STUDENT SOLUTIONS MANUAL
DIFFERENTIAL EQUATIONS
and
BOUNDARY VALUE PROBLEMS
Computing and Modeling
3E
EDWARDS
&PENNEY
Digitized by the Internet Archive
in 2018 with funding from
Kahle/Austin Foundation
https://archive.org/details/studentsolutionsOOOOedwa
STUDENT SOLUTIONS MANUAL
DIFFERENTIAL EQUATIONS
and
BOUNDARY VALUE PROBLEMS
Computing and Modeling
3E
EDWARDS
&PENNEY
PEARSON
Prentice
Hall
Upper Saddle River, NJ 07458
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PEARSON
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Hall
© 2004 Pearson Education, Inc.
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ISBN
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CONTENTS
1
FIRST-ORDER DIFFERENTIAL EQUATIONS
1.1
Differential Equations and Mathematical Modeling
1
1.2
Integrals as General and Particular Solutions
5
1.3
Slope Fields and Solution Curves
9
1.4
Separable Equations and Applications
16
1.5
Linear First-Order Equations
24
1.6
Substitution Methods and Exact Equations
28
Chapter 1 Review Problems
35
2
MATHEMATICAL MODELS
AND NUMERICAL METHODS
2.1
Population Models
37
2.2
Equilibrium Solutions and Stability
45
2.3
Acceleration-Velocity Models
53
2.4
Numerical Approximation: Euler's Method
57
2.5
A Closer Look at the Euler Method
62
2.6
The Runge-Kutta Method
69
3
LINEAR EQUATIONS OF HIGHER ORDER
3.1
Introduction: Second-Order Linear Equations
76
3.2
General Solutions of Linear Equations
79
3.3
Homogeneous Equations with Constant Coefficients
83
3.4
Mechanical Vibrations
87
3.5
Nonhomogeneous Equations and the
Method of Undetermined Coefficients
92
3.6
Forced Oscillations and Resonance
98
3.7
Electrical Circuits
106
3.8
Endpoint Problems and Eigenvalues
110
4
INTRODUCTION TO SYSTEMS
OF DIFFERENTIAL EQUATIONS
4.1
First-Order Systems and Applications
115
4.2
The Method of Elimination
120
4.3
Numerical Methods for Systems
131
5
LINEAR SYSTEMS OF
DIFFERENTIAL EQUATIONS
5.1
Linear Systems and Matrices
137
5.2
The Eigenvalue Method for Homogeneous Linear Systems
142
5.3
Second-Order Systems and Mechanical Applications
159
5.4
Multiple Eigenvalue Solutions
165
5.5
Matrix Exponentials and Linear Systems
173
5.6
Nonhomogeneous Linear Systems
178
6
NONLINEAR SYSTEMS AND PHENOMENA
6.1
Stability and the Phase Plane
184
6.2
Linear and Almost Linear Systems
188
6.3
Ecological Applications: Predators and Competitors
198
6.4
Nonlinear Mechanical Systems
207
6.5
Chaos in Dynamical Systems
213
7
LAPLACE TRANSFORM METHODS
7.1
Laplace Transforms and Inverse Transforms
219
7.2
Transformation of Initial Value Problems
222
7.3
Translation and Partial Fractions
227
7.4
Derivatives, Integrals, and Products of Transforms
231
7.5
Periodic and Piecewise Continuous Forcing Functions
235
7.6
Impulses and Delta Functions
242
8
POWER SERIES METHODS
8.1
Introduction and Review of Power Series
247
8.2
Series Solutions Near Ordinary Points
251
8.3
Regular Singular Points
257
8.4
Method of Frobenius: The Exceptional Cases
265
8.5
Bessel's Equation
270
8.6
Applications of Bessel Functions
273
9
FOURIER SERIES METHODS
9.1
Periodic Functions and Trigonometric Series
277
9.2
General Fourier Series and Convergence
283
9.3
Fourier Sine and Cosine Series
290
9.4
Applications of Fourier Series
297
9.5
Heat Conduction and Separation of Variables
300
9.6
Vibrating Strings and the One-Dimensional Wave Equation
303
9.7
Steady-State Temperature and Laplace's Equation
306
10
EIGENVALUES AND
BOUNDARY VALUE PROBLEMS
10.1
Sturm-Liouville Problems and Eigenfunction Expansions
312
10.2
Applications of Eigenfunction Series
319
10.3
Steady Periodic Solutions and Natural Frequencies
324
10.4
Cylindrical Coordinate Problems
332
10.5
Higher-Dimensional Phenomena
339
APPENDIX
Existence and Uniqueness of Solutions
340
PREFACE
This is a solutions manual to accompany the textbook DIFFERENTIAL EQUATIONS AND
BOUNDARY VALUE PROBLEMS: Computing and Modeling (3rd edition, 2004) by C.
Henry Edwards and David E. Penney. We include solutions to most of the odd-numbered problems
in the text.
Our goal is to support learning of the subject of elementary differential equations in every way that
we can.
We therefore invite comments and suggested improvements for future printings of this
manual, as well as advice regarding features that might be added to increase its usefulness in
subsequent editions.
Additional supplementary material can be found at our textbook Web site
listed below.
Henry Edwards & David Penney
hedwards@math.uga.edu
dpenney@math.uga.edu
www.prenhall.com/edwards
CHAPTER 1
FIRST-ORDER DIFFERENTIAL EQUATIONS
SECTION 1.1 *
DIFFERENTIAL EQUATIONS AND MATHEMATICAL MODELING
The main purpose of Section 1.1 is simply to introduce the basic notation and terminology of
differential equations, and to show the student what is meant by a solution of a differential
equation. Also, the use of differential equations in the mathematical modeling of real-world
phenomena is outlined.
Problems 1-12 are routine verifications by direct substitution of the suggested solutions into the
given differential equations. We include here just some typical examples of such verifications.
3.
If y,=cos2x and y2=sin2x, then y| = -2sin2x and y\-2cos 2x so
y" = -4cos2x = -4 y}
and
y" = -4sin2x = -4 y2.
Thus y" + 4y} = 0 and y" + 4 y2 = 0.
5.
If y = ex-e~xi then y' = ex+e~x so y'-y = (ex +e~x)-(ex-e~x) = 2e"x. Thus
y = y + 2e~x.
11.
If y = y}= x-2 then y' = -2x~3 and y" = 6x~4, so
x2y" + 5xy' + 4y = x2 (6x~4) + 5x(-2x_3) + 4(x“2) = 0.
If y = y2 = x~2 lnx then y' = x-3 -2x“3lnx and y" = -5x~4 +6x-4 lnx, so
x2y' + Sxy' + 4y = x2(-5x-4 + 6x~4 lnx) + 5x(x~3 - 2x~3 lnx)+ 4(x~2 lnx)
= (-5x~2 + 5x~2 ) + (6x-2 - 10x-2 + 4x-2 jinx = 0.
Section 1.1
1
13.
Substitution of y = erx into 3y' = 2y gives the equation 3rerx = 2 erx that simplifies
to 3r = 2. Thus r- 2/3.
15.
Substitution of y = erx into y" + y' - 2y = 0 gives the equation r2erx + r erx -2erx = 0
that simplifies to r +r- 2 = (r + 2){r -1) = 0. Thus r = —2 or r— 1.
The verifications of the suggested solutions in Problems 17-26 are similar to those in Problems
1-12. We illustrate the determination of the value of C only in some typical cases. However,
we illustrate typical solution curves for each of these problems.
17.
C = 2
19.
If ^(x) = Cex -1 then y(0) = 5 gives C-l = 5, so C = 6.
x
2
Chapter 1
21.
C = 7.
23.
If y(x) = }x5+Cx2 then y(2) = 1 gives the equation j-32 + C-{ = 1 with
solution C = -56. See the figure on the left below.
25.
If y(x) = tan(x2+C) then _y(0) = 1 gives the equation tan C = 1. Hence one value
of C is C = n /4 (as is this value plus any integral multiple of tz). See the figure on the
right above.
Section 1.1
3
27.
/ = x+y
29.
If m = y' is the slope of the tangent line and m' is the slope of the normal line at
(x,y), then the relation mm'--1 yields m' = 1 /y' = (jy-l)/(x-0). Solution for
y then gives the differential equation (1 - y)y' = x.
31.
The slope of the line through (x,y) and (-y,x) is y' - (x - y) !{-y - x), so the
differential equation is (x + y)y' = y-x.
In Problems 32-36 we get the desired differential equation when we replace the "time rate of
change" of the dependent variable with its derivative, the word "is" with the = sign, the phrase
"proportional to" with k, and finally translate the remainder of the given sentence into symbols.
33.
dv / dt = kv2
35.
dN/dt = k(P-N)
37.
The second derivative of any linear function is zero, so we spot the two solutions
y(x) = 1 or y(x) = x of the differential equation y" - 0.
39.
We reason that if y = kx , then each term in the differential equation is a multiple of x .
The choice k = 1 balances the equation, and provides the solution y(x) = x2.
41.
We reason that if y - kex, then each term in the differential equation is a multiple of ex.
The choice k - j balances the equation, and provides the solutiony(x) = \ex.
43.
(a)
We need only substitute x(7) = 1 /(C - kt) in both sides of the differential
equation x' = kx2 for a routine verification.
(b)
The zero-valued function x(7) = 0 obviously satisfies the initial value problem
x' = kx2, x(0) = 0.
45.
Substitution of P'= 1 and P = 10 into the differential equation P' = kP2 gives k = so Problem 43(a) yields a solution of the form P(t) = 1/(C-/7100). The initial
condition P(0) = 2 now yields C = y, so we get the solution
4
Chapter 1
100
50-/
2
100
We now find readily that P = 100 when / = 49, and that P = 1000 when / = 49.9.
It appears that P grows without bound (and thus "explodes") as t approaches 50.
47.
(a)
y(10) = 10 yields 10 = 1/(C-10), so C = 101/10.
(b)
There is no such value of C, but the constant function y(x) = 0 satisfies the
conditions y' = y
and y(0) = 0.
It is obvious visually (in Fig. 1.1.8 of the text) that one and only one solution
curve passes through each point (a,b) of the xy-plane, so it follows that there exists a
(c)
unique solution to the initial value problem y = y , y(a) = b.
SECTION 1.2
INTEGRALS AS GENERAL AND PARTICULAR SOLUTIONS
This section introduces general solutions and particular solutions in the very simplest situation
— a differential equation of the form y' = /(x) — where only direct integration and evaluation
of the constant of integration are involved. Students should review carefully the elementary
concepts of velocity and acceleration, as well as the fps and mks unit systems.
1.
Integration of y’- 2x + l yields y(x) = J(2x + 1 )dx = x2+x + C. Then substitution
of x = 0, y = 3 gives 3 = 0 + 0 + C= C, so j;(x) = x2+x + 3.
3.
Integration of yf = Vx yields y(x) = jVvJx = j-x3/2+C. Then substitution of
x = 4, y = 0 gives 0 = -y + C, so y(x) = j(x3/2 - 8).
5.
Integration of y' = (x + 2)_1/2 yields y(x) = j(x + 2)_,/2 dx = 2\jx + 2 + C. Then
substitution of x = 2, y = -1 gives -l = 2-2 + C, so y(x) = 2Vx + 2-5.
7.
Integration of y' = 10/(x2+l) yields y(x) = Jl0/(x2+l)Jx = 10tan_1x + C. Then
substitution of x = 0, y = 0 gives 0 = 10*0 + C, so y(x) = 10tan_! x.
Section 1.1
5
9.
Integration of / = 1/ J\-x2
yields y(x) = Jl / Vl -x2 dx = sin 1x + C. Then
substitution of x = 0, y = 0 gives 0 = 0 + C, so y(x) = sin-1 x.
11.
If a(t) = 50 then v(/) = J*50d/ = 50/ + v0 = 50^ + 10. Hence
x(0 = J(50 / +I0)dt = 25/2+10/ + x0 = 25/2 +10/ + 20.
13.
If a(t) - 3/ then v(/) = |3tdt = f/2+v0 = f/2+5. Hence
x(/) = j*(4^ +5)dt = -jt + 5 / + x0 — y/ +5/.
15.
If a(t) = 4(/ + 3)2. then v(/) = j4(/ + 3)2d/ = y(/ + 3)3 + C = -f(/ + 3)3-37 (taking
C = -37 so that v(0) = -l). Hence
x(/) = J[j(/ + 3)3 -
17.
If a(t) = (/ + 1)"3 then v(/) -
- j(/ + 3)4 -37/ + C - }(/ + 3)4 -37/-26.
J(/ + l)"3 dt = -}(/ + l)"2+C = -T(/ + l)-2+±
(taking C = \ so that v(0) = 0). Hence
x(/) = J[-|(/+i)_2+\\dt= \(t+\yx+\t+c = i[(/ + ir1+/-i]
(taking C = - \ so that x(0) = 0).
19.
Note that v(/) = 5 for 0 < / < 5 and that v(/) = 10 - / for 5 < / < 10. Hence
x(/) = 5/ + Cj for 0</<5and x(/) = 10/-y/2 + C2 for 5</<10. Now Cx - 0
because x(0) = 0, and continuity of x(/) requires that x(/) = 5/ and
x(/) -10/ - \ t2 + C2 agree when / = 5. This implies that C2 = --y, and we get the
graph on the left at the top of the next page.
21.
Note that v(/) = / for 0 < / < 5 and that v(/) = 10 — / for 5 < / < 10. Hence
x(/) = jt2 + Cs for 0 < / < 5 and x(/) = 10/-\t2 +C2 for 5</<10. Now C, = 0
because x(0) = 0, and continuity of x(/) requires that x(/) = |/2 and
x(/) = 10/ - \ /2 + C2 agree when / = 5. This implies that C2 = -25, and we get the
graph on the right at the top of the next page.
6
Chapter 1
Problem 19
23.
v = -9.8/ + 49, so the ball reaches its maximum height (v = 0) after / = 5 seconds. Its
maximum height then is _y(5) = -4.9(5)2 + 49(5) = 122.5 meters.
25.
a = -10 m/s2 and vo = 100 km/h « 27.78 m/s, so v = -10/+ 27.78, and hence
x(/) = -5t2 + 27.78/. The car stops when v = 0, / « 2.78, and thus the distance
traveled before stopping is x(2.78) « 38.59 meters.
27.
a = -9.8 m/s2 so v - -9.8/-10 and
y = -4.9 t2 - 10 / + ^o*
The ball hits the ground when y = 0 and
v = -9.8 /- 10 = -60,
so /« 5.10 s. Hence
_y0 = 4.9(5.10)2+ 10(5.10) » 178.57 m.
29.
Integration of dv/dt = 0.12 /3 + 0.6 /, v(0) = 0 gives v(/) = 0.3 /2 + 0.04 /3. Hence
v(10) = 70. Then integration of dxldt = 0.3 t2 + 0.04 /3, x(0) = 0 gives
x(/) = 0.1 /3 + 0.04 /4, so x(10) = 200. Thus after 10 seconds the car has gone 200 ft and
is traveling at 70 ft/sec.
31.
If a = -20 m/sec
by
and xo = 0 then the car's velocity and position at time / are given
v = -20/ + vo,
2
x = -10 t + vo/.
Section 1.2
7
It stops when v = 0 (so Vo = 20/), and hence when
x = 75 — -10 /2 + (20/)/ = 10 /2.
Thus / = VT5 sec so
vo = 20VT5 « 54.77 m/sec « 197 km/hr.
33.
If vo = 0 and yo = 20 then
v = -a/ and jy = -{a/ +20.
Substitution of / = 2,
jvo = 200 then
= 0 yields a = 10 ft/sec . If vo = 0 and
v = -10/ and y = -5/2 + 200.
Hence y = 0 when / = V40 = 2VTo sec and v = -20Vl() « -63.25 ft/sec.
35.
If vo = 0 and yo = h then the stone's velocity and height are given by
v = -gt,
y = -0.5 gt2 + h.
Hence y = 0 when t = ^2hig
so
v = -gyjlhlg
37.
= -yjlgh.
We use units of miles and hours. If xo = vo = 0 then the car's velocity and position
after / hours are given by
2
v = at,
x = y/ .
Since v = 60 when / = 5/6, the velocity equation fields a = 72 mi/hr . Hence the
distance traveled by 12:50 pm is
x = (0.5)(72)(5/6)2 = 25 miles.
39.
Integration of y' = (9/v,v)(l — 4x2) yields
y = (3/v,-)(3x - 4x3) + C,
and the initial condition j/(-l/2) = 0 gives C = 3/yv. Hence the swimmer's trajectory
is
8
Chapter 1
y(x) = (3/y9)(3x - 4x3 + 1).
Substitution of y( 1/2) = 1 now gives vv = 6 mph.
41.
The bomb equations are a = -32, v = -32, and sB=s = -\6t2 + 800, with t = 0 at the
instant the bomb is dropped. The projectile is fired at time t — 2, so its corresponding
equations are a = -32, v = -32(7-2) + v0, and
sP = s = -16(/-2)2 + v0(/-2)
for t > 2 (the arbitrary constant vanishing because sp{2) = 0). Now the condition
sB(0 = -16/2 + 800 = 400 gives t = 5, and then the requirement that sp(5) = 400 also
yields v0 = 544/3 « 181.33 ft/s for the projectile's needed initial velocity.
43.
The velocity and position functions for the spacecraft are vs(t) = 0.00981 and
xs(t) = 0.0049/ , and the corresponding functions for the projectile are
vP(/) = -j^c = 3x107and xp(t) = 3x 1071. The condition that xs = xp when the
spacecraft overtakes the projectile gives 0.004912 = 3x1011, whence
t =
3x10'
6.12245xlO9 sec
0.0049
6.12245 x 109
194 years.
(3600)(24)(365.25)
Since the projectile is traveling at
the speed of light, it has then traveled a distance of
about 19.4 light years, which is about 1.8367 x 1017 meters.
SECTION 1.3
SLOPE FIELDS AND SOLUTION CURVES
The instructor may choose to delay covering Section 1.3 until later in Chapter 1. However,
before proceeding to Chapter 2, it is important that students come to grips at some point with the
question of the existence of a unique solution of a differential equation — and realize that it
makes no sense to look for the solution without knowing in advance that it exists. It may help
some students to simplify the statement of the existence-uniqueness theorem as follows:
Section 1.2
9
Suppose that the function /(x,y) and the partial derivative df / dy are both
continuous in some neighborhood of the point (a, b). Then the initial value
problem
~ = f(x,y),
dx
y{a) = b
has a unique solution in some neighborhood of the point a.
Slope fields and geometrical solution curves are introduced in this section as a concrete aid in
visualizing solutions and existence-uniqueness questions. Instead, we provide some details of
the construction of the figure for the Problem 1 answer, and then include without further
comment the similarly constructed figures for Problems 2 through 9.
1.
The following sequence of Mathematica commands generates the slope field and the
solution curves through the given points. Begin with the differential equation
dyldx- f(x,y) where
f[x_, y_J
:= -y - Sin[x]
Then set up the viewing window
a = -3; b = 3; c = -3; d = 3;
The components (w,v) of unit vectors corresponding to the short slope field line
segments are given by
u[x_, y_]
v[x_, y_]
:= 1/Sqrt[1 + f[x, y]A2]
:= f[x, y]/Sqrt[1 + f[x, y]A2j
The slope field is then constructed by the commands
Needs["Graphics'PlotField'"]
dfield = PlotVectorField[{u[x, y] , v[x, y] } , {x, a, b), {y, c, d} ,
HeadWidth -> 0, HeadLength -> 0, PlotPoints -> 19,
PlotRange -> {{a, b}, {c, d}} , Axes -> True, Frame -> True,
FrameLabel -> {"x", "y"}, AspectRatio -> 1];
The original curve shown in Fig. 1.3.12 of the text (and its initial point not shown there)
are plotted by the commands
x0 = -1.9; yO = 0;
pointO = Graphics[{PointSize[0.025], Point[{x0, yO}]}];
soln = NDSolve[(Derivative[1][y][x] == f[x, y[x]], y[x0] == yO},
y[x], {x, a, b}];
soln[[1,1,2]] ;
curveO = Plot[soln[[1,1,2]], {x, a, b},
PlotStyle -> {Thickness[0.0065], RGBColor[0, 0, 1]}];
The Mathematica NDSolve command carries out an approximate numerical solution of
the given differential equation. Numerical solution techniques are discussed in Sections
2.4-2.6 of the textbook.
10
Chapter 1
instance the 7th point is (-2.5, 1). It and the corresponding solution curve are plotted
by the commands
xO = -2.5;
yO = 1;
point7 = Graphics[{PointSize[0.025] ,
soln = NDSolve[{Derivative[1][y][x]
y[x],
{x,
a,
soln[[1,1,2]] ;
curve7 = Plot[soln[[1,1,2]],
PlotStyle ->
Point[{xO,
== f[x,
yO}]}];
y[x]],
y[x0]
== yO},
b}];
{x,
a,
b},
{Thickness[0.0065],
RGBColor[0,
0,
Finally, the desired figure is assembled by the Mathematica command
Show[
dffield,
pointO , curveO ,
pointl,curvel, point2,curve2, point3,curve3,
point4,curve4, point5,curve5, point6,curve6,
point7,curve7, point8,curve8, point9,curve9,
,
pointl0 curvel0,
pointl1,curvel1,
pointl2,curvel2];
1]}];
7.
11.
9.
2 2
Because both /(x,jy) =2xy
2
and df / dy =4xy are continuous everywhere, the
existence-uniqueness theorem of Section 1.3 in the textbook guarantees the existence of a
unique solution in some neighborhood of jc = 1.
13.
Both /(x,jy) = y113 and df /dy - (l/3)jy_2/3 are continuous near (0,1), so the
theorem guarantees the existence of a unique solution in some neighborhood of x = 0.
15.
/(x, jy) = {x — y)
1 /9
is not continuous at (2, 2) because it is not even defined if y>x-
Hence the theorem guarantees neither existence nor uniqueness in any neighborhood of
the point x = 2.
17.
Both f(x,y) = (x-l/y and df I dy - -(x-1 )/y
are continuous near (0, 1), so the
theorem guarantees both existence and uniqueness of a solution in some neighborhood of
x = 0.
19.
9
9
Both f(x,y) = ln(l + y ) and df I dy = 2jy/( 1 + y ) are continuous near (0, 0), so
the theorem guarantees the existence of a unique solution near x = 0.
21.
The curve in the figure on the left at the top of the next page can be constructed using the
commands illustrated in Problem 1 above. Tracing this solution curve, we see that
y(-4) « 3. An exact solution of the differential equation yields the more accurate
approximation y(-4) « 3.0183.
12
Chapter 1
■
III
V - /
3
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1-1-
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2
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■
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\ — /
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-
\
\
\
\
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■
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23.
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y / /
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-- / / P)
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/
\
•
\ \ \ ',
-5 _1_1_1_1_
/ / / /--
/
\
-A
/
1
\
-3
1 - III// y ^ -
1
-1 •
-2
////////
////////
II//////
/////// /
Problem 23
Tracing the curve in figure on the right above, we see that j^(2) « 1. A more accurate
approximation is y(2) « 1.0044.
25.
The figure below indicates a limiting velocity of 20 ft/sec — about the same as jumping
off a 6 j-foot wall, and hence quite survivable. Tracing the curve suggests that v(7) = 19
ft/sec when t is a bit less than 2 seconds. An exact solution gives t« 1.8723 then.
> 20
Section 1.3
27.
If 6 < 0 then the initial value problem y' = 2yfy, y{0) = 6 has no solution, because the
square root of a negative number would be involved. If b > 0 we get a unique solution
curve through (0,6) defined for all x by following a parabola — in the figure below —
down (and leftward) to the x-axis and then following the x-axis to the left. But starting at
(0,0) we can follow the positive x-axis to the point (c,0) and then branching off on the
parabola y - (x - c)2. This gives infinitely many different solutions if 6 = 0.
X
29.
Looking at the figure below, we see that we can start at the point (a, 6) and follow a
branch of a cubic up or down to the x-axis, then follow the x-axis an arbitrary distance
before branching off (down or up) on another cubic. This gives infinitely many
solutions of the initial value problem y’ = 3y
, y(a) = 6 that are defined for all x.
However, if 6^0 there is only a single cubic y = (x-c) passing through (a, 6), so
the solution is unique near x = a.
14
Chapter 1
31.
The function y(x) = sin(x - c), with y'(x) = cos(x - c), satisfies the differential
equation y’ = yj 1 - y2 on the interval c-tt/2<x<c + 7t/2 where cos(x - c) > 0, so it
follows that
y2 = yj\ - sin2 (x - c) = ^/cos2(x -c) = -sin(x-c) = y.
If |Z>| > 1 then the initial value problem y' =
yj\- y2, y(a) = b has no solution because
the square root of a negative number would be involved. If \b\<l then there is only one
curve of the form y = sin(x - c) through the point (a, b); this give a unique solution.
But if b = ±1 then we can combine a left ray of the line y = -1, a sine curve from the
line y = -1 to the line y = +1, and then a right ray of the line y = +1. Looking at the
figure below, we see that this gives infinitely many solutions (defined for all x)
through any point of the form (a,± 1).
33.
Looking at the figure provided in the answers section of the textbook, it suffices to
observe that, among the pictured curves y = x /(cx -1) for all possible values of c,
•
•
•
there is a unique one of these curves through any point not on either coordinate axis;
there is no such curve through any point on the y-axis other than the origin; and
there are infinitely many such curves through the origin (0,0).
But in addition we have the constant-valued solution y(x) = 0 that "covers" the x-axis.
It follows that the given differential equation has near (a,b)
•
•
•
a unique solution if a ^ 0;
no solution if a - 0 but b ^ 0 ;
infinitely many different solutions if a = b - 0.
Section 1.3
15
SECTION 1.4
SEPARABLE EQUATIONS AND APPLICATIONS
Of course it should be emphasized to students that the possibility of separating the variables is
the first one you look for. The general concept of natural growth and decay is important for all
differential equations students, but the particular applications in this section are optional.
Torricelli's law in the form of Equation (24) in the text leads to some nice concrete examples and
problems.
dy
1.
- - J2xdx\
Iny — -x2+c;
y(x) = e
- Jsinxdx;
In y = -cosx + c;
y(x)
x +c
Ce
-x
y
j
dy
3.
-COSX+C
y
r
dy
5.
= Ce
-COS*
sin 1 y - Vx+C;
.y(x) = sin^Vx+cj
jy2n = 3x4/3+|C;
y(x) = (2x4/3+C)
;
J
dy
7.
- j*4x1/3 dx;
T,/3
Y 1
"dy
2dx
. y
i-y “ J V 1 + x
9.
1
+
^
dx
(partial fractions)
1-xJ
ln^ = ln(l + x)-ln(l-x) + lnC;
y(x) = C
1+x
1-x
dy
11.
j
13.
15.
16
y
1/2
-
xdx;
3
-j-j- =
^ y dy
4 - = Jcosxdx;
y +1
( 2
2
1 3
4 dy =
y )
f\
J ^X
y^ = (
2)
C~x
|ln(y4+l) = sinx+C
1 'j
2 dx:
X J
2
1
1
-1-- = In x H-b C
X
y 3 y-
Chapter 1
17.
y = 1 + x + y + xy = (1 + x)(l + y~)
r
dy
J(1 + x)dx;
\+y
dy
19.
=
JV dx\
In |l + y\ = x + |x2+C
ln>> = ex +lnC;
y(x) = Cexp(V)
y
X0) = 2e
r
.
21
2 y dy =
y(5) = 2
r
x dx
implies
C=l
C=e 2
r
dy
Vi1
A
y
Vx
J
jf(1) = 1
implies
—+ 2x ;
C-e
-i
2x
Xx) = Cxexp(x )
y(x) = xexp(x“-l).
ey = 3e2x+C;
C=-2
2y-l = Ce
y(x) = y(l + e2*2).
so
so
y(x) =ln(3e2x+C)
y(x) = ln(3e2jc-2).
Separation of variables gives the general solution
11
2
J l
y )
x dx;
i
f
ii
(b)
so
lnj; = lnx + x +lnC;
\ey dy = j6e2x dx;
(a)
f = 1 + Vx2 -16
yln(2^-l) = x + {lnC;
implies
implies
Xx) = 2exp(V).
so
.y(l) = 1
^(0) = 0
so
2 - Vx2-16+C
Vx2 -16
2y
25.
29.
C=2
dy
— = jdx;
23.
27.
implies
j
1
— = -x + C;
y
1
XX =-.
x-C
Inspection yields the singular solution XX = 0 that corresponds to no value of
the constant C.
(c)
In the figure at the top of the next page we see that there is a unique solution
curve through every point in the xy-plane.
Section 1.4
17
31.
The formal separation-of-variables process is the same as in Problem 30 where, indeed,
we started by taking square roots in (/) = 4y to get the differential equation
y' = 2y[y. Separate of variables then gives
dx\
yfy =
x-C;
y(x) = (x-C).
This general solution provides the parabolas illustrated in Fig. 1.4.5 in the textbook.
But the equation y' = ^4y requires that y; be nonnegative. This means that only the
right half of each parabola y - (x - C)“ qualifies as a solution curve. We spot also the
singular solution y(x) = 0 that corresponds to no value of the constant C.
_i_i_i_i___i_i_i_
-15
-10
-5
0
X
18
Chapter 1
5
10
15
Inspecting the figure above, we therefore see that through the point (a,b) there passes
(a)
No solution curve if b < 0,
(b)
A unique solution curve if b > 0,
(c)
Infinitely many solution curves if b = 0, because in this case we can pick any
c > a and define the solution y(x) = 0 if x < c, y(x) = (x- c)2 if x > c.
33.
The population growth rate is
k = ln(30000/25000)/10 « 0.01823, so the population
of the city t years after 1960 is given by P{t) - 25000 e0 01823'. The expected year
2000 population is then P(40) = 25000 e001823x40 « 51840.
35.
As in the textbook discussion of radioactive decay, the number of 14C atoms after t
years is given by N(t) = N0 e"°'0001216 /. Hence we need only solve the equation
±N0 = N0 g_0-000,2I6/ for t = (In 6)/ 0.0001216 « 14735 years to find the age of the
skull.
37.
The amount in the account after t years is given by A(t) = 5000e° 08/. Hence the
amount in the account after 18 years is given by ^4(18) = 5000e°08x18 « 21,103.48
dollars.
39.
To find the decay rate of this drug in the dog's blood stream, we solve the equation
j = e~5k (half-life 5 hours) for k = (In2)/5 « 0.13863. Thus the amount in the dog's
bloodstream after t hours is given by A(t) = A^ e~0A3863'. We therefore solve the
equation ^4(1) = Ade~0 ]3863 = 50x45 = 2250 for
« 2585 mg, the amount to
anesthetize the dog properly.
41.
Taking t = 0 when the body was formed and t = T now, the amount Q(t) of
U in
the body at time t (in j^ears) is given by Q(t) - Qoe~kt, where k = (In 2)/(4.51xl09).
The given information tells us that
.- = 0.9.
Q,;,-Q(T)
After substituting Q{T) = Qoe~kT, we solve readily for ekT = 19/9, so
T = (1 /A:)ln(l 9/9) « 4.86x109. Thus the body was formed approximately 4.86 billion
j^ears ago.
43.
Because A = 0 the differential equation reduces to T = kT, so T(t) = 25e~kt. The
fact that 7(20) = 15 yields k = (l/20)ln(5/3), and finally we solve the equation
Section 1.4
19
5 = 25e kt
45.
(a)
for
t = (In 5)/k « 63 min.
The light intensity at a depth of x meters is given by I(x) = I0e ]Ax. We solve
the equation /(x) = IQe~]Ax = \I0 for x = (In2)/1.4 « 0.495 meters.
(b)
At depth 10 meters the intensity is/(10) = I0e~]Axl° «(8.3*2 x 10~7) /0.
(c)
We solve the equation /(x) = I0e~]Ax = 0.01/0 for x = (In 100)/1.4 « 3.29
meters.
47.
If N(t) denotes the number of people (in thousands) who have heard the rumor after t
days, then the initial value problem is
N' =*(100 -AO,
N(0) = 0
and we are given that N(7) = 10. When we separate variables (dN/(\00- N) = k dt)
and integrate, we get ln(100 - N) = - kt + C, and the initial condition N(0) = 0 gives
C = In 100. Then 100-N = 100e"4', so JV(O = 100(l-e-*'). We substitute t = 7,
N= 10 and solve for the value k = ln(l00/90)/7 « 0.01505. Finally, 50 thousand
people have heard the rumor after t = (\n2)/k& 46.05 days.
49.
The cake's temperature will be 100° after 66 min 40 sec; this problem is just like Example
6 in the text.
51.
(a)
-ki
A(t) = l5e~kl-, 10 = ^(5) = 15e“K', so
3
kl
- = ek;
2
,
1. 3
k = -In-.
5
2
Therefore
J
1/5
f3l
A/
(b) After 8 months we have
A( 8) = 15
(2\
8/5
K^J
(c)
20
1/5
-
II
t, 3)
A(t) = 15 exp --ln5 2
A(t) = 1 when
Chapter 1
7.84 su.
= 15* f2l
vT
Mil
A(t) = 15-
33.3944.
In®
Thus it will be safe to return after about 33.4 months.
53.
If L(t) denotes the number of Native America language families at time t (in years),
then L{t) = ek‘ for some constant k. The condition that Z(6000) = e6000k = 1.5 gives
1
3
k =-In — . If "now" corresponds to time t = T, then we are given that
6000 2
L(T) = ekI = 150,so T = — lnl50= ^001nl50 » 74146.48. This result suggests that the
k
ln(3/2)
ancestors of today's Native Americans first arrived in the western hemisphere about 74
thousand years ago.
55.
With A = /r(3)2 and a = tt(1 /12)2, and taking g = 32 ft/sec2, Equation (20)
reduces to 162/ = -sjy . The solution such that y = 9 when t = 0 is given b^
324 yjy - -t + 912. Hence y - 0 when t = 972 sec = 16 min 12 sec.
57.
The solution of / = -kjy is given by
2 >Jy = -kt + C.
The initial condition y(0) = h (the height of the cylinder) yields C = 2 yfh . Then
substitution of t = T, y = 0 gives k = (2 yfh )/T. It follows that
y = h(\-t/T)2.
If r denotes the radius of the cylinder, then
V(y) = nr2y = 7rr2A(l -r/T)2 = V0(l-t/T)2.
59.
(a)
Since x
= by, the cross-sectional area is ^4(jv) - nx
- nby. Hence the
equation ^(>0/ = -a-Jlgy reduces to the differential equation
ymy’ = -k = ~(a/xb)j2g
with the general solution
(2/3 )ym = -kt + C.
Section 1.4
21
The initial condition y(0) = 4 gives C = 16/3, and then y( 1) = 1 yields k = 14/3.
It follows that the depth at time t is
y(t)
=
(8
-
itf\
(b)
The tank is empty after t = 8/7 hr, that is, at 1:08:34 p.m.
(c)
We see above that k = (alnb)-^2g — 14/3. Substitution of a = nr2, b = 1,
g = (32)(3600)2 ft/hr2 fields r = (1/60)77/12 ft * 0.15 in for the radius of the
bottom-hole.
61.
A(y) = n(Sy-y ) as in Example 7 in the text, but now a =
/144 in Equation (24),
so the initial value problem is
18(8*-yV = -^y,
y(0) = 8.
We seek the value of t when y = 0. The answer is
63.
(a)
869 sec = 14 min 29 sec.
As in Example 8, the initial value problem is
n(%y ~y2)~y =
at
y{0) = 4
where k = 0.6r2j2g = 4.8 r2. Integrating and applying the initial condition just in
the Example 8 solution in the text, we find that
—//2-- y/2'
3
= ~kt+—.
5
15
When we substitute y = 2 (ft) and r = 1800 (sec, that is, 30 min), we find that
A: « 0.009469. Finally, y = 0 when
448
t = - « 3154 sec = 53 min 34 sec.
15 k
Thus the tank is empty at 1:53:34 pm.
(b)
The radius of the bottom-hole is
r = y/k/4.8 «0.04442 ft « 0.53 in, thus about a half inch.
22
Chapter 1
65.
Let t — 0 at the time of death. Then the solution of the initial value problem
T = £(70 - T),
T(0) = 98.6
is
T(t) = 70 + 28.6e“*'.
If t = a at 12 noon, then we know that
7X0 = 70 +28.6 e"*" = 80,
T(a + 1) = 70 + 28.6e"*l“+l) = 75.
Hence
28.6
= 10
and
28.6e'*V* = 5.
It follows that e~k = 1/2, so k = In 2. Finally the first of the previous two equations
yields
a = (In 2.86)/(ln 2) « 1.516 hr « 1 hr 31 min,
so the death occurred at 10:29 a.m.
67.
Let t = 0 when it began to snow, and t — to at 7:00 a.m. Let x denote distance along
the road, with x = 0 where the snowplow begins at 7:00 a.m. If y = ct is the snow
depth at time t, w is the width of the road, and v = dx/dt is the plow's velocity, then
"plowing at a constant rate" means that the product wyv is constant. Hence our
differential equation is of the form
j^dx _ 1
dt
t
The solution with x = 0 when t = to is
t
=
toe1*.
We are given that x = 4 when t = to + 1 and x = 7 when t = to + 2, so it follows
that
to + 1
= A)
and
/o + 2 =
at 8 a.m. and 9 a.m., respectively. To eliminate to we write
t0(e4k-1) = 1,
Section 1.4
= 2.
23
It follows that
elk -1 = 2(e4k -1), so
elk -2e4k +1 - 0.
Solving this equation numerically — using a calculator or Newton's method — we find
the single positive solution k = 0.08276. Using this value, we finally solve one of the
preceding pair of equations for to = 2.5483 hr « 2 hr 33 min. Thus it began to snow at
4:27 a.m.
69.
Substitution of v = dy I dx in the differential equation for y - y(x) gives
dv
adx
= VT + U
and separation of variables then yields
r
dv
dx
Vi + u
a
sinh
x
v = —+ Ci ’
a
dy_
- sinh
dx
fx
\a
A
+G
J
The fact that /(0) - 0 implies that C, = 0, so it follows that
dy_
y(x) = <2 cosh
sinh
dx
+ C.
\a)
Of course the (vertical) position of the x-axis can be adjusted so that C - 0, and the units
in which T and p are measured may be adjusted so that a - 1. In essence, then the
shape of the hanging cable is the hyperbolic cosine graph y - cosh x.
SECTION 1.5
LINEAR FIRST-ORDER EQUATIONS
1.
p = exp^ Jldxj = ex;
Dx (y • ex^j = 2ex\
y • ex =2ex +C;
y(x) = 2 + Ce~x
3^(0) = 0 implies C--2 so y(x) = 2-2e~x
3.
24
p - exp( j3Jxj = e3x;
Dx (3 • e3x) = 2x;
Chapter 1
y • eix = xz +C;
jy(x) = (x2+C)e~3x
5.
p = exp(^ J(2 / x)ckc^ = e2l"x = x2;
y(x) = x + C/x2;
7.
y(l) = 5
Dx (y-x2) = 3x2;
implies C-4
p = exV(\(\l2x)dx) = e,-"'x)l2 =4^-
so
y-x2=x3+C
y(x) = x + 4/x2
D,(yjZ) = 5;
y-^ = 5x + C
y(x) = 5 \[x +C / \Tx
9.
p = exp( J(-l/x)<&) = e~lnA = 1/x;
y{x) = xlnx + Cx;
11.
jy(l) = 0
p - expQld^j = ex\
implies C = 7
implies C- 0
jy(O) = 1
p = exp^j*2x£/xj = eA ;
so
^ -1/x = lnx + C
^(x) = xlnx + 7x
Dx (y • x<T3a) = 0;
Dx (y-ex^J -e2x\
= y-eA+Ce“A;
15.
-1/x) = 1/x;
p = expQ(1 /x -3)<ixj = e]nx~3x = xe~3x;
.y(x) = CxV";
13.
>>(1) = 7
Z)v
so
^-xe“3jc =C
jy(x) = 0 (constant)
y -ex = \e2x +C
implies C = y so
Dx(^y-ex^ = xe:
^(x) = \ex+\e~x
.y-<rv =±ex +C
2
y(x) = y + Ce A ;
17.
^(0) = -2 implies C = -f so .y(x) = j-fe x
p = exp( Jl/(1 + x)dxj = e'"(,+x> = 1 + x;
,(*) =
Dx(y-( l + x)) = cosx;
K0) = l implies C = 1 so y(x) = 1 + SinX
1+X
1+X
19.
p - exp(^ jcotxJxj = eln(smJf) = sinx;
y • sinx = ysin2 x + C;
21.
j-(l + x) = sinx + C
y(x) = ysinx + Ccscx
p - expQ(-3/x)<ixj = e~3lnx = x“3;
3
Dx (y • sinx) = sin x cos x
3
y(x) = x'sinx + Cx ;
_y(2^) = 0
Dx(y-x~3^j = cosx;
implies C = 0
Section 1.5
so
j;-x-3 = sinx + C
3
j/(x) = x' sinx
25
23.
p = exp( J(2 - 3/ x) dx] = e2^3'11" = x’V *;
_y-x“3e2* =2e2jr + C;
25.
£>, (>- • x'3e2x) = 4e2*
>>(*) = 2x3 + CxV2i
First we calculate
r o
3x 3 dx
3x -
x2 +1
3x
dx = — £x2 - ln(x2 + 1)J .
x" +1
It follows that p = (x2 +1)~3/2 exp(3x2 /2) and thence that
Dt(y(x2+l)"3/2exp(3x2/2)) = 6x(x2+4)-5/2,
y-(x2+1)”3/2 exp(3x212) = -2(x2+4 yV2+C,
y(x) = - 2 exp(3x2 / 2) + C (x2 +1)3/2 exp(-3x2 / 2).
Finally, X^) = 1 implies that C = 3 so the desired particular solution is
y(x) = -2exp(3x2/2) + 3(x2 +1)3/2 exp(-3x2/2).
27.
With x' = dx I dy, the differential equation is x' - x = j ey. Then with y as the
independent variable we calculate
p{y) = exp(J(-l)Jj) = e-3';
x ■ e~y = }y2+C;
29.
p = exp^ J(-2x)<ix j = e x ;
j/(x) =
x(y)
Dy[x-e~y) = y
= [\y2+C)ey
Dx[y - e x ^ = e x \
+ yV/r erf(x)j
dx
31.
(a)
= Ce -J' (-P) = -P^., so y'c+Pyc = 0.
P dx
(b)
/„ = (-P)e
-\Pdx
I dx + e 1
\Pdx
-Qe1
= -Py„+Q
\
26
Chapter 1
i
33.
The amount x(t) of salt (in kg) after t seconds satisfies the differential equation
x' = -x/200, so x(t) = 100c“'/20°. Hence we need only solve the equation
10 = 100c~//2°° for / = 461 sec = 7 min 41 sec (approximately).
35.
The only difference from the Example 4 solution in the textbook is that V = 1640 km3
and r — 410 km /yr for Lake Ontario, so the time required is
V
t = — In4 = 4 In 4 « 5.5452 years.
r
37.
The volume of brine in the tank after t minis V(t) = 100 + 2t gal, so the initial value
problem is
— = 5-——,
dt
100 + 21
x(0) = 50.
The integrating factor p(t) = (100 + 21)3/2 leads to the solution
50000
x(0 = (100 + 20
(100 + 2O372 ’
such that x(0) = 50. The tank is full after t = 150 min, at which time
x(150) = 393.75 1b.
The initial value problem
dx
x
x(0) = 100
dt
for Tank 1 has solution x{t) = 100^ //10. Then the initial value problem
+ =JL_Ji = !(),-""> _21,
dt
10
10
y{ 0) = o
10
for Tank 2 has solution y(f) = 10te //10.
(b)
The maximum value of y occurs when
y\t) = 10 e-'no-te-,no = 0
and thus when t=10. We find that ymax = y(10) = 100c_1 « 36.79 gal.
Section 1.5
27
41.
(a)
A\t) = 0.06/4 + 0.125* - 0.06/4 + 3.6e005'
(b)
The solution with A(0) = 0 is
A(t) = 360(e0Mt-e005 t\
so A(40) « 1308.283 thousand dollars.
43.
The solution of the initial value problem y' = x-y,
y(-5) = yQ is
;d» = x-\ + (y0+6)e~x~5.
Substituting x = 5, we therefore solve the equation 4 + (y0 + 6)e~w = y,
with^i = 3.998, 3.999, 4, 4.001, 4.002 for the desired initial values
yo = -50.0529, -28.0265, -6.0000, 16.0265, 38.0529, respectively.
45.
With the pollutant measured in millions of liters and the reservoir water in millions of
cubic meters, the inflow-outflow rate is r = j, the pollutant concentration in the inflow
is co =10, and the volume of the reservoir is V = 2. Substituting these values in the
equation x' = rc0 - (r / V)x, we get the equation
dx
1
— = 2-x
dt
10
for the amount x(t) of pollutant in the lake after t months. With the aid of the
integrating factor p = e//10, we readily find that the solution with x(0) = 0 is
x(t) = 20(l — e-"10).
Then we find that x = 10 when t = 10 In 2 « 6.93 months, and observe finally that, as
expected, x(t) -> 20 as t -> oo.
SECTION 16
SUBSTITUTION METHODS AND EXACT EQUATIONS
It is traditional for every elementary differential equations text to include the particular types of
equations that are found in this section. However, no one of them is vitally important solely in
its own right. Their main purpose (at this point in the course) is to familiarize students with the
technique of transforming a differential equation by substitution. The subsection on airplane
28
Chapter 1
flight trajectories (together with Problems 56-59) is included as an application, but is optional
material and may be omitted if the instructor desires.
The differential equations in Problems 1-15 are homogeneous, so we make the substitutions
y
v = —,
x
dy
dv
— = v + x—.
dx
dx
y = vx,
For each problem we give the differential equation in x, v(x),and v' = dv I dx that results,
together with the principal steps in its solution.
1.
/*
dv
xv' = 2 Vv;
ln(V + 2v-l) = -21nx + lnC
y2+2xy- x 2 = C
x2(V+2v-l) = C;
3.
2 (v +1) dv
— = -j2x<ix;
v2 + 2v
x(v + l)v' = — (v2 + 2v-l)
dx
Vv = lnx + C;
y = x(lnx + C)'
2 Vv
5.
x (v + l)v' = - 2v2;
%)
r
1 >
2 dx
+ 2 dv — — -;
J x
lv vj
fl
lny - lnx - — = -2 lnx + C;
ln(xy) = C + —
y
7.
y
XV2 V ' = 1;
[3v2 dv =
;
v3 = 3lnx + C;
y2 = x3(31nx + C)
—;
x
— = -lnx + C;
v
x = y(C-lnx)
1 — v
dx
J
v
r
xv' = v ;
j
11.
dv
v
r
x(l - v2) v' = v + v3;
dv -
x
V +V
lnv-ln(V+l) = lnx + lnC;
13.
xv ' =
7771
.
1
_
mv— = -2 lnx + C
v
;
J
r
1
2v ^
VV
V2 + ly
r
v = Cx(v2+l);
dv
f dx
Vv2 +1
x
J
x
y = C(x2+y2SJ
In |v + Vv2 + 1 j
Section 1.6
dx
dv =
In x + In C
29
v + yjv +1 — C x\
y + -\fx + y~ — C x2
r
15.
x(v + l)v' = -2(V + 2v)
/*
2 (v +1 )dv
4<ix
;
ln(V+2v) = -41nx + lnC
v2 + 2v
v^+2v = C/x4;
x2_y2+2x3j/= C
r
17.
v = 4x + y;
v' = v2 + 4;
v = 2tan(2x-C);
x
1
—- = —tan
v2 + 4
2
C
— +—
2
2
_J V
y = 2tan(2x -C) -4x
Problems 19-25 are Bernoulli equations. For each, we indicate the appropriate substitution as
specified in Equation (10) of this section, the resulting linear differential equation in v, its
integrating factor p, and finally the resulting solution of the original Bernoulli equation.
19.
v = y~2;
v'-4v/x = -10/x2;
p = l/x4;
21.
v = y~2;
v' + 2v = -2;
23.
v = _y-1/3;
v'-2v/x = -1;
25.
v = y3;
v' + 3v/x = 3/Vl + x4;
27.
The substitution v = y3 yields the linear equation xv'-v = 3x4 with integrating
factor p = 1/x. Solution: y = (x4 + Cx)1/3
29.
The substitution v = siny yields the homogeneous equation 2xvv' = 4x +v .
2
2
Solution: sin y = 4x - C x
p = e2x;
y2 = xl{Cx5 + 2)
y2 = \/[Ce~2x-l)
p-x~2\
y - (x + Cx2)
p = xJ;
y3 =
+ 3\/l + x4 j/(2x3 )
Each of the differential equations in Problems 31-42 is of the form M dx + N dy = 0, and the
exactness condition dM! dy = dN / dx is routine to verify. For each problem we give the
principal steps in the calculation corresponding to the method of Example 9 in this section.
31.
30
F = J(2x + 3j>)dx = x2+3 xy + g(y);
Chapter 1
Fv = 2x + g'(y) = 3x + 2.y = N
g\y) = 2y;
33.
g(y) = y2;
F = J(3x2 + 2y2)dx = x*+xy2+g(y);
gXy) = 6/;
35.
g(j/) - 2/;
jx3 + |/ + ^inx - c
g(y) =
Fy = x/y + g\y) = x/y+ey= N
sinx + xlnj^ + e^ = C
F = J(3xy+/)*& = xV + x / + g(y)-,
=
3x3y2+4xyi + g'(y) = 3x3y2 + y4 +4xy3 = N
gXy) =/;
/•
F =
J
gW = I/;
r 2x
3_y
2T
t+
x
t
gXy)
2\
V T
X
43.
Fy = \nx + g\y) = y2+lnx = N
F = J(cosx + ln>y)<ix = sinx + xlny + g(y);
Fy
41.
x3+2x/+2/ = C
g(y) = i/;
gXy) = ey\
39.
Fy = 4 xy + g'(y) = 4xy+6j>2 = TV
F = JO3 + y/x)dx = }x4 + y\nx + g(y);
gXy) = y2\
37.
x2+3xy + y2 = c
xy+x/+iy = c
2
2
dx = — + ^T + g(>;);
T
x*
7
,( X
+ & (t)
x2
2y
1
= TV
— + —+
T
x
Vr
g(T) = 2,/y;
X
J/
— + —t+2Vt
T
x
= c
The substitution y = p, y = p' in xy = y yields
(separable)
xp = p.
dp
f dx
J p
J x
In p = Inx + lnC,
y = p = Cx,
y(x) = yCx2 4- T? — ytx +
45.
The substitution y = p, y = p p' = p(dp/ dy) in the equation y + 4y = 0 yields
Section 1.6
31
pp’ + Ay = 0,
(separable)
\pdp = - |4ydy
=>
^-p2 = -2j2+C,
p2 = -4/+2C = 4 (|C-/),
1
dy
1 . _i T ~
dy =
= —sin —hZ).
2 --2
2
A:
P '
J 2yfkl-yy(x) = ksin[2x - 2D] = &(sin2xcos2D-cos2xsin2D),
y(x) = ^cos2x + 5sin2x.
2
47.
The substitution y' = p, y = p' in y = (j/)“ yields
(separable)
P = P\
r dp
2
= j*x<Zx
j P
x + 5,
P
dy_ =
dx
x + i?
y(x) = v4-ln x + yt|.
2
49.
The substitution y = p, y" = p p' = p{dpl dy) in j>y + (y) = xy' yields
yp' + p = y
ypp' + p2 = yp
(linear in p),
Dy[yp] = y,
1 2 +—C
yp - —y
2
2
/*
x =
dy =
p
y2 +C
=>
P =
2 ydy
= ln(/+C)-ln£,
y +C
1/2
y2+C = Be;
51.
y(x) = ±(A + Bex)
The substitution y = p, y" = p p’ = p(dp/ dy) in /' = 2.y(/) yields
= jXc/y
/>/?' =
J D2
^ 1
X
=
dy = -~y -Cx + D,
J P
3
y + 3x + 4y + -# = 0
32
Chapter 1
= T2+C,
p
53.
The substitution y' = p, y" = pp' = p(dpldy) in y" = 2yy' yields
'-iy,
P
X
jl ydy
jdp =
PP' = 2yp
p = y2 +A2,
U2
2 = —7 tan 1 — + C,
A
y2+A
J
tan 1 — = A(x - C)
— = tan (Ax-AC),
A
y(x) = Atan(Ax + B).
55.
The substitution v = ax + by + c, y = (v - ax- c)l b in y' = F(ax + by + c) yields the
separable differential equation (dv/dx-a)!b = F(v), that is, dv/dx = a + bF(v).
57.
If v = lny then y = ev so y' = eV. Hence the given equation transforms to
eV + P(x) ev = £?(x) vev. Cancellation of the factor e then yields the linear
differential equation vf-Q(x)v = P(x).
59.
The substitution x = u- 1, y = v - 2 yields the homogeneous equation
dv
u-v
du
u+v
The substitution v = pu leads to
(P +1) dp
In u =
In yp2 + 2/>-l)-lnC
(p2+2p-l)
We thus obtain the implicit solution
u2[p2+2p-\) = C
,.2
u
V
—— +
u
..
~v
2
—
u
\
1
=
+ 2 uv-u
= C
J
(y + 2)2 + 2(x + l)(y + 2) - (x +1)2 = C
y2 + 2xy - xz + 2x + 6y = C.
61.
The substitution v = x-y yields the separable equation v' = 1 - sin v. With the aid
of the identity
1
1 + sinv
2
- _ -— = sec v + sec v tan v
1-sinv
cos v
Section 1.6
33
we obtain the solution
x = tan(x - y) + sec(x - y) + C.
63.
If we substitute y = yx+ 1/v, y' = y[-v'/v2 (primes denoting differentiation with
respect to x) into the Riccati equation y' = y4y + By + C and use the fact that
y[ =
+ By] + C, then we immediately get the linear differential equation
v' + (B + 2Ayt) v = —^4.
In Problem 65 we outline the application of the method of Problem 63 to the given
Riccati equation.
65.
The substitution y = x +1 / v yields the trivial linear equation v' = -1 with immediate
solution v(x) = C - x. Hence the general solution of our Riccati equation is given by
j(x) = x + l/(C-x).
67.
Clearly the line y = Cx - C2/4 and the tangent line at (C/2, C2/4) to the parabola
= x both have slope C.
69.
With a = 100 and k = 1/10, Equation (19) in the text is
y = 50[(x/100)9/i° - (jc/100)1 l/10].
The equation y'(x) = 0 then yields
(jc/ 100)1/10 = (9/1 l)l/2,
so it follows that
Jmax = 50[(9/l 1)9/2 - (9/1 l)ll/2] * 3.68 mi.
71.
(a)
With a = 100 and k = w/v0 = 2/4 = 1/2, the solution given by equation (19) in
the textbook is y(x) = 50[(x/100)1/2 - (x/100)3/2]. The fact that jy(0) = 0 means that
this trajectory goes through the origin where the tree is located.
(b)
With k = 4/4 = 1 the solution is _y(x) = 50[1 - (x/100)2] and we see that the
swimmer hits the bank at a distance y(0) = 50 north of the tree.
(c)
With k = 6/4 = 1 the solution is y(x) - 50[(x/l00)_1/2 — (x/100)572]. This
trajectory is asymptotic to the positive x-axis, so we see that the swimmer never reaches
the west bank of the river.
34
Chapter 1
CHAPTER 1 Review Problems
The main objective of this set of review problems is practice in the identification of the different
types of first-order differential equations discussed in this chapter. In each of Problems 1—36 we
identify the type of the given equation and indicate an appropriate method of solution.
I.
If we write the equation in the form y' - (3 / x)y = x2 we see that it is linear with
integrating factor p = x~\ The method of Section 1.5 then yields the general solution
y = x3(C + In x).
3.
This equation is homogeneous. The substitution y = vx of Equation (8) in Section 1.6
leads to the general solution y = x/(C - In x).
5.
We write this equation in the separable form y' / y2 = (2x - 3) / x4. Then separation
of variables and integration as in Section 1.4 yields the general solution
y = C exp[(l -x)/x3].
7.
If we write the equation in the form y' + (2 / x)y = 1 / x' we see that it is linear with
integrating factor p - x . The method of Section 1.5 then yields the general solution
y = x~2(C + lnx).
9.
If we write the equation in the form y' + (2 / x)y = 6xjy we see that it is a Bernoulli
equation with n = 1/2. The substitution v = y~
2
I/O
of Eq. (10) in Section 1.6 then
2
yields the general solution y = (x + C/x) .
II.
This equation is homogeneous. The substitution y = vx of Equation (8) in Section 1.6
leads to the general solution y = x / (C - 3 In x).
13.
We write this equation in the separable form y' / y2 = 5x4 - 4x. Then separation
of variables and integration as in Section 1.4 yields the general solution
y = 1 / (C + 2x2 - x5).
15.
This is a linear differential equation with integrating factor p =eix. The method of
Section 1.5 yields the general solution
17.
y = (x 3 + Qe~ 3xx.
We note that Dy[ex + yexy>) = Dx(ey +xexy^ = exy+xyexy, so the given equation is
exact. The method of Example 9 in Section 1.6 yields the implicit general solution
e* +
= C.
Review Problems
35
19.
We write this equation in the separable form y' / y2 = (2- 3x5) / x3. Then separation
of variables and integration as in Section 1.4 yields the general solution
y = x2 / (x5 + Cx2 +1).
21.
If we write the equation in the form y' + (l /(x +1)) y = 1 /(x2 -1) we see that it is linear
with integrating factor p - x +1. The method of Section then 1.5 yields the general
solution y = [C + ln(x - 1)] / (x + 1).
23.
We note that Dy(ey+ycosx) = Dx(xey +sinx) = e^+cosx, so the given equation
is exact. The method of Example 9 in Section 1.6 yields the implicit general solution
x ey + y sin x = C
25.
If we write the equation in the form yf + (2 /(x +1)) y = 3 we see that it is linear with
integrating factor p = (x +1)“. The method of Section 1.5 then yields the general
solution y = x + 1 + C (x + l)-2.
27.
If we write the equation in the form y' + (1 / x)y = - x2y4 / 3 we see that it is a Bernoulli
equation with n = 4. The substitution v =
of Eq. (10) in Section 1.6 then yields
the general solution y = x (C + In x)
29.
If we write the equation in the form j/ + (l/(2x + l))y = (2x + l)1/2 we see that it is
linear with integrating factor p = (2x +1)
1/2
. The method of Section 1.5 then yields
the general solution^ = (x2 + x + C)(2x + 1)_1/2.
31.
dy/(y + l) = 3x dx is separable; y' + 3xy = 2 lx is linear.
33.
(3x2 + 2y2) dx + 4xy dy = 0 is exact; y' = -}(3x/y + 2y/x) is homogeneous.
35.
dy/(y + l) = 2x dx/(x2 +1) is separable; y' -(2x/(x2 +l))y = 2x/(x2 +1) is linear.
36
Chapter 1
CHAPTER 2
MATHEMATICAL MODELS
AND NUMERICAL METHODS
SECTION 2.1
POPULATION MODELS
Section 2.1 introduces the first of the two major classes of mathematical models studied in the
textbook, and is a prerequisite to the discussion of equilibrium solutions and stability in Section 2.2.
In Problems 1-8 we outline the derivation of the desired particular solution, and then sketch some
typical solution curves.
1.
Noting that x>l because x(0) = 2, we write
rr
dx
dx - Jl dt
= Jl dt\
vx
x(l - x)
x -1
lnx-ln(x-l) = t + \nC;
=
Ce'
x -1
x(0) = 2
implies
C - 2;
x = 2(x-l)e'
2e‘
x(0 =
2ef -1
2-e
-i
t
Section 2.1
37
3.
Noting that x > 1 because x(0) = 3, we write
r
dx
(1 + x)(l - x)
= J*1 dt;
1
1
V x -1
x +1
f
x -1
ln(x-l)-ln(x + l) = -2r + lnC;
^
J(—2) dt
dx =
2/
-
= Ce
x +1
x(0) = 3
x(0 =
implies
2+e
2-e
-21
-it
C = —;
2(x-l) = (x + \)e
2
2/
-
2e2' +1
2e2' -1
t
5.
Noting that x > 5 because x(0) = 8, we write
dx
x(x - 5)
dx
= J(-3 )dt-
x-5 j
lnx-ln(x-5) = 15r + lnC;
—
=
Ce'5'
x-5
x(0) = 8
x(t) =
implies
-40 e
C = 8/3;
15/
40
15/
8-3e
3-8e
3x = 8(x-5)e15'
-15/
Typical solution curves are shown in the figure at the top of the next page.
38
Chapter 2
7.
Noting that x > 7 because x(0) = 11, we write
r
dx
x(x -7)
= J(-4 )dtJ
fl
1
lx
X - 7y
X
lnx-ln(x-7) = 28/ + lnC;
^
dx =
= Ce
[28 dt
28/
x-7
x(0) = ll
x(0 =
implies
C7 = 11/4;
-77 e28'
77
28/
11 — 4 e
4 —lie
4x = 1 l(x -17)e28/
28/
-
Section 2.1
39
9.
Substitution of P(0) = 100 and P'(0) = 20 into P'= k\[P yields k = 2, so the
differential equation is P' = 2 yfp. Separation of variables and integration,
jdP/2y[P = j*dt, gives J~P = t + C. Then P(0)=100 implies C=10, so
P(t) = (r+10)2. Hence the number of rabbits after one year is P(12) = 484.
11.
(a)
Starting with dP / dt = kjp]
dPIdt =
we separate the variables and
integrate to get P(f) = (A//2 + C)2. Clearly P(0) = Po implies C =
(b)
If P(t) = (kt/2 + 10)2, then P(6) = 169 implies that k = 1. Hence
P(f) = (/V2+10)2, so there are 256 fish after 12 months.
13.
(a)
If the birth and death rates both are proportional to P and J3 > S, then Eq. (1) in
this section gives P' -kP
with k positive. Separating variables and integrating as in
Problem 12, we find that P(t) = 1 /(C - kt). The initial condition P(0) = P0 then gives
C = 1 / P0, so P{t) = 1/(1 /P0-kt) = P0/(l-kP0t).
(b)
If 7^= 6 then P(t) = 6/(1 -6kt). Now the fact that P(10) = 9 implies that
k= 180, so P(t) = 6 /(I — / / 30) = 180/(30-0- Hence it is clear that
P —> oo as r -» 30 (doomsday).
15.
If we write P’ = b P(a / b- P) we see that M = a//?. Hence
M = (aW = — = M
A>
bP‘1
b
Note also (for Problems 16 and 17) that a = B(> / P(l and b = D0/ P^ = k.
17.
The relations in Problem 15 give k = 1/2400 and M= 180. The solution is
P{t) = 43200/(240-60e“3//8°). We find that P = 1.05M after about 44.22 months.
19.
If we write P' = a P{P -bl a) we see that M = b/a, so
DqPq = (KW = b = M
B0
aP(:
a
as in Problem 18. Note also that b = D0/P0 and a = B0/P0
= k. These relations give
k = 1/1000 and M= 90. The solution is P(t) = 9000/(100-10e9'/10°). We find that
P = 10M after about 24.41 months.
40
Chapter 2
21.
