SCHAUMS
SOLVED PROBLEMS SERIES
2500 SOLVED PROBLEMS IN
DIFFERENTIAL
EQUATIONS
•
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solutions for college and university students.
• Solutions are worked out step-by-step, are easy
to follow, and teach the subject thoroughly.
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SCHAUM'S SOLVED
PROBLEMS SERIES
2500 SOLVED PROBLEMS IN
DIFFERENTIAL
EQUATIONS
by
Richard Bronson, Ph.D.
Fairleigh Dickinson University
SCHAUM'S OUTLINE SERIES
McGRAW-HILL PUBLISHING COMPANY
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#
Richard Bronson, Ph.D., Professor of Mathematics and Computer
Science at Fairleigh Dickinson University.
Dr. Bronson, besides teaching, edits two mathematical journals and has
written numerous technical papers. Among the books he has published
are Schaum's Outlines in the areas of differential equations, operations
research, and matrix methods.
Other Contributors to This Volume
# Frank Ayres, Jr., Ph.D., Dickinson College
I James Crawford, B.S., Fairleigh Dickinson College
# Thomas M. Creese, Ph.D., University of Kansas
f Robert M. Harlick, Ph.D., University of Kansas
f Robert H. Martin, Jr., Ph.D., North Carolina State University
I George F. Simmons, Ph.D., Colorado College
I Murray R. Spiegel, Ph.D., Rensselaer Polytechnic Institute
I
C. Ray Wylie, Ph.D., Furman University
Project supervision by The Total Book.
Library of Congress Cataloging-in-Publication Data
Bronson, Richard.
2500 solved problems in differential equations / by Richard
Bronson.
p.
cm.
— (Schaum's solved problems
series)
ISBN 0-07-007979-X
1.
Differential equations
— Problems,
exercises, etc.
I.
Title.
II. Series.
QA371.B83 1988
88-17705
515.3'5'076—dc 19
CIP
2 3 4 5 6 7 8 9
SHP/SHP 8 9
*
ISBN D-D7-DD7T7T-X
Copyright
© 1989 McGraw-Hill, Inc. All rights reserved. Printed in the United
Except as permitted under the United States Copyright Act
of 1976, no part of this publication may be reproduced or distributed in any form
States of America.
or by any means, or stored in a data base or retrieval system, without the prior
written permission of the publisher.
CONTENTS
Chapter 1
BASIC CONCEPTS
1
Classifications / Formulating proportionality problems / Problems involving
Newton's law of cooling / Problems involving Newton's second law of motion /
Spring problems / Electric circuit problems / Geometrical problems / Primitives /
Chapter 2
SOLUTIONS
19
Validating solutions / Primitives / Direction fields / Initial and boundary
conditions / Particular solutions / Simplifying solutions /
Chapter 3
SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS
37
Solutions with rational functions / Solutions with logarithms / Solutions with
transcendental functions / Homogeneous equations / Solutions of homogeneous
equations
Chapter 4
/
Miscellaneous transformations
/
Initial-value problems /
EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS
66
Testing for exactness / Solutions of exact equations / Integrating factors
Solution with integrating factors / Initial-value problems /
Chapter 5
/
LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
92
Homogeneous equations / Nonhomogeneous equations / Bernoulli equations /
Miscellaneous transformations
Chapter 6
/
Initial-value problems /
APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS
110
Population growth problems / Decay problems / Compound-interest problems /
Cooling and heating problems / Flow problems / Electric circuit problems /
Mechanics problems / Geometrical problems /
Chapter 7
LINEAR DIFFERENTIAL EQUATIONS^THEORY OF SOLUTIONS
149
Wronskian / Linear independence / General solutions of homogeneous
/ General solutions of nonhomogeneous equations /
equations
Chapter 8
LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH
166
CONSTANT COEFFICIENTS
Distinct real characteristic roots
/
Distinct complex characteristic roots
Distinct real and complex characteristic roots
Characteristic roots of various types
Chapter 9
/
/
Repeated characteristic roots /
Euler's equation /
/
THE METHOD OF UNDETERMINED COEFFICIENTS
191
Equations with exponential right side / Equations with constant right-hand side /
Equations with polynomial right side / Equations whose right side is the product
of a polynomial and an exponential / Equations whose right side contains sines
and cosines / Equations whose right side contains a product involving sines and
cosines / Modifications of trial particular solutions / Equations whose right side
contains a combination of terms /
Chapter 10
VARIATION OF PARAMETERS
232
Formulas / First-order differential equations / Second-order differential
equations
Chapter 11
/
Higher-order differential equations
/
APPLICATIONS OF SECOND-ORDER LINEAR DIFFERENTIAL
255
EQUATIONS
/ Mechanics problems / Horizontal-beam problems / Buoyancy
problems / Electric circuit problems /
Spring problems
iii
iv
CONTENTS
Chapter 12
LAPLACE TRANSFORMS
283
Transforms of elementary functions / Transforms involving gamma functions /
Linearity / Functions multiplied by a power of the independent variable /
Translations / Transforms of periodic functions /
Chapter 13
INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING
DIFFERENTIAL EQUATIONS
Inverse Laplace transforms by inspection
translations
/
Chapter 14
/
MATRIX METHODS
Finding e At
Chapter 15
/ Completing the square and
Convolutions
/
/ Solutions using
Linearity
/
Partial-fraction decompositions
Laplace transforms
/
306
337
Matrix differential equations / Solutions
/
INFINITE-SERIES SOLUTIONS
354
Recursion formulas /
/ Ordinary and singular points
Solutions to homogeneous differential equations about an ordinary point /
Solutions to nonhomogeneous differential equations about an ordinary point /
Initial-value problems / The method of Frobenius / Bessel functions /
Analytic functions
Chapter 16
EIGENFUNCTION EXPANSIONS
Sturm-Liouville problems / Fourier series
functions / Sine and cosine series /
415
/
Parseval's identity
/
Even and odd
To the Student
This collection of solved problems covers analytical techniques for solving differential equations.
It is
meant to
be used as both a supplement for traditional courses in differential equations and a reference book for engineers
and scientists interested in particular applications. The only prerequisite for understanding the material in this
book is calculus.
The material within each chapter and the ordering of chapters are standard. The book begins with methods
for solving first-order differential equations and continues through linear differential equations. In this latter
category we include the methods of variation of parameters and undetermined coefficients, Laplace transforms,
matrix methods, and boundary-value problems. Much of the emphasis is on second-order equations, but
extensions to higher-order equations are also demonstrated.
Two chapters are devoted exclusively to applications, so readers interested in a particular type can go directly
to the appropriate section.
chapters.
Problems in these chapters are cross-referenced to solution procedures in previous
By utilizing this referencing system, readers can limit themselves to just those techniques that have
value within a particular application.
CHAPTER 1
Basic Concepts
CLASSIFICATIONS
1.1
Determine which of the following are ordinary differential equations and which are partial differential equations:
+ 2y =
^+3^
dx
dx
= + x—
—
ox
oy
<«)
2
dz
dz
z
(b)
I Equation (a) is an ordinary differential equation because it contains only ordinary (nonpartial) derivatives;
{b) is
1.2
a partial differential equation because it contains partial derivatives.
Determine which of the following are ordinary differential equations and which are partial differential equations:
=3
(a)
xy' + v
(b)
/" + 2(y") 2 + y = cos x
2
2
/x
d z
d z
Ix-
8?
+
2
(C)
= X 2 +y
I Equations (a) and (b) are ordinary differential equations because they contain only ordinary derivatives; (c) is
a partial differential equation because it contains at least one partial derivative.
1.3
Determine which of the following are ordinary differential equations and which are partial differential equations:
dx
(a)
-f-
= 5x + 3
dx
(c)
I
1.4
d3 v
d2 y
dx
dx
=
4-4i + (sinx)-4+5xv
1
All three equations are ordinary differential equations because each contains only ordinary derivatives.
Determine which of the following are ordinary differential equations and which are partial differential equations:
d 2 y\ 3
„
2
Jdy\,2
2
d y
(c)
(dy\
d y
xy 2 + 3xy — 2x 3 y = 1
I Equation (a) is an ordinary differential equation, while (b) is a partial differential equation. Equation (c) is
neither, since it contains no derivatives, it is not a differential equation of any type.
It is
an algebraic equation
in x and y.
1.5
Determine which of the following are ordinary differential equations and which are partial differential equations:
(a)
(b)
(c)
+ 2y = 3x 3 — 5
xy
- 2x + 3y 2 =
e
(2x-5y) 2 = 6
(sin x)y
2
f None of these equations is a differential equation, either ordinary or partial, because none of them involves
derivatives.
.
CHAPTER 1
2
1.6
Determine which of the following are ordinary differential equations and which are partial differential equations:
dy
ax
(y")
(b)
+ (y') 3 + 3y = x 2
2
I Both are ordinary differential equations because each contains only ordinary derivatives.
1.7
Define order for an ordinary differential equation.
# The order of a differential equation is the order of the highest derivative appearing in the equation.
1.8
Define degree for an ordinary differential equation.
I
If an
ordinary differential equation can be written as a polynomial in the unknown function and its
derivatives, then its degree is the power to which the highest-order derivative is raised.
1.9
Define linearity for an ordinary differential equation.
I An nth-order ordinary differential equation in the unknown function y and the independent variable x is
linear if it has the form
x
d"~ v
d"v
bJLx)
The functions bj(x)
(j
^+
— 0, 1,2, ... ,n)
b„- ,(x)
j^ +
dv
+ 6,(x) -£ + b (x)y = g(x)
and g(x) are presumed known and depend only on the variable x.
Differentiil equations that cannot be put into this form are nonlinear.
1.10
Determine the order, degree, linearity, unknown function, and independent variable of the ordinary differential
y" — 5xy' — e x +
equation
I
1.
Second order: the highest derivative is the second. The unknown function is y, and the independent variable
First degree: the equation is written as a polynomial in the unknown function y and its derivatives,
with the highest derivative (here the second) raised to the first power. Linear: in the notation of Problem
is
x.
b 1 (x)=-5x,
b 2 (x)=l,
1.11
b o(x) = 0,
and
g(x)
Determine the order, degree, linearity, unknown function, and independent variable of the ordinary differential
y'"
equation
— 5xy' — e x + 1.
I
Third order: the highest derivative is the third. The unknown function is y, and the independent variable is
x.
First degree: the equation is a polynomial in the unknown function y and its derivatives, with its highest
derivative (here the third) raised to the first power.
6,(x)
1.12
= -5x,
b 2 (x) = b (x) = 0,
and
g(x)
Linear: in the notation of Problem
1.9.
b 3 (x) = 1,
= e* + 1.
Determine the order, degree, linearity, unknown function, and independent variable of the differential equation
y - 5xy' = e
x
+
1
I
First order: the highest derivative is the first.
x.
First degree: the equation is a polynomial in the unknown function y and its derivative, with its highest
The unknown function is y. and the independent variable is
derivative (here the first) raised to the first power.
b (x) = 1,
1.13
1.9,
= e*+l.
and
#(x)
Linear: in the notation of Problem
1.9,
b^x) = — 5x.
= e x + 1.
Determine the order, degree, linearity, unknown function, and independent variable of the differential equation
y-5x(y') 2 = e*+l.
t
First order: the highest derivative is the first. The unknown function is y, and the independent variable is x.
Second degree: the equation is a polynomial in the unknown function y and its derivative, with its highest
derivative (here the first) raised to the second power.
Nonlinear: the derivative of the unknown function is
raised to a power other than the first.
1.14
Determine the order, degree, linearity, unknown function, and independent variable of the differential equation
y - 5x(y')
I
4
= e x + 1.
First order: the highest derivative is the first.
The unknown function is y. and the independent variable is x.
Fourth degree: the equation is a polynomial in the unknown function y and its derivative, with its highest
BASIC CONCEPTS
derivative (here the first) raised to the fourth power.
D
Nonlinear: the derivative of the unknown function is raised
to a power other than the first.
1.15
Determine the order, degree, linearity, unknown function, and independent variable of the differential equation
y" - 5x(y') 4 = e x +
1.
I
Second order, the highest derivative is the second. The unknown function is y, and the independent variable
First degree: the equation is a polynomial in the unknown function y and its derivatives, with its highest
is x.
derivative (here the second) raised to the first power.
is
1.16
Nonlinear: one of the derivatives of the unknown function
raised to a power other than the first.
Determine the order, degree, linearity, unknown function, and independent variable of the differential equation
(y")
3
- 5x(y') 4 = e*+l.
I Second order: the highest derivative is the second. The unknown function is y, and the independent variable
is x.
Third degree: the equation is a polynomial in the unknown function y and its derivatives, with the highest
derivative (here the second) raised to the third power. Nonlinear: one of the derivatives of the unknown function
is
1.17
raised to a power other than the first.
Determine the order, degree, linearity, unknown function, and independent variable of the differential equation
(y'")
3
- 5x(y') 4 = e x +1.
I
Third order: the highest derivative is the third.
x.
Third degree: the equation is a polynomial in the unknown function y and its derivatives, with the highest
The unknown function is y, and the independent variable is
Nonlinear: one of the derivatives of the unknown function
derivative (here the third) raised to the third power.
is
1.18
raised to a power other than the first.
Determine the order, degree, linearity, unknown function, and independent variable of the differential equation
y" - 5x 2 y' =ex + 1.
I Second order: the highest derivative is the second. The unknown function is y, and the independent variable
is x. First degree: the equation is written as a polynomial in the unknown function y and its derivatives, with its
Linear: in the notation of Problem
highest derivative (here the second) raised to the first power.
2
b^x) = 5x
1.19
,
b (x) = 0,
I
1.21
b 2 (x) = 1,
and
Determine the order, degree, linearity, unknown function, and independent variable of the differential equation
y" - 5t 2 y' = e' +
1.20
1.9,
g(x) = e* + 1.
1.
This problem is identical to Problem 1.18, with the single exception that now the independent variable is t.
Determine the order, degree, linearity, unknown function, and independent variable of the differential equation
z"
- 5xV = ex +l.
I
This problem is identical to Problem 1.18, with the single exception that now the unknown function is z.
Determine the order, degree, linearity, unknown function, and independent variable of the ordinary differential
equation
c
5x
&y + 3x dy
—
dx
dx
=
.
-»
2
,
^
(sin x)y
=n
I Second order: the highest derivative is the second. The unknown function is y, and the independent variable
is x. First degree: the equation is written as a polynomial in the unknown function y and its derivatives, with
the highest derivative (here the second) raised to the first power.
b 2 (x) = 5x,
1.22
frjfx)
— 3x 2
,
b (x) =
— sinx,
and
g(x)
Linear: in the notation of Problem
1.9,
— 0.
Determine the order, degree, linearity, unknown function, and independent variable of the differential equation
A
d v
dv
—
4 + 3x -/- - (sin x)y =
dx
dx
,
5x
I
2
Fourth order: the highest derivative is the fourth. The unknown function is y, and the independent variable
is x.
First degree: the equation is a polynomial in the unknown function y and its derivatives, with the highest
derivative (here the fourth) raised to the first power.
b 3 (x) = b 2 (x) = 0,
b^x) = 3x 2
,
b (x) = —sin x,
Linear: in the notation of Problem
and
#(x) = 0.
1.9,
b 4 (x) = 5x,
3
CHAPTER 1
4
1.23
Determine the order, degree, linearity, unknown function, and independent variable of the differential equation
4
d v
,
dv
5r-^ + 3r 2 ^-(sinriy =
f
This problem is identical to Problem 1.22, with the single exception that now the independent variable is t
rather than x.
1.24
Determine the order, degree, linearity, unknown function, and independent variable of the differential equation
I
t.
Fourth order, the highest derivative is the fourth. The unknown function is y, and the independent variable is
First degree: the equation is a polynomial in the unknown function y and its derivatives, with its highest
Nonlinear: one of the derivatives of the unknown function is
derivative (here the fourth) raised to the first power.
raised to a power other than the first.
1.25
Determine the order, degree, linearity, unknown function, and independent variable of the differential equation
'V
I
t.
,, 2
/M
.....
Fourth order: the highest derivative is the fourth. The unknown function is y, and the independent variable is
First degree: the equation is a polynomial in the unknown function y and its derivatives, with its highest
Nonlinear: one of the derivatives of the unknown function
derivative (here the fourth) raised to the first power.
(as well as the
1.26
3
unknown function itself) is raised to a power other than the first.
Determine the order, degree, linearity, unknown function, and independent variable of the differential equation
3f
I
2
First order: the highest derivative is the first.
(^Y-(sinf)y 6 =
The unknown function is y, and the independent variable is t.
Third degree: the equation is a polynomial in the unknown function y and its derivative, with its derivative
raised to the third power.
Nonlinear: one of the derivatives of the unknown function y (as well as y itself) is raised
to a power other than the first.
1.27
Determine the order, degree, linearity, unknown function, and independent variable of the differential equation
f
t.
Second order: the highest derivative is the second. The unknown function is y, and the independent variable is
Sixth degree: the equation is a polynomial in the unknown function y and its derivatives, with the highest
derivative (here the second) raised to the sixth power.
Nonlinear: at least one of the derivatives of the unknown
function is raised to a power higher than the first.
1.28
Determine the order, degree, linearity, unknown function, and independent variable of the differential equation
2
3
'
f
d 3y
#-
2 x*
,_.__Jd y
(sin
VJ =°
Third ordpr: the highest derivative is the third. The unknown function is y, and the independent variable is t.
First degree: the equation is a polynomial in the unknown function y and its derivatives, with its highest
derivative (here the third) raised to the first power.
is
1.29
Nonlinear: one of the derivatives of the unknown function y
raised to a power higher than the first.
Determine the order, degree, linearity, unknown function, and independent variable of the differential equation
3t
I
2
d'y
d 2y
..
A -^
- (sin
- (cos i)y =
t)
-pf
Third order: the highest derivative is the third. The unknown function is y, and the independent variable is r.
First degree: the equation is a polynomial in the unknown function y and its derivatives, with the highest
BASIC CONCEPTS
derivative (here the third) raised to the first power.
b 2 (t) = -sin t,
1.30
6,(r)
= 0,
b (t) =
-cos t,
and
Linear, in the notation of Problem
g(t)
1.9,
b 3 (t) = 3t
5
2
,
= 0.
Determine the order, degree, linearity, unknown function, and independent variable of the differential equation
**y
d2y
, 2
3r
—y - (sin t) —-y - cos fy =
I
,
Third order, the highest derivative is the third. The unknown function is y, and the independent variable is t.
No degree: the equation cannot be written as a polynomial in the unknown function and its derivatives, because
the unknown function y is an argument of the transcendental cosine function; degree is therefore undefined.
Nonlinear, the unknown function is an argument of a transcendental function.
1.31
Determine the order, degree, linearity, unknown function, and independent variable of the differential equation
5y + 2e'y -3y = t.
I
t.
Second order, the highest derivative is the second. The unknown function is y, and the independent variable is
First degree: the equation is a polynomial in the unknown function y and its derivatives, with the highest
derivative (here the second) raised to the first power.
b 1 (t) = 2e\
1.32
b (t)=-3,
and
g{t)
=
Linear: in the notation of Problem
1.9,
b 2 (t) = 5,
t.
Determine the order, degree, linearity, unknown function, and independent variable of the differential equation
5y + 2e
I
t.
-3y =
,j!
t.
Second order: the highest derivative is the second. The unknown function is y, and the independent variable is
No degree: the equation cannot be written as a polynomial in the unknown function y and its derivatives,
because one of its derivatives (namely, y) is an argument of the transcendental exponential function; degree is
therefore undefined.
Nonlinear: at least one derivative of the unknown function is an argument of a transcendental
function.
1.33
Determine the order, degree, linearity, unknown function, and independent variable of the differential equation
5y - 3yy = t
I
.
First order: the highest derivative is the first.
The unknown function is y, and the independent variable is t.
First degree: the equation is a polynomial in the unknown function y and its derivative, with its derivative raised
to the first power.
1.34
Nonlinear: the unknown function y is multiplied by its own derivative.
Determine the order, degree, linearity, unknown function, and independent variable of the differential equation
5y - 3(y) y = t.
7
I
First order: the highest derivative is the first.
The unknown function is y, and the independent variable is t.
Seventh degree: the equation is a polynomial in the unknown function y and its derivative, with the highest power
of its derivative being the seventh.
Nonlinear: the unknown function y is multiplied by its own derivative; in
addition, the derivative of y is raised to a power other than the first.
1.35
Determine the order, degree, linearity, unknown function, and independent variable of the differential equation
5y - 3yy
I
7
=
t.
First order: the highest derivative is the first.
The unknown function is y, and the independent variable is t.
First degree: the equation is a polynomial in the unknown function y and its derivative, with the derivative raised
to the first power.
Nonlinear: the unknown function y is raised to a power other than the first (as well as being
multiplied by its own derivative).
1.36
Determine the order, degree, linearity, unknown function, and independent variable of the differential equation
5z - liz
I
1.37
1
=
t.
This problem is identical to Problem 1.35, with the single exception that now the unknown function is z.
Determine the order, degree, linearity, unknown function, and independent variable of the differential equation
ty
+ t 2 y - (sin t)y/y - t2 — t + 1.
I
Second order: the highest derivative is the second. The unknown function is v, and the independem variable is
t.
No degree: because of the term <Jy, the equation cannot be written as a polynomial in y and its derivatives.
in this case the one-half power.
Nonlinear: the unknown function y is raised to a power other than the first
—
CHAPTER 1
6
1.38
Determine the order, degree, linearity, unknown function, and independent variable of the differential equation
+b -b = p
7
5(tt) + 1 \--\
I
Fourth order. The unknown function is b\ the independent variable is p. Fifth degree: the highest (fourth)
derivative is raised to the fifth power.
1.39
5
Nonlinear,
Determine the order, degree, linearity, unknown function, and independent variable of the differential equation
,
s
d
2
dt
t
+ st
d?
Ts=
s
I Second order. The unknown function is the independent variable is s. First degree: the equation is a
polynomial in the unknown function t and its derivatives (with coefficients in s), and the second derivative is
t;
Nonlinear: in the notation of Problem
raised to the first power.
1.40
b,
1.9,
= sf,
which depends on both s and t.
Determine the order, degree, linearity, unknown function, and independent variable for the differential equation
d 2x
dy l
I Second order. The unknown function is x; the independent variable is y.
2
of Problem 1.9,
b 2 (y) = y,
b l (y) = 0, b o(y) = 0, and g{y) = y + 1.
1.41
First degree.
Linear: in the notation
Determine the order, degree, linearity, unknown function, and independent variable for the differential equation
(/')
2
-
W
+ xy = 0.
I Second order because the highest derivative is the second, and second degree because this derivative is raised
to the second power. The unknown function is y, and the independent variable is x. Nonlinear because one of
the derivatives of y is raised to a power other than the first; in addition, the unknown function is multiplied by
one of its own derivatives.
1.42
Determine the order, degree, linearity, unknown function, and independent variable for the differential equation
x4y(4) + xyl3) =
I
-
Fourth order because the highest derivative is the fourth, and first degree because that derivative is raised to
the first power.
Problem 1.9,
1.43
x
<'
The unknown function is y. and the independent variable is x.
b 4 (x) = v
4
.
b 3 (x) = x,
b 2(x) = b 1 (x) = b o(x)
= 0,
and
Linear: in the notation of
g(x) = e
x
.
Determine the order, degree, linearity, unknown function, and independent variable for the differential equation
y
+ xy (3) + x 2 y" — xy' + sin y = 0.
(4)
I
Fourth order: the highest derivative is the fourth. The unknown function is y, and the independent variable is
x.
No degree and nonlinear because the unknown function is the argument of a transcendental function, here the
sine function.
1.44
Determine the order, degree, linearity, unknown function, and independent variable for the differential equation
2
t
's
— ts =
1
— sin
f.
I Second order: the highest derivative is the second. The unknown function is s, and the independent variable
is
First degree: the equation is a polynomial in the unknown function s and its derivatives, with its highest
t.
Linear: in the notation of Problem
derivative (here the second) raised to the first power.
b (t)=-t,
l
1.45
b (t) = 0,
and
g(t)
=
1
- sin
1.9,
h 2 (t) = t
2
,
t.
Determine the order, degree, linearity, unknown function, and independent variable for the differential equation
d 2 r\ 2
d?)
2
+
d r
d?
+y
dr
n
=°
dy
I Second order: the highest derivative is the second. The unknown function is r, and the independent variable
Second degree: the equation is a polynomial in the unknown function r and its derivatives, and the highest
power of the highest derivative is the second. Nonlinear: one of the derivatives of the unknown function is raised
is y.
to a power other than the first.
.
BASIC CONCEPTS
1.46
Determine the order, degree, linearity, unknown function, and independent variable for the differential equation
n
d x/dy
n
= y2 +
1.
f For the derivative to make sense, n must be a nonnegative integer.
If n is positive,
nth order and first degree because this derivative is raised to the first power.
the independent variable is y.
i>i(y) = b (y) = 0,
1.47
7
and
Linear, in the notation of Problem
2
g(y) = y + 1.
If
n = 0,
The unknown function is x, and
b n (y) —
1.9,
then the equation is of
1,
bn
,(>)
= b„
2 (}')
=
'
'
'
=
the equation is algebraic.
Determine the order, degree, linearity, unknown function, and independent variable for the differential equation
(d
2
y/dx 2 ) 3 2 + y = x.
'
f Second order: the highest derivative is the second. The unknown function is y, and the independent variable is x.
No degree because the equation cannot be written as a polynomial in the unknown function and its derivatives;
the 3/2 power precludes such a possibility. Nonlinear: a derivative of the unknown function is raised to a power
other than the first.
1.48
Determine the order, degree, linearity, unknown function, and independent variable for the differential equation
7
d b/dp
/
= 3p.
Seventh order since the highest derivative is the seventh, and first degree since that derivative is raised to the
first
power. The unknown function is b, and the independent variable is p. Linear: in the notation of Problem
b n (p) = 1,
1.9,
1.49
7
b 6 (p) = b 5 (p) =
•
•
•
= b (p) = 0, and
g(p)
= 3p.
Determine the order, degree, linearity, unknown function, and independent variable for the differential equation
(dp/db)
I
7
= 3b.
First order since the highest derivative is the first, and seventh degree since that derivative is raised to the
seventh power.
The unknown function is p, and the independent variable is b. Nonlinear because one of the
derivatives of the unknown function is raised to a power other than the first.
1.50
Must a linear ordinary differential equation always have a degree?
1
1.51
Yes, and the degree is always
If an
1
because the highest-order derivative is always raised to the first power.
ordinary differential equation has a degree, must it be linear?
I Nc Counterexamples are provided by Problems 1.45 and 1.49.
FORMULATING PROPORTIONALITY PROBLEMS
1.52
Radium decomposes at a rate proportional to the amount present. Derive a differential equation for the amount
of radium present at any time t
I
Let R(t) denote the amount of radium present at time t.
proportional to R. Thus,
1.53
dR/dt = kR,
The decomposition rate is dR/dt, which is
where k is a constant of proportionality.
Bacteria are placed in a nutrient solution at time
t
=
and allowed to multiply. Under conditions of plentiful
food and space, the bacteria population grows at a rate proportional to the population. Derive a differential
equation for the approximate number of bacteria present at any time t.
I
Let N(t) denote the number of bacteria present in the nutrient solution at time t. The growth rate is dN/dt,
which is proportional to N. Thus, dN/dt = kN, where k is a constant of proportionality.
1.54
One hundred grams of cane sugar in water is being converted into dextrose at a rate which is proportional
to the unconverted amount.
I
Find a differential equation expressing the rate of conversion after t minutes.
Let q denote the number of grams converted in f minutes. Then
1.55
Bacteria are placed in a nutrient solution at time
100 — q
dq/dt = /c(100 — q),
unconverted, and the rate of conversion is given by
t
=
is
the number of grams still
k being the constant of proportionality.
and allowed to multiply. Food is plentiful but space is
M. Derive a differential equation
known that the growth rate of the bacteria is
limited, so ultimately the bacteria population will stabilize at a constant level
for the approximate number of bacteria present at any time t if it
is
jointly proportional to both the number of bacteria present and the difference between
population.
M and the current
CHAPTER 1
8
f Denote the number of bacteria present in the nutrient solution at time t by N{t). The growth rate is dN/dt.
Since this rate is jointly proportional to N and (M — N),
we have dN/dt = kN(M — N), where k is a
constant of proportionality.
1.56
Express the following proposition as a differential equation: the population P of a city increases at a rate which
is jointly
f
proportional to the current population and the difference between 200,000 and the current population.
Let P(t) denote the current population; then the rate of increase is dP/dt. Since this rate is jointly
proportional to both P and
(200,000 - P),
we have
dP/dt = /cP(200,000 - P),
where k is a constant of
proportionality.
1.57
A bank pays interest to depositors at the rate of r percent per annum, compounded continuously. Derive a
differential equation for the amount of money in an existing account at any time t, assuming no future
withdrawals or additional deposits.
I
Let P(t) denote the amount in the account at time t. Then dP/dt, the change in P, is the interest received,
which is the interest rate (converted to a decimal) times the current amount. Thus, dP/dt = (r/l00)P.
1.58
When ethyl acetate in dilute aqueous solution is heated in the presence of a small amount of acid, it decomposes
according to the equation
CH COOC H + H
3
2
(Ethyl acetate)
5
2
(water)
CH3COOH + C H OH
2
(acetic acid)
5
(ethyl alcohol)
Since this reaction takes place in dilute solution, the quantity of water present is so great that the loss of the
small amount which combines with the ethyl acetate produces no appreciable change in the total amount.
A chemical
Hence, of the reacting substances only the ethyl acetate suffers a measurable change in concentration.
reaction of this sort, in which the concentration of only one reacting substance changes, is called a first-order
reaction.
is
It is
a law of physical chemistry that the rate at which a substance is used up in a first-order reaction
proportional to the amount of that substance instantaneously present.
Find an expression for the
concentration of ethyl acetate at any time f.
I
Let Q be the amount of ethyl acetate present in the solution at time t, let
V be the (constant) amount of
water in which it is dissolved, and let C be the instantaneous concentration of the ethyl acetate. Then
Q — CV,
and, from the law governing first-order reactions,
or finally
dC/dt = -kC.
PROBLEMS INVOLVING NEWTON'S LAW OF COOLING
1.59
Newton's law of cooling states that the rate at which a hot body cools is proportional to the difference in
temperature between the body and the (cooler) surrounding medium. Derive a differential equation for the
temperature of a hot body as a function of time if it is placed in a bath which is held at a constant temperature
of 32 °F.
I Denote the temperature of the hot body at time t by T{t), and assume it is placed in the bath at
rate at which the body cools is dT/dt.
Since this is proportional to
(T — 32),
we have
t
— 0.
The
dT/dt = k(T — 32),
where k is a constant of proportionality.
1.60
C
A red-hot steel rod is suspended in air which remains at a constant temperature of 24 C. Find a differential
equation for the temperature of the rod as a function of time.
f
Denote the temperature of the steel rod at time t by T{t), and assume it is placed in the cooler medium at
The rate at which the rod cools is dT/dt. By Newton's law of cooling (see Problem 1.59), this rate is
proportional to (T — 24). Therefore, dT/dt — k(T — 24). where k is a constant of proportionality.
t
= 0.
PROBLEMS INVOLVING NEWTON'S SECOND LAW OF MOTION
1.61
Newton's second law of motion states that the time rate of change of the momentum of a body is equal to the
net force acting on that body. Derive the differential equation governing the motion of a body when the only
force acting on it is the force of gravity.
.
.
BASIC CONCEPTS
9
f
Denote the mass of the body by m, and let y{t) be the vertical distance to the body from some fixed reference
Then the velocity of the body is dy/dt, the time rate of change of position. Its momentum
dy
d 2y
d ( dy\
The time rate of change
= m -jy, if we
is its mass times its velocity, or m
( m-£
° of its momentum is
height at any time t.
,
— _
•
—
.
.
dt\
dt
]
dtj
dt'
assume its mass remains constant. The force of gravity is the only force acting on the body; it is given by mg,
2
where g denotes the acceleration due to gravity (a constant 32 ft/s or 9.8 m/s 2 close to the surface of the earth).
Thus, the required equation is
2
d y
^
m
1.62
= m9
d 2y
lT2=g
or
Derive the differential equation governing the motion of a body that is subject to both the force of gravity and
air resistance (which exerts a force that opposes and is proportional to the velocity of the body).
I
This problem is similar to Problem 1.61, except now two forces act on the body in opposite directions. The
dy
force of gravity is mg, while the force due to air resistance is
— k — where k is a constant of proportionality.
,
dt
Thus the net force on the body is
2
—
dt
dt
mg — k
dy
—
,
and it follows from Newton's second law of motion that
d y
dy
= mg — k
—tt
2
1.63
Redo Problem 1.62 if the air resistance is replaced by a force that is proportional to the square of the velocity
of the body.
I The new force is —k(dy/dt) 2 so the net force on the body is
,
now yields
—k
m -j-j = ne
mg -d7
1.64
mg — k(dy/dt) 2
.
Newton's second law of motion
[jt
A particle of mass m moves along a straight line (the x axis) while subject to (1) a force proportional to its
displacement x from a fixed point
its
I
velocity.
in its path and directed toward
and (2) a resisting force proportional to
Write a differential equation for the motion of the particle.
dx
The first force may be represented by k x, and the second by — k 2 —, where k and k 2 are factors of
x
Y
dt
—
2
d x
proportionality.
1.65
dx
m -yj- = kl
k x-k
x — k 2 —,
Newton's second law then yields
x
A torpedo is fired from a ship and travels in a straight path just below the water's surface. Derive the
differential equation governing the motion of the torpedo if the water retards the torpedo with a force
proportional to its speed.
I
Let x(r) denote the distance of the torpedo from the ship at any time t. The velocity of the torpedo is dx/dt.
The only force acting on the torpedo is the resisting force of the water, k
dx
—
where
,
k is a constant of
dt
proportionality.
If
we assume the mass of the torpedo remains constant, then its time rate of change of
d x
momentum is m —-,.
It follows from Newton's second law of motion (see Problem 1.61) that
2
dt
1.66
Inside the earth, the force of gravity is proportional to the distance from the center.
drilled through the earth from pole to pole, and a rock is dropped into the hole.
d x
dx
—
dt
dt
m —T2 = k
Assume that a hole is
Derive the differential equation
for the motion of this rock.
Let s(t) denote the distance from the rock at any time t to the center of the earth. The force of gravity is ks,
where k is a constant of proportionality, so Newton's second law of motion (see Problem 1.61) yields
i
d2 s
m — 2 — ks.
,
-r-
dt
1.67
A boat is being towed at the rate of 12 mi/h. At
t
=
the towing line is cast off and a man in the boat begins
to row in the direction of motion, exerting a force of 20 lb.
The combined weight of the man and the boat is
CHAPTER 1
10
480 lb. The water resists the motion with a force equal to 1.75r lb, where r is the velocity of the boat in feet per
second. Derive a differential equation governing the velocity of the boat.
I The boat moves along a straight line, which we take to be the x axis, with the positive direction being the
direction of motion.
Then
v
— dx/dt.
For constant mass, Newton's second law (Problem 1.61) gives us
m — = net force = forward force — resistance
dv
dt
so that
480 dv
_ _
= 20-1.75i>
,
dv
7
4
—
+— =60
or
v
32 dt
dt
We are also given the initial velocity of the boat,
= 0.
need to find the velocity at times after
v(0)
3
= 12 mi/h = 12(5280)/(60) 2 = 17.6 ft/s,
which we would
f
1.68
A mass is being pulled across the ice on a sled with a constant force. The resistance offered by the ice to the
runners is negligible, but the resistance (in pounds) offered by the air is five times the velocity of the sled.
Derive a differential equation for the velocity of the sled if the combined weight of the sled and the mass is 80 lb.
I We assume that the motion of the sled is along a straight line; we designate that line as the x axis, with
v = dx/dt.
From Newton's
the positive direction being the direction of motion. The velocity of the sled is then
m — = forward force - resistance.
dv
second law of motion (see Problem 1.61), we have
dt
m = 80/32 = 2.5 slugs. The differential equation is then
We denote the constant forward force by F, and
2.5
dv
—
= F-5v
dv
—
+ 2v = - F
2
or
dt
dt
5
SPRING PROBLEMS
1.69
Hooke's law states that the restoring force of a spring is proportional to the displacement of the spring from its
normal length. Use Hooke's law along with Newton's second law of motion to derive the differential equation
governing the motion of the following system: A spring with a mass m attached to its lower end is suspended
vertically from a mounting and allowed to come to rest in an equilibrium position.
The system is then set in
motion by releasing the mass with an initial velocity v at a distance x below its equilibrium position and
simultaneously applying to the mass an external force F(t) in the downward direction.
