direction specified by the line y = x.
(b)
6. (a) All solutions become unbounded and y = x/2 serves as the asymptote.
(b)
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.
(b)
611
11.2 Stability of Linear Systems
8. (a) The solutions are ellipses which encircle the origin.
(b)
9. Since ∆ = −41 < 0, we can conclude from Figure 11.18 that (0, 0) is a saddle point.
10. Since ∆ = 29 and τ = −12, τ 2 − 4∆ > 0 and so from Figure 11.18, (0, 0) is a stable node.
11. Since ∆ = −19 < 0, we can conclude from Figure 11.18 that (0, 0) is a saddle point.
12. Since ∆ = 1 and τ = −1, τ 2 − 4∆ = −3 and so from Figure 11.18, (0, 0) is a stable spiral point.
13. Since ∆ = 1 and τ = −2, τ 2 − 4∆ = 0 and so from Figure 11.18, (0, 0) is a degenerate stable node.
14. Since ∆ = 1 and τ = 2, τ 2 − 4∆ = 0 and so from Figure 11.18, (0, 0) is a degenerate unstable node.
15. Since ∆ = 0.01 and τ = −0.03, τ 2 − 4∆ < 0 and so from Figure 11.18, (0, 0) is a stable spiral point.
16. Since ∆ = 0.0016 and τ = 0.08, τ 2 − 4∆ = 0 and so from Figure 11.18, (0, 0) is a degenerate unstable node.
17. ∆ = 1 − µ2 , τ = 0, and so we need ∆ = 1 − µ2 > 0 for (0, 0) to be a center. Therefore |µ| < 1.
18. Note that ∆ = 1 and τ = µ. Therefore we need both τ = µ < 0 and τ 2 − 4∆ = 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 ∆ = µ + 1 and τ = µ + 1 and so τ 2 − 4∆ = (µ + 1)2 − 4(µ + 1) = (µ + 1)(µ − 3). It follows that
τ 2 − 4∆ < 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 − 4∆ < 0. This condition reduces to
−1 < µ < 3.
20. τ = 2α, ∆ = α2 + β 2 > 0, and τ 2 − 4∆ = −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 8.6 that X(t) = Xc (t) + X1 is the general solution to X = AX + F. If τ < 0 and ∆ > 0
then Xc (t) approaches (0, 0) by Theorem 11.1(a). It follows that X(t) approaches X1 as t → ∞.
22. If bc < 1, ∆ = adx̂ŷ(1 − bc) > 0 and τ 2 − 4∆ = (ax̂ − dŷ)2 + 4abcdx̂ŷ > 0. Therefore (0, 0) is a stable node.
612
11.2 Stability of Linear Systems
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 ∆ = −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, ∆ = 64, and τ 2 − 4∆ = 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, ∆ = 0.03, and τ 2 − 4∆ = −0.08, (0, 0) is an unstable spiral point.
26. (a) The critical point is X1 = (1, 1).
(b) From the graph, X1 appears to be a center.
(c) Since τ = 0 and ∆ = 1, (0, 0) is a center.
613
11.2 Linearization
Stability of Linear
Systems
and Local
Stability
11.3
EXERCISES 11.3
Linearization and Local Stability
1. Switching to polar coordinates,
dr
1
1
dx
dy
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 2.8, (4) and (5).] It
follows that
5
r=
and θ = −t + c2 .
1 + c1 e−5αt
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
11.2.
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 11.2.
5. The only critical point is T = T0 . Since g (T ) = k, g (T0 ) = k > 0. Therefore T = T0 is unstable by
Theorem 11.2.
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 11.2.
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),
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
2y
614
−2x
2x − 1
.