Starting with the differential equation dP / dt = AP(200 - P), we separate variables and
integrate, noting that P < 200 because P0 = 100:
r
dP
P(200-P)
/»
=
dt
(1
1
+
yP
200- P y
P
In- = 200Ay + lnC
200-P
3
200-P
dP -
J200k dt;
Cc200*'
Now P(0) = 100 gives C-1, and P'(0) = 1 implies that 1 = k -100(200 -100), so
we find that k- 1/10000. Substitution of these numerical values gives
P
e
200//10000
200-P
and we solve readily for P{t) = 200/(l + e //5°). Finally, P(60) = 200/(l + e 6/5153.7
million.
23.
(a)
x' = 0.8x-0.004x2 = 0.004x(200 - x), so the maximum amount that will dissolve
is M = 200 g.
(b)
With M = 200, Po = 50, and k = 0.004, Equation (4) in the text yields the
solution
, x
X(0 =
10000
50 +150 e
0.08/ *
-
Substituting x = 100 on the left, we solve for t = 1.25 In 3 « 1.37 sec.
25.
Proceeding as in Example 3 in the text, we solve the equations
25.00k(M-25.00) = 3/8,
47.54£(M-47.54) = 1/2
for M = 100 and k = 0.0002. Then Equation (4) gives the population function
2500
Pit) =
25 +75c
-0.02/
We find that P = 75 when t = 50 In 9 « 110, that is, in 2035 A. D.
27.
We are given that
F = kP^ - 0.0IP.
Section 2.1
41
When we substitute P(0) = 200 and P'(0) = 2 we find that k = 0.0001, so
dP_
dt
= 0.0001P2-0.01P = 0.0001 P(P-100).
We separate variables and integrate, noting that P > 100 because P0 = 200:
r
dP
P(P-100)
In
P -100 _
P
(
= Jo.oooi 4*
100
1
VP-100
P-100
■t + InC
= Ce
n dP = Jo.01 dt;
P)
i/m
Now P(0) = 100 gives C = l/2, hence
2(/>-100) = P e"m
29.
=;>
P(t) = -22^
2—e
(a)
P = 1000 when t = 1001n(9/5) « 58.78.
(b)
P —> oo as ^->100 In 2 « 69.31.
Here we have the logistic equation
dP
= 0.03135P-0.0001489P = 0.0001489P(210.544-P)
dt
where k = 0.0001489 and P = 210.544. With P0 = 3.9 also, Eq. (7) in the text gives
p(0 (a)
(210.544)(3.9)
821.122
(3.9) +(210.544-3.9)e"(0000H89)<210'544)'
3.9 + 206.644e 0 03135
- .
/
This solution gives P(140) «127.008, fairly close to the actual 1930 U.S. census
population of 123.2 million.
(b)
The limiting population as t-> oo is 821.122/3.9 = 210.544 million.
(c)
Since the actual U.S. population in 200 was about 281 million — already exceeding
the maximum population predicted by the logistic equation — we see that that this model
did not continue to hold throughout the 20th century.
31.
If we substitute P(0) = 106 and P'(0) = 3xl05 into the differential equation
no =
42
fo~a'p>
Chapter 2
we find that /% - 0.3. Hence the solution given in Problem 30 is
P(t) = P0Qxp[(0.3/a)(\-e-al)\.
The fact that P(6) = 2Pq now yields the equation
/(a) = (0.3)(1 - e”60) - aln 2 = 0
for a. We apply Newton's iterative formula
1.8 c 6" - In 2 and initial guess oco = 1, and find that a « 0.3915.
with
Therefore the limiting cell population as / -» oo is
P0 exp(/?0 la) = 106 exp(0.3/0.3915) « 2.15 xlO6.
Thus the tumor does not grow much further after 6 months.
33.
(a)
We separate the variables in the extinction-explosion equation and use absolute
values to allow for both possibilities P0 < M and P0> M :
r
J
C(
dP
V P-M
P(P-M)
dP =
dt;
kMt + In C
If P0 < M then P <M and P - M\ = M - P, so substitution of t = 0, P = P0 in (*)
gives C = (M - P0) / P0. It follows that
M-P
P
M-P0 pkMt
c
P
* f\
But if P0> M then P> M and \P - M\ - P - M , so substitution of / = 0, P = P0 m
(*) gives C = (P0- M) / P0, and it follows that
P-M _ P0- M ^kMf
c
•
Po
P
1 r
We see that the preceding two equations are equivalent, and either yields
Section 2.1
43
MP0
P.+iM-fye™1'
(P-M)P0 = (P0 - M)P e kMi
(b)
If P„ < M then the coefficient M - P0 is positive and the denominator increases
without bound, so P{t) -> 0 as t —» oo. But if P0 > M, then the denominator
P0 - (P0 - M) ekMl approaches zero — so P(t) -> +oo — as t approaches the value
(1 /kM)\n([P0 /(P0 - M)\ > 0 from the left.
Any way you look at it, you should see that, the larger the parameter k > 0 is, the faster the
logistic population P{t) approaches its limiting population M.
The formulas referred to in the statement of Problem 36 are
M = P.jiP^-p^-PA)
PA-P"
and
,
* = —Llnpo(M-pi).
Mt
P\{M - P0)
Substitution of the given population values in these formulas gives
k - 0.0000668717 and M = 338.027, so P(t)
_25761.7_
76.212 + 261.815 ^-° 0226045'
The latter formula predicts that P = 192.525 in the year 2000.
We readily separate the variables and integrate:
k + bcos27rt)dt
In P
kt h-sin 2nt + In C.
2n
(
Clearly C = P0, so we find that P(t) = P0 exp kt H-sin2;r/t . The oscillatory curve
\
27T
J
in the figure at the top of the next page shows the graph that results with the typical
numerical values P0 = 100, k- 0.03, and b = 0.06. It oscillates about a non-oscillatory
curve which represents natural growth with P0 and k = 0.03. We see that the two agree
at the end of each full year.
Chapter 2
p
120
ii
SECTION 2.2
EQUILIBRIUM SOLUTIONS AND STABILITY
In Problems 1-12 we identify the stable and unstable critical points as well as the funnels and spouts
along the equilibrium solutions. In each problem the indicated solution satisfying x(0) = xo is
derived fairly routinely by separation of variables. In some cases, various signs in the solution
depend on the initial value, and we give a typical solution. For each problem we show typical
solution curves corresponding to different values of x0.
t
1.
Unstable critical point: x = 4
Spout: Along the equilibrium solution x(t) = 4
Solution: If x0 > 4 then
Section 2.1
45
dx
j
x-4
- |dt;
ln(x-4) = t + C;
x-4 = (x0-4)U;
C - ln(x0-4)
x(7) = 4 + (x0-4)e'.
Typical solution curves are shown in the figure above (on the previous page).
Stable critical point: x = 0
Unstable critical point: x = 4
Funnel: Along the equilibrium solution x(t) = 0
Spout: Along the equilibrium solution x(t) = 4
Solution: If xn > 4 then
j*4 dt =
r
4 dx
x(x - 4)
J
.
„
. x-4
4 t + C = In-;
x
41 = In
x(t) =
f
n dx
l
vx -4
X)
_
. xn -4
C = In-2-
X.
x0(x-4)_
x(x0 - 4)’
x0(x-4)
^4,
x(x0 -4)
4 Xr
x0+(4-x0)e 4
/
Typical solution curves are shown in the figure below.
0
12
3
t
Stable critical point: x = -2
Unstable critical point: x = 2
Chapter 2
4
5
Funnel: Along the equilibrium solution x(t) = -2
Spout: Along the equilibrium solution x(7) = 2
Solution: If x0> 2 then
"4 dx
II
>
x2 -4 ~
J
4 t + C = In———;
x+2
f
i
1
kx-2
)
<ix
x+ 2j
C = ln^-2
Xq + 2
(x -2)(x0 + 2)
4r = In (X ~ 2)(X° + 2);
(x + 2)(x0 -2)
(x + 2)(x0 - 2)
(x0 +2) + (x0 -2)e4'
(x0 +2)-(x0 -2)e4'
Typical solution curves are shown in the figure below.
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
t
7.
Critical point: x = 2
This single critical point is semi-stable, meaning that solutions with xo > 2 go to infinity
as t increases, while solutions with xo < 2 approach 2.
Solution:
If x0 > 2 then
r
J
-dx
(x-2 y
r
(-1 )dt\
= -r + C;
x
Section 2.2
C =
1
x0-2
47
1
1
—t+
x -2
*o
/x
^
x(7) = 2+
1 - t(x0 - 2)
x0-2
2
x0(2/-l)-4f
xn — 2
0
l-/t(x0-2)
/Lx0-2/-l
Typical solution curves are shown in the figure below.
-i_I_i_I_i_i_i_
0
1
2
3
4
5
t
9.
Stable critical point: x = 1
Unstable critical point: x = 4
Funnel: Along the equilibrium solution x(t) = 1
Spout: Along the equilibrium solution x(t) = 4
Solution:
If x0 > 4 then
r
i
—>
.
^
. x-4
3 t + C = In-;
x -1
„
.xn-4
C = In-5x0 -1
l
■>
II
mcn
3 dx
I
1
31 = ln (x ~ 4)(yo
kx-4
=
(x-l)(x0 -4) ’
i
X
^
1y
(x~4)(x0-l)
(x-l)(x0—4)
= 4(1 -x0) + (x0 -4)e3'
(l_x0) + (x0-4)e3'
Typical solution curves are shown in the figure at the top of the next page.
48
Chapter 2
0
1
2
3
4
5
t
11.
Unstable critical point: x = 1
Spout: Along the equilibrium solution x(t) = 1
Solution:
—2 dx
r
-- - K-2)dt,
(x-1)3
•>
1
— — 21 +
(x-iy
1
(X„ - 1)‘
Typical solution curves are shown in the figure below.
In each of Problems 13 through 18 we present the figure showing a slope field and typical solution
curves, and then record the visually apparent classification of critical points for the given differential
equation.
Section 2.2
49
The critical points x- 2 and x =-2 are both unstable. A slope field and typical solution
curves of the differential equation are shown below.
x' = (x + 2) (x - 2)'
[1—1—1—I—ri—i—i—np—i—i—np—i—r
III Mil Mil III
1 1 II1 II III M Ml1 1 II
IIII 1 1 1 1 III! 1 M
/ / /
// / / /
/
/
/
/
/
/
/
/
/
/
/
/ // /
/
/
/ 1 /
Ij' 1 /
// 1 /
/
/ /
/ /
/ / // / /
/ ll' 1 / /
/ 1 1 / /
/
I / /
/ / /
/ / /
/
/
/
/ / /
/,(l / / II
1 1 1 /
1 1 / 1
/ I / /, 1
Z.
MM
\ \ M
MM MM
MM
II
II
II
II
II
II
II
IIM
II
j_i
II
II II
i i l_1_1_L
II II
1 1_1_L 1
\\
I
I
I
\
\
\
\
\
\
II
\
\
\
I
1
1
III!
I
1
'
1
1-L
The critical points x = 2 and x =-2 are both unstable. A slope field and typical solution
curves of the differential equation are shown below.
x' = (x2 - 4) 2
n-1-nn-1-1-1
n—r
I
I
I II
I
I
I
I
I
I
I
* / / J/ / /
X 0
MM
MM
II II
II II
f / / /
IIII IIII
II II II II
II II II II
Mil II II
iiii
III! _J_1_1_L -
f
-4 -
/
/
/
/
7
Chapter 2
7
7
17.
The critical points x = 2 and x = 0 are unstable, while the critical point x = -2 is stable.
A slope field and typical solution curves of the differential equation are shown below.
x' = x2 (x2 - 4)
19.
The critical points of the given differential equation are the roots of the quadratic equation
jQx(\0-x)-h = 0,
that is,
x2-10x + 10h = 0.
Thus a critical point c is given in terms of h by
c = io±Vkk>3^ = s±7___
2
It follows that there is no critical point if h> 2j, only the single critical point c = 0 if
h = 2j, and two distinct critical points if h < 2j (so 10 - 25h > 0). Hence the bifurcation
diagram in the he-plane is the parabola with the (c-5) =25-10h that is obtained upon
squaring to eliminate the square root above.
21.
(a)
If k=-o' where a > 0, then kx-x3 = -a2x-x3 = -x(a2+x2) is 0 only if
x = 0, so the only critical point is c- 0. If a > 0 then we can solve the differential
equation by writing
a ax
x(a2 +x2)
U
a2 h
Section 2.2
51
lnx -^Tn(a2 + x2) =
x
2
+X
2
= Ce
a2t + — InC
a2Ce-2a2‘
-2 a2/
X2
=
l-Ce
-2a2/
It follows that x -> 0 as t —> 0, so the critical point c = 0 is stable.
(b)
If k = +a2 where a > 0 then Joc-x3 = + a2x - x3 = — x(x + a)(x - a) is
0 if either x = 0 or x = ±a = ±V&. Thus we have the three critical points c = 0, ±
,
and this observation together with part (a) yields the pitchfork bifurcation diagram shown in
Fig. 2.2.13 of the textbook. If x * 0 then we can solve the differential equation by writing
r
2 a2 dx
r(
x(x -a)(x + a)
V
-2 lnx + ln(x
~/2
=
1 +1 ^
—
dx =
2
— +
x
x-a
|2a2 dt,
x + aJ
a) + ln(x-a) = -2a t + \nC,
c
1 -Ce
It follows that if x(0) ^ 0 then x -»
+y[k
a
x*
^2"2'
X =
-2a2/
if x > 0, x —>
VI-
-2 a2/
if x < 0. This implies that
the critical point c - 0 is unstable, while the critical points c = ±V^ are stable.
23.
(a)
If h < kM then the differential equation is x' = kx((M - hi k)-x), which is a
logistic equation with the reduced limiting population M - h / k.
(b)
If h > kM then the differential equation can be rewritten in the form
f
2
x' = -ax-bx~ with a and b both positive. The solution of this equation is
a xQ
(a + bx0)eal -bx0
so it is obvious that x(t) —» 0
25.
(i)
as t -> go t
In the first alternative form that is given, all of the coefficients within parentheses are
positive if H< xo < N. Hence it is obvious that x(t) —> N
as / —> oo .
(ii)
In the second alternative form that is given, all of the coefficients within parentheses
are positive if xo < H. Hence the denominator is initially equal to N - H> 0, but decreases
as t increases, and reaches the value 0 when
52
Chapter 2
1
, N - x0
In-- > 0.
N-H H-xr
t
27.
Separation of variables in the differential equation x' = - k ((x - a)2 + b2) yields
(
n-x
^
x(t) = a-btan bkt + tan 1-J
V
It therefore follows that x{f) goes to minus infinity in a finite period of time.
29.
This is simply a matter of analyzing the signs of x' in the cases x< a, a <x<b, b <x <c,
and c > x. Alternatively, plot slope fields and typical solution curves for the two differential
equations using typical numerical values such as a =-1, b = 1, c = 2.
SECTION 2.3
ACCELERATION-VELOCITY MODELS
This section consists of three essentially independent subsections that can be studied separately:
resistance proportional to velocity, resistance proportional to velocity-squared, and inverse-square
gravitational acceleration.
1.
Equation:
Solution:
v' = *(250-v),
(-l)dv
v(0) = 0,
- jkdt;
v(10) = 100
ln(250-v) = -kt + lnC.
250-v
v(0) = 0 implies C = 250;
v(t) = 250(1-e-*)
v(10)=100 implies * = ^-ln(250/150) « 0.0511;
3.
Answer:
v = 200 when t- -(In50/250)/A: « 31.5 sec
Equation:
v' = -kv,
Solution:
v(t) = 40 e~kt with k = (1/10)ln 2
v(0) = 40;
v(10) = 20
x' = v,
x(0) = 0
x(t) = (40/k)(\-e^‘)
Answer:
x(oo)
= lim(40/£)(l-e~*') = 40/k = 400/ln2 * 577 ft
/—> OO
Section 2.2
53
5.
Equation:
v’ = -kv9
Solution:
v =
v(0) = 40;
40
v(10) = 20
x' - v,
x(0) = 0
(as in Problem 3)
1 + 40 kt
v(10) = 20 implies 40& = 1/10, so
v(t)
400
10 + t
x(t) = 4001n[(10 + 0/10]
7.
Answer:
x(60) = 400 In 7 « 778 ft
Equation:
v' - 10-0.lv,
r
(a)
x(0) = v(0) = 0
-0.1 dv
— = f(-0.1 )dt;
J
10-0 .1 v
ln(10-0.lv) - -r/10 + lnC
v(0) = 0 implies C = 10;
v(0 - 100(l-e , lu);
(b)
ln[(10-0.1v)/10] = -t! 10
v(oo)
= 100 ft/sec (limiting velocity)
x(t) = 100/ — 1000(1 — e-"10)
v = 90 ft/sec when t = 23.0259 sec and x = 1402.59 ft
9.
The solution of the initial value problem
1000 v' - 5000- 100 v,
v(0) - 0
is
v(t) = 50(1 -e_'/l0).
Hence, as t —> oo, we see that v(/) approaches vmax = 50 ft/sec « 34 mph.
11.
If the paratrooper's terminal velocity was 100 mph = 440/3 ft/sec, then Equation (7) in
the text yields p- 12/55. Then we find by solving Equation (9) numerically with
yo = 1200 and vo = 0 that y - 0 when t « 12.5 sec. Thus the newspaper account is
inaccurate.
Given the hints and integrals provided in the text, Problems 13-16 are fairly straightforward (and
fairly tedious) integration problems.
•*>
17.
To solve the initial value problem v' = - 9.8 - 0.0011 v ,
dv
0.010595 dv
TT = - \dt:
9.8 + 0.001 1V
54
J
1+ (0.010595 v):
Chapter 2
v(0) = 49 we write
= - JO. 103827 dt
tan"1 (0.010595v) = -0.103827/ + C;
v(0) = 49 implies C = 0.478854
v(t) = 94.3841 tan(0.478854-0.103827/)
Integration with jr(0) = 0 gives
y(t) = 108.468 + 909.052 ln(cos(0.478854 - 0.1038270) •
We solve v(0) = 0 for ^ = 4.612, and then calculate y(4.612) = 108.468.
19.
Equation:
v' = 4-(l/400)v2,
Solution:
f
f (1/40) civ
dv
_ u5
„ 4-(l/400)v2
'
tanh ](v/40) = /710 + C;
21.
v(0) = 0
C = 0;
Answer:
v(10) «30.46 ft/sec,
Equation:
v = -g-pv2,
_
f
J 1 -(v/40)2 “ J 10
dt
v(t) = 40tanh(/710)
v(oo) = 40 ft/sec
v(0) = v0,
y(0) = 0
r
dv
Solution:
sjplg dv
- |dt;
g + pv‘
1 + (•Jp/gv)
tan"1 [gpTg v) = -Jgpt + C;
v(0) = v0 implies C = tan'1 {gplg v0)
f
v(0 = -
- jjgpdt;
f
IP
— tan pgP “tan"1
V°\
IP
\ ]Is) J
We solve v(7) = 0 for t =
JL_- tan 1
Jgp
and substitute in Eq. (17) for y(t)\
vo,
g
V
cos(tan"1 v0 Jp/g - tan"1 v0 Jp/g
y
max
y max
—In
P
cos (tan"1 v0yfp7g
— In (sec ( tan-1 voyJp/g
P V
V
2^
(
Pv0
In 1 +
2P
{
g )
= —In , ll +
g
P
Section 2.3
55
23.
Before opening parachute:
v' = -32 +0.00075v2,
v(0) = 0,
y(0) = 10000
v(0 = - 206.559 tanh(0.154919 0
v(30) = -206.521 ft/sec
y(t) = 10000-1333.33 ln(cosh(0.1549190),
.y(30) = 4727.30 ft
After opening parachute:
v' - -32 +0.075v2,
v(0) = -206.521,
.y(O) = 4727.30
v(t) = - 20.6559 tanh(l .54919/ + 0.00519595)
y(t) = 4727.30-13.3333 ln(cosh(l .54919/+ 0.00519595))
y = 0 when
t = 229.304
Thus she opens her parachute after 30 sec at a height of 4727 feet, and the total
time of descent is 30 + 229.304 = 259.304 sec, about 4 minutes and 19.3 seconds.
25.
(a)
The rocket's apex occurs when v = 0. We get the desired formula when we set
v — 0 in Eq. (23),
(i
i \
= V2 + 2 GM-
V2
\r
RJ
and solve for r.
(b)
We substitute v = 0, r = R +10" (100 km = 105 m) and the mks values
G = 6.6726x10'", M = 5.975xl024, R = 6.378xl06 in Eq. (23) and solve for
v0 = 1389.21 m/s « 1.389 km/s.
(c)
When we substitute v0 = (9 /10)V2GM / R in the formula derived in part (a), we
find that r
27.
(a)
= 100R /19.
Substitution of v2 = 2GM / R = k2 / R in Eq. (23) of the textbook gives
We separate variables and proceed to integrate:
/
o/9\2/3
(using the fact that r = R when / = 0). We solve for r(t) = \jkt + R~
r(t) —> oo as t -» oo.
56
Chapter 2
j
and note that
(b)
If v0 > 2GM IR then Eq. (23) gives
dr
\2GM
dt ~ V " V
r
( 2
2 GM\
+[V°
R
+a >
J
•n/t
Therefore, at every instant in its ascent, the upward velocity of the projectile in this part is
greater than the velocity at the same instant of the projectile of part (a). It's as though the
projectile of part (a) is the fox, and the projectile of this part is a rabbit that runs faster.
Since the fox goes to infinity, so does the faster rabbit.
29.
Integration of v
dv
GM
dy
(y + R)2'
1
2
2
3/(0) = 0, v(0) = v0 gives
GM
GM
1
2
y+R
R
2 0
which simplifies to the desired formula for v2. Then substitution of
G = 6.6726 xKT11, M = 5.975 x 1024 kg, R = 6.378 xl06m v = 0,and v0=l
yields an equation that we easily solve for y = 51427.3, that is, about 51.427 km.
SECTION 2.4
NUMERICAL APPROXIMATION: EULER'S METHOD
In each of Problems 1-10 we also give first the explicit form of Euler's iterative formula for the
given differential equation y' = f (x, y). As we illustrate in Problem 1, the desired iterations are
readily implemented, either manually or with a computer system or graphing calculator. Then we
list the indicated values of y(j) rounded off accurate to 3 decimal places.
1.
For the differential equation / = /(x,y) with /(x,y) = -y, the iterative formula of
Euler's method is y„+\ = yn + h(-yn). The TI-83 screen on the left shows a graphing
calculator implementation of this iterative formula.
0. 1+H:0-^:2*V
2.0000
ft+H+X! V+H+C -vm1
1.8000
1.6200
1.4580
Section 2.3
X
1.8000
1.6200
1.4580
1.3122
1.1810
.5000
57
After the variables are initialized (in the first line), and the formula is entered, each press of
the enter key carries out an additional step. The screen on the right shows the results of 5
steps from x = 0 to x = 0.5. with step size h- 0.1 — winding up with y(0.5) « 1.181.
Approximate values 1.125 and 1.181; true value y{\) « 1.213
The following Mathematica instructions produce precisely this line of data.
f[x_,yj
g[x ]
= -y;
=2 Exp[-x] ;
h = 0.25;
x = 0;
Do[
k = f[x,yl];
yl = yl + h*k;
x = x + h,
U,l,2}
h = 0.1;
Do[
(
the left-hand slope
*)
(
Euler step to update y
*)
(
update x
*)
]
x = 0;
y2 = yO;
k = f [x,y2];
y2 = y2 + h*k;
(
(
the left-hand slope
*)
Euler step to update y
*)
x = x + h,
(
update x
*)
U,l,5}
]
Print[x,"
0.5
3.
yl = yO
Iterative formula:
M,yl,"
1.125
",y2,"
1.18098
",g[0.5]]
1.21306
y„+\ = yn + h(yn + 1)
Approximate values 2.125 and 2.221; true value y{\) « 2.297
5.
Iterative formula:
yn+\ = yn + h(yn -xn- 1)
Approximate values 0.938 and 0.889; true value y(y) « 0.851
7.
Iterative formula:
2
yn+\ = yn + h(-3xnyn)
Approximate values 2.859 and 2.737; true value y(j) « 2.647
9.
Iterative formula:
2
yn+\ = yn + h( 1 +y„)/4
Approximate values 1.267 and 1.278; true value y(j) « 1.287
The tables of approximate and actual values called for in Problems 11-16 were produced using the
following MATLAB script (appropriately altered for each problem).
58
Chapter 2
% Section 2.4, Problems 11-16
xO = 0;
yO = 1;
% first run:
h = 0.01;
x = xO;
y = yO;
yl = yO;
for
n = 1:100
y = y + h*(y-2);
yl = [yl,y];
x = x + h;
end
% second run:
h = 0.005;
x = xO;
y = yO;
y2 = yO;
for
n = 1:200
y = y + h*(y-2);
y2 = [y2,y];
x = x + h;
end
% exact values
x = xO : 0.2 : x0+l;
ye = 2 - exp (x) ;
% display table
ya = y2 (1:40:201) ;
err = 100*(ye-ya)./ye;
[x; yl(1:20:101); ya; ye;
11.
The iterative formula of Euler's method is y^\ = yn + h(y„ - 2), and the exact solution is
y(x) = 2 - ex. The resulting table of approximate and actual values is
X
y(h=o.oi)
y (h=0.005)
y actual
error
13.
err]
0.0
1.0000
1.0000
1.0000
0%
0.2
0.4
0.5111
0.5097
0.5082
-0.29%
0.7798
0.7792
0.7786
-0.08%
0.6
0.8
1.0
0.1833
0.1806
0.1779
-1.53%
-0.2167
-0.2211
-0.2255
1.97%
-0.7048
-0.7115
-0.7183
0.94%
1.6
1.8
2.0
3.8084
3.8117
3.8149
0.09%
4.2924
4.2967
4.3009
0.10%
4.8890
4.8940
4.8990
0.10%
3
Iterative formula:
yn+\ = yn + 2Jvcn /yn
Exact solution:
y(x) = (8+x4)l/2
X
1.0
1.2
y ( /f=0.01)
y (7z=0.005)
y actual
error
3.0000
3.0000
3.0000
0%
3.1718
3.1729
3.1739
0.03%
1.4
3.4368
3.4390
3.4412
0.06%
Section 2.4
59
15.
Iterative formula:
Exact solution:
yn+\ = yn + h{3 - 2yn/xn)
y(x) = x + 4/x
X
2.0
2.2
y(h=0.01)
y (h=0.005)
y actual
error
3.0000
3.0000
3.0000
0%
3.0253
3.0259
3.0264
0.019%
2.4
3.0927
3.0936
3.0944
0.028%
2.6
2.8
3.1897
3.1907
3.1917
0.032%
3.3080
3.3091
3.3102
0.033%
3.0
3.4422
3.4433
3.4444
0.032%
The tables of approximate values called for in Problems 17-24 were produced using a MATLAB
script similar to the one listed preceding the Problem 11 solution above.
17.
X
y(h=0A)
y (7z=0.02)
y (h=0.004)
y (k=0.0008)
0.0
0.0000
0.0000
0.0000
0.0000
0.2
0.0010
0.0023
0.0026
0.0027
0.4
0.0140
0.0198
0.0210
0.0213
0.6
0.8
1.0
0.0551
0.0688
0.0717
0.0723
0.1413
0.1672
0.1727
0.1738
0.2925
0.3379
0.3477
0.3497
These data indicate that y( 1) « 0.35, in contrast with Example 5 in the text, where the
initial condition is jy(0) = 1 •
In Problems 18-24 we give only the final approximate values of y obtained using Euler's method
with step sizes h = 0.1, h = 0.02, h = 0.004, and h = 0.0008.
19.
With x0 = 0 and yo = 1, the approximate values of y(2) obtained are:
h
v
21.
With xo =
h
v
23.
0.02
6.3653
0.004
6.4022
0.0008
6.4096
and yo = 2, the approximate values of y(2) obtained are:
0.1
2.8508
0.02
2.8681
0.004
2.8716
0.0008
2.8723
With xo = 0 and yo = 0, the approximate values of y( 1) obtained are:
h
y
60
0.1
6.1831
0.1
1.2262
0.02
1.2300
Chapter 2
0.004
1.2306
0.0008
1.2307
25.
Here f(t,v) = 32-1.6v and t0= 0, v0 = 0.
With h = 0.01, 100 iterations of v#l+1 = v„ + h f(tn, vj yield v(l) « 16.014, and 200
iterations with h = 0.005 yield v(l) « 15.998. Thus we observe an approximate velocity of
16.0 ft/sec after 1 second — 80% of the limiting velocity of 20 ft/sec.
With h = 0.01, 200 iterations yield v(2) « 19.2056, and 400 iterations with h = 0.005
yield v(2) « 19.1952. Thus we observe an approximate velocity of 19.2 ft/sec after 2
seconds — 96% of the limiting velocity of 20 ft/sec.
27.
Here /(x, y) = x2 + y2 -1 and x0 =0, y0 = 0. The following table gives the
approximate values for the successive step sizes h and corresponding numbers n of steps.
It appears likely that y(2) = 1.00 rounded off accurate to 2 decimal places.
29.
h
n
0.1
20
0.01
200
0.001
2000
0.0001
20000
0.00001
200000
y{ 2)
0.7772
0.9777
1.0017
1.0042
1.0044
With step sizes h = 0.15, h = 0.03, and h = 0.006 we get the following results:
X
-1.0
-0.7
-0.4
-0.1
+0.2
+0.5
y with
/z=0.15
y with
/z=0.03
y with
/2=0.006
1.0000
1.0000
1.0000
1.0472
1.1213
1.2826
0.8900
0.7460
1.0512
1.1358
1.3612
1.4711
1.2808
1.0521
1.1390
1.3835
0.8210
0.7192
While the values for h = 0.15 alone are not conclusive, a comparison of the values of y
for all three step sizes with x > 0 suggests some anomaly in the transition from negative to
positive values of x.
31.
With step sizes h = 0.1 and h = 0.01 we get the following results:
X
0.0
0.1
0.2
y with
h — 0.1
y with
h = 0.01
1.0000
1.2000
1.0000
1.2200
1.4428
1.4967
Section 2.4
61
0.7
4.3460
5.8670
8.3349
0.8
0.9
6.4643
11.8425
39.5010
Clearly there is some difficulty near x = 0.9.
SECTION 2.5
A CLOSER LOOK AT THE EULER METHOD
In each of Problems 1-10 we give first the predictor formula for un+\ and then the improved Euler
corrector for yn+\. These predictor-corrector iterations are readily implemented, either manually or
with a computer system or graphing calculator (as we illustrate in Problem 1). We give in each
problem a table showing the approximate values obtained, as well as the corresponding values of
the exact solution.
0. 1-»H:0+X:2*V
2.0000
V-H*V+U:V+CH/2>*
( -V-U)+V
1.8100
1.6381
1.4824
1.
un+1
V-H*V-MJ: V+£hk2>*
< -v-u>+v
1.8100
1.6381
1.4824
1.3416
1.2142
yn + h( yn)
T«+i — yn
(/z/2)[—yv? — ^«+i]
The TI-83 screen on the left above shows a graphing calculator implementation of this
iteration. After the variables are initialized (in the first line), and the formulas are entered,
each press of the enter key carries out an additional step. The screen on the right shows the
results of 5 steps from x = 0 to x = 0.5 with step size h = 0.1 — winding up with
jz(0.5) « 1.2142 — and we see the approximate values shown in the second row of the
table below.
X
y with h=0.1
y actual
62
0.0
2.0000
2.0000
0.1
0.2
1.8100
1.8097
1.6381
1.6375
Chapter 2
0.3
1.4824
1.4816
0.4
1.3416
1.3406
0.5
1.2142
1.2131
3.
Un+\
~
y,7 + h(yn + 1)
yn+1 =
+ (/z/2)[(y„ + 1) + (wrt+i + 1)]
X
y with h=0.1
y actual
5.
Uyt+x
+ (/z/2)[(y„ - xn - 1) + (^+i —
— /z — 1)]
xn
0.3
1.6985
1.6997
0.4
1.9818
1.9837
0.5
2.2949
2.2974
0.3
0.9508
0.9501
0.4
0.9091
0.9082
0.5
0.8526
0.8513
0.3
2.9156
2.9201
0.4
2.8082
2.8140
0.5
2.6405
2.6475
0.3
1.1625
1.1625
0.4
1.2230
1.2231
0.5
1.2873
1.2874
1)
0.0
1.0000
1.0000
y with h=0A
y actual
9.
0.2
1.2103
X
7.
0.1
1.2100
1.4421
1.4428
yn + h(yn
yn+1 =
0.0
1.0000
1.0000
0.1
0.2
0.9950
0.9948
0.9790
0.9786
^«+i
3xnynh
yn+1
(/z/2)[3xw<yw + 3(xn + h) zz^+i]
X
0.0
0.1
0.2
y with A=0.1
y actual
3.0000
3.0000
2.9955
2.9970
2.9731
2.9761
w„+i - yn + h{\ +.y«)/4
T«+i = yn + Ml + T« + 1 + (Wah-i )2]/8
X
y with/2=0.1
y actual
0.0
1.0000
1.0000
0.1
0.2
1.0513
1.0513
1.1053
1.1054
The results given below for Problems 11-16 were computed using the following MATLAB script.
% Section 2.5, Problems 11-16
xO = 0;
yO = 1;
% first run:
h = 0.01;
x = xO;
y = yO;
yl = yO;
for
n = 1:100
u = y + h*f(x,y);
y = y + (h/2)* (f(x,y)+f(x+h,u)) ;
Section 2.5
%predictor
%corrector
63
yi = [yi,y];
x = x + h;
end
%
second run:
h = 0.005;
x = xO;
for
y = y0;
y2 = yO;
n = 1:200
u = y + h*f(x,y);
%predictor
y = y +
%corrector
y2 =
(h/2)*(f(x,y)+f(x+h,u));
[y2,y];
x = x + h;
end
% exact values
x = xO
:
0.2
:
x0+l;
ye = g(x);
% display table
ya = y2(1:40:201);
err = 100*(ye-ya)./ye;
x = sprintf('%10.5f', x),
sprintf('\n');
yl = sprintf('%10.5f’,yl(1:20:101)),
sprintf(f\n');
ya = sprintf('%10.5f’,ya),
sprintf('\n');
ye = sprintf('%10.5fye),
sprintf('\n');
err = sprintf(1%10.5f1,err),
table =
[x;
yl;
ya;
ye;
sprintf(’\n’);
err]
For each problem the differential equation y' = f (x, y) and the known exact solution y - g(x)
are stored in the files f. m and g. m — for instance, the files
function
yp = f(x,y)
yp = y-2;
function ye = g(x,y)
ye = 2-exp(x);
for Problem 11. (The exact solutions for Problems 11-16 here are given in the solutions for
Problems 11-16 in Section 2.4.)
X
y(h=0.0l)
y (h=0.005)
y actual
error
64
0.0
1.00000
1.00000
1.00000
0.000%
0.2
0.77860
0.77860
0.77860
-0.000%
0.4
0.50819
0.50818
0.50818
-0.001%
Chapter 2
0.6
0.8
1.0
0.17790
0.17789
0.17788
-0.003%
-0.22551
-0.22553
-0.22554
0.003%
-0.71824
-0.71827
-0.71828
0.002%
X
y(h=0.01)
y (h=0.005)
y actual
error
X
y(h=0.01)
y (h=0.005)
y actual
error
1.0
1.2
3.00000
3.00000
3.00000
0.0000%
1.6
1.8
2.0
3.17390
3.17390
3.17389
-0.0001%
1.4
3.44118
3.44117
3.44116
-0.0001%
3.81494
3.81492
3.81492
0.0001%
4.30091
4.30089
4.30088
-0.0002%
4.89901
4.89899
4.89898
-0.0002%
2.0
2.2
2.4
2.6
2.8
3.0
3.000000
3.026448
3.094447
3.191719
3.310207
3.444448
3.000000
3.026447
3.094445
3.191717
3.310205
3.000000
3.026446
3.094444
3.191716
3.310204
3.444445
3.444444
0.00000%
-0.00002%
-0.00002%
-0.00002%
-0.00002%
-0.00002%
With h = 0.1:
With h = 0.02:
With h = 0.004:
With h = 0.0008:
y(l) * 0.35183
XI) * 0.35030
XI) * 0.35023
XI) * 0.35023
The table of numerical. results is
y with
h = 0.1
y with
h = 0.02
y with
h = 0.004
y with
h = 0.0008
0.0
0.2
0.00000
0.00000
0.00000
0.00000
0.00300
0.4
0.02202
0.6
0.8
1.0
0.07344
0.17540
0.35183
0.00268
0.02139
0.07249
0.17413
0.35030
0.00267
0.02136
0.07245
0.17408
0.35023
0.00267
0.02136
0.07245
0.17408
0.35023
X
In Problems 18-24 we give only the final approximate values of y obtained using the improved
Euler method with step sizes h = 0.1, h = 0.02, h = 0.004, and h = 0.0008.
19.
With h = 0.1:
With h = 0.02:
With h = 0.004:
With h = 0.0008:
X2) « 6.40834
X2) » 6.41134
x2) « 6.41147
X2) « 6.41147
21.
With h = 0.1:
With h = 0.02:
With h = 0.004:
With h = 0.0008:
X2) * 2.87204
X2) « 2.87245
x2) « 2.87247
X2) « 2.87247
Section 2.5
65
23.
With h = 0.1:
With h = 0.02:
With /i = 0.004:
With h = 0.0008:
25.
Here f(t9v) = 32-1.6v and f0=0, v0 = 0.
XI) «
XI) *
XI) *
XI) *
1.22967
1.23069
1.23073
1.23073
With h = 0.01, 100 iterations of
k\ = /(/>v„)>
h
v„+1 = v„+-(*,+fc2)
k2= f(t + h,vn+hk^),
yield v(l) « 15.9618, and 200 iterations with h = 0.005 yield v(l) « 15.9620. Thus we
observe an approximate velocity of 15.962 ft/sec after 1 second — 80% of the limiting
velocity of 20 ft/sec.
With h = 0.01, 200 iterations yield v(2) « 19.1846, and 400 iterations with h = 0.005
yield v(2) «19.1847. Thus we observe an approximate velocity of 19.185 ft/sec after 2
seconds — 96% of the limiting velocity of 20 ft/sec.
27.
Here /(x, y) = x2 + y2 -1 and x0 =0, y0 = 0. The following table gives the
approximate values for the successive step sizes h and corresponding numbers n of steps.
It appears likely that y(2) = 1.0045 rounded off accurate to 4 decimal places.
0.1
20
1.01087
h
n
y( 2)
0.01
200
1.00452
0.001
2000
1.00445
0.0001
20000
1.00445
In the solutions for Problems 29 and 30 we illustrate the following general MATLAB ode solver.
function
[t,y]
= ode(method,
%
[t,y]
= ode(method,
%
calls
the method described by
%
ODE
'yp'
yp,
yp,
tO,b,
,
t0,b,
yO,
yO
n)
n)
'method'
for the
with initial
(column)
with function header
%
%
y' = yp(t,y)
%
66
%
on the interval
%
vector
%
'impeuler',
%
Results are saved at the endPoints of n subintervals,
%
that is,
%
result
%
while
yO.
Choices for method are
'rk'
(Runge-Kutta),
in steps of length
t
y
[tO,b]
is an
h =
'euler1,
'ode23?,
(b -
'ode45'.
tO)/n.
The
(n+1)-column vector from b to tl,
is a matrix with
n+1
Chapter 2
rows
(one for each
%
t-value)
and one
h =
(b
-
tO) /n ;
t =
tO
:
h
t =
t' ;
:
column
%
i
=
2
:
n+1
tO
=
t(i-1) ;
tl
=
t (i) ;
yO
= y(i-1,:)';
[T,Y]
y =
step
size
%
col.
b;
y = yO';
for
for each dependent variable.
= feval(method,
vector of
t-values
%
1st row of
result matrix
%
for
i=n+l
%
old t
ypf
i=2
%
new t
%
old y-row-vector
tO,tl,
%
[y;Y'];
to
yO) ;
adjoin new y-row-vector
end
To use the improved Euler method, we call as ’method’ the following function.
function
[t,y]
= impeuler(yp,
tO,tl,
yO)
%
%
[t,y]
=
impeuler(yp,
%
Takes
one
tO,tl,
improved Euler
yO)
step
for
%
%
y'
= yprime(
t,y
),
%
%
from tO
%
column vector
h =
tl
tl
with
initial value
the
yO.
tO ;
kl
=
feval(
yp,
tO,
yO
k2
=
feval(
yp,
tl,
yO
k
=
(kl
t =
);
+ h*kl
);
+ k2)/2;
tl;
y = yO
29.
-
to
+ h*k;
Here our differential equation is described by the MATLAB function
function
vp =
vp = vpboltl(t,v)
-0.04*v
-
9.8;
Then the commands
n =
50;
[tl,vl]
n =
= ode('impeuler’,'vpboltl',0,10,49,n);
100;
[t2,v2]
= ode(’impeuler’vpboltl',0,10,49,n);
t = (0:10) ' ;
ve =
[t,
294*exp(-t/25)-245 ;
vl(1:5:51),
v2(1:10:101),
ve]
generate the table
Section 2.5
67
t
with yi — 50
with n
0
1
2
3
4
5
6
7
8
9
10
49.0000
37.4722
26.3964
15.7549
5.5307
4.2926
13.7308
22.7989
31.5115
39.8824
47.9251
49.0000
37.4721
26.3963
15.7547
5.5304
4.2930
13.7313
22.7994
31.5120
39.8830
47.9257
-
-
-
-
-
-
=
actual v
100
-
-
-
-
-
-
49.0000
37.4721
26.3962
15.7546
5.5303
4 2932
13.7314
22.7996
31.5122
39.8832
47.9259
-
.
.
-
-
-
-
-
We notice first that the final two columns agree to 3 decimal places (each difference being
than 0.0005). Scanning the n = 100 column for sign changes, we suspect that v = 0 (at the
bolt's apex) occurs just after / = 4.5 sec. Then interpolation between / = 4.5 and t = 4.6
in the table
[t2(40:51),v2(40:51)]
3.9000
4.0000
4.1000
4.2000
4.3000
4.4000
4.5000
4.6000
4.7000
4.8000
4.9000
5.0000
6.5345
5.5304
4.5303
3.5341
2.5420
1.5538
0.5696
0.4108
1.3872
2.3597
3.3283
4.2930
-
-
-
-
-
indicates that t = 4.56 at the bolt's apex. Finally, interpolation in
[t2(95:96) ,v2(95:96) ]
9.4000
9.5000
43.1387
43.9445
-
-
gives the impact velocity v(9.41) « -43.22 m/s.
68
Chapter 2
SECTION 2.6
THE RUNGE-KUTTA METHOD
Each problem can be solved with a "template" of computations like those listed in Problem 1. We
include a table showing the slope values kv k2, k3, kA and the
values at the ends of two
successive steps of size h = 0.25.
1.
To make the first step of size h = 0.25 we start with the function defined by
f[x__,
y_]
:= -y
and the initial values
x = 0;
y = 2;
h = 0.25;
and then perform the calculations
kl
k2
k3
k4
y
x
=
=
=
=
=
=
f [x, y]
f[x + h/2, y + h*kl/2]
f[x + h/2, y + h*k2/2]
f[x + h, y + h*k3]
y + h/6*(kl + 2*k2 + 2*k3 + k4)
x + h
in turn. Here we are using Mathematica notation that translates transparently to standard
mathematical notation describing the corresponding manual computations. A repetition
of this same block of calculations carries out a second step of size h = 0.25. The
following table lists the intermediate and final results obtained in these two steps.
k{
k2
h
k\
X
-2
-1.55762
-1/75
-1.36292
-1.78125
-1.38725
-1.55469
-1.2108
0.25
0.5
k\
ki
h
h
X
2
2.56803
2.25
2.88904
2.28125
2.92916
2.57031
3.30032
0.25
0.5
k\
h
h
k\
X
0
-28402
-0.125
-0.44452
-0.14063
-0.46458
-0.28516
-0.65016
0.25
0.5
Section 2.6
Approx, y
1.55762
1.21309
Actual y
1.55760
1.21306
Approx, y
1.56803
2.29740
Actual y
1.56805
2.29744
Approx, y
0.96598
0.85130
Actual y
0.96597
0.85128
69
j
7.
kl
ki
h
&4
X
0
-0.55378
-0.14063
-1.21679
-0.13980
-1.18183
-0.55595
-1.99351
0.25
0.5
k\
ki
h
k/\
X
0.5
0.57122
0.53223
0.61296
0.53437
0.61611
0.57126
0.66444
0.25
0.5
Approx, y
2.95347
2.6475
Actual y
2.95349
2.64749
Approx, y
1.13352
1.28743
Actual y
1.13352
1.28743
The results given below for Problems 11-16 were computed using the following MATLAB script.
%
Section 2.6,
xO = 0;
Problems
11-16
yO = 1;
% first run:
h = 0.2;
x = xO;
for
y = yO;
yl = yO;
n = 1:5
f (x,y) ;
kl =
k2 = f(x+h/2,y+h*kl/2);
f (x+h/2,y+h*k2/2) ;
k3 =
k4 = f(x+h,y+h*k3);
y = y +(h/6)*(kl+2*k2+2*k3+k4);
yl = [yi,y];
x = x + h;
end
%
second run:
h = 0.1;
x = xO;
for
y = yO;
y2 = yO;
n = 1:10
kl = f(x,y);
k2 = f(x+h/2,y+h*kl/2);
k3 = f(x+h/2,y+h*k2/2) ;
k4 = f(x+h,y+h*k3);
y = y +(h/6)*(kl+2*k2+2*k3+k4);
y2 = [y2,y];
x = x + h;
end
% exact values
x = xO
:
0.2
:
xO+1;
ye = g(x);
% display table
y2 = y2(1:2:11);
err = 100*(ye-y2)./ye;
70
Chapter 2
x = sprintf( '%10.6f', x),
sprintf('\n');
yl = sprintf(1%10.6f',yl),
sprintf('\n’);
y2 = sprintf(' %10.6f',y2),
sprintf('\n');
ye = sprintf(’%10.6f’ ,ye),
sprintf(’\n');
err = sprintf(’%10.6ferr),
table = [x;y1;y2;ye;err]
sprintf('\n’);
For each problem the differential equation / = f (x, y) and the known exact solution y - g(x)
are stored in the files f. m and g. m — for instance, the files
function
yp = f(x,y)
YP = Y-2;
and
function ye = g(x,y)
ye = 2-exp(x);
for Problem 11.
X
0.0
0.00000%
0.2
0.778600
0.778597
0.778597
-0.00002%
0.4
0.508182
0.508176
0.508175
-0.00009%
0.6
0.177894
0.177882
0.177881
-0.00047%
0.8
-0.225521
-0.225540
-0.225541
-0.00061%
y (6=0.2)
y (6=0.1)
y actual
1.000000
1.000000
1.000000
-0.718251
-0.718280
-0.718282
-0.00029%
error
X
1.0
3.000000
3.000000
3.000000
0.00000%
1.2
3.173896
3.173894
3.173894
-0.00001%
1.4
3.441170
3.441163
3.441163
-0.00001%
1.6
3.814932
3.814919
3.814918
-0.00002%
1.8
4.300904
4.300885
4.300884
-0.00003%
2.0
4.899004
4.898981
4.898979
-0.00003%
y actual
2.0
3.000000
3.000000
3.000000
2.2
3.026448
3.026446
3.026446
2.4
3.094447
3.094445
3.094444
2.6
3.191719
3.191716
3.191716
2.9
3.310207
3.310204
3.310204
3.0
3.444447
3.444445
3.444444
error
0.000000%
-0.000004%
-0.000005%
-0.000005%
-0.000005%
-0.000004%
y (6=0.2)
y (6=0.1)
y actual
error
X
y (6=0.2)
y (6=0.1)
17.
With h = 0.2:
With h = 0.1:
With h = 0.05:
With h = 0.025:
1.0
XI) * 0.350258
XI) * 0.350234
XI) * 0.350232
XI) * 0.350232
Section 2.6
71
The table of numerical results is
A = 0.1
y with
h = 0.05
y with
h = 0.025
0.000000
0.002667
0.021359
0.072448
0.174081
0.350234
0.000000
0.002667
0.021359
0.072448
0.174080
0.350232
0.000000
0.002667
0.021359
0.072448
0.174080
0.350232
y with
y with
X
A = 0.2
0.0
0.2
0.4
0.6
0.8
1.0
0.000000
0.002667
0.021360
0.072451
0.174090
0.350258
In Problems 18-24 we give only the final approximate values of y obtained using the Runge-Kutta
method with step sizes h = 0.2, h
19.
With h = 0.2:
With h = 0.1:
With h = 0.05:
With h = 0.025:
M2)
M 2)
M2)
21.
With h = 0.2:
With h = 0.1:
With h = 0.05:
With h = 0.025:
M2)
M 2)
M 2)
M 2)
23.
With h = 0.2:
With h = 0.1:
With h - 0.05:
With h = 0.025:
Ml)
Ml)
MD
MO
25.
Here f(t,v) = 32-1.6v and ^=0, v0=0.
y(2)
6.411474
6.411474
1.230735
1.230731
1.230731
1.230731
With h = 0.1, 10 iterations of
K = /(4>v„)>
k3 = f(t„
K = f(t„ +t h,v„+±hkt),
+ 2 h- V„ + 2 hk2 )»•
k = i(kl+2k2 + 2k3 + k4),
K = /('„ +h’v„+ hk3 )’
vll+l =vn+hk
yield v(l) « 15.9620, and 20 iterations with h = 0.05 yield v(l) « 15.9621. Thus we
observe an approximate velocity of 15.962 ft/sec after 1 second — 80% of the limiting
velocity of 20 ft/sec.
72
Chapter 2
With h - 0.1, 20 iterations yield v(2) « 19.1847, and 40 iterations with h = 0.05
yield
v(2) « 19.1848. Thus we observe an approximate velocity of 19.185 ft/sec after 2
seconds — 96% of the limiting velocity of 20 ft/sec.
27.
Here /(x, y) = x2 + y2 -1 and x0 =0, yQ = 0. The following table gives the
approximate values for the successive step sizes h and corresponding numbers n of steps.
It appears likely that y(2) = 1.00445 rounded off accurate to 5 decimal places.
0.1
20
1.00447
1
2
1.05722
h
n
y(2)
0.01
200
1.00445
0.001
2000
1.00445
In the solutions for Problems 29 and 30 we use the general MATLAB solver ode that was listed
prior to the Problem 29 solution in Section 2.5. To use the Runge-Kutta method, we call as
’method’ the following function.
function
[t,y]
%
[t,
= rk(yp,
%
Takes one Runge-Kutta step for
y]
= rk(yp,
tO,
tO,tl,
tl,
yO)
yO)
%
%
y’
= YP(
t,y ),
%
%
from tO
%
column vector
h = tl
-
to
tl
with initial value
the
yO.
tO ;
kl = feval(yp,
k2 = feval(yp,
tO
,
tO + h/2,
yO
yO +
(h/2)*kl
) ;
);
k3 = feval(yp,
tO + h/2,
yO +
(h/2)*k2
);
k4 = feval(yp,
tO + h
,y0 +
h *k3
);
k
=
(1/6)*(kl + 2*k2 + 2*k3 + k4);
t = tl;
y = yO + h*k;
29.
Here our differential equation is described by the MATLAB function
function
vp = vpboltl(t,v)
vp = -0.04*v -
9.8;
Then the commands
n = 100;
[tl,vl] = ode ( 'rk',’vpboltl',0,10,49,n) ;
Section 2.6
73
n = 200;
[t2 ,v] = ode('rk','vpboltl' ,0,10,49
t = (0:10)•;
i
0)
[t.
*
CM
II
ve :
t/25)-245;
vl(1:n/20: l+n/2),
v(1:n /10:n+1)
generate the table
t
with ft = 100
with n = 200
actual v
0
1
2
3
4
5
6
49.0000
37.4721
26.3962
15.7546
5.5303
4.2932
13.7314
22.7996
49.0000
37.4721
26.3962
15.7546
5.5303
4.2932
13.7314
22.7996
49.0000
37.4721
26.3962
15.7546
5.5303
4.2932
13.7314
22.7996
7
8
9
10
-
-
-
-
-
-
-
-
-
31.5122
39.8832
47.9259
31.5122
39.8832
47.9259
31.5122
39.8832
47.9259
-
-
-
-
-
-
-
-
-
We notice first that the final three columns agree to the 4 displayed decimal places.
Scanning the last column for sign changes in v, we suspect that v = 0 (at the bolt's apex)
occurs just after r = 4.5 sec. Then interpolation between t = 4.55 and ^ = 4.60 in the table
[t2 (91: 95) , v (91: 95) ]
4.5000
4.5500
4.6000
4.6500
4.7000
0.5694
0.0788
0.4109
0.8996
1.3873
-
-
-
indicates that t = 4.56 at the bolt's apex. Now the commands
y = zeros(n+1,1);
h = 10/n;
for j
= 2:n+1
y(j)
= Y(j“l)
+ v(j-l)*h +
0.5*(-.04*v(j-1)
-
end
ye = 7350*(1
[t,
- exp(-t/25))
y(1:n/10:n+1),
ye]
generate the table
74
Chapter 2
- 245*t;
9.8)*hA2;
t
Approx y
Actual y
0
1
2
3
4
5
6
7
8
9
10
0
43.1974
75.0945
96.1342
106.7424
107.3281
98.2842
79.9883
52.8032
17.0775
26.8540
0
43.1976
75.0949
96.1348
106.7432
107.3290
98.2852
79.9895
52.8046
17.0790
26.8523
-
-
We see at least 2-decimal place agreement between approximate and actual values of y.
Finally, interpolation between t-9 and t = 10 here suggests that y = 0 just after t = 9.4.
Then interpolation between t - 9.40 and t = 9.45 in the table
[t2(187:191),y(187:191)]
9.3000
9.3500
9.4000
9.4500
9.5000
4.7448
2.6182
0.4713
1.6957
3.8829
-
-
indicates that the bolt is aloft for about 9.41 seconds.
Section 2.6
75
CHAPTER 3
LINEAR EQUATIONS OF HIGHER ORDER
SECTION 3.1
INTRODUCTION: SECOND-ORDER LINEAR EQUATIONS
In this section the central ideas of the theory of linear differential equations are introduced and
illustrated concretely in the context of second-order equations. These key concepts include
superposition of solutions (Theorem 1), existence and uniqueness of solutions (Theorem 2),
linear independence, the Wronskian (Theorem 3), and general solutions (Theorem 4). This
discussion of second-order equations serves as preparation for the treatment of «th order linear
equations in Section 3.2. Although the concepts in this section may seem somewhat abstract to
students, the problems set is quite tangible and largely computational.
In each of Problems 1-16 the verification that y\ and y2 satisfy the given differential equation
is a routine matter. As in Example 2, we then impose the given initial conditions on the general
solution y = c\y\ + c^. This yields two linear equations that determine the values of the
constants c\ and c^.
1.
Imposition of the initial conditions jy(0) = 0, .y'(O) = 5 on the general solution
y(x) = cxex + c2e~x yields the two equations cx + c2 = 0, c] - c2 = 0 with solution
cx =5/2, c2-~ 5/2. Hence the desired particular solution is X*) = 5(g* - e~x)/2.
3.
Imposition of the initial conditions j^(0) = 3, j/(0) = 8 on the general solution
y(x) = cx cos2x + c2 sin2x yields the two equations cx = 3, 2c2 = 8 with solution
cx =3, c2 - 4. Hence the desired particular solution is y(x) = 3 cos 2x + 4 sin 2x.
5.
Imposition of the initial conditions y(0) = 1, .y'(O) = 0 on the general solution
jy(x) = cxex + c2e v yields the two equations cx + c2 = 1, cx + 2c2 = 0 with solution
cx- 2, c2 = — 1. Hence the desired particular solution is ^(x) =
7.
- e .
Imposition of the initial conditions j^(0) = - 2, /(0) = 8 on the general solution
j/(x) = cx + c2e~x yields the two equations cx + c2 = - 2, - c2 = 8 with solution
cx = 6, c2 = - 8. Hence the desired particular solution is j^(x) = 6 - Se~x.
76
Chapter 3
9.
Imposition of the initial conditions y(0) = 2, y'(0) = -1 on the general solution
Xx) = cxe x +c2xe ' yields the two equations c] = 2, - cx + c2 = -1 with solution
Cj = 2, c2 = 1. Hence the desired particular solution is y(x) = 2e~* + xe~*.
11.
Imposition of the initial conditions y(0) = 0, y'(0) = 5 on the general solution
y(x) = c}ex cosx + c2ex sinx yields the two equations Cj-0, c, + c2 = 5 with solution
cx = 0, c2=5. Hence the desired particular solution is y{x) = 5exsinx.
13.
Imposition of the initial conditions y( 1) = 3, y'(l) = 1 on the general solution
y(x) = cxx + c2x“ yields the two equations c, + c2 - 3, c, + 2c2 = 1 with solution
q = 5, c2 = - 2. Hence the desired particular solution is y(x) = 5x - 2x2.
15.
Imposition of the initial conditions y(l) = 7, y'(l) = 2 on the general solution
y(x) = cxx + c2xlnx yields the two equations cx - 7, Cj + c2 = 2 with solution
c\ = 7, c2 = - 5. Hence the desired particular solution is y(x) = 7x - 5x In x.
17.
If y = c/xthen y' + y2 = -c/x2+c2/x2 = c(c-l)/x2 ^ 0 unless either c = 0
or c = 1.
19.
.
If y = \ + 4x then yy" + {y'f = (\ + Jx)(-x-vl/4) + (x~m/2)2 = -x~vl/4 * 0.
3
2
3
2
21
Linearly independent, because x = + x x
23.
Linearly independent, because fix) = +g(x) if x > 0, whereas fix) = -g(x) if x < 0.
25.
fix) = exsinx and g(x) = excosx are linearly independent, because fix) = kg(x)
would imply that sin x = k cos x, whereas sin x and cos x are linearly independent.
27.
Let L[y] = y" + py' + qy. Then L[yc] = 0 and L[yp] = / so
if x > 0, whereas x = - x x
ifx<0
LLyc+Tp] = L[yc] + [yP] - 0 +/ = /
29.
There is no contradiction because if the given differential equation is divided by x to
get the form in Equation (8) in the text, then the resulting functions p(x) = -4/x and
q(x) = 61 x are not continuous at x = 0.
Section 3.1
77
31.
W(yi,y2) = -2x vanishes at x = 0, whereas if y\ andy2 were (linearly independent)
solutions of an equation y" + py' + qy = 0 with p and q both continuous on an open
interval / containing x = 0, then Theorem 3 would imply that W * 0 on /.
In Problems 33-42 we give the characteristic equation, its roots, and the corresponding general
solution.
33.
r2 - 3r + 2 = 0;
35.
r2 + 5r = 0;
37.
2r2 -r-2 = 0;
r = 1,-1/2;
39.
4r2 + 4r +1 = 0 ;
r = -1/2, -1/2;
y(x) = (c\ + c2x)e~xl2
41.
6r2-7r-20 = 0;
r = -4/3, 5/2;
y(x) = cxe~Ax,2> + c2eSxl2
r = 1, 2;
r = 0,-5;
y(x) = c\ex + c2elx
y(x) = c\ + c2e~5x
y(x) = c\e~x/1 + c2ex
In Problems 43-48 we first write and simplify the equation with the indicated characteristic
roots, and then write the corresponding differential equation.
(r + 10)(r + 10) == r2+20r + 100 = 0;
47.
49.
II
45.
i
o
(r-0)(r + 10) = r1 +\0r = 0;
i
o
43.
r2 = 0;
/' + 10/ = 0
y" + 20y' + 100y = 0
y" = 0
The solution curve with y(0) - T /(0) = 6 js y(x) = &e x -1 e 2x. We find that
y\x) - 0 when x = ln(7/4) so e~x =4/7
and
e~2x = 16/49 . It follows that
y(ln(7/4)) - 16/7 5 so the high point on the curve is (ln(7/4)), 16/7) « (0.56, 2.29) ?
which looks consistent with Fig. 3.1.6.
51.
(a)
The substitution v = In x gives
, _ dy _ dy dv _ 1 dy
dx
dv dx
x dv
Then another differentiation using the chain rule gives
78
Chapter 3
y
tt
_ d2y _ d f dy^ _ d (1
dx2
1
dx \dx j
dy
1
1
dy^
dx\x dv j
d fdy} dv
1
dy
-— —
x
dv
x dv \dv j dx
x2 dv
+
1
x'
Substitution of these expressions for y' and y" into Eq. (21) in the text then yields
immediately the desired Eq. (22):
+ (b-a)
dy_
+ cy
dv
(b)
If the roots rx and r2 of the characteristic equation of Eq. (22) are real and
distinct, then a general solution of the original Euler equation is
y(x) = cxer'v +c2eriV = c, (V)'1 + c2 (V)'2 = cxxr' + c2xri.
53.
The substitution v = In x yields the converted equation d2y / dv2 + dy / dv -12y = 0
whose characteristic equation r + r -12 = 0 has roots rx~-4 and r2- 3. Because
e - x,the corresponding general solution is
—4 v
y
55.
~ ^1^
3v
^ ^2^
—4
~ C\X
3
+ C2X .
The substitution v = lnx yields the converted equation d2y/dv2 = 0 whose
characteristic equation r - 0 has repeated roots rvr2= 0. Because v = lnx, the
corresponding general solution is
y - c,+c2v = c,+c2lnx.
SECTION 3.2
GENERAL SOLUTIONS OF LINEAR EQUATIONS
Students should check each of Theorems 1 through 4 in this section to see that, in the case
n = 2, it reduces to the corresponding theorem in Section 3.1. Similarly, the computational
problems for this section largely parallel those for the previous section. By the end of Section
3.2 students should understand that, although we do not prove the existence-uniqueness theorem
now, it provides the basis for everything we do with linear differential equations.
The linear combinations listed in Problems 1-6 were discovered "by inspection" — that is, by
trial and error.
Section 3.1
79
1.
(5/2)(2jc) + (-8/3)(3x2) + (-l)(5x - 8x2) = 0
3.
(1)(0) + (0)(sin x) + (0)(ex) - 0
5.
(1 )(17) + (-34)(cos2x) + (17)(cos 2x) = 0, because 2 cos2x = 1 + cos 2x.
7.
W =
1
x
0
1
0
0
is nonzero everywhere.
9.
W = e*(cos2x + sin2x) = ex ^ 0
11.
W = x3e2x is nonzero if x ^ 0.
In each of Problems 13-20 we first form the general solution
y(x) = C]y\(x) + ciy2(x) +
then calculate y'(x) and y"(x), and finally impose the given initial conditions to determine the
values of the coefficients ci, C2, C3.
13.
Imposition of the initial conditions y(0) = 1, y'(0) = 2, y”(0) = 0 on the general solution
y(x) = cxex + c2e~x + c3e~ Y yields the three equations
c, +c2 +c3 = 1,
c, - c2 - 2c3 =2,
cx + c2 + 4c3 - 0
with solution cx - 4/3, c2= 0, c3 = -113. Hence the desired particular solution is
given by y(x) = (4ex - e~2x)/3.
15.