For convenience, we choose the downward direction as the positive direction and take the origin to be the
Furthermore, we assume that air resistance
1.1).
is present and is proportional to the velocity of the mass. Thus, at any time r, there are three forces acting on
the system: (1) F(r), measured in the positive direction; (2) a restoring force given by Hooke's law as
F s = —kx,
where k >
is a constant of proportionality known as the spring constant; and (3) a force due to air resistance
given by
F a — —ax, where a >
is a constant of proportionality. Note that the restoring force F s always
acts in a direction that will tend to return the system to the equilibrium position: if the mass is below the
equilibrium position, then x is positive and —kx is negative; whereas if the mass is above the equilibrium position,
then x is negative and —kx is positive. Also note that because a >
the force F a due to air resistance acts in
the direction opposite the velocity and thus tends to retard, or damp, the motion of the mass.
It now follows from Newton's second law that
mx = —kx — ax + F(t), or
f
center of gravity of the mass in the equilibrium position (see Fig.
-
x J.
-\
Since the system starts at
(7) the initial conditions
t
—
x(0)
a
--l* x =
xH
m
m
with an initial velocity r
=x
and
x(0)
= v
m
F{t)
(/)
m
and from an initial position x
,
we have along with
.
The force of gravity does not explicitly appear in (/), but it is present nonetheless. We automatically
compensated for this force by measuring distance from the equilibrium position of the spring. If one wishes to
exhibit gravity explicitly, then distance must be measured from the bottom end of the natural length of the spring.
That is, the motion of a vibrating spring can be given by
—mm
xH— =
a
x
if the
origin,
x = 0,
is
-\
k
x
F(t)
g H
m
the terminal point of the unstretched spring before the mass m is attached.
BASIC CONCEPTS
Equilibrium Position
Initial Position at
t
11
—
x =
F(t)
Positive x direction
1.70
Fig. 1.1
Derive the differential equation governing the motion of the spring system shown in Fig. 1.1 if the vibrations are
free and undamped.
I The vibrations are free if no external force is applied to the spring, and they are undamped if air resistance
is zero. With
F(t) —
and a — 0, (7) of Problem 1.69 reduces to x + (k/m)x = 0.
1.71
A steel ball weighing 128 lb is suspended from a spring, whereupon the spring stretches 2 ft from its natural
What is the value of the spring constant?
length.
/ The applied force responsible for the 2-ft displacement is the weight of the ball, 128 lb. Thus,
Hooke's law then gives
1.72
- 128 = -k(2),
or
F = — 128 lb.
k = 64 lb/ft.
A 32-lb weight is attached to a spring, stretching it 8 ft from its natural length. What is the value of the spring
constant?
f The applied force responsible for the 8-ft displacement is the 32-lb weight. At equilibrium, this force is
balanced by the restoring force of the spring, so by Hooke's law
1.73
— 32 = — /c(8),
or
k
= 4 lb/ft.
A mass of 1/4 slug is attached to a spring, whereupon the spring stretches 1.28 ft from its natural length. What
is
the value of spring constant?
f The applied force responsible for the 1.28-ft displacement is the weight of the attached body, which is
= 8 lb.
-8 = -fc(1.28),
(l/4)(32)
1.74
At equilibrium, this force is balanced by the restoring force of the spring, so by Hooke's law
or
k = 6.25 lb/ft.
A 10-kg mass is attached to a spring, stretching it 0.7 m from its natural length. What is the value of the spring
constant?
I The applied force responsible for the 0.7-m displacement is the weight of the attached body, which is
10(9.8)
= 9.8 N.
-98 = k{0.1),
1.75
At equilibrium, this force is balanced by the restoring force of the spring, so by Hooke's law
from which
k = 140 N/m.
A steel ball weighing 128 lb is suspended from a spring, whereupon the spring stretches 2 ft from its natural
length.
The ball is started in motion with no initial velocity by displacing it 6 in above the equilibrium position.
Derive a differential equation governing the subsequent vibrations of the spring if there is no air resistance.
f
This is an example of free, undamped motion. The spring constant was determined in Problem 1.71 to be
k = 64 lb/ft;
With these values, the result
so m — 128/32 = 4 slugs.
x -I- 16x = 0. In addition, we have the initial conditions
(the minus sign is required because the ball is initially displaced above the equilibrium position,
the weight of the ball is
of Problem 1.70 becomes
x(0)
= — 1/2 ft
mg — 128 lb,
x + (64/4)x = 0,
which is in the negative direction) and
x(0)
or
= 0.
12
1.76
CHAPTER
D
1
A 32-lb weight is attached to a spring, stretching it 8 ft from its natural length. The weight is started in motion
by displacing it 1 ft in the upward direction and giving it an initial velocity of 2 ft/s in the downward direction.
Derive a differential equation governing the subsequent vibrations of the spring if the air offers negligible
resistance.
I
This is an example of free, undamped motion. The spring constant is 4 lb/ft (see Problem 1.72), and
m = 32/32 =
conditions
1.77
slug.
The result of Problem 1.70 becomes
x(0) = -
x(0) = 2 ft/s.
1
1
ft
and
x + 4x = 0.
In addition, we have the initial
A mass of 1/4 slug is attached to a spring, whereupon the spring stretches 1.28 ft from its natural length. The
mass is started in motion from the equilibrium position with an initial velocity of 4 ft/s in the downward direction.
Derive a differential equation governing the subsequent motion of the spring if the force due to air resistance is
-2xlb.
/
This is an example of free (no external force is applied to the body) but damped (there is air resistance)
motion. Here
m = 1/4,
a = 2,
k = 6.25
2
(see Problem 1.73), and
6.25
x H
x H
1/4
In addition,
x(0)
1.78
x(0)
= 0,
x =
F(t)
— 0,
sothat(7)of Problem 1.69 becomes
x + 8x + 25x =
or
1/4
because initially the body is not displaced at all from its equilibrium position, and
= 4 ft/s.
A 10-kg mass is attached to a spring, stretching it 0.7 m from its natural position. The mass is started in motion
from the equilibrium position with an initial velocity of 1 m/s in the upward direction. Derive a differential
equation governing the subsequent motion of the spring if the force due to air resistance is — 90x N.
I Here
m = 10,
a = 90,
k = 140
Problem 1.74). and
(see
90
x H
140
x H
10
In addition,
x(0)
=
x =
or
F(t)
x
:
= 0,
so that (7) of Problem 1.69 becomes
+ 9x + 14x =
10
(the mass starts at the equilibrium position) and
x(0)
= —
1
(the initial velocity is in
the upward, or negative, direction).
1.79
Redo Problem 1.78 if, in addition, an external force 5 sin f (in newtons) is applied to the system.
I The constants m, a, and k remain as before, but now
x +
140
5sinf
90
—
x + —— x = ———
F(t)
— 5 sin
or
x
:
t
and (7) of Problem 1.69 becomes
+ 9x + 14x = jsmt
Vibrations subject to external forces are called forced vibrations.
1.80
A 128-lb weight is attached to a spring having a spring constant of 64 lb/ft. The weight is started in motion
with no initial velocity by displacing it 6 in above the equilibrium position and by simultaneously applying to
the weight an external force
F(t)
= 8 sin 4f.
Derive a differential equation governing the subsequent vibrations
of the spring if there is no air resistance.
I
This is an example of forced (there is an applied external force) but undamped (there is no air resistance)
motion. Here
m = 128/32 = 4 slugs,
k = 64 lb/ft,
a = 0,
and
F(t)
= 8 sin 4r lb,
so (7) of Problem 1.69
becomes
..
xH
64
x =
8sin4r
4
The initial conditions are
x(0)
= —\ ft
or
„
= 2sin4f
x+16x
4
and
x(0)
= 0.
ELECTRIC CIRCUIT PROBLEMS
1.81
Kirchoffs loop law states that the algebraic sum of the voltage drops in a simple closed electric circuit is zero.
Use this law to derive a differential equation for the current 7 in a simple circuit consisting of a resistor, a
capacitor, an inductor, and an electromotive force (usually a battery or a generator) connected in series.
f The circuit is shown in Fig. 1.2, where R is the resistance in ohms, C is the capacitance in farads, L is the
inductance in henries, E(t) is the electromotive force (emf) in volts, and 7 is the current in amperes. It is known
BASIC CONCEPTS
13
E(t)
Fig. 1.2
that the voltage drops across a resistor, a capacitor, and an inductor are respectively RI,
— q, and L — where
,
C
dt
The voltage drop across an emf is — E(t). Thus, from Kirchhoff's loop law,
q is the charge on the capacitor.
we have
dl
L+ -q-E(t) =
1
RI +
The relationship between q and /is
obtain
R
dl
2
d I
— + L —T + —
dt
_
dt
1
2
dE(t)
.
/
C
/
= dq/dt.
— 0,
Differentiating (7) with respect to t and using :his relation, we
which may be rewritten as
dt
d
2L
Rdl
1
1
dE(t)
2
Ldi
LC
L
dt
dt
1.82
(I)
(2)
Derive a differential equation for the charge on the capacitor in the series RCL circuit of Fig. 1.2.
I From the last problem we have
/
= dq/dt
dl/dt — d q/dt
2
and so
2
Substituting these equalities into (/)
.
of Problem 1.81 and rearranging, we obtain
2
d q
-d^
1.83
C = 1/280 F,
R = 180 Q,
A simple series RCL circuit has
Rdq
1
=
+
LTt LC q L
1
+
m
V)
L = 20 H,
and applied voltage
E(t)
= 10 sin
Derive a differential equation for the charge on the capacitor at any time t.
I
Substituting the given quantities into (7) of Problem 1.82, we get
d2q
dt
1.84
180 dq
+
2
q =
+
20 dt
'
20(1/280)
C = 10
R = 10 Q,
A simple series RCL circuit has
or
q + 9q + I4q = \ sin r
L = 1/2 H,
and applied voltage
(10sint)
20
2
F,
£ = 12 V.
Derive a differential equation for the amount of charge on the capacitor at any time t.
I
Substituting the given quantities into (7) of Problem 1.82, we obtain
d2q
10 dq
1
1
2+ T/2dt + (1/2)(10~
lt
1.85
1/2
)
Find a differential equation for the current in the circuit of Problem 1.84.
f
Substituting the values given in Problem 1.84 into (2) of Problem 1.81, we obtain
d2I
dt
1.86
q + 20q + 200q = 24
or
(12)
2
+
2
10 dl_
l72^
+
A simple series RCL circuit has
(l/2)(10-
R = 6 Q,
/
2
)
=
1
d
d(l2)
or
1/2
dt
C = 0.02 F,
2
dl
-T + 20 — + 200/ =
dt
L = 0.1 H,
I
2
dt
and no applied voltage. Derive a
differential equation for the current in the circuit at any time t.
I
Substituting the given quantities into (2) of Problem 1.81, we obtain
d
2
l
T+
dr
6
dl
0T^
2
1
+
or
(0.1)(0.02)
0.1
dt
dl
d l
—=
+ 60 — + 500/ =
dt
2
dt
t.
CHAPTER 1
14
1.87
Use Kirchoff's loop law to derive a differential equation for the current in a simple circuit consisting of a resistor,
an inductor, and an electromotive force connected in series (a series RL circuit).
I The circuit is similar to the one in Fig. 1.2, but without the capacitor. The voltage drops are given in
Problem 1.81, so it follows from Kirchoff's law that
RI + L
dl
—
-E(t) =
dl
R
—
+ -/ = -£(f)
L
L
1
or
at
1.88
(7)
at
A simple series RL circuit has an emf given by 3 sin 2f (in volts), a resistance of 10 Q, and an inductance of 0.5 H.
Derive a differential equation for the current in the system.
f
Substituting the given quantities into (7) of Problem 1.87, we obtain
dl
10
—
+— =—
dl
—
+ 207 = 6sin2r
1
7
0.5
dt
1.89
(3 sin 2:)
or
dt
0.5
Derive a differential equation for the current 7 in a series RL circuit having a resistance of 10 Q, an inductance
of 4 H, and no applied electromotive force.
Here
£(0 = 0,
R = 10,
L = 4,
and
—+—
so (/) of Problem 1.87 becomes
1.90
7
= 0.
4
dt
Use Kirchoff's loop law to derive a differential equation for the charge on the capacitor of a circuit consisting
of a resistor, a capacitor, and an electromotive force (emf) connected in series (a series RC circuit).
I The circuit is similar to the one in Fig. 1.2, but without the inductor. The voltage drops are as given in
Problem 1.81, so it follows from Kirchoff's law that RI + q/C — E(t) = 0. Since 7 = dq/dt, this may be
rewritten as
dq
1.91
1
1
A series RC circuit has an emf given by 400 cos 2f (in volts), a resistance of 100 Q, and a capacitance of 0.01 F.
Find a differential equation for the charge on the capacitor.
I
Substituting the given quantities into (/) of Problem 1.90, we obtain
dq
1
dt
100(0.01)
— + —r^
1.92
q =
H
—— (400 cos 2f
1
)
q + H
q = 4cos2t
H
or
100
Derive a differential equation for the current in the circuit of the previous problem.
I
d (dq\
Differentiating (/) of that problem with respect to time, we obtain
—
{
—
relationship
7
= dq/dt,
we find that
dl dt + I =
dq
H
dt \dt)
1.93
(7)
= — 8 sin It.
Using the
dt
— 8 sin 2f.
Derive a differential equation for the charge on the capacitor of a series RC circuit having a resistance of 10 Q,
3
a capacitance of 10" F, and an emf of 100 sin 1207rf.
f
Substituting these quantities into (7) of Problem 1.90, we obtain
q + lOOq = 10 sin 1207tf.
GEOMETRICAL PROBLEMS
1.94
Derive a differential equation for the orthogonal trajectories of the one-parameter family of curves in the xy
plane defined by
F(x, y, c) = 0,
where c denotes the parameter.
I The orthogonal trajectories consist of a second family of curves having the property that each curve in this new
family intersects at right angles every curve in the original family. Thus, at every point of intersection, the slope
of the tangent of each curve in the new family must be the negative reciprocal of the slope of the tangent of each
curve in the original family.
To get the slope of the tangent, we differentiate
respect to x, and then eliminate c by solving for it in the equation
F(x, y. c) =
F(x, y, c) —
implicitly with
and substituting for it in the
derived equation. This gives an equation connecting x, y, and y', which we solve for y' to obtain a differential
BASIC CONCEPTS
dy/dx = f(x, y).
equation of the form
D
15
The orthogonal trajectories are then the solutions of
dy
1
dx
(/)
f(x, y)
For many families of curves, one cannot explicitly solve for dy/dx and obtain a differential equation of the form
dy/dx = f{x, y). We do not consider such curves in this book.
1.95
y = ex 2
Derive a differential equation for the orthogonal trajectories of the family of curves
.
I The family of curves is a set of parabolas symmetric about the y axis with vertices at the origin. In the notation
of Problem 1.94, we have
dy/dx = 2cx.
x, we obtain
dy/dx = 2y/x.
1.96
F(x, y, c) = y — ex
= 0.
2
Implicitly differentiating the given equation with respect to
We have found
f(x, y)
= 2y/x,
I The family of curves is a set of circles centered at (e/2, 0).
2
+ y 2 — ex.
found
have
2x + 2yy' =
2
— x 2 )/2xy,
so (7) of Problem 1.94 becomes
F(x, y, e) = x
or
2
+ y2 — e2
dy/dx = —x/y.
\
2
.
We have
,,2'
x2 + y1 = c2
.
Implicitly differentiating the given equation with respect to x, we get
.
Since
f{x,y)=—x/y,
In the notation of Problem 1.94, we have
f(x, y) = y,
(7) of Problem 1.94
y'
— ce — 0.
x
Since
so (7) of Problem 1.94 becomes
= y — ce x
F(x, y,c)
x
y = ee
dy/dx —
it
,
.
In the notation of Problem 1.94, we have
respect to x, we get
y + xy' =
or
y'
y — ee
x
.
Implicitly differentiating this equation with
follows that
y'
—y=
or
y'
— y.
Here
— l/y.
F(x, y, C)
— —y/x.
dy/dx = y/x.
becomes
Derive a differential equation for the orthogonal trajectories of the family of curves
ff
—
dy
=y
—
dx
2xy
gives
x2 — y
dx
respect to x, we obtain
1.99
x 2 + y 2 — ex —
Derive a differential equation for the orthogonal trajectories of the family of curves
I
+ y 2 = ex.
This family of curves is a set of circles with centers at the origin and radii c. In the notation of Problem
1.94, we
1.98
2
In the notation of Problem 1.94, we have
Derive a differential equation for the orthogonal trajectories of the family of curves
I
x
.
Implicitly differentiating the given equation with respect to x, we obtain
Eliminating c between this equation and
f(x, y) = (y
hence,
;
2
dy
2x + 2y -— — e.
dx
1.97
=
—
dx
2y
so (7) of Problem 1.94 becomes
2
—x
dy
Derive a differential equation for the orthogonal trajectories of the family of curves
F(x, y, c) — x
c = y/x
To eliminate c, we observe, from the given equation, that
= xy — C.
Here
xy = C.
Implicitly differentiating this equation with
f(x, y) = —y/x,
so (7) of Problem 1.94 becomes
dy/dx — x/y.
1.100
p — C{\ + sin 6),
Derive a differential equation for the orthogonal trajectories of the cardioid
expressed in
polar coordinates.
Differentiating with respect to 6 to obtain
—
=C
dd
cos 0,
solving for
C—
cos
dp
in the given equation lead to the differential equation of the given family:
—
du
,
6
= p —
—
da
6
+
and substituting for C
cos
:
1
-.
In polar coordinates,
sin
the differential equation of the orthogonal trajectories is obtained by replacing dp/d9 by
—p 2 d6/dp, which gives
us
dd
dp
1.101
cos0
=
p{\
+ sin 0)
or
(sec 6
_ ,„
+ tan 0)dO =
p
A curve is defined by the condition that at each of its points (x, y), its slope dy/dx is equal to twice the sum of
the coordinates of the point.
Express the condition by means of a differential equation.
# The differential equation representing the condition is
1.102
dp
—
+
dy/dx — 2(x + y).
A curve is defined by the condition that the sum of the x and y intercepts of its tangents is always equal to 2.
Express the condition by means of a differential equation.
CHAPTER 1
16
/
dy
Y — y = —- (X — x),
The equation of the tangent at (x, y) on the curve is
X — x—y—
dx
respectively,
Y=y—x
and
dy
and the x and y intercepts are,
ax
dy
—
The
dx
.
differential equation representing the condition is
PRIMITIVES
1.103
Define essential constants in the context of a relationship between two variables.
I
If a
relationship between two variables involves n arbitrary constants, then those constants are essential if
they cannot be replaced by a smaller number of constants.
1.104
y = x
Show that only one arbitrary constant is essential in the relationship
2
+A+ B
involving the variables
x and y.
I
1.105
Since
A + B
is
no more than a single arbitrary constant, only one essential constant is involved.
y = Ae
Show that only one arbitrary constant is essential in the relationship
x+B
involving the variables x and
y-
1.106
I
Since
is
required.
y — Ae
x+B = Ae x e B
and Ae B is no more than a single arbitrary constant, only one essential constant
,
y = A + In fix
Show that only one arbitrary constant is essential in the relationship
involving the variables x
and y.
I
Since
y — A + In Bx — A + In B + In x,
(A + In B)
and
is
no more than a single constant, only one
essential constant is involved.
1.107
Define primitive in the context of a relation between two variables.
f
A primitive is a relationship between two variables which contains only essential arbitrary constants.
4
2
y = x + C and y = Ax + Bx, involving the variables x and y.
Examples are
1.108
Describe a procedure for obtaining a differential equation from a primitive.
f
In general, a primitive involving n essential arbitrary constants will give rise to a differential equation of
order n, free of arbitrary constants. This equation is obtained by eliminating the n constants between the n + 1
equations consisting of the primitive and the n equations obtained by differentiating the primitive n times with
respect to the independent variable.
1.109
y = Ax
Obtain the differential equation associated with the primitive
f
2
+ Bx + C.
Since there are three arbitrary constants, we consider the four equations
y = Ax
2
d2 v
dv
-f = 2Ax + B
+ Bx + C
dx
i
=
—
dx
d
~r^ = 2A
dx
\
\
The last of these, being free of arbitrary constants and of the proper order, is the required equation.
Note that the constants could not have been eliminated between the first three of the above equations. Note
also that the primitive can be obtained readily from the differential equation by integration.
1.110
x 2 y 3 + x 3 y 5 = C.
Obtain the differential equation associated with the primitive
I
Differentiating once with respect to x, we obtain
2xy 3 + 3x 2 y 2
+ ( 3x y + 5x y 4 ^
^\
dx
dx)
2
3
5
\
or
dy
dy\
.. 2 /V.
c..
2y + 3x j- + xy
3y + 5x j.
{
I
,
1
)
=
for xy
^
BASIC CONCEPTS
D
17
Written in differential notation, these equations are
as the required equation.
3
2
(2xy dx + 3x y 2 dy) + (3x 2 y 5 dx + 5x 3 y 4 dy) =
(1)
2
(2y dx + 3x dy) + xy (3y dx) + 5x dy) =
and
(2)
Note that the primitive can be obtained readily from (/) by integration but not so readily from (2). To obtain
2
which was removed from (1).
the primitive when (2) is given, it is necessary to determine the factor xy
1.111
y = A cos ax + B sin ax,
Obtain the differential equation associated with the primitive
constants and a being a fixed constant.
dy
—
= — Aa
dx
Here
d y
—
= —Aa
A and B being arbitrary
+ Ba cos ax
sin ax
2
12
and
2
2
The required differential equation is
1.112
d y/dx
+ a 2 y = 0.
2
y = Ae
Obtain the differential equation associated with the primitive
2
d
2iJ2iAAe
Zx
dx
2
— — + —=
—
ax
dx
dx
d y
The required equation is
-5
3
„
— 2Ae
,
dx
d y
d y
~{d 2 y
dy
3
2
2
dx
dx
2
dx
\dx
dy
d y
-=
3
r
J
ax
dy
d y
L
and
3
-4y = %Ae 2x + Be x
dx 1
dx
+ Be x + C.
d3
-4y = 4Ae 2x + Be'
2x
+ Be"
-r- = 2Ae
Then
2x
d2
dy
Here
1.113
—a 2 {A cos ax + B sin ax) = — a 2 y
cos ax — Ba sin ax =
2
y
dx 2
2
z
0.
y — C e
Obtain the differential equation associated with the primitive
3x
{
+ C 2 e 2x + C 3 e*
I Here
dv
-f-
= 3C e
3jc
t
dx
d 2y
+ 2C 2 e 2x + C 3 e x
Ix 2
= 9C e 3x + 4C 2 e 2x + C 3 ex
v
—
? = 27C^ 3x + SC e 2x + C e
ax
J
2
3
The elimination of the constants by elementary methods is somewhat tedious. If three of the equations are
solved for C l9 C 2 and C 3 using determinants, and these are substituted in the fourth equation, the result may
,
be put in the form (called the eliminant):
e
3x
e
3x
2e
7>e
3x
9e
3x
21e
4e
Se
2x
2
2x
*
2x
e
y
e*
y
y
e*
3
2
1
y'
y'
9
4
1
y"
y
27
8
1
y'"
,6x
2x
The required differential equation is
1.114
Illy
d 3y
2
dx
dx
= e 6x (-2y'" + 12y" - 22y' + 12y) =
—-^ — 6 d—-=y + 11-dy
6y = 0.
dx
Obtain the differential equation associated with the primitive
I
dy/dx — 2Cx,
Since
y = Cx
2
+ C
2
we have
C=
—
—
2x dx
The required differential equation is
and
I
—
-
I
\dxj
y
Lx
+ 2x 3
dx
+L
2xdx
+
4x 2 {dx)
4x 2 y = 0.
{Note: The primitive involves one arbitrary constant of degree two, and the resulting differential equation is
of order
1.115
1
and degree 2.)
Find the differential equation of the family of circles of fixed radius r with centers on the x axis.
CHAPTER 1
18
P(x.y)
Fig. 1.3
# The equation of the family (see Fig. 1.3) is
(x
— C) 2 + y 2 = r 2
dy
x — C — —y
so
ax
and the differential equation is
1.116
2
y
(dy\ 2
1
—
I
+ y2 — r2
C being an arbitrary constant. Then
,
—
ax
.
Find the differential equation of the family of parabolas with foci at the origin and axes along the x axis.
I The equation of the family of parabolas is
A = \yy',
and
2
y
= 2yy'(\yy' + x).
y
2
= 4A(A + x).
The required equation is
(See Figs 1.4 and
fdy\
y
(
—
I
2
+ 2x
V dx I
dy
dx
1.5.)
Then
vv'
= 2/4,
y = 0.
(-A.0)
x
or
2
+y
y
2
2
=
=
(2A + x)
2
y*
44 (M + x)
=
Fig. 1.4
1.117
Fig. 1.5
Form the differential equation representing all tangents to the parabola
I
By — x + \B
Eliminating B between this and
By' = 1.
2
x, we get as the required differential equation
2x{y') — 2yy' +1=0.
.
2
y
= 2x.
y — B - (x — A)/B or, since A = \B
which is obtained by differentiation with respect to
At any point {A, B) on the parabola, the equation of the tangent is
2
4A(A +x)
2
,
CHAPTER 2
Solutions
VALIDATING SOLUTIONS
2.1
Determine whether
I
y(x)
= 3e x
Differentiating y(x), we get
is
a solution of
y'(x)
= 3e x
Then
.
+ y = 0.
y'
y'
+ y = 3ex + 3e x = 6ex ^ 0.
Since y(x) does not satisfy
the differential equation anywhere, it is not a solution.
2.2
Determine whether
I
y(x)
— 5
a solution of
is
Differentiating y(x), we get
y'{x)
= 0.
+ y — 0.
y'
Then
+ 5 = 5 ^ 0.
+y=
y'
Since y(x) does not satisfy the
differential equation anywhere, it is not a solution.
2.3
Determine whether
y{x)
= cos x
is
y'
a solution to
+ y = 0.
Differentiating y(x), we get
Then y' + y = — sin x + cos x, which is not identically zero
y'(x) = — sin x.
on any interval. Because y(x) does not satisfy the differential equation on any interval, it is not a solution.
Note that y + y is zero wherever sin x = cos x, which occurs at infinitely many discrete points. There
sin x = cos x,
so there is no interval on which the differential equation is
is, however, no interval on which
#
satisfied.
2.4
Determine whether
x
y — 3e'
is
a solution of
+ y — O.
y'
Then y' + y = -3e~ x + 3e~ x = 0. Thus y(x) = 3e~ x
satisfies the differential equation for all values of x and is a solution on the interval ( — oo, oo).
I
2.5
y'{x)=—3e~ x
Differentiating y(x), we get
Determine whether
x
y = 5e~
is
.
a solution of
+ y = 0.
y'
Then y' + y = — 5e~ x + 5e~ x = 0. Thus y(x) = 5e~ x
satisfies the differential equation for all values of x and is a solution on the interval ( — oo, oo).
#
2.6
Show that
f
2.7
y(x)
= Ce~ x
is
a solution of
Differentiating y(x), we get
Determine whether
f
y'(x)=—5e~ x
Differentiating y(x), we get
y(x)
y'(x)
=2
is
Differentiating y(x), we get
.
= — Ce~ x
Then
.
a solution of
y'(x)
= 0.
on the interval ( — oo, oo) for any arbitrary constant C.
+y=
y'
y'
Then
+ y = -Ce' x + Ce' x =
y'
for all real values of x.
+ y 2 = 0.
y'
+ y2 =
+ (2) 2 = 4^0.
Thus y(x) does not satisfy the
differential equation anywhere and is not a solution.
2.8
Determine whether
f
y = e
x
Differentiating y(x), we get
is
a solution of
y'(x)
= ex
.
Then
satisfy the differential equation anywhere and
2.9
Determine whether
f
y — —x
Differentiating y(x), we get
is
is
a solution of
y'(x)
+ y 2 = 0.
y'
+ y 2 = e x + (e x 2 = e x + e 2x ^ 0.
y'
)
Thus y(x) does not
not a solution.
y'
+ y 2 = 0.
= — 1. Then
y
+ y 2 = - 1 + (-x) 2 = x 2 - 1,
which is not identically
zero on any interval. Since y(x) does not satisfy the differential equation on any interval, it is not a solution.
Note that x 2 — 1 is zero at ± 1; but for y(x) to be a solution,
interval
and that is not the case.
x2 — 1
would have to be zero on some
—
2.10
Determine whether
y = 1/x
is
a solution of
y'
+ y 2 = 0.
— + /A_ =
2
f
Differentiating y(x), we get
y'(x)
= - 1/x 2
the differential equation is satisfied whenever
.
Then
x # 0,
y'
+ y2 =
1
j
(
)
°
for all nonzero x.
Since
y(x) is a solution on any interval that does not include
the origin.
19
—
20
2.11
CHAPTER 2
U
Determine whether
f
y = 2/x
is
Differentiating y(x), we get
2.12
Determine whether
Here
y'(x)
Then
y = l/(x — 2)
= - l/(x - 2) 2
.
a solution of
y'
+ y2 =
so
,
x # 2,
Differentiating y(x), we get
I
2x
is
+ y2 =
/2\ 2
2
j
I
)
2
~z * °-
Since >* x ) does not
y'
+ y 2 = 0.
— + \x -
+ v2 =
y'
(
)
=
for all
x # 2.
Since the differential
(x
+ k) 2
x = 2.
on any interval that does not include the point
x = k,
Then
.
m + \x + kjr) =0
-
forallx^fc
l
y" — Ay — 0.
a solution of
Differentiating y twice, we find
y'
- 2) 2
(x
= — l/(x + k) 2
y'(x)
y'
y = e
—+~ =
+ y2 =
2/
y(x) is a solution on any interval that does not include
y — l/(x + k) is a solution to
where k denotes an arbitrary constant.
Determine whether
y'
is
Show that
f
2.14
= -2/x 2
anywhere, it is not a solution.
equation is satisfied whenever
2.13
y'
y'
satisfy the differential equation
+ y 2 — 0.
a solution of
= 2e 2x
and
y" = 4e 2x
.
y" — 4y = 4e 2x — 4(e 2x ) — 0,
Then
so y is a
solution to the differential equation everywhere.
2.15
Determine whether
y = e
2x
is
a solution of
y" = 4e
I As in the previous problem,
2.16
Determine whether
f
is
y'
—4 sin 2x + 4(sin 2x) = 0,
Determine whether
y = 2 sin x
then
;
so y is not a solution.
= 2cos2x and f = — 4 sin 2x. Then
y = sin 2x is a solution to the differential equation everywhere.
so
a solution of
is
y" + 4y = 4e 2x + 4{e 2x ) — Se 2x =£ 0,
y" + y = 0.
a solution of
Differentiating y twice, we find
y" + 4y =
2.17
y = sin 2x
2x
y" + 4y = 0.
y" + y = 0.
y' = 2cosx
and y"=— 2sinx. Then
—2 sin x 4- 4(2 sin x) = 6 sin x, which is zero only for integral multiples of n. Since 6 sin x is not
identically zero on any interval,
y = 2 sinx is not a solution to the differential equation.
f
Differentiating y, we obtain
y" + 4y =
2.18
Determine whether
I
is
a solution of
y'=— 4sin2x
Differentiating y, we find
—8 cos 2x + 4(2 cos 2x) = 0,
y" + y =
2.19
y = 2cos2x
Determine whether
y(x)
=
is
so
and
y'
y"=— 8cos2x.
y = 2 cos 2x
Then
a solution to the differential equation everywhere.
is
y" + 4y = 0.
a solution of
I For the identically zero function,
y" + y = 0.
= y" = 0;
y" + 4y =
hence
+ 4(0) = 0.
It
follows that y(x) is a
solution to this differential equation everywhere.
2.20
Show that
y(x)
= Cj sin 2x + c 2 cos 2x
Ci
and c 2
I
Differentiating y, we find
is
a solution of
y" + 4y =
for all values of the arbitrary constants
.
y'
= 2c cos 2x — 2c 2 sin 2x
x
y" + 4y = (
Hence,
y" =
and
— 4c sin 2x — 4c 2 cos 2x
x
4c v sin 2x — 4c 2 cos 2x) + 4{c x sin 2x + c 2 cos 2x)
= (— 4c + 4c )sin2x + — 4c 2 + 4c 2 )cos2x =
j
Thus,
y = Cj sin 2x + c 2 cos 2x
interval (
2.21
x
(
satisfies the differential equation for all values of x and is a solution on the
— oo, oo).
Determine whether
y = e~
2t
is
a solution of
y — 4y — 4y + 16y = 0.
SOLUTIONS
I
y= — 2e~ 2
Differentiating y, we obtain
y = 4e~
',
2
y- 4y - 4y + I6y = -8e~ 2 - 4(4<T
'
2
— Se~ 2
y =
and
',
D
21
Then
'.
- 4(-2e~ 2t + 16{e" 2 =
')
')
)
Thus, y is a solution of the given differential equation for all values of t on the interval ( — 00, 00).
2.22
y = e
Determine whether
I
2'
y — 4y — 4y + 16y = 0.
a solution of
is
y — 2e
Differentiating v, we obtain
y — 4e \
2
2t
,
y = 8e
and
2 '.
y-4y-4y + \6y = Se - 4(4e 2 - 4(2e
2t
')
Then
+ 16(e 2 =
21
')
)
Since y satisfies the differential equation everywhere, it is a solution everywhere.
2.23
v
= e 3f
y(t)
— e3
Determine whether
f
Differentiating
y — 3e ',
y = 9e ',
3
we obtain
',
y — 4y — 4y + 16v = 0.
a solution of
is
3
y = 21e
and
3t
Then
.
y-4y-4y + ify = 27e 3 - 4(9<? 3 - 4(3e 3 + 16(e 3 = -5e 3 #
'
')
'
')
')
Therefore, v is not a solution.
2.24
y = e
Determine whether
#
Differentiating
y(r)
4'
= e4
',
y — 4y — 4y + 16y = 0.
a solution of
is
we obtain
3;
= 4e 4t
y = 16e
,
4f
y = 64e
and
,
4 '.
y- 4y>- 4y + 16y = 64e 4 - 4(16e 4f - 4(4e 4 + 16(e
'
4r
')
)
Then
)
=
Thus y(t) is a solution everywhere.
2.25
I
Differentiating
— 0.5e 4
y =
Determine whether
y(t)
'
is
= -0.5e 4
y=-2e 4
we obtain
',
y — 4y — 4y + 16y = 0.
a solution of
y=-8e 4
',
and
',
y= -32e 4
'.
Then
y_ 4y - 4y + 16y = -32e 4r - 4(-Se 4 - 4(-2e 4 + 16(-0.5e 4 =
')
')
')
Thus y(t) is a solution everywhere.
2.26
Show that
y{t)
= c e 2t + c 2 e~ 2 + c 3 e A
'
arbitrary constants c x c 2
,
f
'
is
v
,
and c 3
y — 4y — 4y + 16y =
a solution of
for all values of the
.
Differentiating y{t), we get
-2c 2 e" 2 + 4c 3 e 4
2
2
+ 16c 3 e 4
y = 4c^ + 4c 2 e~
2
2,
+ 64c 3 e 4
y = Sc e '-Sc 2 e~
-2
2t
2t
4
2
+ 16c 3 e 4
y - 4y - 4y + 16y = 8c e - Sc 2 e~ + 64c 3 c - 4(4c,e + 4c 2 e
- 4[2c e 2t - 2c 2 e~ 2 + 4c 3 e 4 + 16(c,e 2 + c 2 e~ 2 + c 3 e*')
=
y = 2c!e
and
2'
'
'
'
'
'
-'
1
Then
'
'
'
')
x
'
'
'
')
l
Thus y(t) is a solution for all values of t.
2.27
Determine whether
1
x(i)
— — 2t
Differentiating x(t), we get
x — 2x — t.
a solution of
is
x — —\.
Then
x — 2x — — 2 — 2(—\t) = t — \,
which is never equal to t, the
right-hand side of the differential equation. Therefore, x(t) is not a solution.
2.28
Determine whether
I
x(t)
= —\
Differentiating x(t), we get
is
a solution of
is
2.29
t
x — 2x =
Since x(t) does not satisfy the differential equation on any interval, it
— 2(— |) = \.
This is equal to t, the right-hand side of
not a solution.
Determine whether
I
so
— \.
x = 0,
the differential equation, only when
x — 2x = t.
x(t)
= — jt — {
Differentiating x(r), we obtain
for all values of t in the interval (
is
a solution of
x = —\. Then
— 00, 00).
x — 2x = t.
x — 2x = -\ — 2(— {h — i) = r.
Therefore, x(f) is a solution
~
CHAPTER 2
22
2.30
Determine whether
I
= Ae 2
x(t)
'
is
x = 2Ae
Differentiating x(t), we get
x —2x = t
a solution of
2
side of the differential equation, only at
x - 2.x = 2Ae
so
',
= 0.
f
for any value of the arbitrary constant A.