11.3 Linearization and Local Stability
Computing g ((1/2, 1)) we find that τ = −2 and ∆ = 2 so that τ 2 − 4∆ = −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, ∆ = 4, and so τ 2 − 4∆ = 0. We can conclude that (1, 0) is unstable but we are unable to
classify this critical point any further. At X = (−1, 0), ∆ = −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
g (X) =
2y 2x − 1
√
√
to classify these three critical points. For X = ( 2 , 0) or (− 2 , 0), τ = −1 and ∆ < 0. Therefore both critical
points are saddle points. For X = (1/2, −7/4), τ = −1, ∆ = 7/2 and so τ 2 − 4∆ = −13 < 0. Therefore
(1/2, −7/4) is a stable spiral point.
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
2 −2y
g (X) =
y x−1
√
to classify these critical points. For X = (0, 0), ∆ = −2 < 0 and so (0, 0) is a saddle point. For either (1, 2 )
√
√
√
or (1, − 2 ), τ = 2, ∆ = 4, and so τ 2 − 4∆ = −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
−3
2y
g (X) =
2x −2y
to classify these four critical points. For X = (1, 1), τ = −5, ∆ = 2, and so τ 2 − 4∆ = 17 > 0. Therefore (1, 1) is
a stable node. For X = (1, −1), ∆ = −2 < 0 and so (1, −1) is a saddle point. For X = (2, 2), ∆ = −4 < 0 and
so we have another saddle point. Finally, if X = (2, −2), τ = 1, ∆ = 4, and so τ 2 − 4∆ = −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
y
x−3
g (X) =
−2x
2y
to classify these two critical points. For X = (4, 4), τ = 12, ∆ = 40, and so τ 2 − 4∆ < 0. Therefore (4, 4) is
an unstable spiral point. For X = (−1, 1), τ = −3, ∆ = 10, and so x2 − 4∆ < 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
615
11.3 Linearization and Local Stability
to classify these three critical points. For X = (0, 0), τ = 1 and ∆ = 0 and so the test is inconclusive. For
X = (0, 1), τ = −4, ∆ = 4 and so τ 2 − 4∆ = 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), ∆ = −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 11.1. The
Jacobian is
−6xy
1 − 3x2 − 3y 2
g (X) =
.
−2xy
3 − x2 − 9y 2
For X = (0, 0), τ = 4, ∆ = 3 and so τ 2 − 4∆ = 4 > 0. Therefore (0, 0) is an unstable node. Both (0, 1) and
(0, −1) give τ = −8, ∆ = 12, and τ 2 − 4∆ = 16 > 0. These two critical points are therefore stable nodes. For
X = (1, 0) or (−1, 0), ∆ = −4 < 0 and so saddle points occur.
19. We found the critical points (0, 0), (10, 0), (0, 16) and (4, 12) in Problem 11, Section 11.1. Since the Jacobian is
10 − 2x − 12 y
− 12 x
g (X) =
−y
16 − 2y − x
we can classify the critical points as follows:
τ
∆
τ 2 − 4∆
(0, 0)
26
160
36
unstable node
(10, 0)
−4
−60
–
saddle point
(0, 16)
−14
−32
–
saddle point
(4, 12)
−16
24
160
stable node
X
Conclusion
20. We found the sole critical point (10, 10) in Problem 12, Section 11.1. The Jacobian is


−2
1
,
g (X) = 
15
2 −1 −
(y + 5)2
g ((10, 10)) has trace τ = −46/15, ∆ = 2/15, and τ 2 − 4∆ > 0. Therefore (0, 0) is a stable node.
21. The corresponding plane autonomous system is
1
y = (cos θ − ) sin θ.
2
θ = y,
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) =
cos 2θ − 12 cos θ
0
and so at (0, 0), τ = 0 and ∆ = −1/2. Therefore (0, 0) is a saddle point. For X = (±π/3, 0), τ = 0 and ∆ = 3/4.
It is not possible to classify either critical point in this borderline case.
22. The corresponding plane autonomous system is
x = y,
y = −x +
1
2
− 3y y − x2 .
