Elementary diff eqs 8e sol earl rainville

Student Solutions Manual
Phillip E. Bedient * Richard E. Bedient

Eighth Edition

Elementary
Differential
Equations

Earl D. Rainville

Phillip E. Bedient
Richard E. Bedient


Student Solutions Manual
Phillip E. Bedient ¢ Richard E. Bedient

Eighth Edition

Elementary
Differential
Equations
Earl D. Rainville
Late Professor of Mathematics
University of Michigan

Phillip E. Bedient
rofessor Emeritus of Mathematics
Franklin and Marshall College

Richard E. Bedient
Professor of Mathematics
Hamilton College

PRENTICE HALL, UPPER SADDLE RIVER, NJ 07458
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Assistant Editor: Audra Walsh
Production Editor: Carole Suraci
Special Projects Manager: Barbara A. Murray
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Manufacturing Buyer: Alan Fischer

Copyright © 1997 by Prentice-Hall, Inc.
A Pearson Education Company
Upper Saddle River, NJ 07458

All rights reserved. No part of this book may be
reproduced in any form or by any means,
without permission in writing from the publisher.

Printed in the United States of America

ISBN

O-13-592783-6

Prentice-Hall International (UK) Limited,London

Prentice-Hall of Australia Pty. Limited, Sydney
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Definitions; Families of Curves
12
Definitions.
2.

peak
frank

Contents

Numerical

2000.

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The General Solution of a Linear Equation

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Ch bt

2.6

Miscellaneous Exercises

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Equations of Order One
2.1 Separation of Variables...
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2.2 Homogeneous Functions .. 0...
2.3 Equations with Homogeneous Coefficients...
2.4 Exact Equations 2... 0.

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Methods

3.2

Euler’s Method

3.3.
3.4
3.5
3.6
3.7
3.8

A Modification of Euler's Method. . 0... 0
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A Method of Successive Approximation .. 2.6.0.0...
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AnImprovement 2.0.0
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The Runge-Kutta Method...
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A Continuing Method 2.00.0. ee
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Elementary Applications
4.3
Simple Chemical Conversion. 2... 00
4.4
Logistic Growth and the Price of Commodities

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Additional Topics

on Equations of Order One
5.1 Integrating Factors Found by Inspection...
...-.00 00.00.0020
5.2 The Determination of Integrating Factors .......0.00 2.00000.
5.4 Bernoulli's Equation... 0.
5.5 Coefficients Linear in the Two Variables 2... 6
5.6 Solutions Involving Nonelementary Integrals...
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21
22
22
22
23
24
24
25
25

26


CONTENTS

6

Linear Differential Equations
6.2 An Existence and Uniqueness Theorem
...........0..00000000-00.
6.4 The Wronskian 2.0...
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6.8 The Fundamental Laws of Operation... 2.0000
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6.9 Some Properties of Differential Operators .......000..
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39
39
39
40
41

7

Linear Equations

with Constant Coefficients
7.2 The Auxiliary Equation: Distinct Roots ........0..00...0. 0200005 cas
7.3 The Auxiliary Equation: Repeated Roots .......0..00.000. 0.00000008

42
42
43

7.6

A Note on Hyperbolic Functions

Miscellaneous Exercises
8

9

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20.

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44

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45

Nonhomogeneous Equations: Undetermined Coefficients
8.1 Construction of a Homogeneous Equation from a Specific Solution ..........
8.3 The Method of Undetermined Coefficieents 2.2.
0.0 0.0. ee
8.4 Solution by Inspection...
2.1.

48
48
49
54

Variation of Parameters

56

9.2

Reduction of Order...

0.

9.4 Solution of y’+ y= f(@)

0.

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10 Applications
10.3 Resonance

70
60.

70

10.4 Damped Vibrations. ©. 0. ee
10.5 The Simple Pendulum ©. 1... ee
11 Linear Systems of Equations
11.2 First-Order Systems with Constant Coefficients
11.4 Some Matrix Algebra 20
11.5 First-Order Systems Revisited. 2. 0
11.6 Complex Eigenvalues 0 00 0
11.7 Repeated Eigenvalues 2. 0.
11.8 The Phase Plane

12 Nonhomogeneous

. 0.0

13 The Existence

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73
77
78
78
78
79
81
83
85

0

Systems of Equations

12.1 Nonhomogeneous Systems
19.2 Arms Races...

12.4 Simple Networks

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87

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and Uniqueness

of Solutions
13.2 An Existence and Uniqueness Theorem...

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90
94
94


CONTENTS
14 The
14.3

Laplace Transform
‘Transforms of Elementary Functions

14.6
14.10

FunctionsofClassA
Periodic Functions...

2...

95
.......00000.0.......0004..,

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15 Inverse Transforms
15.1
15.2

101

Definition of an Inverse Transform
Partial Fractions ©...

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15.3 Initial Value Problems .. 0.0.0...
15.4 A Step Function «2.0...
15.5

A Convolution Theorem

16.2
16.56
16.7
16.9

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15.6 Special Integral Equations...
15.8 The Deflection of Beams...
15.9 Systems of Equations...
16 Nonlinear

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102
108

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14
115

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120

2...

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Series Solutions

18 Solutions Near Regular

....

Singular Points

Regular Singular Points

2.2...

Difference of Roots Nonintegral

184

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143

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Equal Roots...

18.7

Equal Roots, an Alternative...

0.

18.8

Nonlogarithmic Case...

18.9

Logarithmic Case.

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18.10
18.11

Solution for Largez oe
Many-Term Recurrence Relations 2...