Imposition of the initial conditions y(0) = 2, y\0) = 0, y"(0) = 0 on the general
2
3x
#
solution y(x) = cxex +c2xex +c3x e x yields the three equations
Cj — 2,
cx+ c2= 0,
C| + 2c2 + 2c3 — 0
with solution cx- 2, c2 = - 2, c3 = 1. Hence the desired particular solution is given by
y(pc) = (2 - 2x + x2)ex.
80
Chapter 3
17.
Imposition of the initial conditions y(0) = 3, /(0) = -1, y"(0) = 2 on the general
solution y(x) = c] + c2 cos3x + c3 sin3x yields the three equations
cx + c2 = 3,
3c3=-l,
-9c2=2
with solution cx - 29/9, c2=- 2/9, c3 =-1/3. Hence the desired particular solution is
given by y(x) = (29-2 cos 3x - 3 sin 3x)/9.
19.
Imposition of the initial conditions y(l) = 6, y'(l) = 14, y"(\) = 22 on the general
solution y(x) = c,x + c2x +c3x' yields the three equations
cx+c2+c3=6,
cx + 2c2 + 3c3 = 14,
2c2+6c3=22
with solution c, = 1, c2-2, c3 = 3. Hence the desired particular solution is given by
y(x) = x + 2x2 + 3x3.
In each of Problems 21 - 24 we first form the general solution
X*) = fcW+JpW = ciyi(x) + C2y2(x)+yp(x),
then calculate y'(x), and finally impose the given initial conditions to determine the values of
the coefficients c\ and c^.
21.
Imposition of the initial conditions y(0) = 2, y'(0) = - 2 on the general solution
y(x) = c, cosx + c2 sinx + 3x yields the two equations c} =2, c2 + 3 = - 2 with
solution cx = 2, c2 = - 5. Hence the desired particular solution is given by
y(x) = 2 cos x - 5 sin x + 3x.
23.
Imposition of the initial conditions y(Q) = 3, j/(0) = 11 on the general solution
y(x) = cxe~x + c2e3x - 2 yields the two equations cx +c2 - 2 = 3, - c, + 3c2 = 11 with
solution c, = 1, c2 =4. Hence the desired particular solution is given by
y(x) = e~x + 4e3x - 2.
25.
L [y\ = L[yi +y2] = L[yi] + L[y2] =f+g
27.
The equations
C\ + C2X + C3X
=0,
C2 + 2cyX + 0,
Section 3.2
2c3 = 0
81
(the latter two obtained by successive differentiation of the first one) evidently imply
by substituting x = 0 — that c\ = C2 = C3 = 0.
If coerx + C\xerx + ••• + cnxnerx = 0, then division by erx fields
c 0 + C\x + ••• + cnxn — 0,
so the result of Problem 28 applies.
(a)
Substitution of x = a in the differential equation gives y\a) = - py'(a) - q(a).
(b)
If y(0) = 1 and /(0) = 0, then the equation y" - 2y' - 5y = 0 implies that
/'(0) - 2/(0) + 5y(0) - 5.
This follows from the fact that
1
1
1
a
b
c
2
b
7 2
c
a
(b - a)(c - b)(c - a).
2
0
When we substitute y - vx in the given differential equation and simplify, we get the
separable equation xv" + v' = 0 that we solve by writing
v"
1
— =-=>
v'
x
lnv' = -lnx + ln^4,
A
v' = —
x
=>
v(x) = A\nx + B.
With A- 1 and B- 0 we get v(x)-lnx and hence y2{x)-x lnx.
When we substitute y - vexl2 in the given differential equation and simplify, we
eventually get the simple equation v" = 0 with general solution v(x) = Ax + B.
With A- 1 and B - 0 we get v(x) = x and hence y2(x) = xex/\
When we substitute y = vex in the given differential equation and simplify, we get the
separable equation (1 + x)v" + x V = 0 that we solve by writing
Chapter 3
=
v'
1+x
-1 +
v' = A(l + x)e~x
1
=>
1+x
=>
lnv' = -x + ln(l + x) + ln^,
v(x) = A |(1 + x)e~x dx = - A(2 + x)e~x + B.
With A = -1 and B = 0 we get v(x) = (2 + x)e x and hence y2(x) = 2 + x.
43.
When we substitute y = vx in the given differential equation and simplify, we get the
separable equation x (x2 -1) v" = (2 - 4x2) v' that we solve by writing
_
2
2 -4x2
1
1
+
v'
x(x2-l)
1+x
X
1-x
lnv' = -21nx-ln(l + x)-ln(l-x) + ln^,
v =
x\\-xA)
(
1
= A
( 1
1
+
yxA
1
A
1
+
2(1+ x)
2(1 -x)J
1
A
v(x) = A — + —ln(l + x)—ln(l-x) + B.
x 2
J
With A = -1 and B = 0 we get
y2(x) = 1
v(x) = - - hn(l + x) + hn(l - x)
x 2
2
2
^
1-x
SECTION 3.3
HOMOGENEOUS EQUATIONS WITH CONSTANT COEFFICIENTS
This is a purely computational section devoted to the single most widely applicable type of
higher order differential equations — linear ones with constant coefficients. In Problems 1-20,
we write first the characteristic equation and its list of roots, then the corresponding general
solution of the given differential equation. Explanatory comments are included only when the
solution of the characteristic equation is not routine.
1.
r2 -4 = (r-2)(r + 2) = 0;
r =-2,2;
3.
r2+3r-10 = (r + 5)(r-2) = 0;
y(x)= c\e2x + C2e~Ix
r =-5,2;
Section 3.2
y(x) - C\e2x + C2e~5x
83
y{x) = cje~3x + C2xe~3x
r -- 3,-3;
5.
r2 +6r + 9 = (r + 3)2 = 0;
7.
4r2-12r + 9 = (2r-3)2 = 0;
9.
r2+8r + 25 = 0;
11.
r4 -8r3 + 16r2 = r2(r-4)2 = 0;
r = 0,0,4,4;
13.
9r3 +\2r2+4r= r(2r + lf = 0;
r = 0,-2/3,-2/3
r = -3/2,-3/2;
r = (-8±>/-36 )/2 =-4 + 3/;
XX = c\e3x/2 + C2xe3x'2
y(x) = e~4x(ci cos 3x + C2 sin 3x)
XX — c i + C X + C2>e4x + cyxe4x
2
_y(x) = Ci + cje~lxn + c^xe~2xl3
15.
4r4 -8r2 +16 = (r2-4)2 = (r - 2)2(r + 2)2 =0;
/ \
XX = c\e
17.
x i
2
+ cjxe
x ,
2
+ <23^
-2.x ,
+ Czpce
r =2,2,-2,-2
-2x
6/+llr2+4 = (2r2 + l)(3r2 + 4) = 0;
r = ±//V2, ±2//V3,
XX = ci cos(x/ V2 ) + C2sin(x/ V2 ) + C3Cos(2x/ >/3 ) + C4sin(2x/ V3 )
19.
X+r2-r-l = r(V2-l) + (X2-l) = (r-l)(r + l)2 =0;
r =1,-1,-1;
XX = CieX + C2C_X + C3X£-X
21.
Imposition of the initial conditions
XX = 7, j/(0) = 11 on the general solution
XX = cxex + c2e3x yields the two equations cx + c2 = 7, q + 3c = with solution
q = 5, c2= 2. Hence the desired particular solution is XX = 5ex + 2e .
2
23.
Imposition of the initial conditions
11
XX = 3, y'(0) = 1 on the general solution
XX = e3x (cx cos4x + c2 sin4x) yields the two equations cx = 3, 3cx + 4c =
2
1
with
solution cx = 3, c2 = - 2. Hence the desired particular solution is
XX) = c3x(3 cos 4x - 2 sin 4x).
25.
Imposition of the initial conditions
solution
XX - cx + c2x +
cx + c3 = -1,
84
2
/3
XX = - , /( ) = , X(X - 1 on the general
1
0
0
yields the three equations
c2 - 2c3 / 3 = 0,
Chapter 3
4c3 / 9 = 1
with solution c, — — 13/4, c2 = 3 / 2, c3 = 9 / 4. Hence the desired particular solution is
y(x) = (-13 + 6x + 9e-2x/3)/4.
First we spot the root r = 1. Then long division of the polynomial r3 + 3r2 - 4
by r - 1 yields the quadratic factor r2 + 4r + 4 = (r + 2)2 with roots
r = -2, -2. Hence the general solution is y(x) = c\ex + c2e-2* + c?,xe~2x.
First we spot the root r = -3. Then long division of the polynomial r J + 27 by
r + 3 yields the quadratic factor r2 -3r + 9 with roots r = 3^1±/V3^/2. Hence the
general solution is y(x) = c\e~3x + ^3a/2[c2Cos(3x V3 /2) + C3 sin(3x V3 /2)].
The characteristic equation r +3 r +4r-8 = 0 has the evident root r = 1, and long
2
2
division then yields the quadratic factor r + 4r + 8 = (r + 2) +4 corresponding to the
complex conjugate roots -2 ± 2 i. Hence the general solution is
y(x) = c\ex + c~2a(c2COs 2x + C3sin 2x).
Knowing that y — e is one solution, we divide the characteristic polynomial
r +3 r - 54 by r-3 and get the quadratic factor
r1 + 6r + 18 = (r + 3)2 + 9.
Hence the general solution is Xx) = c\e
+ e~ fecos 3x + C3sin 3x).
The fact that y = cos 2* is one solution tells us that r +4 is a factor of the
characteristic polynomial
6
/ + 5 ? + 25r2 + 20r + 4.
Then long division yields the quadratic factor
r + 5r +1 = (3r + l)(2r +1) with roots
6
r = —1/2, —1/3. Hence the general solution is
y(x) = c\e~xl2 + C2e~x/3 + C3C0S 2x + C4sin 2x
The characteristic equation is ^ -r3 =r3(r-1) = 0 5 so the general solution is
jy(x) = A + Bx + CxL + Dex. Imposition of the given initial conditions yields the
equations
v4 + D = 18,
B + D = 12,
2 C + D = 13,
Section 3.3
D = 7
85
with solution ,4 = 11, B = 5, C- 3, Z) = 7.
y(x) = 11 4- 5x + 3x
39.
(r - 2)
Hence the desired particular solution is
-\-l &
= r - 6r +12r - 8, so the differential equation is
y"' -6y" + \2y' -Sy = 0.
41.
(r2+4)(r2-4) - r4-16, so the differential equation is
y(4)-\6y = 0.
45.
The characteristic polynomial is the quadratic polynomial of Problem 44(b). Hence the
general solution is
y(x) = c1e_,v + c2e3lx = c,(cosx-zsinx)+ c2(cos3x+ zsin3x).
47.
The characteristic roots are r = ±y-2 + 2zV3 = ±(1 + z’V3) so the general solution is
y(x) = c1e(1+,^)v + c2e_(1+/^)x = c,ex(cosV3x + zsinV3 xj + c2e~x^cosV3x-zsinV3x)
49.
The general solution is y = Ae x + 5<TX + C cos x + D sin x. Imposition of the given
initial conditions yields the equations
A+ B+C
2A-B
= 0
+D = 0
4.4 + 5-C
8,4-5
- 0
-Z) - 30
that we solve for A = 2, B = -5, C = 3, and Z) = -9. Thus
y(x) = 2c2a - 5e~x + 3 cos x - 9 sin x.
51.
In the solution of Problem 51 in Section 3.1 we showed that the substitution v = Inx
gives
and
y”
A further differentiation using the chain rule gives
86
Chapter 3
dy
1
d2y
x2 dv
x2
dv2
1
Substitution of these expressions for y\ y", and ym into the third-order Euler equation
3
fff
2
ax ym + bx~y" + cx y' + d y = 0 and collection of coefficients quickly yields the desired
constant-coefficient equation
3
2
a^-^- + (b-3a)^-^- + (c-b + 2a) — + dy = 0.
dv
dv~
dv
In Problems 52 through 58 we list first the transformed constant-coefficient equation, then its
characteristic equation and roots, and finally the corresponding general solution with v = In x
and ev = x.
53.
^ + 6^ + 25 y = 0;
dv2
dv
r2+6r + 25 = 0;
r = -3±4i
y(x) = e~3v [c, cos(4v) + c2 sin(4v)] = x“3 [c] cos(41nx) + c2 sin(41nx)]
3
55.
2
^-^--4^—^- + 4— = 0;
dv3
dv2dv
r3-4r2+4r = Q;
r= 0, 2, 2
y(x) = c, +c3e2r +c3ve2r = ct+x2(c2+c3 lnx)
57.
+
dv3
dv2
= 0;
r}-4r2+4r = 0;
r = 0, 3±S
dv
y(x) = c,+c2e(3_'|/2)v+c3ve(3+'^)v = c, + x3 ^CjX-^3 + c3x*^ j
SECTION 3.4
Mechanical Vibrations
In this section we discuss four types of free motion of a mass on a spring — undamped,
underdamped, critically damped, and overdamped. However, the undamped and underdamped
cases — in which actual oscillations occur — are emphasized because they are both the most
interesting and the most important cases for applications.
1.
Frequency: co0 = \fkim - V16/4 = 2 rad/ sec = l/7T Hz
Period: P-lnlco^-lnll- n sec
Section 3.3
87
3.
The spring constant is k = 15N/0.20m = 75 N/m. The solution of 3x" + 75x = 0
with x(0) = 0 and x'(0) = -10 is x(t) = -2 sin 5t. Thus the amplitude is 2 m; the
frequency is co0 = yjk/m = ^75/3=5 rad/ sec = 2.5In Hz ; and the period is 2zr/5 sec.
5.
The gravitational acceleration at distance R from the center of the earth is g = GM/R .
According to Equation (6) in the text the (circular) frequency co of a pendulum is given
by co2 = g!L = GM/R2L, so its period is p = 2n!co - 2nR^L! GM.
7.
The period equation p = 3960 Vl00.10 = (3960 + x) Vl00 yields x « 1.9795 mi «
10,450 ft for the altitude of the mountain.
11.
The fact that the buoy weighs 1001b means that mg = 100 so m = 100/32 slugs.
The weight of water is 62.4 lb/ft , so the F = ma equation of Problem 10 is
(100/32)x" = 100 - 62.4^r2x.
It follows that the buoy's circular frequency co is given by
®2
= (32)(62.4/r)r2/100.
But the fact that the buoy's period is p = 2.5 sec means that co = 2zr/2.5. Equating
these two results yields r « 0.3173ft « 3.8 in.
13.
(a)
The characteristic equation 1 Or2 + 9r + 2 = (5r + 2)(2r +1) = 0 has roots
r - -2/5, -1/2. When we impose the initial conditions x(0) = 0, x'(0) = 5 on the
general solution x{t) = cy~2"5 + c2e~"2 we get the particular solution
x(0 = 5b[e~2il5-e-112).
(b)
The derivative x'{t) = 25e~'12 -20e~2t/5 = 5e~2t,s(5e~tno-4) = 0
when t = 10 ln(5/4) « 2.23144 . Hence the mass's farthest distance to the right
is given by x(101n(5/4)) = 512/125 = 4.096.
15.
With damping The characteristic equation (1 / 2)r + 3r + 4 = 0 has roots r = - 2,-4.
When we impose the initial conditions x(0) = 2, x'(0) = 0 on the general solution
x(0 = cxe~2' + c2e~41 we get the particular solution x(/) = 4e~2t - 2e~4t that describes
overdamped motion.
Without damping The characteristic equation (1 /2)r2 +4 = 0 has roots r = ± 2i\[2.
When we impose the initial conditions x(0) = 2, x'(0) = 0 on the general solution
88
Chapter 3
u(t) = A cos(2\/2 /) + B sin(2V2 t) we get the particular solution u(t) = 2 cos(2V21).
The graphs of x(t) and u{t) are shown in the following figure.
t
With damping The characteristic equation r + 8r +16 = 0 has roots r = - 4, - 4.
When we impose the initial conditions x(0) = 5, x'(0) = -10 on the general solution
x(t) = (cx + c2t)e~Al we get the particular solution x(t) = 5e~4'(2t +1) that describes
critically damped motion.
Without damping The characteristic equation r +16 = 0 has roots r - ± 4i. When
we impose the initial conditions x(0) = 5, x'(0) - -10 on the general solution
u(t) = ^4cos(4/) + Z?sin(4/) we get the particular solution
u{t) = 5cos(4/)+—sin(4/) «— a/5 cos(4/-5.8195).
2
2i
The graphs of x(t) and u(t) are shown in the following figure.
Section 3.4
89
19.
With damping The characteristic equation 4r2 + 20r +169 = 0 has roots
r = -5/2±6/. When we impose the initial conditions x(0) = 4, x'(0) = 16 on the
general solution x(t) - e~57/2 ( Acos6t + Bsin 6/) we get the particular solution
e 5/ 2 [ 4 cos 6t + — sin 6/ ]
3
e“5'72 cos(6f - 0.8254)
that describes underdamped motion.
Without damping The characteristic equation 4r +169 = 0 has roots r = ± 13/72.
When we impose the initial conditions x(0) = 4, x'(0) = 16 on the general solution
u(t) - A cos(l 3t / 2) + B sin(l 31 / 2) we get the particular solution
4 cos
32 . (13,)
+ —sin
2 j
13
( 2 J
13,)
— 7233 cosf— f-0.5517
13
12
J
The graphs of x(t) and u(t) are shown in the following figure.
21.
With damping The characteristic equation r2 +1 Or +125 = 0 has roots r = - 5 ± 10/.
When we impose the initial conditions x(0) = 6, x'(0) = 50 on the general solution
x(t) = e~51 (A cos 1 Ot + B sin 1 Of) we get the particular solution
x(t) = e~5t(6 cos lOf + 8 sin lOf) « 10 e"5/cos(10f - 0.9273)
that describes underdamped motion.
Without damping The characteristic equation r2 +125 = 0 has roots r = ± 5^5 i.
When we impose the initial conditions x(0) = 6, x'(0) = 50 on the general solution
u{t) = Acos(5y/51) + Bsin(5yf51) we get the particular solution
u(t) = 6cos(^5V5
90
+ 2\/5 sin^5V5 f) « 2>/l4 cos{5^51- 0.6405j.
Chapter 3
The graphs of x(t) and u(t) are shown in the following figure.
t
(a)
With m = 100 slugs we get co = yjkl100 . But we are given that
co = (80 cycles/min)(2;r)(l min/60 sec) = 8tt/3,
and equating the two values yields k » 7018 lb/ft.
(b)
With co\ = 2^(78/60) sec"1, Equation (21) in the text yields c « 372.31
lb/(ft/sec). Hence p = c!2m « 1.8615. Finally
= 0.01 gives z1 « 2.47 sec.
The binomial series
(l + x)a = l + ax + ^Tx2 + g(a~1)(a~2h3 + v
7
2!
3!
converges if x < 1. (See, for instance, Section 11.8 of Edwards and Penney, Calculus,
6th edition, Prentice Hall, 2002.)
With a = 1 /2 and x = -c2 /4mk in Eq. (21) of
Section 3.4 in this text, the binomial series gives
co, =
7
^o
-/>
4
m V
8m/:
2 7.2
128m k
If x\ = x(^i) and *2 = xfe) are two successive local maxima, then co\t2 = co\t\+In
so
x\ = C Qxp(-pt\) cos(co\t\ - a),
\
X2 = C Qxp(-pt2) cos(a>\t2 - cc) = C exp(-^) cos(m^i - a).
Section 3.4
91
Hence x\/x2 = exp[-p(t\ - t2)\, and therefore
ln(xi/x2) = -piti - t2) = 27ip/(0\.
35.
The characteristic equation r2 + 2r +1 = 0 has roots r = -1, -1. When we impose the
initial conditions x(0) = 0, x'(l) - 0 on the general solution x(t) = (c} + c2t)e~' we get
the particular solution x^r) = te~'.
37.
The characteristic equation r2 + 2r + (1 +10 2,1) = 0 has roots r = -1 ± 10 "i. When we
impose the initial conditions x(0) = 0, x'(l) = 0 on the general solution
x(t) = e~‘ r^cos(l0”"?) + 5sin(l0-"^"
we get the equations c} = 0,
- c, +10”" c2 = 1 with solution cx = 0, c2 = 10". This
gives the particular solution x3(/) = 10"e sin(10 "/).
SECTION 3.5
NONHOMOGENEOUS EQUATIONS AND
THE METHOD OF UNDETERMINED COEFFICIENTS
The method of undetermined coefficients is based on "educated guessing". If we can guess
correctly the form of a particular solution of a nonhomogeneous linear equation with constant
coefficients, then we can determine the particular solution explicitly by substitution in the given
differential equation. It is pointed out at the end of Section 3.5 that this simple approach is not
always successful — in which case the method of variation of parameters is available if a
complementary function is known. However, undetermined coefficients does turn out to work
well with a surprisingly large number of the nonhomogeneous linear differential equations that
arise in elementary scientific applications.
In each of Problems 1-20 we give first the form of the trial solution ytriai> then the equations in
the coefficients we get when we substitute ytriai into the differential equation and collect like
terms, and finally the resulting particular solution yp.
1.
yM = Aeix;
25^4 = 1;
3.
ytrial = ^4cos3x + £sin3x;
= (l/25)e3*
-15^4-35 = 0, 3^4-155 = 2;
yp = (cos 3x - 5 sin 3x)/39
92
Chapter 3
5.
First we substitute sin2x - (1 - cos 2x)/2 on the right-hand side of the differential
equation. Then:
Ttrial = A+ Bcos2x + Csin2x;
,4 = 1/2, -35 + 2C = -l/2, -25-3C = 0;
yP = (13 + 3 cos 2x - 2 sin 2x)/26
7.
First we substitute sinh x = (ex - e~x)/2 on the right-hand side of the differential
equation. Then:
Ttdai = Aex + Be~A;
9.
-3A = 1/2, -3B = -1/2;
yp = (e~x - ex)/6 = -(l/3)sinhx
First we note that ex is part of the complementary function yc = c\ex + C2e~3x. Then:
y’t^ai = A + x(B + Cx)ex\
-3,4 = 1, 45 + 2C = 0, 8C = 1;
= -(1/3) + (2x2 - x)exl\6.
11.
First we note the duplication with the complementary function
yc = CjX + c2 cos 2x + c3 sin 2x. Then:
y*trial = x(A + Bx);
13.
y’trial =
4A = -1, 8B = 3;
(^cosx +^sinx);
yp = (3x2 - 2x)/8
7^1 + 45 = 0, -4^+ 75 = 1;
yp = ex(7 sin x - 4 cos x)/65
15.
This is something of a trick problem. We cannot solve the characteristic equation
r5 + 5r4 -1 = 0 to find the complementary function, but we can see that it contains no
constant term (why?). Hence the trial solution _ytrjal = A leads immediately to the
particular solution yp = -17.
17.
First we note the duplication with the complementary function yc = c} cos x + c2 sin x.
Then:
Ttriai “ *[(^ + 5x)cosx + (C + Z)x)sinx];
2B + 2C = 0, 4D = 1,-2,4 + 2D = 1,-45 = 0;
yp = (x sinx - x cos x)/4
19.
First we note the duplication with the part c, + c2x of the complementary function
(which corresponds to the factor r2 of the characteristic polynomial). Then:
Section 3.5
93
ytrial = x2(A +Bx + Cx2);
4A +125 = -1, 125 + 48C = 0, 24C = 3;
yp = (10x2 - 4x3 +x4)/8
In Problems 21-30 we list first the complementary function yc, then the initially proposed trial
function yh and finally the actual trial function yp in which duplication with the
complementary function has been eliminated.
.
21
yc =
(c, cosx+ c2 sinx);
y- = ex (4cosx + #sinx)
yp = x*ex(4cosx + Z?sinx)
23.
yc = c, cosx + c2 sinx;
= (4 + f?x)cos2x + (C + .Dx)sin2x
yp = x-[(4 + i?x)cos2x + (C +Z)x)sin2x]
25.
yc = cy
+ c2c
,
y{ = (A + Bx)e~x +(C + Dx)e~2x
yp = x-(A + Bx)e~x+x-(C + Dx)e~2x
27.
yc =
cosx + c2 sinx) + (c3 cos2x + c3 sin2x)
y{ = (A cosx + i?sinx) + (Ccos2x + Z) sin 2x)
yp = x-[(4cosx + 2?sinx) + (Ccos2x + Z)sin2x)]
29.
yc = (c, + c2x + c3x2 ^ex +c4e2x +c5e~2x;
y{ = (A + Bx)ex +Ce2x +De~2x
yp = x3-(A + Bx)ex +x-(Ce2x) + x\De-2x)
In Problems 31-40 we list first the complementary function yc, the trial solution ytr for the
method of undetermined coefficients, and the corresponding general solution yg = yc + yp where
yp results from determining the coefficients in ytr so as to satisfy the given nonhomogeneous
differential equation. Then we list the linear equations obtained by imposing the given initial
conditions, and finally the resulting particular solution y(x)-
94
Chapter 3
31.
yc = cx cos2x + c2 sin2x;
ytr = A + Bx
yg = cx cos2x + c2sin2x + x/2
cx — 1,
2c2 +1/2 — 2
.y(x) = cos2x + (3/4)sin2x + x/2
33.
yc = c} cos 3x + c2 sin 3x;
.y
ytT = ^cos2x + Fsin2x
= c, cos3x + c2 sin3x + (l/5)sin2x
cx = 1,
3c2 + 2/5 = 0
y(x) = (l5cos3x-2sin3x + 3sin2x)/15
35.
yc = ex (cx cosx + c2 sinx);
ytT = A + Bx
yg = ex (cx cosx + c2 sinx) + 1 + x/2
Cy
+ 1
— 3,
Cy
+ C2 + 1 / 2
—
0
y(x) = ex(4cosx-5sinx)/2 + l + x/2
37.
yc = Cj + c2ex + c3xex\
ytT = x-(A ) + x2 -(B + Cx)ex
yg = Cy + c2£ + c3x 6 + x — x<F/2 + x£?/6
Cj + c2 — 0,
c2 + c3 +1 = 0,
c2 + 2c3 —1 = 1
y(x) = 4 + x + ex (-24 + 18x -3x2 + x3 )/6
39.
yc = Cy + c2x + c3e~x;
yn = x2 •(,4 + i?x) + x-(Ce~j:)
yg = Cy + c2x + c3e~x - x2 !2 + x3 /6 + xe~x
Cy+c3
= 1,
c2 - c3 +1 = 0,
c3 -3 = 1
j/(x) = (-18 + 18x-3x2+ x3)/6 + (4 + x)e_*
41.
The trial solution ytr = A + Bx-\- Cx2 + Dx3 + Fx4 + Fx5 leads to the equations
Section 3.5
95
2A-B-2C-6D + 24E = 0
-2B -2C- 6D - 24E + \20F = 0
-2C-3D-12E-60F = 0
-2D-4E -20F = 0
-2E-5F = 0
-2 F = 8
that are readily solve by back-substitution. The resulting particular solution is
y(x) = -255 - 450x + 30x2 + 20x3 + 10x4 - 4x5.
43.
(a)
cos3x + /sin3x = (cosx + zsinx)3
= cos x + 3 i cos x sin x - 3 cos x sinA x - i sin x
When we equate real parts we get the equation
cos3 x-3(cosx)(l -cos2 x) = 4cos3x-3cosx
and readily solve for cos3 x = |cosx + {cos3x. The formula for sin3 x is derived
similarly by equating imaginary parts in the first equation above.
(b)
Upon substituting the trial solution yp = A cos x + B sin x + C cos 3x + D sin 3x
in the differential equation y" + 4y - f cosx + |cos3x, we find that A — 1/4, B — 0,
C = -1/20, D = 0. The resulting general solution is
y(x) = c\ cos 2x + C2sin 2x + (l/4)cos x - (l/20)cos 3x.
45.
We substitute
sin4x = (1 - cos 2x)2/4
= (1 - 2 cos 2x + cos22x)/4 = (3 - 4 cos 2x + cos 4x)/8
on the right-hand side of the differential equation, and then substitute the trial solution
yp = ^cos2x + £sin2x + Ccos4x + Dsin4x + E. We find that A = -1/10, B = 0,
C = -1/56, D = 0, E= 1/24. The resulting general solution is
y — ci cos 3x + C2sin 3x + 1/24 - (l/10)cos 2x - (l/56)cos 4x.
In Problems 47-49 we list the independent solutions yx and y2 of the associated homogeneous
equation, their Wronskian W = W(y}, y2), the coefficient functions
96
Chapter 3
r
ux(x) = -
r
*(*)/W dx
and
«,(a) =
J
iMM*
W(x)
J
in the particular solution yp = uxyx + u2y2 of Eq. (32) in the text, and finally yp itself.
47.
W = e
Ei =
E2 =
U\ = -(4/3)e3*,
U2 — 2c2a,
-3x
EP = (2/3)ex
49.
Ei
— ^2x
y2 = xe2x.
U\
U2 = 2x,
Ep = xV*
51.
yx = cos2x,
y2 = sin2x,
IE = 2
Liberal use of trigonometric sum and product identities yields
ux = (cos5x-5cosx)/20,
ux = (sin5x-5sinx)/20
yp = -(l/4)(cos 2x cos x - sin 2x sin x) + (l/20)(cos 5x cos 2x + sin 5x sin 2x)
= -(l/5)cos 3x (!)
53.
yi = cos 3x,
W
y2 = sin 3x,
u[ = -(2/3)tan 3x,
u'2 = 2/3
yp = (2/9)[3x sin 3x + (cos 3x)ln|cos3x ]
55.
yi = cos 2x,
y2 = sin 2x,
W = 2
w( = -(l/2)sin2x sin 2x = —(1/4)(1 -cos2x)sin2x
u2 = (l/2)sin2x cos 2x = (1/4)(1 - cos 2x)cos 2x
yp = (1 - x sin 2x)/8
57.
With jq = x, y2 = x~\ and fix) = 72x3, Equations (31) in the text take the form
xu\ + x"1 u2 = 0,
/
ux - x
—2
f
'vo 3
u2 = 12x .
Section 3.5
97
Upon multiplying the second equation by x and then adding, we readily solve first for
u[ = 36x'
so
u\ = 9x4
and then
u2 = -x2M' = -36r
SO
U2
—
-6x6.
Then it follows that
yP = y\U\+yiU2 = (x)(9x4) + (x'')(-6x6) = 3x5.
59.
2
y\ = x,
y2 = x2lnx,
W = X3,
fix) = X2
X
II
^ CN
3
u[ = -x In x,
X
■'t
ll
Q.
61.
y\ = cos(lnx),
y2 = sin(lnx),
W = 1/x,
X*) = (lnx)/x2
= (In x)cos(ln x) - sin(ln x)
^2
= (In x)sin(ln x) + cos(ln x)
TP = Inx (!)
63.
This is simply a matter of solving the equations in (31) for the derivatives
u\ =
1
and
W(x)
if' =
2
mm
W(x)
’
integrating each, and then substituting the results in (32).
SECTION 3.6
FORCED OSCILLATIONS AND RESONANCE
1.
98
Trial of x = A cos 21 yields the particular solution xp = 2 cos 21. (Can you see that because the differential equation contains no first-derivative term — there is no need to
include a sin It term in the trial solution?) Hence the general solution is
Chapter 3
x(/) = ci cos 3/ + c2sin 3/ + 2 cos 2/.
The initial conditions imply that ci = -2 and c2 = 0, so x(t) = 2 cos 2/ - 2 cos 3/.
The following figure shows the graph of x(/).
2jx
3.
First we apply the method of undetermined coefficients — with trial solution
x = A cos 5t + B sin 5/ — to find the particular solution
xp = 3 cos 5/ + 4 sin 51
3
—008
5
4
5/ + — sin 5t
5
= Scos^-/?)
where /? = tan '(4/3) « 0.9273. Hence the general solution is
x(t) = ci cos 10/ + c2 sin 10/ + 3 cos 5/ + 4 sin 5t.
The initial conditions x(0) = 375, jc'(O) = 0 now yield ci = 372 and c2 = -2, so
the part of the solution with frequency co - 10 is
xr = 372 cos 10/ - 2 sin 10/
= V138388
372
cos 10/-
V138388
sin 10/
V138388
= Vl 38388 cos(lO/ -a)
I
_
where a = 2n- tan" (1/186) « 6.2778 is a fourth-quadrant angle. The figure at the
top of the next page shows the graph of x(/).
Section 3.6
99
jt/5
t
5.
Substitution of the trial solution x = Ceos cot gives C = F§!(k - mco). Then
imposition of the initial conditions x(0) = x0, x'(0) = 0 on the general solution
x(t) = c, coscoQt + c2 sinco0t + Ccos cot
(where co0 = yfkJm)
gives the particular solution x(t) = (xo - Qcos coot + C cos cot.
In Problems 7-10 we give first the trial solution xp involving undetermined coefficients A and
B, then the equations that determine these coefficients, and finally the resulting steady periodic
solution xsp.
In each case the figure shows the graphs of x (t) and the adjusted forcing
function F] (t) = F(t) / mco.
t
100
Chapter 3
7.
x
= A cos 3/ + B sin 3/;
xjt)
50
-5,4 + 12# = 10,
10 (
120
cos 3/ +-sin
3/
169
169
„
•
13 V
oc = n - tan-1 (12/5) « 1.9656
12,4 + 5# = 0
5
12
^
—cos3/ +—sin 3/
13
13
y
10
13
cos(3/L-a)
(2nd quadrant angle)
The figure is shown at the bottom of the preceding page.
9.
xp = ^4cosl0/t +lOsinS/1;
60
,A
199A + 20B = 0,
597
•
20,4 + 199# = -3
ia
10/
xs„(o = - 40001 cos 10/-sin
40001
3
f
20
19.9
cos10/
V40001 V
V40001
a = n + tan-1 (199/ 20) « 4.6122
. ln^
sin 10/
J
COS(10/-6/)
740001
(3rd quadrant angle)
The figure is shown below.
X sp
t
Each solution in Problems 11-14 has two parts. For the first part, we give first the trial solution
xp involving undetermined coefficients A and #, then the equations that determine these
coefficients, and finally the resulting steady periodic solution xsp. For the second part, we give
first the general solution x(/) involving the coefficients c\ and <?2 in the transient solution,
then the equations that determine these coefficients, and finally the resulting transient solution
xtr so that x(/) = xtr(/) + x (/) satisfies the given initial conditions. For each problem, the
graph shows the graphs of both x(/) and x (/).
Section 3.6
101
11.
xp = A cos 3/ + B sin 3/;
-4A + UB = 10,
3 .
x(t) = —cos 3/ + — sin3/ =
Vio (
1
4
Vio
1
spv }
4
4
a = /r-tan ’(3) « 1.8925
\
cos3/ +
Vio
sin 3/
Vio
COS (31 -a)
J
(2nd quadrant angle)
x(t) = c~2'(c, cos/ +c2 sin/) +xsp(/);
xJt) = e -it
X2A + AB = 0
(l
7 . ^
cos t—sin t
4
V4
J
C\
^
0,
2c, “t £"2
4
- 0
V50 _2,f 1
7 . ^
-£ 1 ,-=COS/-F=Sin/
V50
V50
y
= — V2 c 2/cos(/-/?)
P = 2^-tan '(7) » 4.8543
(4th quadrant angle)
t
13.
xp = A cos 1 0/ + B sin 10/;
-74,4 + 205 = 600,
20,4 + 745 = 0
11100
1A
3000 . 1A
coslO/+-sin
10/
*sP(0 = 1469
1469
300 f
37
1A
10
. 1A
-7= —y
cos 10/ + ,
smlO/
Vl469 v V1469
Vl469
y
a - ;r-tan '(10/37) « 2.9320
102
c, + 5cn
Vl469
(2nd quadrant angle)
x(/) = e~' (c, cos5/ + c2 sin5/) +xsp(/);
c, -11100/1469 = 10,
300
-30000/1469
Chapter 3
cos (10/- or
-/
(25790 cos 51 - 842 sin 5t)
*tr(0
1469
2^166458266
12895
1469
v Vl 66458266
0 113314
= 2J-e
1469
cos5t -
421
Vl 66458266
A
sh^
J
cospr-Z?)
V
7
/? = 2;r -tan 1 (421 /12895) « 6.2505
(4th quadrant angle)
i
t
In Problems 15-18 we substitute x{t) = A(co) cos cot + B(co) sin cot into the differential equation
mx" + ex' + kx = F0 cos cot with the given numerical values of m, c, k, and F0. We give first
the equations in A and B that result upon collection of coefficients of cos cot and sin cot, and
then the values of A(co) and B(co) that we get by solving these equations. Finally,
c = VZ + B2 gives the amplitude of the resulting forced oscillations as a function of the
forcing frequency co, and we show the graph of the function C(co).
15.
(2-co2)A + 2coB
= 2,
-2coA + (2-co2)B = 0
2(2-co1)
.
4 +co
4
5
C(co) = 2/v4 + co4
A
4 + co
4
begins with C(0) = 1 and steadily decreases as co increases.
Hence there is no practical resonance frequency. The graph of C(co) appears at the top
of the next page.
Section 3.6
103
c
CO
17.
(45-co2)A + 6coB = 50,
-6co A + (45-co2)B = 0
50 (45 — <x>2)
A =
B
2025 - 54co2 + co
300co
2025 - 54co2 + co
C(c6) - 50/^2025-54co2 + co4
so, to find its maximum value, we calculate the
derivative
C'(co) =
-100ft>(-27 + co2)
(2025-54<y2 W)3/2
Hence the practical resonance frequency (where the derivative vanishes) is
co = V27 = 3>/3. The graph of C(co) is shown below.
C
co
104
Chapter 3
19.
m - 100/32 slugs and k = 1200 lb/ft, so the critical frequency is co0 = -Jkl m
=
21.
V384 rad/sec = V384/2;r « 3.12 Hz.
If 0 is the angular displacement from the vertical, then the (essentially horizontal)
displacement of the mass is x = LO, so twice its total energy (KE + PE) is
mix')2 + kx2 + 2mgh = mL1{Or)2 + kL2$ + 2mgL{\ - cos 0) — C.
Differentiation, substitution of 0 « sin 6\ and simplification yields
0"+ {k/m + g/L)0 = 0
so
=
23.
(a)
F/m + gIL.
In ft-lb-sec units we have m = 1000 and k = 10000, so co0 = VlO rad/sec
« 0.50 Hz.
(b)
We are given that co = In 12.25 « 2.79 rad/sec, and the equation
mx" + kx = F(t) simplifies to
x" + 10* = (V 4) co2 sin cot.
When we substitute x(t) = A sin cot we find that the amplitude is
A = ®2/4(l0-®2) * 0.8854 ft « 10.63 in.
25.
Substitution of the trial solution x = A cos cot + B sin cot in the differential equation, and
then collection of coefficients as usual yields the equations
[k -mco2^jA + (cco)B = 0,
- (cco) A + (k - mco2 ^B = F0
with coefficient determinant A = (k-mco2 j +(cco)
and solution A = —(cco)FqIA,
Z? = (k-mco2^F0/A. Hence
x{t) =
F
0
Va
k - mco2 .
sin cot -—r= cos cot
VA
Va
where C = F0/Va and sin or =
= Csin(W
/ Va, cosa = (*
Section 3.6
a),
mco )/7a.
105
27.
The derivative of C(co) = F0/^j{k - mco2) +(cco)2
coF0
C\co)
is given by
(c2 - 2km) + 2(mo?y
—13 / 2 •
\ 2
/
Ik-mco2 j
(a)
Therefore, if c2 > 2km, it is clear from the numerator that C'(co) < 0 for all co,
so C(&>) steadily decreases as co increases.
(b)
But if c2 <
then the numerator (and hence C\co)) vanishes when
co = com - y[kl m-c212m2 < yjkl m = co0. Calculation then shows that
=
v
m/
-2k)
. /
c‘
_ \ 3/2
5
-c J
so it follows from the second-derivative test that C{com) is a local maximum value.
29.
We need only substitute E0 = acco and F0 = ak in the result of Problem 26.
SECTION 3.7
ELECTRICAL CIRCUITS
1.
With E{t) = 0 we have the simple exponential equation 5/'+ 251 = 0 whose solution
with 7(0) = 4 is 7(0 = 4e~5'.
3.
Now the differential equation is 5/'+ 25/ = 100 cos 60/ Substitution of the trial
solution
/p = A cos 60/ + B sin 60/
yields
Ip = 4(cos 60/ + 12 sin 60/)/145.
The complementary function is Ic = ce 5/; the solution with 7(0) = 0 is
/(/) = 4(cos 60/ + 12 sin 60/ - e 5/)/l 45.
106
Chapter 3
5.
The linear equation 7'+ 10/ = 50 e~10'cos 60/ has integrating factor p = e[0>. The
resulting general solution is 7(/) = e~]0‘ [(5/6)sin60/ + C].] To satisfy the initial
condition 7(0) = 0, we take C= 0 and get I(t) = (5/6>f 10'sin 60/.
7.
(a)
The linear differential equation RQ'+ (l/C)Q = E0 has integrating factor
p - e,IRC. The resulting solution with 0(0) = 0 is Q(t) = £0C(1 - e~t/RC). Then
lit) = Q'it) = (E0/R)e~t/RC.
(b) These solutions make it obvious that lim0(/) = £oC and lim7(/) = 0.
f—>oO
9.
/—>00
Substitution of the trial solution 0 = A cos 120 / + B sin 120/ into the differential equation
_ 200 0' + 40000 = 100 cos 120/ yields the equations
4000,4 + 240007? = 100,
-24000,4 + 40005 = 0
with solution A = 1/1480, B = 3/740. The complementary function is 0c = ce~20'., and
imposition of the initial condition 0(0) = 0 yields the solution
0(/) = (cos 1207 + 6 sin 1207 - e~20/)/1480. The current function is then
I(t) = Q'it) - (36 cos 120/ - 6 sin 120/ + e~m)HA. Thus the steady-periodic current is
(6 cos 120/ - sin 120/)
74
6-Jyi ( 6
74
\
sin 120/
cos 120/-
V V37
V37
cos(120/-a)
y
(with a - 2n - tan 11), so the steady-state amplitude is 3/ VTT
In Problems 11-16, we give first the trial solution I
- A cos cot + B sin cot, then the equations
in A and B that we get upon substituting this trial solution into the RLC equation
L I" + R 7' + (1 /C) 7 = E'{t), and finally the resulting steady periodic solution.
11.
7p = A cos 2/ + B sin 2/;
A + 6B = 10,
-6A + B = 0
r
,
10
„
60 . .
10
I At) = —cos 2/ H-sin 2/ = —= ,
sp
37
1
.
6
. . A
,— cos 2/+ -7= sin 2/
V37 1V37
'Jyi
j
37
8 = 2^--tan 1 (1 /6) « 6.1180
10
sin(2/-c>)
V37
(4th quadrant angle)
Section 3.7
107
13.
/
= A cos 5/ + B sin 5/;
_ , .
40
p
13
r
3,4-2# - 0,
60 . c
20 f 2
13
VTnVn
5 = 2^-tan '(2/3) « 5.6952
15.
/
2A + 3B = 20
c
3
20
.
Vb
J
sin(5 t — S)
Vl3
(4th quadrant angle)
= ,4 cos 60;zt + # sin 60;z7;
(1000-36n2)A + 30;r# = 33;r,
A =
33;r(250-9;r2)
15^-(500-18^2)# - 0
250000-17775^+324^
7sp(0 * 70sin(60;rf-£);
495;r‘
# -
2(250000 -17775;r2 + 324;r4)
33;r
70 =
0.1591
2^250000 -1 7775+ + 324+
# = 2# -tan
500-18+^1
-i
V
15;r
4.8576
In each of Problems 17-22, the first step is to substitute the given RLC parameters, the initial
values 7(0) and 0(0), and the voltage E{t) into Eq. (16) and solve for the remaining initial
value
I'm = |[£(O)-i?/(O)-(l/C)0(O)].
(*)
17.
With 7(0) = 0 and 0(0) = 5, Equation (*) gives 7'(0) = -75. The solution of the
RLC equation 27"+167'+ 507 = 0 with these initial conditions is I{t) = -25e~4'sin 37
19.
Now our differential equation to solve is
27" + 607' + 4007 = -1000e~10'.
-i
We find the particular solution 7P = -50te~
At e~XOt; the general solution is
r\ a
by substituting the trial solution
I(t) = cxe-m + c2e-m-50te-10'.
The initial conditions are 7(0) = 0 and 7'(0) = -150, the latter found by substituting
L = 2,R = 60, 1/C = 400, 7(0) =0, 0(0) = 1, and E(0) = 100 into Equation (*).
Imposition of these initial values on the general solution above yields the equations
cx+c2 = 0,
— 1 Ocj -20c2 -50 = -150 with solution c, = — 10,
get the solution
108
Chapter 3
c2=10. Thus we
7(0 - \0e~m - \Oe~]0t - 50te~l0t
21.
The differential equation 10/" +20/'+1007 = -1000 sin 5t has transient solution
7tr(0 = e~l (c, cos 3t + c2 sin 3t),
and in Problem 13 we found the steady periodic solution
20
Isp(t) = —(2008 5^ +3 sin 5/).
When we impose the initial conditions 7(0) = 0, 7'(0) = -10 on the general solution
7(0 = 7tr(0 + Isp(t), we get the equations
cx +40/13 = 0,
-c] +3c2 +300/13 = -10
with solution c] = -40/13,
c2 = -470/39. The following figure shows the graphs
of 7(0 and 7sp(0-
I
23.
The LC equation 7,7" + (l/C)7 = 0 has general solution 7(0 = c, cos co0t + c2 sin (o0t
with critical frequency co0 = 1 / VZc.
25.
According to Eq. (8) in the text, the amplitude of the steady periodic current is
+ (coL -1 / coC)2. Because the radicand in the denominator is a sum of squares,
it is obvious that the denominator is least when coL-\lcoC = 0, that is, when
a)
= i/VZc.
Section 3.7
109
SECTION 3.8
ENDPOINT PROBLEMS AND EIGENVALUES
The material on eigenvalues and endpoint problems in Section 3.8 can be considered optional at
this point in a first course. It will not be needed until we discuss boundary value problems in the
last three sections of Chapter 9 and in Chapter 10. However, after the concentration thus far on
initial value problems, the inclusion of this section can give students a view of a new class of
problems that have diverse and important applications (as illustrated by the subsection on the
whirling string). If Section 3.8 is not covered at this point in the course, then it can be inserted
just prior to Section 9.5.
1.
If X = 0 then y" = 0 implies that y(x) = A + Bx. The endpoint conditions
j/(0) = 0 and y(l) = 0 yield B = 0 and A = 0, respectively. Hence X = 0 is
not an eigenvalue.
If X = a2 > 0, then the general solution of y" + cfy — 0 is
=
A cos ax + B sin ax,
so
y\x) - -A a sin ax + Ba cos ax.
Then j/(0) = 0 yields B = 0, so y(x) = A cos ax. Next y{ 1) = 0 implies that
cos a = 0, so a is an odd multiple of nil. Hence the positive eigenvalues are
{(2n - \)27?/A) with associated eigenfunctions {cos(2« - l);zx/2} for n = 1, 2, 3, • • •.
3.
Much as in Problem 1 we see that X - 0 is not an eigenvalue. Suppose that
X = o'2 > 0, so
j/(x) = A cos ax + B sin ax.
Then the conditions y(-7f) = y(zr) = 0 yield
A cos azr+ B sin an = 0,
A cos an - B sin an = 0.
It follows that
A cos an - 0 = B sin an.
Hence either A = 0 and B ^ 0 with an an even multiple of n/2, or A & 0 and
B = 0 with an an odd multiple of n/2. Thus the eigenvalues are {n /4} for n a
positive integer, and the ^th eigenfunction is yn(x) = cos(rac/2) if n is odd,
yn(x) = sm{nx/2) if n is even.
110
Chapter 3
5.
If X = a2 > 0 and
y(x) =
A cos ocx + 5 sin ax,
y'(x) = -A a sin ocx + Ba cos ax
then the conditions y(-2) = j/(2) = 0 yield
^4 cos 2a - B sin 2a = 0,
-A sin 2a + B cos 2a = 0.
It follows either that A = B and cos 2 a = sin 2 a, or that A = -B and
cos 2a = -sin 2a. The former occurs if
2a = n!4, 57r/4, 9/t/4, • • •,
the latter if
2a = 37r/4, Itc!A, 11^/4, • • •.
Hence the nth eigenvalue is
A„ = a2„ = (In - l)V/64
for n = 1, 2, 3, • • •, and the associated eigenfunction is
yn(x) = cos anx + sin anx (n odd)
or
yn(x) — cos anx - sin anx (n even).
7.
(a)
If X — 0 and y(x) = A + Bx, then y(0) = A = 0, so jy(x) =
j>(l)+y(l) = 25 = 0, so A = B = 0 and Z = 0 is not an eigenvalue.
But then
(b)
0 so
If X - a2 > 0 and ^(x) = v4 cos ax + B sin or, then jv(0) = A
y(x) = B sin ax. Hence
=
j/(1)+j/(1) = B(sm a + a cos a) = 0.
so a must be a positive root of the equation tan a = -a, and hence the abscissa of a
point of intersection of the lines y = tanz and y = -z. We see from the figure at the top
of the next page that att lies just to the right of the vertical line z = (2n -1)^/2, and
lies closer and closer to this line as n gets larger and larger.
Section 3.8
111
y
9.
If y” + Ay = 0 and X = -a2 < 0, then
y(x) = Ae°* + Be'0*.
Then y(0) = A + B = 0, so B = -A and therefore
y(x) = Aie^-e-™).
Hence
y'(L) = Aoieal + e~aL) = 0.
But a^O and eaL + e~aL>0, so ^4 = 0. Thus X = -a2 is not an eigenvalue.
11.
If T = - a2 < 0, then the general solution of y" + Xy = 0 is
y(x) = A cosh ax + B sinh ax. Then j/(0) = 0 implies that B = 0, so y(x)= cosh ax
(or a nonzero multiple thereof). Next,
y(l) + y(l) = cosha + asinha = 0
implies that tanha = -1 la. But the graph of y = tanha lies in the first and third
quadrants, while the graph of y = -1 / a lies in the second and fourth quadrants. It
follows that the only solution of tanh a = -1/a is a = 0, and hence that our
eigenvalue problem has no negative eigenvalues.
112
Chapter 3
13.
(a)
With A = 1, the general solution of y" + 2y' + y = 0 is
y(x) = Ae~x + Bxe~x.
But then 3/(0) = A = 0 and XI) = e~l(A + B) = 0. Hence A - 1 is not an
eigenvalue.
(b)
If A < 1, then the equation y" + 2y' + Xy = 0 has characteristic equation
r2 + 2r + A = 0. This equation has the two^distinct real roots (-2± V4-4T j/2 ; call
them r and s. Then the general solution is
y(x) = Aerx + Besx,
and the conditions X0) = y( 1) = 0 yield the equations
A +B = 0,
If A, B ^ 0, then it follows that e
(c)
Aer + Bes — 0.
= es. But r ^ s, so there is no eigenvalue A< 1.
If A > 1 let A - 1 = a2, so A = 1 + a?. Then the characteristic equation
r2 + 2r + A = (r + 1 )2 + a2 = 0
has roots
-1 ± a i, so
X*) = e~x(A cos ax + B sin ax).
Now X0) = A = 0, so Xx) = Ae~xsm ax. Next, XI) = ^e_1sin a = 0, so
a - nn with n an integer. Thus the ndn positive eigenvalue is An = n2T?+ 1.
Because A - c? + 1, the eigenfunction associated with An is
yn(x) = e~xsinnnx.
15.
(a)
The endpoint conditions are
X0) = /( 0) = y"(L) = y3)(i) = 0.
With these conditions, four successive integrations as in Example 5 yield the indicated
shape function Xx).
(b)
The maximum value ymax of y(x) on the closed interval [0, L] must occur either
at an interior point where X(x) = 0 or at one of the endpoints x = 0 and x = L. Now
y'(x) = k(4x3 - 12Tx2 + 12Z2x) = 4Ax(x2 - 3Lx + 3Z2)
Section 3.8
113
where k = w/24EI, and the quadratic factor has no real zero. Hence x = 0 is the only
zero of y'(x). But y(0) = 0, so it follows that ymax = y(L).
17.
If y(x) = k(x4 -2Lx3 + L3x) with k = w/24EI, then
y\x) = A:(4x3 -6Lx2 +1?) = 0
has the solution x = LI2 that we can verify by inspection. Now long division of the
cubic 4x -6Lx + L
by 2x-L yields the quadratic factor 2x -2Lx -L
whose
zeros 12L ± Vl2Zr j/4 = ^l±V3^Z,/2 both lie outside the interval [0, L\. Thus
x = L/2 is, indeed, the only zero of j/(x) = 0 in this interval.
114
Chapter 3
CHAPTER 4
INTRODUCTION TO SYSTEMS
OF DIFFERENTIAL EQUATIONS
This chapter bridges the gap between the treatment of a single differential equation in Chapters 1-3
and the comprehensive treatment of linear and nonlinear systems in Chapters 5-6. It also is
designed to offer some flexibility in the treatment of linear systems, depending on the background
in linear algebra that students are assumed to have — Sections 4.1 and 4.2 can stand alone as a very
brief introduction to linear systems without the use of linear algebra and matrices. The final Section
4.3 of this chapter extends to systems the numerical approximation techniques of Chapter 2.
SECTION 4.1
FIRST-ORDER SYSTEMS AND APPLICATIONS
1.
Let xl = x and x2 = x\ - x, so x'2 = x" - -lx - 3x' + t2.
Equivalent system:
x[
3.
=
X2,
x'2 = -7x\ - 3X2 + t
Let x, = x and x2 = x\ - x, so x2 = x" - (l-^jx-rxj/?2.
Equivalent system:
x[ = X2,
5.
t2 X2
T
= (1 - t2)xi - tX2
/
/
/
= X3,
X3
n
m
/
(
/\2
,
Let Xj = x, x2 = Xj = x , x3 = x2 = x , so x3 = x = (x J + cosx.
Equivalent system:
x[
_
7.
= X2,
X2
T
= X2 + COS Xj
//
/
/
/
n
y
//
2
,
2\3/2
Let Xj = x, x2 — x, — x, y{ = y, y2 = y{ = y so x2 = x =-kx/(x + y )
/
tt
i
//
2
,
2\3/2
y2 = y =-kyl(x +y )
,
.
Equivalent system:
/
t
(
1
= X2,
X2
= -kxt /[x; + >' J
^ = )>2,
^2
=
X,
+^1
)
Section 4.1
115
9.
Let x, = *, x2 = Xj = x, y, = y, y2 = y' = y', zl = z, z2 = z[ = z, so
x2 = x" -3x- y + 2z, y2 = y*= x+ y -4z, z2 = z" = 5x - y - z.
Equivalent system:
11.
x\ - x2,
x2 = 3x\ - yi + 2z\
y[ = yi,
y'2
= *1 + y\ - 4zi
Zl
Z2
=
= Z2,
- Jl “ Zl
The computation x" = y’ = -x yields the single linear second-order equation
x" + x = 0 with characteristic equation r2 + 1 =0 and general solution
x(t) = A cos t + B sin t.
Then the original first equation y = x' gives
y(t) = B cos t - A sin t.
The figure below shows a direction field and typical solution curves (obviously
circles?) for the given system.
X
13.
The computation x" = -2y' = -4x yields the single linear second-order equation
x" + 4jc = 0 with characteristic equation r2 + 4 = 0 and general solution
x(t) = A cos 2t + B sin 2t.
116
Chapter 4
Then the original first equation y = -x72 gives
y{t) = -B cos 2r + A sin 21.
Finally, the condition x(0) = 1 implies that A = 1, and then the condition y(0) = 0
gives B = 0. Hence the desired particular solution is given by
x(t) = cos 21,
y(t) = sin 21.
The figure on the left below shows a direction field and some typical circular solution
curves for the given system.
The computation x” = y72 = -4x yields the single linear second-order equation
x" + 4x = 0 with characteristic equation r2 + 4 = 0 and general solution
x{t) = A cos 21 + B sin 21.
Then the original first equation y = 2x' gives
y(t) = 4B cos 21 - 4A sin 21.
The figure on the right above shows a direction field and some typical elliptical
solution curves.
The computation x" = y' = 6 x - y = 6 x - x' yields the single linear second-order
equation x" + x' - 6 x = 0 with characteristic equation r + r - 6 = 0 and characteristic
roots r = -3 and 2, so the general solution
x(t) - A e~3t + B e2t.
Section 4.1
117
Then the original first equation y = x' gives
y(t) = -3Ae~2' + 2Be2‘.
Finally, the initial conditions
x(0) = A + B = 1,
y(0) = -3A + 2B = 2
imply that A = 0 and B = 1, so the desired particular solution is given by
x(t) = e~3t,
y(t) = 2 e2'.
The figure below shows a direction field and some typical solution curves.
19.
The computation x" = -y' = -13x-4y = -\3x + Ax' yields the single linear secondorder equation x" - 4x'+ \3x = 0 with characteristic equation r2-4r+13 = 0 and
characteristic roots r - 2 ±3 i, hence the general solution is
x(t) = e2t(A cos 3t + B sin 3t).
The initial condition x(0) = 0 then gives A = 0, so x(t) = B e2lsm 31. Then the original
first equation y = -x' gives
y(t) = -e2t{3B cos 31 + 2B sin 31).
Finally, the initial condition y(0) = 3 gives B = -1, so the desired particular solution is
given by
x{t) = -e2/sin 31,
118
y(t) = e2t(3 cos 3t + 2 sin 3t).
Chapter 4
The figure below shows a direction field and some typical solution curves.
21.
(a)
Substituting the general solution found in Problem 11 we get
x2 + y2 = (A cos t + B sin t)2 + (.B cos t - A sin t)2
= (A2 + £2)(cos2r + sin20 = A2+ B2
2,2
n2
x +y = C ,
2
2 1/2
the equation of a circle of radius C = (A + B )
(b)
.
Substituting the general solution found in Problem 12 we get
X2 -y2 =
CAe' + Be'')2 - (Ae‘ - Be'1)2 = 4AB,
the equation of a hyperbola.
23.
When we solve Equations (20) and (21) in the text for
2x-y = 3Ae-/and
and e2t we get
x + y = 3 Be2t.
Hence
(2x-yf(x + y) = (3Ae~')2(3Be2') = 27A2E = C.
Clearly y = 2x or y = -x if C = 0, and expansion gives the equation
Ax - 3xy2 + y3 = C.
Section 4.1
119
25.
Looking at Fig. 4.1.12 in the text, we see that
my"- -Tsin 0, + Tsin 02 ~ - T tan <9, + T tan 02 = -Ty] /L + T(y2 - y})/L,
my" = - Tsin 02 - T sin 03 ~ - 7 tan 02 - T tan
= -T(y2- yx)lL- Ty2 /L.
We get the desired equations when we multiply each of these equations by LIT and set
k = niL/T.
27.
We apply Kirchhoff s law to each loop in Figure 4.1.14 in the text, and immediately get the
equations
2( I[ -12) + 501 \ = 100 sin 601,
29.
2(I2 -I[) + 2512 = 0.
If 6 is the polar angular coordinate of the point (x, y) and we write
F = ^/ (x2 + y2) = /:/r2, then Newton's second law gives
mx" = -Fcos# = - (k / r2)(xl r) = - kx/r3,
my/r = -Fsin^ = -(k / r2)(y / r) = -ky / r3,
31.
If r = (x, y, z) is the particle's position vector, then Newton's law rar" = F gives
i
rar* = ^vxB = q x
j
/
0
y
k
/
0
z
/
= +(75y'i-^5xj = qB(—y\x\ 0).
5
SECTION 4.2
THE METHOD OF ELIMINATION
1.
The second differential equation y = 2y has the exponential solution
y(t) = c2e2'.
Substitution in the first differential equation gives the linear first-order equation
x' + x = 3c2e2t with integrating factor p-e1. Solution of this equation in the usual way
gives
x(t) = e~J (c, + c2£3r) = c]e~t+c2e2t.
120
Chapter 4
The left-hand figure below shows a direction field and some typical solution curves.
X
3.
From the first differential equation we get y = (3x + jO/2, so y' = (3x' + x")/2.
Substitution of these expressions for y and y' into the second differential equation yields
the second-order equation
x" -x' + 6x = 0
with general solution
x(t) = cxe~2t + c2e2t.
Substitution in y = (3x + x')/2 now yields
y(t) = \cxe~2t +3c2e3t.
Imposition of the initial conditions x(0) = 0, y(0) = 2 now gives the equations
q + c2 = 0, cx /2 + 3c2 = 0 with solution cx = - 4/5, c2 = 4/5. These coefficients give the
desired particular solution
x(t) = 4(e3' - e"2,)/5,
y(0 = 2(6e3' - e‘2')/5.
The right-hand figure above shows a direction field and some typical solution curves.
5.
Substitution of y - -(x' + 3x)/4 and y = - (x"+3x')/4 —from the first equation —
into the second equation yields the second-order equation xn + 2x' + 5x = 0 with general
solution
x(t) - e~] (q cos2t + c2 sin2/).
Section 4.2
121
Substitution of this solution in 3; = — (x' + 3x)/4 gives
y(t) = —e 1 [-(c, + c2)cos2r + (c]-c2)sm2t~\.
2
>
The figure below shows a direction field and some typical solution curves.
-3
7.
-2
-1
1
2
3
4
5
Substitution of y = x' - 4x - 2r and y — x" - Ax' -2 — from the first equation —
into the second equation yields the nonhomogeneous second-order equation
x"-5x' + 6x - 2-21. Substitution of the trial solution xp=A + Bt yields
A = 1/18, B = -1/3 so x=l/18-f/3. Hence the general solution for x is
x(t) = cxe2t + c2e2t -f/3 + 1/18.
Substitution in y = x'-4x-21 now yields
y(t) = -2cxe2t-c2e^ — 2//3 — 5/9.
9.
Substitution of y = (-x' +2x + 2sin2/)/3 and y = (-x/r + 2x/ + 4cos20/3—from
the first equation — into the second equation yields the second-order equation
x" — x — 1 cos2t + 4sin2f with general solution
x(t) = cxe 1 + c2e'
-(7cos2r+ 4sin2f)
Substitution in y = (-x' + 2x + 2 sin 2t) / 3 now yields
122
Chapter 4
y(0 = Cxe
^■(2cos2r+ 4sin2r)
H—c1et
3 2
Imposition of the initial conditions x(0) = 3, y(0) = 2 now gives the equations
cx = 3, - c, / 3 + c2 / 3 = 2 with solution q = 3, c2 = 9. These coefficients give the
desired particular solution
x(t) = e 2t (3cos3t + 9sin3r),
11.
y(r) = e 2' (2cos3?-4sin3f).
First we solve the given equations for the normal-form first-order equations
x = 3x-9 y + e r + 2er,
y = Ix-ly + e'1 ll + le* !2.
Substitution of y = (-x' + 3.x:-l-e_r + 2e')/9 and y' = (-x'+ 3x'- e-' + 2el)/9 —from
the first equation — into the second equation yields the nonhomogeneous second-order
equation x” + 9x = (5e~r +tie1) / 2. Substitution of the trial solution xp = Ae~! + Be1
yields A = -1/4, B = -11/20 so xp = -e r/4-lle'/20. Hence the general solution for
x is
x (t) = Cj cos3r + c2 sin3r-—e
Substitution in y = (~x' + 3x + e r+2e‘)/9 now yields
1
1
i
y(t) = — (cj - c2)cos3t + — (cj + c2) sin 3f + — e 1
3
3
2
13.