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
0
1
g (X) =
−1 − 2x 12 − 9y 2
616
11.3 Linearization and Local Stability
to classify these critical points. For X = (0, 0), τ = 1/2, ∆ = 1, and τ 2 − 4∆ < 0. Therefore (0, 0) is an unstable
spiral point. For X = (−1, 0), τ = 1/2, ∆ = −1 and so (−1, 0) is a saddle point.
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
1
g (X) =
,
2x + 3x2 y x3 − 1
τ = −1 and ∆ = 0, and we are unable to classify the critical point in this borderline case.
24. The corresponding plane autonomous system is
x = y,
y = −
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
(1 + x2 )2
τ = −2, ∆ = 4, τ 2 − 4∆ = −12, and so (0, 0) is a stable spiral point.
25. In Problem 5, Section 11.1, we showed that (0, 0), ( 1/ , 0) and (− 1/ , 0) are the critical points. We will
use the Jacobian matrix
0
1
g (X) =
−1 + 3x2 0
to classify these three critical points. For X = (0, 0), τ = 0 and ∆ = 1 and we are unable to classify this critical
point. For (± 1/ , 0), τ = 0 and ∆ = −2 and so both of these critical points are saddle points.
26. In Problem 6, Section 11.1, 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
2|x| − 1
1
0
.
For X = (0, 0), τ = 0, ∆ = 1 and we are unable to classify this critical point. For (±1/, 0), τ = 0, ∆ = −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 ∆ = β. Since β < 0, we can conclude that (0, 0) is a saddle point.
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
617
11.3 Linearization and Local Stability
τ = −α − β +
1
1 − αβ
= −β +
< 0 and ∆ = αβ − 1 > 0. Therefore (0, 1/β) is a stable critical point.
β
β
29. (a) The graphs of −x + y − x3 = 0 and −x − y + y 2 = 0 are shown in the
figure. The Jacobian matrix is
1
−1 − 3x2
g (X) =
.
−1
−1 + 2y
For X = (0, 0), τ = −2, ∆ = 2, τ 2 − 4∆ = −4, and so (0, 0) is a stable
spiral point.
(b) For X1 , ∆ = −6.07 < 0 and so a saddle point occurs at X1 .
30. (a) The corresponding plane autonomous system is
x = y,
1
y = y − y3 − x
3
and so the only critical point is (0, 0). Since the Jacobian matrix is
0
1
g (X) =
,
−1 (1 − y 2 )
τ = , ∆ = 1, and so τ 2 − 4∆ = 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 − 4∆ = 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.
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
y2
x3 − 3x2 + 2
(x − 1)(x2 − 2x − 2)
=
=
.
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
2y 2x
g (X) =
−2x 2y
and so τ = 0 and ∆ = 4 at either X = (1, 0) or (−1, 0). We obtain no information about these critical
points in this borderline case.
618
11.3 Linearization and Local Stability
(b) The differential equation is
dy
1 − x2 + y 2
y
= =
dx
x
2xy
or
dy
= 1 − x2 + y 2 .
dx
Letting µ = y 2 /x, it follows that dµ/dx = (1/x2 ) − 1 and so
µ = −(1/x) − x + 2c. Therefore y 2 /x = −(1/x) − x + 2c which
2xy
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.
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 differential 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
y2 =
1 4
(x2 − 1)2 + 2v02 − 1
x − x2 + v02 =
.
2
2
The x-intercepts on this graph satisfy
x2 = 1 ±
1 − 2v02
√
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 ≤
1
4
1±
√
1 − 4
.
2
for real solutions to exist. We will use the Jacobian matrix
0
1
g (X) =
2x − 1 0
√
√
to attempt to classify ((1 ± 1 − 4 )/2, 0) when ≤ 1/4. Note that τ = 0 and ∆ = ∓ 1 − 4 .
√
√
For X = ((1 + 1 − 4 )/2, 0) and < 1/4, ∆ < 0 and so a saddle point occurs. For X = ((1 − 1 − 4 )/2, 0),
∆ ≥ 0 and we are not able to classify this critical point using linearization.