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205

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143

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18.6

Miscellaneous Exercises

120
123
124
126
134

17.5 Solutions Near an Ordinary Point... ... . —

18.4

101
102

Equations

Miscellaneous Exercises

18.1

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17 Power

95

96
98

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20 Partial Differential Equations
20.3 Method of Separation of Variables
21 Orthogonal Sets of Functions

208

21.6 Other Orthogonal Sets...

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22 Fourier Series
22.3 Numerical Examples of Fourier Series
22.4 Fourier Sine Series 20

2...

0.2

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208

210
210
213


CONTENTS
23 Boundary Value Problems
23.1 The One-Dimensional Heat Equation... .........0...
23.4 Heat Conduction ina Sphere .. 0...
2. ee
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23.5 The Simple Wave Equation ........0.-.
205. ee ue
23.6 Laplace’s Equation in Two Dimensions. .............
24 Additional Properties of the Laplace Transform
24.1 Power Series and Inverse Transforms
24.2 The Error Function. 2. 0. 0
24.3 Bessel Functions

. 2... 0

.... 0.0.0.0...

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25 Partial Differential Equations:
Transform Methods

25.1 Boundary Value Problems...
25.2 The Wave Equation

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25.5 Diffusion in a Slab of Finite Width...
25.6 Diffusion in a Quarter-Infinite Solid

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Chapter

1

Definitions; Families of Curves
1.2

Definitions

All answers in this section are determined by inspection.

1 . The equation is ordinary, linear in x, and of order 2.

cn

3. The equation is ordinary, nonlinear, and of order 1.
The equation is ordinary, linear in y, and of order 3.

7. The equation is partial, linear in u, and of order 2.
9. The equation is ordinary, linear in x or y, and of order 2.
il. The equation is ordinary, linear in y, and of order 1.
13. The equation is ordinary, nonlinear, and of order 3.
15. The equation is ordinary, linear in y, and of order 2.

1.3.

Families

of Solutions

1. Rewriting the equation yields y = / a? + Idx +c. Integrating, we have y = gx! + 2% +46,
Rewriting the equation yields y = 4 / cos 6a dx + ¢. Integrating, we have y = 2 sin Oz +c.
Rewriting the equation yields y = 2 /

1

ye

dz+c. Integrating, we have y = arctan (2/2)+c.

. Rewriting the equation yields y = 3 / e* dz. Integrating, we have y = 3e* +c,
the initial conditions gives 6 = 34+ corc=

Substituting

3 so y = 3e7 +3.

As in Example 1.2, y = ce**. Substituting the initial conditions gives 3 = ce® = c so y = 3e%*,
11. Rewriting the equation yields y = 4 / sin 2a dx. Integrating, we have ~2cos2¢+c.
ing the initial conditions gives 2 = -2cos7 +e = 2+corc=0s0

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y = —2c0s 2u.

Substitut-


CHAPTER 2.

Chapter

EQUATIONS OF ORDER

ONE

2

Equations of Order One
2.1

Separation of Variables

1. The variables may be separated to give

[tana
few
r
Integrating both sides we have
Inr = ~-2¢ +corr=e =2t? +e orn
=

nt?

677 6,

To simplify this we replacing, the constant e° by another constant. We could use a different
name like k to get, r = ke~?*', but the convention is to reuse the name c to obtain

r= ce

oe

2t

BF

or To = €or Tr = To exp (—2t").

The variables may be separated to give
y
wee dy =

lies,

¥

dz
fm.

x

Integrating both sides we have

din(l+y*) =Ina+cor In(i+y*)/? =Ing+e.

Exponentiating both sides yields

(1 y2)¥2 = ell 49 op (1 $y?)

ax elt tee,

As in Exercise 1 replacing e° with ¢;
(1+y?)\/? = ex or y* = cz? -lory= Vex? — 1.
From the given conditions, 3=

/4¢~-1,

9=4e-1lorc=

y= 3 V 100? - 4.

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42. Thus


2.1.

SEPARATION OF VARIABLES

3

. Separating the variables we have,

2{Z=3

ay

xr

y

Integrating both sides yields, 2Ina +¢ = 3iny or y? = cz.

conditions to obtain, 1 = 4c, or c = 4. Thus y = (x/2)?/8.

Finally, we substitute the initial

. Separating the variables we have,

[eva

= [oe

dz.


Integrating both sides yields, -e7~¥ = —ten
-+-c. Finally, we substitute the initial conditions
to obtain, -1 = -i =corc= —}. Thus 2e7Ơ = e~đ +1 or In2- y = In (e-™” +1) or
y = In2 —In[1 + exp (—2")).
. Separating the variables we have,
2a? [r-8dr—

fran

= f

sins,

Integrating both sides yields, ~a?r~? —Inr = —cos@+.c.

Finally, we substitute the initial

conditions to obtain, ~a?a~? — Ina = ~—cos0 +c or -Ina = c. Thus ~a2r7? ~Inr
—cos@ ~ Ina or r*(Inr — Ina) = r? cos@ — a? or r* In(r/a) = r? cos8 — a?.

=

11. Separating the variables we have,

[a-/

da:

yf

i-«

Integrating both sides yields, ~y~! = ~In(1 ô)+c. Replacing  by Inc, yjln|1 el+ine] = 1.

Thus yin |e(1 — z)j = 1.
13. Separating the variables we have,
dy _

adz

Integrating both sides yields, —}y~? = hers” +c. Replacing —2e by c leaves e~™ + y7? = c.
16. Separating the variables we have,

mde _ fndy
z

¥

Integrating both sides yields, mlng = niny +c or 2™ = cy”.
17. Separating the variables we have,

[%-Vv

aP
Po

Integrating both sides yields, nV = ~ln P+corin(PV)=cor PV =e,

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CHAPTER 2.