- 2Ae 2 = 0.
2'
'
This is equal to t, the right-hand
Since x(t) satisfies the differential equation only at a single
point, it is not a solution anywhere.
2.31
Determine whether
= -\t - \ + Ae 2
x(t)
'
x - 2x = t
a solution of
is
for any value of the arbitrary
constant A.
I
x = - \ + 2Ae 2t
Differentiating x(t), we obtain
Then
.
x - 2x = -\ + 2Ae
- 2(— |f - | + Ae 2 =
2'
')
t
so x(t) is a solution everywhere.
2.32
Determine whether
1
— 2e x + xe~ x
y(x)
a solution of
is
y" + 2/ + y = 0.
Differentiating y(x), we obtain
y'(x)
= — 2e~ x + e~ x — xe~ x — —e~ x — xe~ x
y"(x) — e~
and
x
— e~ x + xe~ x — xe~ x
Substituting these values into the differential equation gives
y" + 2/ + y = xe~ x + 2{-e' x - xe~ x ) + (2e
x
+ xe' x =
)
Thus, y(x) is a solution everywhere.
2.33
Determine whether
I From
y(v) =
Show that
y(x)
is
=
and y"(x) = 0. Substituting these values into the differential
is not a solution.
+ 2(0) + 1 = 1 # x. Thus, y(x) =
y'(x)
y" + 2y' + y —
=
is
1
the only solution of
I By direct substitution, we find that
in (
y" + 2y' + y = x.
a solution of
follows that
it
1
equation, we obtain
2.34
=1
y(x)
r(\)
(y')
+ y 2 —0
2
=
on the entire interval
satisfies the differential
(
— oo, oo).
equation identically for all values of x
— oo, oo) and is, therefore, a solution. Any other function must be nonzero at some point (otherwise
it
would
not be different from the given function), and at such a point its square must be positive. Therefore, for such a
function, the left side of the differential equation must be positive at that point, because it is a sum of squares;
it
then cannot equal zero, the right side of the differential equation.
satisfy the differential equation at
2.35
y = x
Determine whether
2
—
1
some point on
is
(
It
follows that any nonzero function cannot
— oo, oo) and thus cannot be a solution over the entire interval.
a solution of
(
y') 4
+ y 2 = — 1.
I Note that the left side of the differential equation must be nonnegative for every real function y(x) and any x,
since it is the sum of terms raised to the second and fourth powers, while the right side of the equation is
negative.
2.36
Show that
Since no function y(x) will satisfy this equation, the given differential equation has no solution.
y = In x
a solution of
is
I On (0, oo) we have
y'
= 1/x
and
xy" + y
y" —
—
— 1/x
on
2
J = (0, x)
but is not a solution on
J = — oo, oo)r
(
Substituting these values into the differential equation, we
.
obtain
^ + y _,(_»,) + I_0
y = In x is a solution on (0, x). However, y = In x
logarithm is undefined for negative numbers and zero.
Thus,
2.37
Show that
y = l/(x
2
- 1)
is
a solution of
y'
+ 2xy 2 =
cannot be a solution on
on
J = (—
1, 1),
(
— x, x) because the
but not on any larger interval
containing J.
I On (-1,1),
y = l/(x
2
- 1)
and its derivative
y'
= — 2x/(x 2 — l) 2
are well-defined functions.
these values into the differential equation, we have
" +2x" i=
-(^w +2x (i?yf =o
Substituting
SOLUTIONS
— (—1,1). However, l/(x 2 - 1) is not defined at
y = l/(x - 1) is a solution on
therefore cannot be a solution on any interval containing either of these two points.
J
2
Thus,
x-±l
23
and
PRIMITIVES
2.38
Explain what is meant by a primitive associated with a differential equation.
f A primitive associated with a differential equation of order n is a primitive (see Problem 1.107) that contains n
arbitrary constants and is a solution of the differential equation.
2.39
y = C sin x + C 2 x
2
d y
dy
— x - h y — 0.
(1 — x cot x) ^-j
2
Show that
x
is
a primitive associated with the differential equation
—
dx
dx
y = Cj sin x + C 2 x, y' = C cos x + C 2
# We substitute
x
,
and
= — C, sin x
>'"
in the differential equation
to obtain
(1
— xcotx)( — C\ sinx) — x(C cosx + C 2 + {C sinx + C 2 x) =
— C] sin x + C x cos x — C x cos x — C 2 x + C sin x + C 2 x =
)
t
x
x
y = C e
Show that
t
In addition, the order of the differential equation (2) equals the number of arbitrary constants.
so y is a solution.
2.40
x
x
x
+ C 2 xe x + C 3 e~ x + 2x 2 e x
is
a primitive associated with the differential equation
^l3 - ^-1 - Q. + = Se x
dx 2
dx
dx
y =
C e x + C 2 xe x + C i e~ x + 2x 2 e x
y = (C, + C 2 )e x + C 2 xe x - C 3 e~ x + 2x 2 e x + 4x^
y" = (C + 2C2 )ex + C 2 xe x + C i e~ x + 2x 2 e x + 8xe x + 4e x
/" = (C, + 3C 2 )f x + C 2 xe x - C 3 ^" x + 2x 2 e x + 12xe x + 12e x
I We have
x
t
and
and
y'"
— y" — y' + y = 8e x
Also, the order of the differential equation and the number of arbitrary constants
.
are both 3.
2.41
Show that
y — 2x 4- Ce
x
dy
is
y = 2(1 — x).
a primitive of the differential equation
dx
1 We substitute y = 2x + O* and y' — 2 + Ce* in the differential equation to obtain
x
2 + Ce* — (2x + Ce = 2 — 2x.
Furthermore, the order of the differential equation equals the number of
)
arbitrary constants (2).
2.42
Show that
y = C e
x
x
d2y
+ C 2 e 2x + x
y = C e* + C 2 e
differential equation to obtain
I We substitute
is
2x
x
a primitive of the differential equation
/ = C e x + 2C 2 e 2x + 1,
+ x,
dy
y" = C^e" + 4C 2 e
and
x
—^ - 3 — + 2y = 2x - 3.
2jc
in the
C ex + 4C 2 e 2x - 3(C e x + 2C 2 e 2x + 1) + 2{C e x + C 2 e 2x + x) = 2x - 3
x
x
x
Moreover, the order of the differential equation and the number of arbitrary constants in y are both 2.
2.43
Show that
(y
— C) = Cx
M
is
a
Here
2(y — C)
4X
—
=
dx
a
C,
4x
a primitive of the differential equation
so that
I
2
1
y = 0.
r*
—
—
=—
— C)
dx
C2
C
2+2X
4(y-C)
2(y-C)
—
dy
fdy\
+ 2x
\dxj
dx
-.
Then
4x(y')
2
+ 2xy' - y
becomes
2(y
~
'
C 2 x + Cx(y - C) - y(y - C) 2
2
(y - C)
y[_Cx - (y - C)
(y
2
]
- C) 2
Furthermore, the order of the differential equation (1) is the same as the number of arbitrary constants in the
proposed primitive.
Y
24
2.44
CHAPTER 2
D
Determine whether
y — c l e~
x
+ \e 2x
y" — y' — 2y — e 3x
a primitive of
is
.
One can show by direct substitution that y is a solution of the differential equation. However, since y
contains only one arbitrary constant whereas the order of the differential equation is 2, y is not a primitive of
#
the differential equation.
2.45
Determine whether
y = c l xe
x
+ c 2 x 2 e x + \x 2 e x -
— 3y" + 3/ — y = e* +
y'"
a primitive of
is
1
1.
I By direct substitution we can show that y is a solution of the differential equation. However, since y contains
only 2 arbitrary constants whereas the order of the differential equation is 3, y is not a primitive.
2.46
Determine whether
f
y = 3e
2x
y" — 2y' + y = 3e 2x
a primitive of
is
.
Since y contains no arbitrary constants while the order of the differential equation is 2, y cannot be a
primitive of the differential equation.
2.47
Determine whether
I
v'
a primitive of
+ y = 0.
y'
into the left side of the differential equation, we obtain
y = A and its derivative y' =
+ A — A, which equals 0, the right side of the differential equation, only if A — 0. If A # 0,
is not a solution. Since
y = A
y — A is not a solution for arbitrary A, y is not a primitive.
Substituting
Determine whether
f
is
+y=
then
2.48
y — A
Substituting
y = Ax
y = /lx
is
and
a primitive for
y'
= ,4
y
- 3y = 0.
into the left side of the differential equation, we obtain
— 3y — A — 3 Ax = A(l — 3.x). If y = Ax is a primitive, must satisfy the differential equation for all
is zero only when
must be zero for all A, but
x = j. That means y = Ax
A(l — 3x)
values of A. Thus,
y'
it
it
does not satisfy the differential equation on any interval; for that reason it is not a solution and. therefore, not
a primitive.
2.49
Determine whether
v
= Cx + 2C 2
is
a primitive for
2|
—
+x
)
\dxj
m
The derivative of y is
\
= C.
2
—
- =
dx
y
0.
Then
+*-^--y = 2C + xC - (Cx + 2C =
(V)
\flx/
ax
2
2
)
so y is a solution for all values of the arbitrary constant C.
Since y contains only the one arbitrary constant C
and the differential equation is of order 1, y is a primitive for the differential equation.
2.50
Show that
y
— — f.v 2
is
a particular solution of
2
——
+x
=
1
J
V dx J
I
Here
y'
= — $x,
= 0.
so
2
r*y + ,* ,_ 2 f-ix + «(-i«W4*'i=o
4
dx
4
dx)
\
2.51
y
dx
\
J
Use the results of Problems 2.49 and 2.50 to show that not every particular solution of a differential equation
can be generated from a primitive of that differential equation by selecting specific values for the arbitrary
constants.
#
We have shown that a primitive for
2
—
2
— =
dx
fdy\
+x—
\dxj
dx
:
y
is
y — Cx + 2C
with arbitrary constant C,
2
The primitive represents a family of straight lines, and
y = — ^x
clearly the equation of a parabola cannot be obtained by manipulating the arbitrary constant C. (A solution that
while a particular solution is the parabola
.
cannot be generated from a primitive is called a singular solution of the differential equation.)
2.52
Determine graphically a relationship between the primitive
of the differential equation
I
y = xy' — (y)
Referring to Fig. 2.1, we see that
y = x /4.
2
y — Cx — C
2
and the singular solution
y = x /4
2
2
.
y = Cx — C
2
represents a family of straight lines tangent to the parabola
The parabola is the envelope of the family of straight lines.
SOLUTIONS
D
25
Fig. 2.1
DIRECTION FIELDS
2.53
Construct a direction field for the first-order differential equation
y'
—y—
t.
I At each point in the (t, y) plane, we compute dy/dt by substituting y and t into the right-hand side of the
differential equation; then through the point, we graph a short line segment having the derivative as its slope.
In particular, at (0, 0) we have
(1,1),
is
/ = 1 — 1 = 0;
and at
= - = 0; at (0, 1), / = 1 - = 1; at (1, 0), / = - = - 1; at
— 1, — 1), y' — — — (— 1) = 0. The direction field for these and other points
y'
(
1
1
shown in Fig. 2.2.
/
/
,'/—
/
\
\
2.54
\
—
\
\
\
\
\
Fig. 2.2
Graph the solution curves that pass through the direction field obtained in the previous problem.
I The curves are shown in Fig. 2.3.
Fig. 2.3
\
26
2.55
CHAPTER 2
D
Construct a direction field for the first-order differential equation
f
Since y' in this case is independent of t, for any given >'
all
t.
— 5y(y — 1).
y'
the slopes of the solutions at (r, y ) are the same for
Noting that the right-hand side of this equation is zero when y is or 1, positive when y is in
(—oo, 0) u (1, oo), and negative when y is on (0, 1), we can readily verify that Fig. 2.4 gives a reasonable
indication of the direction field. For example, if y = \ then
y' = 5(\)(-j) =
-f, so the solutions have slope
y = \.
— | when they cross the line
y =
Also, if
then
v
=
15
16'
and if
y = —\
then
y
— 25
16
(these
values are indicated in Fig. 2.4).
\ \ \
\ \ \
//////
Fig. 2.4
2.56
Sketch a direction field for
y'
— y — I.
f The derivative is independent of
y'
= 0-l= -1;
(2, 2),
y'
at (2,0),
= 2 — 1 = 1.
y'
t
and depends only on y. At
= 0-l = -l;
at (1,
1),
>•'
=
1
(t,
—
1
= (0, 0), y' = — = — 1; at (1, 0),
= 0; at (2, 1), / = 1-1=0; and at
y)
1
The direction field at these points and others is shown in Fig. 2.5.
y-y-Y-Y—?—/\ \
N
\
s
\
\
r-A-^- Jr- Jr--V
J
Fig. 2.5
X— \— \- -\~\~
/ / /
/
C
7^
7^
-t—t—t—t—t — iFig. 2.6
—
1
SOLUTIONS
2.57
(t,y)
= (0,0),
/ = l _ 1 = 0;
y'
= 1-0=1;
- = 0; at (2,
y' =
/ = 1 - = 1; at (1,
- 1), / = 1 -(-1) = 2. The direction field at these
at (1,0),
y = 1 - 2 = -1;
1
at (2, 2),
27
/ = 1 — y.
Sketch a direction field for
I At
D
1),
1
1
1),
at (-2,
points and others is shown in Fig. 2.6.
2.58
Sketch a direction field for
y'
= y3 — y2
I At (t, y) = (0, 0), / =
/ = (-2) 3 -(-2) 2 = -12,
3
-
2
.
/ = l 3 - l 2 = 0; at (0, 2), / = 2 3 - 2 2 = 4; at (0, -2),
/ = (-1) - (-1) 2 = -2. The direction field at these points and
= 0;
at (0,
1),
3
at (1,-1),
others is shown in Fig. 2.7.
y
/--/-->'
/ / / / /
\ \ \ \ \
-*t
Fig. 2.7
2.59
/ = 1 — y2
Sketch a direction field for
I At
/=
(t,y)
= (0,0), / =
- (- 1) = 0;
2
1
-
= 1;
and at (— 1, — 2),
1
2
.
at (1,
y'
=
1
/ = 1 - 2 = 0, at (1, 2), /=l-2 2 =-3; at (0,-1),
- — 2) 2 = — 3. The direction field at these points and others is
l
1),
(
shown in Fig. 2.8.
y
"\-^K~^\
^-^-^
—
i—i-—
—H-H-H-^-^ —
y
/
-/-
i-
*-
Fig. 2.8
28
2.60
CHAPTER 2
Sketch a direction field for
y'
= 2x,
along with some of the solution curves that pass through it.
# The direction field and three curves are shown in Fig. 2.9.
slope = 4
slope = 2
Fig. 2.9
INITIAL AND BOUNDARY CONDITIONS
2.61
Determine whether the conditions on y(x) given by
y(0)
= 1,
y'(0)
= 2
are boundary conditions or initial
conditions.
f They are initial conditions because they are given at the same value of the independent variable, here at
x = 0.
2.62
Determine whether the conditions on y(x) given by
y(l)
= 0,
y'{2)
—
are boundary or initial conditions.
I They are boundary conditions because they are not both given at the same value of the independent variable.
One is given at x = 1, and the other at x = 2.
2.63
Determine whether the conditions on y(r) given by
y(3)
= 0,
y'(3)
= 0,
y"(3) =
1
are boundary or initial
conditions.
f They are initial conditions because they are all given at the same value of the independent variable, here at
r = 3.
2.64
Determine whether the conditions on x(f) given by
x(7r)
= 1,
x\n) = 2,
x"(n)
= 3,
x'"(n)
=4
are boundary
or initial conditions.
I They are initial conditions because they are all given at the same value of the independent variable, here at
t = n.
2.65
Determine whether the conditions on x(t) given by
x(0)
= 0,
x'(0)
= 0,
x"(n) =
are boundary or initial
conditions.
I They are boundary conditions because they are not all given at the same value of the independent variable.
Two are given at f =
t — n.
while the third is given at
SOLUTIONS
2.66
Determine whether the conditions on s(t) given by
s(5)
= s(10) =15
29
are boundary or initial conditions.
f They are boundary conditions because they are not both given at the same value of the independent variable.
= 5, and the other at
= 10.
One is given at
t
2.67
t
Determine whether a single condition is a boundary or initial condition.
I A single subsidiary condition is an initial condition because it satisfies the criterion that all conditions are
prescribed at the same value of the independent variable.
2.68
Determine whether the conditions on y(x) given by y( — 7.5) = 0, y'( — 7.5) = 1,
(4)
<5)
y ( — 7.5) = 0, and y ( — 7.5) = 1 are boundary or initial conditions.
y"(
— 7.5) = 0,
y
(3,
(
— 7.5) = 1,
I They are initial conditions because they are all specified at the same value of the independent variable, here
x = -7.5.
2.69
Determine C so that
I When
2.70
and
Determine C so that
/ When
(
2.71
x =
y(x)
x = 1
(y
and
= 2x + Ce x
y - 3,
y(0)
= 2(0) + Ce°
C = 3.
we have
— C) 2 = Cx
y = 2,
= 3.
will satisfy the condition
3
and
will satisfy the condition
we have
(2
y = x + C,c
x
y(l)
— C) 2 = C and
y = 2x + 3c*.
Then
= 2.
C = 1,4.
Thus,
(y
— l) 2 = x
and
2
y - 4) = 4x.
Determine C, and C 2 so that
+ C 2 e 2x
will satisfy the
boundary conditions
y(0)
=
and
y0) = o.
f
When
x =
and
C,e + C 2 e 2 =—l.
y = 0,
Then
C,
C, + C 2 = 0.
we have
e
2.72
Determine c, and c 2 so that
——
= — C 2 = -=
y(x)
When
x = 1
y = 0,
and
1
and the required equation is
)»
we have
= xH
_ e 2*
e*
=
—e
e
e
= c sin 2x + c 2 cos 2x +
t
will satisfy the conditions
1
.
y(n/S) =
and
y'(rr/8) = ^2.
# Note that
To satisfy the condition
y(rc/8)
*
*
= c sin - + c 2 cos - +
•
y(7r/8)
= 0,
y'(x)
—
—
= c A/2\ + c 2 A/2\
x
(
I
l
c^y/2) + c^v^) +1=0,
we require
c
Since
1
x
(1)
= 2c cos 2x — 2c 2 sin 2x,
1
1
To satisfy the condition
y'(7t/8)
= y/2,
y/2c { — sflc 2 - yJ2,
we require
Cl
Solving (7) and (2) simultaneously, we obtain
Determine Cj and c 2 so that
I
Because
= 0,
sin
y(0)
y(x)
ct
- c2 =
= c e 2x + c 2 e x + 2 sin x
Y
.
y'(x)
require
2c,
2x
= 2c e + c 2 e + 2 cos x,
+ c 2 + 2 — 1, or
l
x
Solving (7) and (2) simultaneously, we obtain
c,
2
or equivalently,
(2)
To satisfy the condition
we have
Ci
and
c2
— — 2 (\J2 + 1).
will satisfy the conditions
t
= c + c2
^ ~ ^°
1
= — \(\J2 — 1)
cx
From
or equivalently,
l+ c 2 =-j2
yW8) = 2c cos-- 2c 2 sin- = 2c i(-y) _ 2c2 \Y) =
2.73
,
y(0)
= 0,
y(0)
=
and
y'(0)
we require
+ c2 =
y'(0)
(7)
= 2c, + c 2 + 2.
2c, + c 2
=-\
= —
and
1
= 1.
To satisfy the condition
y'(0)
=1,
we
(2)
c2
= 1.
CHAPTER 2
30
PARTICULAR SOLUTIONS
2.74
I
y(x) = c,c
_x
,
Since y(x) is a solution of the differential equation for every value of c u we seek that value of c, which will
also satisfy the initial condition.
sufficient to choose C, so that
we obtain
2.75
y' + y = 0;
if the general solution to the differential
y(3) = 2,
where C, is an arbitrary constant.
Find the solution to the initial-value problem
equation is known to be
y(x)
Note that
c x e~
— 2e i e~ x = 2e i
~x
f
To satisfy the initial condition
.
l
3
c, = 2e
that is, to choose
y(3)
= 2,
it is
Substituting this value for c, into y(x),
.
as the solution of the initial-value problem.
y" + Ay = 0;
Find a solution to the initial-value problem
differential equation is known to be
— c e' i
y(3)
= 2,
2
y(x) — c, sin 2x
y(0) = 0,
+ c 2 cos 2x.
y'(0)
= 1,
Since y(x) is a solution of the differential equation for all values of C, and c 2
,
general solution to the
if the
we seek those values of c, and
To satisfy the first
c 2 that will also satisfy the initial conditions. Note that
y(0) = c, sinO + c 2 cosO = c 2
=
—
Furthermore. y'(x) = 2c, cos 2x — 2c 2 sin 2x; thus,
initial condition.
we choose c
0.
0,
y(0)
y'(0) = 1,
we choose 2c, = 1,
y'(0) = 2c, cosO — 2c 2 sinO = 2c,. To satisfy the second initial condition.
—
=
^sin 2x as the solution of the
y(x)
or c,
j. Substituting these values of c, and c 2 into y(x), we obtain
.
2
initial-value problem.
2.76
Find a solution to the boundary-value problem y" + Ay = 0;
y(x) = c, sin 2x + c, cos 2x.
= 0,
3(71/8)
yin/6)
— 1,
if the
general solution
to the differential equation is
m
f
n
Note that
y(rc/8)
n
—
= c, sin - + c 2 cos - = c, /v2\
I
(yJ2
c
-I-
4
To satisfy the condition
y(n/8)
= 0,
we require
V2\
_
.
^-c 2
A/2
[^-j^0
n
n
y/3
\imb) — c, sin - -I- c 2 cos - = c,
.
Furthermore,
To satisfy the second condition,
y(7r/6)
= 1,
Solving (7) and (2) simultaneously, we find
2
=
>*(x)
(sin 2x
v3 —
2.77
- cos 2x)
c,
1
(2)
= -c 2 = 2/(^3 - 1).
Substituting these values into y(x), we
as the solution of the boundary-value problem.
1
y" + Ay = 0: y(0) = 1,
= c, sin 2x + c 2 cos 2x.
y(n/2) = 2,
Find a solution to the boundary-value problem
the differential equation is known to be
f
Since
y(n/2)
\(0)
= c, sin
satisfy both
c
—
2
1-
we require
W3c, + \c 2 -
obtain
(1)
y(x)
if the
general solution to
= c, sinO + c 2 cos0 = c 2 we must choose c 2 =
to satisfy the condition
Since
v(0) = 1.
+ c 2 cos n = — c 2 we must choose c 2 = -2 to satisfy the condition \inj2) = 2. Thus, to
1
,
71
,
boundary conditions simultaneously, we must require c 2 to equal both
1
and -2, which is
impossible. Therefore, this problem does not have a solution.
2.78
Find a solution to the initial-value problem
differential equation
I
At
x = 0,
Furthermore,
At
y\n/2) —
condition
2.80
y'(0)
= 2,
if
the general solution to the
= A sinO + ScosO = B, so we must choose B —
to satisfy the condition
y(0) = 1.
= /I cos x — B sin x, so y'(0) = .4 cos — B sin = .4. To satisfy the condition y'(0) = 2,
A = 2. Then y(x) = 2 sin x
cos x is the solution to the initial-value problem.
y(0)
x = 7t/2,
1.
= 1,
where A and B are arbitrary constants.
1
-I-
Rework Problem 2.78 if the subsidiary conditions are
f
y(0)
y'(x)
we must choose
2.79
y" + y = 0;
is y(x) = A sin x + B cos x.
Also
y'(n/2)
y(rr/2)
y(7r/2)
= 1,
y'(7r/2)
= 2.
= A sin (rc/2) + B cos (tc/2) = A, so we must choose .4 —
to satisfy the condition
= A cos(7r/2) — Bsin(7r/2) = — B, so we must choose B — — 2 to satisfy the
Then y(x) = sin x — 2 cos x.
1
y'(7t/2)
= 2.
Rework Problem 2.78 if the subsidiary conditions are
y(0)
= 1,
y{n/2)
= 1.
SOLUTIONS
31
f The problem is now a boundary-value problem because the subsidiary conditions are specified at different
values of the independent variable x. At
x = 0, y(0) = A sin + B cos = B, so we must choose B =
to
satisfy the first subsidiary condition. At
x = n/2, y(n/2) = A sin (n/2) + B cos (n/2) = A, so we must choose
to satisfy the second subsidiary condition. Then
A —
y(x) = sin x + cos x.
1
1
2.81
Rework Problem 2.78 if the subsidiary conditions are
— 1,
y'(0)
y'(n/2)
— 1.
I The problem is now a boundary-value problem. For the given y(x), we have y'(x) — A cos x — B sin x.
x = 0, y'(0) = A cos — B sin = A, so we must choose A —
if we are to satisfy the first boundary
condition. At
x = n/2, we have y'(n/2) = A cos (7r/2) — B sin (n/2) = — B, so we must choose B=—
At
1
satisfy the second boundary condition.
2.82
Then
y(x)
Rework Problem 2.78 if the subsidiary conditions are
y(0)
= 1,
Then
2.83
y(x)
B = 1;
Since
Since
To
A = — 1.
to satisfy the second condition we must choose
= — sin x + cos x.
Rework Problem 2.78 if the subsidiary conditions are
I
to
= 1.
y'(n)
I The problem is now a boundary-value problem. With y(x) = A sin x + B cos x and
we have y(0) = /I sinO -f BcosO = B and y'(n:) = Acosn — Bsinn = — A.
y'(x) = A cos x — B sin x,
satisfy the first condition we must choose
1
= sin x — cos x.
y(0)
y(0)
= 0,
y(n)
= 2.
= A sin + B cos = B, we must choose B =
to satisfy the subsidiary condition
y(0) = 0.
—
—
—2
=
Bcos
n
—B,
we
must
choose
B
to
satisfy
the
condition
y(7r)
2.
Thus, to
+
y(n) — A sin n
satisfy both conditions simultaneously, we must require
B to equal both
and —2, which is impossible.
Therefore, this boundary-value problem does not possess a solution.
2.84
Rework Problem 2.78 if the subsidiary conditions are
I At
x = 0,
Furthermore,
y(0)
y'(0)
= A sin
= A cos
y(0)
= y'(0) = 0.
— B,
so we must choose
— B sin = A,
so we must choose
B—
A —
=
sin x
+ B cos
Substituting these values into the general solution, we get
y(x)
to satisfy the first initial condition.
to satisfy the second initial condition.
cos x =
+
as the solution to the
initial-value problem.
2.85
Rework Problem 2.78 if the subsidiary conditions are
I
At
x = 7i/4,
condition
we have
y{n/4) = 0,
y(n/4) - A sin(7r/4) + B cos (tt/4)
y(7r/6)=l.
= A^Jl/l) + B(yf2/2).
Thus, to satisfy the
we require
y(n/4) = 0,
Furthermore,
y(7r/6)
= ^ sin (tt/6) + B cos (tc/6) = A{{) + B(y/3/2).
To satisfy the condition
y(n/6) =
1,
then,
we require
A
^3
-2
Solving (7) and (2) simultaneously, we determine
A — —=
and
>/3into the general solution, we obtain
y(x)
= ——
2
(
— sin x
B — —=
Substituting these values
.
y/3- 1
1
-(-
2
cos x).
>/3- 1
2.86
Rework Problem 2.78 if the subsidiary conditions are
/
At
x = 0,
condition.
At
y(0)
= A sin
x = n/2,
y(0)
= 0,
+ B cos = B, so we must choose
= A cos(7i/2) — B sin (n/2) = — B,
y'(n/2)
second boundary condition. Thus, we must have B equal to both
y'(n/2)
= 1.
B—
to satisfy the first boundary
so we must choose
B— —
1
to satisfy the
and — 1 simultaneously, which is
impossible. Therefore, the boundary-value problem does not have a solution.
2.87
Rework Problem 2.78 if the subsidiary conditions are
f At
x = 0,
condition.
At
y(0)
= 1,
y(n)
= — 1.
= A sin + B cos = B, so we must choose B —
to satisfy the first boundary
—
—
=
A
sin
B
n
—B,
we
x
n,
y(n)
n +
cos
so
must choose B —
to satisfy the second
y(0)
1
1
CHAPTER 2
32
boundary condition. Thus, B — 1 is sufficient to satisfy both boundary conditions, with no restrictions
placed on A. The solution is y = A sin x + cos x, where A is an orbitrary constant.
2.88
A general solution to a certain differential equation is
are arbitrary constants.
I
= c e~ x + c 2 e 2x — 2x 2 + 2x — 3,
y(x)
where c t and c 2
x
Find a particular solution which also satisfies the initial conditions
y = Cjfi
Since
we have
y'
-
*
y(0)
= 1,
= 4.
y'(0)
+ c 2 e 2x — 2x 2 + 2x — 3
(7)
= — c^"* + 2c 2 e 2x — 4x + 2
(2)
Applying the first initial condition to (7), we obtain
Cl e~
w + c e 2m - 2(0) + 2(0) -3 =
2
2
+ c2 = 4
(5)
-c, + 2c 2 = 2
(4)
or
1
c,
Applying the second initial condition to (2), we obtain
-Cie- (0) + 2c 2 e
2(0)
- 4(0) + 2 = 4
cx = 2
Solving (5) and (4) simultaneously, we find that
and c 2 = 2. Substituting these values into (7), we
x
=
2e~
+ 2e 2x — 2x 2 + 2x — 3.
y
obtain the solution of the initial-value problem as
2.89
A general solution to a certain differential equation is
arbitrary constants.
#
or
= c,^ + c 3 xe? + xe*ln |x|,
y(x)
where c, and c 3 are
Find a particular solution which also satisfies the initial conditions
= 0,
y(l)
y'(l)
= 1.
+ xex In |x|
x
x
x
x
y' = c e + c 3 e + c 3 xe + e In |x| + xe" In |x| + e*
y = c,?* + c 3 xe
Since
we have
x
(7)
(2)
x
Applying the first initial condition to (7), we obtain
cxe
l
+ c^e + (l)e In
1
1
1
=
or (since
In 1
= 0),
c x e + c3e =
Applying the second
initial
condition to (2), we obtain
c,?
c x e + 2c 3 e
Solving (3) and (4) simultaneously, we find that
=
(3)
+ c3e +
x
c 3 {l)e
i
+e
1
In
+
1
(l)e
Inl + e = 1,
1
1
or
—e
1
(4)
— —c 3 — (e — l)/e. Substituting these values into (7), we
x ~
x
(e — 1)(1 — x) + xe In |x|.
ct
y — e
obtain the solution of the initial-value problem as
2.90
1
l
2x
y = e
A general solution to a certain differential equation is
{c l cos 2x
7
4
65
65
+ c 2 sin2x) -f — sinx — — cosx,
where c, and c 2 are arbitrary constants. Find a particular solution which also satisfies the initial conditions
/(0) = 0.
y(0)=l,
74
I For y as given, we have
y'
= -2e
2x
(c l cos2x
2x
+ c 2 sin2x) + e
(
-2c, sin 2x + 2c 2 cos 2x) + —cosx +
65
Applying the first initial condition to y, we obtain
— 2c,
-1-
2c 2 = —7/65,
so that
c2
= 131/130.
c{
— 69/65.
— sinx
65
Applying the second initial condition to y' gives
Substituting these values for c, and c 2
,
we obtain the solution
of the initial-value problem as
69
y
2.91
131
7
4
. \
= e -2xfZ7 cos 2x + T^ sin 2x
+ T7 sin x ~ 77 cos X
-,
\od
13U
)
J
65
65
The general solution to a certain third-order differential equation is
y = c^e? + c 2 e
2x
+ c 3 e 3x
where c u c 2
and c 3 are arbitrary constants. Find a particular solution which also satisfies the initial conditions y{n) = 0,
y\n) = 0,
y"(n) =
,
,
1.
# We have
y = Cl e
and
+ 2c 2 e 2x + 3c 3 e 3x
y" = Cl e x + 4c 2 e 2x + 9c e 3x
3
y = Cl e
'
x
+ c 2 e 2x + c 3 e 3x
x
Applying each initial condition separately, we obtain
+ c 2 e 2n + c 3 e in =
K
2
3n
=
c e + 2c 2 e * + 3c 3 e
2n
3*
c e" + 4c 2 e
+ 9c 3 e =
cxe
n
x
{
1
(7)
SOLUTIONS
n
2n
c 2 = -e~
and
~
{x ~ K)
2{x ~ n)
—
—
\e
e
y
+ 2\e3ix K)
Solving these equations simultaneously, we find
c 1 =^e~
values into the first equation of (7), we obtain
2.92
differential equation is
x(f)
= c e~ + c 2 e + \e
x(0)
';
3
2t
l
= 1,
= 2,
x'(0)
= c e~° + c 2 e 2(0) + |e 3<0) = 1,
x(0)
= \e~ 3 *.
Substituting these
general solution to the
if the
where Cj and c 2 are arbitrary constants.
',
x
f The first initial condition yields
c3
,
33
.
x" - x' - 2x = e 3
Solve the initial-value problem
,
D
which may be rewritten as
x
c,+c 2 = l
Furthermore, x'(f) = — c x e~
-0
2{0)
x'(0)
= — Cje
4-
2'
(1)
3r
so the second initial condition yields
+ 2c 2 e + |e
+ |e 3(0) = 2, which may be rewritten as
2c 2 e
l
,
-c,+2c 2 =l
Solving (7) and (2) simultaneously, we find
cx
=^
and
(2)
= §.
c2
Thus,
x(r)
= j^e' + \e + \e
2t
1
3t
is
a
solution to the initial-value problem.
2.93
Rework Problem 2.92 if the initial conditions are
x(0)
= 2,
= 1.
x'(0)
# With x(t) = c e~ + c 2 e 2 + \e 3i we have x'(t) = — c e "' + 2c 2 e 2t + \e 3x The initial conditions then yield
m) + 3(0) = 2 and x'(0) = -c e~° + 2c 2 e 2{0) + |e 3(0) = 1, which may be rewritten as
x(0) = c e~° + c 2 e
i<?
l
'
.
Y
x
Y
x
Cy+C 2 =i
—c x +2c 2 =l
Solving system (7), we obtain
= ^e- + le 2t + {e 3t
ct
= y§ and
c2
= f,
U)
and the solution to the initial-value problem is
,
x{t)
2.94
.
Rework Problem 2.92 if the initial conditions are
x(l)
= 2,
x'(l)
= 1.
I Applying the initial conditions to the expressions for x(t) and x'(f) as determined in the previous problem,
we obtain x(l) = c e~ + c 2 e 2{1) + ie 3(1) = 2 and x'(l) = -c e~ + 2c 2 e 2(1) + f<? 3(1) = 1, which may be
1
x
x
1
rewritten as
c x e~
-c e~
+ c 2 e 2 = 2 — \e 3
+ 2c 2 e 2 = - \e 3
x
l
= e + ^e 4 and
2
- \e)e 2t + \e 3t
x(f) = (e + ^e^e'' + (e~
Solving system (7), we obtain
is
2.95
then
(1)
1
r
c,
c2
= e~ 2 - \e.
The solution to the initial-value problem
.
x(f) = c e' + c 2 e~' + 4 sin t,
where c x and
Find a particular solution which also satisfies the initial conditions x(0) = 1,
A general solution to a certain second-order differential equation is
c 2 are arbitrary constants.
{
x(0)=-l.
f For x(r) as given, we have
= c e — c 2 e~' + 4 cos
l
x(t)
x
t.
The initial conditions then yield
= c e° + c 2 e~ l0) + 4sin0 =
x(0) = c ^°-c 2 e" (0) + 4cos0= -1
x(0)
1
x
1
= — 2 and c 2 = 3.
— 2e' + 3e~' + 4 sin
=
x(r)
Solving system (7), we obtain
get the particular solution
2.96
or
Cl -c =-5
t.
x(t)
= c + 2t.
t
x(r)
= c t + c 2 + 2 — 1,
t
x
c —
and c 2 = 1.
2
- 1 = 2
= (0)f + 1
Solving system (7), we obtain
x(r)
x
-I- 1
where c x and c 2
x(l)
= 1,
x(l)
= 2.
The initial conditions yield
= Cl (l) + c 2 + (l) 2 - 1 = 1
x(l) = c t + 2(1) = 2
t
or
Cl
or
+ c2 =
c, =
1
Substituting these values into the general solution, we get
.
A general solution to a certain second-order differential equation is
arbitrary constants.
2
Substituting these values into the general solution, we
cx
x(l)
the particular solution
1
Find a particular solution which also satisfies the initial conditions
f For x(r) as given, we have
2.97
cx
A general solution to a certain second-order differential equation is
are arbitrary constants.
+ c2 =
or
z(t)
— Ae' + Bte'
-I-
Find a particular solution which also satisfies the initial conditions
fV,
z(l)
where A and B are
= 1,
z(l)
= — 1.