37. The corresponding plane autonomous system is
x = y,
y = −
α
R
β
x − x3 − y
L
L
L
619
11.3 Linearization and Local Stability
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 ∆ = 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 ∆ = −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
dy/dt
x3 + x
=
=−
.
dx
dx/dt
y
Separating variables and integrating we obtain
y dy = − (x3 + x) dx
and
1 2
1
1
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
(x20 + x2 )(x20 − x2 ) + 2(x20 − x2 )
(x2 + 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) =
and so
A1 = g = ((π/6, 0)) =
0
1
√
− 3/2 0
0
1
− cos x 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
dy
1/2 − sin x
=
.
dx
y
Separating variables and integrating we obtain
1
y dy =
− sin x dx
2
and
620
11.3 Linearization and Local Stability
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
dy
2x3 y − y 4
= 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
1
2u3 − 1
dx +
du = 0
x
u(u3 + 1)
1
1
1
2u − 1
dx +
− + 2
du = 0.
x
u+1 u u −u+1
Integrating gives
ln |x| + ln |u + 1| − ln |u| + ln |u2 − u + 1| = c1
or
u+1
x
(u2 − u + 1) = c2
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.
621
11.4
as Stability
Mathematical Models
11.3 Autonomous
LinearizationSystems
and Local
EXERCISES 11.4
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
2g
1
l 2
y2 =
cos x − +
w0
l
2 2g
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 ,
dθ
2g cos θ − cos θ0 .
=−
dt
l
Therefore
dt
l
1
√
=−
dθ
2g cos θ − cos θ0
and so
T =−
l
2g
θ=−θ0
θ=θ0
1
√
dθ =
cos θ − cos θ0
l
2g
θ0
−θ0
√
1
dθ.
cos θ − cos θ0
3. The corresponding plane autonomous system is
x = y,
and
∂
∂x
β
f (x)
−
−g
y
2
1 + [f (x)]
m
y = −g
= −g
f (x)
β
−
y
1 + [f (x)]2
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
1
−gf (x1 ) 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.
622
11.4 Autonomous Systems as Mathematical Models
5. (a) If f (x) = x2 /2, f (x) = x and so
1
x
dy
y
= = −g
.
2
dx
x
1+x 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
1 + x2
y 2 = v02 − g ln
.
1 + x20
Now
v02 − g ln
1 + x2
1 + x20
≥0
2
if and only if x2 ≤ ev0 /g (1 + x20 ) − 1.
2
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.
(b) Since z = x2 /2, the maximum height occurs at the largest value of x on the cycle. From (a), xmax =
2
[ev0 /g (1 + x20 ) − 1]1/2 and so
zmax =
1 2
x2max
= [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
dy
sinh x 1
y
= = −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
2g
2g
y2 =
−
+ v02 .
cosh x cosh x0
Now
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) =
c0
F (xm )G(a/b)
=
< G(a/b).
F (x1 )
F (x1 )
Therefore from Property (2) in the discussion preceding Example 3 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 (1) in the discussion preceding Example 3 in this section of the text, when y = a/b, xn is taken
on at some time t. From Property (3), 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.
623
11.4 Autonomous Systems as Mathematical Models
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).
(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.
10. (a) Solving
x,y
x(−0.1 + 0.02y) = 0
10
x(t)
y(0.2 − 0.025x) = 0
5
y(t)
in the first quadrant we obtain the critical point
(8, 5). The graphs are plotted using x(0) = 7
and y(0) = 4.
t
20
40
80
60
100
(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
0.08(20 − 0.8x − 0.3y)
−0.024x
g (X) =
−0.018y
0.06(10 − 0.2y − 0.3x)
and so
1.6 0
0 0.6
−1.6 −1.2
A3 = g ((50, 0)) =
0
−0.3
A1 = g ((0, 0)) =
A2 = g ((0, 100)) =
A4 = g ((20, 40)) =
−0.8
−1.8
−0.64
−0.72
0
−0.6
−0.48
−0.24
.