EQUATIONS

OF ORDER

ONE

19. Separating the variables we have,
[*

bsin é dé

r

|

1—bcosé’

Integrating both sides yields, In |r| = In|1 — bcos @| + ¢ or r = c(1 ~ bcos @).
ai. Separating the variables we have,
[uev=-[

Zyae=-f

(+045)

dz.

Integrating both sides yields, dy? = —(a+1)?—In (x — 1)+c or (x+1)? +y? +2 In |e(x — 1)| = 0.
23. Separating the variables we have,
[a

Integrating by parts yields, / (i +
Thus -

— i

=~

| xe* de.

| dy = —ze* + / e” dz.

= ~—ze* +e" +c or e*(a — 1) = (2y + 1)/(2y") +e.

25. Separating the variables we have,
fue

dy = [eo

dz.

Integrating both sides yields, —ye~¥ + / e-8 dy = —a7' +c. Thus ~ye7¥ ~ e74 =~

+6

or c(y +1) = (1+ czr)e¥.
27. Separating the variables we have,
J secudy = [cost xa.
Integrating both sides yields, In|secy + tany| = da + ; sin 2n + c.

Thus 4In]secy+ tanyj] = 2x + sin 2x + ¢.
29. Separating the variables we have,
[cost xa

= fe

+t?) dt.

Integrating both sides yields, 42 + 4sin 2x = 4(1+¢)? +c. Thus 2z+sin22 =e+ (1407).

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2.2.
31.

HOMOGENEOUS FUNCTIONS

5

Separating the variables we have,

[tBan- [has
or

[ (E43) aa=- f (G41) a8
Integrating both sides yields, Ina + 3a = ~Inf—§8+c
caf = exp{—3a — ~).

or lnaf = ~3a~ B8+c.

Thus

33. Separating the variables we have,

rdz

/ dy

va? — x3
Integrating both sides yields, -/a? — x? = y +c. Thus y—-c¢ = ~Va? — 2”, the lower half of

the circle x? + (y — c)? = a?.
35.

Separating the variables we have,

a? dx

aye
=f
aVvx* — a?

/

LY,
y+e

Integrating both sides yields, a aresec(t/a) = y +e. Thus x = asec y
37. Separating the variables we have,
2
_fyti,

x

_

lane: [tae

/(v

_

1+

2

ty) dy

Integrating both sides yields, § In (2* +1) = $y? —y+2Inly+1j+e.
Thus In (x? + 1) = y? — 2y + 4 In |e(y + 1).

2.2

Homogeneous

Functions

All functions are homogeneous except those of Exercises 2, 5, 6, and 19 by examination.

2.3.

Equations

1. Substituting

y

with Homogeneous
=

av,

we

have

3(327

Coefficients

+ v*z*)dr

—

(9 + v) da = Qrvdv. We first separate the variables to obtain
[e-2/

udu

x

9 + 2"

Integrating yields Inc = In(9+ v7) +c.

x = [9 + (y/x)*] or 2° = c(9x? + y*),

22?u(ude

+ xdv)

=

O

or

Replacing ¢ by Inc and substituting for v leaves

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CHAPTER 2.

EQUATIONS OF ORDER

3. Substituting y
= av, we have 2(227 + v?x?)dx — x*v(ude
(4+v7)dx = xcudu. We first separate the variables to obtain
/ dx

/

gc
Integrating yields Inz = 3 In 1 (4 +v*) +c.

f

+ adv)

ONE

=

0

or

vdy

442!

Replacing ¢ by Ine and substituting for v leaves

x? = c7[4 + (y/x)*] or ct = ce? (da? + y?).
. Substituting

y

=

xv,

we have

(4a* + Tux? + Quta*)dz — a?(udz + xdv)

=

0, or

(4+ 6v + Qu") dx = xdu. We first separate the variables to obtain
[e-

loss
du

4+6u+

-4/

202

1

+

1

2Q+u

d

itv

Integrating yields Ina = }(In(2+v) In(1+v)] +<Â or
a
leaves Â[1 + (y/x)] = ô7[2 + (y/z)] or x?(y + 2x) = c(y + 2).

°-

= c*z*. Substituting for v

. Substituting y = cv, we have (2 ~ xv)(4z + av)da + x(5a — xv)(vdx + xdv)
(—2u? + 2u + 4) dx = (—5a + xv) du. We first separate the variables to obtain

dx

ve 5

1

[¢ =| sete’

1

0 or

2

-1f (=5+sh)
e(v
— 24/2

Integrating yields Inz = —}In(v ~ 2) —In(v+1) +c or

=

vt

dv.
= z, Substituting for v

leaves e[(y/x) — 2]!/? = a[(y/x) +1] or z(y + z)* = c*(y — 22).

. Substituting z = yu, we have (v2y? + 2y?v — 4y*)(udy + y du) — (uy? — 8y?u — 4y”) dy = 0 or
(v? + v? + 4u + 4) dy = ~y(v? + 2u — 4) dv. We first separate the variables to obtain

[*--[
yo

i

he-/

we+vrtdu+4 0 —

Integrating yields Ina = In (v + 1) ~ In(v? + 4) +c, or

1
v+t1

vt

ed

ay

vi+4

=: y. Substituting for v leaves

yl(y/a)? + 4] = c[(y/x) + 1] or 2? + 4y? = e(a + y).
i. Substituting

y

=

av,

we

have

(2?

+

v*z*)de

/

vw
1+2v?

+

z*u(udz

(1 + 2v”) dz = —zu dv. We first separate the variables to obtain
/ dx
ce

J

+

cdv)

=

0

or

Integrating yields ~Inz = 41In(1+?) +e, or (1 + 2u*)/4z = c. Substituting for v leaves
[1 + 2(y/x)?]a4 = ef or 2?(2* + 2y?) = ct.

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2.3.