The first equation yields y = (x" + 5x)/2, so y" = (x(4) + 5x,r)/2. Substitution in the
second equation yields
x(4) + \3x" + 36x = 0.
The characteristic equation is r4 + 13r2 + 36 = (r2 + 4)(r2 +9) = 0, so the general
solution for x is
x(t) = a, cos 2t + a2 cos2t + bx cos 3t + b2 sin 3t.
Substitution in y = (xn + 5x)/2 now gives
1
1
y(t) - — ax cos2t +—a2cos2t -2bx cos 3? - 2&2 sin 3/.
Section 4.2
123
15.
If we write the given differential equations in operator notation as
(D2 -2)x-3Dy = 0
3Dx + (D2 — 2)y = 0,
we see that the system has operational determinant
(D2 -2)2 + 9D2 = D4 +5D2 +4 = (D2 + 1)(D2 +4). ‘
Therefore (as in Example 3) we see that x satisfies the fourth-order differential equation
(Z)2 +1 )(D2 + 4)x = 0 with characteristic equation (r2 + l)(r2 + 4) = 0 and general
solution
x(t) = a, cos t + a2 sin t + b{ cos 2t + b2 sin 2t.
Similarly, the general solution for y is of the form
y(t) = c]cost + c2sint + dlcos2t + d2sin2t.
Now, substitution of these two general solutions in the first equation xn - 7>y - 2x = 0
and collection of coefficients gives
(-3a, - 3c2)cos£ + (3c, -3a2)sinr + (-6Z?, -6J2)cos2r + (3^/, -3Z?2)sin2r = 0.
Thus we see finally that c, =a2, c2=-al, dx-b2, d2=-bv Hence
y(t) = a2 cost-a{ sint + b2cos2t-b]sm2t.
17.
If we write the given equations in the operational form
(D2-3D-2)x+
(D2-D + 2)y = 0,
(2D2-9D-4)x + (3D2-2D + 6)y = 0
we see (thinking of the operational determinant) that x satisfies a homogeneous fourthorder equation with characteristic equation
(r2 —3r-2)(3r2-2r + 6)-(r2 —r + 2)(2r2—9r-4)
= r4 — 3r2 — 4 = (r2 + l)(r2-4) = (r2 + l)(r + 2)(r-2) = 0.
Hence the general solution for x is
124
Chapter 4
x(t) = ax cos t-\-a2 sin t + bxe~2t+b2e2\
and, similarly, the general solution for y is
y(t) = q cosr + q, sin? + dxe~2t +d2e2t.
To determine the relations between the arbitrary constants in these two general solutions, we
substitute them in the first of the original differential equations and get
(-a, cos t - a2 sin t + 4bxe~2' + 4b2e2t) + (-q cos t - c2 sin t + 4dxe~2t + 4d2e2t ) +
(3a, sin? — 3a2 cost + 6b]e~2’ - 6b2e2,^j + (q sin?-c2 cos? + 2dxe~2t -2d2e2'^ +
(-2a, cos t - 2a2 sin t - 2bxe~2t - 2b2e2t) + (+2q cos t + 2c2 sin t + 2d]e~2t + 2d2e2t)
= 0
If we collect coefficients of the trigonometric and exponential terms we get the equations
-3a, - 3a2 + q - c2 =0,
3a, - 3a2 + q + c2 =0
8bx + 8dx = 0,
and
-4£>2 + 4d2 = 0.
The first two of these equations imply that q = -3a2 and c2 = 3a,, while the latter two
give dx = —bx, d2-b2 We therefore see finally that
y(t) = - 3a2 cos ? + 3a, sin t-bxe~ +b2e .
The operational determinant of the given system is
D-4
2
0
4
D-4
2
0
4
D-4
= Z)3 - 12D2 + 32D,
so x, y, and z all satisfy a third-order homogeneous linear differential equation with
q
o
characteristic equation r — 12r +32r = r(r-4)(r — 8) = 0. The corresponding general
solutions are
x(t) = a, +a2e4t + a3e8',
y(t) = bx +b2e4t +b3eSt,
z(t) = q + c2e4t +c3eSt.
If we substitute x(t) and y(t) in the first differential equation x = 4x - 2y and collect
coefficients of like terms, we find quickly that bx = 2a,, b2= 0, and Z?3 = -2av
Similarly, we find by substitution in the other two equations that q = 2a,,
c2 - -2a2,
and c3=2a3.Thus y and z are given by
Section 4.2
125
y(t) = 2al-2a3eSt
21.
LjLj —
and
z(t) = 2a, -2a2e4t +2a3e8t.
because both sides simplify to the same thing upon multiplying out and
collecting terms in the usual fashion of polynomial algebra. This "works" because different
powers of D commute — that is,DlDj=DjDl because Dl(Djx) = Dl+jx = Dj(Dlx).
23.
Subtraction of the two equations yields x + y = e~2t - e~3t. We then verify readily that any
two differentiable functions x(t) and y(t) satisfying this condition will constitute a solution
of the given system, which thus has infinitely many solutions.
25.
Infinitely many solutions, because any solution of the second equation also satisfies the first
equation (because it is D + 2 times the second one).
27.
Subtraction of the second equation from the first one gives x + y =
in the second equation yields
Then substitution
x{t) = D2(x + y) =
It follows that y(t) = 0, so there are no arbitrary constants.
29.
Addition of the two given equations yields D2x = e \ so x(t) = e 1 + a\t + #2- Then the
second equation gives D2y = a\t + a2, so
y(t) = (\/6)a[t3 + (\/2)a2t2 + ayt + a\.
Thus there arofour arbitrary constants.
31.
Substitution of Z2 = (Z1' + 25ZI -50)/25 and I'2 = (l"+251[)/25 — from the first
equation — into the second equation yields the second-order equation
31* + 301[ +125Zj = 250 with general solution
I At) — 2 + e
51
-
c, cos (5rV6/3) + c2 sin (5tj6/3)
Substitution in I2 = (Z' + 25Z, -50)/25 now yields
I2(t) = —e 51
2
15
-
12c, + V6 c2 jcos^5^V6/3j + ^12c2 - V6c, jsin^5f\/6/3j
Imposition of the initial conditions Z, (0) = 0, Z2 (0) = 0 now gives the equations
126
Chapter 4
c, + 2 — 0, 4c, / 5 + V6 c2 /15 = 0 with solution c, = — 2, c2 = 4a/6 . These coefficients give
the desired particular solution
/. (4) — 2 + c -5f -2 cos (5^/6/3) + W6 sin (5/V6/3)
j
/
\
20
72(r) = —j=e * sin (5fV6/3).
v6
33.
To solve the system
U = -20(7, - /2),
1{ = 40(7, - 72)
we first note that 7' = - 27', so I2 = - 27, + 77. Then 77 = 27, (0) +12(0) = 2(2) + 0 = 4,
so 72 = - 27', so I2 - - 27, + 4. Substitution of this into the first equation gives the
simple first-order linear equation 7' + 607, = 80 with general solution
7,(0 = 4/3 + cc_60f. The initial condition 7,(0) = 2 gives c = 2/3, so
7,(0 = |(2 + e-60'),
35.
/2(0 = |(l-e-60').
The two given equations yield
mi(3) = qBy" = -q2B2x'/m,
so x(3) + ft?x' = 0. The general solution is
x(t) = A cos cot + B sin cot + C.
Now x'(0) = 0 implies B = 0, and then x(0) = ro gives A + C = r0. Next,
ft>y' = x" = -Aft?cos cot,
so y'(0) =
is the circle
implies A = r0, hence C = 0. It now follows readily that the trajectory
x{t) = rocos cot,
37.
(a)
If we set m, = 2, ra2 = 1 / 2,
y(t) = -rosin cot.
= 75, fc2 = 25 in Eqs. (3) of Section 4.1, we get the
system 2x" = -100x + 25y, \yn — 25.x-25y with operational determinant
Da + 1007)2 +1875 = (D2 + 25)(D2+75j. Hence the general form of the solution is
x{t) = a, cos5^ + a2sin5r + Z?, cos5rV3+ /?2sin5^V3,
y(0 = c, cos5t + c2 sin 5^ + <7, cos 5t-\/3 + d2 sin 5^V3.
Section 4.2
127
Upon substitution in either differential equation we see that c, = 2a,, c2 = 2a2 and
dx = -2/?,, d2 = -2b2. This gives
x(0 =
a, cos5^ + a2sin5f +
Z?,cos5rV3 + Z?2sin5r%/3,
y(0 = 2a, cos5^ + 2a2 sin 5r-2/?, cos5+s/3 -2/?2 sin 5rV3.
(b)
In the natural mode with frequency C0[ = 5 the masses move in the same direction,
while in the natural mode with frequency a>i - 5 yfe they move in opposite directions. In
each case the amplitude of the motion of mi is twice that of m\.
39.
The system has operational determinant is 8D4 + 40D2 + 32 = 8(jD2 + l)(T>2 +4).
Hence the general form of the solution is
x(t) = a, cost + a2 sinr + Z?, cos2r + /?2 sin2r,
y(t) = c, cos t + c2 sin t + dx cos 2t + d2 sin 2t.
Upon substitution in either differential equation we see that c, = 2a,, c2 = 2a2 and
dx = -/?,, d2 = -b2. This gives
x(t) =
a,cosr + a2 sinr + /?, cos2t + b2 sin2r,
y{t) = 2alcost + 2a2sint-blcos2t-b1sin2t.
In the natural mode with frequency o\ = 1 the masses move in the same direction, with
the amplitude of motion of the second mass twice that of the first mass (figure on the left
below. In the natural mode with frequency C0i = 2 they move in opposite directions with
the same amplitude of motion (figure on the right below).
t
128
t
Chapter 4
41.
The system has operational determinant is D4 +10D2 +9 = (D2 4 1 ) (D2 + 9). Hence the
-
general form of the solution is
x(t) = ax cost + a2 sin t + bxcos3t + b2sin3t,
y(t) = cx cos t + c2 sin t + dx cos 3r + d2 sin 3t.
Upon substitution in either differential equation we see that cx — ax, c2 = a2 and
dx - —bx, d2 = -b2. This gives
x(t) = ax cost-\-a2 sin t + bxcos3t-\-b2sin3t,
y(t) = a, cost + a2 sinr-Z?, cos3t-b2 sin3t.
In the natural mode with frequency a\ = 1 the masses move in the same direction (lefthand figure below), while in the natural mode with frequency Oh - 3 they move in
opposite directions (right-hand figure below). In each case the amplitudes of motion of the
two masses are equal.
43.
The system has operational determinant is D4 + 6D1 +5 = [D2 +1) (T)2 + 5). Hence the
general form of the solution is
x(t) - axcost + a2sint + bxcosty[5+b2sint^5,
y{t) — c, cost + c2 sin t + dx costy/5 +d2 sin tyfs.
Upon substitution in either differential equation we see that cx =ax, c2= a2 and
dx = -bx, d2 - -b2. This gives
Section 4.2
129
x(t) = ax cos t + a2 sin t + bx cos ty/5 + b2 sin t\[5,
y(t) = a, cost + a2 sint-^ cos?>/5 — b2 sin tS.
In the natural mode with frequency (0\ = 1 the masses move in the same direction (lefthand figure below), while in the natural mode with frequency C0i = V5 they move in
opposite directions (right-hand figure below). In each case the amplitudes of motion of the
two masses are equal.
45.
The system has operational determinant is 2D4 + 20D2 + 32 = 2^D2 + 2)(Z)2 + 8).
Hence the general form of the solution is
x(t) = alcost^/2 + a2 sin t^2 +bx cos
+b2 sin t\J&,
y(t) = c{cost\f2 + c2 sin t\f2 + dx cos tyfs +d2sinfV8.
Upon substitution in either differential equation we see that c} = ax, c2- a2 and
dl - -bx / 2, d2 - -b2 / 2. This gives
x(t) = ax cos^V2 Jra2 sinrV2 +
bx cos ^V8 + Z?2sinrV8,
y(t) - ax costal +a2sinrV2 -\bx cos?V8 -\b2sint\fe.
In the natural mode with frequency ax = V2 the two masses move in the same direction
with equal amplitudes of oscillation (left-hand figure on next page). In the natural mode
with frequency C0i = V8 = 2>/2 the two masses move in opposite directions with the
amplitude of m2 being half that of m\ (right-hand figure on next page).
130
Chapter 4
t
47.
(a)
t
Looking at Fig. 4.2.7 in the text, we see that the first spring is stretched by x, the
second spring is stretched by y-x, the third spring is stretched by z-y, and the fourth
spring is compressed by z. Hence Newton's second law gives mx" = - k(x) + k(y- x),
my" = -k(y - x) + k(z- y), and mz" = -k(z- y)-k(z).
(b)
The operational determinant is
(D2 + 2)[(D2 + 2)2 - 1] + [ -(Z)2 + 2)] = (D2 + 2)[(£»2 + 2)2 - 2],
and the characteristic equation (r2 + 2)[(r2 + 2)2 - 2] = 0 has roots
±i-Jl and ±i\/2±a/2 .
SECTION 4.3
NUMERICAL METHODS FOR SYSTEMS
In Problems 1-8 we first write the given system in the form x = f(t,x,y),
y = g(t,x,y).
Then we use the template
h — 0.1;
tx = t0 + h
xx — x0 + hf(t0,x0,y0),
y{ — y0 + h g(t0,x0, y0)
x2 — xx + h f (tx,xx, yx),
y2 — yx +h g(tx,xx,yx)
Section 4.2
131
(with the given values of t0, xQ, and y0) to calculate the Euler approximations xx ~ x(0.1),
yx ~ y(0A)
and
x2 ~ x(0.2),
actual values jcact = x(0.2),
h = 0.2;
~ y(0.2) in part (a). We give these approximations and the
yact - y(0.2) in tabular form. We use the template
f, = t0 + h
m, = x0 + hf(t0,x0,y0);
v, = y0+hg(t0,x0,y0)
x, = x0+±h[f(t0,x0,y0) + f(t i,«|,v,)]
)>i = yo + ihUit^x^y^ + git^u^vy
to calculate the improved Euler approximations w, ~ *(0.2), ux ~ y(0.2) and
x, ~ *(0.2),
« _y(0.2) in part (b). We give these approximations and the actual values
*act = *(0.2),
yact = ;y(0.2) in tabular form. We use the template
h = 0.2;
= f(t0,x0,y0)-,
G, = g(t0,x0,y0)
F2 = f(t0+yh,x0 + yhF1,y0+^hGl);
G2= g(t0+±h,x0+±hFl,y0 + -LhGl)
F} = f(t0+±h,x0+ %hF2,y0+±hG2);
G, = g(t0 +jh,x0 +\hF2, y0 + \hG2)
F4 = f(t0+h,x0+hF},y0 + hG});
G4 = g(t0 + h,x0 + hF3,y0+hG})
xt = x0+-(F{+2F2 + 2F3 + Fty,
6
y, = y0+ —(G,+2G2+2G,+ G4)
6
to calculate the intermediate slopes and Runge-Kutta approximations x, = x(0.2), y, = y(0.2) for
part (c). Again, we give the results in tabular form.
1.
(a)
2.5
-^act
1.0034
3^act
2.6408
y\
^act
^act
2.6
1.0034
2.6408
X\
y\
*2
J2
0.4
2.2
0.88
U\
Vl
0.8
2.4
*i
0.96
(b)
(c)
132
F,
4
G\
f2
Gi
2
X\
y\
1.0027
2.6401
4.8
-^act
1.0034
3
^act
2.6408
Chapter 4
f3
5.08
G3
Fa
3.26
6.32
G4
4.684
3.
(a)
1.5
*2
2.81
J2
2.31
■Vact
3.6775
^act
2.9628
U\
V\
X\
y\
2.4
2
3.22
2.62
•^act
3.6775
^act
2.9628
x\
1.7
y\
(b)
(c)
5.
Fi
7
Gi
5
f2
g2
11.1
8.1
*i
3.6481
3>i
2.9407
-Vact
3.6775
^act
2.9628
*1
0.9
yi
*2
3.2
-0.52
Wl
-0.2
Vl
3.4
*1
f3
13.57
G3
9.95
f4
23.102
G4
17.122
(a)
J2
2.92
•^act
-0.5793
3^act
2.4488
3>i
2.44
■^act
-0.5793
^act
2.4488
(b)
-0.84
(c)
7.
F\
G\
F2
G2
f3
G3
f4
G4
-11
Xi
-0.5712
2
-14.2
-2.8
-12.44
-3.12
-12.856
-6.704
Ji
2.4485
-Vact
-0.5793
^act
2.4488
-^act
3.2820
3^act
1.7902
^act
3.2820
Jact
1.7902
G3
4
Fa
(a)
*1
yi
1.3
2.5
X2
^2
3.12
1.68
X\
y\
3.24
1.76
(b)
Vl
1.6
U\
3
(c)
F\
5
Xi
3.2816
G,
3
6.2
G2
3.8
yi
1.7899
-Vact
3.2820
}^act
1.7902
f2
Section 4.3
F,
6.48
8.088
G4
5.096
133
In Problems 9-11 we use the same Runge-Kutta template as in part (c) of Problems 1-8 above, and
give both the Runge-Kutta approximate values with step sizes h - 0.1 and h = 0.05, and also
the actual values.
9.
With h = 0.1:
With h = 0.05:
Actual values:
x( 1) - 3.99261,
x( 1) - 3.99234,
x( 1) ~ 3.99232,
y(l) - 6.21770
y( 1) - 6.21768
y(l) - 6.21768
11.
With h = 0.1:
x( 1) - -0.05832,
y(l) ~ 0.56664
With h = 0.05:
x( 1) - -0.05832,
y(l) ~ 0.56665
Actual values:
x(l) - -0.05832,
y(l) ~ 0.56665
13.
With y — x’ we want to solve numerically the initial value problem
y = -32 - 0.04y,
o
M
o
x' = y.
40) = 288.
When we run Program RK2DIM with step size h = 0.1 we find that the change of sign in
the velocity v occurs as follows:
t
7.6
7.7
xv
1050.2
+2.8
1050.3
-0.4
Thus the bolt attains a maximum height of about 1050 feet in about 7.7 seconds.
15.
With y = x', and with x in miles and t in seconds, we want to solve numerically the
initial value problem
x' = y
y’- -95485.5/(x2 + 7920jc+ 15681600)
40) = 0,
40) = 1.
We find (running RK2DIM with h - 1) that the projectile reaches a maximum height of
about 83.83 miles in about 168 sec = 2 min 48 sec.
17.
We first defined the MATLAB function
function xp
=
fnball(t,x)
%
Defines the baseball system
%
134
xl'
=
x'
=
x3,
Chapter 4
x3'
=
—cvx'
%
%
x2'
=
=
x4,
x4'
=
-cvy'-
g
with air resistance coefficient c.
32 #•
g
c
=
0. 0025;
xp
=
x;
V
y'
=
sqrt(x(3).A2)
xp(l)
=
x(3);
xp(2)
=
x(4) ;
xp (3)
—
-c*v*x(3);
xp(4)
=
-c*v*x(4)
+ x(4).A2);
- g;
Then, using the ^-dimensional program rkn with step size 0.1 and initial data
corresponding to the indicated initial inclination angles, we got the following results:
Angle
40
45
50
Time
5.0
5.4
5.8
Range
352.9
347.2
334.2
We have listed the time to the nearest tenth of a second, but have interpolated to find the
range in feet. These data indicate that the range increases when the initial angle is decreased
below 45°. The further data
Angle
41.0
40.5
40.0
39.5
39.0
35.0
Range
352.1
352.6
352.9
352.8
352.7
350.8
indicate that a maximum range of about 353 ft is attained with a ~ 40°.
First we run program rkn (with h- 0.1) with Vo
results:
t
5.0
6.0
250 ft/sec and obtain the following
x
y
457.43
503.73
103.90
36.36
Interpolation gives x = 494.4 when 3; = 50. Then a run with Vo = 255 ft/sec gives the
following results:
Section 4.3
135
t
5.5
6.0
x
486.75
508.86
y
77.46
41.62
Finally a run with vo = 253 ft/sec gives these results:
t
5.5
6.0
x
484.77
506.82
y
75.44
39.53
Now x ~ 500 ft when y = 50 ft. Thus Babe Ruth's home run ball had an initial velocity
of 253 ft/sec.
21.
136
A run with h = 0.1 indicates that the projectile has a range of about 21,400 ft ~ 4.05 mi
and a flight time of about 46 sec. It attains a maximum height of about 8970 ft in about 17.5
sec. At time t ~ 23 sec it has its minimum velocity of about 368 ft/sec. It hits the ground
(,t« 46 sec) at an angle of about 77° with a velocity of about 518 ft/sec.
Chapter 4
CHAPTER 5
LINEAR SYSTEMS OF
DIFFERENTIAL EQUATIONS
Along with Chapter 4, this chapter is designed to offer considerable flexibility in the treatment of
linear systems, depending on the background in linear algebra that students are assumed to have.
Sections 4.1 and 4.2 of the previous chapter can stand alone as a brief introduction to linear
systems without the use of linear algebra and matrices. But this chapter employs the notation
and terminology of elementary linear algebra. For ready reference and review, Section 5.1
includes a complete and self-contained account of the needed background of determinants,
matrices, and vectors. The additional linear algebra that is needed in subsequent sections is
introduced along the way.
SECTION 5.1
MATRICES AND LINEAR SYSTEMS
The first half-dozen pages of this section are devoted to a review of matrix notation and
terminology. With students who've had some prior exposure to matrices and determinants, this
review material can be skimmed rapidly. In this event serious study of the section can begin
with the subsections on matrix-valued functions and first-order linear systems. About all that's
actually needed for this purpose is some acquaintance with determinants, with matrix
multiplication and inverse matrices, and with the fact that a square matrix is invertible if and
only if its determinant is nonzero.
1.
(a)
2A + 3B =
(b)
3A-2B -
9
-12"
14
15
3
-9"
"6
-8“
"4
-6“
8
"6
+
—
"13
-18“
23
17
"0
-l"
2
19_
—
12
21_
2
-37
-8
"-1
3.
AB =
8 “
e
s
46
' 11
-12
14 '
-14
0
7
8
-13
BA =
-1
L
Section 5.1
0
137
(a)
-T
“0
-12
8 "
0
28
21 + 4
16
-12
-35
14
49
8
20
-4
' 9
6
-3"
'0
-15
10 "
0
12
9
5
20
-15
-15
6
21
10
25
-5
0
-6
1
10
31
-15
16
58
-23
7A + 4B =
3A-5B =
—
' 21
2
1
4
44
9
-27
34
45
‘ 9
21
-13"
-5
-8
24
-19
26
—
—
"
i
(b)
14
i
5.
' 21
(c)
(d)
(e)
7.
AB =
-10
-8
5
18
12
-10
11
22
6
' 3
2
-1“
0
4
3
-5
2
7
BA =
A-tl =
—
~t
0
0
0
t
0
0
0
t
"
"3-t
2
-1 "
0
4-t
3
-5
2
1-t
—
det(AB) = 0 = 0-0 = det( A) • det(B)
t
t - At2 + 6 r3
t + t2 -At3 + 8tA
3t + t}-tA
A12 + t3 + tA
3t2
4
A'B + AB' =
r.
1—r
1+r
u>
2 '
*■>*.
1_
" 1
K)
(AB)' =
i—
OJ
9.
_3 + 311
+
1 + 2/-12/2 +32+
~\-%t+m2
f
2/—1
3
r
1
-
At3
8r + 3?2 + At3
'-1
1
6f
12/2
t
1 +1 + 8^
_-3 + 312 - 3t3
-4/ + 3/2 + 3+
l-8r + 18r2
\ + 2t-\2t2 +32t3
3 + 312 - At2
8r+ 3/2 +4/3
+
-7? + 12/2
r — 12r2 + 24+
6-t3
+
(N
1 - r + 6r2
1
138
Chapter 5
x =
X
x =
4'
5
-1
"0
1
f
1
0
1
1
1
0
P(0 =
y
y
P(0 =
"0“
m =
y
1
0
-3
X]"
"0
1
0
0"
“0"
x0L
0
0
2
0
0
x3
0
0
0
3
0
_4
0
0
0_
_0_
i
1
o
t3
z
II
1__1
-1
to
LO
1__
to
l
1
1
LO
l
1__1
to
O
1
—H i
1
T
i
1
1
(N
rn
1
—
’ 2e2' "
' e2' '
+
1
"o
m
C,1
" 2e' "
C1l
2 cxe* +c2e2'
—
-e2' _
-3cxe' -c2e21
i
_i
x(/) = CjXj + c2x2
Axj
Ax 2
i
(N
o
<N
1
_l
" e2' '
f
X,2 =
-e2'
r
i__
1
i
L
ii
19.
t2
m =
o
1_
m =
_i
1
x =
" t~
X
17.
U>
X
0
0
i
z
3e‘
m
1
15.
2
1-
13.
In most of Problems 22-30, we omit the verifications of the given solutions. In each case, this is
simply a matter of calculating both the derivative x' of the given solution vector and the product
Ax, (where A is the coefficient matrix in the given differential equation) to verify that
xj = Ax, (just as in the verification of the solutions x, and x2 in Problem 21 above).
Section 5.1
139
23.
2t
e
W(t) =
e2t
-21
= 4^0
5e
T 2/
T -it
c,x1+c2x2 = c,
<r + c2
e
x(0
25.
e
1_
W(t) =
3e21
e~5‘
2e21
3e~5'
27.
it
e
21
W(t) =
e2t
_5_
cxe2‘ +5 c2e~2'
" e-5'
3 cxe2t + c2e -5/
- 7e~3' * 0
~3e2/'
x(t) = c1x,+c2x2 = cC,1
e
-t
+ C02
_2e2t _
_3e~5'_
-t
e
-e~l
-e~l
" 1 "
x(r) = CjXj + C2X2 + C3x, = Cj 1 e2i + c2
1
1
'0
29.
140
=
-1
l
>' = Ax,
l
"0
1
1
■ 1"
0 e ' = 1
0
1
0 e 1 = Ax.
1
1
0
-1
1
'0
1
f
"0"
-1 e' = 1
0
1
1
1
1
0
-1
"0"
X3
0
f T
x; = 2 e = 1 0 1
2
1 1 0
"-1“
2 cxe 2t + 3c2e -5/
= 3^0
T
X2 =
—
0
0
~2
cxe2i +c2e 2‘
1
3e'2/
e'
W(t) = -2e'2/
—c
2<T2'
e'
e 1 = Ax,
e3'
/
-e
3/
= e2/ * 0
0
Chapter 5
“O'
e ' +C3
1
-1
e ' =
C^6
+ C2C
C\6
+ 0^6
cxe2t -c2e~‘ -c2e
" 3 "
x(t) = CjX, + c2x2 + c3x3 = c,
-2
" 1 "
3c,c
" 1"
e 21 +c2 -1 c + c3 -1 e3' =
1
2
+ c2c + c3c
-2c,c-2' - c2c' - c3c3'
0
2 c,c-2' + c2c'
In Problems 31-34 (and similarly in Problems 35-40) we give first the scalar components
xx(t) and x2(t) of a general solution, then the equations in the coefficients c, and c2 that are
obtained when the given initial conditions are imposed, and finally the resulting particular
solution of the given system.
x,^) = c,c3'+2 c2c-2',
c, + 2 c2 = 0,
3c, + c2 = 5
x,(f) = 2c3'-2c-2',
x2(t) - 6c3'-c-2'
xx(t) = c,c3'+c2c2',
x2{t) = -c,c3'-2c2c2'
c, + c2 = 11,
- c, - 2c2 = - 7
x,(f) = 15c3'-4c2',
35.
x2(0 = -15c3'+ 8c2'
x,(0 = 2c,c'-2c2c3' + 2c3c5',
2c, - 2c2 + 2c3 = 0,
2c, - 2c3 =0,
x,(/) = 2c' -4c3' + 2c5',
37.
I
1
1
ON
II
x,(0 = c,c-'+c4c',
c,+c4 = 1,
x,(0 = 3c-'-2c',
41.
x2(0 = c',
x3(0 = 6c-2' -3c'
x3(0 - c2c-'+3c4c',
c2+3c4 = 1,
2c,c
2c, + c2 = 3
x2(/) = —6c 2' + 3c' + 5c ,
x2(0 = c3c',
c3 = 1,
x3(/t) = c'+2c3'+c5'
x2(t) = -2 c,c2'-c2c'- c3c3',
- 2c, - c2 - c3 = 2,
5c3',
x3(0 = c,c' + c2c3' + c3c5'
c, + c2 + c3 = 4
x2(t) = 2c'—2c5',
x,^) = 3c,c 2' + c2e' + c3c3',
3c, + c2 + c3 — 1,
39.
x2{t) = 2c,c'-2c3c5',
II
33.
x2(t) = 3c,c3' + c2c-2'
U>
31.
x4(f) = c,c-'-2c3c'
c, — 2c3 = 1
x3(0 = 7c-'-6c',
x4(f) = 3c-'-2c'
(a)
X2 = t\\, so neither is a constant multiple of the other.
(b)
W(xi, X2) = 0, whereas Theorem 2 would imply that
if xi and X2 were
independent solutions of a system of the indicated form.
Section 5.1
141
43.
Suppose W(a) = xu(a)x22(a) - xX2(a)x2X(a) = 0.
Then the coefficient determinant of
the homogeneous linear system cxxxx(a) + c2xX2(a) = 0, cxx2X(a) + c2x21(a) - 0
vanishes. The system therefore has a non-trivial solution {cx,c2} such that
cxxx(a) + c2x2(a) = 0. Then x(t) = cxxx(t) + c2x2(t) is a solution of x' = Px such
that x(#) = 0. It therefore follows (by uniqueness of solutions) that x(t) = 0, that is,
cxxx(t) + c2x2(t) = Owith cx and c2 not both zero. Thus the solution vectors x\ and
X2 are linearly dependent.
45.
Suppose that cxxx(t) + c2x2{t) +—h cnxn(t) = 0. Then the ith scalar component of this
vector equation is
cxxjX (/) + c2xj2(/) + ••• + cnxin(/) = 0. Hence the fact that the scalar
functions xn(t), xi2(t),
xjn(t) are linear linearly independent implies that
cx = c2 = • • • cn = 0. Consequently the vector functions xx{t), x2(/L), • •
xn(t) are linearly
independent.
SECTION 5.2
THE EIGENVALUE METHOD FOR
HOMOGENEOUS LINEAR SYSTEMS
In each of Problems 1-16 we give the characteristic equation, the eigenvalues X\ and /I2 of the
coefficient matrix of the given system, the corresponding equations determining the associated
eigenvectors v, = [ax
bx]
T
and v2 = [a2
TT
b2] , these eigenvectors, and the resulting scalar
components x\(t) and X2(t) of a general solution x(/) = cx\xe^' +c2v2e/l2/ of the system.
1.
Characteristic equation
Eigenvalues
A - 2/1 - 3 = 0
A\ = -1 and A2 = 3
x\(t) =
X2 (t) = -c\e~l + C2e3t
cie'1 + C2e3t,
and V2 = [1
1]
—
-1]
1
Eigenvectors vi = [1
-1
<N
1
_0_
<N
and
CN
T
'o'
—
(N
2
,1
_1 _1
Eigenvector equations
a
2_ A.
2"
1
2
a
A.
.1
"0"
—
_0_
T
The left-hand figure on the next page shows a direction field and some typical solution
curves for the system in Problem 1.
142
Chapter 5
yy/vy^/
,1 j i m\T iv/y^r
j Jj 1 j i J\ /r r / /
\Y ?7 t 1/ r
f\U( it r
Ivl/y
J
vZ//C<Xl^'t
-5
-4
-3
-2
-1
^
>-1-1--7-;
^
^
^ ^ xyy'yr x.
"4.
ij
J y\y
//V /V 1 X
/
1
2
3
4
7
T
^
^
1-1-1_1_1_L_J
5
-5
-4
-3
-2
-1
0
1
2
3
4
5
X2 -51-6 = 0
Characteristic equation
1\ = -1 and Z2 = 6
'4
—
Eigenvector equations
4" a,1
1
LO
UO
1_
Eigenvalues
S\ / y ///
r
:_1_1_1_1_
0
yyX
T
'O'
and
_
_0_
A
-1]
and V2 = [4
3]
xi(7) =
X2(t) = -c\e * + 3c2e
4' a,Z
"O'
—
_ 3
Eigenvectors Vi = [1
c\e / + 4c2^ ,
"-3
-4_ A.
_0_
T
61
The equations
xi(0) =
c\ + 4c2 = 1
*2(0) = ~c\ + 3c2 = 1
yield c\ = -1/7 and C2 = 2/7, so the desired particular solution is given by
x,(0 = j(-e_' + 8e6'),
x2(0 =
+ 6e6').
The right-hand figure above shows a direction field and some typical solution curves.
Characteristic equation
Eigenvalues
1~ -41-5 = 0
1\ = -1 and I2 = 5
“7
Eigenvector equations
L11
-1
a.1
-1 A
_
0"
and
0_
_!
_
Eigenvectors \\ = [1
1]T and V2 = [7
xi(0 = c\e~‘ + 7c2^5/,
X2(0 = c\e~‘ + C2e5t
Section 5.2
"1
-7" a,Z
-1 _ A.
'O'
—
0_
_
1]
143
The figure below shows a direction field and some typical solution curves.
7.
Characteristic equation
Eigenvalues
X + 8T - 9 = 0
X\ = 1 and X2 = -9
I— —i
I
VO
,n|- VO
l
i_ _1
Eigenvector equations
ax
—
A
"o'
_0_
Eigenvectors vj = [1
1]T and v2 = [2
x\(t) = C\e + 2c2e 9‘,
xiU) = C\e‘ - jCie 91
and
"6
4“ a.1
"0“
_6
4_ _b2_
_0_
-3]T
The figure below shows a direction field and some typical solution curves.
-1—r—i—r-1-r-- -^-,-i-1-r—3
A\v\\
\ Af
V/T ^
\\\\ ^ V/i
T
x
.
4
X UV\ y
-
AAmx
■ ^ V Vx\ \vvv
'^
X ^vVvV
lx
\m\\
^_i_i_i_i_i_A_ _\_i\
-5
-4
-3
-2
-1
0
1
2
144
Chapter 5
, 1_X—I_A_i_
3
4
5
9.
Characteristic equation A2+16 = 0
Eigenvalue X = 4i
4
a
^f1
1
1
(N
l_
Eigenvector equation
-5
_b_
"0"
0_
_
i
Eigenvector v = [5
2-4z]T
The real and imaginary parts of
x(0 = \eAn =
5 cos 4/1 + 5/sin 4/1
(2 cos 4^ + 4 sin 4/) + i (2 sin 4f - 4 cos 4/)
yield the general solution
x\(t) = 5cjcos 4t + 5c2sin 4/
X2(0 = ci(2 cos 4/ + 4 sin 4/) + ^(2 sin 4f - 4 cos 4/).
The initial conditions xi(0) = 2 and *2(0) = 3 give c\ = 2/5 and C2 = -11/20, so
the desired particular solution is
xi(7) = 2 cos At - -y-sin4f
*2(0 = 3 cos At + jsin4t.
The figure below shows a direction field and some typical solution curves.
Section 5.2
145
11.
Characteristic equation
A -2/1 + 5 = 0
Eigenvalue A = 1-2/
Eigenvector equation
~2i
"O'
—
_2
Eigenvector v = [1
-2~ a
2i_ _b
0_
_
z]T
The real and imaginary parts of
x(t) = [1
i]T e'(cos 2t - i sin 2/)
= el [cos 2t
sin 2/L]T + ie [-sin 2t
cos 2t\T
yield the general solution
x\(t) = c^cicos 21 - C2sin 2t)
X2(t) = e'(cisin 21 + C2COS 2t).
The particular solution with xi(0) = 0 and X2(0) = 4 is obtained with c\ = 0 and
C2 = 4, so
x\{t) = -4e'sin2/L,
*2(0 = 4e'cos 2+
The figure below shows a direction field and some typical solution curves.
146
Chapter 5
13.
Characteristic equation 22 - AA + 13 = 0
Eigenvalue A - 2-3 /
Eigenvector equation
"3 + 3/
2
-9
a
"0"
-3 + 3/_ _b_
0_
'T'
Eigenvector v = [3
1+/]
3 cos 3t - 3/ sin 3/
x(0 = ve(2 3/)' = e1'
(cos 3t + sin 30 + / (cos 31 - sin 30
xi(0 = 3e2/(cicos 3/ - C2sin 30
*2(0 = e2t [(ci + ^2)cos 31 + (ci - C2)sin 30].
The figure below shows a direction field and some typical solution curves.
15.
2
Characteristic equation K - 10/1 + 41 = 0
Eigenvalue A = 5-4/
Eigenvector equation
Eigenvector v = [5
x(0 = ve
(5-4/)/
"2 + 4/
4
a
0
-2 + Ai_ _b_
0
-5
2+4/]T
5 cos At — 5/ sin At
(2 cos 4/ + 4 sin 4/) + / (4 cos At-2 sin At)
Section 5.2
147
xi(7) = 5e5'(ciCOS A1 - c2sin 41)
x2(t) — e5t [(2ci + 4c2)cos At + (Ac\ - 2c2)sin 4/)]
The figure below shows a direction field and some typical solution curves.
; \\ r—r
A /K T 'jf
1] 7/7
\
- 4— <— +—• ^ r
Ay]
x\X)J7A/
AftA
^ -> ^
y c/i f l >
^ ~~~~A -+
/ // i L \ ^
/ At ty/t
J i iX'y
17.
IL
,1
-5
-4
,\ A\\
-3
-2
-1
0
1
2
3
4
5
Characteristic equation -A3 + 15/12 -5AA = 0
Eigenvalues
A\ - 9, A2 = 6, A3 = 0
Eigenvector equations
_T>
1_
1
-2
_C2
v2 = [1 -2
+ C2£6/ + C3
c\e9t - 2C2e6t
x3(t) = c\e9t + c2e6t - c3
18.
Characteristic equation -A3 +15A2 - 5AA = 0
Eigenvalues
148
A\ =9,
_
T2 = 6,
A3 = 0
Chapter 5
1-
1
r-H
4
—
1_
b2
1
1
0 5
1
7
1
b3
1
0
_1
1
1]T,
O
l__
1
<N
1_
1-
x2(0 =
-H
X\(t) =
1
1
T—
Eigenvectors vj = [1
(N
1
-5
1_
1
0 J
1
0
_1
4
1
0
1_
1
II
-2
1
1_
1'sT
1
1_
1
~a~
4
1
4
_C3
1]T,
V3 = [1
—
_
0 -1]T
Eigenvector equations
"-8
2
2 " a,
2
-2
1
2
1
-2 .ci.
\
—
Eigenvectors \\ = [1
x\(t) =
0
“-5
2
2
a2
0
2
1
1
b2
0
2
1
1
_C2
v2
—
"
c\e9t
"
2 2]T
"0"
0
—
0
_
1 - 1]T
[0
5
"1
2
2“
a3
2
7
1
b3
2
1
7
_C3 _
v3 - [4
5
"0"
—
0
0
-1 --1]T
+ 4c3
X2(f) — 2c\e9t + c'2^6/ - C3
*3(7) = 2c\e9t - C2e6t - C3
Characteristic equation - A3 + \2A2 -451 + 54 = 0
Eigenvalues
A\ = 6,
,
A2 = 3, A3 = 3
Eigenvector equations
'-2
1
1 " a
1
-2
1
1
1
-2 _ci_
"O'
"1
1
f ~a2~
0 5
1
1
1
b2
0
1
1
1
_C2_
=
bi
1
1]T,
"1
1
f
Cl^
0 5
1
1
1
b3
0
1
1
1
_C3_
V2 = [1 -2
"0"
—
0
0
1]T,
V3 = [1
0 -1]T
^0
Eigenvectors vi = [1
—
"0"
x\(t) = C\e6t + C2e3t + c3e3/
X2(t) = c\e6t - 2c2e3t
X3(t) = c\e6t + C2e3t - C3e3t
•
•
^
•
Characteristic equation -A' + A = 0
Eigenvalues
A\ = 0,
A2 — 1, A3 = -1
Eigenvector equations
4
-2
-5
6ci
+ 3c2^f + 2c3^
X2 (0 =
2ci
+
_C2 _
v2 = [3
C3£ 1
X3 (t) = 5ci + 2c2^ + 2c3e"'
Section 5.2
=
1
0
2
0-2
b.
0
4
-2
-3
2]t,
v3 = [2
1
1
b2
1
-2
5]t,
xi(Y) =
C2^ +
-2
~a~
0
0
2
2
1_
5
1
O
1_
Eigenvectors Vl = [<3
11
1
.ci_
0
0
<N
1
—
b\
L
cf
1_
11
_1
1
0
1_
0
-2
1
in
-1
—1
1_
2
a2
2]T
=
23.
Characteristic equation -A3 + 3A2 + 4A -12 = 0
Eigenvalues
A\ = 2, Ai — -2, A3 = 3
Eigenvector equations
1
1
1
-5
-5
-1
0 5
-5
-1
-1
5
5
1
0
5
5
5
Eigenvectors vi = [1 -1
0]T,
xi(f) =
X2(t)
'O'
=
_ci_
"
"
o
'
a2
—
.C2_
V2 = [0
'
'
0
0
0
5
1 -1]T,
1
1
'
-1
b3
0
_C3
5
V3 = [1 -1
0
a3
—
_
1]T
+ c^e31
C\e2t
= -cic2/ + C2e~2t - c^e31
X3(t) =
25.
a]
"
1
1
On
1
1
5
"
- C2e~2t + c^e31
Characteristic equation -A3 +4A2 -13A = 0
Eigenvalues A = 0 and 2 ± 3 /
Ut
-5
6
6
10
1
O
1_
-6
1
-6
cf
5
(N
1_
1—
1_
jTi
0
—
[1 -1
gives eigenvector vj
0]T
1
0
_1
—1
b\
1—
1
With A = 1 the eigenvector equation
With A = 2 + 3/ we solve the eigenvector equation
"3-3/
5
2
a
-6
-8 - 3/
-5
b
6
6
3-3/
c
"O'
—
0
0
to find the complex-valued eigenvector v = [1+/
complex-valued solution is
-2
T
2] . The corresponding
(cos 3t - sin 31) + / (cos 3t + sin 31
x(0 = xe
(2+3/)/
-2 cos 3t-2i sin 31
2cos3t + 2/sin3/
The scalar components of the resulting general solution are
X\(t) =
c 1 + e2t [(c2 + C3)cos 3/ + (-C2 + C3)sin 3/]
9/
*2(0 = -c 1 + 2e (-C2COS 3t - (?3sin 3/)
X3(/) =
150
2c2/(c2Cos 3t + C3sin 3/).
Chapter 5
0
0
27.
The coefficient matrix
0.2
0
0.2
-0.4
-
A
-
has characteristic equation A2 + 0.6/1 + 0.08 = 0 with eigenvalues A\ = -0.2 and
A2 = -0.4. We find easily that the associated eigenvectors are Vi = [1
V2 = [0
1]T, so we get the general solution
xx(t) = c^-°-2/,
1]T and
*2(7) = c^'+c^-4'.
The initial conditions x,(0) = 15, v2(0) = 0 give c,=15 and c2 = -15, so we get
jci(0 = \5e~02t,
x2(t) = \5e-02t - \5e~0At.
To find the maximum value of x2(t), we solve the equation x'2(t) = 0 for t = 5 In 2,
which gives the maximum value *2(5 In 2) = 3.75 lb. The following figure shows the
graphs of xx{t) and x2(t).
t
29.
The coefficient matrix
A
-0.2
0.4
0.2
-0.4
has eigenvalues A\ = 0 and A2 = -0.6, with eigenvectors vj = [2
Section 5.2
1]
T
and
151
T
\2
= [1
-1]
that yield the general solution
x,(f) = 2 c,+c2e~0'6',
x2(t) =
The initial conditions x,(0) = 15, x2(0) = 0 give cx=c2= 5, so we get
x,(0 = 10 + 5e-0'6',
x2(t) = 5-5e-0'6'.
The figure below shows the graphs of x^/1) and x2(f).
31.
The coefficient matrix
A
=
-10
0
1
-2
0
0
2-3
has as eigenvalues its diagonal elements X\ = -1, T2 = -2, and T3 = -3. We find
T
T
readily that the associated eigenvectors are vi = [1 1 1] ,
- [0 1 2] , and
T
V3 = [0 0 1 ] . The resulting general solution is solution is given by
Xj(7) = cxe~'
x2(0 = Cj e~' + c2 e~2t
x3(t) = cx e~‘ + 2c2 e~2< + c3 e~31.
152
Chapter 5
The initial conditions x,(0) = 27, x2(0) = x2(0) = 0 give c} = c3 = 27, c2 = -27, so we
get
x,(7) = 27 e~'
x2{t) = 21 e~' -21 e~2'
jc3(0 = 27
- 54 e~2/ + 27 e~3'.
The equation x3(t) = 0 simplifies to the equation
3e~2t -Ae~l +1 = [3e~l - l)(ef' -1) = 0
with positive solution tm = In 3. Thus the maximum amount of salt ever in tank 3 is
x3(In3) = 4 pounds. The figure below shows the graphs of xfa), x2(t), and x3(t).
t
The coefficient matrix
A =
-4
0
0
4
-6
0
0
6
-2
has as eigenvalues its diagonal elements fa = -4, fa — -6, and fa = -2. We find
XX
readily that the associated
eigenvectors
are
Vi
=
[-1
-2
6],
V2 = [0 -2 3],
rp
and V3 — [0 0 1 ] . The resulting general solution is solution is given by
Section 5.2
153
x,(/) = - c, e 41
x2(t) = -2c, c"4' -2c2 c"6/
x3(t) = 6c, c“4' + 3c2 c"6' + c3 c~2/.
The initial conditions x,(0) = 45, x2(0) = x2(0) = 0 give c, =-45, c2=45, c3=135, so
we get
x, (t) = 45 c"4'
x2(t) = 90 c"4'-90 c"6'
x3(0 = -270c~4'+ 135c"6'+ 135c"2'.
The equation x3(/) = 0 simplifies to the equation
3c"4' -4c~2' +1 = (3c"2'-l)(c"2'-l) = 0
with positive solution tm = \ In 3. Thus the maximum amount of salt ever in tank 3 is
x3(jln3) = 20 pounds. The figure below shows the graphs of x,(/), x2(7), and x3(t).
o
35.
2
The coefficient matrix
A =
154
-6
0
6
-20
0
20
Chapter 5
has characteristic equation -T3 -29/12 -198/1 = - A(A-18)(A-11) = 0 with
eigenvalues Aq = 0, fa = -18, and T2 = -11. We find that associated eigenvectors
are v0 = [10
3
20]T, v, = [-1 -3
4]T, and v2 - [-3 -2 5]T. The
resulting general solution is solution is given by
x,(7) = 10c0
e~w -3c2 e~Ut
x2(t) = 3c0 -3c, c~l8/ -2c2 e~Ul
x3(t) = 20c0 + 4c, c~18' + 5c2 e~Ul.
The initial conditions x,(0) = 33, x2(0) = x2(0) = 0 give c, = 1, c2- 55/7, c3 = - 72/7,
so we get
x,(r) = 10-y(55e-18' -216e""')
x2(/) = 3-y(l65e'18' -144e”"')
x3(0 = 20 + y(220e"18' -360e“"').
Thus the limiting amounts of salt in tanks 1, 2, and 3 are 10 lb, 3 lb, and 20 lb. The'
figure below shows the graphs of x,(7), x2(f), andx3(7).
The coefficient matrix
A
=
-1
0
2
1
-3
0
0
3
-2
Section 5.2
155
has characteristic equation -A3 -6A2 -l\A = 0 with eigenvalues Ao - 0, A\ =
-3 - i V2 , and A2 = -3 + i y/2. The eigenvector equation
"-1
0
2 ' a
1
-3
0
b
0
3
-2
c
'O'
=
0
0
associated with the eigenvalue Ao = 0 yields the associated eigenvector
Vo = [6
2
3] and consequently the constant solution x0(t) = v0. Then the
eigenvector equation
LC,
+
(N
1_
a
1
/V2
0
b
0
3
associated with A\ = -3 - i V2
vi =
(-2 + iy/l)/3
&
1_
2
+
0
'O'
—
c
0
0
yields the complex-valued eigenvector
(—1 — /V2)/3
1
. The corresponding complex-valued solution
is
Xi(0 = Vj<?
(-3-/V2)/
(-2cos(rV2) + V2 sin(rV2)j + i{^2 cos(tyfl) + 2sin(/s/2)j
-e~3'
3
(-cos(7a/2) - V2 sin(7V2)) + i ^-V2 cos(/>/2) + sin (/%/2) j
3cos(/V2)
-
3/sin(rV2)
The scalar components of resulting general solution x = c0x0 + cx Re(Xj) + c2 Im(Xj) are
given by
X\(t) = 6c0+\e~31 ^-2Cj + V2 c2 j cos(/s/2) + ^ V2 c, + 2c2 j sin(f\/2)
x2(t) = 2c0+je_3/ h-v^c2) cos (f%/2) + (-V2 c, + c2) sin (/V2)
x3(7) = 3 cQ+e~31 c, cos 0V2 )-c2 sin (/V2)|.
When we impose the initial conditions x, (0) = 55, x2(0) = x2(0) = 0 we find that
c0 = 5, Cj =—15, and c2 = 45/ y/l. This finally gives the particular solution
156
Chapter 5
Xj( t) = 30+ e-3' 25 cos(f%/2) +10V2 sin(f>/2) J
x2(0 = 10-<?~3/ 10 cos(/V2)-fV2 sin(rV2)
x3(0 = 15 -e“3/ 15 cos(f>/2) + ^V2 sin(/s/2) .
Thus the limiting amounts of salt in tanks 1, 2, and 3 are 30 lb, 10 lb, and 15 lb. The
figure at the top of the next page shows the graphs of x,(/), x2(t), and x3(t) .
In Problems 38-41 the Maple command with (linalg) : eigenvects (A), the Mathematica
command Eigensystem[A], or the MATLAB command [V,D] = eig (A) can be used to
find the eigenvalues and associated eigenvectors of the given coefficient matrix A.
39.
Characteristic equation:
(A2 - 1)(T2 - 4) = 0
Eigenvalues and associated eigenvectors:
A =
1,
A = -1,
V
A =^
V
= [3
= [0
= [0
V
=
2,
A =:-2,
V
[i
-2
0
1
-1
4
1
0
0
i]'
0]T
0]T
0]T
Section 5.2
157
Scalar solution equations:
3c\el
+ C4e~2t
X2(t) = -2c\el
+ c3e2t - c\e~2t
x\(t) =
41.
x3 (t) =
4c\et + c2e~‘
X4 (t) =
c \e‘
The eigenvectors associated with the respective eigenvalues X\ = -3, X2 = -6,
X3 = 10, and X4 = 15 are
v, =
v2 =
v3 =
v4 =
[ 1
[ 0
[-2
[ 1
0
0
1-1
1
1
2
2
-1]T
0]T
-2]t
1]T.
Hence the general solution has scalar component functions
jti(f) =
c\e~3t
- 2c3el0t +
C4el5t
X2 (t) =
C2C~6t +
C3el0t +2C4^15/
X3(t) =
- C2e~6t +
C3^10/ + 2c4^15/
X4(t) = -c\e
- 2c3e
+
C4e
The given initial conditions are satisfied by choosing c\ = C2 = 0, C3 = -1, and
C4 = 1, so the desired particular solution is given by
xi(7) = 2e10/+ e]5t = X4 (t)
x2(t) = -eWl + 2e>5' = x3(t) .
In Problems 42-50 we give a general solution in the form x(t) = clvleA‘l + c\v2e'1’-' h— that
exhibits explicitly the eigenvalues X[9X2,... and corresponding eigenvectors v,, v2,... of the
given coefficient matrix A.
T
" 3 "
or
11
-1
5
158
e 2> + C2
1
1
~ 1 “
6
41
+
-1
3
Chapter 5
1
'
■
1
~
1
45.
2
C
~h C7L
1
49.
x(0
2
1
3
- /
£
,
+
2“
'
"
2
£*2
_-l_
_
1 “
"
2
1
_
Q
3/
-1
1
,
4~ C4
"
-1
e31 + c4
1
1
~ 2 "
x(7)
+ c^
1
“
1
-1
_-l_
47.
"2“
'
1
e
61
1 "
-1
+ c4
2
-1
1
~\
'0“
"f
0
3
7
3 e 31 + c2
0
+ c4 1
1
-1
1
1
2
1 _
1
_ _
1
_ _
1
_ _
_
1
-1
0
"2"
1
0
~
~
e3' + c4 0 e61 +c5 5
1
SECTION 5.3
SECOND-ORDER SYSTEMS
AND MECHANICAL APPLICATIONS
This section uses the eigenvalue method to exhibit realistic applications of linear systems. If a
computer system like Maple, Mathematica, MATLAB, or even a TI-85/86/89/92 calculator is
available, then a system of more than three railway cars, or a multistory building with four or
more floors (as in the project), can be investigated. However, the problems in the text are
intended for manual solution.
Problems 1-7 involve the system
m\x\" = ~(k\+k2)xi+
m2x2
=
k2x2
k2xi - (k2 + k2)x2
with various values of m\,m2 and k\, k2, k2. In each problem we divide the first equation by m\
and the second one by m2 to obtain a second-order linear system x" = Ax in the standard
Section 5.2
159
form of Theorem 1 in this section. If the eigenvalues A, and
are both negative, then the
natural (circular) frequencies of the system are cox = ^/-A, and co2 = Ay/-A2, and — according to
Eq. (11) in Theorem 1 of this section — the eigenvalues v, and v2 associated with A, and ^
determine the natural modes of oscillations at these frequencies.
1.
The matrix A =
-2
has eigenvalues A0 = 0 and A, = -4 with associated
2
rp
eigenvalues v0 = [1
1]
'T'
and v, = [1
-1] . Thus we have the special case described
in Eq. (12) of Theorem 1, and a general solution is given by
x^/) = ax+a2t + bx cos 2t + b2 sin 2t,
x2(t) = a} + a21 - bx cos2t - b2 sin2t.
The natural frequencies are co\ = 0 and coi = 2. In the degenerate natural mode with
"frequency" co\ = 0 the two masses move by translation without oscillating. At
frequency coi = 2 they oscillate in opposite directions with equal amplitudes.
3.
The matrix A =
-3
2
1
-2j
iT
eigenvalues v, = [1
1]
„
has eigenvalues A, = —1 and ^ = -4 with associated
i
n
and v2 = [2
nT
-1] . Hence a general solution is given by
xx(t) = ax cost + a2 sint + 2bx cos2t + 2b2 sin2t,
x2(t) = ax cost + a2 sint -b^oslt -b2sin2t.
The natural frequencies are co\ = 1 and cc>2 — 2. In the natural mode with frequency
co\, the two masses m\ and m2 move in the same direction with equal amplitudes of
oscillation. In the natural mode with frequency coi they move in opposite directions
with the amplitude of oscillation of m\ twice that of m2.
5.
The matrix A =
-3
1
eigenvalues v, = [1
x,(7) =
has eigenvalues A, = -—2 and ^ = -4 with associated
1]T and v2 = [1
- 1]T. Hence a general solution is given by
cos t^2 +a2 sin t^2 + bx cos2^ + b2 sin 2/,
x2(t) = ax cos tV2 + a2 sin tyjl - bx cos 2t - b2 sin It.
160
Chapter 5
The natural frequencies are cox - a/2 and co2 = 2. In the natural mode with frequency
co\, the two masses m\ and m2 move in the same direction with equal amplitudes of
oscillation. At frequency 02 they move in opposite directions with equal amplitudes.
7.
The matrix A
eigenvalues v, = [1
-10
6
6
-10
has eigenvalues \--4 and ^ =-16 with associated
1]T and v2 = [1
-1]T. Hence a general solution is given by
x,(7) = a] cos 2/ + a2 sin2f + bx 008 4/ + b2 sin 4t,
x2(t) = a{cos2t + a2sm2t - bxcos4t - b2sin4t.
The natural frequencies are co\ = 2 and coi = 4. In the natural mode with frequency
cdi, the two masses m\ and m2 move in the same direction with equal amplitudes of
oscillation. At frequency 0)2 they move in opposite directions with equal amplitudes.
9.
Substitution of the trial solution xx = c, cos 3t, x2 = c2 cos 31 in the system
x" = -3x,+2x2,
2x" = 2xj-4x2 + 120cos31
yields cx =3, c2 = -9, so a general solution is given by
xx(t) =
cost + a2 sint + 2b^ cos2t + 2b2 sin2t + 3cos3/,
x2{t) = ax cos/ + a2 sint -/?,cos2/1 -b2 sin2^ — 9cos3t.
Imposition of the initial conditions Xj(0) = x2(0) = xf(0) = x2(0) = 0 now yields
ax =5, a2= 0, bx = -4, b2 - 0. The resulting particular solution is
Xj(7) = 5 cos t- 8 cos 2t + 3 cos 3f,
x2(t) = 5 cos t + 4 cos 2t - 9 cos 3t.
We have a superposition of three oscillations, in which the two masses move
•
•
•
in the same direction with frequency a>\ = 1 and equal amplitudes;
in opposite directions with frequency o>2 = 2 and with the amplitude of
motion of m\ being twice that of m2;
in opposite directions with frequency co3 = 3 and with the amplitude of
motion of m2 being 3 times that of m\.
Section 5.3
161
11.
(a)
The matrix A =
-40
8
12
-60
has eigenvalues T, = -36 and
rr
associated eigenvalues v2 = [2
= -64 with
rp
and v2 = [1
1]
-3] . Hence a general solution is
given by
x(t) = 2a, cos 6t + 2a2 sin 6r + bx cos 8/ + b2 sin 8/,
y(t) =
a, cos 6/ + a2 sin 6/ - 36, cos^ - 3b2 sin 8/.
The natural frequencies are co\ = 6 and coi = 8. In mode 1 the two masses oscillate in
the same direction with frequency co\ = 6 and with the amplitude of motion of m\ being
twice that of m2. In mode 2 the two masses oscillate in opposite directions with
frequency CO2 = 8 and with the amplitude of motion of m2 being 3 times that of m\.
(b)
Substitution of the trial solution x-cx cos It, y = c2 cos 71 in the system
x" = -40x + 8y-195cos7r,
y" = 12x-60y -195COS7/1
yields c, = 19, c2 = 3, so a general solution is given by
x(t) = 2a, 0056^ + 2^2 sin6t+ bx cos8t + b2 sin8/t + 19cos7/,
y(t) =
a, 0056/ + a2 sin 6/ -3/?, cos 8/- 3b2 sin 8/ + 3 cos It.
Imposition of the initial conditions x(0) = 19, xf(0) = 12, y(0) = 3, y'(0) = 6 now yields
a, = 0, a2 = 1, Z?, = 0, Z?2 = 0. The resulting particular solution is
x(7) = 2sin6f + 19cos7f,
y{t) =
sin 6^ + 3 cos It.
Thus the expected oscillation with frequency 02 - 8 is missing, and we have a
superposition of (only two) oscillations, in which the two masses move
•
•
13.
in the same direction with frequency co\ = 6 and with the amplitude of
motion of m\ being twice that of my,
in the same direction with frequency 0^ = 1 and with the amplitude of motion
of m\ being 19/3 times that of m2.
The coefficient matrix
A =
-4
2
0
2
-4
2
0
2
-4
has characteristic polynomial
-A3 -12 A2- 40 A - 32 = -(A + 4 )(A2 + 8 A+ 8).
162
Chapter 5
Its eigenvalues A, - — 4, /^ = -4 - 2 V2, Ag = -4 + 2>/2 have associated eigenvectors
vi=[l
0
-1]T, v2=[l
-V2
1]T, v3=[l
V2
1]T. Hence the system's three
natural modes of oscillation have
•
Natural frequency &>, = 2 with amplitude ratios 1 : 0 : -1.
•
Natural frequency co2 = ^4 + 2^2
with amplitude ratios 1: - V2 :1.
•
Natural frequency co2 = V4-2V2
with amplitude ratios 1: V2 : 1.
First we need the general solution of the homogeneous system x" = Ax with
A
-
-50
25/2
50
-50
The eigenvalues of A are A\ = -25 and X2 = -75, so the natural frequencies of the
system are co\ = 5 and
and V2 = [1
(02
= 5>/3 . The associated eigenvectors are vi = [1
2]T
-2] , so the complementary solution xc(/) is given by
x,(/) =
a] cos5/ + a2 sin5/ + bx cos5\/3/ + b2 sin5>/3/,
x2(/) = 2^, cos5/ + 2a2 sin5/- 2bx cos5V3/-2Z?2 sin 5 St.
nr
When we substitute the trial solution xp(/) = [c\
C2] cos 10/ in the nonhomogeneous
system, we find that c\ = 4/3 and <?2 = -16/3, so a particular solution xp(/) is
described by
x\(/) = (4/3)cos 10/,
xi(t) = -(16/3)cos 10/.
Finally, when we impose the zero initial conditions on the solution x(/) = xc(/) + xp(/)
we find that a\ = 2/3, a2 = 0, Zq = -2, and 62 = 0. Thus the solution we seek is
described by
xi(/) = fcos 5/ - 2 cos 5\/3/ + ^cos 10/
X2(/) = |cos5/ + 4cos 5V3/ + -j-cos 10/.
We have a superposition of two oscillations with the natural frequencies co\ = 5 and
o>2 = 5a/3 and a forced oscillation with frequency ^ = 10. In each of the two natural
oscillations the amplitude of motion of m2 is twice that of m\, while in the forced
oscillation the amplitude of motion of m2 is four times that of mi.
Section 5.3
163
17.
With c\ = C2 = 2, it follows from Problem 16 that the natural frequencies and
associated eigenvectors are
= 0, vj = [1
Hence Theorem 1 gives the general solution
1]
and a>i - 2, V2 = [1
-1] .
xi(f) = a\ + b\t + r^cos 2t + b2sin 2t
x2(t) = ai + b\t - a2cos 2r - ^sin 2t.
The initial conditions x[ (0) = vo, xj(0) = X2(0) = x' 0) = 0 yield a\ = a2 = 0 and
b\ = v0/2, b2 = v0/4, so
xi(Y) = (vo/4)(21 + sin 2/)
x2(f) - (vo/4)(2/- sin 2/)
while X2 - xi = (vo/4)(-2 sin 2/) < 0, that is, until r = /r/2. Finally, xi'(tt/2) = 0 and
X2\7r/2)
19.
= v0.
With ci = 1 and C2 = 3, it follows from Problem 16 that the natural frequencies and
associated eigenvectors are co\ = 0, vi = [1
Hence Theorem 1 gives the general solution
1]T and
- 2, V2 = [1
-3]T.
xi(0 = a\+b\t+ a2cos 2t + b2sm2t
x2(t) — a\ + b\t - 3cf2cos 2t - 2b2s\n 21.