34
CHAPTER 2
f For the given z(t), we have
z(l) = Ae + Be + e = 1
and
= Ae' + B(e' + te') + (2te' + t 2 e'). The initial conditions then yield
=
i(l)
Ae + B(e + e) + (2e + e) = — 1. which may be rewritten as
z(t)
A+ B = e-' -
1
A + 2B= -e~ -3
l
A —
Solving system (7), we obtain
1
4-
3e"
solution, we get the particular solution
2.98
1
B = —2 — 2e
and
Substituting these values into the general
.
— (1 + 3e~ )e* + — 2 — 2e' )te' + rV.
z(t)
1
(
A general solution to a particular third-order differential equation is z(t) — A + Bt + Ct 2 + 2r 3 where /I, B,
and C are arbitrary constants. Find a particular solution which also satisfies the initial conditions
z(l) = z(l) = z(l) = 0.
,
f For z(r) as given, we have
z(t)
= B + 2Ct + 6t 2
and
z(t)
= /I + 5(1) + C(l) 2 + 2(1) 3 =
2
z(l) = B + 2C(1) + 6(1) =
z(l) = 2C + 12(1) =
A = — 2,
Solving system (7), we obtain
The initial conditions yield
A + B+
C=
-2
B + 2C = -6
2C = -12
or
or
U)
and C = — 6. Substituting these values into the general
= — 2 + 6f — 6f 2 4- 2r 3
5 = 6,
solution, we get the particular solution
z(f)
.
A general solution to a third-order differential equation is
arbitrary constants.
= 2C + 12r.
or
z(l)
2.99
_1
l
z(t)
= Ae 2 + Be~ 2t + Ce' 3 \
where A, B. and C are
'
Find a particular solution which also satisfies the initial conditions
z(0)
= 0,
z'(0)
= 9,
z"(0)= -5.
I For
initial
z(f)
as
given,
we have
r'M = 2Ae
2'
- 2Be' 2 - 3Ce" 3
'
+ fl<?- 2(0) + Ce- 3i0) =
2(0)
- 2Be" 2<0) - 3Ce- 3,0) =
z'(0) = 2/le
z"(0)
/1<?
= 4/le
Solving system (7), we obtain
2.100
z"(t)
= AAe 2 + ABe~ 2 + 9Ce' it
'
'
.
The
conditions yield
z(0)=
z(t)
and
'
2,0)
2(0)
+ 4Be
A = 2,
" 2(0)
+ 9Ce
B = — 1,
" 3,0)
and
/1+B+C=0
or
9
or
2/1
- 2B - 3C =
=-5
or
4/l
+ 4fl + 9C=-5
C = — 1,
so the particular solution is
9
(7)
= 2e 2 '-e- 2 -e-*'.
'
A general solution to a fourth-order differential equation is yis) = Ae + Be' + Ce 2s + De is + s 2 + 2, where
A, B, C, and D are arbitrary constants. Find a particular solution which also satisfies the initial conditions
y(0)=l, y'(0) = y"(0) = 4, y"'(0)= 10.
1
5
f For y(5) as given, we have
y\s) = Ae*- Be-* + 2Ce
2s
+ 3De 3i + 2s
= Ae* + Be~* + 4Ce 2s + 9De 3s + 2
2s
y'"(s) = Ae* - Be'* + 8Ce
+ 21De is
y"(s)
Consequently, the initial conditions yield
= Ae w + Be~ w + Ce 2(0) + De M0) + (0) 2 + 2=1
2m + 3De 3{0) + 2(0) = 4
y'(0) = Ae m - Be~ w + 2Ce
w + Be~ m + 4Ce 2l0) + 9De 3<0) + 2 = 4
y"(0) = Ae
y-(0) = Ae (0) - Be~ {0) + 8O? 2,0) + 21De M0) = 10
jlO)
which may be rewritten
A + B+ C +
D = -I
A-B + 2C+3D=4
A + B + 4C+9D= 2
A - B + 8C + 27D =
System (7) has as its solution A = D = 0. B = -2,
2s
initial-value problem is y(s) = — 2e~* + e
+ s 2 + 2.
and
(J)
10
C = 1,
so the particular solution to the
1
SOLUTIONS
2.101
35
A general solution to a fourth-order differential equation is y(0) = A + B9 + CO 2 + DO 3 + 4 where A, B, C,
and D are arbitrary constants. Find a particular solution which also satisfies the initial conditions y(- 1) = 0,
/(-1)=1, y"(-l) = 2, y"'(-l) = 0.
,
I For the given function y(9), we have y'(0) = B + 2C6 + 3D9 2 + 40 3
y'"(6) = 6D + 129. Applying the initial conditions, we obtain
y"(0)
,
= 2C + 6D9 + \26 2
,
and
y(-l) = ,4 + fl(-l) + C(-l) 2 + D(-l) 3 + (-l) 4 =
y'(-l) = B + 2C(- 1) + 3D(- 1) 2 + 4(- 1) 3 = 1
y"(-l) = 2C + 6D(-l) + 12(-1) 2 = 2
y'"(-l) = 6D + 12(-1) =
which may be rewritten as
B+ C- D =
B-2C + 3D =
-1
5
(')
2C -6D = -10
=
6£>
System (7) has as its solution A = B = C = 1,
problem is y(0) = 1 + 9 + 2 + 20 3 + 4
and
D = 2,
12
so the particular solution to the initial-value
.
2.102
Find a particular solution to the initial-value problem
y'
solution to the differential equation is given implicitly by
2
2
=x +y
y(l)
;
= — 2,
xy
2
=
x 2 In x 2 + kx 2
y
if it is
known that a general
where k is an arbitrary constant.
,
I Applying the initial condition to the general solution, we obtain — 2) 2 = l 2 In
= 0). Thus, the solution to the initial-value problem is
(Recall that
In
(
2
(
)
+ /c(l 2
),
or
k = 4.
1
2
y
= x 2 In x 2 + 4x 2
y = — Vx 2 lnx 2 + 4x 2
or
The negative square root is taken so as to be consistent with the initial condition. That is, we cannot choose the
positive square root, since then y(l)
2.103
= yf\ 2 ln(l 2 + 4(1 2 = 2,
Find a particular solution to the initial-value problem
y'
solution to the differential equation is given implicitly by
2
f Applying the initial condition, we obtain
l
which violates the initial condition.
)
)
= e x /y; y(0) = 1, if it is known that a general
x
2
y = 2e + k, where k is an arbitrary constant.
= 2e° + k,
k =
or
— 1.
Thus, the solution to the initial-value
problem is
y = 2e —
,2
x
or
1
y = >j2e
[Note that we cannot choose the negative square root, since then
x
y(0)
—
1
= — 1,
which violates the initial
condition.]
To ensure that y remains real, we must restrict x so that 2e x — 1 > 0. To guarantee that y' exists [note that
x
we must restrict x so that 2e x — 1 # 0. Together these conditions imply that
y'(x) = dy/dx = e /y\
2e
2.104
x
-
1
> 0,
or
x > In \.
y — y sin t; y(0) = 1,
4
solution to the differential equation is given implicitly by
1/y = 4cosr + c u
Find a particular solution to the initial-value problem
f Applying the initial condition, we obtain
l/l
4
5
= 4cos0 + c lt
the general solution and solving explicitly for y, we obtain
y(t)
=
or
ct
l
if it is
known that the general
where c t is an arbitrary constant.
= —3.
\
I
\4 cos t — 3/
Substituting this value into
1/4
.
This equation makes sense
4 cos t — 3 > 0.
(a solution to an initial-value
Also, since y must be defined on an interval containing
we
that
if
is
the number in (0, n/2) such
initial
point),
see
contains
the
on
an
interval
that
problem is defined
= Arccos |), then the solution y is indeed defined on ( — 0, 0). Moreover,
that
cos = | (that is,
+
0"
->
and as t ->• -0
y(r)->+oo as t
only if
.
2.105
Find a particular solution to the initial-value problem x cos x + (1 - 6y )y' =
x sin x + cos x + y — y 6 = c,
to the differential equation is given implicitly by
5
y(n)
= 0;
if the general
solution
where c is an arbitrary constant.
2
= c, or
f Applying the initial condition to the general solution, we find n sin n + cos n + +
6
A particular solution is then x sin x + cos x + y - y = 1. We can rearrange this equation to
c
= — 1.
CHAPTER 2
36
x sin x + cos x + 1 = y 6 — y,
but since we cannot solve either equation for y explicitly, we must be content with
the solution in implicit form.
SIMPLIFYING SOLUTIONS
2.106
Verify and reconcile the fact that
y = c x cos x + c 2 sin x
y = A cos(x + B)
and
are primitives of
d2 y
+ ^ = o.
t4
d?
m
From
y = c { cos x + c 2 sin x,
we obtain first
y'
= —c sinx + c 2 cos x and then
1
y" = — c t cosx — c 2 sinx =
From y = A cos (x + B), we obtain first
To reconcile the two primitives we write
y'
—y
^y
-j-^ + y =
or
= — A sin (x + B)
dx1
and then, again,
y" —
— A cos (x + B) = — y.
y = A cos (x + B) — A{cos x cos B — sin x sin B) — (A cos B) cos x + ( — A sin B) sin x = c, cos x + c 2 sin x
2.107
Show that
In x
+ In (y 2 /x 2 = /I + x
2
2
#
Since we have
2.108
Show that
In x
2
I
sin (Arcsin x
x 2 —^
Show that
In (1
(1
2.110
Show that
In (1
= Be1
.
we may write
y
2
= e^ + x = e A e x = Be*.
xVl — y 2 — y Vl — ^ 2 = B-
Then, for a difference of angles, we have
— cos (Arcsin x) sin (Arcsin y) = xyjl — y 2 — y-Jl — x 2 = B
may be written as
xy + x + y = c.
+ y) + In (1 + x) = ln[(l + y)(l + x)] = A.
+ y)(l + x) = xy + x + y +
sinh y + cosh y — cx
I By definition,
2
= In y 2 = A + x,
— Arcsin y) = sin A — B.
+ y) + In (1 + x) — A
I We first note that
J
may be written as
sin (Arcsin x) cos (Arcsin y)
2.109
y
v2 \
/
v
+ In —^ = In
Arcsin x — Arcsin y = A
f We first let
may be written as
)
1
= eA = B
may be written as
and
Then
xy + x + y = B-l=c.
y = In x + A.
sinh y + cosh y = j(ey — e~ y ) + \{ey + e~ y ) = e y = cx.
Then
y = In c + In x = A + In x.
,
CHAPTER 3
Separable First-Order
Differential Equations
SOLUTIONS WITH RATIONAL FUNCTIONS
3.1
Define separable as applied to differential equations.
A first-order differential equation is separable if it can be expressed in the form
is
dy
A(x)
dx
B(y)
where A(x)
a function only of x, and B(y) is a function only of y. Such equations have the differential form
A(x) dx + B(y) dy = 0.
The variables x and y may be replaced by any two other variables without affecting
separability.
3.2
Prove that every solution of the separable differential equation A(x) dx + B(y) dy =
+ B(y) dy = c, where c represents an arbitrary constant.
j A(x) dx
I
given by
is
j"
Rewrite the differential equation as
identically in x; hence,
A{x) + B(y)y' — 0.
A{x) + B[y(x)]y'(x) = 0.
a solution, it must satisfy this equation
If y{x) is
Integrating both sides of this last equation with respect to x,
we obtain
§A(x)dx + JB[y(x)~]y'(x)dx = c
y = y(x),
In the second integral, make the change of variables
dy — y'{x) dx.
so that
The result of this
j A(x) dx + J B{y) dy = c.
substitution is
The two integrals may be, for all practical purposes, impossible to evaluate.
In such a case, numerical
techniques may have to be used to obtain an approximate solution. Even if the indicated integrations can be
performed, it may not be algebraically possible to solve for y explicitly in terms of x. In that case, the solution
is left
3.3
in implicit form.
Solve
x dx — y 2 dy = 0.
and B(y) — — y 2
The solution is j x dx + j ( — y 2 )dy — c, which,
2
after the indicated integrations are performed, becomes
x /2 — y 3 /3 — c. Solving for >' explicitly, we obtain the
2
1/3
solution as y — (fx + /c)
where k — — 3c.
i For this differential equation,
A(x) = x
.
,
3.4
Solve
y'
= y2x 3
.
x 3 dx — {l/y 2 )dy = 0.
I We first rewrite this equation in the differential form
B(y) —
— l/y
2
The solution is
.
j x dx + j ( — \/y ) dy — c
2
3
Solving explicitly for v, we obtain the solution as
3.5
Solve
y =
= x3
and
x 4 /4 -I- l/y — c.
-4
x
A
where
+ k'
k
Ac.
x dx + v dy — 0.
explicitly for y, we obtain the two expressions
Solve
,4(x)
or, after the indicated integrations,
f The variables are separated, so integrating each term gives
3.6
Then
y = \Jk — x
2
\
xdx + j y dy = c
and
v
= ->fk - x
or
2
,
\x
where
2
4-
\y
2
= c.
Solving
k = 2c.
x dx — y 3 dy = 0.
b
2
4
= c. Solving
f The variables are separated, so integrating each term gives J x dx — J y 3 dy = c or ±x
2X
2 1/4
2 1/4
and y = —{k + 2x
where k = -4c.
explicitly for y, we obtain the two expressions
y = {k + 2x
)
)
3.7
Solve
f
4
y / = x + 1.
4
y dy — (x + 1) dx = 0, which is separable. The
5
2
or, after the indicated integrations are performed,
jy — \x — x = c
This equation may be rewritten in the differential form
solution is
J y
4
dy — J (x + 1) dx = c
This may be solved explicitly for y to yield
3.8
,
Solve
y'
y = (fx
2
+ 5x + k)
1 ' 5
,
where
k = 5c.
= (x 4- l)/y.
37
—
38
CHAPTER 3
D
f
This equation may be rewritten as
=x+
yy'
dy — \(x +
J y
differential equation is separable and has the solution
integrations, we get
y = Jx
3.9
2
I
+ 4x + k
y = y
Solve
2
y~ 2 dy — dt.
j
y~ 2 dt =
J
dt
+c
or
2 2
c 1
/•
\
\
-jdy —
J
for y,
.
2
y
J
we get
y
=c
dt
t
-= dy — t 2 dt — 0,
y
—+
=
dz/dt = z
Solve
or, after the indicated integrations are
\
Z
dz —
J
,
—c
3
t
Solving explicitly
.
3
/c
3 2
t
.
=c
2
f
11
performed,
k = 3c.
where
r
-r dz — t
This equation may be rewritten in the differential form
J
which is separable. The solution
y
3
f
t/f
dt = 0,
2
which is separable. The solution
— —^
or, after the indicated integrations are performed,
J
— =
22T
/
explicitly for z, we obtain
3.12
This has the solution
y = - l/(r + c).
Thus,
dy/dt = y t
Solve
is
The
Performing the indicated
.
This equation may be rewritten in the differential form
3.11
dx = c.
— \x 2 — 2x = c. Solving for y explicitly, we obtain the two expressions
2
y = —y/x + Ax + k, where k = 2c.
Separating variables gives, in differential form,
is
1 )
2
- 1/}' = t + c.
3.10
\y
and
ydy - (x + 1) dx = 0.
or, in differential form, as
1
r
= +1 —
i
^
y
3
f
Solving
c.
3
:
k — —2c.
where
,
J
~7dx + dt = 0.
Solve
x
I
This equation is separable.
(assuming it is the unknown function), we obtain
3.13
Solve
s ds
+ s 3 (0 2 - 3) d0 =
-\x
Integrating term by term, we obtain
- (3t
\(M
+ k)
'
3
where
\
+
t
— c.
k
= —3c,
Solving explicitly for x
for s\0).
— 3)dd = 0, which has as its solution j s~ 2 ds + J(0 2 — 3) d0 = c.
— s~ + ^0 3 — 30 — c. Solving explicitly for we find
Performing the indicated integrations, we obtain
We can rewrite this equation as
I
2
s
ds + (9
2
'
s
3.14
= j6
l
3
(
Solve
-30-cr
.x
= x2 + \
s,
1
.
:
t
.
This equation may be rewritten as
dx dt = x 2 (t + 1)
or, in differential
form, as
—r dx —
(f
+ \)dt — 0.
x
Integrating term by term, we obtain
\t
2
— = c.
which may be explicitly solved for x(t). giving
t
x
x= -(\t +t +C)2
3.15
xx =(t- l) 2
Solve
I
1
.
.
This equation may be rewritten as
x
dx
—
= —
((
2
l)
,
x dx
which has the differential form
— (t — I) 2 dt = 0.
at
Integrating term by term, we obtain
3.16
x" 2 x = (f + 3) 3
Solve
\x 2 — j(t — l) 3 = c
or, explicitly,
x= ±[k + f(t — l) 3 ] 2
1'
.
I
This equation may be rewritten as
x
_ 2,
dx
—
= +
(t
3)\
which has the differential form
<//
v
:
dx - (t + 3) 3 dt = 0.
x= -
1
[c
4
+ i(/ + 3) ].
Integrating term by term, we obtain
-x
1
- ^(t + 3) 4 = c.
Thus
,
where
k = 2c.
'
SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS
3.17
2
x 3 dx + (y + l) dy = 0.
Solve
f
The variables are separated.
y =
we obtain
3.18
Integrating term by term, we get
— + (k - |x 4 1/3
1
)
dx
Solve
x
dt
2
Integrating term by term, we get
y = 3 -I- (k — fx )
3 1/5
for y, we obtain
— 2)
Integrating term by term, we get
x = 2 + rj(f- l)
2
Solve
+ /c] 1/3
.
.
(f
..„
.
.
— l)
.
— (s + 3) _1 + t~ = c.
l
=
$
dt
s
2
=c
.
(x
where
k — 3c.
„
dr
r
Solve
dt
which is separable.
Solving explicitly for x, we obtain
.
dt
3
\(s
+ 3) — jt = c.
3
+ 3r 2 + 3r +
x 2 - 2x +
i
{s
+ 3) 2 ds —
= — 3 + t/(\ — ct)
s
— 0,
dt
which is separable. Integrating
s
= — 3 + (f 3 + k) lli
,
1
dx
dr
5-
(r+1)
—\{r + l)" 2 + (x — 1)
Integrating term by term, we get
- 2c(x - 1)"
2
t
Solving explicitly for s, we obtain
This equation can be written in the differential form
"2
which is separable. Integrating
0,
2
3)
Solving explicitly for s, we obtain
m
1
Solving explicitly
— 2) 2 dx — (t — \)dt = 0,
—+ds^-^ —
j =
~~t
„
This equation can be written in the differential form
,
- = c.
+ 6s + 9
term by term, we get
3.22
y _ 3)5
t
term by term, we get
f
(
h -
2
This equation can be written in the differential form
Solve
x3
k = 3c.
where
,
3
(s
3.21
Solving explicitly for y,
2
dt
~.
3
— = c.
t-l
- 4x + 4
This equation may be rewritten in the differential form
/
l)
—
k = 5c.
where
,
(x
3.20
(v +
1-
4
The variables are separated.
f
4
k = 3c.
where
,
x
x 2 dx + (y - 3) dy = 0.
Solve
f
3.19
39
_1
= c.
3
:
(x-1)
2
Solving explicitly for r(x), we obtain
1/2
+
1
3.23
Solve
>•'
= x/(y + 2).
(y + 2)y' = x,
I We may rewrite this equation as
and in differential form as
(v -I- 2)dy
— xdx = 0.
The variables are separated, so term-by-term integration produces the solution \y 2 + 2y — \x 2 — c
2
2
= 0, where k = —2c. To solve explicitly for y we use the quadratic formula,
or y + 4y + (k — x
)
getting
y=
3.24
Solve
y'
4±v/ *6
4(fc
x2 )
^_ 2± ^4 _
(fe
_ x 2 ^_ 2±
)
^^2
= x 2/(y - 1).
I We rewrite this equation first as
(y
— l)y' — x 2 =
y
— 2y + (k — |x 3 = 0.
)
y =
2 ± v4
where
4{k
k = —2c.
3x3)
=
1
— \)dy — x 2 dx = 0.
— y — 5X — c or
and then in the differential form
Integrating term by term, we obtain as the solution of this separable equation
2
d = 4 _ k
where
\y
2
(y
3
To solve explicitly for y we use the quadratic formula, getting
+ 71 _ (k - |x 3 =
)
1
± JdTJx1
where
d = 1
—
CHAPTER 3
40
3.25
Solve
— = y-\
dt
I
y =
Solve
f
|y
2 + y/4 - 4(k -=!
2
dz
=
—
t
dt
z
t
+2
+3
2
t
- 2r)
z
/
r-,
Solve
k = —2c.
=
^
4
+
y
f
\z
2
+ 3)dz — (r 2 + 2)dr = 0, which is separable.
+ 3z — \t 3 — 2t = c or z 2 + 6z + {k — ft 3 — 4r) = 0,
= _ 3±V9 _
(z
_ jf3 _ 4t) = _ 3±
(fc
^ +
t3
4r
+
^ = 9 _ fc
where
<f
1
This equation may be written in differential form as
The solution is
d= 1 -k
To solve explicitly for z we use the quadratic formula, getting
-6±V36-4(fc-jt3_4t)
y'
where
.
Integrating term by term, we obtain the solution
=
t
= 1 ± Vl - (k - t 2 - 2f) = 1 ± Vt 2 — 2t + d
This equation may be rewritten in the differential form
where
3.27
2
— 2c. To solve explicitly for y we use the quadratic formula, getting
k =
where
3.26
which is separable.
(y - \)dy — (f + \)dt = 0,
— y — \t 2 — = c or y 2 - 2y + (k — 2 — 2t) = 0,
This equation may be rewritten in the differential form
Integrating term by term, we obtain the solution
(
x2
+ \)dx + J (-y 4 - lMy = c
j (x
+ 1) dx + — v 4 — \)dy = 0,
(x
after integration,
or.
which is separable.
5
—- + x —
y
—
y = c.
Since it is
algebraically impossible to solve this equation explicitly for y, the solution must be left in implicit form.
3.28
Solve
y'
x2 + 7
=
y
I
— 3y 4 dx — (x 2 + l)dx — 0. which is separable. Its solution is
10
5
3
or, after integration,
^y — }y — \x + Ix = c. Since it is algebraically
In differential form this equation is
9
J"
-3/'
9
(y
— 3y
4
)
dy — J (x
2
+ 7) dx — c
(y
9
)
impossible to solve this equation explicitly for y, the solution must be left in implicit form.
SOLUTIONS WITH LOGARITHMS
3.29
Solve
f
= 5y.
y"
5dx — (l/y)dy — 0.
This equation may be written in the differential form
5 dx
—
+
dy = c
5x — In |y| = c.
or, after integration,
To solve for y explicitly, we first rewrite the solution as
inM = e 5x ~ c
e
both sides. Thus,
.
Then the solution is
Noting that
5x
inM —
e
= 5x — c
In |y|
we obtain
|y|,
\y\
and then take the exponentials of
= e 5x e~ c or y = ±e~ c e 5x The
,
.
k — ±e~
y = ke
Note that the presence of the term — 1 y in the differential form of the differential equation requires the
solution is given explicitly by
c
,
.
since
y ^0 in our derivation of the solution. This restriction is equivalent to the restriction k # 0,
5x
However, by inspection, y =
is a solution of the differential equation as originally given.
y = ke
Thus,
y = ke
Solve
y'
restriction
3.30
I
.
5x
is
= A\\
the solution for all k.
where A denotes a constant.
In differential form this equation is
integration.
— Ax — c,
In \y\
sides of this last equation and noting that
y = ±e e
Ax =
Solve
= xy.
c
3.31
y'
ke
Ax
where
,
solution is
(l/y)dy — J A dx = c or, after
J
= Ax + c. Taking the exponentials of both
we obtain |y| = eAx+c = e e Ax Thus,
(\/y)dy — A dx — 0.
which may be rewritten as
k —
±e
e
ln|y|
= |y|,
Its
In \y\
c
.
c
.
Integrating term by term, we obtain the solution
(1/y) dy — x dx = 0.
— \x 2 — c, which we rewrite as In \y\ — c + jx 2 Taking the exponentials of both sides of this equation,
we get lyl = e c+x2 2 = e e xZ 2 so y = ke x2 2 where k — ±e
I The differential form of this equation is
In \y\
.
'
c
c
.
.
.
SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS
3.32
41
dy/dt = y(t - 2).
Solve
- {t - 2) dt = 0. The solution is j (1/y) dy - \ (t - 2) dt = c.
2
The indicated integrations result in In \y\
|(r
which may be rewritten as ln|y| = c + |U - 2) 2
2) = c,
Taking the exponentials of both sides of this equation, we get \y\ = ^+c-2) /2 _ ^^t-2^12^ so ^ at y _ ^u-2^/2^
= +e
where
I
In differential form this equation is
(1/y) dy
.
2
t
c
/c
3.33
'
n |y| + f'
3
=c
In \y\
>
(l/y)dy + 2f dt = 0.
2
.
c
so
.
(l/y)dy — (l/t)dt = 0.
In differential form this equation is
\y/t\
Integrating term by term, we obtain the solution
= c — ft
Taking the exponentials of both sides of this last equation, we
2til
\ where k=±e
y = ke~
3
dy/dt = y/t.
In |y| — In
—e
c
,
|f|
= c,
y/t
z
+ 1
= ±e
c
Integrating term by term, we obtain the solution
In \y/t\ = c.
Taking the exponentials of both sides of this last equation, we get
y = kt, where k=±e
or, rewritten,
so
dz
3.35
.
or rewritten,
= ec {2l3) ' 3 = e c e- 2,3 i\
\y\
Solve
I
-2yr 2
In differential form this equation is
get
3.34
Jy/dr =
Solve
f
.
c
and
.
Solve
dt
In differential form this equation is
In \z 4-
— In
1|
|f|
= c,
Solve
#
+ 1
which we rewrite as
In
z+
equation, we get
3.36
z
1
=e
z+
c
1
dt
Integrating term by term, we obtain the solution
0.
t
z+
= ±e
or
— —
dz
c
.
1
= c.
Then
Taking the exponentials of both sides of this last
z
= kt — 1,
uy
dy/dx — 8 — 3y.
We have
Separating the variables gives
— |ln |8 — 3y| = x + c,
Taking the exponentials of both sides gives us |8 — 3y| = e
3x
that
where k = ±\e~ 3c
y = f + ke~
The indicated integrations produce
'
\ In |3 + 5y| — x = c,
5c + 5x
= e 5c e 5x
|3 + 5y| = e
Solve
(5
3
ay
c
J
+ 5y
(5
- t) dx + (x + 3) dt =
8
term, we obtain the solution
In |x
t
,
where
k —
x + 3
—- dt = 0,
dx t
+ 3| — In |5 —
x + 3
t-5
±e
x + 3
=e
c
-5
t\
x + 3
=c
Hence,
.
which is separable. The solution,
or
In
t-5
x = — 3 + k(t — 5),
c.
Taking the exponentials
where
k =
±ec
.
c
.
dx + -
—^ dt = 0,
t-5
which is separable. Integrating term by
— 5\ = c or In |(x + 3)(r — 5)| = c. Taking the exponentials
Hence, (x + 3)(f — 5) = ±e
and
|(x + 3)(r - 5)| = e
+ 3| + In
In |x
of both sides of this last equation, we get
k
.
,
for x(f).
We rewrite the differential equation as
- 5
Integrating term by term
for x(r).
- t) dx - (x + 3) dt =
—3 H
dy — dx = 0.
.
of both sides of this last equation, we get
x —
with solution
In |3 + 5y| — 5c + 5x.
Taking the exponentials of both sides gives
+ 5y - ±e 5c e 5x Then y = -| + ke 5x where k = ±\e 5c
obtained by integrating term by term, is
Solve
= dx,
so that
or
We first rewrite the differential equation as
3.39
— 3v
dy/dx - 5y = 3.
3
3.38
c
= p\dx + c
— 3v J
which were written as In |8 — 3y| = — 3x — 3c.
— e 3c e~ 3x or 8 - 3y = ±e~ 3c e„- 3x SO
Separating the variables, we obtain the differential form
gives
±e
.
,
Solve
k =
dy/dx + 3y = 8.
8
3.37
where
\t
c
c
.
\
42
3.40
CHAPTER 3
D
Solve
dy/dt = -y/t.
#
where
Integrating term by term, we get
(1/y) dy + (1/r) dt = 0.
= c. Taking the exponentials of both sides gives us |yf| = e or yt — ±e Thus
k — +e
Solve
ydy + (y 2 + l)dx = 0.
In differential form this equation is
or
3.41
c
In \yt\
c
.
2
y
|ln(l + y 2 ) + x — c
+y =e
2c ' 2x
2
.
f
In (1
)
2
y x dy + (y
Solve
)
1
.
,
.
- 1) dx = 0.
3
We rewrite this equation as
-=
y
2
.
dx — 0,
which is separable. Integrating term by term, we get
x
1
+ 3 In |x| = 3c and then as In |(y 3 — l)x 3 = 3c.
3c
3
Taking exponentials gives |(y — l)x = c
The solution in implicit form is
so that (y — l)x = + c
ic
3
3
= kx~
Solving explicitly for y, we obtain y = (1 + kx~ 3 113
then y —
where k = ±e
3
I In |y
1
H- In |x|
1
= c,
—
1
dy h
y —
1
—
which is separable. Integrating term by term, we get
1
+ y 2 = 2c — 2x. Taking the exponentials of both sides, we obtain
Then y 2 — — + ke~ 2x where k — e 2c Solving explicitly for y, we obtain as
or, rewritten,
2x
2c
dy + dx = 0,
+
— e e~
=
±(-1 + ke~ 2x 112
y
the solution
3.42
=c
|r|
y = k/t,
.
We rewrite this equation as
1
+ In
In \y\
c
which we may rewrite as
3.43
2
Solve
(y
k —
3.44
±e ic
3
3
+ 3y| — In |x + 1| = c,
^
+ 3y
.
)
+ 3y
= e 3c
(x+ l) 3
3
—
+ 3y
t= ±e
(x+ 1) J
so that
,
dx — 0,
which is separable. Integrating term by term, we
x + 1
In
3
+ 3y
3
(x + l)
y
from which we find
y
y
dy -
This yields
.
Taking the exponentials of both
3c.
y
3
+ 3y = k(x + l) 3
,
where
as the solution in implicit form
(4x + xy ) dx + (y + x y) dy =
2
Solve
f
,
|
.
— —— 4±Li,— —
3
sides gives
1|
3
+ l)(x + 1) dy = (y 3 + 3y) dx.
y
j In |y
—
.
,
• ~
..
We rewrite this equation as
get
3
,
|
1
In |y
3c
3
3
2
for y(x).
x(4 + y )dx + y(l + x )dy =
2
2
This equation can be written as
= H
= 0. Integrating term by term gives
+ xz 4 +
2c
2
2
2
2
In [(1 + x )(4 + y )] = 2c
or (1 + x )(4 + y = e
2c
where k = e
j In (1
=-
1
and then separated into
+ x 2 + j In (4 + y 2 = c,
)
)
which we rewrite as
v
Thus the general solution is
.
)
(1
+ x 2 )(4 + y 2 = k.
)
.
3.45
Solve
(y
3
+ y)(t 2 + 1) dy = (fy 4 + 2y 2 t) dt
^
This equation can be written as
iln (y
4
+ 2y 2 - \ In (t 2 + 1) = c,
)
for y(f).
—+—
y
-£
3
y
-^ dy
t
- -^
df
which we rewrite as
After taking the exponentials of both sides, we have
In (y
2
3.46
k(t
2
+ l) 2
k = e
where
,
Solve
xv dv - (1 + v 2 dx = 0.
I
„
...
4
which is separable. Integration yields
+ 2y 2 - In (f 2 + l) 2 = 4c
)
y* + 2y 2
—
2
(t
/ + 2y =
= 0,
—j = c 4c
+ 1)
,
y
or
4
In -^
+ 2y 2
-y = 4c.
which yields the implicit solution
Ac
.
)
....
—_v_
—j ^
We first separate the variables and rewrite the differential equation as
dx — 0).
-
+v
2
1 + V
-
t-
Then,
1
integrating term by term yields
Thus, in implicit form,
1
+
2
t'
^ In (1 + v
= /ex 2
,
2
)
— In |x| = c,
where
k — e
2c
.
from which
In
2
—=
_
2c,
1
so that
+ V2
—=
^
e
2c
==
<
SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS
3.47
2xv dv + (v
Solve
#
- 1) dx = 0.
2
2
We first separate the variables and rewrite the differential equation as
-r
v
integrating term by term yields
\x(v
3.48
— 1)| = e
2
from which
,
3xv dv + (2v
Solve
I
c
2
k= ±c 4c/3
#
1
+ - dx = 0.
Then,
x
Taking exponentials gives
2
2
1|
c
= ±e
c
where
1
.
/c
.
3d
—
2v —
- + In |x| = c,
3 4
4c
(2v — l) x = ±c
2
so that
1|
2
1
dv + - dx = 0,
2
2v
In
|
which is separable. Integrating term by
x
1
-
+ In |x| 4 = 4c,
3
l|
and
and the solution, in implicit form, is
,
2v
In \(2v
2
—
1
- l) 3 x 4 = 4c.
2
|
= k/x 413
,
where
.
4xo dv + {3v
Solve
1
di;
—
- 1) dx = 0.
fin \2v
Exponentiation gives
3.49
x(v
—
2
+ In |x| = c, so that In \x(v 2 — 1)| = c.
— 1) = ±e Thus, in implicit form, v 2 — = /c/x,
In \v
This differential equation may be rewritten as
term, we obtain
43
2
-l)dx = 0.
4o
—
dv — dx —
x
3o —
1
This differential equation may be rewritten as
A
which is separable. Integrating term by
0,
1
term yields
|
3v
2
where
— l) x = 3c,
3c/2
k — ±e
Solve
x(v
In \{3v
3.50
§ In
I
2
2
3
|
+ In |x| = c, which we rewrite as In \3v 2 — 1| 2 + In |x| 3 = 3c. Then
3c
2
2 3
so that (3v — l) x = ±e
and the solution, in implicit form, is 3i; 2 —
—
1|
2
yln \v 3 — 3v\ + In |x| = c,
term yields
3
— 3o)| = 3c,
3
(i;
Solve
I
3
so that
|x (y
v
2
-=
+ x 3 )dy = x 2 ydx
(1
In
1
+ x 3 = 3c,
3
2
Solve
(t
y
3c
In |t>
— k{\ + x 3
In
+ 4) dx + 3f(l - 2x) dt =
+ 4) 3
1
-2x
= ±e 2c = k
so that
'
t
3f
= 0.
+ 4) 3 = k{\ — 2x),
2
2
(t
x =
Solve
I
zz = (z
2
+ l)/(t + l) 2
+ 4)-
(t
+ 4)
1
-2x
In
3
2c.
Taking exponentials, we get
and in explicit form
1
A(t
2
+ 4)
3
where
A —-
.
We rewrite this equation as —2
z
— dt =
T dz-
2
+ \
0.
Integrating term by term yields
1
\\n{z 2 + 1) +
Then
and
z
2
_|_
^
so that, in
;
2k
3.53
,
or
Integrating term by term, we obtain
+4
which is equivalent to
(t
= ±e
|
x(t).
—2
-\
dx
"
- 2x
+X
+ x3 = c
.
2
2
x 3 (v 3 — 3v) = ±e 3c
- | In |1
1
±e 3c
for
1
— 2x| + |ln(f 2 + 4) = c,
(t
Then
In implicit form, the solution is then
.
+x 3
k =
where
),
1
|1
— 3o| + In |x| 3 = 3c.
dx — 0. Integration yields In
+ x Jf
y
= 3c. Exponentiation gives us
1
so that
We rewrite this equation as
— | In
3
which is separable. Integrating term by
x2
1
- dy
1
implicit form,
1
dv + — dx = 0,
— 3v
x
for _y(x).
We rewrite this equation as
3 In
-1
which we rewrite as
— 3o)| = e
3
y
3.52
,
- 1) dv + (v 3 - 3o) dx = 0.
v
3.51
= /cx" 3/2
.
This differential equation may be rewritten as
In |x
1
2
=c
ln(z + l) = 2c
or
t+ 1
_ e 2c-2/(» + l) _ e 2c e -2/(t+l) _ ^ e -2/(t+l)
z= ±(-l+/cc- 2/(I+1) 1/2
)
.
t
+
1
.
1
44
3.54
CHAPTER 3
Q
Solve
+ l)i = 4r(z 2 + z).
(2z
-=
This equation has the differential form
+z
z
\z
2
'
2
c
'
2
'
—+
v
1
— dx
Solve
I
4
X
and
3y
- dx
Solve
x
f
In -r
In |x|
2
+ In jl - 2u
X4
+1
= 4c.
By taking exponentials we get
Solve
y
|
m
du
—
=
—
d
v
1
-r
i
so that
= e**,
+
which gives us
dv — 0.