EQUATIONS

WITH HOMOGENEOUS

COEFFICIENTS

7

13. Substituting 2
=
vy,
we have v*(udy + ydv) +
(y? + 2y) du = —udy. We first separate the variables to obtain

- [=

at

ve

=a

Jf w+oy 2

yu(yy

il
yo

\,

y+2

+

v)dv

=

O

or

y.

1/2
}

=>

Substituting for y leaves

a

Integrating yields — Inv = g(Iny —In(y+2)}+eor € 2

c{(z/v) + 2] = (x/v)u* or xv? = efx + Qu).
16. Substituting yo = rv, we have 3 + susin 2) dx — asin Foy dx + xdv)
=

(2 + zusin®
v — usin? v) dz = x* sin? udu. We first separate the variables to obtain
dx
x

O

or

= | sin* vav.

1
Integrating yields Ing = 3Ui sin2v +c. Replacing c by Inc and substituting for v leaves
In |z/el= 4(y/x) — 4 sin (2(y/x)] or 4a in |x/e| ~ 2y + vsin (2y/x) = 0.

17, Substituting y = xv, we have (« — zvarctanu)dr + xarctanv(vdr + xdv) = 0 or
—(i — varctanv + varetanv) dz = xarctanudv. We first separate the variables to obtain
-{2
x
Integrating

yields

2
2
In eee)

~Inz

=

+c

=

= f asctanv do.

varetany

Qvarctanyv.

Then

—

4 In(l

+7).

substituting

for

Replacing

vu gives

¢ by

Inc

leaves

=

0 or

2
2
In (Eege*)
x

= 2(y/x) arctan (y/x) or 2y arctan (y/x) = x In [c?(x? + y*)/24}.
19. Substituting

t = sv, we have vs(s* + v%s?)ds — s(s? — v?s?)(uds + sdv)
(v +3 —v +") ds = s(1 — v*) dv. We first separate the variables to obtain

ds
seme
/ 8

SE

fi-v?
1
—————1) =
“3 yn
we
9 ~ 3 / (eno)

de

.

Integrating yields In |s| tes= 4(—4u7?—In[u}) or Injesv!/?|= ~jv*. Substituting for v leaves

—2t? in |es?(t/s)|= s? or s? = ~2#? In jest}.
al. Substituting

y

=

xv,

we

have

(3x?

— 227y

+ 327v")
de + 4z*u(udz

+ zdv)

(3— 2u — v*) dx = drvdv.

We first separate the variables to obtain

dz

/?

du

-/[sE=3
god
i”

—3

1

=-/[(St)

do.

Integrating yields Ing = ~3In(34+v)—-In(l—v)4+e

Substituting for v leaves z[3 + (y/x)P{1 ~ (y/x)] = c or (y — x)(y + 3x)? = ex.

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=

0

or


CHAPTER 2.

EQUATIONS OF ORDER

23. Substituting

y = «xv, we have (x — zu)dz + (382 + zv)(ude + adv)
(l—v+3u +07) dz = —(3+ v)zdv. We first separate the variables to obtain
dz

-{¢

3+4

-/ ase

1

=f

for v leaves

Gs

or

2

ino™

dy

In[z(1 +v)| +c. Substituting

= Infa(1 + [y/2])] + ¢ or 22 = (2 + y) In (aw + y) + e(a + y). The initial

conditions give 6 = (1) In@ij)+e=c,
25,

0

2

(+a)

Integrating yields —Inx = In(1+v)~2(1+u)7!+cor

=

ONE

Thus 2(2r + 3y) + (@ + y) Infe+y) = 0.

Substituting y =
zu, we have (rv + Ja? +a%v?)dx — a(udr
(v+ V1 + vu? — v)dz = adv. We first separate the variables to obtain

+ xdv)

=

0

or

dy
Jitu?
Integrating

27.

yields

Inn

=

xz = ef(y/x) +
zg? =Qyt+l.

/1+ (y/z)*].

Substituting y

=

In(vu+Vl+v7)

+

«

Substituting

for

vu

leaves

The initial conditions give 3 = c(1+/3+1) orc = 1. Thus

xv, we have (x*v? + 7x*v + 16z*)dr + 2°(udz + 2dv)

=

0 or

(vu? + 7v +164 u)dx = ~—xdv. We first separate the variables to obtain

[¢Integrating yields Inz = —

ees

rs ie -/ oe

+c. Substituting for v leaves [(y/x) + 4]Ing = 1+ ¢[(y/x) + 4).

The initial conditions give 0 = 1+ 5¢ or c = ~—(1/5).
a2-y=5(yt+4z) ing.

Thus (y+ 4c) Ing = x — By + 4x) or

29. Substituting

c =
yv, we have yv(ydu + udy) + 2(y7u?
(u? + Qu? + 4) dy = —vy du. We first separate the variables to obtain

+ 2y2)dy

=

O

or

dy
v
-{% -f a pat

Integrating yields ~Iny = 3 In(3v? + 4)+c. Substituting for v leaves c = In ([3(x/y)? + 4]y®).
The initial conditions give C== 4. Thus y4(32? + 4y?) = 4,
31. Substituting

y

=

xv,

we

have

au(Qx — 2xv)dx

— 2(64 ~ cu)(vdz + adv)

(Qu ~ 2u? — 6v + vu?) dz = (6 ~v)adu. We first separate the variables to obtain

[2- [arene [qo=ae=
[Go sea) *
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=

0 or


2.4.

EXACT EQUATIONS

9

Integrating yields Ing = 2lnv ~In(v —3)+c¢.

_

2*(y

—

32)

Th

c= ———~;-——. The

Seated

‘a2

.

Substituting for v leaves c = al(y/z) = 3]

(y/z)?

a

_

initial conditions give c = {1(1 ~3)|/1 = —2.

rn

2_.