The initial conditions x[ (0) = vo, xi(0) = X2(0) = x\ (0) = 0 yield a\ = a2 = 0
and b\ = 3vo/4, b2 = vo/8, so
x\(t) = (vo/8)(6/+
sin 2/)
x2(t) = (vo/8)(6/ - 3 sin 2t)
while X2 - xi = (vo/8)(-4 sin It) < 0; that is, until t = ;z/2. Finally, xj (;r/2) = vo/2
and x^ (n!2) = 3vq/2.
21.
(a)
The matrix
A
-160/3
320/3
8
-116
has eigenvalues X\ « -41.8285 and h2 « -127.5049, so the natural frequencies are
« 6.4675 rad/sec « 1.0293 Hz
02 « 11.2918 rad/sec « 1.7971 Hz.
164
Chapter 5
(b)
Resonance occurs at the two critical speeds
vi = 20(O\/7r « 41 ft/sec « 28 mi/h
V2 = 20co2/tt « 72 ft/sec « 49 mi/h.
23.
lOOx" = -4000x,
800#" = 100000#
-40
Obviously the matrix A =
0
25.
Up-and-down:
CO\
Angular:
coi
=
0
has eigenvalues X]=-40 and ^ = —125.
-125
V40 ,
vi « 40.26 ft/sec « 27 mph
Vl25,
V2 « 71.18 ft/sec « 49 mph
100 x" = -3000x-5000#
800#" - -5000x-75000#
The matrix A
« 5.0424,
C02 ~ 9.9158,
-30
-50
-25/4
-375/4
has eigenvalues Xl,X2 = —(^—99± V3401).
vi « 32.10 ft/sec « 22 mph
V2 ~ 63.13 ft/sec « 43 mph
SECTION 5.4
MULTIPLE EIGENVALUE SOLUTIONS
In each of Problems 1-6 we give first the characteristic equation with repeated (multiplicity 2)
eigenvalue X. In each case we find that (A - /II) = 0. Then w = [1
0]
is a generalized
eigenvector and v = (A - X I)w ^0 is an ordinary eigenvector associated with X. We give
finally the scalar component functions x\(t), X2 (t) of the general solution
x(t) = c\\eAt + ciiyt + w)eA/
of the given system x' = Ax.
I.
Characteristic equation
T2 + 6 A + 9 — 0
Repeated eigenvalue
Generalized eigenvector
X = -3
w = [1
v = (A-Tl)w =
0]T
■ 1
rT
■ 1 “
-1
-i_ _0_
-1
Section 5.3
165
X\(t) = ( Cl + c2 + C2t)e 3/
x2(0 = (-ci
- c2t)e~3t.
The left-hand figure below shows a direction field and typical solution curves.
o
Characteristic equation
X - 6/1 + 9 = 0
Repeated eigenvalue
Generalized eigenvector
X = 3
T
0]T
1
r
<N (N
1
l__i
(N
r—-1
<N (N
1
(N
1
1__1
v = (A-Tl)w =
w — [ 1
i—
o
1_
3.
3/
x\(t) = (-2ci + c2 - 2c2t)e'
x2{t) = ( 2c i +
2c2t)e3t.
The right-hand figure above shows a direction field and typical solution curves.
X2 - 10T + 25 - 0
Characteristic equation
Repeated eigenvalue
Generalized eigenvector
w = [1
'2
i
i
xi(/) = ( 2ci + c2 + 2c2t)e
x2(t) = (-4ci
i_
r T
i
v = (A-Tl)w =
X = 5
to
5.
_o_
0]T
"
2"
-4_
5/
- 4c2t)e5t.
The figure at the top of the next page shows a direction field and typical solution curves.
166
Chapter 5
In each of Problems 7-10 the characteristic polynomial is easily calculated by expansion along
the row or column of A that contains two zeros. The matrix A has only two distinct
eigenvalues, so we write Al, A^ X, with either A, or
= A3. Nevertheless, we find that it
has 3 linearly independent eigenvectors v,, v2, and V3. We list also the scalar components
x\(t), X2(0, *3(0 of the general solution x(f) = cx\xe^‘ + c2\2e^ + c3\3e^' of the system.
7.
Characteristic equation
-A3 + 13A2 -40A + 36 = - (A - 2)2 (A - 9)
Eigenvalues
Eigenvectors
A = 2, 2, 9
[1
1
0]T, [1
0
1]T, [0
1
0]T
xi(f) = c\e2t + C2e2t
X2 (t) = c\e2t
x3(0 =
9.
+ C2,e9t
C2e2t
Characteristic equation
-A3 +19A2 -115A + 225 - -(A-5)2(A-9)
Eigenvalues
Eigenvectors
A - 5, 5, 9
[1 2 0]T, [7
xi(0 =
0
2]t, [3
0
1]T
C\e^ + 7c2e5/ + 3c3e9/
X2 (0 = 2c\e5t
x3(0 =
2c2e5/ + C2>e9t
Section 5.4
167
In each of Problems 11-14, the characteristic equation is-A3 - 3A2 -3A-1 = - (A +1)3.
Hence A = -1 is a triple eigenvalue of defect 2, and we find that (A - AT)3 = 0. In each
problem we start with v3 = [1
0
0]
and then calculate v2 = (A - Al)x3 and
v, = (A - AT)\2 ^ 0. It follows that (A - AT)\] = (A - Al)2v2 = (A - TI)3v3 = 0, so the vector
(if nonzero) is an ordinary eigenvector associated with the triple eigenvalue A. Hence
{vi, V2, V3} is a length 3 chain of generalized eigenvectors, and the corresponding general
solution is described by
Vi
x(^) = e~' [c\V\ + ci(y\ t + x2) + c*3(vi t2l2 + V21 + V3)].
We give the scalar components xi(/% X2(t), X3(t) of x(/).
11.
v, = [0
1
0]T,
v2 = [-2 -1
1]T,
v3 = [1
0 0]T
x\(t) = e~t(-2c2 + c3 - 2c3 t)
X2(0 = e~\c\ ~ C2 + C2 t - C3 t + C3 t2/2)
x3(0 = e~l{c2 + C3t)
13.
Here we are stymied initially, because if V3 = [1
0 0]
then (A-/lI)v3=0 does not
qualify as a (nonzero) generalized eigenvector. We there make a fresh start with
V3 = [0 1 0] , and now we get the desired nonzero generalized eigenvectors upon
successive multiplication by A - Al.
V,
= [1
0 0]T,
v2 = [0 2
1]T,
V3
= [0
1
0]T
x\(t) = e~'{c\ + c2 / + c312/2)
x2(t) = e~‘(2ci + C3 + 2c3 l)
x2(t) = e'\c2 + C31)
In each of Problems 15-18, the characteristic equation is -A3+3A2-3A + l = -(A-1)3.
Hence A - 1 is a triple eigenvalue of defect 1, and we find that (A - Al)2 = 0. First we find
the two linearly independent (ordinary) eigenvectors ui and 112 associated with A.
start with v2 = [1
0
0]
Then we
and calculate v, = (A - Al)x2 ^ 0. It follows that
(A - TI)Vj = (A - Al) v2 = 0, so vi is an ordinary eigenvector associated with A. However,
Vi is a linear combination of ui and U2, so v1e/ is a linear combination of the independent
solutions u,e' and u2e''.
But {vi, V2} is a length 2 chain of generalized eigenvectors
associated with A, so (yxt + y2)e' is the desired third independent solution. The corresponding
general solution is described by
168
Chapter 5
x(0 = el [ciu, + c2u2 + c3(vi t + v2)]
We give the scalar components
15.
X2(t), x3(/) of x(7).
u, = [3
-1
0]T
u2 = [0
0
1]T
v, = [-3
1
1]T
v2 = [1
0
0]T
u, = [2 0 -9]t
u2 = [1 -3
0]T
vi = [0
v2 = [0
0]T
x\(t) = e'(hc\ + c3 - 3c3t)
x2(t) = e'(-ci + c31)
x2(t) = e\c2 + c31)
17.
6 -9]t
1
(Either v2 = [1
0 0]T or v2 = [0
0 if can be used also, but they
yield different forms of the solution than given in the book's answer section.)
x\(t) = e\2c\ + c2)
X2(t) = e{(-3c2 + C3 + 6C3 t)
X3(0 = e\-9c\ - 9c3 t)
19.
Characteristic equation
A4 - 2/12 +1=0
Double eigenvalue A = -1 with eigenvectors
Vj = [1
0 0
1]T and v2 = [0 0
1
0]T
0 3
0]T.
Double eigenvalue A = +1 with eigenvectors
v3 = [0
1
0 -2]t and v4 = [1
General solution
x(0 = e~l(c\\\ + c2v2) + ^(c3v3 + C4V4)
Scalar components
xi(/) = c\e~l + ctf1
x2(t) = C3el
x3(0 = C2e~t + 3c4et
x4(t) = c\e~t - 2c3et
Section 5.4
169
21.
Characteristic equation
A4 -4A3 + 6A2 -4/1 + 1 = (A-l)4 = 0
Eigenvalue A = 1 with multiplicity 4 and defect 2.
We find that (A - Al)2 & 0 but (A - AI)3 = 0. We therefore start with
V3 = [1
0
0 0]T and define v2 = (A-AI)v3 and v, = (A-AI)v2 ^ 0, thereby
obtaining the length 3 chain {vi, v2, V3} with
V) — [0 0
0
1]T,
v2 = [-2 11 0]T,
v3 = [1
0 0 0]T
T
Then we find the second ordinary eigenvector V4 = [0
general solution
0 10]. The corresponding
x(t) = el [civi + c2(vi t + v2) + c3(vi t2/2 + v21 + v3) + c4v4]
has scalar components
xi(f) = e\-2C2 + c3 - 2c3 t)
x2(0 = e\c2 + c31)
x3(t) = e\c2 + c4 + c31).
x4(f) = e\c\ + c21 + c312/2.)
23.
A\ = -1:
{vj} with vi = [1
-1
A2 = 3:
{v2} with v2 = [4
{V3} with v3 = [0
0
2
2]T
9]T and
1]T
Scalar components
xi(0 =
ci^_/ + 4c2e3/
x2(f) = -c\e~{
+ 2 c3e3t
x3(t) = 2c\e~l + 9c2e3t + c3e3t
In Problems 25, 26, and 28 there is given a single eigenvalue A of multiplicity 3. We find that
(A - AI)2 & 0 but (A - AI)3 = 0. We therefore start with \3 = [1 0 0]T and define
v2 = (A - AI)v3 and v, = (A - AI)v2 ^ 0, thereby obtaining the length 3 chain {vi, v2, V3} of
generalized eigenvectors based on the ordinary eigenvector vi. We list the scalar components of
the corresponding general solution
x(t) = c\\\eAl + c2(vif + \i)eXl + c3(\\t2/2 + v2r + y3)e21.
170
Chapter 5
25.
{vi, v2, v3} with
vi = [-1
0 -1]T,
v2 = [-4 -1 0]T,
v3 = [1
0 0]T
Scalar components
x,(0 = e2l(-c] -4c2 + c3 - c2t - Ac3t - c3t212)
x2(t) = e2‘(-c2-c3t)
x3(t) = e2l(-c]-c2t-c3t2 /2)
27.
We find that the triple eigenvalue k- 2 has the two linearly independent eigenvectors
[1
1
0]T and [-1
0
1]T Next we find that (A -/II) ^0 but (A-/lI)2=0. We
therefore start with \2 = [1
0 0]
v, = (A - kl)\2
=
T
and define
[-5
3
8]t
*
0,
thereby obtaining the length 2 chain {vj, V2} of generalized eigenvectors based on the
ordinary eigenvector \\. If we take V3 = [1 1 0] , then the general solution
x(7) = e2t [civi + C2(\\t + V2) + C3V3] has scalar components
x{(t) = e2'(-5c\+c2+c3-5c2t)
x2(t) = e1,(3c] + 3c2t)
x3(t) = e2,(Sc] +%c2t).
In Problems 29 and 30 the matrix A has two distinct eigenvalues \ and k^ each having
multiplicity 2 and defect 1. First, we select V2 so that v, = (A-/l,I)v2 ^ 0 but
(A - k[l)\l = 0, so {vi, V2} is a length 2 chain based on vj. Next, we select U2 so that
u, = (A - A,I)u2 ^ 0 but (A - 21I)u1 = 0, so {ui, U2} is a length 2 chain based on uj. We give
the scalar components of the corresponding general solution
x(0 = eAl' [cjvi + c2(\\t + v2)] +
29.
[c3ui + c4(u\t + u2)].
k =-l: {vi,V2} with \\ = [1 -3 -1 -2]T and V2 = [0
1
0
0]T,
k—
0
2
1]T
2: {ui, U2} with ui = [0 -1
1
0]T and 112 = [0
Scalar components
x,(7) = e~'(cl+c2t)
x2(t) = e~'(-3c, +c2 -3c2t) + e21 (~c3 -c4t)
x3(t) = 6 \—Cy—c1t) + e2 (c3 + 2c4 + c4/)
x4(0 = e~\-2cx -2c2t) + e2l(c4)
Section 5.4
171
31.
We have the single eigenvalue X - 1 of multiplicity 4. Starting with
V3 = [1
0
0
0]T, we calculate v2 = (A-/LI)v3 and v, = (A-/lI)v2 ^ 0, and find
that (A - /II)v, = 0. Therefore {vj, V2, V3} is a length 3 chain based on the ordinary
eigenvector vj. Next, the eigenvector equation (A - Xl)\ = 0 yields the second linearly
T
independent eigenvector V4 = [0
1
3
0] . With
v, = [42
7 -21 -42]r,
v2
V3 = [1
0
V4 = [0
0
0]T
and
= [34 22 -10 -27]T,
1
3
0]
the general solution
x(t) = el [civi + c2(\\t + V2) + c3(vi/2/2 + \2t + v3) + c4v4]
has scalar components
XjiT) = e'(42c, + 34c2 + c3 + 42c2/+ 34c3/+ 21c3/2)
x7(t) = c (7cj + 22c2 +
+ 7c2t + 22c3t + 7c3t / 2)
x3(t) = e'(~2\C\ -\0c2+3cA -2\c2t -I0c3t-2\c3t2 /2)
x4(t)
e'(-42c, 21c2-A2c2t-21c3t-2\c3t2).
-
'4 i
-4
1
0"
"1
i
0
O'
4
4i
0
1
0
0
1
0
0
0
4i
-4
0
0
0
1
_0
0
4
4i
_0
0
0
0_
-32
-32 i
8/
-8
32/
-32
8
0
0
_ 0
0
II
HH
1
The chain {vi, V2} was found using the matrices
>
33.
=
—>
and
o'
8/
0
0
1
i
-32
-32 i
0
0
0
0
32/
-32_
_0
—1
0
0
0
=
1
0
(A-/II)2
1
‘
where
signifies reduction to row-echelon form. The resulting real-valued solution
vectors are
172
xi(7) = e3t [ cos 41
sin 41
0
0
]T
x2(t) = e3t [-sin 41
cos 4t
0
0
]T
X3(t) = e3t [ t cos 41
t sin 4t
cos 41
sin 41 ]T
x4(f) = e3t [-t sin 41
t cos 41
-sin 41
cos 41 ]T.
Chapter 5
35.
The coefficient matrix
A
0
0
1
0
0
0
0
1
-1
1
-2
1
1
-1
1
-2
=
has eigenvalues
X =
0 with eigenvector
X = -1
with eigenvectors
X = -2 with eigenvector
Vi = [1
1
0
v2 = [1
0-1
0]
T
0]T and v3 = [0
v4 = [1 -1 -2
1
0-1]
2]T.
When we impose the given initial conditions on the general solution
t
t
x(7) = C\\\ + C2\2C~ +
we find that c\ = vo, C2 =
masses are given by
i
+ C4V4^"
= -vo, c\ = 0. Hence the position functions of the two
*i(0 = X2(t) = v0(l - e~).
Each mass travels a distance vo before stopping.
SECTION 5.5
MATRIX EXPONENTIALS AND LINEAR SYSTEMS
In Problems 1-8 we first use the eigenvalues and eigenvectors of the coefficient matrix A to
find first a fundamental matrix O(t) for the homogeneous system x' = Ax. Then we apply the
formula
x(0 = 0(00(0)-’x0,
to find the solution vector x(7) that satisfies the initial condition x(0) = xo. Formulas (11) and
(12) in the text provide inverses of 2-by-2 and 3-by-3 matrices.
1.
Eigensystem:
A, = 1,
T
v, = [1
A, =3,
-1] ;
v2=[l
If
3/
0(0
x(0
eA,lv
e^'v-
M
-e
e‘
ev' 1 '1
-f " 3 "
1 ’ 5e‘ +ev ’
—e'
e31 *2 _1
1 _
-2
” 2 _-5e' +e3'_
Section 5.4
173
3.
Eigensystem:
0(0
5.
[Re(v^')
=
v = [1 + 2 i
2]T
cos 4t-2 sin 4t
2 cos 4t + sin 4t
2 cos41
2 sin 41
Im(v^/l/)] =
cos 4/-2 sin 4/
2 cos 4t + sin 4t
1 "0
2“ "0”
2 cos4/
2 sin 4/
4 2
-1
Eigensystem:
<t(0
7.
=
A = 4 /,
A = 3i,
v = [-!+/
[Re(veA')
-5 sin 4/
1
_1_ ” 4 4 cos 4/ -2 sin 4t
3]T
Im(veA')
— cos 3/ — sin 3/
cos 3t - sin 3t
3 cos 31
3 sin 31
-cos 3/-sin 3/
cos 3/-sin 3/
1 "0
f " 1 "
3cos 3/
3 sin 3/
*3 _3
1
3 cos 3/-sin 3/
1
_-l_ ~ 3 -3 cos 3/+ 6 sin 3/
Eigensystem:
Al= 0,
0(0
Vj = [6
2
5]t;
A, = 1,
e^'x2
=
e'
e~'
5
2e'
2e~'
3e‘
2e~'~
2
e‘
e"
5
2e‘
2e'
e^,x3
1
2
-2
-13
0
•
A3=-l,
v3 = [2
1
2]T
i
2
6
1
2er
2]T;
ro
3e'
1
o
i_
~6
v2=[3
~-12 + 12e' +2e~'~
~2
1
•
—
0
_4 + 4e' + e~‘
-10 + 8e' +2e~‘
In each of Problems 9-20 we first solve the given linear system to find two linearly
independent solutions xi and X2, then set up the fundamental matrix 0(0 = [x,(0
and finally calculate the matrix exponential eAl = 0(0 O(O)-1.
9.
Eigensystem:
0(0
Al = 1,
e
—*
31
-1
2
3/
1
-1
2e
174
T
v, = [1
1] ;
e1
i
e
h = 3,
v2 = [2
2ei‘~
3/
e
_
3/
2e' - 2e3'
3/
2e' -e3‘
-e + 2e
j
■6
.
+6
Chapter 5
If
x2(0]>
[eVv,
>
c
1]T;
X, = 3,
e21
3e3'"
_e2‘
2e3' _
v2 = [3
2]T
—
3e3'" '-2
e2'
II
3"
-2e2' + 3e3'
3e2' -3e3'"
_-2e2' + 2e3'
3e2'-2e3'
—
/'
13.
v, = [1
to
II
©
Hi
A, =2,
_1
Eigensystem:
1
11.
2e3'.
_
1
1
-
_
rp
A, - 1,
cD(/)
\
v, = [1
1] ;
A, =3,
v2 = [4
3]t
i
Eigensystem:
~-3e'
>3'
4e' - 4e3'
3e3/
-1
-3ef
.3/
4e' - 3e3'
Eigensystem:
A, = 1,
v, =[2
1]T;
r ;,
j,
[^'v,
2e2'
"2_
-2e‘+2e21
i
e'
=
_ 1
A, =2,
[e^'v,
' e2'
v, = [1
-1]T;
1 "1
-l"
5e'-4e2' _
^ = 4,
e4'_ * 2 _1
1 _
1 " e2' + e4'
v2 = [1
If
—e
2/ .
4/1
+ 2
1
ii
to
10e'-10e2'’
evv2
e4'"
-e2' + e4'
e
2/ .
4/
+e
1
1
2]T
2,
L/i
_1
evv2] =
v2=[5
5e2‘
4e' +5e2'
Eigensystem:
<D(0
2e/
Oi
to
_1
i
I
to
=
-i
^=2,
to
cD(0
17.
e'
4"
-3
4e3'
15.
ns
U>
-
1_
o
-i
C ^21 v
ii
©
e
A/
=
A, = 5,
_<f
Eigensystem:
i——i
19.
' e5'
-2e5'
2e10'
e10'
v,
[1
-2]T;
^ =10,
v2 = [2
If
e5'
2e10'
-2e5/
e10/
-2"
1 ‘ e5' + 4e10'
-2e5' + 2el0'~
1
5 -2e5' +2e10'
4e5' + e10'
J
Section 5.5
175
A = 2I + B
A = I+B
where
33.
\+t
-t
-t-t2
t
1 -t
t-t2
0
0
1
2,2
where
so
eA'= e2,'eB'= (e2'I)(I + Bt).
5 te2'
"4“
e2'
1_
B3=0,
so
2 te'
(3/ + 2/V"
0
e'
2 te'
0
0
e'
2/ 4 + 35/"
7
= e
5 + 12/
6
e
At
It
= e e
Bt
6
3,3 •
(e'I)(I + B/+iBV+£BV).
T
2/
3/ + 6/2
4/ + 6/2 + 4/3
0
1
6/
3/ + 6/
0
0
1
2/
/ \
A/ 1
x(0 = eA
1
_0
0
0
1
_1_
,
cosh/1
sinh/1
sinh/
cosh/1 5
x(/) = eA'c
Hence
"4 + 28/ + 12/2’
"l
eAl = I cosh/ + A sinh/
Hence
eA/ = elteBt = (<?' I)(I + /+}B2/2).
e'
where B4 = 0. so
= e
1
i_
A = I+B
/
1 ~t
B2=0,
e2'
A/
eA' =
e
t
1
0
29.
—t
so eA‘ = 1 + At + -AZ/
e
27.
1+/
>
25.
A'=0
so eAl = I + A/ =
II
23.
A2 =0
*
.
21
Hence
1 + 9/ + 12?2 + 4+
= e
/
1 + 9/ + 6/2
1 + 2/
1
so the general solution of xF = Ax is
c} cosh/ + c2 sinh/1
cx sinh / + c2 cosh /
In Problems 35-^40 we give first the linearly independent generalized eigenvectors up u2, • • •, u/;
of the matrix A and the corresponding solution vectors x,(7), x2(/1), • ••, xfl(t) defined by Eq.
(34) in the text, then the fundamental matrix ®(/) = [xj(/)
calculate the exponential matrix eA/ = ®(/)®(0)~\
176
Chapter 5
x2(/)
•••
xw(/)]. Finally we
35.
A = 3:
u, = [4
Of,
u2 = [0
if
(up u2} is a length 2 chain based on the ordinary (rank 1) eigenvector up so
U2 is a generalized eigenvector of rank 2.
x2(t) = eXl (u2 + (A-AI)u2t)
<D(0 = [x,(r)
x2(f)] = e3' 4
4^
1
4^
_1
x,(0 = eA'u„
O
_1
1
= e
\ = 2:
u, = [1
0
Of,
f = 1:
u, = [9
-3
_0
37.
0"
1 '1
•—
1 _ 4
4‘
3/
"1
40
_0
1 _
x,(0 = evu,
Of,
u3 = [10
1 -If
{u2, u3} is a length 2 chain based on the ordinary (rank 1) eigenvector 112, so
113 is a generalized eigenvector of rank 2.
x2(0 = eA2‘u2,
0(0 = [xj(0
0
x3(0 = eAlt (u3 + (A -/l2I)u3/t)
x2(0
90
-30
x3(0]
(10 + 900
(1 - 3t)el
0
39.
f =1:
e21
90
(10 + 900
0
-3e'
(1 - 3t)e‘
0
0
-0
"3
9
13"
0
-1
-1
0
0-3
e2t
-3el+3e2t
(-13-9f)0 + \3e21
0
0
3 te'
0
0
0
u, = [3
0
0
Of,
u2 =[010
Of
{lip u2} is a length 2 chain based on the ordinary (rank 1) eigenvector ui, so
U2 is a generalized eigenvector of rank 2.
Xj(0 = evU[>
x2(0 = eAl' (u2 + (A -A,I)u2f)
Section 5.5
177
^ = 2:
u3 = [144
36
12
0]T,
u4 = [0
27
17
4]T
{u3, u4} is a length 2 chain based on the ordinary (rank 1) eigenvector 113, so
114 is a generalized eigenvector of rank 2.
x3(0 = evu3,
x4(0 =
(u4 + (A -/ljl)u4f)
3e'
3 te‘
144e2'
- 144? e2'
0
e'
36e2'
(27 + 36?) e
0
0
12e2'
(17 + 12?)e
0
0
0
4e2'
0(0 = [x,(0 x2(0 x3(0 x4(0] =
At
0
-192
816"
1
0
48
-144
288
48
0
0
4
-17
_0
0
0
<N
e
"16
3 te‘
144e2'
144?e2'
0
e‘
36e2'
(27 + 36?)e2'
0
0
I2e21
(17 + 12?)e2'
_ 0
0
0
4e2'
e'
3te‘
(-12-9/y +12te2‘
(51 + 1 $t)e‘ + (-51 + 36t)e21
0
e'
-3e‘ + 3e21
6e‘ +(-6 + 9t)e21
0
0
e21
3 te21
0
0
0
1
~3e'
SECTION 5.6
NONHOMOGENEOUS LINEAR SYSTEMS
1.
Substitution of the trial solution x (t) = a, yp(t) = b yields the equations
<2 +26+ 3 = 0, 2a + 6-2 = 0 with solution a = 7/3, 6 = -8/3. Thus we obtain the
particular solution x(0 = 7/3, y(0 = - 8 / 3.
3.
When we substitute the trial solution
xp — £?| + b^t + Cj/^ ,
= #2 + b2t + c2^
and collect coefficients, we get the equations
178
Chapter 5
3tf]+4a2 = bx
3b}+4b2 = 2c,
3c,+4c2 = 0
3^j -f- 2a2 — b2
3/>, + 2b2 — 2c2
3c, + 2c-, +1 — 0.
Working backwards, we solve first for c, = -2/3, c2 = 1/2, then for
= 10/9, b2 = -7 /6, and finally for a, =-31/27, #2 =41/36. This determines the
particular solution x/,(/L),
Next, the coefficient matrix of the associated
homogeneous system has eigenvalues 2, = -1 and /^ = 6 with eigenvectors
v, = [1
-1]
and v2 = [4
3] , respectively, so the complementary solution is given
by
xc(t) = c,e“' +4c2e6',
xc(t) = -c,e“' +3c2e61.
When we impose the initial conditions x(0) = 0, y(0) = 0 on the general solution
x(0 = xc(0 + xp(0>
y(0 = yc(0 + yp(0 we find that c, = 8/7, c2 = 1/756. This
finally gives the desired particular solution
x(t) =
J_( 864e"' +4e6' -868 + 840?-504^2)
756
y(t) = -J_(-864e"' +3e6' +861-882/ + 378/2)
756
5.
The coefficient matrix of the associated homogeneous system has eigenvalues \ - -1
and ^2=5, so the nonhomogeneous term e~‘ duplicates part of the complementary
solution.
We therefore try the particular solution
xp(t) = a] + bxe~' + c/c_/, yp(t) = a2 +b2e~' +c2te~‘.
Upon solving the six linear equations we get by collecting coefficients after substitution
of this trial solution into the given nonhomogeneous system, we obtain the particular
solution
x(t) = I(-12-e"'-7 te~‘\
3
7.
y(t) = -(-6-7te").
3
First we try the particular solution
x(t) - a} sin t + bx cos T
r
yM) = a2 sin t + b2 cos t.
a
Upon solving the four linear equations we get by collecting coefficients after substitution
of this trial solution into the given nonhomogeneous system, we find that a, = -21 / 82,
bx = -25 / 82, a2 = -15 / 41, b2 = -12 / 41. The coefficient matrix of the associated
Section 5.6
179
homogeneous system has eigenvalues \ - 1 and
T
v, = [1
1]
= -9 with eigenvectors
'T'
and v2 = [2
- 3] , respectively, so the complementary solution is given
by
xc(t) = cte' +2c2e~9',
yc(t) = cte‘-3c2e~9'.
When we impose the initial conditions x(0) = 1, y(0) = 0, we find that c, =9/10 and
c2 = 83/410. It follows that the desired particular solution x = xc + x , y = yc + yp is
given by
x(t) - —(369e' + 166e~9' - 125cos/- 105sinf)
410
y(t) = -1—(369^'-249e”9/- llOcos^-lSOsin/1).
410
9.
Here the associated homogeneous system is the same as in Problem 8, so the
nonhomogeneous term cos 21 term duplicates the complementary function. We
therefore substitute the trial solution
xp(t) = ax smlt + fy cos2t + cjsin2t + d\tcos2t
yp(t) = a2 sin2t + b2 cos2t+ c2tsin2t + d2tcos2t
and use a computer algebra system to solve the system of 8 linear equations that results
when we collect coefficients in the usual way. This gives the particular solution
x(t) = —(sin2t + 2tcos2t + tsin2t),
4
11.
y(t) = — /sin2/\
4
The coefficient matrix of the associated homogeneous system has eigenvalues \ - 0
and A? =4, so there is duplication of constant terms.
We therefore substitute the
particular solution
xp(t) = al+blt,
yp(t) = a2 + b2t
and solve the resulting equations for a, = -2, a2 =0, b}= -2, b2 = 1. The eigenvectors
of the coefficient matrix associated with the eigenvalues \ - 0 and A^-4 are
v, = [2
-1]
and v2 = [2
1] , respectively, so the general solution of the given
nonhomogeneous system is given by
x(t) = 2cx +2c2e4' -2 -2t,
180
Chapter 5
y(t) = -c, +c2e4' +t.
When we impose the initial conditions x(0) = 1, y(Q) = -1 we find readily that
C\ =5/4, c2 = 1/4. This gives the desired particular solution
x(r) = }(l-4 t + e4l\
13.
y(t) = K-5 + 4 t + e4t).
The coefficient matrix of the associated homogeneous system has eigenvalues \ = 1 and
^2= 3, so there is duplication of el terms. We therefore substitute the trial solution
xP(t) = (Px+bfye',
yp{t) = (a2+b2t)e‘.
This leads readily to the particular solution
x(t) = ±(1 + 5 f)e\
y(t) = -|fe\
In Problems 15 and 16 the amounts x}(t) and x2(t) in the two tanks satisfy the equations
x[ = rc0-k}x p
x^ = k]xl-k2x2
where ki=rl Vj in terms of the flow rate r, the inflowing concentration co, and the volumes
Vx and V2 of the two tanks.
15.
(a)
We solve the initial value problem
xj = 20-x]/10,
*i(0) = 0
x2 = Xj/10-x2/20,
^2(^) = 0
for xj(0 = 200(1 -e-,/w),
(b)
x2(t) = 400(1 + e-'/i0 -2e~,/20).
Evidently x, (t) -> 200 gal and
x2 (t) -> 400 gal as
t -> oo.
(c)
It takes about 6 min 56 sec for tank 1 to reach a salt concentration of 1 lb/gal, and
about 24 min 34 sec for tank 2 to reach this concentration.
In Problems 17-34 we apply the variation of parameters formula in Eq. (28) of Section 5.6. The
answers shown below were actually calculated using the Mathematica code listed in the
computing project for Section 5.6. For instance, for Problem 17 we first enter the coefficient
matrix
Section 5.6
181
A = {{6,
-V},
-2}};
{1,
the initial vector
xO = {{0},
{0}>;
and the vector
f[t_]
:= {{60},
{90}};
of nonhomogeneous terms. It simplifies the notation to rename Mathematical exponential
matrix function by defining
exp [A_]
: = MatrixExp [A]
Then the integral in the variation of parameters formula is given by
integral =
Integrate[exp[-A*s]
.
f[s],
{s,
0,
t}]
// Simplify
'-102 + 7e~5' + 95e'~
-96+ e"5' +95e'
Finally the desired particular solution is given by
solution =
exp[A*t] .
(xO + integral)
// Simplify
~102-7*T5'-95e'"
_ 96-e~51 -95e‘
(Maple and MATLAB versions of this computation are provided in the applications manual that
accompanies the textbook.)
In each succeeding problem, we need only substitute the given coefficient matrix A, initial
vector xO, and the vector f of nonhomogeneous terms in the above commands, and then reexecute them in turn. We give below only the component functions of the final results.
17.
182
r/O - 102 -95e-/ -7e5/,
x2(t) = 96 - 95 e" - e51
Chapter 5
19.
x,(?) = — 70 — 60/ + 16e_3/ + 54e2',
21.
x,(?) = -<T'-14e2'+15e3',
23.
Xj(?) = 3 + ll? + 8? ,
25.
^,(0 = -1+ 8/1+ 008^-88111 z1,
27.
x,(?) = 8?3+6?4,
29.
x,(?) = ?cos?-ln(cos?)sin?,
31.
x, (?) = (9?2+4?3)e',
33.
x,(?) = 15?2+60?3+95?4+12?5,
x3(?) = 15?2+20?3,
x2(t) = 5-60?-32e-3' +27e2'
x2(?) = -5e"-10e2' + 15e3'
x2(?) = 5 + 17? + 24?2
x2(/) = -2 + 4zL + 2coszL-3siiU
x2(?) = 3?2-2?3+3?4
x2(?) = ?sin?-ln(cos?)cos?
x2(?) = 6?V,
x2(?)
x3(?) = 6?e'
= 15?2 + 55?3 +15?4,
x4(?) = 15?2
Section 5.6
183
CHAPTER 6
NONLINEAR SYSTEMS AND PHENOMENA
SECTION 6.1
STABILITY AND THE PHASE PLANE
1.
The only solution of the homogeneous system 2x- y = 0, x - 3y = 0 is the origin
(0, 0). The only figure among Figs. 6.1.11 through 6.1.18 showing a single critical point
at the origin is Fig. 6.1.13. Thus the only critical point of the given autonomous system
is the saddle point (0, 0) shown in Figure 6.1.13 in the text.
3.
The only solution of the system x -2y + 3 = 0, x-y + 2 = 0 is the point (-1,1). The
only figure among Figs. 6.1.11 through 6.1.18 showing a single critical point at (-1, 1) is
Fig. 6.1.18. Thus the only critical point of the given autonomous system is the stable
center ( -1, 1) shown in Figure 6.1.18 in the text.
5.
The first equation 1 - y2 =0 gives y = 1 or y = -1 at a critical point. Then the
second equation x + 2y - 0 gives x = -2 or x = 2, respectively. The only figure
among Figs. 6.1.11 through 6.1.18 showing two critical points at (-2, 1) and (2, -1) is
Fig. 6.1.11. Thus the critical points of the given autonomous system are the spiral point
(-2, 1) and the saddle point (2, 1) shown in Figure 6.1.11 in the text.
7.
The first equation 4x - x =0 gives x = -2, x = 0, or x = 2 at a critical point.
Then the second equation x-2y = 0 gives y = -l9 y = 0, or y = l, respectively.
The only figure among Figs. 6.1.11 through 6.1.18 showing three critical points at
(-2, -1), (0, 0), and (2, 1) is Fig. 6.1.14. Thus the critical points of the given
autonomous system are the spiral point (0, 0) and the saddle points (-2, 1) and (2, 1)
shown in Figure 6.1.14 in the text.
In each of Problems 9-12 we need only set x' = x" = 0 and solve the resulting equation for x.
9.
The equation 4x - x3 = x(4 - x2) = 0 has the three solutions x = 0, ± 2. This gives the
three equilibrium solutions x(t) = 0, x(/) = 2, x{t) = -2 of the given 2nd-order
differential equation. A phase plane portrait for the equivalent 1 st-order system
184
Chapter 6
x> —
y — —4x + x3 is shown in the figure below. We observe that the critical point
(0,0) in the phase plane appears to be a center, whereas the points (±2,0) appear to be
saddle points.
X
The equation 4sinx = 0 is satisfied by x-nn for any integer n. Thus the given 2ndorder equation has infinitely many equilibrium solutions: x(t) = me for any integer n.
A phase portrait for the equivalent lst-order system x' = y, yr = -3y - 4 sin x is shown
below. We observe that the critical point (wr, 0) in the phase plane looks like a spiral
sink if n is even, but a saddle point if n is odd.
Section 6.1
185
In Problems 13-16, the given x- and ^-equations are independent exponential differential
equations that we can solve immediately by inspection.
13.
Solution:
x(t) = xoe~2t, y(t) = y§e~2t
Then y = (y0 /x0)x = kx, so the trajectories are straight lines through the origin. Clearly
x(t), y(t) —> 0 as t —> +oo, so the origin is a stable proper node like the one shown in
the left-hand figure below.
-7V
\j WA/ n!/ st vy/1/ y
\ 1 i / //V
^
\u i UX^ ^
^\i V J/ A K'
^ 1/<—
V•
X* s* X?h n\\X
+ X/ / f t \ \\^
T
V\\
\\
X/
T
T
T
*1
/
_\ _ _
-5
15.
Solution:
0
-5
5
t
x(t) = x§e~ , y{t) = y0e~
0
5
X
t
Then x = (x0/ y0 )(y0e~ ) = ky , so the trajectories are parabolas of the form
x = ky , and clearly x(t), y(t) -» 0 as t -» +oo.
Thus the origin is a stable improper
node like the one shown in the right-hand figure above.
17.
Differentiation of the first equation and substitution using the second one gives
x" = y' = x, so
x" + x = 0.
We therefore get the general solution
x(t) = A cos t + B sin t
y(t) = B cos t - A sin t
(y = xf).
Then
x2 + y2 = (Acost + Bsint)2 + (Boost - Asint)2
= (A2 + B2)cos21 + (A2 +B2)sin21 = A2 + B2.
186
Chapter 6
Therefore the trajectories are clockwise-oriented circles centered at the origin, and the
origin is a stable center as in the figure below.
0
19.
Elimination of y as in Problem 17 gives x" + 4x = 0, so we get the general solution
x{t) = A cos 2t + B sin 21,
y{t) = B zos2t - A sm2t
OO
0
(y = jxr).
0
Then x + y = A + B , so the origin is a stable center, and the trajectories are
clockwise-oriented circles centered at (0, 0), as in the figure above, which serves equally
well for both Problem 17 and 19 — the circular orbits are the same, but are parameterized
differently in the two problems.
21.
We want to solve the system
-ky + x(l - x2 - y2) = 0
be + y(l- x2 - y2) = 0.
If we multiply the first equation by -y and the second one by x, then add the two
results, we get k(x2 + y2) = 0. It therefore follows that x = y - 0.
23.
The equation dy/dx = -x/y separates to xdx + ydy = 0, so x2 + y2 = C. Thus
2
2
the trajectories consist of the origin (0, 0) and the circles x + / = c> 0 (looking,
again, like those shown in the figure above).
•
Section 6.1
187
25.
The equation dy / dx = -x/4y separates to xdx + 4ydy = 0, so x2+4y2 = C.
Thus the trajectories consist of the origin (0, 0) and the ellipses x + 4y
shown in the figure below.
i__^
-5
0
= C> 0
>
5
X
27.
If (ffat) = x{t + y) and y/j) = y(t + y) then
<j)'(t) = x’(t + f) = y(t + y) = yKt),
but
V<t) = y\t + y) = x(t + r)it +y) = t fit) + yfit)*t fit).
SECTION 6.2
LINEAR AND ALMOST LINEAR SYSTEMS
In Problems 1-10 we first find the roots X\ and fa of the characteristic equation of the
coefficient matrix of the given linear system. We can then read the type and stability of the
critical point (0,0) from Theorem 1 and the table of Figure 6.2.9 in the text.
1.
The roots fa = -1 and fa = -3 of the characteristic equation T2 + 4/1 + 3 = 0 are
both negative, so (0,0) is an asymptotically stable node as shown on the right at the top
of the next page.
3.
The roots fa = -1 and fa = 3 of the characteristic equation A2 - 21-3 = 0 have
different signs, so (0,0) is an unstable saddle point as shown on the left at the top of the
next page.
188
Chapter 6
5.
The roots A\ = 12 = -1 of the characteristic equation A2 + 2 A +1 = 0 are negative and
equal, so (0,0) is an asymptotically stable node as in the right-hand figure below.
>s o
7.
The roots A\, A2 = 1 ± 2 i of the characteristic equation A2 - 2A + 5 = 0 are complex
conjugates with positive real part, so (0,0) is an unstable spiral point as shown in the
left-hand figure above.
Section 6.2
189
9.
The roots A\, Ai = ±2 i of the characteristic equation A + 4 = 0 are pure imaginary, so
(0,0) is a stable (but not asymptotically stable) center as in the left-hand figure below.
^
11.
The Jacobian matrix J =
1
3
J
2
-
0
'
has characteristic equation A + 3A + 2 = 0
-4
—H
and eigenvalues A\ = -1, Ai = -2 that are both negative. Hence the critical point
(2, 1) is an asymptotically stable node as in the right-hand figure above.
13.
The Jacobian matrix J =
2
-1
3
-2
has characteristic equation A~ ~ 1 = 0
and eigenvalues A\ — -1, I2 = +1 having different signs. Hence the critical point
(2, 2) is an unstable saddle point as in the figure below.
0
190
Chapter 6
15.
The Jacobian matrix
has characteristic equation X2 + 2A + 2 = 0
and eigenvalues X, X — ~ 1 ± i that are complex conjugates with negative real part.
Hence the critical point (1,1) is an asymptotically stable spiral point as shown in the
figure below.
0
-5
The Jacobian matrix
-
-5
-1
has characteristic equation X +4 = 0 and pure
imaginary eigenvalues X\, X. = ±2/. Hence (5/2,-1/2) is a stable (but not
asymptotically stable) center as shown below.
Section 6.2
191
In each of Problems 19-28 we first calculate the Jacobian matrix J and its eigenvalues at (0,0)
and at each of the other critical points we observe in our phase portrait for the given system.
Then we apply Theorem 2 to determine as much as we can about the type and stability of each of
these critical points of the given almost linear system. Finally we a phase portrait that
19.
J=
1 + 2y
-3 + 2x
4-y
-6-x
At (0,0): The Jacobian matrix J
1
-3
4
-6
has characteristic equation T2 +5T+ 6 = 0
and eigenvalues X\ = -3, X2 = -2 that are both negative. Hence (0,0) is an
asymptotically stable node of the given almost linear system.
9/5
-5/3
At (2/3, 2/5): The Jacobian matrix J =
has characteristic equation
18/5 -20/3
X2 + jjX-6 = 0 and approximate eigenvalues Xl « -5.89, X1 « 1.02 with different
signs. Hence (2/3, 2/5) is a saddle point.
The figure below shows both these critical points.
-202
*
21.
1 + 2x
2 + 2y
2-3 y
-2 - 3x
At (0,0): The Jacobian matrix J
1
2
has characteristic equation X + X - 6 = 0
and eigenvalues X\ - -3, X2 = 2 with different signs. Hence (0,0) is a saddle point of
the given almost linear system.
192
Chapter 6
At (-0.51,-2.12): The Jacobian matrix J
-0.014
-2.236
/-*W/
8.354
-0.479
has complex conjugate
eigenvalues A, « -0.25 + 4.32/, ^ « -0.25 + 4.32/ with negative real parts. Hence
(-0.51,-2.12) is a spiral sink.
The figure on the left below shows these two critical points.
0
23.
0
J=
2 + 3x2
4
-6 + 4y1
At (0,0): The Jacobian matrix J
2
-5
4
-6
has characteristic equation A~ + 4 A + 8 = 0
and complex conjugate eigenvalues A, - -2 + 2z,
= -2 - 2/ with negative real part.
Hence (0,0) is a spiral sink of the given almost linear system.
["5.495
-5 1
At (-1.08,-0.68): The Jacobian matrix J«
has eigenvalues
v
4
-7.276
Aj « -5.45, Aj « 3.67 with different signs. Hence (-1.08,-0.68) is a saddle point.
The figure on the right above shows these two critical points.
15.
1 + 3_y
-2 + 3x
2-2x
-3-2 y
has characteristic equation
At (0,0): The Jacobian matrix
+ 2A + 1 — 0
and equal negative eigenvalues Aj = -1, A2 = -1. Hence (0,0) is either a nodal sink or a
spiral sink of the given almost linear system.
Section 6.2
193
At (0.74,-3.28): The Jacobian matrix J
—8.853
0.226"
_ 0.516
3.568
hhas real eigenvalues
^ «-8.86, fa & 3.58 with different signs. Hence (0.74,-3.28) is a saddle point
At (2.47,-0.46): The Jacobian matrix J
"-0.370
5.410
-2.940
-2.087
has complex conjugate
eigenvalues fa « -1.23 + 3.89/, fa « -1.23 + 3.89/ with negative real part. Hence
(2.47,-0.46) is a spiral sink.
At (0.121,0.074): The Jacobian matrix J
1.222
-1.636'
1.758
-3.148
has real eigenvalues
fa « -2.34, fa -0.42 with different signs. Hence (0.121,0.074) is a saddle point.
The left-hand figure below shows clearly the first three of these critical points. The righthand figure is a close-up near the origin with the final critical point now visible.
/
i
l
/ /1 /
/ i// aV / y / if
ifai /// fa fa fa/ \l /Jj
////A / / / Lfa?
/ y/y fa j / far y
j f /A + 'Zf iT r /
'r/i t/ y/y
Ji {
i 4 fafay fafar faff
fa fa fa 7/T
fafay far?
fafa? fa fa T/T 7.
/
0.2
-
l + 4x3
-l-2y
2-2x
-1 + 4>’3
0.2
X
X
27.
/-/_i_
0
At (0,0): The Jacobian matrix
has characteristic equation T2+1 = 0
and equal positive eigenvalues fa = -/, fa = +/. Hence (0,0) is either a center or a
spiral point, but its stability is not determined by Theorem 2.
0.934
0.014
At (-0.254,-0.507): The Jacobian matrix
has real eigenvalues
2.508 -1.521
fa « —1.53, fa -0.95 with different signs. Hence (-0.254,-0.507) is a saddle point.
194
Chapter 6
At (—1.557,—1.637): The Jacobian matrix J
-14.087
-4.273'
5.113
16.532
has real eigenvalues
4 ~ -13.36, Aj » 15.80 with different signs. Hence (-1.557,-1.637) is a saddle point.
At (-1.070,-1.202): The Jacobian matrix J
-3.905
1.403 '
^ 4.141
-7.940
has unequal negative
eigenvalues A, « -9.07, A, « -2.78 . Hence (-1.070,-1.202) is a nodal sink.
The left-hand figure below shows these four critical points. The close-up on the right
suggests that the origin may (but may not) be a stable center.
0
0.6
-
X
29.
J=
1
-1
2x
-1
At (0,0): The Jacobian matrix J =
real eigenvalues X] « -1,
1
0
1
-
'
has characteristic equation T2 -1 = 0 and
-1
« +1 with different signs. Hence (0,0) is a saddle point.
At (1,1): The Jacobian matrix J =
1
2
1
-
'
-1
has characteristic equation T2 +1 = 0 and
pure imaginary eigenvalues X] « +i, X^ « -i. Hence (1,1) is either a center or a spiral
point, but its stability is not determined by Theorem 2.
The figure at the top of the next page suggests that (1,1) is a stable center.
Section 6.2
195
31.
J=
' 0
2y
3x2
-1
At (1,1): The Jacobian matrix J
0
2
3
-1
has characteristic equation X + X - 6 = 0
and real eigenvalues Xx = -3, X1=+2 with different signs. Hence (1,1) is a saddle point.
At (-1,-1): The Jacobian matrix J =
0
-2
l3
-1
has characteristic equation
X~ + X + 6 = 0 and complex conjugate eigenvalues Xl « -0.5 + 2.398/,
with negative real part. Hence (-1,-1) is a spiral sink.
These two critical points are shown in the figure below.
196
Chapter 6
« -0.5 - 2.398/
The characteristic equation of the given linear system is
(A- sf + 1 = 0
with characteristic roots A\, A2 = e± i.
(a)
So if £< 0 then A\, X2 are complex conjugates with negative real part, and
hence (0, 0) is an asymptotically stable spiral point.
(b)
If s = 0 then Ai, A2 = ±i (pure imaginary), so (0,0) is a stable center.
(c)
If £> 0, the situation is the same as in (a) except that the real part is positive, so
(0, 0) is an unstable spiral point.
(a)
If h = 0 we have the familiar system x' = y, / = -x with circular trajectories
about the origin, which is therefore a center.
(b)
The change to polar coordinates as in Example 6 of Section 6.1 is routine,
yielding r' = hr
and 6' = -1.
(c)
If h = -1, then r' = -r integrates to give 2r = 1 /(t + C) where C is a
positive constant, so clearly r -> 0 as t —> +oo, and thus the origin is a stable spiral
point.
(d)
If h = +1, then r' - r3 integrates to give 2r2 = -1 /(t + C) where C = -B
is a positive constant. It follows that 2r = 1 /(B - t), so now r increases as t starts at
0 and increases.
The substitution y - vx in the homogeneous first-order equation
dy_ = y(2x3 -y3)
dx
x(x3-2 y3)
yields
dv _
v4 + v
dx
2v3 -1
Separating the variables and integrating by partial fractions, we get
U
i
^
V
+
2v -1
+
V + l
V
^
— V + 1y
ln((v + l)(v2 - v +1))
=
Section 6.2
dv
=
dx
x
lnv-lnx + lnC
197
Cv
(v + l)(v2-v + l)
X
X
Finally, the replacement v = y/x yields x3+y3 = Cxy.
SECTION 6.3
ECOLOGICAL APPLICATIONS:
PREDATORS AND COMPETITORS
1.
200-4y
-Ax
2y
-150 + 2x
At (0,0): The Jacobian matrix
has characteristic equation
(200-/l)(-150-/L) = 0 and real eigenvalues \ =-150, ^ =200 with different signs.
Hence (0,0) is a saddle point of the linearized system x' = 200x, y' = -150y. Seethe
left-hand figure below.
-5
At (75,50): The Jacobian matrix J =
0
iOO
0
U
5
-300
0
has characteristic equation
T2 + 30000 = 0 and pure imaginary eigenvalues Tj, /^ = +100/V3. Hence (75,50) is a
198
Chapter 6
stable center of the linearization u' — —300v, V = 100u. See the right-hand figure on the
preceding page.
3.
The effect of using the insecticide is to replace b by b +f and a by a - f in the
predator-prey equations, while leaving p and q unchanged. Hence the new harmful
population is (b+f)/q > blq- xe, and the new benign population is
(a -f)/p < alp = yE.
Problems 4-7 deal with the competition system
x' = 60x-4x2-3xy,
that has Jacobian matrix J =
5.
y' - 42y -2y2 -3xy
60 -8x-3y
-3x
-3 y
42-4y-3x
At (0,21) the Jacobian matrix J =
-3
0
-63
-42
(2)
has characteristic equation
(-3 - A)(-42 - A) = 0 and negative real eigenvalues \ - -42, /^ = -3. Hence (0,21) is
a nodal sink of the linearized system u' = - 3u, vf = - 63u - 42v .
7.
-18
At (6,12) the Jacobian matrix
-24
has characteristic equation
(-24 - A)2 -(-36)(-18) = 0 and real eigenvalues Xx =-24 + 18a/2>0,
^ = - 24 -18 V2 < 0 with different signs. Hence (6,12) is a saddle point of the
linearized system u' = - 24u -18v,
v' = -36w - 24v . The figure on the left below
illustrates this saddle point. The figure on the right shows all four critical points of the
system.
>
-5
0
5
U
0
5
10
15
20
X
Section 6.3
199
Problems 8-10 deal with the competition system
x' = 60x-3x2-4xy,
that has Jacobian matrix
9.
j/ = 42y-3y2-2xy
60 - 6x - 4 y
-4x
-2y
42-6y- 2x
-80
At (20,0) the Jacobian matrix
2
has characteristic equation
(-60 - A)(2 - A) = 0 and real eigenvalues A, = - 60,
=2 with different signs. Hence
(20,0) is a saddle point of the linearized system u' = - 60u - 80v, v' = 2v. The figure
below shows the four critical points of the system in (3).
\
TIT
-
TV 7 / V £/
i.u i/
l i
/ A kA
0
/
\
N
V
i/ /
15
7
-
/
^
/
/
/
y ^
i/
AT
f i/
S<>(12,6) ^
1 /y
-"AAr-—^
i
<—
^—
(20^^
-J_
15
10
<—
<r:
20
Problems 11-13 deal with the predator-prey system
x' = 5x - x2 - xy,
that has Jacobian matrix
11.
5-2x-y
-x
y
-2 + x
At (0,0) the Jacobian matrix
0
-2
y' = -2y + xy
has characteristic equation
(5 - A)(-2 - A) = 0 and real eigenvalues A, = - 2, A2 = 5 with different signs. Hence
(0,0) is a saddle point of the linearized system x' = 5x, y' = -2y.
200
Chapter 6
13.
-2
At (2,3) the Jacobian matrix
has characteristic equation
0
(~2 — A)(—A) — (3)(—2) — A2 + 2A + 6 = 0 and complex conjugate eigenvalues
4, Aj = -1 ±iyjs with negative real part. Hence (2,3) is a spiral sink of the
linearized system u' = - 2u - 2v, v' = 3u (illustrated below).
Problems 14-17 deal with the predator-prey system
x’ = x~ -2x- xy,
that has Jacobian matrix
15.
y' - y -4y + xy
2x-2- y
-x
y
2y-4+x
0
At (0,4) the Jacobian matrix
has characteristic equation
4
(-6 - A)(4 - A) = 0 and real eigenvalues A, = - 6, /^ = 4 with different signs. Hence
(0,4) is a saddle point of the linearized system u' = - 6w, v' = 4w + 4v.
17.
At (3,1) the Jacobian matrix J =
3
-3
1
1
has characteristic equation
(3 - A)(l - A) - (l)(-3) = A2' - 4A + 6 = 0 and complex conjugate eigenvalues
Section 6.3
201
\= 2 ± / V2
with positive real part. Hence (3,1) is a spiral source of the
linearized system u' = 3u - 3v,
v' = u + v (illustrated below).
>■
u
Problems 18 and 19 deal with the predator-prey system
x’ = 2x-xy,
that has Jacobian matrix
y' = - 5y + xy
—x
-5 + x
u
202
Chapter 6
(7)
19.
-5
At (5,2) the Jacobian matrix
has characteristic equation
0
- (2)(—5) = A2 +10 = 0 and pure imaginary roots A - ±i
VlO, so the origin
is a stable center for the linearized system u' = — 5v, v’ = 2u. This is the indeterminate
case, but the figure shown at the bottom of the preceding page suggests that (5,2) is also
a stable center for the original system in (7).
Problems 20-22 deal with the predator-prey system
x' = - 3x + x" - xy,
that has Jacobian matrix
21.
y' = -5y + xy
-3 + 2 x-y
-x
y
-5 + x
-3
At (3,0) the Jacobian matrix
-2
has characteristic equation
(3 - A)(~2 -A) = 0 and real eigenvalues A[=- 2,
= 3 with different signs. Hence
(3,0) is a saddle point of the linearized system u' = 3u- 3v, V = - 2v.
Problems 23-25 deal with the predator-prey system
x' = 7x-x -xy,
that has Jacobian matrix
23.
yf = -5y + xy
7-2x-y
-x
y
-5 + x
0
At (0,0) the Jacobian matrix
has characteristic equation
-5
(7 - A)(-5 - A) = 0 and real eigenvalues A[=- 5, A^-l with different signs. Hence
(0,0) is a saddle point of the linearized system x' = 7x, y' = -5y.
25.
At (5,2) the Jacobian matrix J
-5
-5
2
0
has characteristic equation
(-5 - A)(-A) - (2)(-5) = Az + 5A +10 = 0 and complex conjugate eigenvalues
Al,A2 = |(-5 ± i Jl5^j with negative real part. Hence (5,2) is a spiral sink of the
linearized system u' = — 5u — 5v, v' — 2u (illustrated in the left-hand figure at the top
of the next page).
Section 6.3
203
2x
27.
x-3
At (0,0): The Jacobian matrix
has characteristic equation
A +7/1 + 12 = 0 and negative real eigenvalues Aj = -4, /^ = -3. Hence (0,0) is a nodal
sink.
At (3,2): The Jacobian matrix J =
0
6
2
0
has characteristic equation A -12 = 0 and
real eigenvalues Al,A2= ±2>/3 with different signs. Hence (3,2) is a saddle point.
If the initial point (x0, j/0) lies below the northwest-southeast separatrix through (3,2),
then (x(t),y (7)) -» (0,0) as t —» oo. But if (x0,y0) lies above this separatrix, then
(x(r), >>(?))-> (oo,oo) as t —^ oo. See the right-hand figure above.
29.
4 - 2x
At (0,0): The Jacobian matrix J =
3
0
0
4
has characteristic equation
/l2 - 7A +12 = 0 and positive real eigenvalues Al = 3, /^ = 4 . Hence (0,0) is a nodal
source.
204
Chapter 6
At (3,0): The Jacobian matrix
has characteristic equation
T2 + 5X + 6 = 0 and negative real eigenvalues A, = -3, ^ = -2 . Hence (3,0) is a nodal
sink.
-1
At (2,2): The Jacobian matrix
0
X + 2X-4 = 0 and real eigenvalues Xl
has characteristic equation
-3.2361, Xj =1.2361 with different signs.
Hence (2,2) is a saddle point.
If the initial point (x0, y0) lies above the southwest-northeast separatrix through (2,2),
then (x{t),y (/)) -» (0,oo) as t -» oo. But if (x0,y0) lies below this separatrix, then
(x(0? y (7)) —> (3,0) as t —» oo. See the figure below.
J=
-2x-\y + 3
-\x
y
x—2
At (0,0): The Jacobian matrix J =
3
0
0
-2
'"N
has characteristic equation X -X-6 = 0
and real eigenvalues Xl = -2, /^ = 3 of opposite sign. Hence (0,0) is a
At (3,0): The Jacobian matrix J =
2
-3
u 0
saddle point.
4
1
has characteristic equation
X2 + 2X - 3 = 0 and real eigenvalues Xl = -3, X^ = 1 of opposite sign. Hence (3,0) is a
saddle point.
Section 6.3
205
At (2,4): The Jacobian matrix
has characteristic equation
A2 + 2A + 2 = 0 and complex conjugate eigenvalues A,, ^ = -1 ± i with negative real
part. Hence (2,4) is a spiral sink.
As t —> oo, each solution point (*(/), t(/)) with nonzero initial conditions approaches
the spiral sink (2,4), as indicated by the direction arrows in the figure below.
0
5
X
33.
-6x + y + 30
X
4y
4 x-6y + 60
0
At (0,0): The Jacobian matrix
has characteristic equation
80
A2 -110A + 2400 = 0 and positive real eigenvalues A = 30, A2 = 80. Hence (0,0) is a
nodal source.
At (0,20): The Jacobian matrix
10
0
40
-80
has characteristic equation
A2 + 70A - 800 = 0 and real eigenvalues A, = -80, \ = 10 of opposite sign. Hence
(0,20) is a saddle point.
15
At (15,0): The Jacobian matrix
110
has characteristic equation
A2 - 80A - 3300 = 0 and real eigenvalues A, = -30, ^ = 110 of opposite sign. Hence
(15,0) is a saddle point.
206
Chapter 6
At (4,22): The Jacobian matrix J =
-8
-4
44
-88
has characteristic equation
A2 + 96/1 + 880 = 0 and negative real eigenvalues fa » -85.736, fa = -10.264 . Hence
(4,22) is a nodal sink.
As t —> oo, each solution point {x{t\y (7)) with nonzero initial conditions approaches
the nodal sink (4,22). See the figure below.
SECTION 6.4
NONLINEAR MECHANICAL SYSTEMS
In each of Problems 1-4 we need only substitute the familiar power series for the exponential,
sine, and cosine functions, and then discard all higher-order terms. For each problem we give
the corresponding linear system, the eigenvalues fa and fa, and the type of this critical point.
1.
x' — 1 — ^1 + x + y x2 4- • • • j + 2y ~ —x + 2y
/ =
-x-4(y~ly2 h—) ~-x-4y
The coefficient matrix A =
-1
2
-1
-4
has negative eigenvalues fa - -2 and
fa = -3 indicating a stable nodal sink. Alternatively, we can calculate the Jacobian
matrix
Section 6.3
207
J(x,y) =
-ex
2
-1
-4cosx
so
J(0,0)
-1
2
-1
-4
This sink at the origin is illustrated in the left-hand figure below.
0
-3
3.
-
jt' = (\ + x + jX2-\—) + 2y-l « x + 2y
y = 8x + (l + y + ±y2 + ---)-l » 8x + ^
1
The coefficient matrix A
8
2
1
has real eigenvalues fa = -3 and fa = 5 of
opposite sign, indicating an unstable saddle point as illustrated in the right-hand figure
above. Alternatively, we can calculate the Jacobian matrix
J(x,f) =
5.
so
8
J(0,0) =
e'
1
2
8
1
The critical points are of the form (0, mi) where n is an integer, so we substitute
x = u, y = v + nn. Then
u! - x* = -u + sin(v + n7r) = -u + (cosn7r)v = -u + {-l)”v.
Hence the linearized system at (0, rut) is
u' = -u ± v,
vr = 2u
where we take the plus sign if n is even, the minus sign if n is odd. If n is even the
eigenvalues are fa = 1 and fa = -2, so (0, mi) is an unstable saddle point. If n is odd
the eigenvalues are fa,fa = (-l±/V7)/2, so (0, nn) is a stable spiral point.
208
Chapter 6
-IV3pi :.VsA
'A
/
A
A
2pi
V\
Pi
V
\x
/
VssV
AVA
Vs
ASS
' Vs
VssV .
A r\A
v\.a
-pi
A
V V •
V
sV
0
v V
A +\V\ \\
- V V
Vs
-2pi -
V
V
V Vv
V
V
-3pi -
/xSsQA
sVr 5a
-5
^
Vss<\ __
0
X
Alternatively, we can start by calculating the Jacobian matrix J(x,y) =
At (0 ,nn), n even: The Jacobian matrix
cosy
2
0
1
0
A2 + A - 2 = 0 and real eigenvalues A, = -2,
-1
has characteristic equation
= 1 of opposite sign. Hence (0,nn) is a
saddle point if n is even, as we see in the figure above.
At (0 ,nn), n odd: The Jacobian matrix
-1
has characteristic equation
0
A2 + A + 2 = 0 and complex conjugate eigenvalues Zl,A7= y(-l ± iyfl) with negative
real. Hence (0,nn) is a spiral sink if n is odd, as indicated in the figure.
7.
The critical points are of the form (nn, nn) where n is an integer, so we substitute
x = u + nn, y = v + nn. Then
u' = x' =
1 -eu~v
= l-(\ + (u-v) + j(u-v)2 + •••)
« -m + v,
v' = y' = 2sin(w + ^27r) = 2 sin wcos ^ « 2(—1)"m.
Hence the linearized system at (nn, nn) is
u'
=
-m + v,
Section 6.4
v' = ±2m
209
and has coefficient matrix A =
-1
±2
where we take the plus sign if n is even, the
minus sign if n is odd. With n even, The characteristic equation X2 + X - 2 = 0 hasreal roots X = 1
and X = “2 of opposite sign, so inn, ruf) is an unstable saddle point.