= 2c.
3
— jln(r 4 + 1) = c,
4
This equation is separable. Integrating term by term, we get
In x
3.57
.
2
- 2r 3j
1
We may solve for z explicitly,
k = e**.
where
1
3.56
.
1
— In (y* + 1) = Ac
= /c(r 4 + 1),
c
dv — 0.
v*
v
4
— ±e
1
=
This equation is separable. Integrating term by term, we get
In x
x
k
3
-.
x
z
Then exponentiation gives
.
2
,
using the quadratic formula to obtain
3.55
Integrating term by term, we get
+ z\ — 2t 2 — c, which we may write as In \z 2 + z\ — c + 2t 2
from which z 2 + z + ke 2 — 0, where
+ z\ = ec+2 = e e 2
— + yj\ — 4ke 2t2
In \z
2
dz — 4t dt — 0.
2
Writing this as
In |x (l
- 2r
3
)|
In |x|
+ }ln |1 — 2i> 3 = c,
|
= 2c.
we then find that
from which we obtain
x 2 (l - 2r 3 - ±e 2c = k.
)
u
\-u
1
-^— du
In differential form, this equation is
u
dy = 0,
which is separable. Integrating term by term, and
y
noting that
J
we have as the solution
In \u\
u
A
J \u
— In \y\ — c,
uj
u
which tnav be simplified to
u
3.58
Solve
x(l
- =
— c — In \uv\.
u
+ y/v)dv + vjvdx = 0.
—
Separating the variables, we obtain
dv +
— dx — 0.
Integrating term by term, after noting that
x
VyJV
+^
w +"-H
/^*-/(^^)*-*1
—— +
2
we get
In \v\
2
+ In |x| = c,
p + In |i"x| = c.
so that, in implicit form,
yJV
3.59
Solve
I
yJV
xv dv - (1 + 2r 2 + iA ) dx =
_
Separating the variables, we obtain
—^
_v_
V
-^—
(1
+ r2
I
dv
dx — 0.
Term-by-term integration yields
)
— In |x| = c. which we rewrite as — (1 + r
= In x + In \k\, where
—{(1 -I- u 2)
2
2
so that, in implicit form, (1 + r )ln|fcx = — 1.
Then -(1 + r 2 )" = In \kx 2
2
1
2
)
'
k — e
2c
r
!
.
1
\,
3.60
Solve
f
dy/dx = y- y
|
2
.
We rewrite this equation as
1
y — y
T dy — dx = 0.
which is separable. The solution is
J
y ~ y
Now, using partial-fractions techniques, we have for the leftmost term.
y
/A-Js£rJG iy*-w-M'->i-* i-y
+
= dx -
r
\dx = c.
J
SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS
The solution thus becomes
—
y
= e + x — e ex
c
-
\-yy
ke
y —
which
y
so that
,
—x=c
\-y
or, after rearrangement and exponentiation,
= ±e e x — kex
c
1-y
To solve explicitly for y, we write
.
4xdy — ydx = x 2 dy.
Solve
/•
The solution is
1
—
J x(x
y dx + (x
— 4x) dy =
2
- 4)
J
-4
x —
4
(x
= 4x.
1
J
— dx
+ - dy = 0,
x(x — 4)
dx = - In x — 4
x
J
4
5 In |x — 4| — \ In Ixl + In lyl = c
The solution then becomes
In
1
- 4)y
or
'
In |x
1411x
In
— 4| — In Ixl + In y 4 = 4c,
4
+e
Exponentiation then yields
y =
from which
,
+
k= +e*
x(y — 3)
y- 3
We rewrite this equation in the differential form
4
(>
dy —
J
which has as its solution
x
4
- dx = c.
By the method of partial fractions,
x
/
I 1
7
" 3,,,w
1
*'"J'( "j)*'" ,
so the solution becomes
y — 3 In |y| — 41n|x| = c.
~
c
3 4
3 4
y c =
=
that
e
e~
e\
so
x
y x = ke\ where
|y
3
Thus
fc
|
Solve
dx = 0,
dy
y
3.63
ln|y x
= ±e~
4
|
=y — c
and, after exponentiation,
c
.
x 2 (y 4- 1) dx + y 2 (x - 1) dy = 0.
x
This equation may be rewritten as
x-1
dx H
y
y+1
dy = 0,
with solution
c
x
J
x - 1
Since
S^
and
$JT[
dx
=${ x + ,+
d
> = S{r-
l+
^)
ih)
{
dx == -x 2 + x + lnlx - ll
'
2
dy =
1
2
y2 - y + ,nly+,
\
the solution becomes
2
lv-2
x
4-
x + In |x - 1| + \y 2 - y + In \y + 1| = c
x 2 + 2x + y 2 -2y + 21n|x- 1| + 21n|y + 1| = 2c
or
or
(x
or
3.64
where
.
dx
Jy v
,
c
J- =
—3
so that
1/4
kx
x~^~4
x
/
which is separable.
y
For the leftmost term, the method of partial fractions gives
1/4
(x - 4)y
—
or
1
— dx + r -dy = c.
— 4)
J v
J x(x
c(x
Solve
from
3
This equation may be rewritten as
3.62
y = /ce*(l - y),
+ ke x
1
3.61
c
In
45
where
k = 2c + 2.
Solve
-f = y
dt
2
- y3
.
2
+ 2x + 1) + (y 2 - 2y + 1) + 2 In |(x - l)(y + 1)| = 2c + 2
2
2
(x + l) + (y - l) + 2 In |(x - l)(y + 1) = k
dx +
r
y
J
y+ 1
dy
=
CHAPTER 3
46
#
This equation may be rewritten in the differential form
~2
dy — j dt = c.
3
— dt = 0.
_ 3,3 dy
3,2
with solution
By the method of partial fractions.
1
1
3,2(l_j,)
3,3
1
1
y
y
1
= -5-2 + --
so after the indicated integrations the solution becomes
— In ll — vl — — c
In Id
1-
!__v
t
or.
rearranged.
y
1
y
In
3.65
- = t + c.
y
-y
1
Solve
(f
+ \)yy =
1
- y2
.
I
This equation may be rewritten in the differential form
y
-,
1
dy H
-l
2
)'
— 0.
dt
t+
with solution
1
2
Integrating directly, we have
^ In v — ll + In
+ ll = c, so that
dv + f
dt = c.
J
—
+
2
2
2
2
Exponentiation then yields |(y 2 - l)(f + 1) 2 = e
ln|y - 1| + ln(f + l) = 2c or In \(y - l)(t + 1) = 2c.
2
2
2
2
= ±e \ from which y = + k/(t + l) 2 where k = ±e 2c
or (y
l)(f + l)
-=
I
J
-
v
1
If
I
1
t
|
1
Solve
y
'2
d~y
+U
4
dy + -
This equation may be rewritten in the differential form
y
r
J
'
,
(/\+
P 2
+ "
rdu — c.
,
L
u
r
r
By* the
method
of *partial fractions.
r,
•
becomes
the indicated integrations, the solution
4 In lyl + 8 In |»| — In (1
+ u 4 = 4c.
)
,
—- = -2 + "*4— = 2
u + ir
4
1
3.67
AuH
\
which may be written as
In
+u
)
+ u
= c.
= 4c.
A
u
)
"
3
-.
+ w4
1
Thus, after
Multiplication by 4 gives
Exponentiation yields
8
+u
c
4
Solve
I
u(\
4
+ 2 In |u| — | In (1 + u
In \y\
which has as its solution
2 + "*
1
y u
—rdu = 0,
u + ir
'
4
u + ir
J
y
.
.
U + IT
du
3.66
r
4 '.
+ x
4
which we write as
y i<
8
= k{ + u 4
1
k = e
where
).
4
—=
This equation may be rewritten as
dv
=
x—
dx
r(r
2
5
v
l
+1)
— or
-
v
1
dx H
2
-
—
I
5
dr = 0.
Mr- + 1)
1
Using the method of
partial fractions, we can expand this to
2v
1
dx +
+ T
r TT
r
dv =
2
— In
The solution to this separable equation is
In |.v|
—x(n +
\)
where
Solve
v
2
3.68
2
|
= ±e
c
.
Then
—
dy
x(u
2
+ 1) = kv,
|r|
k -
+ In (c 2 + I) = c,
±e
\.\U-
so that
+ 1)
In
-5
In differential form, this equation is
c
.
w+u
du + - dv = 0.
y
which is separable. Using partial-fraction
techniques, find that
>
(^±^du= J(-"^^du^ J\u
|f?-— L- M = 21n|u|-ln|u+
J
u+lj
2
)rf
+
and
for n( v).
u + 2
u
=c
11
u(u +
'
1)
'
'
If
'
= In
u
+ 1
SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS
so the solution to the differential equation becomes
= c,
+ In \y\
11
In
u + 1
k =
u
2
yu 2 = k(u + 1).
c
±e or
- (k/y)u — (k/y) =
where
,
2
y{u
1
To solve for u explicitly, we rewrite this last equation as
2
+ A(k/y)
A 2 ± A^/l + Ay
=
A = ±yfk
where
2
Solve
= k,
u +
and then use the quadratic formula to obtain
k/y ± J(k/y)
3.69
from which we obtain
47
2y
+ 2) du - (u 3 - u) dy = 0.
dy = 0.
-j— - du
Separating the variables gives us
—u
u
By the method of partial fractions, we have, for the
y
leftmost term,
u
J
u
+ 2
3/2
Ju(u-l)(u+l)
i{u - \){u + 1)
-u
3
2
J\
7 +
u-\
u
\
—3/2—r)du
= —2
3
L
3
+-ln|u-
In |m|
U+\J
ll
+-ln|u+ ll
Then the solution to the differential equation is
3
— 2 In
I
*S
I
I
I
-4 In |u| 4- 3 In \u -
or
(u
so that
- l) 3 (u + l)
In
u*y
This yields
3.70
3
\u +
+ - In \u — ll + -In
T
\u\
Solve
y(u
(u
2
2
— l) 3 = ku*y 2
+ 2) du + (u 3 + u)
where
rfj;
= 0.
I
u
1
+ 3 In \u +
3
(u
and
2c
2
,
1
— In \y\ — c
I
I
1
1
= +e 2c
/c
2
I
- l) 3 (u + l)
3
= +e 2c
2
.
+2
+u
1
By the method of partial fractions, we have, for the
-
u
•*
- 2 In lyl = 2c
u*y
—Jj— - du + - dy — 0.
Separating the variables gives us
ll
I
y
leftmost term,
I
-i
J u
+u
du =
\
J
^
—
u(w +
i{u
du
2
Jly +
1)
U V
In-
to
ir +
3.71
Solve
y(u
4
2
2
- = 2c.
U V
—
In (u
2
2
+ 1) + 2 In "
|i/|
which may be simplified
2
— e 2c
-=
2
u +
Exponentiation gives
1
wj
— ^ln(u 2 + 1) + 2 In lul + In lyl = c,
Then the solution to the differential equation is
4
Ju =
+
i
A
so that
,
u y
2
= k(u 2 + 1),
where
k — e
2c
.
1
+ \)du + (u 3 - 3u) Jy = 0.
I
m
Separating the variables gives
2
+
du + - dy = 0.
-3u
y
—
1
—3i
u
u
2
+
J"U"
•^
1
1
Using the method of partial fractions, we find that
u
du =
3u
2
+
u(u
- V3)(u + 73)
1/3
=
1
du
J'
/H
\
2/3
2/3
- 73
u + 73/
du
u
2
2
= — - In \u\ + - In |u - 73| + z In |u + 73|
Then the solution to the differential equation is
—3 In \u\ + fin \u — \J3\ + § In \u + V3| + hi
multiplication by 3 and rearrangement, this solution becomes
(u
exponentiation gives
k =
3.72
+e ic
Solve
- 73) 2 (u + 73)V
.
x(v
2
- 1) dv + (v 3 - Av) dx = 0.
— +c 3f
,
(u-73) 2 (u + 73) 2 V 3
In -
which we may write as
-
(ir
|.V'|
= 3c,
— 3) 2 y = /cu,
= c.
After
and
where
1
CHAPTER 3
48
*
•
,
,
2
-1
dv H
dx = 0.
—
v*
4v
x
v
•
—
1
-=
Separating the variables gives us
Using the method of partial fractions, we find that
2
~
v - {
*
rA/4 + —3/8 + —3/8 \)dv = -1 In
—
= C
dv
dv = \[
3
J v - 4v
J v{v - 2)(v + 2)
J\v
4
v-2 v + 2J
C v
— — —
2
l
,
.
,
3,
3,
„.
„.
+ - In \v - 2 '8
+ - In \v + 2
.
.
t;
.
.
1
8
'
'
'
\\n \v\ + |ln \v — 2\ + fin \v + 2\ + In |x| = c.
3 8
2
3
= 8c, and exponentiation gives
In \v (v — 2) (v + 2) x
Then the solution to the differential equation is
Multiplication by 8 and rearrangement yield
2
v {v
3.73
- 2) 3 {v + 2) 3 x 8 = ±e 8c
Solve
x(v
2
,
|
2
which may be written as
v {v
- 4) 3 x 8 = k,
2
k =
where
±e Sc
.
+ I) do + (v 3 - 2v) dx = 0.
f
——+ — + — =
2
v
1
1
-=
Separating the variables gives us
dx
dv
+1
f£±i»-f
-2v
w _ r
J
v
3
J
r
- y/2)(V
J2)( V + 72)
J2)
.
V{V
t
Using the method of partial fractions, we find that
0.
x
2v
D
».r(Jg + -*U-*U»
J \
_
v
V-yJl
v
V
V
V
+ yfi
= -^\n\v\+l\n\v-y/2\+l]n\v + j2\
— \ In + f In |r — v2| + f In \v + v2| + In |x| = c.
3
3 A
(v
J2) (v + j2) x
= 4c, and exponentiation gives
In
The solution to the differential equation is, therefore,
Multiplication by 4 and rearrangement yield
(v
3.74
— >/2) 3
Solve
(i>
x(t<
2
+ v2) 3 * 4 = kv 2
,
where
— ±c 4c
k
|r|
which we may write as
,
(v
2
— 2) 3 x 4 = kv 2
.
- I) do + (v 3 + 2v) dx = 0.
—- —
2
#
o
1
-=
- dv + - dx = 0.
tr + 2v
x
Separating the variables gives us
c v
J
r
2
3
-\
+ 2r
r
r
,
-l
2
J i^c
2
r/
,
J\
+ 2)
(v
+ 2) x
(3/2)p\
v
|t;|
2
+ 2)
1
..
2
'
'
+ |ln(f 2 + 2) + In |x| = c.
3,
_
.
4
After multiplication by 4
4
— — - = 4c,
In -
and rearrangement, this becomes
where
3
v
— \ In
Thus, the solution to the differential equation is
2
— 1/2
Then, by the method of partial fractions,
5
so that exponentiation gives
(v
2
+ 2) 3 x 4 = kv 2
.
k — e^.
SOLUTIONS WITH TRANSCENDENTAL FUNCTIONS
3.75
Solve
dy/dx = y 2 + 1.
I
1
^dy — dx = 0,
By separating the variables we obtain
The integrations yield
3.76
Solve
arctan y — x = c,
from which
—
dy - [dx = c.
Then
yf y = tan v/6(x + c),
3.77
Solve
dy/dx - y 2 + 2y + 2.
#
j
!
^ dy
— r dx — c.
\
y = tan (x + c).
dy/dx = 2v 2 + 3.
— —- dy — dx =
By separating the variables we obtain
f
r
which has the solution
^
—
The integrations yield
so that
.
which has the solution
arctan >/§ y — x = c,
or
arctan v |.v = \ 6(x + c).
y = Vf tan *j6(x + c).
1
.
We separate the variables to obtain
2
y
+ 2y + 2^
i..
dy
=
^-^
+
+2
f
J y2
0,
2y
the solution to the differential equation is
^—- =
r
f
J
dy - dx = 0.
1
+ (y + l) 2
arctan (y + 1) - x = c;
Now, since
arctan (y + 1)
hence,
y = — 1 + tan (x + c).
SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS
3.78
Solve
dy/dx - y 2 + 6y = 13.
dy — (y 2 - 6y + 13)dx
We first write the equation as
dy =
Sy 2 -6y + l3
5(y-3) 2 + 4
1
3.79
the solution to the differential equation is
- arctan
\{y ~ 3) = tan (2.x + /c),
or
dy =
y —
l$i + H\-3)
3
dy - dx = 0.
-=
2dy = \* TCl * n
y - 3
x = c
or
^
= 2(x + c).
arctan
Since
Thus,
y = 3 + 2 tan (2x + /c).
3
dy
y
dx
2
Solve
*
k = 2c,
where
and then as
x(y
+ 4y
+ 2y + 4)
—+ + + dy-2
o
L
•
,.
4 J
2y
y
,
Separating the variables gives us
3
y
1
dy
A
dx = 0.
Then, by the method of partial fractions,
x
4y
we obtain
J
y
3
+ 4y
J
y(y
2
+ 4)
Then the solution to the differential equation is
arctan (y/2) — In |/cx/y|,
dy
3.80
y
Solve
2
x (2y
dx
#
i
k =
where
±e
2y
..„„_.
+ lOy + 9 J
y3 + 9 y
'H
or,
rearranged,
—+ + 9 dy
—+10y
2
1
J
j
3
^ dx
= 0.
And, by the method of partial fractions,
9y
2
+ lOy + 9 J
y + 10\ ,
P /l
dy
+
=
2
^
J
y(y + 9)
y^9 )
[y
+
1
^ + tW
/G t
dy = ln |y| +
m
2
ln (y + 9) +
+ -1,_,.2
10
y
—
arctan
,
,
ln \y\
+ 9) +
ln {y2
^
1
t arctan
= c,
which we write
as
2 ln \y\ + ln (y
and then as
Solve
/
+ arctan (y/2) — In |x| = c
In \y\
The solution to the differential equation is, then,
3.81
2
.
r 2y
=
first
+ arctan -
+ 4/
c
y
J
y
+ 9y
+ lOy + 9)
.__ 4t
2
\y
3
2
Separating the variables gives us
r 2y
2
J
(v
2
2
ln |fcy (y
2
+ 9) + ln \k\ + — arctan ^ = -
k =
where
±e~ 2c
+ 9)1+^ arctan (y/3) = 2/x.
2
+ 2v + 2) dx + x(v-l)dv = 0.
—
17—1
...
o
Sepa
Separating
the variables gives us
1
-z
dv + — dx — 0.
tr + 2v + 2
x
.
Then, by the method of partial fractions,
we have
v-1
J
2
l)-2
J (u + l) 2 +
r (i>+
,
+ 2u + 2
t;
O+
l)
2
1
J
2
J
1
= - In
y+
r
,
(i>
+ l) +
1
1
„ r
J
i
+(v + iy
dr
+ 1] - 2 arctan (y + 1)
The solution to the differential equation is thus \ ln (v 2 + 2v + 2) — 2 arctan (u + 1) + In |x| = c, which we
2
2
multiply by 2 and rearrange to obtain ln [x (t> + 2v + 2)] - 4 arctan (t; + 1) = k, where k = 2c.
3.82
Solve
(u
+ 2e u dy + y(l + 2e u du = 0.
)
)
#
1
Separating the variables gives us
+ ln |u + 2e u = c,
u
where
y(w + 2e = k,
ln |y|
\
)
,
1
which we rearrange to
k —
±e
c
.
—
+ 2e u
J
n
- dy H
dw = 0.
u + 2e
y
ln |y(u
T
Integrating term by term then yields
+ 2e u = c.
)\
Exponentiation then gives the solution as
49
50
3.83
CHAPTER 3
Q
dy/dx = sec y tan x.
Solve
f
Separating the variables gives us
dy sec y = tan x dx.
cos y dy = (sin x/cos x) dx.
which we rewrite as
Then integration yields
r
3.84
rflicos.x)
di cos x)
—
cos v «v =
J
or
sin x
,11,
— —In cos x + c.
cos x
J
dy/dx = tan y.
Solve
f
tan y = sin v cos v.
Since
the differential equation mav be rewritten as
cos v
dx — dx — 0.
Integrating
sin y
term by term, we obtain the solution
sin y
3.85
Solve
I
3.86
k —
where
,
y = ey
±e
c
Separating the variables gives us
y
=k —
e~
Solve
y = te y
t.
y = —In (A:
Solve
y = ye'.
— ^f 2
— c + e'.
Solve
y = y
5
),
where
e~ y dx — dt — 0,
and
r ).
e~ y dy — t dt — 0,
and integrating term by term yields
— e~ y — jt 2 = c.
- dy — e' dt = 0.
y
and integrating term bv term vields
= e + e = e ee
c
\y\
c
'
y = ke '.
e
Thus,
'.
In \y\
where
— e' — c.
k =
±e
sin t.
dx
Separating variables vields
r dx
—r = sin
and integration gives us
dt
t
J
y
— iy -4 = — cosr + c.
Solve
— e~ y — t — c.
and integrating term by term yields
y = — In (k —
so that exponentiation yields
m
3.89
= c 4- x. Then
k — — c.
Separating the variables gives us
3.88
In [sin y\
.
Hence
In \y\
which we rewrite as
y = arcsin ke*.
k — —c,
where
Separating the variables gives us
Then
— x = c,
In |sin y\
so that
.
.
Hence
f
3.87
= kex
or
— = 4cosr +
k =
where
A:.
(•
-^ = sin i dt t- c.
J
y
Therefore.
— Ac.
x dv - y/l-v2 dx = 0.
di
N
arcsin v — In |x| = c
dx = 0.
-
Separating the variables gives us
1
- v2
and then integrating term bv term vields
x
as the solution in implicit form.
HOMOGENEOUS EQUATIONS
3.90
Define homogeneous with regard to first-order differential equations.
I
A first-order differential equation in standard form
dy
—
=
dx
f(x, x)
is
homogeneous if
fitx. tx) = fix.
x)
for every real number r in some nonempty interval.
Note: The word homogeneous has an entirely different meaning in the general context of linear differential
equations. (See Chapter 8.)
3.91
Determine whether the equation
y'
= (y + x)x
is
homogeneous.
I The equation is homogeneous because
fu
f(tx. tx) =
,
*
—+ = — + = + =
tx
ty
tx
3.92
Determine whether the equation
y'
= y2 x
is
x)
t(y
tx
homogeneous.
x
y
X
f(x, VI
c
.
SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS
I The equation is not homogeneous because
tx
3.93
Determine whether the equation
y'
2xye xly
r—
+ y sin (x/y)
— —=z
is
:
x
'
X
tx
homogeneous.
I The equation is homogeneous because
3.94
Determine whether
'
ty)
y'
2
,x " y
x/y
,y
t 2xye?
2xye
_
_
= /( *' j)
~ (tx) 2 + (ty) 2 sin (tx/ty) ~ t 2 x 2 + t
~
2
sin (x/y)
x + 2 sin (x/y)
2(tx)(ty)e
f{tX
V
= (x 2 + y)/x 3
y
homogeneous.
is
I The equation is not homogeneous because
3.95
Determine whether
y =
fx J
rx
(txY
2y* + x 4
r
xy
—
homogeneous.
is
I The equation is homogeneous because
/(fX
3.96
Determine whether
y'
=—
^
+ (txf
= 2(tyf
(rx)(^)
=
3
2fV + 4 x 4 = 2y 4 + x 4 =
f
'~?V" ~^^
/(X
'
'
>)
2xy
=
x
)
—
r
is
homogeneous.
_y
f The equation is homogeneous because
2
(fx)
3.97
Determine whether
/=
x2 + y2
is
z
2xy
2t xy
2(fx)(0')
— (f v)
rlx"
— )^)
1
x
z
— yz
homogeneous.
xy
I The equation is homogeneous because
(r x)
/('*' ty) =
3.98
Determine whether
y'
= (y — x)/x
2
— — —r—
+ (ty) 2 _ t 2 (x 2 + y 2 _ x 2 + y 2
^-7
=
=
= Z( x y)
2
)
/...w..a
(tx)(ty)
xy
t
'
xy
homogeneous.
is
I The equation is homogeneous because
,,,
.
.
=
f(tx, ty)
—ty
tx
—t(y
=
Determine whether
y'
= (2y + x)/x
is
= y-x = f(x, y)
X
tx
tx
3.99
x)
homogeneous.
I The equation is homogeneous because
2(ty)
/(tx, ty) =
3.100
Determine whether
y'
=
x 2 + 2y 2
is
+ rx = t(2y + x) = 2y + x
X
tx
tx
= f(x, y)
homogeneous.
xy
I The equation is homogeneous because
f(tx, ty) =
— + —=
(rx)
2
2(ry)
(tx)(ty)
2
2
t
(x
2
+ 2y 2 = x 2 + 2y 2
)
=
2
t
xy
xy
= f{x, y)
51
CHAPTER 3
52
3.101
Determine whether
y'
= 2x + y
2
homogeneous.
is
xy
I The equation is not homogeneous because
2
2 2
2(tx) + (ty)
2tx +
2x + ty 2
y
= —tt^—s =
=
* fix, y)
is
2
— — xy
—
t
f(tx, ty)
(tx)(ty)
3.102
Determine whether
y'
2xy
= -=z
y
:
txy
t
=
homogeneous.
is
— xL
I The equation is homogeneous because
—
2t
2(tx){ty)
= 7T-2
f(tX> ty)
— (tx)
(ty)
3.103
Determine whether
y'
=
x2 + y2
2
t
xy
2xy
=~
2 =
— x K y 2 — x 2 /(*» y)
= -27—2
-Tj
-
(y
)
homogeneous.
is
2xy
a
The equation is homogeneous because
=
f(tx, ty)
3.104
Determine whether
y'
=
(tx)
2
+ (ty) 2
t
=
^..
W .. A
2(tx)(ty)
V^ +
2t
2
y )
=
2 ...,
xy
x2 + y2
= fix, y)
„____
2xy
homogeneous.
is
x + \fxy
I The equation is homogeneous for
f(tx, ty)
3.105
Determine whether
y
=
t
> 0,
because then
ty
ty
tx + yf(tx)(ty)
tx + \t\yfxy
=
j-jTj
is
xy + (xy 2 )
= f(x,y)
x + y/xy
homogeneous.
f The equation is not homogeneous because
f(tx, ty)
(ty)
=
(tx)(ty)
3.106
/=
Determine whether
2
2
2
t
2
+ [(tx)(ty) ]
1/3
2
t
x 4 + 3x 2 y 2 + y 4
=
is
x3y
2
3
tf
xy + f(xy 2 ) ,/3
txy + (xy )
t
2
xy + (f xy )
1/3
2
t
2
y
y
2 173
* f(x> y)
homogeneous.
i The equation is homogeneous because
v
/(tx, ty)
=
(tx)*
+ 3(tx) 2 (ty) 2 + (tyf = 4 x 4 + 3t 4 x 2 y 2 + t*y4 = x 4 + 3x 2 y 2 + y 4
f
-r-z3
t*x y
3
(tx) (ty)
3.107
3
x 3y
g(tx, ty)
numbers t in some nonempty interval.
Determine whether
g(x, y)
= xy + y 2
is
homogeneous and, if so, find its degree.
I The function is homogeneous of degree 2 because
g(tx, ty)
3.109
t
k
What is a homogeneous function of degree nl
I A function g(x, y) of two variables is a homogeneous function of degree n if
3.108
,
= f(x.
y)
Determine whether
g(x, y)
= x + y sin (y/x) 2
I The function is homogeneous of degree
g(tx, ty)
1
= (tx)(ty) + (ty) 2 =
is
2
t
(xy + y
2
)
homogeneous and, if so, find its degree.
because
= tx + ty sin
=
t
x + y sin
—
[
= t"g(x, y)
for all real
SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS
3.110
Determine whether
= x 3 + xy 2 exly
g(x, y)
is
53
homogeneous and, if so, find its degree.
I The function is homogeneous of degree 3 because
= (tx) 3 + (tx)(ty) 2 e ,xl,y =
g(tx, ty)
3.111
Determine whether
= x + xy
g(x, y)
3
t
(x
3
+ xy 2 e x/y
)
homogeneous and, if so, find its degree.
is
I The function is not homogeneous because
= tx + (tx)(ty) = tx + t 2 xy
g(tx, ty)
3.112
Determine whether
= yjx 2 — y 2
g(x, y)
I The function is homogeneous of degree
g(tx, ty)
3.113
Determine whether
homogeneous and, if so, find its degree.
is
because, for
1
> 0,
t
= J(tx) 2 - (ty) 2 = Jt 2 (x 2 - y 2 = \t\y/x 2 -y 2 = \t\g(x, y) = tg(x, y)
g(x, y)
)
= 2xsinh(y/x) + 3ycosh(y/x)
homogeneous and, if so, find its degree.
is
f The function is homogeneous of degree 1 because
3.114
Determine whether
g(tx, ty)
= 2(tx) sinh
g(x, y)
= sjx + y
ty
h 3(ty) cosh
ty
y
y
—
= 2tx sinh + 3ty cosh - = tg(x, y)
homogeneous and, if so, find its degree.
is
I The function is homogeneous of degree 1/2 because
g(tx, ty)
3.115
Determine whether
g(x, y)
= y/tx + ty = y/t{x + y) = yftjx + y = t l/2 g(x, y)
= Xyfx + y
is
homogeneous and, if so, find its degree.
I The function is homogeneous of degree 3/2 because
= txyjtx + ty = txjt(x + y) = ty/txy/x + y = t 3 2 g(x, y)
'
g{tx, ty)
3.116
Determine whether
#(x, y)
— x sin(y/x 2
is
)
homogeneous and, if so, find its degree.
I The function is not homogeneous because
g(tx, ty)
= tx sin
—=
ty
(*x)
r
j
2
.
t
tx
tX
Sin
sm
—y ^
:
2
r"6'(
x y)
'
tx'
for any real value of n.
3.117
Determine whether
g(x, y)
= x 3 sin (x 2 /y 2
)
is
homogeneous and, if so, find its degree.
f The function is homogeneous of degree 3 because
(rx)
g(tx, ty)
2
2
f
x2
x2
y*
y
= (tx) 3 sin ——= = 3 x 3 sin 1z-T = t 3 x 3 sin -^z = t 3 g(x, y)
t
(ty)
-
t
SOLUTIONS OF HOMOGENEOUS EQUATIONS
3.118
Show that the differential equation
M(x, y)dx + N(x, y)dy =
is
homogeneous if M(x, y) and N(x, y) are
homogeneous functions of the same degree.
I
The differential equation may be rewritten as
M(x,
dy
= ——
—
dx
y)
-.
N(x, y) are homogeneous of
If M(x, y) and
N(x, y)
degree n, then
M(tx,ty)
/(tX
'
ty)
t"M(x,y)
M(x, y)
= - A^Ty) = ~7N{x7i) = -N&J) = /(X y)
'
3.119
Prove that if y' = f(x, y) is homogeneous, then the differential equation can be rewritten as
where g(y/x) depends only on the quotient y/x.
y'
= g(y/x),
.
CHAPTER 3
54
Since this equation is valid for all in some interval, it must be true, in
f(x, y) = f{tx, ty).
= 1/x. Thus, f(x, v) = f(\, y/x). If we now define g(y/x) = f{l, y/x), we then have
/ = /(*> y) = /(!» jV*) = »(y/*) as required.
f We know that
particular, for
3.120
t
t
y = vx;
Show that the transformation
dy/dx = i + x dv/dx
>
converts a homogeneous differential equation into
a separable one.
I From the previous problem, we know that the homogeneous differential equation
written as
= g(y/x).
y'
Substituting for v' and v/x in this equation, we get
+x
v
= f(x, y)
y'
—
=
dx
can be
dv
which may be
#(t;),
rewritten as
[v
— g(v)~\ dx + x dv =
— dx H
or
x
- g(v)
v
dv —
This last equation is separable.
3.121
Prove that if y' = f(x, y) is homogeneous, then the differential equation can be rewritten as
where h(x/y) depends only on the quotient x/y.
= h(x/y),
y'
= f{tx, ty). Since this equation is valid for all in some interval, it must be true in
— 1/y. Thus, f(x,y)=f(x/y,l). If we now define h{x/y) = f(x/y,
we have
y' = f(x, y) = /(x/y, 1) = h(x/y)
as required.
I We have
f(x, y)
particular for
3.122
t
t
1 ),
x — yu;
Show that the transformation
du
—
= u + y — converts homogeneous
ay
ay
dx
a
differential equation into a
'
separable one.
I
From the previous problem, we know that the homogeneous equation
y'
= ——
—
dy
= h{x/y),
dx
which is equivalent to the differential equation
y
— f(x, y)
can be written as
1
-
-.
Substituting for dx/dy and x/y in this
h(x/y)
—= —
du
last
m + y
equation, we get
'
-
dy
,
which may be rewritten as
h(u)
u =
-.
—
dy + v du =
—— du + - dy —
or
u - \/h{u)
h(u)
y
The last equation is separable.
3.123
Solve
f
y'
= (y + x)/x.
This differential equation is homogeneous (see Problem 3.91). Using the substitution
—
= +x—
dx
dx
dy
dv
v
,
,
we obtain
v
xv + x
+x— =dv
.
dx
3.124
Solve
y
v
=—
= y/x,
-^
xy
I
v
= In |x| - c,
or
—
=
dx
v
= In \kx\,
1
J
where
we obtain the solution to the given differential equation as
u
,
- dx — dv = U.
x
k =
±e~ e
.
Finally,
—
dv
v
or
y = x In \kx\.
This differential equation is homogeneous (see Problem 3.95). Using the substitution
dy
—
= +x—
dx
dx
+x—=
dv
.
we- obtain
v
2(xr)
=
—
ax
dv
4
+ v
^
dx
x
y(.\t)
r
4
—
j
+
3—
,
1
which can be simplified to
r
1
or
ir
the solution becomes
x
4
4
= fc[(y/x) + 1],
or
y - vx;
4
3
- Jx - ^~~- dr =
v
r +
1
This last equation is separable; its solution was found in Problem 3.55 to be
3.125
x
x
This last equation is separable; its solution is
substituting
which can be simplified to
.
y = r.v:
dv
.
4
x = k(\ A + x
8
)
x
4
= A:(r 4 + 1).
in implicit form.
Rework Problem 3.124 using the transformation suggested in Problem 3.122.
Since
= y v.
—
SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS
§
dx
xv
—
=—
— Then,
dy
2y + x
We first rewrite the differential equation as
du
dx
—
= + y—
dy
dy
u
u + y
,
we have
,
-
4
—
— = —7(yu)y*
du
2y
4
.
,
-
dy
4
,
,
which can be simplified to
-j-,
+ (yuf
1,2 +
5
du
u + u
dy
2 + 11*
u
u + u
y
4
5
u = x/y,
3.126
y\x/y)
4
= k[\ + (x/y) ]
).
/ = (x 2 + y 2 )/xy.
Solve
I
the solution becomes
4
4
Since
y u* = k(\ + u
or, on simplification,
x 8 = k(y* + x 4 ) as before.
This last equation is separable; its solution was found in Problem 3.66 to be
8
x = yu;
using the substitution
.
55
y — vx;
This differential equation is homogeneous (see Problem 3.97). Using the substitution
dv
dy
—
— +x—
dx
dx
v
dv
.
,
we obtain
-,
v
+ x —- =
x
2
+ (xv)
2
,
.
,
,.„
,
which can be simplified to
,
dx
x{xv)
dv
I
— dx — v dv — n
1
x —— = dx
v
or
,
,
x
The solution to this separable equation is In |x| — v 2 /2 = c or, equivalently, v 2 = In x 2
where
=
k — —2c.
Substituting
equation
is
we find that the solution to the given differential
v
y/x,
-I-
2
)
3.127
= x 2 In x 2 + kx 2
I
—
= +x
=v
x—
dx
dx
dv
v
2
,
we obtain
-3—,,
+ 1) = kv
Solve
I
y'
r
v
+x
—
=———
dx
x —
dv
—^=
y
2
This last equation is separable and has as its solution
-
-j
(xv)
v
.
= y/x,
this solution may be rewritten as
x[()'/x)
+ 1] = k(y/x)
2
or,
+ x = ky.
2
= (y - x)/x.
y = vx;
This differential equation is homogeneous (see Problem 3.98). Using the substitution
—
= +x—
dx
dx
dy
dv
.
v
we obtain
,
v
+x
=
—
dx
dv
vx — x
which may be simplified to
,
x
—
= —
dx
dv
x
and substitute
v
= y/x,
1
n
,
dv + — dx =
,
or
1
x
This last equation is separable and has as its solution
3.129
y — vx;
2x(xv)
(see Problem 3.67). Since
after simplification,
3.128
).
This differential equation is homogeneous (see Problem 3.96). Using the substitution
dy
x(v
.