Thus —2y*

or

2

= 2?(y — 32) or

¥

33 — zy — 2y? = 0.
33. Substituting y = av, we have (16x + 5xu)dx + (3x + rv)(udx + x2dv)
(16 + 5u + 3u + v*) dz = (~3 — v)rdv. We first separate the variables to obtain
-[S-

flatts
v+3

v= |

wtb pie.”
Integrating

yields

—-Ing

=

1

~In(ai(y/2) +4) = Gog

In(v+4)

ter

—In(-3+4)

= 1/(-3+4)+core
yt 3e = (y + 42) In (y + 42).

_t

+

ny

(v + 4)

1

1

v+4

(u4+4)?

+6

=

0

or

dv

‘

Substituting

for

vu

leaves

£
+4) = ae

= —1.

+c. The initial conditions give
Thus —(y + 4x) In(y+42) = xc - y — 42 or

35. Substituting

2 = yv, we have 2y*v(ydu + udy) — (8y2v?
(3u? — 2 — 2v*) dy = Quy dv. We first separate the variables to obtain

— 2y7)dy

=

0

or

dy
Integrating yields Iny = In(v? —2)+

¢.

Substituting for v leaves In (ars)

= ¢ OF

3
Cc=

2.4

gop

orc = 1/2. Thus 2? = 2y?(y +1).

Exact

Equations

OF
1. Let an 72 +y. Integrating with respect to z we have F = x7/2+2y+T(y).

with respect to y yields x
cor 2? +Iry
OF

Let a

¢

Differentiating

= 2+T"(y) = ¢-y so T(y) = —y?/2. Thus F = 2?/2+2y—y?/2 =

~yr=c.
2

= 2xy—32*.

.

.

4

2

3

Integrating with respect to x we have F = 2*y~2°+T(y).

:

sae

Differentiating

with respect to y yields 3 = x? +T7"(y) = 2? +y so that T(y) = y?/2.
Thus F = a2?y— 23 + y?/2 = cor x*y— 23+ $y? =e
. LetoF

ax

= c~2y. Integrating with respect to x we have F = x7/2—2zy+T(y).

with respect to y yields &

= —22+T"(y) = 2y — 2x so that T(y) = y?.

Thus F = 22/2 ~2ry + y? =c or 2? + 2y? = dry te.

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Differentiating

CHAPTER

10

2.

EQUATIONS OF ORDER

ONE

This equation has homogeneous coefficients. Substituting y = av, we have (a — 2av) dx +
av —a)(vdz + adv) = 0 or (1 ~ 2u + 2v? — Qu) dz = (—2u + 2)xdu. We first separate the
variables to obtain
[e-

ze

(~2u+2)du

(2v — 2) dv

jf Q?—4u+1—

(Qu?
— 4u +1)

Integrating yields Ina = —1} In (2u? — du 4+ 1) +c or x7(2v* —4u +1) =c.
Substituting for v leaves z*(2(y/x)? ~ 4(y/x) + 1] = ¢ or 2? + 2y? = dey +e.
. Let i

= y* — Ixy + 6x. Integrating with respect to z we have F = zy” — xy + 32? + T(y).

Differentiating with respect to y yields

PP

a

‘

= Qay~ 224 7"(y) = ~2? + Izy ~ 2 so that

T(y) = —2y. Thus F = xy? ~ xy + 32" -2y =e.
il.

Let a

= cos 2y ~— 327y?. Integrating with respect to a we have F = xcos2y — zy" + T(y).
oF

Differentiating yields By

=

~Qrsin2y — 2x5y + T"(y)

:

= cos2y — 2xsin2y ~ 22%y so that

T(y) = 4sin 2y. Thus F = zcos2y — x*y? + § sin 2y = c or 4} sin 2y + x cos 2y — ay? = c.
,_
OF
13. Let an"

.
:
Integrating with respect to z we have F = x + ry? + tx7y? + T(y).
OF
Differentiating with respect to y yields By = Qry+c*y t+ T’(y) = x?y + y+ 2zy so that
1+y°+ay*.

T(y) = y?/2. Thus F = a+ zy? + $27y? + gy =corQet+y%(l+c)? =e
15. Let x

= 2? —xsec*y.

Integrating with respect to y we have F = x2*y — rtany + Q(z).

OF
Differentiating with respect to z yields —~- = 2xy —~ tany + Q(x)
Ox

= 2xy — tany so that

Q(z) = 0. Thus F = x°y — xtany =e.
17. Let a

= r+sin?—cos@. Integrating we have F = r?/2+r(sin@—cos0)+T7(@).

with respect to @ yields a

Differentiating

= r(cos@+sin 0) +7'(6) = r(cos @ +sin @) so that T(@) = 0. Thus

F = r?/24+r(sin@ — cos@) =c or r* + 2r(sin@ — cos) =e.
19. Let a

Fe

ir

= sin@ — 2rcos*@.

.

Integrating with respect to 2 we have F = rsin@ — r? cos?@ +

T(6). Differentiating yields a

= rcos 6 + 2r? cos @sin@ + 7" (8) = rcos O(2r sin @ +1) so that

T(@) =0. Thus F = rsin# — r? cos? @ =e.
ai. Let a

me

= Qry. Integrating with respect to « we have F = 2*y + T(y).
a

respect to y yields i

Differentiating with

= 2? +T'(y) = y? +2? so that T(y) = y°/3. Thus F = xy + y3/3 =c

or y(3a? + y”) =e.

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2.6.

THE GENERAL SOLUTION OF A LINEAR EQUATION

OF
23. Let a = zy? + y ~ x. Integrating with respect to
OF
Differentiating with respect to y yields a

11

we have F = Za7y* + cy — $2? + Ty).

= ay +ao+T'(y)

= a*y +2

so that Ty)

= 0.