Since det(A1 ) = ∆1 = 0.96 > 0, τ = 2.2 > 0, and τ12 − 4∆1 = 1 > 0, we see that (0, 0) is an unstable node.
Since det(A2 ) = ∆2 = 0.48 > 0, τ = −1.4 < 0, and τ22 − 4∆2 = 0.04 > 0, we see that (0, 100) is a stable node.
Since det(A3 ) = ∆3 = 0.48 > 0, τ = −1.9 < 0, and τ32 − 4∆3 = 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. ∆ = r1 r2 , τ = r1 + r2 and τ 2 − 4∆ = (r1 + r2 )2 − 4r1 r2 = (r1 − r2 )2 . Therefore when r1 = r2 , (0, 0) is an
unstable node.
13. For X = (K1 , 0), τ = −r1 + r2 [1 − (K1 α21 /K2 )] and ∆ = −r1 r2 [1 − (K1 α21 /K2 )]. If we let c = 1 − K1 α21 /K2 ,
τ 2 − 4∆ = (cr2 + r1 )2 > 0. Now if k1 > K2 /α21 , c < 0 and so τ < 0, ∆ > 0. Therefore (K1 , 0) is a stable node.
If K1 < K2 /α21 , c > 0 and so ∆ < 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 11.38(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.
624
11.4 Autonomous Systems as Mathematical Models
15. K1 /α12 < K2 < K1 α21 and so α12 α21 > 1. Therefore ∆ = (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 =
−g
β
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
1
g
β
− cos x −
l
ml
has trace τ = −β/ml and determinant ∆ = g/l > 0 at (0, 0). Therefore
τ 2 − 4∆ =
0
β 2 − 4glm2
β2
g
−4 =
.
2
2
m l
l
m2 l 2
√
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 = −
β
k
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
k
β
−
− 2|y|
m
m
and so τ = 0 and ∆ = 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.
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
cos θ = 1
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
1 + [f (x)]2
1 + [f (x)]2
and so
x = g
µ − f (x)
β x.
−
1 + [f (x)]2
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
625
11.4 Autonomous Systems as Mathematical Models
and so at a critical point X1 ,

0
g (X) =  −gf (x1 )
1 + µ2

1
β .
−
m
Therefore τ = −β/m < 0 and ∆ = gf (x1 )/(1 + µ2 ). When f (x1 ) < 0, ∆ < 0 and so a saddle point
occurs. When f (x1 ) > 0 and
β2
f (x1 )
τ 2 − 4∆ = 2 − 4g
< 0,
m
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,
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).
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
bx
g (X) =
r
−cy
−cx + (K − 2y)
K
and so at X1 = (0, 0), ∆ = −ar < 0. For X1 = (0, K), ∆ = n(Kb − a) = −rb(K − a/b). Since we are
given that K > a/b, ∆ < 0 in this case. Therefore (0, 0) and (0, K) are each saddle points. For X1 = (x̂, ŷ)
−a + by
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 ∆ = 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 − 4∆ =
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
626
11.4 Autonomous Systems as Mathematical Models
Therefore τ 2 − 4∆ < 0 if and only if
ŷ <
4bK
4bK 2
=
.
+ 4b
r + 4bK
r
K
Note that
4bK 2
4bK
=
·K ≈K
r + 4bK
r + 4bK
where K is large, and ŷ = a/b < K. Therefore τ 2 − 4∆ < 0 when K is large and a stable spiral point will
result.
21. The equation
y
x =α
x−x=x
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
1
α
x̂ = α β −
=
[(α − 1)β − 1]
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)) = 
.
2
1
(α − 1)
−
−1
−
α
α2
It follows that
(α − 1)2
(α − 1)2
τ =−
x̂
+
1
<
0,
∆
=
x̂
α2
α2
and so τ = −(∆ + 1). Therefore τ 2 − 4∆ = (∆ + 1)2 − 4∆ = (∆ − 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 critical points
√
√
are (0, 0), (0, −2), (0, 2), (0, − 2 ), and (0, 2 ). The Jacobian matrix
is
0
1
g (X) =
−8 + 18x2 − 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.