/ = 2x>'/(x 2 - y 2
Solve
/c,
v
+ In |x| = c.
y/x = In \k\ — In |x| = In \k/x\,
the solution becomes
c — In \k\
we set
If
k —
(that is,
±e
c
Rework Problem 3.128 using the substitution suggested in Problem 3.122.
I
We first write the diflerential equation as
du
—
= +y—
dy
dy
du
dx
u
,
we obtain
u
+y—=—
dy
separable and has as its solution
u = x/y,
substitute
yu
x
dx
=- —
—
dy
x
y
—
y — yu
.
Then, using the substitution
x — yu;
2
du
This equation may be simplified to
- = — c — In \uy\
the solution becomes
.
y
— = - u— —
dy
(see
—— = In
\k\
Problem 3.57).
If we set
- In |(x/v)y| = In \k/x\,
c
or
= — In \k\
y
.
which is
,
u
1
and
= x In \k/x\
as before.
x/y
3.130
)
y = x In \k/x\.
or
Solve
y'
= yi
2xye ixly)2
+ y 2 e w,)* + 2x 2 e (xly)
I Noting the (x/y) term in the exponential, we shall try the substitution
substitution
x = uy.
u = x/y,
To do so, we rewrite the differential equation as
2
which is equivalent to the
+ y 2 e lxly)2 + 2x 2 e ix,y)2
—=— —
dx
y
(x/y)2
—
,
and
CHAPTER 3
56
—=u+y—
dy
x = uv;
then use the substitution
to obtain
fly
du
\+e ul
y—- = — u 5-
dy
= In \k\. Substituting
equation as y = /c[l + ? Uy)2 ].
3.131
c
2
Solve
y'
I
v
.
we obtain
,
v
dv
2vx + x
dx
x
+ x —- =
Since
respectively).
v
= y/x,
which may be simplified to
,
equation is separable and has as its solution
= kx — I
v
y = vx\
(see
x
dv
—
— +
dx
v
This last
1.
Problem 3.35, with z and t replaced by
2
or y = kx — x.
i;
and x,
y/x = kx — 1
the solution becomes
Rework Problem 3.131 using the substitution suggested in Problem 3.122.
'
dx
x
=
—
dy
2y + x
We first write the differential equation as
du
dx
—
— + —
ay
u
v
'
u + y
,
we obtain
.
du
uy
—
=—
'
fly
.......
2y + uy
fly
u — x/y.
Since
y = Ax
rewrite this as
Solve
2
+u
.
u + 2
yu 2 = k(u + 1) (see Problem 3.68).
= k(x + y). Setting A = l//c, we may
2
y'
= (x 2 + 2y 2
xy.
)
This differential equation is homogeneous (see Problem 3.100). Using the substitution
dy
—
= + x—
ax
dv
r
,
we obtain
r
x +
—=—
2
dv
+x
</\
xvdv — (1 4- v 2 )dx — 0.
Problem 3.46). Since
= y/x,
v
2{vx)
2
—
y = vx;
= +
—
ax
dv
which may be simplified to
,
x
v(!'\)
flx
differential form,
(see
u
u=
y(x \)
or x
which is identical to the solution obtained in the previous problem except for the
— x,
2
= k(x/y + 1)
y
designating the arbitrary constant.
letter
I
2
that solution becomes
du
=
—
2 + u
dy
u
which may be simplified to
.
x = yu;
Then, using the substitution
.
This last equation is separable and has as its solution in implicit form
3.134
y = k(\ + e"\
which can be rewritten as
)
into this result, we obtain the solution of the given differential
This differential equation is homogeneous (see Problem 3.99). Using the substitution
dv
—
= +x—
dx
dx
3.133
j du
= (2y + x)/x.
dy
3.132
=
,
+ eu
1
— In (1 + e" = c,
In \y\
— x/y
u
2ue"
1
- dv
y
y
or
2ue
This equation is separable; its solution is
where
2
—
1
2
1;
or, in
r
This last equation is separable and has as its solution
+ (y/x) = kx
2
the solution becomes
1
2
or
y
= kx* — x
2
+ r 2 = fex 2
1
2
.
Rework Problem 3.133 using the substitution suggested in Problem 3.122.
„
dx
xy
—
—
—^
~
dy
x + 2y
.._
_
.
We first write the differential equation as
du
dx
- — u + v
dy
dy
—
du
y
;
dy
4
,
'
u
—u=
~ -j
u2 + 2
= fc(w 2 + 1)
x = fc(x 2 + y 2
:
i/
u + y
we obtain
,
(see
i
4
).
—+
u
2
^
-
du
=
—
dy
2
(v»)v
'
A — \/k,
i
'
l
s
,
separable and has as its solution in implicit form
u — x/y,
Problem 3.70). Since
Setting
2
^ ms ast ec uat on
2
,._
.
,
which may be simplified to
r-,
+ 2y
x = yu;
Then, using the substitution
.
,
=
(yu)-
j.
the solution becomes
we may rewrite this solution as
(x/y)*y
Ax* = x + y
2
2
— k[(x/y) 2 + 1]
2
or
which is algebraically
,
identical to the solution obtained in the previous problem.
3.135
Solve
f
y'
= (x 2 + y 2 )/2xy.
This differential equation is homogeneous (see Problem 3.103). Using the substitution
dy
—
dx
== v
+x
—
dx
2
dv
,
L
we obtain
differential form,
(see Problem 3.47).
3.136
Solve
f
it
y'
i•
+x
dv
x +
=
—
dx
(vx)
_
2
.
,
.
.
.
Since
v = y/x,
x
1
-
d
2
or, in
2v
2x{vx)
2xvdv + (t' 2 - l)fl"x = 0.
—
=——
dx
di>
which may be simplified to
y = vx;
This last equation is separable and has as its solution
the solution becomes
(y/x)
2
— 1 = k/x
or
y
2
v
2
— 1 = k/x
2
= x + kx.
= (x 2 + y 2 )/3xy.
This differential equation is almost identical to that of the previous problem. The same substitution reduces
to
i;
+x
2
x +
=
—
dx
dv
(t;x)
3x(vx)
2
,
which may be simplified to
l-2t;
—
=—
dx
dy
x
3v
2
or, in differential form,
=
,
SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS
3xvdv + (2v 2 - 1) dx = 0. This last equation is separable and has as its solution
Problem 3.48). Since v — y/x, the solution becomes 2y 2 — x 2 = kx 2 3
2v
-
2
= k/x*' 3
1
57
(see
'
.
3.137
Solve
= (x 2 + y 2 )/4xy.
y'
I
This differential equation is almost identical to those of the two previous problems. The same substitution
2
2
dv
x + (vx) 2
dv
1
.
,. c
= -3v or, in differential form,
=
which may be simplified
reduces it to
v + x
to
x
dx
4x(vx)
dx
4v
— —— — —
.
.
—
.
-
,
4xvdv + (3v 2 — l)dx = 0. This last equation is separable and has as its solution
Problem 3.49). Since v = y/x, the solution becomes 3y 2 — x 2 = kx 112
1
= kx~ 3 2
Then, using the substitution
x = yu;
3r
—
2
'
(see
.
3.138
Rework the previous problem using the substitution suggested in Problem 3.122.
f
4xy
ta
dx
—
=
x +
dy
^
_
iL
J a
We cfirst
write the differential equation as
.
r
.
a
.
,
\
y
—^2
j.
y
du
dx
—
=«+
dy
dy
v -7_
u + y
}
we obtain
i
du
= 4(yu)y
—
dy
,.
=
(yu)
2
+y
,..,.«.,
,
which may
to
J be simplified
y
=-,
2
This last equation is separable and has as its solution in implicit form
3.139
Setting
u — x/y,
then to
3y
Solve
#
y'
2
— x 2 = Ax 112
= 2xy/(y 2 - x 2
—
— +x—
dx
dx
where
,
I
y
+ x —- =
.
x 3 (v 3 — 3v) — k
Solve
dv
we obtain
,
or, in differential
2
— 3) 2 y 3 = ku
which may be simplified to
v
+x
x{v
2x(t>x)
.
1
Problem 3.71).
and
— 3y 2 2 = kx
2
)
y = vx;
=
.
which may be simplified to
=-,
—
=
dx
2y
di?
x
= 3xy/(y 2 - x 2
—v—
-=
— +
—
—
—
y
3
3v
1
).
=
—
dx
(vxf — x
dv
3x(wc)
5
,
.
,.,._,
,
y'
3
— + 4v
dv
—
=—
—
dx
v
which may be simplified to
j,
x
ir
= 3xy/(y 2 + x 2
.
5
1
the solution becomes
y (y
— 4x
2
2 3
dv
3x(vx)
-
r
(vx)
2
^
2
2
+ 1) dv + (v 3 — 2t;) dx = 0.
Problem 3.73). Since v = y/x,
x(v
2
Solve
y'
= 3xy/(x 2 - y 2
2
— 4) 3 x 8 = k
(see
).
=
—
— —
wmcn may
which
ma
T
dx
+x
+x
v
v (v
— k.
)
This equation is similar to that of the previous problem. Using the substitution
we obtain
2
This last equation is separable and has as its solution
2
form,
or, in differential
m
.
-u +
^ =—
—
—
x
the solution becomes
(y
v
2i;
=
-
dx
v
2
+
i
or, in differential form,
1
1
This last equation is separable and has as its solution
2
dy
dv
—
= + x—
dx
dx
3
.....
.
De simplified
to
y be
J
*
'
,
y — vx;
— 2x
2 3
= ky
)
(v
2
— 2) 3 x 4 = kv 2
(see
2
.
).
This problem is similar to Problem 3.139. The same substitution reduces the equation to
f
dv
—
dx
— which may
3x(vx)
—:
+ x -7- = -5—
2
-
x
r-,,
(vx)
2
which
— \)dv + (v 3 + 2v)dx = 0.
Problem 3.74). Since v = y/x,
x(v
3.143
u
2
2
2
Solve
v
(see
(x
3w
=
v
dx
(vx) — x
v — 1
— \)dv + (v 3 — 3v)dx = 0. This last equation is separable and has as its solution
3
2
(see Problem 3.50). Since
the solution becomes y — 3yx = k.
v = y/x,
form,
— \)dv + (v 3 — 4v)dx — 0.
Problem 3.72). Since v — y/x,
3.142
+1
=
This differential equation is similar to that of the previous problem. The same substitution reduces it to
x(v
3.141
— 3] y = k(x/y),
A = ±yfk.
(u
u
u
).
dv
v
v
[(x/y)
3
2
dy
2
-u +
—
+
3
^u
4u
—
= —5
This differential equation is homogeneous (see Problem 3.102). Using the substitution
dy
3.140
we obtain
2
y
y
2
.
to
be simplified
v
dv
=
—
x
x—
dx
v
3
+ 2v
- v j;
1
2
or, in differential form,
This last equation is separable and has as its solution
this solution becomes
2
(y
+ 2x
2 3
)
= ky
(v
2
+ 2) 3 x 4 = kv 2
2
.
Solve
x + -Jxy
i
This differential equation is homogeneous (see Problem 3.104). Using the substitution
dy
=v+x—
-f-
dx
dv
dx
.
,
.
we obtain
v
+x
=
—
dx
dv
vx
.
x + y/x(vx)
y = vx;
-vy/v
=
—
=
dx
dv
which may be simplified to
x
I
+ Jv
or,
(see
.
58
CHAPTER 3
D
in differential form,
3.144
x(l
Problem 3.58). Since
(see
Solve
+ -Jv) dv + Vyjvdx = 0. The solution to this last equation is
— 2yfx/y + In \y\ = c.
v = y/x,
the solution becomes
-2/yfv + In \vx\ = c
4
2 2
4
= x + 3x /y + v
—
y'
x y
I
This differential equation is homogeneous (see Problem 3.106). Using the substitution
—
= +x—
dx
dx
dy
v
we obtain
,
or, in differential form,
+
(1
or
2
i;
y
2
In \kx\
)
= —
2
= -x 2
(
1
1
Solve
I
(x
3
+
dv
dx
x (vx)
dx
v
\
— — x2
)
We use the transformation
y = vx; dy = v dx + xdv, to
3
—
2f
dx — 3v 2 xdv =
(1
and then
which we simplify to
)
2
3v dv
= 0.
" 1 - ^-j
2
~x~
2v
— y/x,
+ 2v 2 + v*
2
x 3 [(l + v 3 )dx — 3v 2 (vdx + xdv)~\ = 0,
write as
1
+ y 3 dx - 3xy 2 dy = 0.
dx
3.146
x 4 + 3x 2 (ux) 2 + (vxf
This equation is homogeneous of degree 3.
obtain
v
dv
xvdv — (1 + 2v 2 + v 4 )dx = 0. The solution to this last equation is
2
2
2
this solution becomes
(see Problem 3.59). Since
v — y/x,
(x + y )ln \kx
In \kx
3.145
y = vx;
dv
we have
The solution to this equation is
x 2 [l — 2(y/x) 3 ] = k
x (l - 2tr) = k
2
(see
Problem 3.56). Since
x 3 — 2y 3 = /ex.
or
x dy - ydx- yjx 2 - y 2 dx = 0.
Solve
I The equation is homogeneous of degree 1. Using the transformation y — vx; dy = vdx + xdv and
dividing by x, we get
vdx + x dv — vdx — sj\ — v 2 dx =
or x dv — y/l — v 2 dx = 0, which we write as
dx
- = 0.
X
dv
y/\
-V
2
arcsin (y/x) = c + In |x| = In \kx\,
we have
3.147
2
3.148
—
3
dv = 0.
x
sinh v
— y/x,
this becomes
Solve
I
c
(see
Problem 3.89). Since
v
= y/x,
= In k.
2 sinh vdx — 3x cosh v dv — 0.
Integrating, we get
2 In x
y = vx;
dy = vdx + xdv
and
Separating the variables yields
— 3 In sinh v = In c,
so that
x 2 = c sinh 3 v.
Since
x 2 — c sinh 3 (y/x).
(2x + 3y) dx + (y - x) dy = 0.
y = vx;
This equation is homogeneous of degree 1. The transformation
(2
+ 3v) dx + (v - l)(v dx + x dv) =
= y/x,
the solution becomes
or
(v
2
dy = vdx + xdv
2
In [x (v
reduces it to
+ 2v + 2)dx + x(v - 1) dv =
+ 2v + 2)] — 4arctan(i; + 1) = k
x + y
2
2
— 4 arctan
= k.
In (y + 2xy + 2x
The solution to this last equation is
v
where
This equation is homogeneous of degree 1. Using the transformation
dividing by x, we obtain
v
arcsin v — In |x| = c
[2x sinh (y/x) + 3y cosh (y/x)] dx - 3x cosh (y/x) dy = 0.
Solve
I
.
The solution to this equation is
2
(see Problem 3.81).
Since
)
x
3.149
Solve
I
(
1
+ 2e xl >) dx + 2e* /y (l - x/y) dy = 0.
This equation is homogeneous of degree zero. The appearance of the quantity x/y throughout the equation
suggests the substitution
x = uy;
—
= +y—
dy
dy
u
or, equivalently,
dx — udy + ydu.
This transforms the
+ 2e u )(u dy + ydu) + 2e u (\ — u)dy = 0, which we simplify to
=
0.
The solution to this last equation is y(u + 2e") = k (see Problem 3.82).
(u + 2e dy + y(l + 2e du
—
the solution becomes
u
x/y,
x + 2ye xly = k.
Since
differential equation into
u
u
)
)
(1
SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS
59
MISCELLANEOUS TRANSFORMATIONS
3.150
dy/dx = ( y - Ax) 2
Solve
y — 4x = v,
I The transformation
dv
dx
— 47
=
it
= 0-
.
A
v
= ce
-2
4jc
2
i;
—-
yields
y
+ 2
- = In c - 4x.
-2
Ax - 2
= ce~* x
2
x + y — v;
dy = dv — dx
— (dv — dx) —
or
f The transformation
tan
2
v dx
reduces this equation to
^— =
dx
x — \v — £sin 2v — c
— y) = c + sin 2(x + y).
Integrating gives
Solve
dx — cos 2 v dv =
or
+ tan 2 v
1
3.152
v
and substitution for
,
In-
so that
or
tan (x + y) dx - dy = 0.
Solve
2(x
v
= c,,
— r2
v
4 dx + dv = v 2 dx
reduces this equation to
lf +
x + -ln
Then integration gives
Exponentiation then yields
3.151
dy = A dx + dv
which, after substitution for v and simplification, becomes
x
+ 2x 2 y 1/2 )y dx + (x 2 y 1/2 + 2)x dy = 0.
(2
#
The transformation
x y
v
(2
'
y
— v;
2
2w
y — —r\
x*
dy — —rdv
x
(2v
Av 2
2
+ 2v) -; dx + x(v + 2)[ -j dv
dx
.
.
2
=-
x3
=
1
dx
reduces the given equation to
or
2 dv
1
3
3 v
v(3
+ v)dx- x{v + 2) dv =
dv
= 0, and integration yields
+ 3
x 3 = c v 2 {v + 3). Finally, substitution gives
Then the method of partial fractions gives
x
— 21n v — ln(i' + 3) = In c
from which
= c xy(x 2 y 2 + 3) or xy(x 2 y 1/2 + 3) = k.
3 lnx
\
r dx
4i>
{ ,
v
{
x '
1
x
3.153
Solve
(2x
2
+ 3y 2 - 7)x dx - (3x 2 + 2y 2 - 8)y dy = 0.
f The transformation x 2 — u; y 2 — v reduces this equation to
(2w + 3i> — 7) du — (3u + 2v — 8) dv — 0.
Then the transformation u — s + 2; u = t + 1 yields the homogeneous equation
and the transformation s = rt; ds = rdt + tdr yields
(2s + 3t) ds — (3s + 2t) dt — 0,
~
>
2(v — 1
„
,
)
dt + (2r
.
~
+ 3)f «r = 0.
,
^
,
•
,
2r + 3
r dr
2
^dt
,
Separating the variables, we get
2
,
h -j
r
t
Then integration yields
4 In t
— In (r + 1) + 5 In (r — 1) = In C.
-
\
^dt
—2
t
\
dr
2 r +
5
h
1
dr
2 r -
1
Exponentiation and successive substitutions
then yield
t\r - l) 5 _ (s - t) 5 _ (u - v - l) 5 _ (x 2 - y 2 - l) 5 _
r r
so that
3.154
Solve
f
(x
2
s
\
+
u + v
t
—3
x 2 + y2 - 3
- y 2 - l) 5 = c(x 2 + y 2 - 3).
x 2 (x dx + y dy) + y(x dy — y dx) = 0.
Here
x Jx + y dy = i^(x 2 + y 2 )
and
x (iy — y Jx = x 2 d(y/x)
suggest the transformation
x2 + y2 = p 2
y = p sin 0; dx = —p sin 9d6 + cos dp; dy = p cos 6 d0 + sin dp.
2
2
2
or dp -(- tan sec dO = 0.
given equation then takes the form
p cos 0(p dp) + p sin 0(p dO) —
y/x = tan fl,
or
Integration gives
(x
3.155
2
x — p cos 0;
p + secfl = c,,
+ y 2 )(x+ l) 2 = cx 2
Solve
so that
yjx
2
+ y2
x+ =
c,,
1
which may be written
.
y(xy + 1) dx + x(l + xy + x y ) dy - 0.
f
The transformation
2
2
dy =
xy = u;
x di> — v dx
-.
,
,
.
reduces this equation to
x^
- (v + 1) dx + x(l + v + v 2
X
)
=
X
= 0,
which can be simplified to
v
3
dx - x(l + v + v 2 ) dv = 0.
The
;
60
CHAPTER 3
D
........
dx
dv
do
x
T
ir
T
£
Separating the variables yields
3.156
— 2v — 1 = cv
2
2v In (v/x)
that
Solve
2
r
The transformation
dy =
xy = v;
x
x = cye xy
3.157
Solve
I
= c,,
In t>
2x 2 y 2 In y — 2xy — 1 = cx 2 y 2
yields
so
v
.
— xy) dx — x{\ + xy) dy = 0.
x
xz
5
=
2 In x — In v
and integration gives
2v dx - x(l + v) dv =
or
— v = In c,
from which
x 2 /v = ce v
and
,
u
.
(1
- xy + x 2 y 2 dx 4- (x 3 y - x 2 dy =
~,
r
or, rewritten,
)
)
v
ay =
xy = u;
t
xdv — vdx
,.
.
,
.
- xy 4 x 2 y 2 dx 4 x(x 2 y 2 - xy) dy = 0.
)
,
xdv — v dx
^
.2
X
=
vdx + x(v 2 - v) dv =
or
X"
Then
dx/x + (v — \)dv = 0,
Solve
(x + y) dx
and integration gives
\n x
+ jv 2 — v = c,
from which
In x
= xy — \x 2 y 2 + c.
+ (3x + 3y - 4) dy = 0.
(x + y)
f The expressions
dy = dt - dx
y(l
reduces this equation to
j
- v + v 2 )dx + x{v 2, - v)
..
(\
3.159
= xy
2
reduces this equation to
=
xz
x dv — v dx
The transformation
3.158
v
y(l
v
dv — 0,
2
r 4-
2v
v
x dv — v dx
-(1 - v)dx -x(l 4 v)
Then
11
In x H
Integration then gives
or, rewritten,
)
-T-.
= 0.
Finally, substitution of
.
— xy 2 dx — (x + x 2 y) dy =
(>•
v
dv
to obtain
t
(3x + 3y)
and
or, rewritten,
It -4
—
dt = 2 dx — 3 dt +
the variables are separable.
We then have
2x — 3r — 2 In (2 — f) = c,,
and after substitution for t we have
2 dx +
which
x + 3y + 2 In (2 — x — y) = c.
Solve
(2x - 5y + 3) dx - (2x + Ay - 6) dy = 0.
2x — 5y + 3 =
I We first solve
2x 4- 4y — 6 =
and
x + y —
We use y = — x;
2t) dx + (3f
in which
(4
4) dt = 0,
suggest the transformation
dx + (3f - 4)(dt - dx) =
t
t.
2
df
= 0.
Integration yields
2x — 3(x + y) — 2 In (2 — x — y) = c ls
simultaneously to obtain
x = h = 1,
from
y = A: = 1.
Then the transformation
x — x' + h — x' +
d.\
1 :
y = y' + /c = y' + 1;
reduces the given equation to
(2x'
= dx'
dy = dy'
— 5y')dx' — (2x' + 4y')dy' = 0,
which is homogeneous of degree 1. (Note
that this latter equation can be written without computing the transformation.)
y' =
x':
—
5v) dx'
(2
(2 + 4v)(v dx' + x' dv) =
Using the transformation
dx
4
</r
1
x'
3
dv
3 v
+2
1
—
3 4r
x
2
1
i
= 0.
dy' = v dx'
or
(2
+ x' dv,
Integrating, we get
In x' 4- 5 In (4r
— 1) + § In (r + 2) — In c,,
or
(4u- l)(r + 2) =c.
yields the primitive
Solve
f
which we separate into
2
2
(4y' — x')(y' + 2x') = c,
2
—
—
—
=
x
c.
(4y
3)
3)(y + 2x
Replacing v by y'/x gives us
3.160
we obtain
— 7v - 4v 2 )dx' - x'(2 + 4v) dv = 0.
(x
and replacing x' by
x —
1
and y' by
y — 1
- y - 1) dx + (4y + x - \)dy = 0.
Solving
x — y — 1 =
and
4y + x — 1 =
simultaneously, we obtain
x = h = 1,
y = k = 0.
The
transformation
reduces the given equation to
this transformation
terms
(x
— y — 1)
(x'
x = x' + ft = x' + 1;
dx — dx'
y = y + k = y';
dy = dy'
— y')dx' + (4y' + x')dy' — 0,
x — 1 = x', y = y'
and (4y 4- x — 1).]
which is homogeneous of degree 1. [Note that
could have been obtained by inspection, that is, by examining the
SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS
Using the transformation
/ = vx';
dy'
dx'
+
= v dx' + x' dv,
we obtain
(1
61
- v)dx' + (4v + l)(v dx' + x dv) = 0.
Then
IT
4t>
1
J
dx
8u
1
dv
4?TT * " T + 2 4^TT * + 47TT = °
+
+ \ In (4v 2 + 1) + 5 arctan 2t> = c, which we rewrite as In (x') 2 (4v 2 + 1) + arctan 2v - c.
2
2
Substitution for v then gives In (4y' + x' ) + arctan (2y'/x') = c, and substitution for x' and y' yields
Integration gives
In [4y
2
In x'
+ (x - l) 2 ] + arctan
= c.
x — 1
INITIAL-VALUE PROBLEMS
3.161
Discuss how to solve the initital-value problem
A(x) dx + B(y) dy = 0;
y(x
)
=y
.
I The solution may be obtained by first solving the separable differential equation for its general solution and
then applying the initial condition to evaluate the arbitrary constant. Alternatively, the solution may be obtained
directly from
x
P
=
( A(x)dx+
B(y)dy
Jxo
Jyo
(7)
This last approach may not determine the solution uniquely; that is, the integrations may produce many solutions
(in implicit form),
3.162
Solve
e* dx
of which only one will satisfy the initial-value problem.
- y dy = 0;
= 1.
y(0)
f The solution to the differential equation is j e x dx + J — y)dy = c or, after evaluation, y 2 = 2e x + k,
Applying the initial co ndition (s ee Problem 2.103), we find that k — — 1 and that the solution to
k = —2c.
x
the initial-value problem is
x > In \.
y — -J2e —
(
1
3.163
,
Use (/) of Problem 3.161 to solve the previous problem.
I Here
x
=
and
= 1,
y
so (7) of Problem 3.161 becomes
^e x dx + \\( — y)dy = 0.
Evaluating these
integrals, we get
-v 2
y
=
3.164
= 2e* —
Thus,
y
Solve
x cos x dx + (1 — 6y 5 ) dy — 0;
I Here
x
= n,
3.161, we obtain
y = \J2e* — 1
and, as before,
1
e
x
2
lo
2
or
-e° + ^_-(-i) =
x > In j.
,
y(n) — 0.
Substituting these values into (7) of Problem
y = 0, A(x) = x cos x, and B(y) — 1 — 6y
5
—
j* x cos x dx + j& (1
6y ) dy = 0. Evaluating these integrals (the first one by integration
5
.
by parts), we find
x sin x\
|«
+ cos x* + (yJ — Jy b )\|0 =0
x sin x + cos x + 1 = Jy 6 — Jy
or
\n
Since we cannot solve this last equation for y explicitly, we leave the solution in implicit form. (See also Problem
2.105).
3.165
Solve
sin x dx
+ y dy — 0;
y(0)=— 2.
# The differential equation is separable, so we have
Jo sm
xdx + \yl 2 ydy = 0.
or
— cos x +
Evaluating these integrals, we
get
2
— cos x| + 1J
?y -2 =0
|o
\
from which
y
2
= 2 + 2cosx,
or
y = —-J2 + 2cosx.
1
2
—2=
+ |y
z
The negative square root is chosen to be consistent
with the initial condition.
3.166
Solve
(x
2
+ \)dx + (l/y)dy - 0;
y(-l)=l.
f The differential equation is separable, so we have
(ix
3
+ln|y|T =0
+ x)|*
i
or
ji , (x
2
+ 1) dx + $\ (1/y) dy = 0,
from which
|x 3 + x - (-^ - 1) + In |y| - In 1 =
CHAPTER 3
62
Then
In \y\
= -(x 3 + 3x + 4)/3,
y = e
and
-(* 3 + 3*+4)/3
The plus sign in front of the exponential is consistent
with the initial condition.
5
xe* dx + (y - 1) dy = 0;
2
3.167
Solve
= 0.
y(0)
f The differential equation is separable, so we have
Jo xe
x2
dx + J (y 5 - \)dy = 0.
The indicated integrations
give
yxT + (b 6 ~ y)L =
from which we obtain
3.168
Solve
y'
6
y
2
— 6y + 3(e* — 1) = 0,
= (x 2 y - y)/(y + 1);
y(3)
y* - i + iv 6 - y - =
or
which is the solution in implicit form.
= -1.
I
v
+ 1
Separating the variables, we find that the differential equation has the form
solution to the initial-value problem then is
(v + In |y|)|
y
3.169
Solve
y'
+ 3y = 8;
- (W - x)\\ =
x
y + In \y\ = ,x
from which we obtain
11+) dy —
'
y(0)
3
(x
2
— \)dx — 0.
y + In \y\ - {- 1 + In 1) - (|x
or
dy — (x 2 — \)dx — 0.
The
The indicated integrations give
3
- x) + 9 - 3 =
— x — 7.
= 2.
I The solution to the differential equation was shown in Problem 3.36 to be y(x) = f + ke' ix Applying the
initial condition, we get
k = — f.
2 = y(0) = \ + ke~ 3l0 \
so that
Thus, the solution to the initial-value
problem is y(x) = f — \e~ ix
.
.
3.170
Solve the preceding problem if the initial condition is
y(0)
= 4.
f The solution to the differential equation remains the same. Applying the new initial condition, we get
-3
4 — y(0) — f + ke~ M0 \ so that k = *. Thus, the solution to this initial-value problem is y(x) — | + fe *.
3.171
y(l) = 0.
Solve Problem 3.169 if the initial condition is
f The solution to the differential equation remains the same. Applying the new initial condition, we get
= y(l) = f
y(x)
3.172
-(-
ke~ Ml)
= f-feV
3*
— fe 3
so that
k =
= f(l- e - 3 «*- , »).
,
.
The solution to this initial-value problem is then
Solve Problem 3.169 if the initial condition is
y(
— 2) =
1.
1 The solution to the differential equation remains as before. Applying the new initial condition, we get
= )'( — 2) = f + kc M 2 \ so that k = - \e 6 The solution to this initial-value problem is then
ix
= }(8 - 5e~ Mx 4 :,
y(x) = f _ ^e^e1
.
:
|.
3.173
Solve Problem 3.169 if the initial condition is
y(
— 2) = — 1.
I The solution to the differential equation remains as before. Applying the new initial condition, we get
— 1 — y( — 2) = I + ke' M ~ 2) so that k — —"e' 6 The solution to this initial-value problem is then
.
,
y(x)
3.174
=f - ^e' b e~
ix
= ^(8 - lle- 3(vf 2)
).
Solve Problem 3.169 if the initial condition is
y(4)
= —3.
# The solution to the differential equation remains as before. Applying the new initial condition, we get
— 3 = y(4) — f + ke~ M4 K so k — —^e 12 The solution to the initial-value problem is then
.
3.175
y(x) =
Z-^e i2 e- ix = (&-ne- 3u " )/3.
Solve
dy/dt - y
4,
s
sin f;
y(0)
- 1.
f The general solution to the differential equation was shown in Problem 3.88 to be
result of Problem 2.104 we have
the initial-value problem.
y(t)
=
/
I
1
V
\4cosf — 3/
/4
3
,
where
1/y
4
= 4 cos + c v
f
As a
3
-arccos - < t < arccos -,
4
4
as the solution to
—
SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS
3.176
Solve the previous problem if the initial condition is
y(0)
63
= {.
I The general solution to the differential equation remains as before. Applying the initial condition, we get
—
+ = 4cos0 + c„
so
c,=2 4 -4=12. The solution to the initial-value problem becomes
/
4
1/y
= 4 cos + 12
or, explicitly,
t
y(t)
^4 cost
4cosf +12
3.177
\i /4
.
+ 12
The solution y is defined fur all t, since
always positive.
is
y + y = 0;
= 2.
1
Solve
—
i
=
y(3)
y = ke~
I The general solution to the differential equation was shown in Problem 3.30 to be
Problem 3.30 is equal to — 1.) As a result of Problem 2.74, we have
y(x)
— 2e
3 ~x
x
.
(Here, A of
as the solution to the
initial-value problem.
3.178
y = y
Solve
= 4.
2
y(0)
;
The general solution to the differential equation was shown in Problem 3.9 to be y(t) = — l/(f + c).
so that
c = —\.
The solution to the
f
Applying the initial condition, we get 4 = >'(0) = — 1/(0 + c),
initial-value problem then is
y(t) = - l/(r - |) = -4/(4f - 1).
3.179
Solve the previous problem if the initial condition is
y(
— 1) = —2.
The solution to the differential equation remains the same. Applying the new initial condition, we get
1
The solution to the initial-value problem is then
c = §
m
— 2 = y(— 1) = — l/( — + c), so that
y(t)= -l/(f + f)= -2/(2f + 3).
.
Observe that this solution is not valid in any interval containing t = — §
Since a solution to an initial-value
problem must be valid in some interval containing the initial time, in this case t = — 1, it follows that the
above solution is valid only on the interval (— f, oo). By similar reasoning, the solution to the previous problem
.
is
3.180
valid only on (
— oo, £).
Solve Problem 3.178 if the initial condition is
y(0)
= — 2.
I The general solution to the differential equation remains as before. Since
is the solution to the initial-value problem.
y(t) = — l/(f + j) = — 2/(2f + 1)
for
f
/ -},
y(0)
= — 2, we have
c
— \. Then
Since this solution is defined only
must be in the interval for which y is defined, it follows that ( — \, oo) is the interval of
and since
definition for y.
3.181
Solve
dz/dt = z
3 2
t
z(2)
;
= 3.
The solution to the differential equation was found in Problem 3.1
Applying the initial condition, we get
explicit form,
3.182
z
/
=
I
^
—^
i
\
— ——^ — -(2) 3 = c,
so that
to have the form
1
c
= -f§
— —^ — -t 3 — c.
and the solution becomes, in
1/2
,
where the positive square root is chosen consistent with the initial condition.
J
Rework the previous problem if the initial condition is
z(2)
= -3.
f The solution to the differential equation remains the same. Applying the initial condition, we get
3
(2)
Z\
J)
= c,
so again
c
= — f§.
Now, however, the solution to the initial-value problem in explicit
j
form becomes
z —
— (^
(
Y
i
jT
)
/2
>
where the negative square root is chosen to be consistent with the initial
condition.
3.183
Solve
dy/dt = y(t
- 2);
y(2)
= 5.
'~
f The solution to the differential equation was found in Problem 3.32 to be >-(f) = ke 2)1/2 Applying the initial
~ 2)2 2
{2
= ke° - k, so the solution to the initial-value problem is y(t) = 5e"" 2)2/2
5 = ke
condition, we get
{
.
'
.
—
CHAPTER 3
64
3.184
dy/dt =
Solve
-2yt 2
y(2)
\
= 3.
I The solution to the differential equation was found in Problem 3.33 to be
condition, we get
3.185
3
= 3e- 16 e 2 = 3e
'
y(t)
3
= y(2) = ke 2{2)3 = ke 16
2{,
16
k = 3e~
so
.
— ke 2
'
3
.
= -1.
y(5)
I The solution to the differential equation was found in Problem 3.61 to have the form
4
(5 — 4)(— l) /5 — k,
Applying the initial condition, we get
problem is
|_5(x
3.186
k =|
so
(x
— 4)y 4 /x — k.
and the solution to the initial-value
/4
- 4) J
where the negative fourth root is taken consistent with the initial condition.
,
x 2 (y +l)dx + y 2 (x - \)dy = 0;
Solve
f
T
x
—T
=
y(x)
Applying the initial
.
Then the solution to the initial-value problem is
- 2)3
4x dy - y dx = x 2 dy,
Solve
,
y(t)
y(-l) = 2.
The solution to the differential equation was found in Problem 3.63 to have the form
+ \) 2 + (y — I) 2 + 2 In |(x — l)(y + 1)| = k. Applying the initial condition, we get
2
2
k = (-l + l) + (2- l) -h 2 In — — 1)(2 + l)| = 1 + 2 In 6.
The solution to the initial-value problem is then
(x
1
|(
(x+ l) 2 +(y- l) 2 + 21n|(x- l)(y + 1)| = 1 + 2 In 6
(x + l)
or
2
+ (y- D 2 + In
~ 1}
(X
(
/
+ 1}
=
1
36
3.187
dy/dt = y
Solve
- y3
2
y(l)
;
= 2.
f The solution to the differential equation was found in Problem 3.64 to be — 1/y + In \y\ — In |1 — y\ — t = c.
— \ + In 2 — In 1 — 1 = c, so c ^ -0.80685. The solution to the
Applying the initial condition, we get
initial-value problem
3.188
is
- 1/y + In \y\ — In
thus
1
j
- y\ — -0.80685.
Solve the preceding problem if the initial condition is
= 0.
y(2)
I The solution obtained in Problem 3.64 is the solution to the differential equation only when y ^
and
partial-fraction
these
two
because
the
decomposition
used
to
generate
the
solution
is
undefined
at
values
1,
y ^
=
of y. Here y„
so we are in one of these special cases. By inspection, we note that two constant solutions
0,
and y = 1. Since the first of these also satisfies the initial condition,
to the differential equation are
y =
it
3.189
is
the solution to the initial-value problem.
Solve
f
dy/dx = y- y 2
;
y(0)
= 2.