Thus F = $2*y? + ay — $27 = cor 2?y? + 2ay — 2? =.
2

25. Let -

= y?+27(1—cy)~*. Integrating we have F = y3/3+2(1—zy)7'!+Q(z). Differentiating
OF

with respect to a yields on = zy(1—zy)7?+(1—zy)7!+Q'(z) = (1~zy)~? so that Q(x) = 0.

Thus F = 3y°+a(1—ay)~* = c. Substituting the initial conditions gives }~2 = core = —8,
Thus 4y° + 2(1 — zy)! = —8 or ayt — y9 + Say — 32 = 5.
OF
27. Let iy

:

= x + 3y* + exp(—2?).

Integrating we have F = yz? + y° + yexp(—2?) + Q(x).

Differentiating with respect to x,

i = Qry — 2zyexp(—z?) + Q'(z) = 6x? + Iry — Ary exp (~x?) so that Q(z) = 223. Thus
Fe=x?y+y* + yexp(—27) +2 =e.
o

2.6

The

General Solution of a Linear Equation

d
1, The equation in standard form is = ~ (3/z)y = x‘. Thus the integrating factor is
exp (-3/<)

=exp(—3inz) = =

d
Multiplying by the integrating factor we have is (5)

= x. Integrating and simplifying yields

y/e®
= x7/2+.c or 2y=
2° + cx,
“

dz

The equation in standard form is dy + [4/(y + ijx = y/(y +1). Thus the integrating factor is

exp (1/4)
Multiplying

= exp (4In(y+ 1] = (y+1)4.

by the integrating factor we have

sl

+ 1)4a]

= y(y +1)*.

Integrating and

simplifying yields 20z(y + 1)4 = (4y — 1)(y+1)4 +c or 2027 = 4y~1+e(y+1)74.
da
The equation in standard form is ——- + [(1 — 3u)/u]z = 3 Thus the integrating factor is
du

exp (/ Co

au)

= exp (Inu — 3u) = uexp (—3u).

d
Multiplying by the integrating factor we have Gy ue")

= 3ue~*". Integrating and simpli

fying yields ue~**z = —ue~*" ~ ge" +c or cu = ce™ ~ uf.

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CHAPTER 2,

12

_ @
The equation in standard form is oy

ultiplying
Multiplyi

P

ONE

(cot z)y = cse x. Thus the integrating factor is

dx

ex

EQUATIONS OF ORDER

/ cota dz) = exp |[—In (sinx)| =
i
=
exp
™ sina’

2= esc”cae

Y)

df

by the integrating factor we have Ge

Gna)

oe:

-

x. Integrating and simplifying

yields y/sinz = ~—cotr-+cor y = csinz — cose.

dy

The equation in standard form is dn + (1/cosx)y = cosx. Thus the integrating factor is
exp (/ secede

= exp [In (secx + tan z)] = sec + tan x.

d
Multiplying by the integrating factor we have 7, lulsec x +tanz)| = 1+sinz. Integrating and
simplifying yields y(secz + tanz) = c+

— cosa.

dy

li, The equation in standard form is ix + (cotz +1/a})y = 1. Thus the integrating factor is
1

exp ( / (cots + *)

az)

.

d
Multiplying by the integrating factor we have az
fying yields zysina = ¢+ sing ~ xcosa.
13. The equation in standard form is

dy
a _

.

= exp [In (sin) + Ing] = vsina.
sing) = zsinz.

Integrating and simpli-

[2x /(1+27)ly = (x? +24)/(1+27). Thus the integrating

factor is

exp (-/ ar) = exp[-In(1+2”)] =(1+2%)"1
;

_,

d

Multiplying by the integrating factor we have qv

simplifying yields y(1 + 2?)~! = x — aretanz +

Bet

+2°)~"]

xe

= ite’

Integrating and

or y = (1 + 2”)(¢ + x — arctanz).

d
.
.
:
15. The equation in standard form is = ~ (ma)y = cie™*. Thus the integrating factor is
exp (/ —ma dz)

= exp (—Mmgr).

d
_m
Multiplying by the integrating factor we have ——(e~™*"y) = cye™~™2)*_
simplifying yields y = cye™” + cge™?*, where cz = e,/(m, — m2).

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:
Integrating and


2.6.

THE GENERAL SOLUTION OF A LINEAR EQUATION

13

17. The equation in standard form is a + [2/(x(a? + 1)ly = (2? + 1)*/x. Thus the integrating
factor is

oo ([ garry) ores
2dx

_

wa:
:

.
Multiplying by the integrating factor we have

2

2

Ul ay

d
2?
is (ay )

=

= a? +2.

Integrating and

simplifying yields x7y/(x? +1) = 3 (0? +1)? +c or 2?y = (x? +1)9 +e(2? +1).
d

+ (2/(x? ~ 1)]y = 1. Thus the integrating factor is

19, The equation in standard form is =
ex

P

fea

gz? —]

dx}

= exp

=

exp

{in

vl

d
Multiplying by the integrating factor we have — (v3
dx

fying yields y ral
a+]

=

_ 27}

e+ij//
-I1
\"ri+l

«+1
bp 1
) =,
g+1

Integrating and simpli-

—-2In|[x+1|+ecor (x — ijy = (a@+1)(e+
¢~ 2injx +1).

d
21. The equation in standard form is = ~ {3tanz)y = 1. Thus the integrating factor is
exp(-3 / tanz ar)

= exp [~3
In (sec x)} = cos? x.

d
Multiplying by the integrating factor we have ay lu(eos” x)| = cos? a. Integrating and simpli-

fying yields ycos* ô = 4(2 + cos* x) sinx + Â or 3ycos*x = c+ 3sinz ~ sin? z.
d
23. The equation in standard form is a
:
dx

[6a/(x? + a)]y = 20(x? + a*). Thus the integrating

factor is

exp(-6 / wept)

= exp{—3ln (2? + a?)] = (x? + a?)7?.