627
11.4 Autonomous Systems as Mathematical Models
11.5 Periodic Solutions, Limit Cycles, and Global Stability
EXERCISES 11.5
Periodic Solutions, Limit Cycles,
and Global Stability
1. y = x − y = 0 implies that y = x and so 2 + xy = 2 + x2 > 0. Therefore the system has no critical points, and
so, by the corollary to Theorem 11.4, there are no periodic solutions.
2. x = 2x − xy = x(2 − y) = 0 implies that x = 0 or y = 2. If x = 0, then from −1 − x2 + 2x − y 2 = 0, y 2 = −1
and there are no real solutions. If y = 2, −x2 + 2x − 5 = 0 which has no real solutions. Therefore the system
has no critical points and so, by the corollary to Theorem 11.4, there are no periodic solutions.
3. For P = −x + y 2 and Q = x − y,
∂P
∂Q
+
= −2 < 0. Therefore there are no periodic solutions by
∂x
∂y
Theorem 11.5.
4. For P = xy 2 − x2 y and Q = x2 y − 1,
∂P
∂P
∂Q
∂Q
+
= y 2 − 2xy + x2 = (y − x)2 . Therefore
+
does not
∂x
∂y
∂x
∂y
change signs and so there are no periodic solutions by Theorem 11.5.
5. For P = −µx − y and Q = x + y 3 ,
∂Q
∂P
+
= −µ + 9y 2 > 0 since µ < 0. Therefore there are no periodic
∂x
∂y
solutions by Theorem 11.5.
6. From y = xy − y = y(x − 1) = 0 either y = 0 or x = 1. If y = 0, then from 2x + y 2 = 0, x = 0. Likewise x = 1
implies that 2 + y 2 = 0, which has no real solutions. Therefore (0, 0) is the only critical point. But g ((0, 0))
has determinant ∆ = −2. The single critical point is a saddle point and so, by the corollary to Theorem 11.4,
there are no periodic solutions.
7. The corresponding plane autonomous system is x = y, y = 2x − y 4 . Therefore (0, 0) is the only critical point.
But g ((0, 0)) has determinant ∆ = −2 < 0. The single critical point is a saddle point and so, by the corollary
to Theorem 11.4, there are no periodic solutions.
8. The corresponding plane autonomous system is
x = y,
and so
y = −x +
1
+ 3y 2 y − x2
2
∂P
∂Q
1
+
= 0 + + 9y 2 > 0. Therefore there are no periodic solutions by Theorem 11.5.
∂x
∂y
2
9. For δ(x, y) = eax+by ,
∂
∂
(δP ) +
(δQ) can be simplified to
∂x
∂y
eax+by [−bx2 − 2ax + axy + (2b + 1)y].
Setting a = 0 and b = −1/2,
1
∂
∂
1
(δP ) +
(δQ) = x2 e− 2 y
∂
∂y
2
which does not change signs. Therefore by Theorem 11.6 there are no periodic solutions.
628
11.5 Periodic Solutions, Limit Cycles, and Global Stability
10. For δ(x, y) = ax2 + by 2 ,
∂
∂
(δP ) +
(δQ) can be simplified to −5ax4 − 3bx2 y 2 + 10(a − b)x2 y. Setting
∂x
∂y
a = b = 1,
∂
∂
(δP ) +
(δQ) = −5x4 − 3x2 y 2
∂x
∂y
which does not change signs. Therefore by Theorem 11.6 there are no periodic solutions.
11. For P = x(1 − x2 − 3y 2 ) and Q = y(3 − x2 − 3y 2 ),
∂P
∂Q
+
= 4(1 − x2 − 3y 2 )
∂x
∂y
and so
∂Q
∂P
+
>0
∂x
∂y
for x2 + 3y 2 < 1. Therefore there are no periodic solutions in the elliptical region x2 + 3y 2 < 1.