With n odd, the characteristic equation X + X + 2 = 0 has complex conjugate
eigenvalues are X\, X2 = (-1 ± i V7 )/2 with negative real part, so (nn, nn) is a stable
spiral point. Alternatively, we can start by calculating the Jacobian matrix
J (x,y)
2cosx
0
1
At (n7r,n7r), n even: The Jacobian matrix
0
has characteristic equation
X2 + X - 2 = 0 and real eigenvalues Xl = -2, X -1 of opposite sign. Hence (nn,nn) is
a saddle point if n is even, as we see in the figure above.
At {rm^rur), n odd: The Jacobian matrix J =
-1
-2
1
0
has characteristic equation
X2 + X + 2 = 0 and complex conjugate eigenvalues Xl,X ~ -0.5 ± 1.3229i with negative
real. Hence (rm,rm) is a spiral sink if n is odd, as we see in the figure.
210
Chapter 6
As preparation for Problems 9-11, we first calculate the Jacobian matrix
r
j (X,y) =
o
r
-co cosx
-c
of the damped pendulum system in (34) in the text. At the critical point {nn, 0) we have
0
1 "
o
r
-co2 cos nn
-c
±co2
-c
J(«;r,0) =
where we take the plus sign if n is odd, the minus sign if n is even.
9.
If n is odd then the characteristic equation X2 + cA - co2 =0 has real roots
.
Q
—c ± yjc2 +4co2
'V'b
- ---
with opposite signs, so (nn,; 0) is an unstable saddle point.
11.
If n is even and c < 4co~ then the two eigenvalues
-c± \[c2 -4ft)2
c i f—2
2
a,,/L = - = — + — \I4cq —c
2
2
2
are complex conjugates with negative real part, so {me, 0) is a stable spiral point.
Problems 12-16 call for us to find and classify the critical points of the first order-system
x'= y, y' = ~f(x,y) that corresponds to the given equation x" + f(x,x') = 0. After finding
the critical points (x,0) where /(x,0) = 0, we first calculate the Jacobian matrix J(x,^).
13.
J (x,y) =
0
1
15x2 -20
-2
At (0,0): The Jacobian matrix J =
0
1
-20
-2
has characteristic equation
T2 + 2T + 20 = 0 and complex conjugate eigenvalues Al,A7=-\± iyj\9 consistent with
the spiral node we see at (0,0) in Fig. 6.4.6 in the textbook.
ro
r
40
~2
At (+2,0): The Jacobian matrix J =
T2 + 2/1 - 40 = 0 and real eigenvalues
has characteristic equation
= -1 ±
of opposite sign, consistent with
the saddle points we see at (+2,0) in Fig. 6.4.6 in the textbook.
Section 6.4
211
15.
o
1
2x-4
0
At (0,0): The Jacobian matrix J =
0
1
has characteristic equation
-4
0
A +4 = 0 and pure imaginary eigenvalues A], ^ = ±2i, consistent with the stable center
we see at (0,0) in Fig. 6.4.13 in the textbook.
At (4,0): The Jacobian matrix J =
0
r
4
0
'•y
has characteristic equation A -4 = 0 and
real eigenvalues Al,A2=± 2 of opposite sign, consistent with the saddle point we see at
(4,0) in Fig. 6.4.13.
17.
0
1
J(x,y) =
-5-jX2
At (0,0): The Jacobian matrix J =
0
-5
1
-2
has characteristic equation
T2 + 2A + 5 = 0 and complex conjugate eigenvalues Al,A1 = -1 ±2i with negative real
part, consistent with the spiral sink we see in the figure below.
.t
T
T
T
T
T
T
7
10 -
5 -
T 1
0
T 1
T 4
-T T
T IT
-10
19.
.T
T
0
J(x,j/) =
L
_5_Jlx2
-4 y
At (0,0): The Jacobian matrix J =
0
1
-5
0
has characteristic equation A +5 = 0 and
pure imaginary eigenvalues Al,A2 = ±/V5. This corresponds to the indeterminate case of
212
Chapter 6
Theorem 2 in Section 6.3, but is not inconsistent with the spiral sink we see in the figure
below.
-
-1T IT \ t It
T '
T
X
V
T
\A
IT
T (
T
T 1 T / T
T
T
T 1 T
\ t \
T
T
T
v
V\
T l T
T
T
.T
l
It
t It
i
A
-T
4/
l
J
ll
4,
4.
t\
T1
T
T-
1 1 1 T
T
^v \ 4 1 ^ 1 T
T
* j ± 1 T
T
^
T
IT
r
-T
f
A
1 -r
'l-r
A
\\
T\
\
f
T\
T \ T \ T\
1
Tl tit
y
T\
\
Vv
,
W. ^ / T
T
tit
_1_
-5
0
5
The statements of Problems 21-26 in the text include their answers and rather fully outline their
solutions, which therefore are omitted here.
SECTION 6.5
CHAOS IN DYNAMICAL SYSTEMS
We list here some programs that may be useful in the projects for this section. Further
discussion of these projects can be found in the applications manual that accompanies this text.
As indicated in Fig. 6.5.1 in the text, you can use the Maple commands
r := 1.5:
x = array(1..200):
x[1] := 0.5:
for n from 2 to 200 do
z := x[n-1]:
x[n] := r*z*(1-z):
od:
the Mathematica commands
Section 6.4
213
r = 1.5;
x = Table[n,{n,1,200}];
x[ [1]] = 0.5;
For[n=2, n<=20 0,
n=n+l,
z = x[[n-l]];
x[[n]] = r*z*(1-z)];
or the MATLAB commands
r = 1.5;
x = 1:200;
x(1) = 0.5;
for n = 2:200
z = x(n-l);
x(n) = r*z* (1-z) ;
end
to calculate and assemble a list of the successive iterates given by xfl+l = rxtl( 1 -xn), as
illustrated in Figures 6.5.2 through 6.5.7 in the text. The following BASIC program can be used
to investigate periodic cycles for this iteration.
100
110
120
130
140
150
160
170
180
190
200
210
220
230
240
250
260
270
280
290
300
310
320
330
340
350
360
370
380
214
'Program PERIODS
i
'The period-doubling iteration
i
i
X
=
rx (1 - X)
i
’
'
’
'
’
'
’
'
’
i
r
r
r
r
r
r
r
r
r
=
=
=
=
=
SS
=
—
—
2.75
3.25
3.50
3.55
3.565
3.57
3.84
3.845
3.848
Period
1
Period 2
Period 4
Period 8
Period 16
CHAOS
Period 3
Period
6
Period 12
DEFDBL R,X
.
INPUT "Value of r"; R
"; K
INPUT "Print in blocks of k =
n
P$
=
"#.####
'Initial seed
X =
.5
i
=
1 TO 500
FOR I
= R*X*(1 - X)
X
NEXT I
'500 initial
'iterations to
’stabilize.
=
1
TO K
I
FOR
X = R*X*(1 - X)
PRINT USING P$; X;
'Final iterations
i
Chapter 6
390
400
410
420
430
440
450
460
470
480
NEXT
IF K <> 8 THEN PRINT
f
'Press any key but Q to continue:
A$
=
INKEY$
IF A$
=
"" THEN GOTO 430
IF A$
=
"q" OR A$
=
"Q” THEN END
GOTO 360
'End of loop
'
END
The next BASIC program below can be used to plot pitchfork diagrams as in Figures 6.5.8 and
6.5.9 in the text. As written, it runs well in Borland TurboBasic (probably now obsolete), but
may have to be fine-tuned to run in other dialects of BASIC.
100
110
120
130
140
150
160
170
180
190
200
210
220
230
240
250
260
270
280
290
300
310
320
330
340
350
360
370
380
390
400
410
420
430
440
'Program PICHFORK
i
'Exhibits the period-doubling toward chaos
'generated by the Verhulst iteration
r
'
i
x
=
rx(l - x)
'as the growth parameter
'in the range from about
»
r
3
is increased
to about
4.
DEFDBL H,K,R,X
DEFINT I,J,M,N,P,Q
INPUT "Rmin,Rmax"; RMIN, RMAX
INPUT "Xmin,Xmax"; XMIN, XMAX
'Try 2.8 and 4.0
'Try
0
and
1
i
KEY OFF
: CLS
'SCREEN 1
:
N
=
319
SCREEN 2
:
N
=
639
M
=
200
H
=
(RMAX - RMIN)/N
K
=
(XMAX - XMIN)/M
i
'For med resolution
'For hi
resolution
'Hor rows for either
LINE (0,0) - (N, 0)
LINE - (N,199)
LINE - (0,199)
LINE - (
0,0)
'Draws a box
'around
'the screen
FOR P
=
Q
=
LINE
LINE
NEXT P
1 TO 9
(PMN+D/10) - 1
(Q, 0) - (Q, 5)
(Q,195) - (Q,199)
'Tick marks on
'top and bottom
'of box
FOR J
R
0 TO
N
RMIN + J*H
'Jth vertical column
'of pixels on screen
i
1
=
=
Section 6.5
215
450
460
470
480
490
500
510
520
530
540
550
560
570
580
590
600
X
=
.5
FOR P
0 TO 1000
'These iterations
X
R*X*(1-X)
'to settle down.
NEXT P
=
FOR Q
0 TO 250
'These iterations
=
X
R*X*(1-X)
'are recorded.
I
INT((X - XMIN)/K)
I
200 - I
IF (0< = I) AND (I<200) THEN PSET (J,I)
NEXT Q
NEXT J
—
—
—
—
i
WHILE INKEY$
=
""
WEND
SCREEN 0
:
CLS
:
END
'Press a key when
'finished looking
KEY ON
A more elaborate construction of these pitchfork diagrams is given by the following
Mathematica program, a slight elaboration of one found on page 102 of T. Gray and J. Glynn,
Exploring Mathematics with Mathematica, Addison-Wesley, 1991.
g[x_J : =
r x (1 - x) ;
Clear[r] ;
a
=
2.8;
b
=
4.0.
(* r-range for Fig 6.5.8 *)
c
=
0;
d
=
1;
(* x-range *)
m
=
250;
(* no of x-points *)
n
=
500;
(* no of r-values *)
ListPlot[
Flatten[Table[
Transpose[{
Table[r, {m+1}],
NestList[g, Nest[q, 0.5, 2m], m] }]
{r, a, b, (b-a)/n> ]
13 ,
PlotStyle -> PointSize[0.001],
PlotRange -> {{a,b},{c,d}},
AspectRatio -> 0.75,
Frame -> True,
AxesLabel -> {"r","x”} ]
This Mathematica program runs slowly, and requires a fast machine with plenty of memory to
finish within a reasonable waiting time. The following MATLAB program (which was actually
used to construct Figs 6.5.8 and 6.5.9) runs much faster on a comparable computer, and may be
easier to understand.
% pitchfork diagram script
% for Figures 6.5.8 and 6.5.9
216
Chapter 6
hold off
%
no
n
=
400;
%
no of x-subintervals
a
=
CM
4.0;
%
r-range
dr
=
b;
%
vector of
%
x-range
%
vector of x-values
• N.
400;
•
=
00
m
(b
b
-
=
=
a+dr/2
:
dr
c
=
0;
d
=
=
X
<d
=
=
:
for
j
1
:
:
d;
% matrices
m
r
=
a
x
=
0.5;
-
dr/2
=
X
1;
meshgrid(R,X)
zeros(m,n);
for k
:
dx
C =
=
6.5.9
r-values
c) /n
c+dx/2
[rr,xx]
for
a) /m;
R
dx
of r-subintervals
+
of
r-
and x-coords
%
of grid points
in rx-rect
%
Cycle
%
Initialize x-value
%
1000
%
1000 more
%
lattice point
through r-values
j*dr;
1:1000
=
r*x*(1-x);
=
1:1000
iterations
to
stabilize
end
for k
x
i
=
iterations
r*x*(1-x);
= ceil(x/dx);
C(i,j)
=
1;
to plot
end
end
C = C
C =
+
1;
flipud(C);
% matrix of points
for
image
image(R,X,C)
colormap([l
axis
1
1;
0
0
0])
%
color
them black or white
square
The following MATLAB function defines the forced Duffing equation for Figures 6.5.13
through 6.5.16.
function
F0
=
yp = ypduffing(t,x)
0.80;
YP = x;
y = x (2) ;
yp (l)
yp(2)
x = x (1) ;
= y;
= F0*cos(t)-y+x-x.A3;
Then the following MATLAB script can be used to construct Fig. 6.5.16.
%
fig6_5_16.m script
options
[t,y]
= odeset('RelTol',le-8,'AbsTol',le-8);
= ode45('ypduffing',
[0
100],
Section 6.5
[1; 0] ,options);
217
n = length(t);
ylOO = y(n,:)';
[t,y] = ode45('ypduffing',
hold off
plot(y(:,1),y(:,2),*b1)
axis([-1.5 1.5 -1.5 1.5])
axis square
hold on
plot([-1.5 1.5],[0 0],'k')
plot([0 0], [-1.5 1.5],'k')
pause
hold off
plot(t,y(:,1),'b')
axis ( [100 300 -1.5 1.5])
axis square
hold on
plot([100 300],[0 0],’k’)
plot([0 0], [-1.5 1.5] ,'k')
218
[100 300], ylOO,options);
% Fig.
6.5.16(a)
% Fig.
6.5.16(b)
Chapter 6
CHAPTER 7
LAPLACE TRANSFORM METHODS
SECTION 7.1
LAPLACE TRANSFORMS AND INVERSE TRANSFORMS
The objectives of this section are especially clearcut. They include familiarity with the definition
of the Laplace transform £{f[t)) = F(s) that is given in Equation (1) in the textbook, the direct
application of this definition to calculate Laplace transforms of simple functions (as in Examples
1-3), and the use of known transforms (those listed in Figure 7.1.2) to find Laplace transforms
and inverse transforms (as in Examples 4-6). Perhaps students need to be told explicitly to
memorize the transforms that are listed in the short table that appears in Figure 7.1.2.
1.
(■u = -st,
£{t} = J^° e s,t dt
du = - s dt)
r— oo
1
1 r
-i-°o
t
ueu du
3.
£{e3,+l) = £ £fV/+I dt = e £ e'(,"3)/ dt =
5.
£{sinh^} = \£{e' -e~1} = y J£° e~sl (el -e~l^dt =
_1_
1 _1
2 s-\
5+1
7.
£{m} =
9.
£{f(t)} = / e~"tdt = 1
e
J0
11.
£ | ^Jt + 3/| —
se
5
r(3/2)
_i
53/2
Section 7.1
219
13.
£{t-2e3'} =
s
s- 3
15.
£{l+cosh5/} = — +
S
s s -25
17.
f1
^
^
£{cos22/j = — £{l+cos4/} = — -1-r2\s s2+\6J
19.
£\(l +
21.
Integration by parts with u = t and dv = e
2! 3!
13
6
6
£{l + 3/ + 3/2 + /3} = - + 3~ + 3~ + 4 - T + “r + 3 +
s
s
s s
s
s
cos 21 dt yields
£[t cos2t] = J^° te~st cos2t dt = —^~v/ (-5 cos 2/+ 2 sin 2/) dt
= —^—[-5£{cos2/} + 2£{sin2/|]
5+4
1
52
52
-
+ 4 52+4
+
5-4
4
52+4
(*2+4)2
[1. 2 1l - £-.J 1
u s5n J ’■
15
25.
27.
£ -l
= 3£
5-4
29.
31.
220
_! [5-35
= 3e
5-4
_! f 105-3
=
10-£
-
1
=
-
i.i
2
~
3aAt
2
i = —sin
3/-3 cos 3/
3
*1 r——
52+9
-1
s2 -25
t3/2 = 1-
8/ 3/2
4/
S.£-'UL. -3£
3
52 +9
52+9
25-52
2
r(5/2)
r(5/2)’ 55/2
i ^— £
5
Chapter 7
5 -25
3
= -10cosh5/ +—sinh5/
5
33.
£{sin kt] = £
(eik' - e~ik'}
l
s + ik j
(because i2 = -1)
s2 + k2
Using the given tabulated integral with a = -s and b = k, we find that
co
/»oo
£{coskt} =
-St
e v/cos ktdt =
s2+k2
= lim
/-> 00
37.
^
2 ik
2i (s - ik)(s - ik)
35.
i
i
J ~~ 2 i Ks -ik
2/
1
1 f
V s2+k2
■(-s cos kt + k sin kt)
J/=0
1
Qw
)
sz+k
(-scosAt +Arsing)—,—— (—^ * 1 + A: • 0) =
2
,
j 2 *
sz+A:
/(O = 1 - ua{t) = 1 - u(t - a) so
£{/(/)} = £{l }-£{«„(/)} =
1
o~as
5
S'
= i-'(l-0.
For the graph of f note that f{a) = 1 - w(a) = 1-1 = 0.
39.
Use of the geometric series gives
oo
oo
—
ns
i
i „-2.v .
•c{/(0} =
= X— = ~(1+e 1 + £
//=o s
s
n-0
+ 6
-3.y
.
+ •**
\
I
= -(i+(0+(02+(03 + - ) = --T-U. = -7-U5
41.
^ 1-e
j(l-c-)
By checking values at sample points, you can verify that g(t) = 2/(7) -1 in terms of
the square wave function f(t) of Problem 40. Hence
2
£{g(0} = £{2/(0-1} =
1 1-e
s l+e
1
-s
e
s/2
1 1 - e~s
\ 1 + e-'
J
s 1+e
1 W'-e-"')
s esl2 + e 'y/2
s' ±(e'«+e-"’)
sinh(s / 2) _ 1
s cosh(.s72)
_p
sl2_e-s,2
1 e
e'/2
2
1 _ If
s
Section 7.1
-s
s
2
221
SECTION 7.2
TRANSFORMATION OF INITIAL VALUE PROBLEMS
The focus of this section is on the use of transforms of derivatives (Theorem 1) to solve initial
value problems (as in Examples 1 and 2). Transforms of integrals (Theorem 2) appear less
frequently in practice, and the extension of Theorem 1 at the end of Section 7.2 may be
considered entirely optional (except perhaps for electrical engineering students).
In Problems 1-10 we give first the transformed differential equation, then the transform X(5*) of
the solution, and finally the inverse transform x(t) of X(s).
1.
[s2X(s) - 5s] + 4{X(s)} = 0
X(s)
5s
s2 + 4
= 5-
^
s2 +4
,-l
x(t) = £ {JT(5*)} = 5 cos 21
3.
[42f(4 - 2] - [sX(s)] - 2[X(s)] = 0
2(
X(s) =
s2-s-2
(.$-2)0 + 1)
1
1
3 \s-2
^
S+\;
x(t) = (2/3)(e2‘ - e~‘)
5.
[,s2X(s)] + [X(s)] = 2!{s2 + 4)
X{s) =
I
(s1 +l)(>y +4)
3 s2 +1
2
3 V+4
x(t) = (2 sin t - sin 2t)l3
7.
[^2X(5) - s] + [Z($)] = s/s2 + 9)
(s2 + 1)X(.$) = 5 + s/(s2 + 9) = (.y3 + 10s)/O2 + 9)
1
s2 + 10s
X(s) =
0+1)0 +9)
9 5+1
8 s +9
x(t) = (9 cos t - cos 3t)/ 8
9.
s2X(s) + 4sX(s) + 3X(s) = 1 Is
X{s) =
1
s(s2+4s + 3)
1
s(s + 1)(s + 3)
x(t) = (2 - 3el + e2t)!6
222
Chapter 7
111
11
1
--1-
3 5*
2 s +1
6 5*4-3
11.
The transformed equations are
sX(s) - 1 = 2X(s) + Y(s)
sY( s) + 2 = 6X(s) + 37(s).
We solve for the Laplace transforms
X(s)
s-5
_ 1
s(s-5)
s
Y(s) = X(s) =
-2s+ 10
s(s - 5)
Hence the solution is given by
x(t) = 1,
13.
y{t) = -2.
The transformed equations are
sX(s) + 2[s7(s) - 1] + X(s) = 0
sX(s) - [s7(s) - 1] + Y(s) = 0,
which we solve for the transforms
X(s) = 3s
-1
X(s) = ^±1
3
s2-\/3
s + 1/3
3i2 -1
Hence the solution is
x(t) = -[21 >/3 j sinh^/7V3 j
y(t) = cosh(V/V3) + (l/sinh(/7V3^.
15.
The transformed equations are
[s2X-s] + [sX- 1] + [sY- 1]+2X- 7=0
[s2Y-s] + [sX- 1] + [sY - 1] + AX - 27 = 0,
which we solve for
Section 7.2
223
X(s) =
1(2
s + 3s + 2
s+3
—+
— +
s + 3s + 3s
f
2
3ys
s + 3s + 3 j
s + 3/2
— +
3Vs
Y(s) =
r2
\
(s + 3/2)2 +(>/3/2)2
r 28
-s — 2s + 2s + 4
9
21
s—1
(s + 3/2)2+(V3/2) J
2s +15 ^
+
s -1
s + 3s.+ 3 J
(s + 3/2) + 3/4 J
+ 2-
21 V. 5
A
2s+ 15
+
28
A
(.? + 3/2)z+(3/4)J
73/2
9
21 ^ s
s + 3s + 3s
1 ( 28
+ 7?
3 Ks
s+3
s-l
s + 3/2
(s + 3/2)2 +(73/2)'
73/2
(.j + 3/2)2 +(73/2)' y
+ 873
Here we've used some fairly heavy-duty partial fractions (Section 7.3). The transforms
S-tf
Jl\eat coskt} =
,
nt
.
,
'\
k
£{eat sin At J =
(s-a)2 + k2
2
.
r_2
(s-a) + A:
from the inside-front-cover table (with a = -3 / 2, k = V3 / 2) finally yield
11
x(0 = +2
: +e
—3// 2
COS
(73f/2) + 73sin(73r/2'
/ ,
-3//2
j(0 = T-j28_9e'+c
2 cos ^ V3// 2 j + 8 V3 sin ^ ^// 2
n /
17
19.
21.
/(0 = ( e3r* =
f |sin2zdz =
A) 2
fit) = £
224
f{t) = £
- —cos2r
= -^j-^1 — cos 2^
4
- r=0
£ sin t dt dz —
23.
1 3r
—e
= ^3'-0
3
_ r=0
(l-cosr)Jr = [r-sinr]
—I
j^sinhrd/1 dz =
= ^ — si
sin t
r=0
(coshr - \)dz = [sinhr-r]^o = sinhr-r
Chapter 7
25.
With 7(0 ~ cos kt and F(s) — s/(s2 + k2), Theorem 1 in this section yields
£{-ksinkt} = £{f'(t)} = sF(s) - 1 = s- — s
-1 = - k
s +k
s +k
so division by -k yields £ {sin kt} = kl (s2 +k2).
27.
(a)
With fit) = feal and /'(t) = ntn~'ea' + at"eat, Theorem 1 yields
£{nf~x eat + atnea‘} = s£{tnea'}
SO
«£{/”-' ea'} = (s - a)£{tnem}
and hence
£\t"e‘“\ = -^—£\t"-'e“'\.
1
}
s-a 1
J
(b)
« = !:£{ te“') -
<N
II
R
[f-e- 1 -
1
1
M i " s -a s-a
1
^
2
£
1
(s-a)2
1
s-a (s--a)2
n = 3:
£|\t3 e°'j! -
J £\
s-a *
3
2!
s-a (s -a)3
_
2!
(s-a)3
3!
(s-a)
And so forth.
29.
Let f(f) = t sinh kt, so /(0) = 0. Then
f'(t) = sinh kt + A:/1 cosh A:/
/"(0 = 2k cosh kt + tft sinh kt,
and thus /'(0) = 0, so Formula (5) in this section yields
£{2k cosh kt + J^t sinh kt} - s2£{sinh kt},
2k- , - T + A^FO) = i2F(i).
/-A2
We readily solve this last equation for
£{tzos\\kt} = F(T) =
Section 7.2
225
31.
Using the known transform of sin kt and the Problem 28 transform of t cos kt, we obtain
k
£< —^(sin&f-ktcoskt) i = —
[2k3 K
;j
2k
2k3 s2+k2
r
33.
v
2k5 ^+k2)
\
1
2t
s2-£2
2k‘
s2+£2
2k‘
J
f(t) - ua(t)-uh(t) = u(t - a) - u(t - b\ so the result of Problem 32 gives
■£{/(0} = £{u(t-a)}-£{u(t-b)} =
35.
s2-k2
-as
e
e
— bs
e
-as
-e
-hs
Let's write g(t) for the on-off function of this problem to distinguish it from the square
wave function of Problem 34. Then comparison of Figures 7.2.9 and 7.2.10 makes it
clear that g(t) = y(l + /(7)), so (using the result of Problem 34) we obtain
2s V
37.
1+
,-v \
1
1+e
esl2_e-sl2
— h-tanh— = — 1 + s/2 . -s/2
2s 2s
2
2s V
e
+e
G(s) = i + L(S)
2s 2
1 f
1 (
11
1
—,v
e-s! 2\
e
-s/2
J
1
2s 1 + e
—s
(1+e~T
We observe that /(0) = 0 and that the sawtooth function has jump -1 at each of the
points t„ = n = 1,2,3, • • •. Also, f\t) = 1 wherever the derivative is defined. Hence
Eq. (21) in this section gives
oo
=
00
sF(s) + Yue~m
=
+
=
n=0
n=]
v }
l-e~ns
using the geometric series ^x" = 1 /(I - x) with x = e~s. Solution for F(s) gives
/»=0
-s
F(s) = \ + ~ —
s
s s
226
s
Chapter 7
s
(1-i
SECTION 7.3
TRANSLATION AND PARTIAL FRACTIONS
This section is devoted to the computational nuts and bolts of the staple technique for the
inversion of Laplace transforms — partial fraction decompositions. If time does not permit
going further in this chapter, Sections 7.1-7.3 provide a self-contained introduction to Laplace
transforms that suffices for the most common elementary applications.
1.
£{t4} = f,
so
£{fV'} =
24
O-O'
3.
3n
£{sin3tU) = —:-so £{e~2tsm37tt} =
(s + 2) +97t‘
s + 9n
5.
F(s) =
3
2s -4
7.
9.
11.
13.
3
1
so fit) =
2 s-29
so fit) = t e 2t
s \
'■y
s-3
7
4
F(s) — 3-^-1-y-,
(s-3) +16 2 (s-3) +16
1111
F(s) =-,
4 s-2 4 s+2
so
F(s) = 3—-5-—so
s ~h 2
s+5
fit)
=
so fit) = e3t[3 cos 41 + (7/2)sin 41]
-(e2'-e 2‘) = 7sinh2t
4V
/(O = 3e‘2' -5e“5'
(
15.
F(s) =
'2
1
1
1 ^
,
-1- — 5 •-+ +
25 V
s
s
s-5 7
so
m
= ^(-l-5t + e5')
/(O = ^ [^' (2 + 5f) + e2' (-2 + 5/)]
Section 7.3
227
17.
K
ns) = -
l
l
8U -4
1 (
5Z+4J
2
3
2
161,5-4
5 +4 y
/(Z1) = — (sinh2/L-sin2zL)
16
19.
F(5) =
(-2s-\
52-25
25 + 4^
—--+ —-
(V +l)(^2 +4)
3 V 5 +1
5
+4
f (t) = —(-2cosZ-sinzL + 2cos2zL + 2sin2/t)
21.
First we need to find A, B, C, D so that
52 + 3
vl5 + B
Cs + D
^+2^ + 2 + (52 +25 + 2)2 ’
(.s2+25 + 2)2
When we multiply both sides by the quadratic factor
coefficients, we get the linear equations
+25 + 2 and collect
5
-2B-D + 3 = 0
-2A-2B-C = 0
-2,4-5 + 1 = 0
-A = 0
which we solve for A = 0, 5 = 1, C - -2, D-1. Thus
5+ 1
2
F(5)
-
(5 + 1)
*
?
+1
(5 + 1)
-I2
+1
+ 3■(5 + l)2+lT
We now use the inverse Laplace transforms given in Eq. (16) and (17) of Section 7.3
supplying the factor c_/ corresponding to the translation
m = e
-l
sinZ-2 • — tsint + 3 - — (sin/-1 cos t)
2
2v
2
5 -a
23.
54+4a
9
2 ^5Z - 2^5 + 2#2
and 5 ± 2as + 2a
228
9
9
+
5+a
A
52 +2a5 + 2azj
9*
= (5 ± a) + a , so it follows that
Chapter 7
5 -» 5 +1
— and get
e ' (Ssinr-^sinz1-3zLcoszt)
ir1
.S’4
— (ec" + e "')cos at = cosh at cos at.
+ 4a
1 (
25.
\
4a\s -2as + 2a2
s*+4al
1 (
s-a
4a
-2as + 2a
s2+2as + 2a‘
a
s+a
s -2as + 2a
s +2as + 2a
+
+
a
\
s + 2as + 2a' J
and s2 ± 2as + 2a2 = (s ± a)2 + a2, so it follows that
1
iT1
r
4a
sA+4a
1
ecjr (cos a/1 + sin at) - e al (cos at - sin at) J
~(ea‘ + £ 0')sina/ + ^-(ea' -e
2a
)cos£tf
= y- (cosh at sin a/1 H-sinhatfcos a/1).
In Problems 27—40 we give first the transformed equation, then the Laplace transform X(s) of
the solution, and finally the desired solution x(t).
27.
[s2X(s) - 2s - 3] + 6[sX(s) - 2] + 25X(s) = 0
2^ + 15
*(*) =
.
=
2
s + 6s + 25
9
4
+(s + 3) +16 4 (j + 3) +16
s+3
-3/1
x(/) = e"J'[2
cos 4f + (9/4)sin 4t]
29.
s2Z(s)-4X(i) =
3
3(
*(*) =
(»’-«)
x(t) = ^-sinh2/
1 \
1
4 v ^-4
s2 J
= -^■(sinh2/L - 2t)
Section 7.3
229
31.
\s3X(s) - s - 1] + [.y2X(i') - 1] - 6[5Z(j)] = 0
X(s) =
s+2
5 +s2 -6s
1 f
5
1
15 V
s
s+3
+
6
^
s-2 J
x(t) = ^(-5-e_3/+6e2/)
33.
[s4X(s)~ l]+X(s) = 0
1
X(s) =
54 +1
It therefore follows from Problem 26 with a =a/1 / 4 = 1 / V2 that
if
. t . t
. . t
t A
cosh -t= sin -T=- - sinh -t= cos —j=
V2
V2
V2
V2 y
V
x(t) =
35.
[j4Ar(,s)-l] +8^(5-)+ 16^(5) - 0
X(s) =
1
s + 85 +16
(by Eq. (17) in Section 7.3)
x(/L) = ^(sin2/L-2/Lcos2/t)
37.
[s2X(s) - 2] + 45X(5) + 13X(5) =
(J + 1)J
2 + l/(5 + l):
252 + 45 + 13
52 + 45 + 13
(5 + I)2 (V + 45 + 13)
=
1
1
50
5+1
1
50
s+1
+
+
(5 + \)
+
(y + l)2
+
5 + 98
(5 + 2) +9
5
+2
(y + 2)2 +9
+ 32-
x(t) = -|+(-l + 5/)e ' +e 2,(cos3t + 32sin3f)J
230
Chapter 7
0 + 2)2+9
39.
x" + 9x = 6cos3l
s2X(s) + 9X(s) =
x(0) = x'(0) = 0
6s
s +9
6s
X(s) =
(r+9)'
1
x(t) = 6-/sin3f = rsin3r
2•3
(by Eq. (16) in Section 7.3)
The graph of this resonance is shown in the figure below.
SECTION 7.4
DERIVATIVES, INTEGRALS, AND
PRODUCTS OF TRANSFORMS
This section completes the presentation of the standard "operational properties" of Laplace
transforms, the most important one here being the convolution property £{f*g} = £{/}-£{g},
where the convolution j*g is defined by
f*gu) =
J[ f(x)g(t-x)dx.
Here we use x rather than r as the variable of integration; compare with Eq. (3) in Section 7.4
of the textbook.
Section 7.3
231
1.
With f{t)-t and g(t) = 1 we calculate
-!-*=/
II
"1 21
—x
_2
1 2
= -r,
2
JC=0
To compute (sin t) * (sin t) = J£ sin x sin(/ - x) dx, we first apply the identity
sin ,4 sin B = [cos(^4 - B) - cos (A + B)\/2. This gives
(sin t) * (sin t) =
sin x sin(t - x) dx
1 r' [cos(2x -t)~ cos t] dx
2
-|JC=/
i sin(2x -t)-x cos t
2
- jc=0
(sint)* (sin/1) = -^-(sinr-rcosr).
5.
eal*eal
= f e^e-^dx = [V'<& = e‘"[x£' = teal
7.
/(/) = 1 * e3' = e3' * 1 = JVm dx = -(e3'-l)
1
1 r<
/(r) = — sin3/*sin3r = —
= ^
sin3xsin3(/t - x)<ix
sin 3x [sin 31 cos 3x - cos 31 sin 3x] dx
1
rt
1
rt
9
Jo
9
Jo
2
= —sin 3m sin 3x cos 3x dx—cos 3m sin 3xdx
X—t
1 . 7
(
A=/
1
—cos3r
= — sm31 —sin2 3x
9
2l
9
|_6
jc=0
y
"l
J
f(t) =
.
11
232
^
6
; Jjr=0
(sin 3t-3t cos 3/)
fU) — cos21 * cos 21 —
=
1
cos 2x cos 2(r — x) dx
cos 2x (cos 21 cos 2x + sin 21 sin 2x) dx
Chapter 7
= (cos 21) J[ cos2 2x dx + (sin 21)
cos 2x sin 2xdx
-\x=l
1 •
,
= (008 20 V
— x H—sin4x
2v
4
-1 X=l
1 • 2„2x
+ (sin 2/) —sin
4
x=0
*=0
1
f{t) = —(sin 2/ +2/cos 2/)
13.
f(t) = e3'*cost =
(cosx)e3('A)Jx
= e■3/ ^ e~3x cos xdx
-i x=t
-3x
.3/
(-3cosx + sinx)
10
(by integral formula #50)
jc=0
fit) = -^(3e3/-3cosf 4-sinf)
15.
17.
6s
£{tsint} = - — (£{sinO) = - —f—^
dsy X
n
ds{s2+ 9 y
£{e2‘cos 31)
(s'+9)'
= (s - 2)/(s2 - 4s +13)
£{te2'cos 31) = ~(d/ds)[(s - 2)/(sz - 4s + 13)] = (s2 - 4s - 5)/(s2 - 4s + 13)'
■ 9.
21.
^ fsin t]
*{-}-
ds
=
/» 00
J
-i ‘ 00
n
-i
s
=-tan s’ = tan
f l 'N
—
2
i
, U-3
23.
m - -i^'(FW) =
25.
/(?) = -ije-'{r(s)} =
L
tan
, s2 +1
£\e3t-l! = —--—,
1
j
s-3 s
e3,-l
r
UJ
so
n
r
ds -
“100
3^i
In
sj
V
s
J
= In
-J.V
'
s
^
fs-3 y
, ^2/
— e
-e
28^2/
-2/
t
2s
V
l s +1
i—-} = -( e
s +f 2 s — 3 j
y
Section 7.4
-2/
.
+e
3f
-2cos/)
233
27.
/(O =
t
= -£■'
t
29.
-2/s3
y,
{F’(s)}
\\+\/s:
—t j——~—
= —(l - cos/)
s +1
t
s +s
~[s2X(s) - x'(0)]' - [jI(j)]' - 2[sX(s)] + X(s) = 0
s(s + 1 )X'(s) + As J^s) = 0
X{s)
A
(separable)
with A ^ 0
(s+iy
x(/) = Ct3e 1 with C & 0
31.
-|YX(s) - x'(0)]' + 4[sX(s)]' - [sX(s)] -4[X(s)]’ + 2X(s) = 0
(.s2 - As + A)X’(s)+(3s - 6)X(s) = 0
(separable)
(s - 2)X’(s)+ 3X(s) = 0
X(s)
A
with A & 0
(s- 2):
x(0 = Ct2e2t with C * 0
33.
-[/I(5) - x(0)]' - 2[sX(s)] - [X(s)]f = 0
(s2 + l)X'(s)+ 4^X(^) = 0
m
A
(separable)
with A ^ 0
(s2+1):
x(/) = C(sin Z1 - t cos /) with C ^ 0
w
35.
ir1
dx
V7Tt
(s-l)\fs
A
„ o si7TX
rd
^ ■e " -2udu =
—
o
37.
u
s2X(s) + 2sXOO + ^X?) = F(s)
X(s) = F(s)-!—5(s + 1)2
x(t) = te~’*f(t) = £Terf{t-x)dT
234
Chapter 7
Yj=Ye “2du = e'erf(>y7j
SECTION 7.5
PERIODIC AND PIECEWISE CONTINUOUS
FORCING FUNCTIONS
In Problems 1 through 10, we first derive the inverse Laplace transform f(t) of F(s) and then
show the graph of /(f).
1.
F(s) = e
so Eq. (3b) in Theorem 1 gives
m = u(t — 3) • (/ — 3)
J 0 if / < 3,
\t- 3 if t >3.
f(t)
3.
F(s) = e“'£{e'2'} so /(f) = u(t- l)-e“2(,_1) = |e_2(,-,) if ^ > j
f (t)
Section 7.5
235
5.
F(s) = e ^{sin t} so
fit) = u(t - n) • sin(7 - n) - -u(t - n) sin t —
0 if t <n,
-sinr if t>7r.
f(t)
7.
F(s) = £{sint}-e 27rx£{sint} so
fit) = sint- n{t-27i)$,m(J-2n) = [l —
— 2tt)] sin / =
fsinr if t< 2n,
0 if t > 2n.
f (t)
t
9.
F(s) - £{cosnt} + e 3v£{cos^} so
fit) = cos;rr + u(t -3) cos 7r(t -3) = [l - u(t - 3)]cos7rr =
[cos 7Tt if t < 3,
[
The graph of /(t) is shown at the top of the next page.
236
Chapter 7
0 if t>3.
f (t)
11.
f(t) = 2-u(t-3)-2 so
13.
f(t) = [\-u(t-2n)~\smt = sint- u(t-27r)sm(t-In) so
F(s) =
1
-e
1
-2 ns
<7+1
15.
f(t) = [1 —
F(s) = - - e~3' - = —(l-e"3').
s
s
s '
'
,7+1
i
7+1
.
1+e
F(s) = -2— + 6
7+1 7+1
-3 ns
s2 +1
f{t) = [u(t-2)-u(t-3)]sin7rt =
n
F(s) = (e"2’ + e“3')
5
19.
-2 ns
— 3tt)] sin / = sint + u(t-37r)] sin (t-37r) so
.-3ns
17.
l-e
2
+ 7T
2
u(t - 2)sinn(t - 2) + u(t- 3)sinn(t- 3) so
_ n(e~2s + e~3')
2 .
2
S +7T
If g(t) = ^ + 1 then /(/) = u(t-\)-t = u(t-1)• g{t-1) so
F(s) = e~sG(s) = e~sL{t +1} = e~s
21.
i
n
e~\s +1)
—+ -
U
S)
If g(f) = ? + l and h(t) = t + 2 then
m = /[i - u(t -1)] + (2 — t) [u(t -1) - u(t - 2]
= Z1 -2tu{t-1) + 2u(t -1) -2u(t-2) + tu(t — 2)
= t - 2u(t - l)g(t -1) + 2u(t -1) - 2u(t - 2) + zz(^ - 2)h(t - 2)
so
F(^)
1
/ 1
1 A 2e*
- 2e"s
—+ - +
\S
Sj
Section 7.5
2e
-2 s
+e
-2 s
( 1
2^1 _ (1-^7
+—
s)
237
23.
With f(t) = 1 and p = 1, Formula (12) in the text gives
25.
With p = 2a and flj) = 1 if 0 < t < a, f[t) = 0 if a <t<2a, Formula (12) gives
A/«» - ~rJV'
_
l ~ g"aV
_ _1
” s(l-e~as)(\ + e-as) ~ ^(l + g-®)’
27.
G(s) = £{ t/a-f(t)} = (Mas2) - F(s). Now substitution of the result of Problem 26
in place of F(s) immediately gives the desired transform.
29.
With p - 2 n/k and fit) = sin kt for 0 <t<n/k while j[t) = 0 for 7t/k<t <2nlk.
Formula (12) the integral formula
eal sin bt dt = e
a! a sin bt — b cos bt
+C
a2 +b2
give
£{/(0} = --ke~"-sinktdt
1-e
^
1
1 - e~2”slk
1
( -s sin kt -k cos kt
C l
S2+k2
1
e-*’,k{k)-{-k)
- e-2KS'k
7+1?
k{ \ + e-**'k)
^
J
k
(l_e-^«)(l + e—/*)(s2+A;2) ~ (j2+jfc2)(l-e-”'*)‘
In Problems 31-42, we first write and transform the appropriate differential equation. Then we
solve for the transform of the solution, and finally inverse transform to find the desired solution.
238
Chapter 7
31.
x" + 4x = 1 - u(t— 71)
1-e
s2X(s) + 4X(s)
X(s) =
1-e
-ns
-ns
+ 4)
if1-'-”)
f1
A
s +4 J
x(t) = (1/4)[1 - u(t- 7T)][ 1 - cos 2(t - 7T)] = (1/2)[1 - w(f- ;r)]sin2/
The graph of the position function x(t) is shown below.
x(t)
t
33.
x"+9x = [1 - u(t — 2^)] sin t
1
^
,,1 :*T ,-hi
-HV ^ 1+1 ,r+9y
8
(V + l)(*s2 +4^
x{t) = -^■[l-w(/1-2^)]^sin/t-^-sin3/L
1 ‘ ^
J
The figure below shows the graph of this position function.
x(t)
Section 7.5
239
35.
x" + 4x' + 4x = [1 - u(t - 2)] t = t - u{t - 2)g(t - 2) where g(t) = t + 2
(s + 2)2X(s) = \-e~2'
(2
1 'l
— +
\s
1
X(s) =
-e
s )
2s + 1
2V
.s'2 (5 + 2)"
(
4
V
1
1
1
S
s2
5+2
A
1
+
(s + 2y
J
--e~2s
4
V
1
1
1
5
52
s+2
\
(J + 2y
y
Jt(0 = (l/4){ — 1 + / + (1 + t)e~2t + u(t-2)[\ -t + (3t - 5)e_2(/_2)]}
x(t)
37.
/f(0+10'
5 7(5)+ 10
1{s)
4/(5)
.~A-e
=
-2 ns
100
= 10t; - = (l-e-2“)£{sinl0°r}
i(t) = sin 100^ - w(/-2;r)sin 100(7 - 2;z) = [1 - u(t - 2;r)]sin 100/
39.
i\t) + 150 i(t) + 5000 \i(t) dt = 100?[1 - u(t- 1)]
sl(s) +150/(s) + 5000
100
I(s)
... _Jl
--100e
s
100
I(s)
-e
5(5 + 50)(5 + 100)
1 f1
2
50 ^5
5 + 50
+
1
-S
\s
I
O
y
s j
100(5 + 1)
5(5 + 50)(5 + 100)
(1
\
1
98
-e
s +100 j 50
—:v
-h
^5
5 + 50
99
^
5 + 100 J
i(t) = (1/50)[1 - 2e~501 + e"100'] - (l/50)u(f- 1)[1 + 98e‘50(M) - 99e-,00(M)]
240
Chapter 7
41.
x"+4x = fit),
x(0) = x'(0) = 0
(^ + 4)IW
(by Example 6 of Section 7.5)
(^2+4)lW
e~nxx
(as in Eq. (16) of Section 7.5)
Now let
4
g(t) =
.s(£2 +4)
l-cos2/ = 2 sin21.
Then it follows that
oo
x(t) = gW + 2^(-l)"w,„r(0g(>-wr) = 2sin2? + 4]T(-l)"Mror(/)sin2f.
n=1
«=1
Hence
x(/) = <
2 sin2/ if 2nn < t < (2n +1 )tt,
-2 sin2/ if (2n - X)n < t < 2nn.
Consequently the complete solution
x(t) = 2 sin/ sin/
is periodic, so the transient solution is zero. The graph of x(t) is shown below.
X(t)
Section 7.5
241
SECTION 7.6
IMPULSES AND DELTA FUNCTIONS
Among the several ways of introducing delta functions, we consider the physical approach of the
first two pages of this section to be the most tangible one for elementary students. Whatever the
approach, however, the practical consequences are the same — as described in the discussion
associated with equations (11)—(19) in the text. That is, in order to solve a differential equation
of the form
cix"{t) + bx\t) + cx(t) = f(t)
where XO involves delta functions, we transform the equation using the operational principle
£{Sa(t)} = e~a\ then solve for X(s), and finally invert as usual to find the formal solution x(t).
Then we show the graph of the inverse transform x(t).
1.
tfX(s) + 4X(s) = 1
1
s2+4
X(s) =
x(t) = —sin2/
2
3.
?X(s) + 4sX(s) + 4X(s) = - + e
X(s)
1
5,(.s, + 2)z
-2 s
(^ + 2)^
-2s
\_
1
4
s+2
-2 s
+
(.s' + 2) j
x(t) = — [l - e'2' - 2/ e~2' ] + m(/ - 2){t - 2)e“2('“2>
The graph appears at the top of the next page.
242
Chapter 7
(^ + 2)
X
t
5.
(sf + 2s + 2 )X(s) = 2e
X(s) =
-
7TS
2e -ns
0 + i)2+i
x(t) = 2u(t - 7i) e {l *} sin(7 - Ti)
0
if
0 <t<7T,
-2e~{,~K) sint if t>7i.
X
t
7.
[sire?) - 2] + 4sX(s) + 5X(s) = e'™ + e~lns
-Ins
X^s) =
2 + e'xs + e
(s + 2)2 +1
Section 7.6
243
x(t) = 2e 2/sin t + ujj)e 2(t
sin(7 - re) + U2jp)e 2(/ 2;r)sin(7 - Ire)
- [2 - el7Tu(t - 7i) + eAnu(t - 2/r)] e_2/sin t
9.
rJTCs) + 4X(s) = F(5)
X(s) =
1
F(^)
.s'2 +4
x(7) = I J[ {s\n2u) f {t - u) du
11.
(s2 + 6s + i)X(s) = F(j)
X(s) =
x(7) =
13.
(a)
(b)
1
(s + 3)2 -1
e~3w(sinhz/)/(7- w) <iw
mx8"(t)
(p/s)[uo(t) - mXO]
ms2X^s)
(/?/£•) [1/s - e_£7s]
mXJjs)
(p/£-)[(l - e “)/^3]
mxjj)
{pl2s)[^ - ujt)(t - ef]
If t_> s then
mxjj) = (p/2s)[t2 - (Z2 - 2st+ s'2)] = (p/2s)(2st - e?).
Hence mxjj) -> pt as s-> 0.
(c)
244
mv = (wx)f = (/?/)' = p.
Chapter 7
15.
Each of the two given initial value problems transforms to
(ms2
17.
(b)
+ k)X(s) = mvQ = po.
/'+ 100/ = Si(/) - Si(t),
0) = 0
i(
e * -e 2s
I(s)
=
I(s)
5 + 100
/(/) = ui(t)e-'m-" - u2(t)e-xm~2)
OO
19.
(52+100)/(5) = 10£(-1)'V"*' /10
n-0
OO
10£(-1)V*”
I(s)
/10
n= 0
=
= S (-!)''
+100
-nns! 10
10
N
s2+1007
'
oo
i(t)
= ^(-l)"i^/10(/)sinlO(/-/7W10) = ^w(/-wr/10)sinl0/
»=0
n-0
because sin(10t - rur) = (-l)nsinl0/. Hence
i(t)
=
(n +
l)sin 10^
if n7r/\0 < t < (n + 1)7r/10.
.
21
oo
(52+605 + 1000)7(5) = 10£(-l)"e —/;^r.v/10
n=0
00
>710
10^(-l)"e‘""v'
l(s) =
i(f)
=
s
r\n
r^r = El (-1)"
+ 60s + 1000
-nns! 10
//=0
10
A
(s + 30)2 +100 J
2 (-l)"iwio(0 g(t - nn! 10)
where g(7) = e 30/sin 10/, and so
g(t - nnl
10) = exp[-30(/ - /77r/10)] sin 10(/ - mc/VS)
= e3"V30'-(-l)',sin 10/
Therefore
Section 7.6
245
i(t) = ^u(t-ri7r/\0)e3,,7re 301 sint.
n=0
If nn! 10 <t <(n + 1) 7r/10 then it follows that
(3/i+l)>r _ i
i(t) = (1 +e3;r+- + ^>'30/sin lot = ——r--<T30'sinlOt.
e3*-\
The graph of i(t) is shown below.
i(t)
246
Chapter 7
CHAPTER 8
POWER SERIES METHODS
SECTION 8.1
INTRODUCTION AND REVIEW OF POWER SERIES
The power series method consists of substituting a series y =
into a given differential
equation in order to determine what the coefficients {cn} must be in order that the power series
will satisfy the equation. It might be pointed out that, if we find a recurrence relation in the form
cn+1 = <t(n)cn, then we can determine the radius of convergence p of the series solution directly
from the recurrence relation,
lim
P
1
lim
»QO
n—»co
'n+\
In Problems 1-10 we give first that recurrence relation that can be used to find the radius of
convergence and to calculate the succeeding coefficients cl9 c2, c3, • • • in terms of the arbitrary
constant cq. Then we give the series itself
1.
c
"+1
= ——
, i
n +1
(,
2
3
XXX
n\
\
4
n-> oo
^
6
V
3c
24
it follows that c
y(x) = cc
V
=
1-
Cr
V
3x
1! 2
+
9x
+
8
16
32x2
3V
2! 2
3! 2
27x'
128
3
3
o4 .4
+
X3
x4
and p - lim
2 nn\
3x 9x'
1-+
(
X2
=
2{n +1)
(
X
'0 1 + —+ — + — + — +
1! 2!
3!
4!
V
1 + X H-1-1-1-
y(x) = c,
Cn+1
and p = \im(n +1) = oo.
it follows that c.
3“x
4! 2
Section 8.1
\
c,e-
>
)
= 00
-
«->oo
A
J
\
J
c0e
-3x/2
247
5.
When we substitute y = Y,cnx into the equation y' = x2y, we find that
oo
cl+2c2x + '£i[(n + 3)cl,+3-c„]x"+, = 0.
/v=0
Hence cj = C2 = 0 — which we see by equating constant terms and x-terms on the two
c
sides of this equation — and c3 = ——. It follows that
n+3
C3k+\ ~ c3k+2 - 0
and
c3k =
c0
k '
k\3
3-6.(3k)
Hence
f,
3
6 .
XXX
A
9
^
x3
x6
x9
= c,o 1 H-1-— H-t + * ■'
1! 3 2!3
3!3:
J
V
J
y(x) = cc 1 + — + — + — +
18 162
V
(P/3)
and p — go.
7.
cw+1 = 2cw;
it follows that cM = 2"c0 and p - lim — = -
2
2
y(x) = c0 (l + 2x + 4x + 8x +16x
= Cn 1 + (2x) + (2x)“ + (2x)3 + (2x)4 h—
1 -2x
9.
(» + 2)c„
Ci+1
n +1
n +1
it follows that cw = (n + l)c0 and p - lim
= 1
n~*°0 n + 2
y(x) = c0 (l + 2x + 3x2 + 4x3 + 5x4 H—j
6V
Separation of variables gives j/(x)
(i-xy
In Problems 11-14 the differential equations are second-order, and we find that the two initial
coefficients Co and c\ are both arbitrary. In each case we find the even-degree coefficients in
terms of cq and the odd-degree coefficients in terms of c\. The solution series in these
problems are all recognizable power series that have infinite radii of convergence.
11.
'n+1
it follows that c2k =
(2k)
(n +1 )(n + 2)
^
x2
x4
x6
y(x) = c,o 1 + — + — + — +
2!
4!
6!
V
248
\
+ c,
J
^
x3
x5
and c 2k+l
x7
\
c0 cosh x + cx sinh x
X H-1-1-h
V
Chapter 8
3!
5!
7!
(2k +1)!
J
13.
o2A:
9c
Cn+\
~
j(x) =
— ; it follows that c2k = V3^0 and c.
_
,
'2A+1
(« + !)(«+ 2)
24
(2Jfc)!
(2£ + l)!
A
A
f. 9x2
27x4 81x‘
3x3 21 x5 8 lx7
^
Cr
1+
+
+ c i x-+-+
8
80
2
40
560
7
V
f
= cf
V
t
P*)2 , (3x)4
2!
(3x)6 t
'
6!
J
4!
c/
+—
3 V
(3„-M+e£>i_Oi):+ A
3!
5!
7!
J
- cn cos3x + — sinx
15.
Assuming a power series solution of the form y = Ec„x", we substitute it into the
differential equation xy' + y = 0 and find that (n+\)cn = 0 for all n > 0. This
implies that c„ = 0 for all n > 0, which means that the only power series solution of
our differential equation is the trivial solution y(x) = 0. Therefore the equation has no
non-trivial power series solution.
17.
Assuming a power series solution of the form y = Hc„xn9 we substitute it into the
differential equation x y' + y = 0. We find that Co = c\ =0 and that cn+ \ = -ncn for
n > 1, so it follows that cn = 0 for all n > 0. Just as in Problems 15 and 16, this
means that the equation has no non-trivial power series solution.
In Problems 19-22 we first give the recurrence relation that results upon substitution of an
assumed power series solution y = Xc^x" into the given second-order differential equation.
Then we give the resulting general solution, and finally apply the initial conditions y(0) = c0
and y(0) = cx to determine the desired particular solution.
22c
*
_ {-X)k2lkcx
(-l)*22*c0
^
for
n
>
0,
so
c2k
=
-and
c2k+x
Cn+2 —
(2k)\
” (2& + 1)!
(n +1 )(n + 2)
f
o2„3
o4„5
06„7
A
A
2V
2V
2 x'
f
22x2
24x4
26x6
+
+
+ C, X-1-h
y(x) = cc 16
!
3!
5!
7!
4!
2
!
V
7
V
c0 = y(0) = 0 and cx = y'(0) = 3, so
22x3
y(x) = 3
6
24x5
2 x
5!
7!
x-1-
V
3!
\
7
+
7
_ 2 {2x)_(2xl+(2xl_(2xl +
2
3!
Section 8.1
7!
3 sin2x.
■ o
= —
2
249
21.
2 YIC„n - cnCn+1
—
n(n +1)
,
^2
i
^3
~j
2
for n > 1; with c0 = jy(0) = 0 and c, = y(0) = 1, we obtain
i
r
^4
6
3!
2
X3
j
i
^5
1
24
X4
4!
5
C6
2
X5
y(x) = x + x +-1-1-1— = x
2!
3!
4!
23.
1 =—. Evidently c„ = -—^777, so
120 5!
(n-1)!
3
4
,
XXX
1 + X H-1-1-h • • •
2!
3!
4!
\
= xe
J
Co = ci = 0 and the recursion relation
(n~ - n+ l)c„ + (n — l)c„_i = 0
for n > 2 imply that c„ = 0 for n > 0. Thus any assumed power series solution
y = Ec„x" must reduce to the trivial solution y(x) = 0.
25.
Substitution of ^J_ocwx" into the differential equation y" = y' + y leads routinely —
via shifts of summation to exhibit x" -terms throughout — to the recurrence formula
(n + 2)0 + l)c,f+2 = 0 + l)c„+]+c„,
and the given initial conditions yield c0 = 0 - F0 and c, = 1 = F]. But instead of
proceeding immediately to calculate explicit values of further coefficients, let us first
multiply the recurrence relation by n!. This trick provides the relation
0 + 2)!c„+2 = (n + \)\cn^+n\cn,
that is, the Fibonacci-defming relation Fn+2 = Fn+l + Fn where Ffl=n\ctt, so we see that
cn - Fn!n\ as desired.
27.
(b) The roots of the characteristic equation r
(-1 + i V3 )/2, and 7*3
P
(1
\
= 1 are r\ = 1,
)/2. Then the general solution is
y(x) = Aex + Beax +Cepx.
Imposing the initial conditions, we get the equations
A +
B +
C =
1
A + ccB + pC =
1
A + c?B + (?C = -1.
250
- a =
Chapter 8
(*)
The solution of this system is A = 1/3, B = (l-z'V3)/3, C = (1 + z'V3)/3.
Substitution of these coefficients in (*) and use of Euler's relation e,6> =
cos 0+ i sin # finally yields the desired result.
SECTION 8.2
SERIES SOLUTIONS NEAR ORDINARY POINTS
Instead of deriving in detail the recurrence relations and solution series for Problems 1 through
15, we indicate where some of these problems and answers originally came from. Each of the
differential equations in Problems 1-10 is of the form
(Ax2 + B)y" + Cxy’ + Dy = 0
with selected values of the constants A, B, C, D. When we substitute y = Sc^\ shift indices
where appropriate, and collect coefficients, we get
O0
y'\An(n-1 )cn + B(n +1 )(n + 2)ctl+2 + Cncn + Dcn ] x" = 0.
n=0
Thus the recurrence relation is
An2 +(C-A)n + D
cn+2
for n > 0.
B(n +1)0 + 2)
It yields a solution of the form
y
CO Teven
C\ JVodd
where yQven and jy0dd denote series with terms of even and odd degrees, respectively. The evendegree series c0 + c2x2 + c4x4 + • • • converges (by the ratio test) provided that
k+2*"+2
oo
cnx"
lim
//—>
Ax2
< 1.
B
Hence its radius of convergence is at least p = ^\B / A\, as is that of the odd-degree series
cxx + c3x3 + c5x4 + • • •. (See Problem 6 for an example in which the radius of convergence is,
surprisingly, greater than ^ B / A\.)
In Problems 1-15 we give first the recurrence relation and the radius of convergence, then the
resulting power series solution.
Section 8.1
251
1.
Cn+2
~
p = i;
Cn \
Ci
C0 ~ C2 ~ C4
C4
^3
00
X*) = c^x2" + c^x
n=0
3.
'/»+2
CQ + C,X
2//+1
1-x2
H=0
p = 0°;
(ft+ 2)
_ (-1)^0.
(-1)^0
ft! 2”
(2«)(2« — 2).4-2
C2/; “
(-iyv,
(2« + l)(2ft-l).5-3
'2«+l
2n
oo
(2ft + 1)!!
oo
^ i x
•A'
v2«+l
»A
^v n
^(x) = c0O^v
y (-1)" ——+
y (-I/ (2« + l)!!
w!2„ c,I^v
/?=0
5.
nc.
Cn+2 ~
2ft -1
'2/i+l
c4 = c6 = • • • = 0
P =3;
3(ft + 2)
2ft -3
1
3(2ft +1) 3(2ft -1)
00
■c,
3(5) 3(3)
=
(2ft+ 1)3"
1
2//+1
A
y(x) = c0+c,X
+ (2n +1)3"
7.
(«-4):
^n+2 —
c'■'n ■
P > V3
?
3(ft + l)(ft + 2)
The factor (ft - 4) yields c6 = c8 = cw =
We find first that
=-cJ 2 and c5 = cx /120, and then for ft > 3 that
f
(2n - 5)2
'2/7 + 1
v
3(2ft)(2ft + l)y V
= ("I)
1
(2n - if
"j
r
i2
3(2ft — 2)(2ft — 1);
{
3(6)(7)J
3"+2« + l)(2«-l).7-6 120
8 2
8
252
~
C< =
(ft + l)(ft + 2)
3”(2« + l)!
oo
1
3
1
X-X
H-X
2
120
(ft + 3)(ft + 4)
Cn+2
\
[(2/7-5)!!]=
77-2
‘
' —X
+ c,
y(x) = cc l--xz+
3
27
y
V
9.
0, so j+ven is a 4th-degree polynomial.
p=i
Chapter 8
5
//=3
1
[(2ft — 5) ! !]2 (— 1)" ^ 2/;+l
(2ft+ 1)! 3"
_ (2k +1)(2« + 2)
2"
(2«-l)(2«)
(2n-l)(2n)
(2/?-3)(2«-2).1-2 C°
_ (2« + 2)(2« + 3)
C2»+l ~
(2n)(2n +1)
1
~
1,
2-3
„
_
_
3
+1)(2« + l)x2" + - c, y (« +1)(2« + 3)x2n+l
«=0
n=0
c ■
1
00
Z
2{yi — 5)
^w+2
_
2(" + )(2” + 1)C°
4-5
(2«-2)(2«-l)
00
11.
1
T~T-—..C, = —(« + 1)(2« + 3)C,
(2n)(2n +1)
+(x) = co
3-4
P — 00
5(n +1)(/7 + 2)
The factor (« - 5) yields c7 = c9 = cn = • • • = 0, so y0dd is a 5th-degree polynomial.
We find first that c2 = -c], c4 = c0 /10 and c6 —cQ! 750, and then for n > 4 that
cn2u =
2{2n-7)
_
2(1)
2(2n - 5)
5(2n)(2n -1) ’ 5(2n - 2)(2n - 3)
2"_3(2« — 7)!!
5(8)(7)
c0
5"~3(2«)(2« -1).(8)(7) 750
53 -6!_2"(2w-7)!!
c, = is
23 -750'5"{2n)(2n).(8)(7)-6!
5” (2/7)!
4x'
4x 5A
x. .^(2n-7)\l2n 2„
X
x-1- + Cr 11 — X 2 H-1+ 151
-x
15
375 J
10 750
ti (2n)\ 5"
V
00
13.
C//+3
p — co
—
72 + 3
When we substitute y = Zcvc" into the given differential equation, we find first that
c2 = 0, so the recurrence relation yields c5 = c8 = cu =• • •= 0 also.
//
(-1)3
y(x) = coZ
7=0
/
15.
Cn+4
~
3//
«! 3"
3/7+1
// .
+ c.
Zr
(-1)3
(3n +1)
w=0 1
p - CO
(n + 3)(n + 4)
When we substitute y = 'Lcnxn into the given differential equation, we find first that
c2 = c3 = 0, so the recurrence relation yields c6 = c]Q = • • • = 0 and c7 =cu = • • • = 0 also.
Then
-1
CAn ~
-1
(4«)(4«-l) (4w-4)(4w-5)
-1
3//+1
-1
-1
4-3
C0 ~
4”«!-(4w-l)(4«-5).5-3
-1
(4w + l)(4w) (4w — 3)(4w — 4)
(-1)%
ci =
5-4 1
4"^!• (4w +1)(4/?-3).9-5
Section 8.2
253
n
00
(-l)"x
An
y(x) = c. 1 +Y—
J—J An
£?4"w!-3-7.(4« -1)
17.
«
00
+ Ci
(-l)"x
4m+1
*+E
t^4^!-5-9.(4/7 + 1)
The recurrence relation
c
=
"+2
(n ~ 2K
(n +1 )(n + 2)
yields C2 = Co = y(0) = 1 and C4 = ce = ••• = 0. Because c\ - y(0) = 0, it
follows also that c\ = C3 = cs = ••• = 0. Thus the desired particular solution is
y(x) = 1 + x2.