Multiplying by the integrating factor we have < [y(a? + a?)3}

Integrating and

* (a? $a)?
simplifying yields y(x? + a7)~* = —(a? + a?)~1 + ¢ or y = (x? + a*)"[e(a? + a?) — 1),
25. If n = 0, the equation is separable so fa

= of

x
zr+a

dx or y = ba +e~-abln |x + al.

If nm = —1, the equation in standard form is a —y/(x@+a) = be/(x+a). Thus the integrating
factor is

exp (-/


= exp[—In(z +4)} = (2 +a)7}.

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CHAPTER 2.

14

d

Multiplying by the integrating factor we have que

EQUATIONS OF ORDER ONE

+a)7}}

_

bx

Integrating and

~ (x-+a)?"

simplifying yields y(x+a)~! = bIn|x + al+ab/(x+a)+c or y = ab+e(x+a)+b(2+a) In |x + al.

dy
27. The equation in standard form is in

[1/(Qx + 3)]ly = (22 +3)-/?.

Thus the integrating

factor is

a

on (-/

+ 3)] = (22 +3)7¥/?,
3) = exp[—4 In (2x
d

Multiplying by the integrating factor we have 5, lv(ant3)-

2) = (20+3)~!. Integrating yields

y(2z +3)~1/? = 4 In (2x + 3) +c. Substituting initial conditions gives 0 = § In(~2+3)+c or

c= 0 so 2y = (2c+ 3)'/?
In (2x + 3).
29, The equation in standard form is “ + (R/L)i = E/L. Thus the integrating factor is
exp (cR/2) fz)

= exp (Rt/Z).

Multiplying by the integrating factor we have < [iexp (Rt/L)] = (£/L) exp (Rt/L). Integrating yields iexp(Rt/L)

= (E/L)(L/R)exp({Rt/L) +c.
R

O= E/R+core=—(E/R) soi =5

Substituting initial conditions gives

E - exp (-F

|

d
31. The equation in standard form is = + 2y = 4x. Thus the integrating factor is
exp (22)

= exp 22.

d
Multiplying by the integrating factor we have qa (ve)

= 4xe**.

.
Integrating yields ye** =

Qne?* — e** + ¢. Substituting initial conditions gives -1 = —-1+corc=0soy=2z—-1.
d.
:
33. The equation in standard form is = + [(2t3)/(1 + t*)|s = 6t(1 + ¢?). Thus the integrating
factor is

exp { 2
.

‘

°
1+
:

dt ) = exp {t? — In(t? + 1)] =
d

Multiplying by the integrating factor we have i
ee

815 si}

= 3e°

2

+c.

et

say

e .
i +1
?

= 6te’.

Substituting the initial conditions gives 2 = 3+c¢

s = (1+ t*)[3 — exp (—¢?)).

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.

:

Integrating yields
orc = ~1 so


2.6.

THE GENERAL SOLUTION OF A LINEAR EQUATION

Miscellaneous

15

Exercises

1. The equation is separable.
Separating the variables we have,
few

ss j[ era.

Integrating both sides yields, eƠ = $e* + Â. Thus 2eƠ = e?* +c.

3. The equation is linear in y.
The equation in standard form is a + [3/(a + 1)]y = 4/(c + 1)”. Thus the integrating factor
1s

exp (3 / =)

= exp (31n (2 + 1)} = (2 +1)°.
d

Multiplying by the integrating factor we have que

+1)5]

= 4(2+ 1).

simplifying yields y(x + 1)? = 2(2 +1)? + cor y = (x41)!

Integrating and

+e(24+1)79.

5. The equation is homogeneous.

Substituting

y

=

xv,

we

have

(v*z*)dz

+ (~22?

— 3x?u)(vde

+ zdv)

=

0

or

(uv? ~ 2u ~ 3u”) dz = (2+ 3v)xdu. We first separate the variables to obtain
de

,

f3v+2,

,

2

1

patfapeentl Gta) &
Integrating yields ~ Ina = Inv + $In(v+1)+c or z*v?(v +1) = c. Substituting for v leaves

a*(y/x)*[(y/z) + 1] = ¢ or yx + y) = on.

7. The equation is linear in y.

a
The equation in standard form is = + (2x)y = x3. Thus the integrating factor is

exp ( / 2x ax) = exp (x),
Multiplying by the integrating factor we have lle)
yields y(e™’) Fe 1 (p22 - ev’) +c,

=e" 23. Integrating and simplifying

= $(x?~1)+ ce~™",

Substituting initial conditions gives 2 = ce~! or c = 2e so 2y = x? ~ 14 4exp(1 — 2”).
9. The equation is homogeneous.
Substituting y
=
xv, we have «v(t + 3ev)dr + a2(vde
(vu + 3u? + v)de = —adv. We first separate the variables to obtain

- [2 -fate-i/
ze

fj 3v2?+22u

?

1
»

3

+

xdv)

=

0

or

NG

Bupa)"

Integrating yields —Inz = 3{Inv—1n (3u + 2)|+¢ or x?v = c(3v +2). Substituting for v leaves

x*(y/x) = c[3(y/x) + 2] or x*y = c(2x + 3y).

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CHAPTER 2.

16

EQUATIONS

OF ORDER

ONE

=

0 or

il. The equation is separable.

Separating the variables we have,

dy

x4

[f-/aee
Integrating both sides yields, ~gy3 =m za% —a-+arctangz +c.

Thus 23y3 + 1 = y3(c + 32 — 3arctan 2).
13. The equation is separable.