12. The corresponding plane autonomous system is x = y, y = g(x, y) and so
∂P
∂Q
∂g
∂g
= 0
+
=
=
∂x
∂y
∂y
∂x
in the region R. Therefore
∂P
∂Q
+
cannot change signs and so there are no periodic solutions by
∂x
∂y
Theorem 11.5.
13. For δ(x, y) =
1
a
r 1
, δP = − + b, δQ = −c +
(K − y) and so
xy
y
K x
∂(δP ) ∂(δQ)
r
+
=−
<0
∂x
∂y
Kx
in the first quadrant. By Theorem 11.6, there are no periodic solutions in the first quadrant.
14. If n = (−2x, −2y),
V · n = 2xy + 2x2 ex+y − 2xy + 2y 2 ex+y = 2(x2 + y 2 )ex+y ≥ 0.
Therefore any circular region of the form x2 + y 2 ≤ r2 is an invariant region by Theorem 11.7.
15. If n = (−2x, −2y),
V · n = 2x2 − 4xy + 2y 2 + 2y 4 = 2(x − y)2 + 2y 4 ≥ 0.
Therefore x2 + y 2 ≤ r serves as an invariant region for any r > 0 by Theorem 11.7.
16. n = −∇t = (−6x5 , −6y). Since the corresponding plane autonomous system is x = y, y = −y − y 3 − x5 ,
n · V = −6x5 y + 6y 2 + 6y 4 + 6x5 y = 6y 2 + 6y 4 .
Therefore n · V ≥ 0 and so by Theorem 11.7, the region x6 + 3y 2 ≤ 1 serves as an invariant region.
≤ x2 + y 2 ≤ 1 is an invariant region for the plane autonomous system. If
the only critical point is (0, 0), this critical point lies outside the invariant region and so Theorem 11.8(ii) is
applicable. There is at least one periodic solution in R.
17. We showed in Example 8 that
1
16
18. The corresponding plane autonomous system is
x = y,
y = y(1 − 3x2 − 2y 2 ) − x
and it is easy to see that (0, 0) is the only critical point. If n = (−2x, −2y) then
V · n = −2xy − 2y 2 (1 − 3x2 − 2y 2 ) + 2xy = −2y 2 (1 − 2r2 − x2 ).
√
2 , 2r2 = 1 and so V · n = 2x2 y 2 ≥ 0. Therefore
Theorem 11.8(ii) there is at least one periodic solution.
If r =
1
2
629
1
4
≤ x2 + y 2 ≤
1
2
serves as an invariant region. By
11.5 Periodic Solutions, Limit Cycles, and Global Stability
19. If r < 1 and n = (−2x, −2y) then
V · n = −2xy + 2xy + 2y 2 (1 − x2 ) = 2y 2 (1 − x2 ) ≥ 0
since x2 < 1. Therefore x2 + y 2 ≤ r2 serves as an invariant region. Now (0, 0) is the only critical point and,
since τ = −1 and ∆ = 1, τ 2 − 4∆ < 0. Therefore (0, 0) is a stable spiral point and so, by Theorem 11.9(ii),
limt→∞ X(t) = (0, 0).
20. Since
∂P
∂Q
+
= −1 − 3y 2 < 0, there are no periodic solutions. If n = (−2x, −2y),
∂x
∂y
V · n = −2xy + 2x2 + 2xy + 2y 4 = 2(x2 + y 4 ) ≥ 0.
Therefore the circular region x2 + y 2 ≤ r2 serves as an invariant region for any r > 0. If (x, y) is a critical point.
y − x = 0 or y = x. From −x − y 3 = 0 we have −y(1 + y 2 ) = 0. Therefore y = 0 and so (0, 0) is the only critical
point. It is easy to check that τ = −1, ∆ = 1, τ 2 − 4∆ = −3 and so (0, 0) is a stable spiral point. By Theorem
11.9(ii), (0, 0) is globally stable. For any initial condition, limt→∞ X(t) = (0, 0).