The substitution t = x - 1 yields (1 -1 )y" - 6ty'- 4y = 0, where primes now denote
differentiation with respect to t. When we substitute y = Ecnf we get the recurrence
relation
n+4
cn+2
Cn
n+2 n
-
•
for n > 0, so the solution series has radius of convergence p = 1, and therefore
converges if -1 <t< 1. The initial conditions give Co = 0 and cj = 1, so ceVen = 0
and
2/7 + 3 2ft +1
7 5
2/7 + 3
C2"+1 ~~ 2/7 + 12/7-1.5?' ”
3
Thus
1
CO
-£(2« + 3)++i
5 «=o
Z(2« + 3)(x - l)2n+1,
n=0
and the x-series converges if 0 < x < 2.
21.
The substitution r = x + 2 yields (4r + 1 )y” = 8y, where primes now denote
differentiation with respect to t. When we substitute y = lLcntn we get the recurrence
relation
_
4(/7 - 2)
0+2
/
I O'v
(/7 + 2)
for n > 0. The initial conditions give Co = 1 and c\ = 0. It follows that c0dd = 0,
C2 = 4 and C4 = C6 = • • • = 0, so the solution reduces to
y = 2 + 4Z2 = 1 + 4(x + 2)2.
In Problems 23-26 we first derive the recurrence relation, and then calculate the solution series
yx(x) with c0 = 1 and cx - 0, the solution series y2(x) with c0 = 0 and cx = 1.
254
Chapter 8
23.
Substitution of y = Zc„xn yields
oo
c0 + 2 c2 +
Zfc-.+‘.
+ (n + \)(n + 2)c,H2]x" = 0,
W=1
SO
C2
+ cn
~ ^O’
'n+2
2
2
3
25.
«>1
X3
X4
X5
6
12
120
(n + \){n + 2)
4
j'iW = 1-+-+
2
for
6
VAX) = x-+-+
24
2
Substitution of y = Z c„xn yields
2c2 + 6c3x + ^T[c„_2 +(n-l)c„_, + (« +1)(« + 2)c„+2]x” = 0,
/;=2
SO
c2 = c3 =
, N
,
X
4
0,
'/»+2
X
7
X
= _^+(»-1K-.
(« +1 ){n + 2)
8
„>2.
x4
x5
x7
VAX) - X-+-+
2
12 20 126
y’j(x) — 1-1-1-f
1
12 126 672
27.
for
Substitution of y = Z c,jcn yields
oo
c0 + 2 c2 + (2 c, + 6c3)x +
Zt
+-2
+ (n + \)cn+(n + \)(n + 2)cfl+2]x" = 0,
n=2
2 c„-2 + (« + l)c„
'n+2
(ft + l)(ft + 2)
for
ft > 2.
With c0 = y(0) = 1 and c, = /(0) = -1, we obtain
x4
y(x) = 1 —x
x5
29x6
13x7
143x8
24 + 30+^20
630
40320 + 22680 +
3 lx9
Finally, x = 0.5 gives
y(0.5) = 1-0.5-0.125 + 0.041667-0.002604 + 0.001042
+ 0.000629 - 0.000161 - 0.000014 + 0.000003 + • • •
y(0.5) » 0.415562 * 0.4156.
29.
When we substitute y = Zcnxn and cosx = ^(-l)"x2" /(2ft)! and then collect
coefficients of the terms involving 1, x, x2, • • •, x6, we obtain the equations
Section 8.2
255
c0+2c2
-
0,
c,4-6c3 = 0,
-^c2-5c4+30c6 = 0,
12c4 = 0,
-2c3 + 20c5 = 0,
^c3-9c5+42c6 = 0,
—— Ci +—cA -14c,; + 56c* - 0.
360
2 4
6
8
Given c0 and c,, we can solve easily for c2, c3, • • *, c8 in turn. With the choices
c0 = 1, Cj = 0 and c0 = 0, c, = 1 we obtain the two series solutions
M*) =
1_ -1-h
33.
2
720
13x8
|
and
40320 +'
x5
y2 (x) =
13x7
60_5040+
Substitution of y = Ec„xn in Hermite's equation leads in the usual way to the recurrence
formula
c
=
2 (a-n)ch
"+2
(n +1 )(n + 2)'
Starting with c0 = 1, this formula yields
c2 —
22a(a-2)
2a
C4 ~ +
2
~
23a(a-2)(a-4)
C6 =
~4\~
!~
~6!~
Starting with cx - 1, it yields
2(a -1)
C3 ~
c5 - +
22(a-l)(a-3)
~3!~
2}(a - l)(q - 3)(a - 5)
ci ~
5!
7!
This gives the desired even-term and odd-term series yx and y2. If a is an integer,
then obviously one series or the other has only finitely many non-zero terms. For
instance, with a - 4 we get
M*) =
2-4 2
22-4-2 4
1_ -X
+-X
2
24
= l-4x2 + —x4
3
—(16x4-48x2+12),
12v
;
and with a - 5 we get
2-4
y2(x)
3
4
22-4-2 5
X-X
H-X
6
120
3
4
= X-X +—X
3
15
5
1
120
(32x5 -160x3 +120).
The figure at the top of the next page shows the interlaced zeros of the 4th and 5th
Hermite polynomials.
256
Chapter 8
y
SECTION 8.3
REGULAR SINGULAR POINTS
1.
Upon division of the given differential equation by x we see that P(x) = 1 - x and
Q(x) = (sinx)/x. Because both are analytic at x = 0 — in particular, (sinx)/x —> 1
as x -> 0 because
1 ^(-l)"x
x
xn (2n + \)\
— it follows that x
3.
n . 2n+\
sinx
CO
n .2 n
(-i yx
,
2
4
XXX
6
1-+-+
„=i (2« +1)!
3!
5!
7!
: 0 is an ordinary point.
When we rewrite the given equation in the standard form of Equation (3) in
this section, we see that p(x) = (cos x)/x and q(x) = x. Because (cosx)/x-»oo as
x —> 0 it follows that p(x) is not analytic, so x = 0 is an irregular singular point.
5.
In the standard form of Equation (3) we have p(x) = 2/(1 +x) and q(x) = 3x /(I +x).
Both are analytic, so x = 0 is a regular singular point. The indicial equation is
r(r - 1) + 2r = r2 + r = r(r + 1) = 0,
so the exponents are r\ = 0 and r2 = -1.
7.
In the standard form of Equation (3) we have p{x) = (6 sin x)/x and q(x) = 6, so
x =0 is a regular singular point with po = qo = 6. The indicial equation is r2 + 5r + 6
= 0, so the exponents are r\ — -2 and V2 = -3.
Section 8.2
257
9.
x
x
The only singular point of the differential equation /' + --/ + --y = 0 is x = 1
1-X
1-X
Upon substituting t = x - 1, x = r + 1 we get the transformed equation
y' _ LLL y - ^ + D - j; = 0, where primes now denote differentiation with respect to t.
t
t
In the standard form of Equation (3) we have p(t) = - (1 +1) and q{t) = - /(I + t) .
Both these functions are analytic, so it follows that x = 1 is a regular singular point of
the original equation.
11.
2x
12
The only singular points of the differential equation y"--y' +--y = 0 are
1 -x'
1 - x'
x = +1 and x = -1.
x = +1; Upon substituting t = x - 1, x = t + 1 we get the transformed equation
y" + ^ -+-- y-—— y = 0, where primes now denote differentiation with respect to
t{t 4- 2)
t{t + 2)
t. In the standard form of Equation (3) we have p{t) = — and q(t) = - ^
t+2
t+2
Both these functions are analytic at t = 0, so it follows that x = +1 is a regular singular
point of the original equation.
1: Upon substituting / = x + 1, x = t- 1 we get the transformed equation
12
„ 2(7-1)
y - 0, where primes now denote differentiation with respect to
y+ ------ y
t(t - 2)
t{t - 2)
2(t -1)
121
t. In the standard form of Equation (3) we have p(t) = ——and q(t) - t-2
t-2
Both these functions are analytic at t = 0, so it follows that x = -1 is a regular singular
point of the original equation.
13.
The only singular points of the differential equation /' +
1
x-2
-y = 0 are
x+2
x = +2 and x = -2.
x = +2: Upon substituting t = x - 2, x = t + 2 we get the transformed equation
y" + —— V + -y = 0, where primes now denote differentiation with respect to t. In the
t+4
t
t
standard form of Equation (3) we have p{t) = - and q{t) = t. Both these
t+4
functions are analytic at t = 0, so it follows that x = +2 is a regular singular point of the
original equation.
258
Chapter 8
x-2: Upon substituting t = x + 2, x = t-2 we get the transformed equation
y" + - / + —~y = 0, where primes now denote differentiation with respect to t. In the
^
/
*
f2
standard form of Equation (3) we have p(t) = 1 and q(t) = -. Both these
t-4
functions are analytic at t = 0, so it follows that x = -2 is a regular singular point of the
original equation.
15.
The only singular point of the differential equation y"
x
4 f
x+2
-TT +7—TUT =
(x - 2Y '
(x - 2)
is
x — 2. Upon substituting t
x 2, x : U2 we get the transformed equation
„ t+4 , t+4
y - 0, where primes now denote differentiation with respect to t. In
y +
y t
V
the standard form of Equation (3) we have p(t) = -(7 + 4) and g(7) = t + 4. Both
these functions are analytic, so it follows that x = 2 is a regular singular point of the
original equation.
Each of the differential equations in Problems 17-20 is of the form
Axy" + By’ + Cy = 0
with indicial equation Ar +{B-A)r = 0. Substitution of y = Zc+x'7 r into the differential
equation yields the recurrence relation
"
A(n + r)2 +(B-A)(n + r)
for n > 1. In these problems the exponents r\ - 0 and r2 = (A -B)/A do not differ by an
integer, so this recurrence relation yields two linearly independent Frobenius series solutions
when we apply it separately with r = r\ and with r = 7*2.
17.
With exponent rx = 0:
7
y,(x) = x
,
cn - -
2
..2
„
3
„3
XX
^
X
1-1-b
2
24 720
v
1
With exponent r2- — \
.12
(„
X
CO
= Z0
y
(-ir (V+)
cos vx
(2 n)\
n=
ctj = -
2
y2(x) = x
4n2 - 2n
"
X
1-+
6 120
V
4m2 + 2«
X'
5040
3
+
00
=Z
7
Section 8.3
(-ir (v^)2
= smvx
(2n +1)!
259
19.
With exponent rx = 0:
^l(x) = x‘
1-X
2
X
2
18
X
07-1
2/7 2 - 3/7
A
X
cn =
3
With exponent r2 = — :
2
3
=i-x-E
^n!(2n-3)U
J
360
c/7 — 1
c/7 =
"
2«2 + 3 7?
3
y2(x) = x
l+ X
3/2
x"
X
X
+
v +I+70+T¥90
A
X
+
= X
83160
3/2
,
x"
l + 3l /?! (2/7 + 3)!!
The differential equations in Problems 21-24 are all of the form
Ax2y" + Bxy’ +(C + Dx2)y = 0
(1)
<Hr) = Ar2 + (B -A)r+C = 0.
(2)
with indical equation
Substitution of y = 'Lcnxn+' into the differential equation yields
0(r)coxr +<f>{r + l)c,x'+l +
+ n)cn + Dcll_2]x"*r = 0.
(3)
77 = 2
In each of Problems 21-24 the exponents r\ and r2 do not differ by an integer. Hence when
we substitute either r = r\ or r = 7*2 into Equation (*) above, we find that Co is arbitrary
because (j)(r) is then zero, that c\ — 0 — because its coefficient </>(r +1) is then nonzero —
and that
Dc 77-20
Dc 0
__77-2_
<f)(r + n)
A(n + r)2 +(B - A)(n + r) + C
for 77 > 2. Thus this recurrence formula yields two linearly independent Frobenius series
solutions when we apply it separately with r = r\ and with r = 7*2.
21.
With exponent rx = 1:
cx = 0,
2c
cH
77-2
77(2/7 + 3)
\
(,
X"
X
X
v
7
154
6930
With exponent n = — :
c3 = 0,
x
1 H-1-1-h
hW = x1
c
260
1/2
2c
=
x
2/7
^77!- 7-11.(4/7 + 3)
7
77-2
77(2/7-3)
2
y2(x) = x'
00
6.
2 x
x
1 + x + — +-+
10 270
V
\
)
Chapter 8
1
CO
2/7
i+y-—
t^77M-5.I(4/7 - 3)
23.
1
c
With exponent r = — : c, = 0, c = -——
2
1
"
n(6n + 7)
A
6i X
X
X
1/2
yt(x) = x
1+-+-+
-v/X
+
38
4712
1215696
V
x
2/j
-
2
= —:
3
With exponent
^(x) = x
25.
-2/3
c, = 0,
1
c
c
= -——
«(6w-7)
\
^
x2
x4
X
1 + — +— +
+
10 680 118320
V
1
c
With exponent rx = —: c/f = ——
2
"
2^
3
6 ^ X ^ X"2
X
X
1/2
X
yx(x)
v
2+Y~ 48 + 384
With exponent r2- 0:
^ 2"«!* 19 • 31.(12« + 7)
00
- x
-2/3
X
1 +Y—
In
t(2"n\-5-n.(12« — 7)
\
y
00
= v*z
= ^/xe
-x/2
n\2"
«=0
'n-l
cw =
2n-\
y2(x)
7
'
X
2
3
XXX
4
\
l-x +-+
3
15 105
V
(-l)“x”
= i+I
tf(2«-l)!!
7
The differential equations in Problems 27-29 (after multiplication by x) and the one in Problem
31 are of the same form (1) above as those in Problems 21-24. However, now the exponents r\
and ri = r\ — 1 do differ by an integer. Hence when we substitute the smaller exponent r — r2
into Equation (3), we find that c0 and c 1 are both arbitrary, and that cn is given (for n > 2)
by the recurrence relation in (4). Thus the smaller exponent r2 yields the general solution
y{x) = cQyx{x) + cxy2(x) in terms of the two linearly independent Frobenius series solutions
T'i(x) and y2(x).
27.
1; with r — —1:
Exponents rx = 0 and r2
c„ = -
9cn-2
n(n -1)
f
y(x) =
X
81x(
, 9x2 21x
1-+
8
V
+
80
f
9x2 81x4 729x6
co ,1-+-+
720
24
x V
y(x)
cos3x
x
^
1
3
C!
3x
27x+ — x-12
40
7 x V
\
(
+
7
jx 3x
27x3
+
8 lx
560
\
+
7
243x5
2187x7
120
5040
+
sin3x
X
The figure at the top of the next page shows the graphs of the independent solutions
, x cos3x
.
, v
sin3x
yx(x) =- and y2(x) =
X
x
Section 8.3
261
y
29.
Exponents r, = 0 and r2 = -1; with r = -1:
r,
X
X
x V
8
384
46080
f
x‘
1-+
y(x) =
£o
x V
y(x)
A
X
1-
+
2 -2
+
X
X
2 -24
2-720
cn
f
-——
4n{n -1)
X
X-b
X'
X
x V
24
1920
322560
^
2c ( x
+—
x V2
+
+
X'
\
+
)
+
2-6
+
2 •120
2 -5040
x 2c, . x
cos —+ —Lsm —
x
2
x
2
'0
The figure below shows the graphs of the independent solutions
, .
cosx/2
.
/ .
sinx/2
y](x) = —- and y2(x) = x
x
y
31.
The given differential equation 4x2y" - 4xyf + (3 - 4x2)y = 0 has indicial equation
4r2 - 8r + 3 = (2r - 3)(2r -1) = 0, so its exponents are rx = 3/2 and r2 =1/2.
262
Chapter 8
With r = 3/2, the recurrence relation c„ = c„_2/«(«-!) yields the general solution
j'(x) = c0x
,
1/2
A
XXX
1 +— +— +-+
V
2
24
f
+ c,x
720
1/2
V
XX
X
X + — +-+-+
6
120
5040
y(x) = c0 Vx cosh x + c, Vx sinh x.
The figure below shows the graphs of the independent solutions
yx (x) = Vx cosh x
and y2 (x) = Vx sinh x.
y
33.
Exponents r} =1/2 and r2 = -1. With each exponent we find that Co is arbitrary and
we can solve recursively for cn in terms of cn~i.
yi(x)
r(, 1 lx llx2
671x3
9577x4
^
Vx 1 +-+-+•••
20
224
24192 387072
f
y2(x)
35.
1 + 1 Ox + 5x +
xV
10x3
~9~
7x4
u
...
+
J
Substitution of y - x; ^c/;x" into the differential equation yields a result of the form
-rc0xr~] +(---)x^ + (--*)x,"+1 h— = 0,
so we see immediately that c0 + 0 implies that r = 0. Then substitution of the power
series y = ^ctlx" yields
(c0 -Cj) + (4c, -2c2)x + (9c? -3c3)xz +(16c3 -4c4)x4 h— = 0
Evidently cn = ncn_x, so if Co = 1 it follows that cn=n\ for n > 1. But the series
Section 8.3
263
]>>!*” has zero radius of convergence, and hence converges only if x = 0. We
therefore conclude that the given differential equation has no nontrivial Frobenius series
solution.
37.
Substitution of px'Vc/ into the differential equation x3y”-y' + y = 0
yields a result of the form
(r - l)2c0xr +(•• -K+l +(■••) xr+2 + --- = 0,
so it follows that r = 1.
But then substitution of y = xYc/' into the differential
equation yields
UjX~ + 4c9x + 9c2x + 16c4x + 25c5x +••• — 0,
so it follows that c{ = c2 = c3 = c4 = • • • = 0. Hence y(x) = c0x,
39.
Exponents r, = 1 and r2 = -1; with r = +l:
7
x4
— +—
8
192
X2
jf(x) = c0x
V
(
X
c0x 1
-
+
2 1! 2!
V
Cj = 0,
9216+ 737280
X
X
2 2! 3!
2°3!4!
2
+ 2)
\
X
X
=
+
7
\
X
2°4!5!
7
If co - 1/2, then
2//
y(x) = J,(x) = 2^n!(« + l)U J
Now, consider the smaller exponent r2 = -1. A Frobenius series with r = -1 is of the
oo
form y = x”1^c/Jxw with c0 ^ 0. However, substitution of this series into Bessel’s
n=0
equation of order 1 gives
-c*, + CqX + (c, + 3c^ )x + (c2 + 8c4 )x +
+ 1 5c5 )x + • • • — 0,
so it follows that co = 0, after all. Thus Bessel's equation of order 1 does not have a
Frobenius series solution with leading term cqx~] . However, there is a little more here
that meets the eye. We see further that c2 is arbitrary and that c, = 0 and
cn - ctl_2 / n(n - 2) for n > 2. It follows that our assumed Frobenius series
00
y = x~'
x" actually reduces to
n=0
264
Chapter 8
f
x2
x4
y(x) = c,x 1-+8
192
V
X
X
\
9216+ 737280
But this is the same as our series solution obtained above using the larger exponent
y — +1 (calling the arbitrary constant C2 rather than cq).
SECTION 8.4
METHOD OF FROBENIUS — THE EXCEPTIONAL CASES
Each of the differential equations in Problems 1-6 is of (or can be written in) the form
xy" + (A + Bx)y'+Cy = 0.
The origin is a regular singular point with exponents r = 0 and r = 1 -A, so if A is an
integer then we have an exceptional case of the method of Frobenius. When we substitute
y = YjCnXn+r in the differential equation we find that the coefficient of xn+r is
[(n + r)2 + (A - \){n + r)]cn + [B{n + r) + C -B]cn-\ = 0.
(*)
Case 1:
In each of Problems 1-4 we have A >2 and B = C, so the larger exponent
r i = 0 and the smaller exponent V2 = 1-A = -N differ by a positive integer. When we
substitute the smaller exponent r = -N in Equation (*) above, it simplifies to
n(n -N)cn + B(n -N)cn-\ = 0.
(1)
This equation determines C\, C2, •••, cam in terms of Co, thereby yielding the solution
jq(x) = X_A/(Co + C\X + ••• + C/V-iX77-1),
(2)
provided it is possible to choose cN = 0. But when n = N, Equation (1) reduces to
0*Cyv + 0'Cyv-l = 0,
so cn may be chosen arbitrarily. With CN = 0 we get the terminating Frobenius series solution
in (2). For n > N, Equation (1) yields the recurrence formula cn = -Bcn-\/n, which if
CN ^ 0 gives a second (non-terminating) Frobenius series solution of the form
yi(x) = cn + c^v+ix + cn+2Xa + • • •.
Section 8.3
(3)
265
Case 2:
If A < 0 then the larger exponent r\ = l-A = N and the smaller exponent
V2 — 0 again differ by a positive integer. In Problems 5 and 6 we have this case with B = -1.
When we substitute the smaller exponent r = 0 in Equation (*), it simplifies to
n(n-N)cn-(n-C- X)cn-\ = 0.
This equation determines c\, C2,
(4)
cam in terms of co. When n = N it reduces to
0-cat-(TV-C- l)Cyv-l = 0.
(5)
If either TV-C-1 = 0 or cam = 0 (the latter happens in Problem 5) then cn can be chosen
arbitrarily, and finally cam, cv+2, ••• are determined in terms of cn- Thus we get two Frobenius
series solutions
yi = C0 + C1X + ••• + camxam,
(terminating)
y2 = cnxn + cn+\xn+{ + • • •.
(not terminating)
On the other hand, if (as in Problem 6) neither N -C - \ = 0 nor cam = 0, then c/v cannot
be chosen so as to satisfy Equation (5), and hence there is no Frobenius series solution
corresponding to the smaller exponent r2 = 0. We therefore find the single Frobenius series
solution by substituting the larger exponent r\ = N in Equation (*) and using the resulting
recurrence relation to determine c 1, C2, C3, • • • in terms of c0.
Problems 1-4 correspond to case 1 above. We give first the indicial roots and the critical index
TV, then the recurrence relation that defines cn in terms of cn-1, for both the TV-term solution
y{(x) in (2) and the non-terminating series solution y2(x) in (3).
1.
r, = 0, r2= -2, TV = 2,
y2(x) ~
2
cn = - n—\
n
J,(x) = x-2 (l + x);
xx2
x3
“
—1-1- +••• = i+2y
3
3-4
n =0, r2 = -4, TV = 4,
3 4-5
„=i (« + 2)!
-
c„=-
2 .2
3
3' x
, .
, 3x 3 x
^2(x) = 1-+
5
5-6
x"
3c 7-1
/
n
V
3
5-6-7
11 nil ..11
+
= 1 + 24 X
266
r, = 5, r2 = 0, TV = 5,
c„
(-l)"3"x
,,=1 (« + 4)!
Problems 5 and 6 correspond to case 2 described above.
5.
l-3x + —x2 - —x3
JiW =
(» ~ 4)c„_|
for 7? 3* 5
n(n - 5)
Chapter 8
2
2
With n = 5 the recurrence relation is 0-c5-c4 = 0. Because c4 = 0 we can choose
c5 = 1 arbitrarily and proceed:
oo
2x6
3x7
4x8
6
6-7
6-7-8
i+i2oy (n + l)x"
y2(x) = x5 + -1-1-h • • •
7.
n=\
The indicial exponents are r = -2, 1.
(n + 5)!
Substitution of T = x“2 II cMx" in the
differential equation leads to the recurrence relation
n{n ~ 3)c„ + 3(« - 3)cw_, - 0
that reduces to 0 • c3 + 0 • c2 = 0 when n - 3 so — having found ci and C2 — C3 can be
chosen arbitrarily. With Co = 2 and C3 = 0 we get the terminating Frobenius series
yi(x) = x~2(2 - 6x + 9x2).
Starting afresh with C3 = 3/3!=l/2, the recurrence relation
= -3cn_xl n for n > 3
yields the second Frobenius series solution
f
y2(x) = x
-2
3 ..5
3w
00
32x4
3jx
-h
4!
5!
V 3!
= 1
7
H=1
(« + 2)
In Problems 11-15, we give first the Frobenius series solution ^(x) corresponding to the larger
indicial exponent r\ of the given differential equation. Then, writing the equation in the form
y" + P(x)yr' + Q(x)y = 0, we apply the reduction of order formula
expj- LP(x)<ix)
-^—3-}-dx
y2(x) =
)\(X?
to derive a second independent solution y2(x).
9.
r, = r2 = 0
,
X2
X4
X6
4
64
2304
y, = 1 + —+—+
P(x) = 1/x;
r
T2 = Ti
X
+
+
147456
exp(-jJP(x)Jxj = 1/x
f
X
2
X
4
1 +-+-+
x
V
4
64
X
6
X
\ -2
dx
2304 +147456 +
Section 8.4
267
ft
=
35x8
\
2
32
V
f , x2
5x
1-+
32
V
576
73728
7
23x6
677x
1 + — +-+-+-+
X
= y\
j
11.
5x6
X
y\
r
y2
3x4
X2
r,
x2
5xz
In
x-+
= y\
128
v
576
23x
6
+
3456
+
dx
A
8
+
dx
73728
7
677x
A
589824
7
r, = r2 = 2
7
3x2
2x3
5xz
2
3
24
1 — 2x + -+
Yi =
V
P(x) = 1-3/x;
7
x3e A • x"
1 -2x +
V
r
J
exp(^-J/>(x)^/xj = x3e
r
y2 = -Vl
A
f
3x
—X
A -2
2x
5xz
-+
24
2
dx
Y
22xA 16xz
dx
x~xe'x 1 - 4x + 7x2- +
y\
3
j
2
3
4
r
f,
XXX
v
46x3 67.x:4
l
+
4x
+
9x"
+
+
+
x_1
l-x
+-+
y\
2
6
24
7V
V
r
a
3
llx2
49x3 87xz
+
+
+
dx
x
1 + 3x +
y\
2
8
J
j
V
A
f
llx2 49x3 87xz
+
+
+
Y2 = Yi lnx + 3x +
18
32
7
v
13.
rx
= 3,
2
dx
7
r2 = 1
/
2x4
+1
2x
+
2x2
Yi = x
3
3
V
4x3
r
= y i x3e 2x •x 6 1 - 2x + 2x
V
j
r
= Yi
\
7
exp(^-|7>(x)(ixj = x3e
P(x) = 2-3/x;
y2
A
2
4x
+
-2x
\ -2
2x
dx
7
1
.
2 32x
32x
x 3e 2a 1 -4x + 8x2-+-
4
A
dx
V
r
= Yi
f
X
-3
4x3
2xz
1 - 2x + 2x2 - -+—
7v
V
268
V
Chapter 8
32x3 32x4
1 + 4x + 8x + - + - +
\
dx
r
r
X
y\
-3
4xJ
1 + 2x + 2x2 +
3
V
r
1
21nx -
y2 = y\
r\
=
V2
=
2
4x
A
+
dx
J
x2
^
—i-1-(- • • •
2?
V
15.
+
2x4
x
3
3
0
. . .
X2
X4
A
x
X
J 0(x) = 1-+-+
4
64 2304 147456
P(x) = 1/x;
exp(^- j/)(x)<ixj = 1/x
r
X
y2(x) = Jo(x)
-i
f
1-
x
J0(x)
-l
j
—
4
X
4
+-
X
+
X
147456
2
3x4
i-^+
2
32
V
5x6
35x8
f
x
,
+
Y
8
2304
f
= JJx)
2
64
V
r
X
576
73728
x2
5x4
23x6
677x8
2
32
576
73728
dx
J
y
dx
J
dx
1 + — +-+-+-+
V
f
rnn.X
,
x“
5x
23x 6
677x
j0M In x + — +-+-+
128 3456 589824
V
J
\
J
= J0(x)lnx +
6
1-
x2
x4
X
++
4
64 2304 147456
vx2
5x4
23x6
677x'
+- +-+
128 3456 589824
7V 4
J
x2
3x4
llx6
yJx) = JJx) lnx + -— — +
2
0
4
128 13284
17.
The given first solution
xe
^
x2 x3
x4
x5
^
x l + x + — + — + — +-+ •••
6
24 120
J
V
can be derived by starting with the single exponent r = 1, substituting y = xU,=oc»x"
into the differential equation, and calculating successive coefficient recursively as usual.
We can verify the alleged second solution by applying the method of reduction of order
as in Problems 9-14:
P(x) = — 1 — 1 / x;
exp(-|P(x)(ixj - xex
y2 = y] ^xex -(xf)
dx = yx Jx”1 e~x dx
Section 8.4
269
r
(
1
X
= y\
2
x+ X
2
l
J
(,
= y\
3
4
X
X
—+
x2
x4
A
5
+ ... dx
120
)
24
6
x3
X
x5
111 X — X H-1-h
v
4
= y, In x +
^
x2
96
x3
600
x4
j
\
x5
1 — X H-1-h
v
^
= Ti1 nx
18
2
6
24
120
X2
X3
X4
X5
—X H-1-h
4
J
96
600
J
\
3x3 llx4
25x5 137x6
+
+
x +— +
36
288 + 7200
V
2
oo
18
7
M+l
y2(x) = xe^lnx-Y^-
„-i
«!
SECTION 8.5
BESSEL'S EQUATION
Of course Bessel's equation is the most important special ordinary differential equation in
mathematics, and every student should be exposed at least to Bessel functions of the first kind.
Though Bessel functions of integral order can be treated without the gamma function, the
subsection on the gamma function is also needed for Chapter 7 on Laplace transforms. The final
subsections on Bessel function identities and the parametric Bessel equation will not be needed
until Section 10.4, and therefore may be considered optional at this point in the course.
1.
h
00
( IV"
y^
y~[) x_
^22'"-'(m-l)!(OT!)
- -V 2 \
3.
(a)
(
\
V
J
°o
2 2m(m\)2
/
i\///+l v2/«+l
y^
*_
2lm+\m)\{m +1!)
*- = - Jj(x)
r m + — - r ( 3m + 2^
3
V
3m- \ 3m - 4
3m -1 3m -4
(3m-A'\
-1
3
\
5 2 J7\
• F
—
2-5-8
3 3
OO
*^-1/3 (x)
270
2/H-1/3
(-1)"'
/L
^ m! Y(m + 2 / 3) ( 2 y
Chapter 8
(2\
=-1
3
(x/2)~1/3
m
J
(-1)w3wx2w
F(2/3) ^ml-2-3-8.(3m-1)
5.
Starting with p = 3 in Equation (26) we get
J4(x) = -J,{x)-J2(x) = - -J2(x)-J (x)
X
X
^ 2
(2A
-^(x)-J0(x)
2LX
x2 - 24 . . .
2
x
8(6-x2) T/ ,
*^o(X) +
X
X
11.
-J,(X)
3
*A(X)
r(p + m + 1) = (p + m)(p + m - 1).(/? + 2)(p + l)r(p +1), so
oo
jp(x) =
I
f x\2,n+p
(-1)"'
m=0m\r(p + m + \){2)
(x/2)p ^_
n> +1) »,=o w!
(-1)
^xV'"
+ i)G? + 2).(/> + m) v 2 j
In Problems 13-21 we use a conspicuous dot • to indicate our choice of u and dv in the
integration by parts formula Ju*dv = uv - Jvdu. We use repeatedly the facts (from Example
l)that
|xJQ(x)dx = xJj(x) + C and
13.
jx2J0(x)dx = ^x*xJ0(x)dx
^Jx{x)dx = -J0(x) + C.
= x2Jx(x)- [x*Jx(x)dx
= x2J](x) + xJ0(x)~ ^J0(x) dx + C
15.
|x4J0(x)Jx = jx3.xy0(x)Jx
= x4J1(x) - 3 [x3»J,(x)(ix
= x4J,(x) + 3xV0(x) - 9^x2J](x) - |x*J,(x)Jxj
= x4J1(x) + 3x3J0(x)-9xV1(x) + 9^-xJ0(x) + jj0(x)Jxj
= (x4 -9x2)J1(x) + (3x3 -9x)./0(x) + 9 jj0(x)<ix + C
Section 8.5
271
17.
jx2J](x)dx = ^x2*Jx(x)dx
= - x2J0(x) + 2 JxJ0(x)dx = - x2J0(x) + 2xJ](x) + C
19.
Jx4J,(x)Jx = jx4 •Jl (x) dx
= - x4J0(x) + 4 Jx2*xJ0(x) dx
= - x4J0(x) + 4(V3J](x) - 2 Jx2«Ji(x)^j
= - x4J0(x) + 4x3J,(x) - 8^-x2J0(x) + 2 JxJc
= (-X4 + 8x2 )J0(x) + (4x3 -16x)Jj (x) + C
21.
With p = 2, Eq. (23) in the text gives
jx 2J3(x)dx = -x 2J2(x) + C. Hence
jj3(x)dx = Jx2*x_2l/3(x)(ix
= x2 (-x_2J2(x)) + 2 Jx_I J2(x)dx
(by Example 3)
= -J9(x)-Jj(x) + C
X
f2
3
2
— t/,(x) — t/0(x) —J1(x) + C
yx
y x
(By Eq. (26) with p= 1)
= J0(x)—Jx (x) + C.
x
23.
This is a special case of the discussion below in Problem 24.
24.
Given an integer n > 1, let us define
gfl(x) = j cos(^6l-xsin^)(i6>.
Differentiation yields
g',(x) = j s\n(nO - xs\n6)s\n0d6.
Integration by parts with u = sin(«0- x sin 0) and dv = sin 6d0 yields
g'nW =
cos <9 cos(nO - xsmO)dO - x
cos2 6 cos(nO -xsin#)dO.
But differentiation of the first equation for g„'(x) yields
g"(x) = - j sin2 0 Qos(nO - xsmO)dO.
272
Chapter 8
It follows that
g"(x) + —gn(x)= -gn(x) + — [ cos6 cos(n0 — xsinQ)d6
X
X
-
=
~gn(x)
8„(x)
-
JC
~
n
X‘
[(n~ xcosO)-n]cos(nO- xsin9)d&
[sin(n6-xsindjl +^gn(x)
X‘
1V
n2 \
xj y
#„(*)•
Upon equating the first and last members of this continued inequality and multiplying by
x2, we see that y = g„(x) satisfies Bessel's equation of order n > 1. The initial values
of g„(x) are
g„(0) = J[ cos(n0)dO = 0
and
g'(0) = £ sin(0)sin(w0)rf0 = 0.
If n * 1 then gf(0) = /r/2, whereas gw'(0) = 0 if n > 1. In either case the values of
g«(0) and g»(0) are n times those of Jn(0) and J„'(0)9 respectively. Now we know
from the general solution of Bessel's equation that gn(x) = c Jn(x) for some constant c.
If n = 1 than the fact that
/r/2 = g„'(0) - cJiX0) = c/2
implies that c = n9 as desired. But if n > 1 the fact that
0 = g„'(0) = cJn\0) - c-0
does not suffice to determine c.
SECTION 8.6
APPLICATIONS OF BESSEL FUNCTIONS
Problems 1-12 are routine applications of the theorem in this section. In each case it is
necessary only to identify the coefficients A, B,C and the exponent q in the differential
equation
x2y" + Axy' + (B + Cxq)y - 0.
(1)
Then we can calculate the values
a=
l-A
p = —-
q
Section 8.5
273
and finally write the general solution
y(x) = xa cxJ P(kxp) + c2J_p{kxp)
specified in Theorem 1 of this section. This is a "template procedure" that we illustrate only in a
couple of problems.
1.
We have A = -1, B - 1, C = 1, q = 2 so
p
a - l-(-l)
Vd-(-l))2-4(1)
Q
2
so our general solution is y(x) = x[c\Jo(x) + C2T0W], using To(x) because p = 0 is an
integer.
3.
y{x) = x[cjJi/2(3x2) + c2J_i/2(3x2)]
5.
To match the given equation with Eq. (1) above, we first divide through by the leading
coefficient 16 to obtain the equation
f
5
x2y” + -xy' +
1
^
-b — X'
36
\
4
y
y
=
0
with A - 5/3, 5 = -5/36, C = 1/4, and g = 3. Then
1-5/3
1
3
3
a =-= —,
„
3
/? = -,
2
,
2VI74
1
J(1 -5/3)2 — 4(—5 / 36)
k =-= -, /? = -33
113
3
3/2,
.3/2,
so our general solution is y(x) = x~ j[ciTi/3(xj/z/3) + £2«7-i/3(x'3/z/3)]
7.
y(x) = * 1 [ciJoM + c2T0(x)]
9.
y(x) = x1/2[c,Ji/2(2x3/2) + c2J_i/2(2x3/2)]
11.
y(x) = xl/2[ClJl/6(x3/3) + c2X1/6(x3/3)]
13.
We want to solve the equation xy" + 2y' + xy = 0. If we rewrite it as
x2yr> + 2xyf + x2y = 0
then we have the form in Equation (1) with A = 2, B = 0, C = 1, and
q = 2. Then Equation (2) gives a = -1/2, p = 1, k = 1, and p = 1/2, so by
Equation (3) the general solution is
274
Chapter 8
T(x)
x
j^Cjt/j/2 (x) + ctl/_y2 (-^)]
1/2
-
= X
2
cosx + c~
7TX
sinx
a] cosx + a2 smx
7TX
(with ai =cjyj2/7r) using Equations (19) in Section 3.5.
The substitution
y =
u'
y
,
u
2
immediately transforms y' = x +y
2
,
(u')2 u"
= —u
u
2
to a" + xa = 0. The equivalent equation
x2z/' + x4w = 0
is of the form in (1) with A = 5 = 0, C = 1, and # = 4. Equations (2) give a =
1/2, p = 2, k = 1/2, and
= 1/4, so the general solution is
w(x) = x1/2[dJi/4(x2/2) + c2J-i/4(x2/2)].
To compute w'(x), let z = *2/2 so x = 2I/2z1/2. Then Equation (22) in Section 8.5
with p = 1/4 yields
-f(^V|/4(x2/2)) = E(2.MzlMJ|/4(z)).^
axv
/
dzy
/ dx
= 21/4zwV.3/4(z).fdx
= 2,/4-E/j3/4(x2/2)-x = x3/2J_3/4(x2 /2).
Similarly, Equation (23) in Section 8.5 with /? = -1/4 yields
2-(x1/2J1/4(x2/2)) = ^(2l/4z,/4J1/4(z))-^ = -x3/V3/4(x2/ 2).
ax v
'
azv
’ dx
Therefore
w'(x) = x3/2[ciJ_3/4(x2/2) - c2J3/4(x2/2)].
2
It follows finally that the general solution of the Riccati equation y' = x + y
;Kx)
X
^3/4(2 X )
2
is
c ^- 3/4
C^l/4('2 X2) +
1/4
where the arbitrary constant is c = C1/C2.
Section 8.6
275
17.
If we write the equation x4y "+ fy = 0 in the form
x2y"+ fx~2y = 0,
then we see that it is of the form in Equation (3) of this section with A = B = 0,
C = /, and q = -2. Then Equations (5) give a = 1/2, /?=-!, k = y, and
p = -1/2, so the theorem yields the general solution
X*) = ^1/2[^i^i/2(y/x) + C2J-i/2(y/x)] = x[A cos(y/x) + B sin(y/x)\,
using Equations (19) in Section 8.5 for Jm(x) and J-m(x). With a and b both
nonzero, the initial conditions y(a) = y{b) = 0 yield the equations
A cos(y/d) + B sin(y/a) = 0
A cos(y/b) + B sin(y/b) = 0.
These equations have a nontrivial solution for A and B only if the coefficient
determinant
A = sm(y/b) cos(y/a) - sin(y/a) cos(y/b)
= sin (ylb-yla) = sin (yL/ab)
is nonzero. Hence yL/ab must be an integral multiple nn of 7T, and then the /7th
buckling force is
P
276
EIjI
EI0 f nnabN
k Z
Chapter 8
= EL
( nn^
V L j
UJ
CHAPTER 9
FOURIER SERIES METHODS
SECTION 9.1
PERIODIC FUNCTIONS AND TRIGONOMETRIC SERIES
The basic trigonometric functions cos© and sin© have period P = In, so the sine or cosine
of cot (as in Problems 1-4) completes its first period when cot = 2n\ hence P -2n I co.
1.
Smallest period P = 2/r/3 (left-hand figure below)
3.
Smallest period P = 4/r/3 (right-hand figure above)
However, the basic tangent and cotangent functions have period re (instead of 2©, so P - n / co
in Problems 5 and 6.
5.
Smallest period P - n\
Section 9.1
277
The hyperbolic sine and cosine functions of Problems 7 and 8 are steadily increasing (for t > 0),
and hence are not periodic.
7.
Not periodic:
9.
Smallest period P = n:
11.
with M
cin — bn
1 the integral formulas of Eqs. (16) and (17) in the text give ao = 2 and
0 for n > 0. Thus the Fourier series of / is the single term series fit) = 1:
t
-2 7T
2 7T
4 7T
In Problems 12-13, 18-19, 23, and 26 the function /(t) is defined by one formula on the
interval (-7T, 0) and by another formula on (0, /r). The coefficient integrals must therefore be
split accordingly, and the appropriate formula substituted in each integral:
a0 = — f f(t )dt +— [f(t)dt,
71
278
*
71
Chapter 9
an = — f f(t)co$ntdt + — f f(t)cosntdt
71 *~n
O>0)
71 Jo
= — f° /(/) sinw/<# + — [ f(t)s\nntdt.
71 *~n
13.
7T ^
a0 = - f (0)dt+— f(l)dt = 1
ft J-n
71
1
a.
71
b„ = —
71
71
„ ,
ji
f (°)
JO
cos ntdt-i—
J-7T
) cos nt dt =
71
sin nn
= 0
Y17Z
f° (0) sinntdt + — Jof (1) sinntdt =
J-n
n
1 -coswr
0 for n even
Y171
2/n7r for n odd
sin/
12
no ~ 2
-+-
/r
sin 3/ sin 5/ sin It
+-+-+- +
3
7
1
The graph of the series:
—'—
15.
-71
7T
tdt - 0,
a.
t
5 71
3 7T
PK
an = — I
— f t cos ntdt = 0
71 *~n
b
2 sin
= — r / sin nt dt =
;r
fit) ~ 2
sin/
sin 2/
1
2
- 2ft/r cos
„2 _
ft ;r
-2In for ft even
+2/ft for n odd
sin 3/ sin 4/
+-+
4
-
See the graph at the top of the next page.
Section 9.1
279
17.
a0 = — f° (-t)dt + — [ (0^ = #
71
=
a.. =
71 J°
— f° (~t) cosntdt + — f (/) cosntdt
7T ')~n
'~TV
71
TT JO
71
2(cos nn + nn sin me -1)
0 for n even
n2 7i
-Ain 7e for n odd
b„ = — f° (-t) sinntdt + — f (t) sinntdt = 0
71 J-TV
n Jo
71
fit)
19.
a0
2
= — f°
cost
cos 31 cos51 cosIt
-+-+-+-+
1
25
49
71
4
(n
+
71 J-tv
a.
-
71 J-TV
2
7T J°
n
Jo
1 - COS n7T
0 for n even
n2n
2 / n 7T for n odd
svnnn -nn
= — f° (n + t) sinntdt + — f (0) sinntdt - SmW71 ^~K
280
)dt
— f° in +1) cosntdt + — f (0) cosntdt
71
h
71
-\— [ (0
t)dt
n ^
Chapter 9
n '71
-i-i
21
.
1 r*
2
r
2;r
a0 = —
t dt = 7T *L;r
3
4ft;rcosft/r -2(ft ;r -2)sinft;r
an = — f t2cosntdt =
3
ft 7T
71
bfl = — I
+4/ft
for « even
-4/ft
for ft even
t2 sin ntdt = 0
cos/
cos 2/
1
cos3/
cos 4/
1
-- -
1
4
9
• • •
16
The graph:
23.
dr
= — r t2 dt =
71 **°
a
71
3
= — f t2 cos ntdt =
7T JO
71
1
6#; = — I
2ft7Tcosrm + (n27T2 -2)sinrm
2(-l)"
n^TT
n‘
/ sin nt dt
71 ^
Section 9.1
281
(2 - n2 7t2) cos rur + 2nn sinnn - 2
_ J
3
(n27r2 - 4) / 7rn3 for n odd
n n
m
71
-2
sin/
+ 71
cost
1
cos 2/ cos 31 cos 4/
-+-+
16
sin 2/ 1 sin 3/
sin At 1
--_
1
2
-n I n for n even
3
(figure below)
4_ sin /
sin 3/
sin 5/
sin 7/
-+-+-4--
71
1
27
125
343
The graph:
The trigonometric identities
2 cos A cos B = cos(y4 + B) + cos(/l - B)
2 sin A cos B = sin(^4 + B) + sin(^4 - B)
2 sin AsmB = cos (A - B) - cos(yt + B)
are needed to evaluate the integrals that appear in Problems 24-26.
25.
In order to evaluate the coefficient integrals in Eqs. (16) and (17) of the text we would
need the trigonometric identity
cos2 2/ = ^-(l + cos4/)
which, however, tells us in advance that the coefficients in the Fourier series of f{t) =
cos 2t are given by ao = 1, a* = 1/2, an = 0 otherwise, and bn = 0 for all n > 1.
'j
282
Chapter 9
SECTION 9.2
GENERAL FOURIER SERIES AND CONVERGENCE
1.
ar
= ij°3(-2)<#+i{(2)<* = 0
3
/•
1 f°
HTTf
1
a,. = —
(-2) cos-dt + —
3 J _3
;
r0
-
bn = —
3
J
3
/
~x
3, o
n7Tt
7
1
=
4(1 - cos wr)
(2) sin-dt =
3J o
3
_3
3
/*3
(-2) sin-dt + —
no -
3
3
3
fO
3
1
5
3
7
f2 n
/IN
no
1
2 7T c -2k
(graph below)
3
an = — f (2)dt+—— [ (-1)* = 1
2n J~2n
2n Jo
<2k
.
1
nt _
1
a.. =
(2) cos—dt +
(-1) cos—dt
- —
271 J -In
2
2/r 0
2
b„ =
4 r
. 7lt 1 . 37Tt 1 . 57T ^ 1 . 7/Z^
sin-i—sin-f- — sin-b —sin-+
71
3.
•
(2) cos-a/1 = 0
1
nt
(-l)sin — dt
0
2
1
. nt 7
sm
—
dt
+
(2)
2
^
r2;r
2/r
sinwr
YITC
3(coswr-l)
mr
U7T
1
6
. t
1.31 1.51 \ . It
-sin— +—sin— + -sin— + — sin—i—
2
n
2325272
i
(graph below)
-
.
.
-2 7T
2 7T
t
6 7T
10 ;
-1
Section 9.2
283
5.
1
a0 =
r2n
rin
r2;r
tdt = 0
a,
'In *~2n
2;r
-2;r
nt 7
/ cos—at - 0
2
'2n
If
. nt
4(sin/7/r -nn cosnn)
4(-l)
—
/sin
—
dt
=
—----=
——-—
K =
2n J -in
2
n~n
n
m
7.
=
. t
1.2/ 1.3/ 1.4/
sin-sin-i— sin-sin-1= 4
2 2
2
3
2
4
2
£(-0 dt + j^(t)dt = 1
a., = |° (—0 COS «7T?
m =
ac =
1
4
2
;r^
+ p COS M7Z? fifr
2[ cos «;r + wrsinwr -1]
2 2
n n
1
0
1
c
1
COS 7T/H—cos37t/+—cos5;r/+ —cos//r/+
9
25
49
f t2 dt = -
J-i
3
i-i 2
.
4«;rcosM;r + 2(« n -2)sinK^
4(-l)"
a „ = j t cosrmtdt = 3 3
2 2
n n
n n
284
2
2
n n
2
((-!)"-(
eft + f t sin rmt dt = 0
j° (-0 sin
9.
(graph below)
Chapter 9
(graph below)
fl 2
bn = J ^ ^ sin rmtdt = 0
14
1
„
1
„
1
cos7rr—cos
2/r
r+
—
cos
Znt--cos
4;^+
/(O = T-i
7V
4
9
16
f. s
x
(graph below)
To calculate the Fourier coefficients in Problems 11-14 we use the trigonometric identities for
sin A cos B and sin^t sin B that are listed above in Section 9.1 (prior to Problems 24-26 there).
11.
an =
rl
4
cos—at = —
J ~i
2
71
r]
a,. =
j-
7Tt
i
cos — cos nnt at =
2
4cosft;r
4(-l)
7r(4n
7r(4n
-l)
n+1
r
h
=
cos—sin rmtdt = 0
i
2
^ ,
2
4 1
1
„
1
„
1
f{t) = - + - — COS 711-COS ATI t H-COS 5711-COS 4 7Tt +
71
71 3
15
35
63
Section 9.2
(graph below)
285
13.
a0 =
a
a
''
2
sin nt dt = —
o
ft
i sin nt cos nntdt
=
1 + cos nn
1+ (-!)"
n{vr -1)
n{n2 -1)
for n > 1
sin nt cos nt dt = 0
sin ft7T
sin nt sin nnt dt = -
7T(yI
~ 1)
= 0 for n > 1
rl . 2
1
6, = j sin ntdt = —
2
„
1
„
1
^
1
11.
2 1
—
cos2;rH-COS
4
7Tt-\
-COS
6
7Tt-\
-cos8;r/l+
j(t) = —+ —sin nt15
35
n 2
;r
63
1
ao = ~
15.
ft J
an
t2 dt
o
8 7T2
3
1 ^ 2
7
4wrcos2/7/r + 2(2/r/r2-l)sin2wr
4
— I
t cos ntdt = --- = —
;r •k
;z77
«
1 r2*- 2 •
,
(2-4n ft ) cos 2wr + Ann sin 2nft - 2
An
— I
t smntdt = ----3- =n Jo
nn
n
m
286
An
.
cos nt
. v—^ sin nt
-r+4X—2—4*Z—
3
„=i
n
w
Chapter 9
(graph below)
(b)
If we substitute t = 0 in the Fourier series of part (a) and note that
/(°) = y[/(0-) + /(0+)] = \ iT.n'f +(0)2] = In1, we get
2
4;r
1
2nl = -+ 4 V —r,
3
tin2
,
00 1
y-
SO
7Z
When we substitute t = n and fin) = nr in the series of part (a) we get
,2 =
+
3
17.
a0 =
(a)
SO
n=\
2
tsvnnTztdt =
12
,
2 ^ sinwr/
/(o =1—Z7i n=x
2
= 0
2
sin 2n7T - 2yitz cos 2mt
2
77 7T
N
(b)
n‘
cos 2wr + 2t77t sin 2t7;t -1
77 /T
&/; =
71
t dt = 2
t cos n7Ztdt =
w =
I
t! n2
(-1)"+I
2
77;T
(graph below)
n
Substitution of t = 1/2, /(f) = 1 / 2 in this series gives
i.i-if
2
i1 —
1 +-+
1
1
3 5 7
^V
\
so
)
111
1-1-1—
3 5 7
The expansions in Problems 19 and 21 are valid on the interval -ti < t < tz rather than the
interval 0 < t < In. When we calculate the Fourier series of the functions fit) and g(t) of
period 2tz defined for -tz < t < tz by fit) = t and g(t) = t2, we find that
Section 9.2
287
sin 2/ sin 3t sin At
t = 2 sin t-+-+
and
r =
cos 2/ cos 31 cos41
-4 cos t-h-h
9
16
71
J
if -n < t < n.
19.
rn
\
an =
0
71 J -n 2
K = —
7i
2
t
oo
2
Z-
21.
t ,
— dt
«=i
~
/
t
—sin ntdt
—
it
rx t
1
an = —
—cos ntdt = 0
71 J
2
sin n7i - n7t cos rm
(-1) «+i
2
n2n
n
1V/ + 1
(-1)"+1 sin nt
{-71 <t < 7T)
(graph below)
n
71
= —
t2 dt 71 J _ it
1 r 2
i
4n7r cos n7r+ 2(n27r2-l)smn7T
4(—1)
an = —
tcos ntdt = -^--- =
—
7i
b
n 7i
- — P t2 sin ntdt = 0
oo
s = *
+ *Z
n=\
7i2
-
3r
12
(-1)" cos nt
n
7ll
_2
1 f 71
12
4v
.2
{-71 <t<7t)
oo
+ *Z
M=1
00
(-1)" cos nt
ft
y
See the graph at the top of the next page.
288
n
Chapter 9
-z
«=i
(~1)"+1 cos nt
ft
25.
Now we want to sum the alternating series
,1111
33+53
73 + 93 +"
.
of reciprocals of odd cubes. Having used a Fourier series of t4 in Problem 24 to evaluate
Z(1 In), it is natural to look at a Fourier series of t3. Let f[t) be the period 2/r function
with flj) = t if -n < t < n. We calculate the Fourier coefficients of j[t), and get
1 rn ,
ar = — I t dt = 0,
;r J-7r
2/77T (n~7r~ — 6) cos U7i — d>(n~ 7T^ — 2) sin n7r
( 6
= —
t sin ntdt ------- 0rr J-n
nA7r
n
n
1
&
1 pr ,
au - — I t cos ntdt = 0
;r J~;r
«r
oo
00
r3 = 2^2Y (-f —-12T (-1)
ti
«
sin/?/
„+1
(graph below)
n
«-i
If we substitute t = n/2 and use Leibniz's series I(-l )"+1/« = 7t/4 of Problem 17 we
find that
,
1
1
1
1
_ 7T3
33
53
73
93
32
Section 9.2
289
There is no value of t whose substitution in the Fourier series of f{t) = t yields the
series E(1 In ) containing the reciprocal cubes of both the odd and even integers.
Indeed, the summation in "closed form" of the series
1+
1
1
1
is a problem that has challenged many fine mathematicians since the time of Euler. Only
in modern times (by R. Apery in 1978) has it been shown that this sum is an irrational
number. For a delightful account of this work, see the article "A Proof that Euler Missed
. . . An Informal Report" by Alfred van der Poorten in the The Mathematical
Intelligencer, Volume 1 (1979), pages 195--203.
SECTION 9.3
FOURIER SINE AND COSINE SERIES
l.
2_
ao -
rn
pTi
\ dt - 2
an — —
71
7t
7
2 sin nn
cosnt dt = - = 0
nn
Cosine series: f(t) = 1
t
— 71
2_
K =
/* n
sin^/1 dt =
n
Sine series:
7T
2(1
cos nn
2 7T
nn L
J
4( .
1
1 .
1 .
sin t + - sin 3/ + - sin 5/ + — sin It +
nV
3
5
7
See the graph at the top of the next page.
290
4 71
1 = A [i _(_!)-]
nn
/(t) =
3 7T
Chapter 9
)
1
—L __
—
3.
an
=
7T
3 7T
71
5 7T
£(1 -t)dt = 0
„
x
met 7
4-4cos/?/r - Irursinrur
(\-t)C0S — dt = -j-;-
a„ =
o
2
/r/r
2
8 (
n n
2 [i-(-i)2]
\
37lt
1
57111
1
Int
Cosine series: /(7) =
COS
+ —rCOS -h—cos — H-7-COS ■
+
/r V
2
32
2
52
2
72
2
Tit
A
J
/*2
K =
„
x . rmt .
2ft/r(l + cos wr)- 2 sin
2 r
(1-0 sin-dt = ——-r—r- = —
o
2
n n
ri7T
Sine series:
/(0
4 f sinTzt
+
sin 2nt
+
71 V
Section 9.3
sin 37it
+
sin 47it
8
+
\
J
291
5.
ar
r2
an
"
= —
n7Zt
COS — dt
3ji
3
Cosine series:
/(7)
-
2 r
-2V3Inn if « = 2, 8,14,--sin
2 nn
sin
nn V
nn \
3 7
<+2V3 Inn if n = 4,10,16, •••
0
otherwise
1 2a/3 [ 1
2n:t 1
Ant 1
8nt
1
10nt
-—cos-cos-+ —cos-cos-i—
3
n
2
34
38
3
10
3
0 for n even
2 r 2 . nnt .
sm-dt
k. =
3
3
f
nn
2 nn ^
cos-cosnn V
3
3 )
+2Inn if n = 1, 7,13, •••
-4Inn if n = 3, 9,15, •••
+2!nn if w = 5,11,17, •••
Sine series:
2
. nt 2 . 3nt 1 . 5;z7 1 . 7^-Z1 2 . 9nt
I . 11 nt
sm-sin-+ —sin-+ —sin-sin-+ — sm-h—
;r
333537393
II
3
/(0
7.
292
2
*0
= —
rn
t(n -1) dt =
0
n
2
^
Chapter 9
r .
x
7
2[wrcoswr +wr-2sinwrl
2 r
[
t(7r-t)cosntdt
=-1—
-i
=
_4[i+
(_!)»
an = “
71
IT
Cosine series: /(/) =-4
r cos2r
+
cosAt
+
cos6/
+
cos8^
8'
V
,
2 r ,
x .
,
J
2[2-2costt;r-2ft;rsin/?/rl
4 r
„-i
= — [ t(7r -t)smntdt = —-3-- = —-| 1 — (—1) I
77 71
;T Jo
Sine series:
9.
3
+
2 f
dr
,, x
4
sin t dt - —,
7T
a.
a,
=
rr
JO
J
8(
2[l + cos n7i]
2 cn
= —
sin t cos nt dt =
7i(\-n2)
71 •k
2 r
L
sin 3? sin5f sin It
\
sin
t
-I--—I--—i--—h
•
•
•
fit) =33
53
73
ft V
7T J o
—
Tin
2 [l+ (-!)"]
if n > 1
7i{ 1 - n2)
sintcostdt = (
Cosine series: f{t)
2
=-
/r
4 A cos2r _i-1cos4/ 0086/
15
7T
35
+
cos8/
63
+
^
/
See the graph at the top of the next page.
Section 9.3
293
K
2sin«;r
smtsmntdt
= 0 if n > 1
7r( 1 - n2)
'i
sin tdt = 1
b\
Sine series:
11.
(graph below)
f{t) =sint
In order to satisfy the endpoint conditions x(0) = x(n) = 0 we substitute the sine series
x(t) = ^bn sinnt and 1 = —
n=\
S-—(from Example 1 in Section 9.1) into the
IT t1 qJj
Tl
differential equation x" + 2x = 1. This gives
00
CO
sin/?/1
-]T n2bfl sin nt + 2^ bn sin nt
n=\
w=l
^
n odd
n
We therefore choose bn-M nn{l-n ) for ^ odd, bn — 0 for n even. This gives the
formal series solution
sin nt
n(2-n2)
294
4( .
sin 3t
— sin tnv
21
Chapter 9
sin 5t
sin It
7l5
329~
13.
In order to satisfy the endpoint conditions x(0) = x(l) = 0 we substitute the sine series
tt'u i
2 ^ (-1),,+1 sinrmt
.
x(t) = 2jb„smrmt and t = ~2—-- (from Example 1 in Section 9.3 with
«=1
//=i
^
L = 1) into the differential equation x" + x = t. This gives
00
E2
00
o 00 / i\m+1
.
L
*
>
2^(—1)
Sin ftTTt
;r bn sin nnt + / bn sin rmt = — > -— 2r
,
n=\
ii=\
IT
„_i
VI
2_2
We therefore choose bn - 2(-l)"+1 /nn{\-nlnL). This gives the formal series solution
2 ^ (-1)" sin rurt
X (t) = -Z,— 2 2—TT of our endpoint value problem.
;r „=1 «(« ^ -1)
15.
In order to satisfy the endpoint conditions x'(0) = x'(2) = 0 we substitute the cosine
series x(t) = — + V an cosnt and t = — - — V ~0S^lt
2
tl
2 7inodd n
(from Example 1 in Section
9.3, with L — 7f) into the differential equation x" + 2x = t. This gives
Z2n an eosnt + a0 + 22^an
71 4
cos nt =-^—-—
oo
/;=1
r—i COS Ylt
//=1
2
71 n
VI
We therefore choose a0 = 7i / 2, a/f = 0 for n > 0 even, and an -M Tin (n -2) for
n odd. This gives the formal series solution
4 v~i
cos nt
_ 7T
= t+tZ
4 n „^dd n2(n2 -2)
4
71
4(
cos3t
cos5t
coslt
63
575
2303
— COS t H-1-1-h
71 V
of our endpoint value problem.
17.
Suggestion: Substitute u = -t in the left-hand integral.
19.
The Erst termwise integration yields
t2 _
2~ "
("I)” cos nt
h
~
n2
co
and substitution of t = 0 gives C, = 2^(-l)"+1 In2 = 7i216, so
//=!
y.2
oo
(-1)" cos??/1
y = 2Z
^
n 1
—
7T'
ft
A second termwise integration gives
Section 9.3
295
jj
00
O
,1-1
(-1)" sin ft/ 7r2t
——+—+C2,
^ = 2Z
ft
and substitution of / = 0 gives C2 = 0. The final termwise integration gives
A
(-l)"C0Sft/ At2
_
—:-+-+
C,
— = -2Y
24
^
«=1
^
00
00
and substitution of / = 0 yields C3 = 2^T(-1)" /ft4.
«=i
21.
We want to calculate the coefficients in the period 4L Fourier sine series
00
sin
m = !>>
tl = \
nnt
21
which agrees with fit) if 0 < t < L. Then
r, , . ftTT/ 7
2
y(/)sin-—-dt +
2L
2L
bn = —
"
2L
2L
f(2L-t)s\n—dt.
2L
The substitution u = 2L - t yields
rL
*" = Z
x . ft/T/ 7
1 ° „ ^ • nn(2L - ft) ^
/ nsm-ft// (ft)sm-aft
21
L, ,
2Z
7
.,
. ft/r/
(-1)"
/ (/)sm-dt~
2L
L
. nnu
f (ft) sin-aft.
2L
Now it is clear that
2 [L r, .
nnt ,
b„ = 7
/(Ocos—dt
L J0
if ft is odd, whereas bn = 0 if ft is even.
23.
b„ = n
. nt 7
/sin—dt -
2
f
2 sin
WT
nn cos
nn \
nn V
nn
It
/
1
3/
1.5/
1
sin--sm
—
+
sin
—
2
sin—b
2
/(O = —
2
72
2
;r V
2 3
2 5
„,
8 r
'
296
8(-l) («-l)/2
Chapter 9
for ft odd
SECTION 9.4
APPLICATIONS OF FOURIER SERIES
.
1
00
We substitute the sine series x(t)-^bft sin nt and F(t) =
/7=1
sin yit
Z— - (from
n n odd n
12
Example 1 in Section 9.1) into the differential equation x" + 5x = F(t). This gives
oo
i o smnt,
Eoo n~bn smnt + 5smnt
=
2_j
?7
.
z-7
•
12 v-'
—
//=1
n=\
-•
n
fl
We therefore choose bn = 0 for n>0 even, and bn = 12/;th(5-n2) for n odd. This
gives the formal series solution
. N
12 ^
sin^
12 ( sin t
xsP(0 = — L -77—7 = —
^ 7,odd n(5-n )
n V
sin3^
sin 5/
sin It
A
12
100
308
)
The graph of xsp (,t):
t
3.
We substitute the sine series x(t) =
sin nt and F{t) = 2V-— - (from
77=1
77=1
n
Example 1 in Section 9.3, with L = 71) into the differential equation x" + 3x = F{t) . This
gives
“ (-1)"-1 sin«7
-Vt72&„ smnt + 3y]bn sinnt = 4V
77=1
77 = 1
1
n
o
We therefore choose b = 4(-l)"” I n{3 - n ). This gives the formal series solution
x sp (0
= 4l
77=1
(•
sm^r
n(3-n )
( sin t sin 21 sin 3/ sin 4/
-+-+52
18
V
\
)
The graph of xsp(/L) is show at the top of the next page.
Section 9.4
297
t
5.
We substitute the sine series x(t) =
sin rurt and F(t) =
n=\
^ sin^-?-^ into the
n odd
^
differential equation x" + lOx = F(t). This gives
v-1
2
2i
i A V1 7 .