Separating the variables we have,
dz

me SE / cos’ tdz.
cosx

Integrating both sides yields, In|seca + tanz| = $¢+ ; sin 2t +c.
Thus 41n| seca + tanz| = 2¢+ sin 2i+c.
15. The equation is homogeneous.
Substituting y = av, we have

(2°

— 2*v*)dx

— (2?v + x*)(ude + adv)

(vy ~v? — v? — v) de = x(v + 1) dv. We first separate the variables to obtain
ax
-{G-

vt
ye

1
aw =3

[

1
1
(3 +53)

Integrating yields —Inc = }(Inv — v7!) + ¢ or 1/v = Incx*v.

t/y = In|cx*(y/z)| or x = yln|ecy).

du

Substituting for v leaves

17. The equation is exact.

Let a

= 7+2y. Integrating with respect to 7 we have F = x?/2+2cy+T(y).

with respect to y yields 3

Differentiating

= 9n+T"(y) = 22 +y so Ty) = 7/2. Thus 2? + 4cy + yi=e.

19. The equation is exact.

Ee
Let a = £°+y3, Integrating with respect to z we have F = x‘/ 4+y%2+T(y).

with respect to y yields %

Differentiating

= 3y2x+T"(y) = 3y22 + ky? so T(y) = ky*/4.

Thus ky! + 4zy? + 24 =.
21. The equation is exact.

Let a

= 3y + 2zy'. Integrating with respect to x we have F = 3xry + xy? + T(y).

0

entiating with respect to y yields %

Differ-

mt

= 3a + 327y? + T'(y) = 327y? + 8x -— 3 so Ty) = —-3y.

Thus 27y? = 3(¢ + y — xy).

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2.6.

THE GENERAL SOLUTION OF A LINEAR EQUATION

17

23. The equation is linear in y.

d
‘The equation in standard form is = + (a)y = 6. Thus the integrating factor is
exp ¢ /az)

= exp ar.

d

Multiplying by the integrating factor we have qe

} = be**.

Integrating and simplifying

yields yeđđ = (b/a)e** + Â or y = b/a + ce7*™.
The equation is also separable.

Q

yo. / dz. Integrating both sides yields —2 In (b — ay) =

Separating the variables gives
x+e.

Thus y = b/a+ce~™*.

25, The equation is exact.
OF
Let a

b-ay

= siny — ysing.

.

.

.

‘

Integrating with respect to z we have f = rsiny + ycosx + T(y).

OF
Differentiating with respect to y yields —T(y) = 0. Thus xsiny + ycos¢ = ce.
aq. The equation is linear in y.

d

The equation in standard form is -

= rcosy + casx + T’(y)

by

— (2cot x)y = sin? x. Thus the integrating factor is

exp(-2 / cot de

= exp[~2In (sinz)| = sin7? «.

Multiplying by the integrating factor we have

plifying yields y(sin~?xz) = —cosxz +c.
~ cos (7/2) +eorc=1s0y

= cosx + xcosy so

dz

[y(sin™? x)|

= sing.

Integrating and sim-

Substituting initial conditions gives, 1/sin? (7/2) =

= 2sin?x sin? $a.

29. The equation is separable.
Separating the variables we have,
/

dz

/

Vi — 22

dy

fi —y

Integrating both sides yields, arcsinz + arcsin y = c, or a part of the ellipse x? + 2eyry + y? +

cf — 1 = 0; where c; = cose.
31. Substituting, arcsinO = — aresin /3/2+c¢ or 0 = —7/34+c. arcsina + arcsiny = RT, or that
arc of the ellipse x? + zy + y? = 3 that is indicated by a light solid line in Figure 2.4.
35. The equation is linear in x.

The equation in standard form is

dz

dy

(1/y)z = 2/y*. Thus the integrating factor is

exp (-[%)

=exp(—Iny) = y7?}.

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CHAPTER 2.

18

d
Multiplying by the integrating factor we have qe)

EQUATIONS

OF ORDER

ONE

= 2y7?. Integrating and simplifying

yields cy7! = -y7? +¢ or zy = cy? — 1.
i.

37. The equation is linear in y.

d
The equation in standard form is = — (tanaz)y = cosz. Thus the integrating factor i
exp(- / tanz az)

= exp [In (cos x)] = cos x.

Multiplying by the integrating factor we have ay cosz) = cos’ x. Integrating and simplifying
yields ycosx = 3(cosz sing
+ x) +c or 2y = sinz + (x +c) seco.
39.

The equation is linear in y.

d

The equation in standard form is = + [2/(1 —2*)iy = (1 ~ 32? + 224)/(1 — 2).

Thus the

integrating factor is

exp (/ ; —2

az)

= exp[~jln(1 ~2*)) = (1- a2)" U2,

ge

d

Multiplying by the integrating factor we have

ve

— gy)

= (1

1 —

ar

2x?

— g2)i/2°

Integrating

and simplifying yields y(1 — £2)~!/? = aresina + 2/1 — x? — arcsinx + ¢ or

yea—a+c(1—2*)/2,

41. The equation is linear in y.

d
.
.
:
The equation in standard form is i +(tanx)y = secz. Thus the integrating factor is

exp (/ tan edz)

= exp [In (sec x)] = seca.
d

Multiplying by the integrating factor we have an

eps

sec x) = sec” x. Integrating and simplifying

yields ycos™' 2 = tanz + ¢ or y = sina + ccosz.
43. The equation is linear in y.

d

:

The equation in standard form is = +[(1/x) ~ tanazly = 1. Thus the integrating factor is

exp (fore) — tan 2] az) = exp (Inz ~— In (sec x)| = a cos =.
d
Multiplying by the integrating factor we have a

cosz) = xcosz.

fying yields ry cosa = c + cosx + rsinz.

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Integrating and simpli-