The solution to the differential equation was found in Problem 3.60 to be
y(x)
=1
initial
condition, we get
I
problem is then
3.190
y(x)
=
ke°
k
+ke°
l+k
—^ = - —-
2 = y(0) =
-2e x
2
\-2e x
2-e~ x
k =
so
Solve the previous problem if the initial condition is
y(0)
— 2.
ke*
Applying the
.
+ ke x
The solution to the initial-value
= 1.
I The solution obtained in Problem 3.60 is the solution to the differential equation only when
y ^ 1,
y.
Here
y
— 1,
so we are in one of these special cases.
y =
the differential equation are
it is
3.191
y ^
and
because the partial-fraction decomposition used to obtain the solution is undefined at these two values of
and
y = 1.
By inspection, we note that two constant solutions to
Since the latter solution also satisfies the initial condition,
the solution to the initial-value problem.
Solve
dy/dt = 2ty
m
If
y
^ 0,
2
;
y(0)
= yQ
.
then by separation of variables we have
V
—= = 2t
/*>•
and hence
y
integrations result in
=
1
y
y
2
t
or
- =
y
y
2
f
,
so that
d\
— = ft 2f dt.
=•
Jyo y
v(r)
=-
——
l
-y t
y
The
J"
as long as
y t
2
#
1.
If
then
y(t)
+ x as t
l/vyo> an ^ so solutions to this equation "blow up" in finite time
y >
and
whenever the initial condition is positive. Note, however, that if y < 0. then y exists for all t >
y(r)
>0
as
f
»+oo.
—
SEPARABLE FIRST-ORDER DIFFERENTIAL EQUATIONS
= 0, then the solution becomes y(t) = 0, which is the solution to
found
above solves the initial-value problem for all values of y
Thus, y(t) as
Note also that if
y(0) = 0.
3.192
Solve
y
65
dy/dt = 2ty
2
:
.
y'
= (y + x)/x;
y(- 1) = -2.
# The solution to the differential equation was found in Problem 3.123 to be y = xln \kx\. Applying the initial
condition, we get
-2 = y(-l) = -1 In \k(- 1)|, so In \k\ = 2 and the solution to the initial-value problem
=
—
y(x)
x In \kx\
x(ln \k\ + In |x|) = x(2 + In |x|).
is
3.193
Solve
y'
= (x 2 + y 2 )/xy;
y(l)
= -2.
I The solution to the differential equation was found in Problem 3.126 to be y 2 — x 2 lnx 2 kx 2 Applying
— 2) 2 = (l) 2 In (l) 2 + /c(l) 2 or k = 4. Thus, the solution to the initial-value
the initial condition, we obtain
2
2
problem is y = x In x 2 + 4x 2 or y = —yfx 2 \nx 2 + Ax 1
The negative square root is taken consistent with
-I-
(
.
,
.
the initial condition.
3.194
Solve
y'
= (x 2 + y 2 )/2xy; y(l)=-2.
i The solution to the differential equation was found in Problem 3.135 to be y 2 — x 2 + kx. Applying the
— 2) 2 = (l) 2 + k(l), or k = 3. The solution to the initial-value problem is then
initial condition, we get
2
2
2
y = x + 3x or, explicitly, y = —Jx + 3x, where the negative square root is chosen consistent with the
(
initial
3.195
Solve
condition.
y'
= 2xy/(y 2 - x 2
><(4)
);
= 0.
I The solution to the differential equation was found in Problem 3.139 to be y 3 — 3yx 2 = k. Applying the
3
2
3
2
initial condition, we get
k — (0) — 3(0)(4) = 0,
so the solution to the initial-value problem is y — 3yx — 0.
— +y
y= x + 3xV
4
4
,
3.196
Solve
x 3y
;
y(2)=l.
I
was found in Problem 3.144 to be
The &
general solution to the differential equation
M
y
J
'
Applying the initial condition, we get
(l)
2
= -(2) 2 (
I
1
1
\
= — x2
1
+
\
4
so
-(- -
2
In |4/c|
= —-. Then the general
J,
solution becomes
-x 2 |Y
2
ln|(4/c)(x /4)|J
or, explicitly,
y =
—x
5
1
+
-4 + 5 ln(x 2 /4)Jj
In \4k\
+ In (x 2 /4)
In \h
+
-4/5 + ln(x 2 /4)J
CHAPTER 4
Exact First-Order
Differential Equations
TESTING FOR EXACTNESS
4.1
Define exact as regards a differential equation.
I A differential equation
dg{x, y) = M(x, y) dx
4.2
M
(x, y) dx
+ N(x, y) dy =
is
Develop a test for determining whether a first-order differential equation is exact.
I
If M(x, y) and
N(x, y) are continuous functions and have continuous first partial derivatives on some
M(x, y) dx + N{x, y) dy =
rectangle of the (x, y) plane, then the differential equation
4.3
dM(x, y)
dN(x, y)
dy
dx
Determine whether the differential equation
I
Here
4.4
exact if there exists a function g(x, y) such that
+ N{x, y) dy.
M(x, y) = 2xy
N(x, y) = 1 + x
and
Determine whether the differential equation
I
Here
2xydx + (1 4- x 2 )dy =
M(x, y) = x + sin y
dM
,
(x + sin y) dx
dN
the differential equation is exact.
+ (x cos y — 2y) dy =
N(x, y) = x cos v — 2 v.
and
exact if and only if
exact.
-r— = —— = 2x,
dy
ox
Since
.
is
is
is
exact.
dM
dN
—— = cos y = -5—,
Since
the differential
ex
dy
equation is exact.
4.5
xy
Determine whether the differential equation
xe
—
d(ye
I
Here
xy
dM d{xe
—
— = —; = x z e xy
)
dy
,
J
and
dN
—=
^— = —dx
dy
dx + ye xy dy —
exact.
is
xy
,
)
y*e y
.
,
.
J
.
Since these two partial derivatives are not equal,
.
dx
the differential equation is not exact.
4.6
Determine whether the differential equation
I
Here,
,
M(x, y) = xy + x'
and
(x>
+ x 2 )dx + (— \)dy =
j
— 1;
N(x, y) =
dM
-^— = x
oy
hence
is
exact.
cN n
—— = 0.
and
„.
dM
cN
5y
dx
^^ # ^—
Since
dx
,
the
equation is nof exact.
4.7
Determine whether the differential equation
#
Here
M(x, v) = 2xy + x
and
(2xy + x) dx + (x
N(x, y) = x^ + y.
+ y) rfy =
2
3M ^
5N
—
— = 2x = ——
Since
Determine whether the differential equation
I
,
Here
M(x, y) = y + 2x1^
and
,
,
.
-I-
dM
,
N(x, y) = 1 + 3x"V + x.
.
the equation is exact.
,
3x 2 y 2 + x)dy =
3
(y + 2xy )cix + (1
exact.
dx
dy
4.8
is
is
exact.
,
cN
^^ = + 6xy^ = ——
Since
1
,
the equation is
ex
dy
exact.
4.9
Determine whether the differential equation
I
Here
4.10
M(x, y) = ye* y
and
N(x, y) = xe* y
Determine whether the differential equation
I
Here
M(x, y) = sin x cos y
and
JV(x, y)
dN
is
not equal to the partial derivative
.
xy
dx + xe xy dy —
Since
dM
=
—
sin x cos y dx
,
e
exact.
is
xy
+ xye xy =
dN
—
.
,
.
the equation is exact.
,
— sin y cos x dy = 0.
= — sin y cos x.
^— = sin v sin x,
dx
66
ye
.
,
,
.
.
Since the partial derivative
.
.
the equation is not exact.
dM
^— = — sin x sin y
EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS
4.1 1
M(x, y) = y
Here
4.12
y dx + x dy =
Determine whether the differential equation
and
= x.
JV(x, y)
Determine whether the differential equation
M(x, y) = x - y
Here
and
exact.
is
—= =—
Since
1
67
the equation is exact.
,
— y) dx + (x + y) dy = 0.
(x
N(x, y) = x + y.
—dy = -
Since
-
—=
and
1
are not equal, the
1
dx
equation is not exact.
4.13
(y sin x + xy cos x) dx
Determine whether the differential equation
M(x, y) = y sin x + x>' cos x
Here
and
+ (x sin x + 1) dy —
N(x, y) = x sin x + 1.
exact.
is
—— = sin x + x cos x =
Since
dy
—
the
,
dx
equation is exact.
4.14
Determine whether the differential equation
M(x, y) = y
# Here
.
2
dM
—— = 2y
.
derivatives
is
exact.
dN
—
— = 2x
J
and
N(x, y) = x 2
and
[recall that M(x, y) is the coefficient of dx~]
Since the partial
.
are not equal, the equation is not exact.
dx
dy
4.15
x 2 dy + y 2 dx =
Determine whether the differential equation
x sin y dy + \x 2 cos y dx =
is
exact.
Pi
M(x, y) = \x 2 cos y
Here
N{x, y) = x sin y.
and
A /f
Since the partial derivatives
dy
^^ = sin y
= — \x 2 sin y
and
are not equal, the equation is not exact.
dx
4.16
Determine whether the differential equation
M(x, y) = 4x 3 y 3 — 2xy
Here
is
4.17
and
4
(3x y
2
— x 2 dy + (4x 3 y 3 — 2xy) dx =
4
N(x, y) = 3x y
— x2
2
—
—— = 12x y — 2x =
3
Since
.
exact.
2
the equation
,
dx
dy
exact.
Determine whether the differential equation
M(x, y) = 3x 2 ye*
Here
3
— x 2 e*
3
and
3
e* (3x
2
y — x ) dx + e
2
x
N(x, y) = e \
Since
xi
dy =
is
exact.
—— = 3x e* = ——
2
3
,
the equation is exact.
dx
dy
4.18
is
)
Determine whether the differential equation given in the previous problem is exact after it is divided by the
nonzero quantity e x \
I The new equation is
dM = „
dN
and ^— =
3x
,
,
2
2
)
dx + dy = 0,
M(x, y) = 3x 2 y — x 2
in which
and
N(x,y)—\.
Since
,._
,
,
,
are not equal, the new differential equation is not exact.
.
.
.
dx
dy
4.19
(3x y — x
Determine whether the differential equation
——
=
dx
r dy
x y
M(x, y) = x" 1 — x
Here
_2 _1
y
and
N(x, y) =
=
is
exact.
xy
— x~ y~ 2
l
.
Since
—— = x _2 y -2 = ——
dy
,
the equation is
dx
exact.
4.20
Determine whether the differential equation
M(x, y) = 3x 2 y 2
f Here
dM
—
—
cy
dN
= 6x _ y = ——
,
ox
t
,
2
2
(2x y + 4y ) dy + 3x y dx =
3
3
[recall that M(x, y) is the coefficient of dx]
the equation is exact.
and
is
exact.
yV(x, y)
= 2x 3 y
-I-
4y
3
.
Since
=
CHAPTER 4
68
4.21
Determine whether the differential equation
M(t, y) = 2f
Here
3
+ 3y and
N(t, y)
(2r
+ 3y)dt + (3r + y — l)dy =
3
= 3f + y - 1.
—— = 3=—-,
Since
dy
4.22
M(f, y) = t
Here
4.23
—y
2
and
AT(f, y)
(t
= —t.
Determine whether the differential equation
I
M(t, y) = v(t — 2)
Here
— y) dt — t dy =
2
Determine whether the differential equation
Since
the equation is exact.
exact.
1
the equation is exact.
,
is
exact.
dM
—
—=t—2
and
t
N(t,y)=-r.
and
is
— 2) dt — 2 dy =
y(t
exact.
dr
—— = — = ——
dt
dy
Since
is
cN
—— = -It
dy
are not equal, the
dt
equation is not exact.
4.24
Determine whether the differential equation
I
4.25
and
N{t, y)
M(f, v) = 3e ')' — 2t
3
Here
and
(3e
3
'_y
= e3
N(t, y)
Since
— 2f) <ir + e 3 dy =
'
—— =
Since
'.
2>e
3'
dy
4.26
I
M(t, y) = cos y + y cos t
Here
and
N(t, y)
= sin —
t
t
is
the equation is not exact.
2
r
exact.
— ——
,
the equation is exact.
dt
(cos >• + >' cos t) dt + (sin t
Determine whether the differential equation
exact when a denotes a constant.
cM
cN
It
^^ = ^ -^— = — 2
dy
dt
Ja _
5-
= —y/a z — r.
Determine whether the differential equation
is
t
r~.
M(t, y) = 1
Here
— yja 2 — 2 dy =
dt
sin y.
—
Since
t
sin y) dy
=
is
exact.
cM
dN
^^ = —sin v + cos = -z—
t
rv
,
the
or
equation is exact.
4.27
Determine whether the differential equation for x(f) denned by
+ 3x + 4) dt + (3f + 4x + 5)dx —
(2r
is
exact.
f With
N(t, x)
6
= 3f + 4x + 5.
M=
^—
Then
and
dN
= -r-
3
CX
4.28
M(f, x) = 2f + 3x + 4
as the independent variable and x as the dependent variable, we have
t
(
,
so the equation is exact.
f
Determine whether the differential equation for x(t) defined by
(6f
5
x 3 + 4t 3 x 5 )dt + (3f 6 x 2 + 5t 4 x 4 )dx =
is
exact.
I With
/V(r, x)
t
= 3t 6 x 2 + 5t 4 x 4
.
—— = 18f x + 20f x 4 = —-,
5
Then
2
3
ox
4.29
)
M
is
t
5
(2r
+ 3x + 4) dx + (3f + 4x + 5) dt =
as the independent variable and x as the dependent variable, we have
dM
dN
—
= 4 and — =
dx
-
-
.
and
so the equation is exact.
is
M(t, x) = 3f + 4.x + 5,
always the coefficient of the differential of the independent variable; then also
Since
+ 4t 3 x 5
3
dt
Determine whether the differential equation for x(f defined by
# With
M(t, x) = 6t x
as the independent variable and x as the dependent variable, we have
N(t, x)
exact.
because
= It + 3x + 4.
are not equal, the equation is not exact.
2
dt
2
4.30
Determine whether the differential equation for x(f) defined by
2t{xe'
- 1) dt + e' dx =
2
f With f as the independent variable and x as the dependent variable, we have
dN
cM
—
N(t, x)
4.31
= e'
2
.
Then
—— = 2te' = —
dt
dx
2
,
f
M(t, x) = 2r(xe'
2
- 1)
and
so the equation is exact.
Determine whether the differential equation for x(r) defined by
f With
exact.
is
2
2f(xe'
2
- 1) dx + e' dt =
as the independent variable and x as the dependent variable, we have
exact.
is
M(t, x) = e'
2
,
because M(t, x)
,
EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS
is
N(t, x) = 2t(xe'
the coefficient of the differential of the independent variable; then also
dM
—
=
dN
=
—
and
- 1).
2
69
Now
- I) + (2t)(2txe')
2
2(xe'
Since these partial derivatives are not equal, the equation is not exact.
4.32
Determine whether the differential equation for x(t) defined by
+ -z2
3
t
V
x2
§
Here
4.33
M(t, x) = 3 + -=-
r
N(t, x)
= -2—.
Here
,
M{t, z) = r + z^
and
N(f, z)
dx
= 2rz - z.
Determine whether the differential equation for z(t) defined by
#
Here
M(t, z) = t + z cos r
and
the equation is exact.
,
dt
+ z cos r) dt -f (sin — 3z 2 + 5) dz =
dM
dN
^— = cos = ——
dz
dt
f
exact.
is
t
Since
f
exact.
is
)
(t
= sin — 3z^ + 5.
N{t, z)
the equation is exact,
,
+ z 2 dt + (2tz — z) dz =
2
(t
dz
4.34
exact.
dt
dM
dN
—
— = 2z = ——
Since
is
t
2
Since
t
Determine whether the differential equation for z(t) defined by
f
— 2 - dx =
dt
dM
dN
—
— = 2xt~ = —-
x
and
1
j
the equation is
,
exact.
4.35
Determine whether the differential equation for u(v) defined by
2(u
2
+ uv — 3) du + (u 2 + 3v 2 — v) dv =
is
exact.
f Here
M(v, u) — u 2 + 3u 2 — v,
independent variable; then
M with the coefficient of the differential of the
because we associate
N(v, u) = 2(tr + uv — 3).
dM
dN
—— = 2u = ——
Since
du
4.36
.
the equation is exact.
dv
3
Determine whether the differential equation for u(v) defined by
.
,
(4u m
3
+ \/v)dv + (3v 4 u 2 — \/u)du =
is
exact.
Here
M(v, u) = 4v u
3
3
+-
and
4
N(v, u) = 3f u
2
— - = 12d u = ——
Since
.
3
-
2
du
u
v
the equation is
,
dv
exact.
4.37
2 u" 2
Determine whether the differential equation for v(u) defined by
(v e
+ 4m 3 du + (2uve uv2 — 3v 2 dv —
)
)
is
exact.
f Here u is the independent variable and v is the dependent variable, so
N(u, v) = 2uve
uv
- 3i> 2
dM
Since
.
= 2ve uv + 2v 3 ue uv 2 = —,
4.39
M{0, p) = 2pe
28
and
N(0, p) =
Determine whether the differential equation
Here
M(t, y) = tsjt
2
+ y2 - y
and
1
+ e 2e
Since
.
(1
+ e 2e )dp + 2pe 2e d6 =
—— = 2e 2e = —-,
dp
(tyjt
N{t, y)
2
Determine whether the differential equation
y =
+ ye"
—
2y — xe"
2
t.
Since
.
is
exact.
the equation is exact.
+ y 2 - y)dt + (y^Jt 2 + y 2 - t) dy =
= yjt 2 + y 2 -
is
du
the equation is exact.
4.40
and
the equation is exact.
Determine whether the differential equation for p(0) defined by
Here
+ 4u 3
du
dv
4.38
2
dN
.
2
M(u, v) = v 2 e u "
exact.
—=
ty(t
2
is
exact.
+ y 2 )~ ,/2 -
1
=
—
CHAPTER 4
70
f
+ ye xy dx + {xexy — 2y) dy = 0. Here,
vM
dN
—
— = exy + xyexy — ——, the differential equation is
Rewriting this equation in differential form, we obtain
M(x, y) — 2 + ye
xy
N(x, y) — xe xy — 2y.
and
Since
(2
)
ex
dy
exact.
4.41
dy/dx — y/x
Determine whether the differential equation
I
In differential form, this equation may be written as
is
exact.
y
dy
dx =
or
y dx — xdy — 0,
The original differential equation also has the differential form
exact. If, however, we write the original differential equation as
dM
N(x, y) — 1/y,
and
so that
dN
= ——
=
y
i
y
—
dx — dy = 0,
which is not exact.
x
x
but this equation also is not
— (1/x) dx + (1/y) dy — 0,
M(x, y) = — 1/x
then
and the equation is exact. Thus, a differential equation has manv
x
<
differential forms, some of which may be exact.
Exactness is a property of differential equations in differential
form (see Problem 4.1).
4.42
dy/dx = —y/x
Determine whether the differential equation
I
y
exact.
•
- dx + dx = 0,
This equation has the differential form
is
which is not exact.
If,
however, we write the original
x
equation as
ydx + x dx = 0,
then we have
M(x, y) = v
and
N(x, y) = x,
so that
cM
—— =
dy
1
dN
= ——
and
ex
the equation is exact.
SOLUTIONS OF EXACT EQUATIONS
4.43
Develop a method for solving an exact differential equation.
#
If the differential
equation
A/(.\\ y)dx
+ N{x, y)dy =
exact, then it follows from
is
Problem 4.1 that there
exists a function g(x, y) such that
dg(x,y) = M{x,y)dx + N(x,y)dy
dgix, y) =
But also
J^ + J^lA
d
d
dx
(
It
= M(x,y)
dy
(2)
= *<*,,)
«
dy
ox
so g(x, y) must satisfy the equations
^
(1)
^
and
dy
a
follows from (/) that the exact differential equation may be written as
dg(x, y) =
= Odx.
Integrating this
with respect to x and noting that y is itself a function of x, we obtain the solution to the exact differential
equation in implicit form as
g(x, y)
=c
(4)
where c denotes an arbitrary constant. The function #(x, y) is obtained by solving (5).
4.44
Solve
I
2.\y dx
+
(
1
+ x 2 dy = 0.
)
We must determine a function g(x, y) that satisfies (i) of
dg/dx = 2xy. Integrating
This differential equation is exact (see Problem 4.3).
Problem 4.43. Substituting
M(x, y) — 2xy
into (3) of Problem 4.43, we obtain
both sides of this equation with respect to x, we find
f^dx = $2xydx
or
g{x, y)
= x 2y + Hy)
(J)
Note that when we integrate with respect to x, the constant (with respect to x) of integration can depend on y.
We now determine h(y). Differentiating #(x, y) of (/) with respect to y, we obtain dg/dy — x 2 + h'(y). Then,
2
substituting this equation along with
into (3) of Problem 4.43, we obtain
N(x, y) = 1 + x
x 2 + h'(y) = 1 + x 2
y(x, y)
= x 2y + y + cv
h'(y)
=
1
h(y) — y + c x
(c t constant). Substituting this expression
Thus, the solution to the differential equation, which is given implicitly
Integrating this last equation with respect to y, we obtain
into (/) yields
or
EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS
by
gix, y)
y =
4.45
= c,
x 2 y + y = c2
is
where
,
c2
=c-c
x
Solving for y explicitly, we obtain the solution as
.
x^T
Solve
f
It
(x
+ sin y) dx + (x cos y - 2y) rfy = 0.
With M(x, y) = x + sin y and
we seek a function #(x, y) that satisfies (J) of Problem 4.43. Substituting Mix, y) in (i) of
follows from Problem 4.4 that this equation is exact.
N(x, y) = x cos y - 2y,
dgfdx = x + sin y.
Problem 4.43, we obtain
Integrating both sides of this equation with respect to .x, we find
—
dx =
dx
dg
I
or
c/(x, y)
I
(x
+ sin y) dx
= jx 2 + x sin y + hiy)
To find hiy), we differentiate (7) with respect to y, obtaining
result
Nix, y) = x cos y — 2y
along with
(J)
dg/dy = xcosy + h'(y),
and then substitute this
Thus we find
into (5) of Problem 4.43.
x cos y + /j'(y) == x cos y — 2y
or
= — 2y
h'(y)
from which it follows that hiy) = —y 2 + c v Substituting this h(y) into (7), we obtain
2
2
The solution of the differential equation is then given implicitly by
g(x, y) = \x + x sin y — y + c
.
x
4.46
2
+ x sin y — y 2 = c 2
or by
\x
Solve
(2xy + x) dx + (x
f
71
2
where
,
c2
2
M(x, y) = 2xy + x and N(x, y) = x + y (see Problem 4.7).
dg/dx = 2xy + x. Integrating this with respect to x, we obtain
so
= c,
+ y) dy = 0.
This equation is exact, with
dg/dx = Mix, y),
gix, y)
— c — c,.
=
g(x, y)
J
£ dx = J(2xy +
x) dx - x y + |x
2
2
We require
+ h(y)
(7)
To find /j(y), we first differentiate (7) with respect to y, obtaining dg/dy = x 2 + h'iy). Since we require
2
2
dd/dy — ^(•x y\ 't follows that x + h'iy) = x + y or h'iy) = y. Upon integration, we find that
»
hiy)
is
4.47
= \y 2 + c,,
then
Solve
I
gix, y)
so (7) becomes
— c,
x y + {x
or
3
(y + 2xy ) dx + (1
dg/dx — Mix, y),
2
+ 3x 2 y 2 + x) dy = 0.
M(x, y) = y + 2xy 3 and W(x, y) = 1 + 3x 2 y 2 + x (see Problem 4.8).
Integrating this with respect to x, we obtain
dg/dx = y + 2xy 3
This equation is exact, with
require
— x 2 y + ^x 2 + |y 2 + c,. The solution to the differential equation
+ \y 2 = c 2 where c 2 = c — c,.
g(x, y)
2
so
= f-£-dx =
gix, y)
J(y
+ 2xy 3 dx - xy + x 2 y 3 + hiy)
(7)
)
To find hiy), we first differentiate (7) with respect to y, obtaining
dg/dy = Nix, y),
We
.
dg/dy = x + 3x y 2 + h'iy).
2
Since we require
follows that
it
x + 3x 2 y 2 + h'iy) =
1
+ 3x 2 y 2 + x
or
h'iy)
=
1
so (7) becomes
h(y) — y + c
2
3
—
differential
equation
is
solution
the
then
or
The
to
xy
x
+ y + y + cv
gix, y) — c,
gix, y)
2
3
—
=
=
where c 2
Cj.
xy + x y + y
c
c2
Upon integrating this last equation with respect to y, we find
4.48
Solve
f
ye
xy
dx + xe
xy
dy = 0.
M(x, y) = ye*y
This equation is exact, with
dg/dx = Mix, y),
dg/dx = ye
so
xy
and
N(x, y) = xe
=
j ||
dx = JV> dx = e
To find hiy), we first differentiate (7) with respect to y, obtaining
it
follows that
with respect to y, we find
differential equation
is
hiy)
#(x, y)
xe
xy
(see
Problem 4.9).
We require
Integrating this with respect to x, we obtain
.
gix, y)
dg/dy = Nix, y),
x ,
xy
+ h'iy) = xe
xy
,
or
h'(y)
xy
+ h(y)
dg/dy = xe
= 0.
xy
(7)
+ h'iy).
Since we require
Upon integrating this last equation
xy
+ c v The solution to the
gix, y) = e
= c,, a constant, so (/) becomes
= c, or e xy = c 2 where c 2 = c — c v
CHAPTER 4
72
4.49
3x 2 y 2 dx + (2x 3 y + 4y 3 ) dy = 0.
Solve
I
M(x, y) = 3x 2 y 2
This equation is exact, with
dg/dx = M(x, y),
so
dg/<?x
= 3x y
2
2
and
= 2x 3 y + 4y 3
7V(x, y)
(see Problem 4.20).
We require
Integrating this with respect to x, we obtain
.
*3
C
2 2
3 2
g (x, y) = f 4- dx = f 3x y dx = x y + h(y)
cg/cy = 2x y + h'(y).
3
To find h(y), we first differentiate (7) with respect to y, obtaining
dy/dy — N(x, y\
it
(7)
Since we require
follows that
2x 3 y + h'(y) = 2x 3 y + 4y 3
or
h'(y)
= 4y 3
Upon integrating this last equation with respect to y, we find h(y) — y 4 + c u so (7) becomes
3 2
4
The solution to the differential equation is g{x, y) = c, or x 3 y 2 + y 4 = c 2
g(x, y) = x y + y + c v
c2
4.50
Solve
I
where
=c-c v
ydx + xdy = 0.
M(x, y) = y
This equation is exact, with
dg/dx = M(x, y),
cg/dx = y.
so
and
7V(x, y)
=x
We require
Problem 4.11).
(see
Integrating with respect to x. we obtain
dg
g(x, y)
= \±dx = \y dx = xy + h(y)
J
dx
(7)
>>
To find /i(y), we first differentiate (/) with respect to y, obtaining cg/cy = x + h'(y). Since we require
Upon integrating this last equation with respect
dg/dy = 7V(x, y), it follows that x + h'(y) — x or h'(y) — 0.
to y, we find
4.51
h(y)
= cu
so (7) becomes
= c,
Solve
(y sin x + xy cos x) dx + (x sin x +
f
or
xy = c 2
where
dg/dx = M{x, y),
—
dx=
dx
It
c2
so
The solution to the differential equation is
1 )
dy = 0.
(
v sin x
J
N(x, y) = x sin x + 1
and
dg dx = y sin x + xycosx.
(see
Problem 4.13).
Integrating this with respect to x, we obtain
+ xv cos x) dx = — y cos x + (xy sin x + y cos x) = xy sin x 4- h(y)
dg dy = x sin x + h'(y).
follows that
= xy + c v
M(x. y) = y sin x + xy cos x
This equation is exact, with
We require
g(x, y)
— c — cv
g(x, y)
We require
cg/dy = 7V(x, y),
x sin x + /j'(v) = x sin x + 1
(7)
so we have
or
h'(y)
=
1
Upon integrating this last equation with respect to y, we find h(y) = y + c,, so (/) becomes
#(x, y) = xy sin x + y + c,.
The solution to the differential equation, #(x, y) = c, may then be written as
=
xy sin x + y + c,
or y = c 2 — xy sin x where c 2 = c — c x
c,
.
4.52
Solve
I
4
(3x y
2
- x 2) Jy + (4xV - 2xy) dx = 0.
M(x, y) = 4x 3 y 3 — 2xy
This equation is exact, with
and
JV(x, y)
= 3x 4 y 2 — x 2
(see
Problem 4.16). Then
we have
g(x, y)
from which
4
cg/cy = 3x y
2
= JM(x, y) dx = f(4x 3 y 3 - 2xy) dx = x*y 3 - x 2 y + h(y)
(7)
Now since dg cy = N(x, y) = 3x 4 y 2 — x 2 we have h'(y) = 0.
4 3
2
The solution to the differential equation is then
g(x, y) = x y — x y + c v
- x 2 + h'(y).
,
- c u and (7) becomes
x 4 y 3 — x 2 y = k, where k is an arbitrary constant.
4.53
Then
h(y)
Solve
e
I
x
\3x 2 y - x 2 dx + e
)
xi
dy = 0.
This equation is exact, with
4( X ,
M(x, y) = 3x 2 ye*
3
- x 2 e xi
and
2
y)=\-f
3
v3
dx = jM(x, y) «ix = |(3x yr
N(x, y) = e
-x
x}
(see
V dx = y^ 3
Problem 4.17). Then
3
)
x
^e
xi
+ h(y)
we have /i'(y) = 0, so that
= c v Then (7) becomes ^(x, y) = ye — y^ + c u and the solution to the differential equation is
xi
x3
=
or
+ \, where k is an arbitrary constant.
ye* — 3f
v = ke~
from which
dg/dy = e*
+ h'(y).
Since this last must equal
xi
h(y)
3
A'
3
N(x, y) = e \
(7)
EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS
4.54
Solve
f
—?— dx
= 0.
= dy
xzy
73
D
xy
g(x, y)
_1
M(x, y) = x
This equation is exact, with
- x~ 2 y~
-£ dx = JM(x, y) dx =
=
J*
1
N(x, y) =
and
JV - x" y~
2
1
l
)<*x
-x~ y~ 2
l
Problem 4.19). Then
(see
= In |x| + x~ y~ l + My)
l
(7)
-2
_1
from which we may write dg/dy = —x~ 1 y~ 2 + h'{y) = N(x, y) = — x y
This gives us h'(y) = 0, from
-1 _1
which h(y) = c v Then (7) becomes g(x, y) = In |x| + x y
+ c u and the solution to the differential
_1 _1
= k, where k is arbitrary. This solution may also be written as
equation is In |x| + x
y
.
In |x|
4.55
Solve
/
+ In \C\ = — x
{It
_1
_1
k=— In
where
y
|C|,
from which
= — 1/xy
In |Cx|
- l/(x In |Cx|).
y =
or
+ 3y) dt + (3t + y - 1) dy = 0.
3
M(t, y) = 2r
This equation is exact, with
d
g(t,
y)
=$
f
dt
t
3
+ 3y
and
= 3r + y —
N(t, y)
= S M{U y) dt = i (2f3 + 3y) dt = \
*
l
1
(see Problem 4.21).
Then
+ 3ty + Hy)
(1)
dg/dy = 3r + h'(y) = N(t, y) = 3r + y — 1. This yields h'{y) = y + 1, from which
+ y + c v Then (7) becomes g(t, y) = \t* + 3ry + \y 2 + y + c lt and the solution to the differential
equation is jt 4 + 3fy + \y 2 + y — k where k is an arbitrary constant. If we rewrite the solution ls
4
2
where C — —2k, then we can use the quadratic formula to solve for y
y + (6f + 2)y + (f + C) =
from which we may write
h(y) = {y
2
explicitly, obtaining
(6r
y =
4.56
Solve
I
2
(t
+ 2) ± V(6t + 2) 2 - 4(t 4 + C)
= — (3t + 1) ± yjyt + ot — t + K
K= —C
I
- y)dt-tdy = 0.
M(t, y) = t
This equation is exact, with
g(t, y)
2
—y
N(t,y)=—t
and
(see
Problem 4.22). Then
X
=
dt
jf
= JM(t, y) dt = J(r 2 -y)dt= -t i -ty + h(y)
(7)
t
from which we may write
Then (7) becomes
g(t, y)
dg/dy = — t + h'(y) = N(t, y) = —t.
= \t 3 — ty + c lt and the solution is
This yields
^r
— ty = k
3
h\y) = 0,
h(y) = c v
2
=
%t
k/t,
y
from which
or, explicitly,
with k arbitrary.
4.57
Solve
I
3
(3e 'y
- It) dt + e 3 dy = 0.
'
M(t, y) = 3e y - It
3,
This equation is exact, with
N(t, y) = e
and
3'
d
2
3
3
g (t, y) = j Jj-dt = JM(t, y)dt = J(3e 'y - 2t)dt = e 'y - r +
%)
(7)
This yields h'(y) = 0, so that
from which we may write dg/dy = e + h'(y) = N(t, y) = e
3t
2
2
3
+ c x and the solution is e 3 'y - t 2 = k or y = {t + k)e~
g(t, y) = e 'y - t
(7) becomes
3
3'
Solve
f
(cos y + y cos f ) dt + (sin t
This equation is exact, with
—
r sin y)
h(y) = c v
'.
,
4.58
Then
(see Problem 4.25).
,
Then
with /c arbitrary.
dy = 0.
M(t, y) = cos y + y cos r
and
7V(r, y)
= sin r - t sin y
(see Problem 4.26).
Then
g (t, y)= f -j-dt = ilM(t, y) dt = f(cos y + y cost) dt = t cosy + y sin t + h(y)
dg/dy =
- sin y + sin f + ft'(y).
h(y) = c v
Then (7) becomes
from which we may write
have
h'(y)
implicitly,
4.59
Solve
I
(tVt
= 0,
t
2
so that
1
Since this must equal
g(t, y)
=
t
N(t, y)
cos y + y sin t + c {
,
(1)
= sin t
t
we
sin y,
and the solution is,
cos y + y sin r = k.
+y 2 -y)dt + (yjt 2 + y 2 -t)dy = 0.
This equation is exact, with
g{t , y)
=
d
j
M(t, y) = rV'
Jl dt =
jM
(t,
2
y) dt
+ y2 - y
and
N
( l>
y)
= y^f^^y 2 -
t
(see Problem 4.39).
= j[t(t 2 + y 2 y> 2 - y] dt = i (r 2 + y 2 3 2 - ty + fc(y)
Then
'
)
(7)
.
CHAPTER 4
74
from which we may write dg/cy = y{t 2 + y 2 ) 1 2 - t + h\y) = N{t, y) = y[t 2 + y 2 ) 1 2 - t. This yields h'(y) = 0,
from which h(y) = c v Then (7) becomes g(t, y) = ^(t 2 + y 2 ) 3 2 - ty + c u and the solution is, implicitly,
'
'
2
(t
4.60
+ y 2 3 2 - 3ty = k.
'
)
Solve (2f + 3.x + 4) dt + (3t + 4x + 5) dx = 0.
f
We presume that
the independent variable and x is the dependent variable, so we are seeking x(t).
is
t
this equation is exact,
M(t, x) = 2r + 3x + 4
with
and
iV(r, x)
= 3r 4- 4x + 5
(see
Then
Now
Problem 4.27).
C
2
g (t, x) = f 4- dt = (M(t, x) dt = Ult + 3x + 4) dt = t + 3tx + 4f + h(x)
(7)
cg/cx — 3t + h'(x) — N(t, x) — 3t + 4x + 5. This yields h'(x) — 4x + 5, from which
= 2x 2 + 5x + Cj. Then (7) becomes 2 + 3tx + 4t + 2x 2 + 5x + c u and the solution is. implicitly,
2
+ 3fx + 4t + 2x 2 + 5x = k. This solution may be rewritten as 2x 2 + (3f + 5)x + (r 2 + 4f - k) =
and
2
2
-(3r
+
8(t
4t
k)
+
+
5)
v(3r
±
5)
f
=
the quadratic formula,
-.
from which we write
h(x)
t
t
,.
.
.
,
.
.
.
.
.
,
then solved explicitly for x with
x
to yield
^
4
4.61
Solve
I
(6f
5
We presume that
is
t
the independent variable, x is the dependent variable, and we want x(f).
M(t, x) = 6f x
5
equation is exact, with
g (t, x) = f
from which we may write
from which
6
f
fc(x)
and
= 3f 6 x 2 + 5f 4 x 4
N(t, x)
^ dt = JM(t, x)dt - f(6f x + 4f x dt =
3
5
3
6
5
)
f
Then this
Now
(see Problem 4.28).
x 3 + r 4 x 5 + h(x)
(1)
+ 5f 4 x 4 + h'(x) = N(t, x) = 3t 6 x 2 + 5f 4 x 5 This yields h'(x) = 0,
Then (/) becomes g(t, x) = 6 x 3 + 4 x 5 + c lt and the solution is. implicitly,
6
dg/dx = 3f x
2
.
f
f
4
Solve
f
= cv
+ 4r 3 x 5
3
x 3 + f x 5 = k.