8
7r~bnsmn7rt + \02_Jbnsmn7rt = — >
/j=1
«=1
sin Y17T t
-3—•
n ocj(j
Yl
.2 _2
We therefore choose bn = 8/ «V3(10 - nlnl). This gives the formal series solution
, x
sp
*sd(0
z
8 ^
=
—
sin rurt
The graph:
n* tZL n3(\0-n2n2)
In Problems 7-12 we are dealing with the equation mx" + kx = F(t) where F(t) is the external
periodic force. The natural frequency is coo = yjk/m . If the Fourier series of F(t) contains a
term of the form cos(Nrt/L) or sin(N7Tt/L) with coq = Nn/L, then pure resonance occurs.
Otherwise, it does not.
7.
The natural frequency is &>o = 3, and
( .
sin 3/^ sin5^ sin 7/
sin
t
h-+-H-+
m = 71 V
3
5
7
298
Chapter 9
A
y
Thus the Fourier series of F(t) contains a sin 31 term, so resonance does occur.
9.
The natural frequency is coq = 2, and
f
m = - sin/ +
nV
sin 3t
+
sin 5t
+
sin It
7
'n
+
J
Because the sin 21 term is missing from the Fourier series of F(t), resonance will not
occur.
11.
The natural frequency is coo = 4. From Equation (15) in Section 9.3 we see that
F(t) =
71
4 (
COS
2
t+
cos3/
+
cos 5/
+
71
^
J
Because the cos 41 term is missing, we see that resonance will not occur.
Problems 13-18 are based on Equations (14)—(16) in the text, according to which the steady
periodic solution of
mx" + ex' + kx =
sin
nnt
~L~
n=i
is given by
xsp(0 = yo„sm(
n=\
where
U7T
T’
tan-1 —CC°n
k - mco„
in the interval [0, ti],
Bn
This calculation is readily automated. The following MATLAB script was written to calculate
the coefficients {b„} for Problem 13. Only the values of m,c,k,L and the calculation of the
force function coefficients {Bn} need to be changed for Problems 14-18.
m = 1;
c = 0.1;
k = 4;
L = pi;
results = ones(0,4);
for n = 1:9
w = n*pi/L;
alpha = atan(c*w/(k-m*wA2));
Section 9.4
299
if k-m*wA2<0
alpha = pi + alpha;
end
B = 12/(pi*n);
% force function coeffs
if floor(n/2)==n/2
% are nonzero if n is odd,
B = 0;
% zero if n is even
end
b = B/sqrt((k-m*wA2)A2+(c*w)A2);
results = [results; n, b, w, alpha];
end
results
13.
Bn = 12/m for n odd, Bn = 0 for n even
xsp(0 « 1.2725 sin(t - 0.0333) + 0.2542 sin(3l - 3.0817) + 0.0364 sin(51 - 3.1178) + *
15.
Bn = 8/rc37? for n odd, Bn = 0 for n even
jcsp(0 * 0.08150 sin(;zf - 1.44692) + 0.00004 sin(3/tf - 3.10176) + • • •
17.
Bn - 60In7r for n odd, Bn = 0 for n even
xsp(f) ~ 0.5687 sin(/rf - 0.0562) + 0.4271 sin(3;tf - 0.3891)
+ 0.1396 sin(5 nt - 2.7899) + 0.0318 sin(7^ - 2.9874) + • • •
xsp(5) « 0.248 ft « 2.98 in.
SECTION 9.5
HEAT CONDUCTION AND SEPARATION OF VARIABLES
1.
From Equation (31) in the text, with L — n and k — 3, we get
oo
u(x,t) - ^bn Qxip(-3n2t)sinnx .
n=\
With 62 = 4 and bn = 0 otherwise we get the solution
w(x, t) = 4e~12/sin 2x.
300
Chapter 9
With L = 1 and k = 2 in Equation (31), we take Zq = 5, b3 = -1/5, and b„ = 0
otherwise. The result is the solution
1 t-18^
2yr2/ sin^x-T^ 167rfsin3;rx.
5
w(x,0 =
From Equation (40) in the text, with k - 2 and L = 3 we get,
.
.
2/7 7T ^
u(x,t) = ~r+2^an exp
2
U7TX
y
V
>/=i
With a0 = 0, a2 = 4, a4 = -2, and
w(x,f) = 4 exp
cos
= 0 otherwise, and a„ = 0 we get the solution
^
8tt2T
2;rx _
f
cos-2 exp
32n2t\
V
J
V
J
cos
47TX
From Equation (40) in the text, with k = 1/3 and L = 2 we get,
n27r2t^
w(X0 = y + Z«„exp
V.
«=1
12
cos
rmx
y
Because of the identity cos 2;zx = (1 + cos 4;zx)/2 , we choose a0 = 1, a8 = 1/2, and
aM = 0 otherwise. This gives the solution
y
,
,
1
1
u(xJ) = —+ —exp
2
2
V
9.
2,\
16/rT
cos4;rx.
y
Because of the zero endpoint conditions u(0,t) = u(5,t) = 0, we use the Fourier sine
series expansion
/ ^
100
1
• xinx
w(x, 0) = -> —sin^ Hodd XI
5
of u(x, 0) = 25 on the interval 0 < x < 5. When we supply the exponential factors in
Eq. (31) with k= 1/10 and 1 = 5, we get the solution
u(x,t) =
100
n
li.
f-n2n2t^
Z ~exp
n odd XI
sin
nnx
V 250 y
Because of the zero-derivative endpoint conditions ux(0,t) = ux(\0,t) = 0, we use the
Fourier cosine series expansion
w(x, 0) = 20-
1
nnx
—cos
I
To~
n odd n
Section 9.5
301
of u(x,0) = 4x on the interval 0<x<10 (from Eq. ( 15) in Section 9.3). When we
supply the exponential factors in Eq. (40) here with k = 1/5 and L = 10, we get
u(x,t)
13.
(a)
=
20
cos
-
nnx
To~
The boundary value problem is
ut = kuxx (0 < x < 40),
*4(0, 0 = *4(40, t) = 0,
u(x, 0) = 100.
By Equation (31) in the text (with L = 40) the solution is of the form
f
oo
u(x,t) = y|ft„exp
n2n2kt^
V
n-\
sin
nnx
40~
1600
We use the Fourier sine coefficients bn = 400for « odd, bn = 0 otherwise, of the
initial value function /(x) = 100 on the interval 0 < x < 100. This gives
u(x,t)
400
n
(b)
I
■n n kt
exp
1
V
// odd
1600
. nnx
sin
40
y
With k = 1.15 for copper we find that
w(20,300) « 15.1591-0.000000204 + ••• « 15.16°C.
With k = 0.005 for concrete, the first term of the series gives
u(20,t)
400
r
exp
n
V
0.00
1600 y
and we solve for t « 66,342 sec « 19 hr 15 min 42 sec. As a check that the first term
suffices for this computation, we find that the next term in the series is then
approximately 0.00000019.
15.
We need only calculate the coefficients in the usual zero-endpoint series
CO
n2n2kt '
u(xj) = ^y„exp
n=\
V
L2
For the function /(x) = A for 0 < x < L/2,
the Fourier sine coefficient
302
Chapter 9
sm
nnx
L
/(x) = 0 for L/2 < x < L we calculate
rl! 2
1/2 for /? odd,
nnx .
A sin ——ax
Z
a;r
4/4
X <
T
a/r
1 for n- 2,6,10, •••,
0 for n = 4,8,12,
SECTION 9.6
VIBRATING STRINGS AND THE
ONE-DIMENSIONAL WAVE EQUATION
In Problems 1-10 we use the general solution
oo
y(x,t) =
nnx
nnat
nnat
sin
4, cos—— +5„sin
L
L
L
(*)
of the string equation ytt = a yxx with endpoint conditions y(0, t) = y(L, t) = 0. This form of
the solution is obtained by superposition of the solutions in Equations (23) and (33) of Problems
A and B in this section. It remains only to choose the coefficients {An} and {Bn} so as to
satisfy given initial conditions
y(x,0) =
= f{x), thus, An = y (' /(x)sin^^<&; and
,
a\
ZJo
Z
H=1
NT nna n
•
n7TX
(
\
2 f1
. nnx
g(x)sin-ax.
nna Jo
Z
n
y,(x,0) = 2_—— = g(x),thus, b„
ti=\
.
1
z
z
Here a = 2 and L = n. To satisfy the condition y(x, 0) = (1/I0)sin2x we choose
A2 = 1/10 in Eq. (*) above, and An = 0 otherwise. To satisfy the condition
yt(x, 0) = 0 we choose Bn = 0 for all /r. Thus
— cos 41 sin 2x.
1)
3.
Z
10
Here a = 1/2 and L = n. Choosing Ai = 1/10 and An = 0 otherwise,
and Bn = 0 otherwise, we get
=
,
If/
A
101
2)
= 1/5
— cos—+ 2sin— sinx.
2
Section 9.5
303
5.
Here a - 5 and L = 3. Choosing ^3 = 1/4 and
Bn = 0 for n ^ 6, we get
= 0 for n ^ 3, Be = l/n and
y{x,t) = — cos 5sin 7zx + —sin 10^ sin 2zdc.
4
;r
7.
Here a = 10 and L = 1. To satisfy the condition y(x, 0) = 0 we choose An = 0 for
all n, so
00
y(x, /) =
Z?/; sin 10«;zt sin nnx.
n=1
To satisfy the condition yt(x, 0) = x we choose
_
1
2(-l)',+1 _ (-l)w+1
10/7/T
77 7T
5^72^2
for 77 > 1 (see Equation (16) in Section 9.3). This gives
n+1
— sin 1Onnt sin nnx.
9.
Here a = 2 and T = 1. To satisfy the condition
all n, so
0) = 0 we choose An = 0 for
co
y(x, t) - ^jBllsin2nntsinnnx.
n=\
To satisfy the condition yt(x, 0) = x(l -x) we choose
1
=
r1
x(l - x)si
sin^xJx =
2-2
cos nn - nn sin nn
4
nn
n n
4
Hence
,
.
4
sin 2nnt sin nnx
y(x,t) = — ^
4
-
^
11.
n odd
-■
^
Substitution of L = 2 ft, T = 321b, and the linear density
l/32oz
P =
2
ft
1 oz
lib
lslug
1
slug
64ft 16oz 321b
323
ft
in Eqs. (2) and (26) in the text yields the velocity a = yjT / p = a/324 = 1024 ft/sec with
which waves move along the string, and its fundamental frequency
v
a
= 256 Hz
2L
304
Chapter 9
(approximately middle C).
13.
If y(x>0 — F(x + Qt) — F(u) with u = x + at, then the chain rule gives
dy
dF du
dx
dy_
du dx
dF du
dt
du dt
d2y
dF' du
dx2
du dx
tfy_
= a
dt2
15.
- F'(u)-\ - F\x + at);
= F'{u)-a = aF'(x + at) = a
dF' du
dx
F"{u) • 1 = F\x + at)\
= a-F"(u)-a = a2F"(x + at) = a2
du dt
dx2
If y(x,0) = 0 then the fundamental theorem of calculus gives
y(x,t) = y(x,t)~ y(x, 0) =
[V(x,r)<ir =
^
19.
dy_
— [G(x + tfr) + G(x-ar)]<ir
Jo 2
The general solution of the second-order ordinary differential equation a2y" = g is a
second-order polynomial in x with leading coefficient g/2a . But the polynomial
0(x) = gx(x - L)/2a2 has this leading coefficient and satisfies the endpoint conditions
y(0) = y(L) = 0.
23.
If 7r/4 < x < 3 7t/4 then
n
2
n
< x H— <n
.
and
_
0
n
n
< x-< —
2
so
y(x, tr/4)
~ [F(x + n!\) + Fix - 7tl4)\
^ [1 - cos 2(x + 7t/4) + 1 - cos 2(x - 7r/4)]
— [1 - cos (2x + n!2) + 1 - cos (2x - tt/2)]
— [2 + sin 2x - sin 2x1
2
y(x, 7t!4) = 1
Section 9.6
305
SECTION 9.7
STEADY-STATE TEMPERATURE AND LAPLACE'S EQUATION
1.
Because 7(0) = Y{b) = 0 we take our separation of variables in the form
X"-AX = 0 = Y" + AY
with A > 0. Then it follows that
n2n
2
and thence that
Xfix) = An cosh^^ + 2^ sinh
b
Y17TX
b
The condition that J7(0) = 0 implies that An = 0 so Xn{x) = ^coshw^x/6, and
hence
/
x
• i nnx . wry
w(*,y) - > C„sinh—- sin——.
tf
b
b
Finally we satisfy the condition u(a9y) = g(y) by choosing Cn = bn /(sinhnna/b)9
where the {/?„} are the Fourier sine coefficients of g(y) on 0 < y < b.
3.
Just as in Example 1 of Section 9.7 we have Xn(x) = sin nnx Ia and
7„(y) = An cosh+ Bn sinhm
a
a
The condition 7(0) = 0 now yields An = 0 so Yfiy) = Bn sinhnny /a, and hence
u(x,y) - > C/7sin-sinh-.
a
a
Finally we satisfy the condition u(x, b) = fix) by choosing Cn - bn /(sinh nnb/a),
where the {bn} are the Fourier sine coefficients of fix) on 0 < x < a.
5.
Now 7'(0) = Y\b) - 0, so we work with the separation of variables
X’-AX = 0 = Y" + AY.
The eigenvalue problem
Y" + AY = 0,
306
Chapter 9
7'(0) = Y\b) = 0,
has eigenvalues and eigenfunctions Ao = 0, Y0(y) = 1 and
2
n n
K =
2
i_2
Y„(y) = cos
5
«;ry
for « - 1,2, 3, • • •. When n = 0, X£(x) = 0 yields Wo(v) = Ax + B. Then
Xo(a) = 0 is satisfied by X${x) = a -x. For n > 0 we have
-rr
x
,
.
,
WTX
_
.
-
Y171X
Xn(x) = An cosh —— + Bn sinh ——,
and Xn{a)
0 is satisfied by the particular linear combination
of cosh nnx/b and sinh n7txlb. Therefore
,
x
^ /
x v- ^
, nn(a - x)
nny
u(x,y) = C0(a -x) + > Cn sinh-cosW=1
Finally we satisy the condition w(0, y) = g(y) by choosing
r
- ao
2a
and
Cn = --sinh nna / b
where the {an} are the Fourier cosine coefficients of g(y) on 0 < y < b.
7.
The eigenvalue problem
X" + AX = 0.
X(0) = X(a) = 0
yields the eigenvalues and eigenfunctions
2
n n
K =
2
X„(x) = sin
a'
nnx
a
for n = 1, 2, 3, • • •. Then
y:+w = o
yields
Y„(y) = Ane"^ + Bne
-nny! a
In order that Y(y) —> 0 as y —> oo we take An = 0, so
u(x,y) =
Section 9.7
sin—
307
Finally we satisfy the condition fix,0) = fix) by choosing the constants {Bn} as the
Fourier sine coefficients of fix) on 0 < x < a.
9.
If in Problem 8 we have /(x) - lOx on 0 < x < 10, then
rio
_2_ r10
=
a
=
To
10x dx = 100,
flu
2 rio
10
nnx .
cos-dx
l 10*
200(cos rm -1 + rm sin n)
2
10
n n
2
so
1
u(x,y) = 50
„-nnyl 10__
e
y
cos-,
10
„odd"
Then
w(0,5) « 50-8.4250-0.0405-0.0006-0.0000-••• « 41.53,
w(5,5) = 50-0-0-0-0-- 50,
w(0,5) » 50+ 8.4250+ 0.0405+ 0.0006 +0.0000+••• « 58.47.
11.
Now the boundary value problem is
Uxx
0
Uyy
(0 < x < a, 0 < y < b)
fia, y) = uy(x, 0) = fix, b) =
0,
u($,y) = g(y)The eigenvalue problem
Y" + AY = 0.
T'(0) = Y(b) = 0
yields (similar to Example 4 in Section 3.8)
2 _2
o
(2«-l) 7C
A.n = ---,
4 b'
Yfiy) = cos
(2n-\)ny
2b
for n = 1,2,3,***. Then
XU-AX
= 0
n
n
n
yields
XSx) = An cosh
+ B. sin(2w ~
2b
2b
Now Xn(a) = 0 is satisfied by the particular linear combination
Xfix) = Cn sinh
308
(2n - \)n {a - x)
Chapter 9
2b
of cosh (2/7-1)7al2b and sinh (2n-\)7ix/2b. Hence
u(x,y) = V C„ sinh(2W ~1Ma ~ X) cos
tT
2b
2b
E.A smh—
. t nn(a-x)
rmy
--cos——.
2b
2b
Finally we satisfy the condition u(0, y) = g(y) by choosing
a.
A
sinh nn a! 2b
where the {an} are the odd half-multiple cosine coefficients of g(y) on [0, b\, as given
by Problem 22 in Section 9.3.
We start with the periodic polar-coordinate solution
_ a0 ^
u(r,6) = —+ 2^r'\an cosnO + bn sinnO)
3
n=\
and choose an = 0 in order to satisfy the conditions u(r, 0) = u{r,n) = 0. Then
00
u(r,0) -
sin n<9
n=]
satisfies the nonhomogeneous boundary condition u(a, 0)-f (0) provided that ancn is
is the nth Fourier sine coefficient of /(0) on the interval 0 < 6< n, that is,
- J^ /(<9)sin nOdO.
na
As in the textbook discussion of the polar-coordinate Dirichlet problem, the substitution
u(r,6) =
in Laplace's equation yields the separated ordinary differential
equations
r2R" + rR'-XR = 0
(25)
©" + 20 - 0.
(26)
and
With 2 = a1 the general solution of (26) is
©(#) = A cos aO + B sin aO,
and the endpoint condition 0(0) = 0'(O) = 0 yields A = 0 and 0 = (2n -1) / 2, so
the nth eigenvalue and eigenfunction are given by
a, =
{2n-\y
Section 9.7
r4 fQx
■ (2n-\)9
®„(&) = sin--
309
As in the discussion of Eqs. (29) and (30) in the text, the bounded solution of
r2R: + rKJ*LJlRn
=
0
Rn(r) = r<2"”')/2
for n = 1, 2, 3, • • ■. We thereby obtain the formal series solution
u(r,6) = X c„r"12 sin^.
It remains only to satisfy the nonhomogeneous boundary condition u(a, 0) = f (<9) by
choosing
rn
2
Cn
so that (for « odd)
17.
=
/o
_ _«/2
/(6») sm~dd,
0
2
equals the ftth odd half-multiple sine coefficient of /(0).
The substitution u(r,6) = R(r)®(6) in Laplace's equation yields the same separated
solution functions
©0(6>) = 1,
R0(r) = C0 + D0 lnr
and
0/;(0) = 4,cos
sin
K(r) = C»r" + —“
r
as in Eqs. (28)-(30) in the text. We choose Bn = 0 to satisfy the boundary condition
u(r,6) = u(r,-6), and n = 1 with Cj = £/0 to satisfy the given limit condition as
A
r-> oo. Then the condition that ur(a,6) = 0 requires that D{=U0a, so
u(r,0) = — (r2+a2)cos0.
r v
'
21.
(a)
Since we cannot simply substitute r = 0, we apply continuity of u(r, t) at
r = 0 and calculate
w(0,0 = limw(rT)
r—»0
v
7
noting that
lim
r-»0
sinnnr la
nn,.
ft;r .. sin
smwzri
wzr / a
— lima
nnr I a
by the elementary fact that (sin 0) / 6 —» 1 as # —» 0.
310
Chapter 9
nn ..
,. sin #
— lima
6
a
(b)
With a — 30 and 70=100 we have
oo
f
u (0,0 = 200^(-l)"+1exp
n=1
V
n27r2kt^
900
,
If A: = 0.15 for iron then after 15 minutes = 900 seconds the center temperature is
w(0,900) * 45.5075-0.5361+ 0.0003-0.0000+••• « 44.97.
If k= 0.005 for iron then after 15 minutes the center temperature is
w(0,900) « 190.37-164.174 + 128.276-90.8081 + 58.2426
- 33.8449 +17.8190 - 8.4998 + 3.6734 -1.4384
+ 0.5103 - 0.1640 + 0.0478 - 0.0126 + 0.0030
- 0.0007 + 0.0001 - 0.00002 + 0.00000 - • • •
w(0,900) * 100.00
Thus the center of the ball has not yet begun to cool. For the center of this concrete ball
to reach 45° (as with the iron ball after 15 minutes) would require (0.15/0.005) x 15 =
450 minutes, that is, seven and a half hours!
Section 9.7
311
CHAPTER 10
EIGENVALUES AND
BOUNDARY VALUE PROBLEMS
SECTION 10.1
STURM-LIOUVILLE PROBLEMS
AND EIGENFUNCTION EXPANSIONS
1.
In the notation of Equation (9) in Section 10.1 of the text we have a\ - J3\ = 0 and
CC2 = fh = 1, so Theorem 1 implies that the eigenvalues are all nonnegative. If X = 0,
then y" = 0 implies that y(x) = Ax + B. Then y'(x) = A, so the endpoint conditions
yield A = 0, but B remains arbitrary. Hence To = 0 is an eigenvalue with
eigenfunction
ToO) = 1.
If A = a2 > 0, then the equation y"+ c?y = 0 has general solution
y(x) = A cos ax + B sin ax,
with
y\x) = -Aa sin ax + Ba cos ax.
Then y'(0) = 0 yields B = 0 so A * 0, and then
y\L) = -Aasin aL = 0,
so aL must be an integral multiple of n. Thus the wth positive eigenvalue is
n2n2
««
=
“F"
and the associated eigenfunction is
yn(x) = cos
3.
312
nnx
~Y
If X = 0 then y” = 0 yields y(x) = Ax + B as usual. But y'(0) = A = 0, and then
hy(L)+y'(L) = h(B) + 0 = 0, so B = 0 also. Thus X = 0 is not an eigenvalue. If
X = d > 0 so our equation is y"+c?y = 0, then
Chapter 10
y(x) = A cos ax + B sin ax,
so
y'(x) = -A a sin ax + Ba cos ax.
Now y'(0) - 0 yields B = 0, so we may write
y(x) = cos ax,
y’(x) = -asinocx.
The equation
hy(L) + y'{L) = h cos aL - asm aL = 0
then gives
T
h
hL
tmaL = — = —
a
aL
so pn = anL is the nth positive root of the equation
tanx =
hL
x
Thus
y„(x) = cos
L
The following figure shows the graphs y = tanx and y = hL/x (with hL- 5 to
illustrate the situation). Each intersection point is labeled with its x-coordinate pn.
It is apparent that pn « (« - 1 )n for n large.
If X = 0 then y" = 0 yields y(*) = Ax + B as usual. Then the endpoint condition
hy(0) - y'(^) = 0 yields A = hB. Substitution for A in the other endpoint condition
hy(L) + y\L) - 0 then gives B(h2L + 2 h) = 0, so it follows that A-B- 0. Thus
X — 0 is not an eigenvalue.
If X = a2 > 0 so our equation is y"+c?y = 0, then
Section 10.1
313
y(x) = A cos ax + B sin ax,
so
y'(x) = -Aa sin ax + Ba cos ax.
Then the endpoint condition hy( 0) - y(0) = hA- Ba = 0 gives
a
Substitution of this value of B in the other endpoint condition hy{L) + y\L) = 0 yields
an equation that simplifies to
(h~-a )sinaL + 2hacosaL = 0.
Substitution of /? = aL, a~piL in this equation then gives the equation
(/?2-/fZ2) sin/? = IhLPzosP,
whence we see that Pn = auL is the nth positive root of the equation
7hT.r
314
Chapter 10
r2 — h2 J2
The figure at the bottom of the preceding page shows the graphs y = 2hL cot x and
y = (x —hL)/x (with hL = 1 to illustrate the situation). Each intersection point is
labeled with its x-coordinate pn. It is apparent that /]„ « (n - \ )n for n large.
The coefficient c„ in Eq. (23) of this section is given by Formula (25) with fix) = r(x)
Bx
B
= 1, a = 0, b = L, and yn(x) = sin-y-. Using the fact that tan /?„ = - — , so
-Lj
hL
sin/?„ =
cos/?„
we find that
J3„
hL
rL
PnX j
sm 2 —^—dx
L
•
r‘-1 (
ip„x \
1 - cos
^ J
o 2v
\_ r
dx
Pn
\
T
. 2 fix
2 P„
L
~iL
x-sin -
2
l-L^JLC0Sp
2
1_
Jo
1 f T T cos B
\
L + L-—cos Pn
hL
V
)
hL + cos2 Pn
2h
rL
and
sin ——dx = ——C°-S
o
L
pn
. Hence the desired eigenfunction expansion isoo
1 - cos /?„
1 = 2hLY- .
^ P„(hL +cos2 /?„)
s.n f^x_
L
for 0 < x < L.
The coefficient c„ in (23) is given by Formula (25) with fix) = r(x) = 1, a - 0,
b = 1, and y„(x) = sin pnx. Using the fact that tan pn = ~PJh, so
h sin pn = -pncos pn, we find that
/*i
/•l
i(l-cos2 /?„x)r& = i x —
sin2 Pnxdx -
sin2/?„x
2A
1 \
2
l
sin/?
'
1
Pn
J
2
)
h J
fi icos2 p*
l
h + cos2 /?„
2/2
and
| xsin/?„xc& = -|t 1 pnxsmp„x-p„dx
rn
A
1 rA
~2 [” usinudu
a;
sin pn - p„ COS pn
2 LSinM-MCOSMJ0
Pi
sin P„~P„ cos pn
(1 + h) sin pn
Pi
Pi
Section 10.1
315
It follows that the desijed expansion is given by
x
2/j(1 + /2)£
n=1
sin /?„ sin Pnx
Pl(h + cos2/?„)
for 0 < x < 1.
11.
If A = 0 then j/'= 0 implies that y(x) = ylx + £. Then yO) ~ 0 gives i? = 0, so
y(x) = ^4x. Hence
hy(L) - y'{L) = /z^Z) - ^ = ,4(/zZ - 1) = 0
if and only if hL = 1, in which case Aq = 0 has associated eigenfunction yo(x) = x.
13.
If A = +c? > 0, then the general solution of /' + c?y = 0 is
y(x) = A cos ax + B sin ax.
But then y0) = A = 0, so we may take y(x) = sin ax. Now the condition hy{L) =
y'(L) yields
h sin aL = a cos aL.
It follows that p = aL must be a root of the equation
x
tanx = —.
hL
So if pn is the n\X\ positive root of this equation, then An - an - pn / L and the
corresponding eigenfunction is yn (x) = sin Pnx / L.
15.
If Ad = 0, then a general solution of y” - 0 is y(x) = Ax + B. The conditions
T(0) + /(0) = B + A = 0,
y( 1) =
A+B = 0
both say that B = -A, so we may take y0 (x) = x -1 as the eigenfunction associated with
Ao=0. If A = +c? < 0, then the general solution of y" + o?y = 0 is
y(x) = A cos ax + B sin ax.
But
y 0) 4- y'(0) = A + Ba = 0, so A - - Ba, and then
yi) = Acosa + Bsina = - B(acosa-sina) = 0.
316
Chapter 10
Thus the possible values of a are the positive roots {/?„} of the equation tanx = x, and
the «th eigenfunction is yn(x) = pn cos Pnx- sin J3nx,
The Fourier sine series of the constant function f(x) = w for 0 < x < L is
4w
1 . nnx
w = — ^ — sin—•
^ n odd n
L
If y = ^ bn sin nnx / L, then
4
47
n n b„
EIy{4) =
n=\
r
sin
nnx
~L~
Upon equating coefficients in these two series and solving for bn, we see that
. ,
Awl
1
.
nnx
y<x) = 77T
L—sm~r
Eln „odd«
L
With X = cc, the general solution of >,l4) - oty = 0 is
y(x) = A cosh ax + B sinh ax + C cos ax + D sin ax,
and then
y'(x) = ar(,4 sinh ax + i? cosh ax - C sin ax + D cos ax).
The conditions y(0) = 0 and j/(0) = 0 yield C = -A and D = -B, so now
y(x) = yf(cosh ax - cos ax) + i?(sinh ax - sin ax).
The conditions y(L) = 0 and j/(Z) = 0 yield the two linear equations
A (cosh aL - cos aL) + Z(sinh aL - sin aL) = 0,
A (sinh aL + sin aL) + Z(cosh aL - cos aL) = 0.
This linear system can have a non-trivial solution for A and B only if its coefficient
determinant vanishes,
2
2
*2
(cosh aL - cos aL) - (sinh aL - sin aL) = 0.
Using the facts that cosh2 A - sinh2 A = 1 and cos2 A + sin2 A-1, this equation
simplifies to
cosh aL cos aL - 1 = 0,
so p = aL = x satisfies the equation
Section 10.1
317
cosh x cos x = 1.
The eigenvalue corresponding to the /7th positive root pn is
K
=
<
f R V
A
=
lLJ
Finally the first equation in the pair above yields
B = -
cosh aL- cos aL
sinh aL - sin aL
so we may take
(
A
y„(x) = (sinh /?„ - sin /?„) cosh —-eos^
I
L
V
\
- (cosh Pn - cos pP) j sinh
- sin
1
L
L
J
as the eigenfunction associated with the eigenvalue An.
21.
As in Problem 19, the solution of y(4) - ay - 0 satisfying the left-endpoint conditions
y(0) = 0 and /(0) = 0 is given by
y(x) = yt(cosh ax - cos ax) + 5(sinh ax - sin ax).
The right-endpoint conditions y(L) = 0 and y"{L) = 0 yield the two linear equations
A (cosh aL - cos aL) + 5(sinh aL - sin aL) = 0,
^4(cosh aL + cos aL) + Z?(sinh aL + sin aL) = 0.
This linear system can have a non-trivial solution for A and B only if its coefficient
determinant vanishes,
(cosh aL - cos oT)(sinh aL + sin aL)
- (cosh aL + cos aT)(sinh aL - sin aL) = 0.
This equation simplifies to 2 cosh aL sin aL- 2 cos aL sinh aL - 0, which is equivalent
to tanh aL = tan aL. Hence p - aL = x satisfies the equation tanhx = tanx,
and the eigenvalue corresponding to the nth. positive root pn is An = aAn = (pn / L)4.
318
Chapter 10
SECTION 10.2
APPLICATIONS OF EIGENFUNCTION SERIES
1.
The substitution u{x,t) = X{x)T(t) yields the separated equations
X"+c^X=0
and
T' = -kXT
with separation constant A = a2. In Problem 3 of Section 10.1 we saw that the SturmLiouville problem
X" + c?X = 0,
X'(0) = hX(L) + X\L) = 0
has eigenvalues Xn = a] = /T /1: and eigenfunctions
Xn(x) = cos PnX
L
for n = 1, 2, 3, • • •, with {A} being the positive roots of the equation tanx = hLIx.
The solution of Tl - -kA„T.
is then
n n
(
Pi to
\
L
T„(t) = exp
n21^ \
so the resulting formal series solution is
oo
(
R2 \
Pn
ft)_PnX
cos
V
L
u (xj) = Xc«exP
n=\
L
The coefficients in the eigenfunction expansion are given by
rL
c„ =
B x
/(x) cos ^—dx
i
L
rL
2 A* 7
cos2 t-A—dx
L
1
2h
f (x) cos
hL + sin Pn J o
Bx
— dx.
L
because
rL
rL
2 A*
T
cos2 —— dx
L
\
2A*
dx =
1 + cos
L )
o 2 V
f
2V
\(
L . 2fix')
x +-sin
L J
2A
\
Jo
u
L .
L h-sin 2 A
2A
x
f
hL + sin A
2h
hL cos P„ '
n
J
hL + sin2 A
2h
Section 10.2
319
In the final step here we use the fact that [hL cos pn) / Pn = sin pn because
tan pn = hL / Pu.
3.
The substitution u(x,y) - X(x)Y(y) yields the separated equations
X"- c?X = 0
and
7" + ^7=0
with separation constant X - a . Problem 3 of Section 10.1 we saw that the SturmLiouville problem
Y"+ c?Y = 0,
Y'(0) = hY(L) + Y\L) = 0
has eigenvalues Xn = a\ = p~ / L2 and eigenfunctions
Y„(y) = cosAz
L
for n = 1, 2, 3, • • •, with {/?„} being the positive roots of the equation tanx = hL/x.
The solution of
K~yx» = °’
xW = 0
is
Xn(x) = sinh /?"(I~X),
1-J
so the resulting formal series solution is
GO
u (x,y) = Xc«sinh
MnXC0SM
L
n=l
L
The coefficients in the eigenfunction expansion are given by
cn =
i
g(y) cos^y dy
2h
L
(sinhyffjfcos
2^dy
COS
0
L
L
g(y) cos
Py
(sinhfi„)(hL +sin A)Jo
L
because
ri
rL
cos 2^Ldy =
L
1 (
1 + COS
V
Mz dy =
L
0
1+
320
\
I
y
Chapter 10
L . 2pny^
y +-sin
2V
L
2 Pn
/zZ, + sin2 pn
sin2/?„
2A
\_
7
2h
jo
dy,
The final step here is the same as in Problem 1, using the fact that
(hL cos /?„) / Pn = sin /?„ because tan /?„ = hL / /?„.
5.
The substitution u(x,t) = X(x)T(t) yields the separated equations
X"+ cPx = 0
T = -kXT
and
with separation constant X - a2. In Problem 4 of Section 10.1 we saw that the SturmLiouville problem
X"+ c?X - 0,
hX(0)-X\0) = X(L) = 0
has eigenvalues Atl = aln - p2} / L2 and eigenfunctions
Xn{x) = Pn cos
L
+/zl sin^nX
L
for n = 1, 2, 3, • • •, with {/?„} being the positive roots of the equation tanx = -x/hL.
The solution of T’ = -kAT
n n is then
P>A
T„(t) = exp
L
V
J
so the resulting formal series solution is
°°
(
u(x,t) = ^]c„exp
/?2 + ^ /
Kkt
V
«=1
i
Pn cos
Jv
£
PnX
L
. Pnx^
+ hL sin
L J
The coefficients in the eigenfunction expansion are given by
nL
\
f (x) pn cos PnX + hL sin PnX dx
J
0
L
V
rL
f0
P„X
L J
1 T . PyjX A2
Pn cos !-JL- + hL sin
dx
L
L J
V
The evaluation of the denominator integral here is elementary, but there seems little point
in carrying it out explicitly.
7.
The boundary value problem here is
Uxx+Uyy
=
0
(0 < X < 1,
7>0)
ux(0,y)= u(l,y) + ux(l9t) = 0,
u(x, 0) = 100.
Section 10.2
321
The substitution u(x,y) = X(x)Y(y) yields the separated equations
X" + dx = 0
Y" - dY = 0
and
with separation constant X = a . In Problem 3 of Section 10.1 we saw (taking h = L =
1) that the Sturm-Liouville problem
X"+ dx = 0,
X\G) = X(\)+X'(\) = 0
has eigenvalues Xn = a] and eigenfunctions
Xfl(x) = cos anx
for n = 1, 2, 3, • • •, with {an} being the positive roots of the equation tanx = 1/x.
The bounded solution of Y”-anYn = 0 is then
Y„ (y) = exp (~a„y),
so the resulting formal series solution is
oo
u(x,y) = ]>y„cosa„xexp (~a„y).
n=1
The coefficients in the eigenfunction expansion are given by
100 .
sin anx
a.
Jo
fioo cos a.xdx
=
j* cos2 a„xdx
Jb
"
1 (
i
2
2«„
l
200 sin a,
1
J
-ii
an + sin an cos an
so
u(x,y) = 200^
n-1
sin an cos anx exp (~any)
an + sin an cos an
The first five positive solutions of tanx = 1/x are 0.8603, 3.4256, 7.4373, 9.5293, and
12.6453, and we find that
m(1,1) « 30.8755 + 0.4737 + 0.0074 + 0.0002 + 0.0000 +••• « 31.4°C.
9.
(a)
With X = 0, the endpoint-value problem in (19) is X" = 0, X(0) = X'(0) = 0,
which has only the trivial solution X(x) = 0. Thus X = 0 is not an eigenvalue.
322
Chapter 10
(b)
With A = -a2 < 0, the endpoint-value problem in (19) is
= 0,
-ma2X(L) = ASX\L).
Ar(0) = 0,
The differential equation and the left-endpoint condition here give X(x) = sinh ax, and
substitution in the right-endpoint condition gives
2
k
-ma sinh aL = ASa cosh aL, that is, tanhaZ, =aL
with k = AS LI m> 0. But the graph y = tanh x lies (aside from the origin) in the first
and third quadrants, while the graph y = -kfx lies interior to the second and fourth
quadrants. Hence the two cannot intersect, and it follows that there cannot be an
eigenvalue of the assumed form X = -a2 < 0.
11.
Xp
m/V
K_
(a)
a = J— =
(b)
K
J
a =
\ypV
m
\ynRT}K
nmr
29
mr
(1.4x8314x273 (
/l.4x8314(273 + rc)
Mr
K
29
“ V
1 i
{
t \
c
273 J
Tr m
Tr miles
331.02J1 + —-* 740.47Jl + -^
273 sec
273 hour
i
i+- f Tc ) 4-...
2 I273J
740.47+ 1.35673-.
sin(P„+P„)
1
>
Pm + Pn
1
1
1
L
"L . Pmx . Pnx
sm^-^sm——ax =
2
Jo
L
L
n
15.
r
. .-H
c
1_
740.47
( Pm + Pn ) (Sin Pm COS /?„ - sin /?„ COS Pm )
L
- {Pm - P„) (sin p,„ cos pn + sin p„ cos p
L
r'Pni ~P
r'n
[P„ sin P„, cos p„ - Pm sin p„ cos /?,„ ]
M cos p
Pn-“ • COS Pn ~ Pm-4^ ' COS /?
Pi - Pi
mP„,
L
_
Mcos
n
LM
Pn
cos pm COS P„
V Pn,
A Pi ~ Pi)
Section 10.2
n
\
Pm
Pn J
Z.M cos/?„, cos/?„
* o
mP,„Pn
323
19.
With the given initial velocity function g(x) with constant value P /2ps concentrated
in the interval L/2-s <x<L/2 + s, the coefficient formula of Problem 18 gives
rL/2+e
2L
C„
=
2
2
n n a
2
3
2
. nnx i
sin-ax
U2-e
L2P
3
P
L
cos
n 7i a ps
f nn nns
-cos — +L
L )
V
f nn
nns ^
V
. nn . nns
-sin
— sin
3
3
2
n n a ps
~L~
2L P
This gives the ^-dependent solution
00
2 UP
1 . nn . nns . n2n2a2t . nnx
sin — sin
sin
sin
V
y(x,t,s) - — 9
L
n~a ps^Txn
2
L
L
/
i
2
Because
L . nns
sin (nns IL)
— sin- = -11 —> 1
nns
L
nns IL
as
s —» 0,
the limit y(x, t) = lim y(x, t, s) has the expansion
£—>0
00
2LP
1i . nn . n2n2a2t . nnx
sin
y(x,t) = —j-j— y—sin—sin—~
n a pyT\ n
2
L
L
SECTION 10.3
STEADY PERIODIC SOLUTIONS
AND NATURAL FREQUENCIES
In Problems 1-6 we substitute u(x,t) = X(x)coscot in
utt -
2
a uxx
(ia
2
= E/S)
and then cancel the factor cos cot to obtain the ordinary differential equation
a2X" + co2X = 0
with general solution
X(x) = A cos
cox
+ 5 sin
a
324
Chapter 10
cox
a
It then remains only to apply the given endpoint conditions to determine the natural (circular)
frequencies — the values of co for which a non-trivial solution exists.
1.
Endpoint conditions:
X(0) = X(L) = 0
With conditions X(0) = 0 in (*) implies that A = 0, so X(x) = sin (cox/a).
Then
X(L) = sin(coL/a) = 0 implies that coLla = nn, an integral multiple of n. Hence
the nth natural frequency is con =
3.
Endpoint conditions:
X(0) = X\L) = 0
The condition W(0) = 0 gives A = 0 in (*), so we have
X(x) = sin
cox
so
a
.... .
co
cox
X (x) = —cos —
a
a
Hence the condition X\L) - 0 implies that coL/a is an odd integral multiple of 7t/2.
Thus the ^th natural frequency is cou
5.
Endpoint conditions:
(2^z — 1)tt^
2L
_
(2^-1)^-
l~E
2L
\J'
“
ux(0, t) = ku(L, t) + AEux(L, t) = 0
The condition vT'(O) = 0 gives B = 0 in (*), so we have
x// x
a(x)
cox
= cos —,
a
so
,
cox
u(xj) = cos—cos cot.
a
Then
,
x
CO . cox
ux(x9t) =-sin—cos cot,
a
a
so the other endpoint condition is
/
at-co . coL
A
k cos-cos cot - AE — sin-cos cot = 0.
a
a
a
Upon canceling the cos cot factor, we find that
,^coL
coL
7T
AE-tan- = kL.
a
a
Thus P = coLI a is a positive root of the equation AEx tan x = kL, and the nth. natural
frequency is given by
Section 10.3
325
®n
=
Pn<*
L
Pn \E
L\S
where pn is the nth positive root of this equation.
7.
Endpoint conditions:
u{0, t) = mun(L, t) + AEux(L, t) + ku(L, /) = 0
The condition w(0, /) = 0 implies that
_ x
. cox
X(x) = sin —,
a
so
.
co
cox
X (x) = —cos —.
a
a
When we substitute u(x9t) - X(x)cos cot in the endpoint condition at x = L and cancel
the cos cot factor we get
-mo?X{L)+AEX\L) + kX(L) = 0.
Next we substitute
z = coL/a,
co = az/L,
X(L) = sinz,
X\L) = (z/L) cos z.
a
= El S,
The result simplifies readily to the frequency equation
(,mEz2 - kSL2)sin z = MEz cos z.
If pn is the /7th positive root, then the /7th natural frequency is con = Prfi
L
Pn \j_
L\S
In Problems 8-14 we substitute y(x, t) = X(x)cos cot in
ytt
ex yXxxx ~ 0
(a4 = Blip)
and then cancel the factor cos cot to obtain the ordinary differential equation
a 4V4) - co2X = 0
with general solution
X(x)
326
a
u#x _t
Ox „
Ox
Ox
A cosh — + B sinh — + C cos — + D sin —
a
a
a
a
Chapter 10
(**)
where 0
yfco . We then get the natural frequencies of vibration by applying the given
endpoint conditions.
9.
Endpoint conditions: y(0, t) = yx{0, t) = 0,
y(L, t) = y^L, t) = 0
Just as in Problem 21 of Section 10.1, the endpoint conditions X(0) = X'(0) = 0 and
X(L) = X”(L) - 0 imply that
where j3tl is the nth positive zero of the frequency equation
tanhx = tan*.
Therefore the ftth natural frequency con is given by
11.
Endpoint conditions:
y(0,t) = yx(0,t) = 0,
yx(L, t) = y^L, t) = 0
Here we have the equation
V4> - xx = o
with X = co11 aA = f?4 / a4 = a4 and endpoint conditions
X(0) = X'(0) = X'(L) = X(3\L) = 0.
The left-endpoint conditions readily give C =-A and D = -B in (**), so
X(x) - Acos\iax + Bsmhax- Atosax- Bsmax.
= A (cosh ax - cos ax) + B (sinh ax - sin ax).
Then the right-endpoint conditions give
74(sinhaZ + sinaT) + i?(coshaZ-cosaZ) = 0,
A(smhaL-sinaL) +B(coshaL + cosaL) = 0.
The determinant of coefficients of A and B must vanish if there is to be a nontrivial
solution, so
Section 10.3
327
(sinh aL + sin aL) (cosh aL + cos aL)
-(sinh aL - sin aL)(cosh aL- cos aL) = 0.
This equation simplifies to 2 sinh aL cos aL + 2 cosh aL sin aL = 0, which upon division
by cosh aL cos aL gives the frequency equation
tanhx + tanx = 0
for (5 = aL. Then the /7th frequency is given as usual by
2
= ana
2
13.
This problem is the special case m = 0 of Problem 14 below.
14.
Endpoint conditions:
y(0, t) = yx(0, t) = yxxx(L, t) = 0
mytt{L, t) = EIyxxx{L, /) - £y(Z, 0
With p = 0/a, <9 = V&S" we may write
X(x) = ^4 cosh px + i? sinh px + C cos px + D sin px.
The conditions X(0) = X'(0) = 0 readily imply that C = -A and D = -5, so
X
=
^4(cosh px - cos px) +
£(sinh px - sin px),
X
— p^4 (sinh px + sin px) + p5(cosh px - cos px),
X"
= p A (cosh px + cos px) + p 2?(sinh px + sin px),
x3) = p3^(sinh px - sin px) + p3i?(cosh px + cos px).
The endpoint conditions at x = L are
X"(L) = 0,
(A: - mco2)X(L) - EIX°\L) = 0.
When we substitute the derivatives above and write z = pL we get
>4 (cosh z + cos z) + B (sinh z + sin z) = 0,
328
Chapter 10
A [(A: - mco2){cosh z - cos z) - EIp3(sinh z - sin z)]
+ B[(k- mco2){sinh z - sin z) - Elp3{cosh z + cos z)] = 0.
If A denotes the coefficient determinant of these two linear equations in A and B, then
the necessary condition A = 0 for a non-trivial solution reduces eventually to the
equation
EIp3{\ + cosh z cos z) - (k - mco2){sinh z cos z - cosh z sin z) = 0.
Finally we substitute p = z/L, M = pL, and
rtf2 - /a4 = (z4/L4){EI/p)
to get the frequency equation
MEIz3{ 1 + cosh z cos z) = (kML3 - mEIz4){sinh z cos z - cosh z sin z).
We may divide by cosh z cos z to write this equation in the form
MEIz3( 1 + sech z sec z) = {kML3 - mEIz4)(Xm\h z - tan z).
If pn denotes the nih. positive root of this equation, then as usual the nth. natural
frequency is
We want to calculate the fundamental frequency of transverse vibration of a cantilever
with the numerical parameters
L = 400 cm
E = 2 • 1012 gm/cm-sec2
/ - (1/12)(30 cm)(2 cm)3 = 20 cm4
p = (7.75 gm/cm3)(60 cm2) = 465 gm/cm.
When we substitute these values and p\ = 1.8751 in the frequency formula
we find that co\ « 6.45 rad/sec, so the fundamental frequency is co\l2n ^ 1.03
cycles/sec. Thus the diver should bounce up and down on the end of the diving board
about once every second.
Section 10.3
329
17.
When we substitute y{x, t) = X(x)cos cot in the given partial differential equation
2y
i
d4y
5/"
^4 dx2dt2
„
d4y
—v-+ El —v
<3x4
and cancel the factor cos cot, we get the ordinary differential equation
£/V4) + PX" - XX = 0
where P = XI / pA and /l = pco . By solving the characteristic equation
£/ / + Pr2 - X = EI(r2-a2){r2 + p2) = 0
we find the general solution
X(x) =
cosh ox + 5 sinh ox + C cos /3x + D sin fix
where
a
-P +Vp2 +4AEI
2
=
-P - VP2 + 4ZEI
n2
B
-,
2 El
=-.
2 El
The endpoint conditions X(0) = W"(0) = 0 imply that A = C = 0, so
X(x) - 5 sinh ox + Z) sin /?x.
Then the conditions X(Z) = X"(Z) = 0 yield the equations
B sinh aL +
D sin PL — 0,
c?B sinh aL - ftD sin PL = 0.
The determinant of these two linear equations in B and D must vanish in order that a
nontrivial solution exist, so
(a2 + /?)sinh aL sin pL = 0.
It follows that sin pL — 0, so PL must be an integral multiple of n. The definitions of
(X2 and ft imply that
P
_
X
Yl ~ pAE ’
X
El'
Hence if pn = nn!L, the corresponding value of an is given by
2
n2n2
Xn
a~ = —--—.
L
pAE
330
Chapter 10
,
2
2
Then or,, J3n = An/EI gives the equation
(22
n n
L
\
that we readily solve for
2
A
El
The resulting value of the n\h natural frequency is
2
=
2
n 7i
pAE J L2
n 71
<°n
\
/L
2
L1
P
f
1+
V
2_2 t \
nLnLI
1/2
-
pAL2 J
]E1_
P
Substitution of y(x, t) = X(x) sin cot in the transverse bar problem
a2y
dt2
+ a
4
d y
= 0
dx4
4
EIA
a =—
P J
V
y(0,0 = yx(0,t) = 0,
yxx(Lp) = EI yxxx (Lj) + F0 sin cot = 0
yields the endpoint problem
X(4)-p4X = 0
(where p2=co/a2),
X(0) - X\0) = 0,
X'\L) = EIX"\L) + F{; = 0.
When we impose the fixed-end conditions X(0) = X'(0) = 0 on the general solution
X{x) = A cosh px + B sinhpx + C cos px + D sin px
we find readily that C = -A and D = -B, so
X(x) = ^4 (cosh px - cos px) + Z?(sinh px - sin px).
It remains only to find A and B. But the free-end conditions yield the linear equations
,4 (cosh pL + cos pL) + i?(sinh pL + sin pL) = 0
^4(sinh pL - sin pL) + £(cosh pL + cos pL) = -Fo/p EI
that can be solved for
A = K(s\rihpL + sinpT),
B = -A^(coshpZ+ cospZ)
where
K =
Fn
2EIp (1 + cosh pL cos pL)
Section 10.3
331
SECTION 10.4
CYLINDRICAL COORDINATE PROBLEMS
1.
Substitution of u(r,t) - R(r)T(t) in the wave equation
d2u
- a
~dF
^ d2u
1 du^
—Y H-
v dr
r dr
yields the separation
R" + -R'
R
The /^-equation has general solution
T(f) = Acosaat + Bsmaat,
and we choose B- 0, so that r'(0) = 0 (because the membrane is initially at rest).
The r-equation can be written in the form
r2R” + rR' + a2r2R = 0,
which is the parametric Bessel equation of order zero, with continuous solution
R(r) = J0(ar). In order that the fixed boundary condition R(c) = 0 be satisfied, we
choose a-yu!c, where yn is the n\Yi positive solution of J0(x) = 0. At this point we
have product functions of the form J0 (jnr / c) cos(j„at / c) that satisfy the wave equation
and the homogeneous boundary conditions, so we form the formal series solution
u(r,t) =
//=]
'7nr' COS 7nat
c
cJ
In order to satisfy the initial position condition u(r,0) = /(x) is suffices that the {cn} be
the Fourier-Bessel coefficients of the function /(x) given by
2
Cn
=
[J>(/„)}
2.
/
rc
\
Ynr
rf{r)J0
V
C
dr.
J
This is the same as Problem 1, except that the membrane has initial position u(r, 0) = 0,
so in the ^-factor T(t) = A cos aat + B sin a at we choose A- 0 so that T(0) = 0. We
then get product functions of the form J0 (yflr / c)sm(ynat / c) that satisfy the wave
equation and the homogeneous boundary conditions, so we form the formal series
solution
332
Chapter 10
(r„r)
■ r„at
—2— sin1-2—
u{r,t) =
\ c J
/>=!
In order that the initial velocity condition u, (r, 0) = v0 we satisfied, we want
= 1
Yna
w=l
' Cn^0
( Ynr^
V
C
and hence
re
2
rv0J0
r„a c-j,(r„y
2cvr
arAir,,)
arlMr„)‘
f YA
dr
c J
[ xJ0(x) dx
mmt =
(with x = y„r / c)
2"-
°r,Mr,,)
This gives the desired solution
u(r [) =^tf Mr„r/c)sm(y„at/c)
r,Mr„)
a
3.
(a)
As in Problem 2,
CO
fr„f) ■ r„at
<r,t) = XcVo LJL— sin ——
n=\
\ c J
c
In order to satisfy the given initial condition we must choose
c„ =
2
r£ f
r„a c J,(y„)
p
ro
rJn
r rA dr
V Pne J
2P0c
C J
C
xJ0[x)dx
Pn£l r],a J\{y,y
c„ =
^
2 P0c
YnS
pne1 y\a J\{y,„f
c
J,
(with x-ynr!c)
' 7 nA
c
J
2aP«
J\(r„£ic)
ncpci-ynJx(ynf
yns!c
(b)
The final formula given in the text for u(r, t) now follows because pa
J\(x)!x —^ 1/2 as x —^ 0.
Section 10.4
7
= T and
333
5.
(a)
We start with the steady-state boundary value problem
d~u 1 du d^u
-— H-1- = 0
dr2
r dr dz~
(r < c,
—
0 < z < L)
u(c, z) = 0
u(r, 0) = 0,
u(r, L) = uq.
The substitution w(r, z) = R(r)Z{z) yields the equations
rR" + -R'+ cfrR = 0,
Z" - c/Z = 0
with separation constant X — of. The homogeneous endpoint conditions are
72(c) = Z( 0) = 0.
If X = of = 0 then r72" + 7? = 0 implies
72(r) = A + 5 In r.
We choose B = 0 for continuity at r = 0, so 72(r) = Z. Then 72(c) = 0, so A = 0
also, and hence 0 is not an eigenvalue.
If X = a2 > 0 then we have the parametric Bessel equation with general solution
R(r) = AJo(ar) + BY(ar).
In order that 72(r) be continuous at r = 0 we choose 5 = 0, so 72(r) = AJo(ar).
Then
72(c) = aAJo(ac) = 0
requires that y — ac be a root of the equation
J0(x) = 0.
If an = yjc where yn is the nth positive root of this equation, then
K(r) = J0 rYnr'
K
C
)
The corresponding function Z(z) of z is
Zn (z) = An cosh
334
+ Bn sinh —c
c
Chapter 10
and we choose An
0 because Z(0) — 0. Thus we get the formal series solution
sinh 7nZ
u{r,z) =
//=!
c J
where My„) = 0. To satisfy the condition u(r, L) = u0, we need (by Eq. (22) in the
text)
re
1
c„ =
r u0J0
sinh(r„L/c) C Jf(y„y
y„r ^
dr
\ c J
2u0
xJQ(x)dx
rUi(rl,fs'mHrl,L/c)
2u0
(with X = r„r / c)
2u0
r;,Mr„)2smh(y„L/c)
Mdr = rA(r„)m
This gives the desired solution
»<M> -
2„„ ±
»»i
r„'/i(r„)sinh(y„Z/c)
We want to solve the boundary value problem
d2u 1 du d2u
.
—- +-+ —- = 0
dr~ r dr dz~
(r < 1, z > 0)
hu{\, z) + wr(l, z) = 0
w(r, z) bounded as z —» oo
w(r, 0) = woWe start with the separation of variables in Problem 5,
rR" + R'+ c?rR - 0,
Z" - cfz = 0
and readily see that a — 0 is not an eigenvalue. When we impose the condition
hR(\) + R'(l) = 0
on R(r) = Jo(ocr), we find that a must satisfy the equation
hJo(x) + xJq'(x) = 0
Section 10.4
335
that corresponds to Case 2 with n - 0 in Figure 10.4.2 of the text. If {/„} are the
positive roots of this equation then
Rn(r) = Jo(ynr).
The general solution of Z" = yn Z is
Z„(z) = A„exp(-ynz) + BnQxp(ynz),
and we choose Bn = 0 so that Zn(z) will be bounded as z -» oo. Thus we obtain a
solution of the form
oo
u(r,z) = £c„exp(-;K„z)J0O„r)
n=\
where
w0(r„)+r„J'o(r„) = o.
so ynJx(jn) = hJ0(yn) because J'0=-Jv Finally, Eq. (23) in the text gives
rc
c..
dr
r w0^0
=
V c J
c2(^ + ^2)j0(r„):
2ur
(r;+h2)J0(r„)2
__
{rl+h2)J0(r„Y
f” xJ0(x)dx
[xJ,(x)]
(with x = y„r/c)
Yn
0
2 hur
2uo
{rl+h2)J0(r„)2
(;r!,+h2)j0(r„)
so
00
u (r,z) = 2hu0 Y
exp(-r„z)J0{r„r)
% (rl+h2)j0{y„) '
.
11
When we substitute w(r, /) = R{r) sin cot in the given partial differential equation and
cancel the factor sin cot, we get the ordinary differential equation
'f co \2
x
R" + -R' +
R =
r
\aj
The associated homogeneous equation is the Bessel equation of order zero with
parameter cola. Hence it follows readily that the solution that is continuous at r = 0 is
336
Chapter 10
R(r) = AJ0
r a>A R
\ a )
CO
2 •
The condition R(b) = 0 yields A = FJor.JJojb/a), so it follows that the desired
steady periodic solution is
u{rj) =
Jo
co-Jacobi a)
13.
^ cor ^
r cob^
Jo
\ a )
V a )
sin cot.
With w(x) = wx and h(x) = h (where w and h on the right are constants) the given
partial differential equation
w(x) d2y
g
dt2
d f
dy^
— Mx)h(x) —
OX V
ox J
(*)
reduces to
x
d2y
dr
f dy
= gh
+
v dx
d2y^
dx' y
When we substitute y(x, t) = W(x)cos cot we get the parametric Bessel equation
2
2
x2X" + xX' + ^-X = o
gh
with bounded solution
(
X(x) = A J0
cox
^
The condition X(L) = yo implies that A = y0! J0{^coLI ^gh\, so
y(x,t) = y0
J^x/Xh)
cos cot.
J0(a>L/yfgh)
15.
With w(x) = wx and h{x) = hx (with w and h being constants on the right) the
partial differential equation in (*) above reduces to
~T
dr
=
gh
„ dy
dV
2 —+ x—4
dx
dx' J
When we substitute y(x, t) = W(x)cos cot we get the ordinary differential equation
x2X" + 2xX' + — x = 0.
gh
Section 10.4
337
#
^
This has the form of Equation (3) in Section 8.6 with A = 2, B = 0, C = co Igh, and
g = 1, so its (bounded) solution is given by
X(x) =
4 J, 2 co g/z
x
-
\
=
V
J
The condition X(L) = go now implies that A = y0 VZ/ J] (icOyjL /g/z j, so
L Jx[2cOyjxlgh)
cos cot.
y(x,t) = y0A—
x J, [icOyjL!gh
17.
Just as in Problem 1 above, substitution of u(r,t) = R(r)T(t) in the wave equation
fd2u
\ du^
-— H-
Kdr
r dr J
yields the separation
R" + -R’
r
R
The /'-equation has general solution
T(t) = A cos aat + B sin aat.
and we choose B- 0, so that T'(0) = 0 (assuming, for instance, that the membrane is
initially at rest). The r-equation can be written in the form
r2R" + rR' + a2r2R = 0,
which is the parametric Bessel equation of order zero. By Problem 16, its solutions
satisfying R(a) = R(b) = 0 are of the form ^(x) = Yo(yna)Jo(ynX) - Jo(yna)Yo(ynx)
with cc-yn being one of the positive roots of Equation (#) there. This leads to a formal
series solution of the form
oo
u(r,t) = y^Rn(x)(An cos ynat + Bn sin ynat),
n=\
where the frequency of the n\Yi term is con = yna = yn
338
Chapter 10
/ p.
19.
We want to solve the boundary value problem
d~u
'
1 du
d~u
7 H-1-r- = 0
dr2
r dr
dz~
, z > 0)
(a < r < 1
u(a,z) = u(b,z) = 0
u(r, z) bounded as z —> oo
u(r, 0) = /(r).
Just as in Problem 5, the substitution u(r, z) = R(r)Z(z) yields the equations
rR" + R' + ctrR = 0,
with separation constant X =
Z" - c^Z = 0
When we impose the conditions
ot.
= R{b) = 0 on
the r-equation here, we have the Sturm-Liouville problem of Problem 16, so R(r) must be
one of the eigenfunctions {^(r)} corresponding to the positive roots [yn] of Equation
(#) there. The general solution of Z" = y„2Z is
Zn{z) - Z„exp(-y„z) + Bnexip(ynz),
and we choose Bn - 0 so that Z„(z) will be bounded as z —» oo. Thus we obtain a
solution of the form
OO
w(r,z) = Xc«exP (-r„z)K(r)
n=1
with the coefficients {c/;} calculated as in Problem 18.
SECTION 10.5
HIGHER-DIMENSIONAL PHENOMENA
This section provides the interested student with an opportunity to study several applications at
greater depth than is afforded by the usual textbook exercises. The problem sets outlined in
Section 10.5 can serve as the basis for several fairly substantial computational projects. Because
these problem sets and projects are rather heavily annotated in the text, further outlines of
solutions are not included in this manual. However, additional discussion — particularly
regarding computer implementations — may be found in the applications manual that
accompanies the text.
Section 10.4
339
APPENDIX
EXISTENCE AND UNIQUENESS OF SOLUTIONS
In Problems 1-12 we apply the iterative formula
y„+1 = b+ £ f(t, yH (t)) dt
to compute successive approximations {yn(x)} to the solution of the initial value problem
/ = f(x, y),
y(a) = b.
starting with yo(x) = b.
1.
y0(x) = 3
= 3 + 3x
y2(x) = 3 + 3x + 3x2/2
^(x) = 3 + 3x + 3x2/2 + x3/2
y^(x) = 3 + 3x + 3x2/2 + x3/2 + x4/8
X*) = 3 - 3x + 3x2/2 + x3/2 + x4/8 + ••• = 3ex
3.
y0(x) = 1
y\(x) = 1 - x2
y2(x) = 1 - x2 + x4/2
jXx) = 1 - x2 + x4/2 - x6/6
y^ix) = 1 - x2 + x4/2 - x6/6 + x8/24
y(x) = 1 - x2 + x4/2 - x6/6 + x8/24 - ••• = exp(-x2)
5.
yoO) = 0
Ji(x) = 2x
y2(x) = 2x + 2x2
33 (x) = 2x + 2x2 + 4x3/3
jXx) = 2x + 2x2 + 4x3/3 + 2x4/3
X*) = 2x + 2x2 + 4x3/3 + 2x4/3 + ••• = e2x - 1
340
Appendix
7.
yo(x) = o
^l(x) = x2
y2(x) = x2 + x4/2
^(x) = x2+x4/2 + x6/6
^(x) = x2 + x4/2 + x6/6 + x8/24
y(x) = x2 + x4/2 +x6/6 + x8/24 + ••• = exp(x2) - 1
9.
y0(x) = 1
y\(x) = (1 + x) + x212
y2(x) = (1 + x + x2) + x3/6
^(x) = (1 + x + x2 + x3/3) + x4/24
y(x) = l+x + x2 + x3/3 + x4/12 + ••• = 2ex - 1 - x
11.
yo(pc) = 1
yi(x) = l +x
^2W = (1 + x + x2) + x3/3
jV3(x) = (1 + x + x2 + x3) + 2x4/3 + x5/3 + x6/9 + x7/63
y(x) = 1 + x + x2 + x3 + x4 + • • • = 1/(1 — x)
13.
1
X0(0
Jo(0_
-1
-
r-
1 + 31
x,(t)
_y,(0.
-
-
1 + 3 t + \t2
*2(0
I .2
—1 + 5t ——t
i-
2
.
1 ,3
x3(0
1 + 3t + — t
_y3( 0_
-l + 5f-++ +
+ j/
Appendix
341
V
,
-
.
'•
I
A
J">
A.
■
r-
9780130475794
01/17/2017 16:34-3
Upper Saddle River, NJ 07458
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