21. (a)
∂P
∂Q
+
= 2xy − 1 − x2 ≤ 2x − 1 − x2 = −(x − 1)2 ≤ 0. Therefore there are no periodic solutions.
∂x
∂y
(b) If (x, y) is a critical point, x2 y = 12 and so from x = x2 y − x + 1, 12 − x + 1 = 0. Therefore x = 3/2 and so
y = 2/9. For this critical point, τ = −31/12 < 0, ∆ = 9/4 > 0, and τ 2 − 4∆ < 0. Therefore (3/2, 2/9) is a
stable spiral point and so, from Theorem 11.9(ii), limx→∞ X(t) = (3/2, 2/9).
2y
2x
y
22. (a) From x
− 1 = 0, either x = 0 or y = 2. For the case x = 0, from y 1 −
−
= 0,
y+2
y+2 8
y
2x
1
y 1−
= 0. Therefore (0, 0) and (0, 8) are critical points. If y = 2, 1 −
− = 0 and so x = 3/2.
8
4
4
Therefore (3/2, 2) is the additional critical point. We may classify these critical points as follows:
X
τ
∆
τ 2 − 4∆
Conclusion
(0, 0)
–
−1
–
saddle point
–
− 35
3
8
–
saddle point
(0, 8)
( 32
, 2)
1
8
− 95
64
unstable spiral point
(b) By Theorem 11.8(i), since there is a unique unstable critical point inside the invariant region, there is at
least one periodic solution.
CHAPTER 11 REVIEW EXERCISES
1. True
2. True
3. A center or a saddle point
630
CHAPTER 11 REVIEW EXERCISES
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 11.25 in the text.
8. False; see Figure 11.29 in the text.
9. True
10. False; we also need to have no critical points on the boundary of R.
11. Switching to polar coordinates,
dx
dy
1
x
+y
= (−xy − x2 r3 + xy − y 2 r3 ) = −r4
dt
dt
r
dx
dθ
1
dy
1
= 2 −y
+x
= 2 (y 2 + xyr3 + x2 − xyr3 ) = 1.
dt
r
dt
dt
r
dr
1
=
dt
r
Using separation of variables it follows that r = √
3
1
and θ = t + c2 . Since X(0) = (1, 0), r = 1 and θ = 0.
3t + c1
It follows that c1 = 1, c2 = 0, and so
1
, θ = t.
3t + 1
As t → ∞, r → 0 and the solution spirals toward the origin.
r= √
3
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, ∆ = 11 > 0 and so (0, 0) is a center.
(b) τ = −2, ∆ = 1, τ 2 − 4∆ = 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
1 + y − 6x
x
g (X) =
−2y
4 − 2x − 2y
to classify these four critical points. The results are as follows:
X
If δ(x, y) =
τ
∆
(0, 0)
5
4
(0, 4)
τ 2 − 4∆
9
Conclusion
unstable node
–
−20
–
saddle point
( 13 , 0)
–
− 10
3
–
saddle point
(1, 2)
−5
10
−15
stable spiral point
1
1
x
4
y
, δP = + 1 − 3
and δQ = − 2 − . It follows that
xy
y
y
x
x
∂
∂
3
1
(δP ) +
(δQ) = − − < 0
∂x
∂y
y x
in quadrant one. Therefore there are no periodic solutions in the first quadrant.
631
CHAPTER 11 REVIEW EXERCISES
15. The corresponding plane autonomous system is x = y, y = µ(1 − x2 ) − x and so the Jacobian at the critical
point (0, 0) is
0 1
g ((0, 0)) =
.
−1 µ
Therefore τ = µ, ∆ = 1 and τ 2 − 4∆ = 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.
16. 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.
dy
y
−2x y 2 + 1
17.
= =
. We may separate variables to show that y 2 + 1 = −x2 + c. But x(0) = x0 and
dx
x
y
y(0) = x