2
4.62
4
4
x 3 + 4f 3 x 5 ) dt + (3f 6 x 2 + 5t x ) dx = 0.
2t(xe'
2
-l)dt + e' dx = 0.
We presume that
t
equation is exact, with
is
the independent variable, x is the dependent variable, and we want x(t).
M(t, x) = 2r(xe'
ig
=
g(t, a)
-
2
— 1)
and
N(t, x)
= e'~
dt - f M(f, x) dt = f2f(xe'
2
(see
Problem 4.30).
-\)dt = xe'
2
Then this
Now
- 2 + h(x)
t
(/)
J*
from which we write
cg/cx = e'~ + h'(x) = N{t, x) = e'
2
g(t, x) = xe' —
(7) becomes
4.63
Solve
f
2
(t
t
+ c,;
from which we have
2
h(z) = -\z + c,.
or. explicitly,
Solve
( 3
the solution is
z(t)
This yields
2
— k
xe' —
h'(x)
= 0,
from which
t
or, explicitly,
h(x)
= cv
= (f 2 + k)e~'
x(t)
Then
2
.
for r(f).
)
M(t, z) = t
This equation is exact, with
.
2
+ z 2 dt + (2tz - z) dz =
g(t,
4.64
2
2
2
+ z2
and
N{t, z)
= 2tz — z
(see Problem 4.33).
Then
z)= (T-dt = §M(t, z) dt = JV 2 + z 2 dt = -t 3 + tz 2 + h(z)
(7)
)
cg/dz = 2tz + h'(z) = N(t, z) = 2tz - z.
Then (7) becomes
This yields
3
g(t, z) = \t
+ tz 2 - \z 2 + c,;
h'(z)
= -z,
the solution is
from which
2f
3
+ 6fr 2 - 3r 2 = k
fk-2t 3 V 2
= ±1
+ ^- j dt - 2- dx = 0.
x2
I
This equation is exact, with
g(t, x)
from which we may write
M(r, x) = 3 + -y
x
and
= j^dt = jM(r, x) dt =
cq
2x
—
= —- +
ex
t
N{t, x)
J
= —2 -
(see
Problem 4.32). Then
h + ^-J A = 3r - y +
X
h'{x)
= N{t, x) = — 2 '-.
t
This yields
h'(x)
h(x)
= 0.
(7)
so that
V
EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS
h{x)
= cv
Then (/) becomes
g(t, x)
= 3t
+ ct
3t -
the solution to the differential equation is
;
l
4.65
Solve
#
(t
x(t)
= ±y/3t 2 - kt.
+ z cos t) dt + (sin - 3z 2 + 5) dz =
for z(t).
t
M{t, z) = t + z cos t
This equation is exact, with
N(z, t) = sin t
and
dg/dz = sin t + h'(z) = N(t, z) = sin t — 3z 2 + 5.
from which we write
— 3z + 5
2
z)= f -~dt = JM(t, z) dt = f (t + z cos t) dt = - t 2 + z sin
g(t,
= -z + 5z + c,. Then (7) becomes
3
2
\t + zsinr - z + 5z =
3
h(z)
x2
- = k
t
,
or, explicitly,
75
This yields
= \t 2 + z sin - z 3 + 5z + c ls
g(t, z)
(see
Problem 4.34). Then
+ hi:)
t
h'(z)
(7)
= - 3z 2 + 5,
so that
and the solution is. implicitly,
t
/c.
4.66
Solve
I
2(u
2
+ ud - 3) du + (u 2 + 3i> 2 - v) dv
M(v, u) = u
This equation is exact, with
for u(v).
+ 3i? 2 — d
2
N(v, u) = 2u
and
+ 2uv - 6
2
(see Problem 4.35).
Then
g(v, u)
=
ydv= { M(v, u)dv = Uu + 3v -v)dv= vu + v - - v + h(u)
2
2
J
2
2
3
(1)
from which we may write dg/du = 2uv + h'(u) = N(v, u) = 2u 2 + 2uv — 6. This yields h'(u) = 2u 2 — 6, from
which h(u) = f u 3 — 6u + c v Then (7) becomes g(v, u) = vu 2 + v 3 — \v 2 + f u 3 — 6u + c ,; the solution is,
after fractions are cleared,
4.67
Solve
4u 3 + 6v 3 + 6i>u 2 — 3i; 2 — 36u — k.
(4pV + l/v) dv + {3v*u 2 - l/u) du =
This equation is exact, with
for u(v).
M(v, u) — Av u
3
3
+-
4
N{v, u) = 3v u
and
2
(see Problem 4.36).
g ( Vy u ) = f j- dv = JM{v, u)dv= f
from which we may write
= — In \u\ + c v
+ In \v/u\ = k.
which
v*u
4.68
3
Solve
I
h(u)
2
(v e
uv2
A
Then
u
v
Uv u +
3
A
3
-J
dv = v u
3
+ In
|o|
+ h(u)
U)
+ h'{u) — N{v, u) — 3v*u 2 — l/u. This yields h'(u) = — 1/w, from
— In \u\ + c x the solution is, implicitly,
Then (7) becomes g(v, u) — v A u 3 + In
dg/du — 3v u
2
|t;|
+ 4u 3 )du + (2uve uv2 - 3v 2 )dv =
M(u, v) = v e
2
This equation is exact, with
;
tor v(u).
ul 2
'
+ 4u 3
N(u, v) = 2uve
and
u" 2
- 3v 2
(see
Problem 4.37).
Then
2
g (u, v) = f j- du = jM(u, y) Ju = JV>
h(v)
4.69
= -v + c,.
Solve
f
(1
dg/dv = 2uve
+ e 2e dp + 2pe 2e d9 =
This equation is exact, with
from which we write
Solve
y=
2 + ve
.
.
for p(0).
)
Then (7) becomes
'
'
'
M(6, p) = 2pe
g(6, p)
4.70
(7)
)
m 2 + h'(v) = N{u, v) — 2uve m 2 — 3v 2
This yields h'(v) = —3v 2
from which
2
4
m
3
+ u — v + c^ the solution is, implicitly, e'"~ + u 4 — v 3 = k.
Then (7) becomes g(u, y) = e
from which we write
3
+ 4u 3 Ju - e uv2 + w 4 + /i(r)
pe
xy
2y — xe
x
dg/dp = e
20
20
+ p + cv
=
2e
and
7V(0, p)
=
1
+ e 2e
(see
Problem 4.38). Then
j^M = $M(0, p)d9 = J2pe d0 = pe +
2e
h(p)
(7)
+ e 20 This yields h'(p) = 1, so that h(p) = p + c v
pe
+ p = k or, explicitly, p = fc/(l + e ie
+ h'{p) = N{0, p) =
The solution
20
1
.
20
is
).
CHAPTER 4
76
I
was shown in Problem 4.40 that this equation is exact in the differential form
Here M(x, y) = 2 + ye" and N(x, y) = xe*> - 2>\
eg ex — M(x, y), we have
(2
It
+ yexy)dx + (xex>- 2y)dy = 0.
g(x, y)
=
J*(2
cg/cy — xexy + h'(y);
from which we may write
from which h'(y) — —2y.
It
follows that
+ yO dx = 2x + e xy + h(y)
dx =
J j-
= —y + c v
2
Then (7) becomes
2x + e
and the solution to the differential equation is given implicitly by
4.71
Solve
I
It
with
(7)
xy
g(x, y)
— y2 — c2
— 2y,
— y 2 + cu
c 2 = c — cv
xe* y + h'{y) — xe
then equating this to N(x, y) yields
h(y)
Since
— 2x + e
where
,
xy
xy
dy/dx — y/x.
— (\/x)dx + [l/y)dy — 0.
was shown in Problem 4.41 that in the differential form
M(x, y) = — 1/x and N(x, y) = 1/y. Then
g(x, y)
= j-JLdx = JM(x, y) dx =
—
this equation
is
exact
dx = -In |x| + h(y)
(7)
J"
from which we may write cg/cy = h'(y) — 7V(x, y) = 1/y. This yields h'(y) — l/y, so that h(y) = In |y| + c v
Then (7) becomes g(x, y) = — In |x| + In \y\ + c,, or g{x, y) = In |y/x| + c,. The solution to the differential
equation is In \y/x\ — k, or y = Cx where C = ±e\
4.72
Solve
I
It
dy/dx = -y/x.
y dx + xdy = 0.
was shown in Problem 4.42 that in the differential form
and N(x, y) — x. Then
this equation
is
exact with
M(x, y) = y
g(x, y)
=
J
-^ dx =
J*
Af (x, y) dx = jy dx = xy + h\ y)
(7)
from which we may write dg/dy = x + h'(y) = N(x, y) = x. This yields h'(y) — 0, so that h{y) — c v Then
so the solution to the differential equation is xy = k
or, explicitly,
g(x, y) = xy + c,,
y = k/x.
(7) becomes
v
4.73
Solve
1
- -^ dx + - d y = 0.
x
I
x
—Ay and
Mix, y) = -
Here
x
,'M
1
= -,
/V(x,
N(x, y)
,
and since
x
—— =
dy
— =—
TV
1
x
r
;
the differential equation is exact.
-r-
cx
Then
9(X
'
PI
=-+
from which we write
-£-
g{x, y)
=J
^ JX = J M(X
= y/x + c u
=
V) d 'X
/(?)
-
dX =
x
+ *°°
(7)
1
= N(x, y) = -.
h'(v)
This yields
h'(y)
= 0,
from which
h{y)
x
x
ay
becomes
>'>
and the solution to the differential equation is
y x = k
or
= cv
Then (7)
y = /ex.
x
- dx - -j dy = 0.
1
4.74
Solve
>'
>'
A/(x, y)
=-
—
dM
cN
—
^— = — = —
T~
x
1
Here
and
= — yy,
?,
2
7V(x,
7V(x,y)=
y)
1
so
-
y
y
r
^
cy
and the equation is exact. Then
cx
y
1
9(x, y)
= f^-dx= fM(x, y) dx = f - dx = - + Hy)
J
from which we can write
-f-
dy
Then (7) becomes
where
C = l/k.
#(x, y)
=
—+
=-
2
y
= x/y + c u
cx
h'(y)
J
J
= 7V(x, y) =
—
5-.
y
This yields
(/)
y
h'(y)
= 0,
so that
Hy) =c v
y
so the solution to the differential equation is
x y = k,
or
y = Cx
—
=
EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS
4.75
~y
Solve
x
,
+ y2
2
I u
Here
dx +
77
* , d>- = 0.
,
x2 + y2
"^
— and
^
-
*«
^
is/
-.—
M(x,
5
y) =
* + y
= -=
*
7v(x, y)
*
dM
-x— —
dN
—
= — = —y
2
2
x
so
_-,
+ yz
dy
(x
2
2
2
+y
.
.
and the equation is exact.
dx
)
Then
— dx = Jc M{x, y) dx = Jc
Jdg
—y
y
2
dx = arctan - + h'(y)
2
(1)
—
da
x
x
/i'(y) = N(x,
7V(x, y) = -^
This yields /i'(y) = 0, from which
+ h'(y)
j- = -5j.
2
x + y
x + y
dy
Hy) — c v Then (/) becomes g(x, y) = arctan (y/x) + c lt and the solution to the differential equation is
arctan (y/x) = k. This may be rewritten as y/x = tan k, or as y — Cx
where C = tan k.
— ~^—^
from which we may write
4.76
Solve
(1/x) dx
+ (1/y) dy = 0.
M(x, y) = 1/x
Here
N(x, y) = 1/y,
and
—— = = ——
so
g(x, y)
=
- dx = In |x| + h(y)
dx = jM(x, y) dx =
J jdx
and the equation is exact. Then
dx
dy
(/)
J*
from which we write dg/dy — h'{y) = JV(x, y) = 1/y. This yields h'(y) = 1/y, from which h(y) — In |y| + c
Then (7) becomes #(x, y) = In |x| + In \y\ + c, = In |xy| + c 1? and the solution to the differential equation is
In |xy|
4.77
Solve
= k,
y = c/x
or
where
— ±e k
c
.
xy 2 dx + x 2 y dy = 0.
M
Here
(x, y)
= xy 2
and
JV(x, y)
= x 2 y,
—— = 2xy = ——
so
=
J jdx
and the equation is exact. Then
dx
dy
g(x, y)
2
dx = ]- x 2 y 2 + h(y)
dx = jM(x, Jy)' dx = Jxy
J
v
J
J
(1)
2
2
from which we may write dg/dy = x 2 y + h'(y) = N(x, y) = x y. This yields h'(y) = 0, so that h{y) — c
2 2
Then (1) becomes g(x, y) = \x y + c u so the solution to the differential equation is \x 2 y 2 = k or,
y = c/x
explicitly,
4.78
Solve
X
"n
Here
where
c
y"" +1 dx + x~ n + y~"dy = 0,
for real numbers
l
M(x, y) = x ~"y~
g(x, y)
and
from which we may write
/i(v)
= cv
r da
iV(x, y)
= x~ n + y' n
l
,
n #
-=
fc.
r
<
dg/dy = x~
n + 1
r
+ '
dx =
g(x, y)
=
This may be rewritten as
—+
(xy)"
"y
"
= —-
and the
dx
—
x"" +1 v" n+1
y~" + h'(y) = N{x, y) = x~ n+l y~".
-(xy)~"
l)x
-
J^-- +
(/)
/j(y)
This yields
h'(y)
= 0,
so
+1
'
and the solution to the differential equation
c l9
= — ~~
1
-,
or as
y =
C
—
where
=
dN
—
and the equation is
l/(n-l)
c=
4.79
Solve
= 0.
-^2
dx + 2,
, dy
2
x + y
x + y2
/
Here
M(x, y) = -^
x
5
.
1.
—dy = {-n +
so
-£ dx = J M(x, y) dx = jx-"y-"
J
Then (/) becomes
— (xv)~" +1
is
=
1
= ± \J2k.
equation is exact. Then
that
.
x
and
iV(x, y)
—
= -j-y
2,
so
8M
—
-=
—2xy
,
2 2
I
x
=
-
CHAPTER 4
78
exact.
Then
= j ~dx = J*M(x, y) dx =
9(x, y)
dg
= -= y T +
—
dy
x*
*
j^
dx = - In (x 2 + y2) + h(y)
A
= N(x, y) = —.L y T
x + v
(1)
This yields h'{ y) = 0, from which
+y
Then (/) becomes g(x, y) = \\n(x 2 + y 2 + c
h(y) = c v
the solution to the differential equation is
2
2
= k, which we may rewrite as x 2 + y 2 = e 2k or explicitly as
= ± yJC — x 2 where C = e 2k
\ In (x + y
from which we may write
h'(y)
)
Y;
)
4.80
Solve
2
Here
v
2
)"
.,,
exact.
2
(x
2
+
>'
,
+ v
dx - /"<*•
= c,.
(x
v
Then (/) becomes
,
2
,
2
—=
-1
^t~ - .w
l)(x
y = ±Vc —
Solve
>'
axa
f
Here
2
t,
2
+ y )"
r
g(x, v)
=
=
2
l
y f
dN
and the equation
dx
y
— -—
,
-——,
\){x
2
+ y
._. + c,.
.
h'(y)
= 0,
so that
and the solution to the differential equation
.
(x
-1
= —— —
,.._.
"
,
2
+ y 2 )"
'
'
,
and then explicitly as
2k(n- 1)
'
where
This yields
.
(/ »
)
_..
This
us n
may be rewritten as
.
k.
-1
2
= N(x, y) = —-=-^j—
2
2
(x + y )"
h'(y)
y )"
2(«
4.81
(x
** - J r?^? * - a .-i^' + ^- + **
OQ
y
—
== j-^
j— +
+
-T<
Hi,
ay
)
2
Then
from which we may write
is
so
+y
{x
cM —
—
=
+
-2nxy
y
.„
,
N(x, y) =
)
«* * - J
/i(y)
n # 1.
for real numbers
and
2
.
)"
*
M(x, y) =
(x
is
>'
,
x
y
=- dx + —.
—
^- dy =
(x +
'
.
l/(n- 1)
c
_2fc(n- 1)
~1 b
a b~ 1
y dx + bx y
M(x, y) = ax°
l
dy =
for nonzero values of the real constants a and 6.
—- =
<\\/
N(x,y) = bx"y b
and
y*
l
,
so
afrx"
l
y*
rv
'=—
—
oy
and the equation is exact.
ox
Then
(
</(.v.
= f / dx = fM(x, y)dx = fax ' V</x = x"/ + /i(y)
y)
This yields /j'(y) = 0, from which
+ />'(>') = N(x, y) = bx a y h ~
g(x, y) = x"y + c x
the solution to the differential equation is
x ay b = k, or
dg/cy = frx'V
from which we may write
Ky) — c \- Then (/) becomes
lb
where C = k h
y = Cx-"
l
'
.
h
\
l
.
INTEGRATING FACTORS
4.82
Define integrating factor for a differential equation of the form
I
M(x, y)dx + N(x, y)dy — 0.
A function I(x, y) is an integrating factor for such a differential equation if
is exact.
+ N(x, y) dy] =
/(x, y)[M(x, y) dx
4.83
Determine whether — 1/x 2 is an integrating factor for
I
ydx — x dy — 0.
Multiplying the given differential equation by —1/x 2 yields
—-1 (y dx — x d\) —
4.84
- 1/x 2 is an integrating factor for the equation.
Determine whether — 1/xy is an integrating factor for
I
Multiplying the given differential equation by
—
xy
This last equation is exact; hence
(
v dx
H
x
x
This last equation is exact (see Problem 4.73); hence
—-yr dx —x dy =
1
or
=-
y dx — x dy = 0.
— 1/xy yields
- x dy) =
or
— +
dx
x
- dy =
y
— 1/xy is an integrating factor for the equation.
(7)
—
EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS
4.85
Determine whether 1/y 2 is an integrating factor for
ydx — xdy — 0.
I
2
Multiplying the given differential equation by 1/y
1
-5- (y dx
- xdy) =
or
This last equation is exact (see Problem 4.74); hence 1/y
Determine whether
f
— l/(x 2 + y 2
is
)
2
—1
y
=
T dy
ydx — xdy — 0.
yields
)
— — y dx
r+y
or
^
2
x
H
^
x* +
^ dy
y
—
— l/(x 2 + y 2 is an integrating factor for the equation.
This last equation is exact (see Problem 4.75); hence
)
y dx + x dy = 0.
Show that 1/xy is an integrating factor for
#
x
an integrating factor for the equation.
is
— l/(x 2 + y 2
— x dy) =
-2
j (y dx
x" + y-
4.87
1
- dx
y
an integrating factor for
Multiplying the given differential equation by
79
yields
y
4.86
D
Multiplying the given differential equation by 1/xy yields
—
xy
(y dx + x dy)
=
— dx + - dy =
or
x
y
Since this last equation is exact (see Problem 4.76), 1/xy is an integrating factor.
4.88
y dx + x dy — 0.
Show that xy is an integrating factor for
I
Multiplying the given differential equation by xy yields
xy(y dx + x dy) —
xy 2 dx + x 2 y dy =
or
Since this last equation is exact (see Problem 4.77), xy is an integrating factor.
4.89
Show that l/(xy) n is an integrating factor for
ydx + xdy — 0,
for any real number n.
n
I
Multiplying the given differential equation by l/(xy) yields
(xy)
n
(ydx + xdy) =
x~"y~" +l dx + x~" +l y~" dy =
or
Since this last equation is exact for all real values of n (see Problems 4.76 and 4.78), l/(xy)
n
is
an integrating
factor.
4.90
Show that
(x
2
+ y 2 )""
is
an integrating factor for
m
Multiplying the given differential equation by
(x
2
xdx + ydy — 0,
for any real number n.
+ y 2 )""
—^
Since this last equation is exact (see Problems 4.79 and 4.80),
that if
n = 0,
then
(x
2
+y
2
)
°
=
1
is
yields
(x
2
+ y 2 )~"
x
is
j— dx + —^
y
^— dy = 0.
an integrating factor. Observe
an integrating factor, which implies that the differential equation is
exact in its original form.
4.91
Show that x a ~ y b
l
I
~ x
an integrating factor for
is
ay dx + bx dy =
Multiplying the given differential equation by x
a' l
b~ 1
yields
y
ax a ~'y b dx + bx y
a
Here
x
4.92
M(x, y) - ax a ~
a-iyb-i
js
Show that
V and
N(x, y) = bx"/' 1
.
Since
for any real-valued constants a and b.
b- 1
dy =
—dy = abx
-
(1)
a~ 1
y
b~ l
-
—
ox
,
(7) is exact;
hence
an integrating factor for the original differential equation.
Mx — Ny
,
for
Mx - Ny
M dx + N dy = yf {xy) dx + xf {xy) dy =
x
2
not identically zero, is an integrating factor for the equation
0.
Investigate the case
Mx - Ny = 0.
x
CHAPTER 4
80
— yields
—
m.x - zvy
Multiplying the given equation by
/^
...
*y[/i(xy) - /2 (*y)]
dx +
,?***,
n ^ = 0- /2 (*y)]
.
r
*y[/i(*>')
This equation is exact because
cy
dylx(f
L-x(/, - h)\
2 )]
i
r/2 \
/s/j
,
5/i
cy
V cy
* 2 (fi ~ h) 2
"
-f
, (/l
-M-l
_ /2l
^_
ox
/2
cy ) _
—
/
<M
-l-Ar_A_i
/i
dy
Mx — Ny = 0,
M
+ N dy =
f(x) = (M y — N x )/N,
If
x(h-f
ex J
M/!-/2
)
ex J
ex
ex
2
v
V"
xyif, - /2 )
—c/2
/
/i
2
)
r/2
x
Sx
o'
2
y cf(xy)/cy = x df(xy)/dx.
M/N = y/x
and the differential equation reduces to
xdy + ydx = 0,
with
xy = C.
solution
4.93
then
o'
IW/, - fa2 )j
Sx
0xb(/i-/
)\
This last is identically zero because
If
V
2
*(/i - f2 )\
u
cy
2
2)
<?/i
and
cy
~
\ex
3/A
+
^+ —
*
a/2
.
,
^_^ ,%-£«!
r(/i-/2
/2 -y
a/i
.
has an integrating factor
rfx
I
By hypothesis,
M y = dM/cy,
f.i
which depends only on x. show that
Nx = 3JV 3x.
and
is
exact. Then
= e1 **** 4*,
where
Write the condition that /< depends only on y.
—dy— = — —
d(jiN)
r(/iAf)
/iM </x + /.iN dy =
//
-
Since /i depends only on x, this last
.
(
equation can be written
CM _
dN
d\i
dfi
dy
dx
dx
dx
M - N dx —
—=—
N
dn
T
Thus
.
y
x J
_
(dM
cN\
\ cy
dx J
f{x) dx
H
and integration yields
In n
= \f(x)dx,
so that
//
= e ifix)dx
.
If
we interchange M and N, x and y, then we
(N x —
)/M = giy) and that in this
y
M
see that there will be an integrating factor n depending only on y if
case
4.94
/<
- e i9Wdy
.
Develop a table of integrating factors.
f From the results of Problems 4.83 through 4.91, we obtain the first two columns of Table 4.1. The last
column follows from Problems 4.71 through 4.81, where in each case we have suppressed the c term in
t
gix, y) for simplicity.
SOLUTION WITH INTEGRATING FACTORS
4.95
Solve
(y
2
- y) dx + x dy = 0.
f No integrating factor is immediately apparent.
differential equation can
Note, however, that if terms are strategically regrouped, the
be rewritten as
~iydx-xdy) + y 2 dx =
(V)
The group of terms in parentheses has many integrating factors (see Table 4.1). Trying each integrating factor
2
Using this
I(x, y) = l/'y
separately, we find that the only one that makes the entire equation exact is
integrating factor, we can rewrite (/) as
.
_ydx-xdy
+ldx = Q
{2)
y
Since (2) is exact, it can be solved by the method of Problem 4.43. Alternatively, we note from Table 4.1
that (2) can be rewritten as
x
y
— x +c
or
y =
x
x + c
-dix/y) + 1 dx = 0,
or as
dix/y) =
1
dx.
Integrating, we obtain the solution
^
EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS
Q
81
TABLE 4.1
Group of Terms
Integrating Factor I(x, y)
Exact Differential
xdy-ydx
Jy\
~x~2
x2
\x)
1
ydx — xdy
j{*\
7
y
1
ydx — x dy
ydx — xdy
2
i
Xdy - ydX
xy
xy
ydx — xdy
2
x
+y
2
x
2
1
ydy + xdx
x 12 + y*2
2
l
ay dx + bx dy
+ yY
'
(x
x a-y-i
2
x a-
(a, b constants)
4.96
Solve
(y
,^_.
ir1
,
2
= d[±\n(x
+ y 2 )]
/
,
-1
ydy + xdx
=
—
—
_2(n+
+
i
(x
2
l
l(n-\)(xyT- \
(xyf
+y
= d(\n \xy\)
ydx + xdy
n >
1
ydy + xdx
+y
xy
1
x
:
2
xy
(xyf
ydy + xdx
y\
— = d (\ arctan —xj
5
2
ydx + xdy
,
X )
\
1
ydx + x dy
ydx + x dy
= d(\n y_
xdy-ydx
1
ydx — xdy
\y)
d
^
2
l)(x
y )"
2
y
2 )"- 1
_
y l^ dx + bx dy _ d x y
-
a
j
b)
(
- xy 2 dx + (x + x 2 y 2 dy = 0.
)
)
I No integrating factor is immediately apparent. Note, however, that the differential equation can be rewritten
as
(ydx + xdy) + (-xy 2 dx + x 2 y 2 dy) =
(1)
The first group of terms has many integrating factors (see Table 4.1). One of these factors, namely
= l/(x_y) 2 is an integrating factor for the entire equation. Multiplying (J) by \/{xy) 2 yields
ydx+xdy — xy 2 dx + x 2 y 2 dy
=
or, equivalently,
+
/(x, y)
,
•
.
(xy)
(xyY
,
2
ydx + xdy
--22
(xy)
From Table 4.1,
ydx + xdy
—
(xyY
so that (2) can be rewritten as
I
d
xy
xy
Solve
(2)
x
(— 1
= d\
both sides of this last equation, we find
4.97
\
= -dx-\dy
= In Ixl — y + c,
I
1
\xy )
= - dx — \dy.
Integrating
x
which is the solution in implicit form.
/ = (y + l)/x.
Rewriting the equation in differential form, we obtain
(y + \)dx
- xdy = 0,
or
(ydx- xdy) + Xdx =
(1)
The first group of terms in (7) has many integrating factors (see Table 4.1); one of them,
7(x, y)
ydx — xdy
an integrating factor for the entire equation. Multiplying (7) by 7(x, y) yields
which we write as
d I - j + d I - j = 0.
Integrating this last equation, we get
as the solution to the differential equation.
—
-=
-+-=c
+
(
or
= - 1/x 2
,
dx = 0.
y = ex — 1
is
CHAPTER 4
82
4.98
Solve
I
y'
= y/(x - 1).
y dx — (x — \)dy — 0,
Rewriting the equation in differential form, we obtain
or
(ydx-xdy)+ \dy =
(7)
The first group of terms in (7) has many integrating factors (see Table 4.1); one of them,
an integrating factor for the entire equation. Multiplying (7) by I{x, v) yields
y — k(x — 1)
4.99
Solve
I
y'
— =
d\ — ) + d\
I
\y)
V
y)
where
k —
1/c.
= r,
or
y
y
= (x 2 + y + y 2 )/x.
We rewrite the equation as
(x
2
+ y + y 2 dx — xdy = 0,
or
)
(
- x dy) + (x 2 + y 2 dx =
v dx
(7)
)
The first group of terms in (7) has many integrating factors (see Table 4.1), one of which,
7(x, y)
— — l/(x 2 + y 2
— y dx + x dy
is
which
y
Integrating this last equation, we obtain, as the solution,
0.
is
,
—= dy = 0,
1-
5
y
we write as
= 1/y 2
I(x, y)
also an integrating factor for the entire equation.
— d(x) = 0.
di arctan —
which we write as
Multiplying (J) by 7(x, y) yields
1
z
x 1 + yl
),
dx = 0,
Integrating this last equation, we obtain as the solution
J
arctan (y/x) — x — c
4.100
Solve
I
y = x tan (x + c).
or, explicitly,
/= -y(l+x 3y3)/x.
We rewrite the equation as
>(1
+ x i y i )dx + xdy — 0,
{ydx + xdy) +
or
xV dx =
(7)
The first group of terms in (7) has many integrating factors (see Table 4.1). one of which.
is
also an integrating factor for the entire equation.
Multiplying (7) by 7(x, y) yields
7(x, >•)
= l/(xv) 3
y dx + x dy
-^
1-
1
,
dx = 0,
Ixy)
which we write as
d
d(x) = 0.
\
Integrating this last equation, we obtain as the solution
2{xyf
+x =c
± l/xV2(x — c).
y =
or, explicitly,
2(xyY
4.101
Solve
+ x 4 dx - x dy = 0.
(>-
)
ydx — xdy + x* dx = 0.
I We rewrite the equation as
integrating factors (see Table 4.1), but only 1/x
ydX ~ Xdy
x2
x
v
Then integration gives
4.102
Solve
(x
3
1
2
+ x>dx =
or, explicitly,
y = 3X
4
V 3
— ex.
+ xy 2 - y) dx + x dy = 0.
We rewrite the equation as x(x 2 + v 2 dx + x dy — ydx =
2
2
I{x, y) = l/(x + y
to obtain
f
)
and multiply by the integrating factor
)
x*c +
Then integration gives
4.103
Solve
suggests several
-d(y-) + dl~\ =
or
V.x /
=c
ydx — xdy
The combination
leads to favorable results, i.e., to
y + tan
^£
" l
3 2
x dy + y dx - 3x y dy = 0.
-=c
=
or
,g) + ,(,a„-i| =
2
or, explicitly,
y = x tan (c - ^x ).
—
EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS
I The terms
xdy + ydx
suggest
I(x, y)
= l/(xy)
fc
y
by the integrating factor l/(xy) 3 the equation becomes
,
-^--31ny=C
2x y
4.104
Solve
61ny = lnC--^
z z
Then
= l/(x 2 + y 2
—
which is exact.
2
2
Solve
y
6
whose primitive is
y
= Ce" 1/(x2y2)
)
I(x, y)
"T"
or
x y
f The last term suggests
A
- - dy = 0,
f
,
xJyJ
Upon multiplication
xdx + y dy + 4y 3 (x 2 + y 2 (/y = 0.
x dx + y dy
4.105
1.
k = 3.
and the last term requires
,
83
•"
^ dy =
0,
Its
as an integrating factor, and multiplication by I(x, y) yields
)
primitive is
\ In (x
2
+ y 2 + y 4 = In C
)
x
or
(x
2
+ y 2 )e 2y = C.
"
_y
x dy -ydx -(I- x 2 ) dx = 0.
xdy — ydx
# Here 1/x 2 is the integrating factor, since all other possibilities suggested by
render the last term
2
inexact. Multiplication by 1/x yields
£«£Z*-(£-l)<x_0
—x
y
Integration yields the solution
1
I
\-
x — C
y + x
or
^-ngJ + ^-O
or
2
+
l
= Cx.
x
4.106
Solve
(x + x
4
+ 2x 2 y 2 + y 4 dx + y dy =
x dx + y dy + (x 2 + y 2 ) 2 dx = 0.
or
)
An integrating factor suggested by the form of the equation is
I{x, y)
= —zz
(x
x dx + v dy
—
+ dx =
-z
(x
4.107
2
z-zr
+ y2 2
y'
1
—-
whose primitive
is
v
z
2
2(x
)
Solve
/
....
0,
z-
+ y2
+ x = C,
ydx
+ xdyy
y
,
h - x
Integrating yields
xy
I
or
3
=c
or, explicitly,
y = (^x
d\
4
— cx)~
)
l
.
= -y/(y 3 + x 2 y - x).
(ydx — xdy) + y(x 2 + y 2 )dy = 0.
2E*£j£-,*-0
Integrating yields the solution
f
.
3
This equation has the differential form
Solve
An integrating factor for the first
2
=
l/(xy)
I(x, y)
Using it, we get
/-l\
(\ \
+ d -x 3 =0
\xyj
\3
J
,
+ x 2 dx =
^2
group of terms that also renders the entire equation exact is
4.109
,
(ydx + x dy) + x 4 y 2 dx = 0.
(xy)
y'
,
(C + 2x)(x^ + y ) = 1.
= -y(l + x 4 y)/x.
This equation has the differential form
Solve
Using it, we get
)
group of terms that also renders the entire equation exact is
4.108
or
zr=.
+ y zY
arctan (y/x) — \y
An integrating factor for the first
2
2
—
=
I(x, y)
l/(x + y
Using it, we get
).
i^iy^.O
or
= c.
2
xy 2 dx + (x 2 y 2 4- x 2 y) dy = 0.
Rearranging gives us
(xy){y dx + x dy) + (xy)
2
dy = 0,
(ydx + x dy) + xydy = 0.
from which we find
integrating factor for the first group of terms that also renders the entire equation exact is
Using it, we get
xd + y dx
>'
+idy =
or
d(ln Ixyl) + d(y)
xy
Integrating yields, as the solution in implicit form,
4.1 10
Solve
(x
3
y
2
-y)dx + (x 2 y 4 -x)dy = 0.
In \xy\
+ y = c.
=
/(x, y)
= 1/xy.
An
y
CHAPTER 4
84
+ y dx) — x 3 y 2 dx — x 2 y 4 dy = 0.
2
that also renders the entire equation exact is
I(x, y) = l/(xy)
f
Rearranging yields
An integrating factor for the first group of terms
(x dy
(xy)
«W<r-o
2
Using it, we get
.
d (=±)- d
or
\xyj
xy
4.111
where
k = 6c.
Solve
y'
I
is
=
\2
J
y 3 — c,
x2
Integrating, we obtain as the solution in implicit form
(lA- d (lf) =
2
\3
3x 3 y + 2xy 4 + kxy = — 6
or
3
3yx 2
x 3 + 2/
'
(3yx )dx + ( — x
— 2y 4 )dy — 0. No integrating factor
—
x (3ydx
xdy) — 2\ A dy = 0. The group in
2
Rewriting the equation in differential form, we have
3
2
immediately apparent, but we can rearrange this equation as
ay dx + bxdy, where a — 3 and b = — 1, which has an integrating factor
x 2 y~ 2 Since the expression in parentheses is already multiplied by x 2 we try /(x, y) = y~ 2 Multiplying by
y~ 2 yields
parentheses is of the form
.
,
.
x 2 y " 2 (3y dx - x dy) - 2y 2 dy =
3
d(x y~
which can be simplified (see Table 4.1) to
— 2y 2 dy.
l
)
x 3 y~
Integration then yields
'
= |y3 + c
as
the solution in implicit form.
4112
Solve
I
y'= -%(y/x).
Rewriting the equation in differential form gives us
with
a — 1
3
2
y dx + 3xy =
4.113
Solve
b — 3.
and
An integrating factor is
d(xy ) — 0.
3
or
y'= -(2y 4 + l)/4xy 3
y dx + 3x dy — 0, which is of the form ay dx + bx dy =
= x 1_1 y 3_1 = y 2 Multiplying by /(x, y), we get
/(x, y)
.
xy 3 = c
Integrating then yields
or, explicitly.
y = (c/x)
113
.
.
4xy 3 dy + (2y 4 + 1) dx — 0, which we rearrange to
y (2ydx + 4xdy) + 1 dx — 0. The terms in parentheses here are of the form ay dx + bx dv with a — 2 and
2
4~
= xy 3 Since the expression in parentheses is
6 = 4, which suggests the integrating factor /(v. y) = x
f
Rewriting the equation in differential form gives
3
'
'
already multiplied by r\ we try
(
2vv
4
/(v. y)
= x.
dx + 4x 2 y 3 dy) + x dx =
Integrating, we obtain the solution in implicit form as
4.114
Solve
I
y'
.
Multiplying by x yields
2
or
d(x y
x 2 y 4 + ^x 2 = c
4
)
+ d{\x 2 =
)
or, explicitly, as
y — ±{cx'
2
— f) 4
1
= 2xy — x.
Rewriting this equation in differential form, we have
(
— 2xy + x)dx + dy — 0. No integrating factor is
M(x, y) = — 2xy + x and N(x. y)=l. so that
immediately apparent. Note, however, that for this equation
1
fdM
N \8y
is
a function of x alone.
dN\ _ (-2x) - (0) _
~ ~dx) ~
1
Then from Problem 4.93, we have
/(x)
= e JI
~ 2jc)dx
= e~*
2
as an integrating factor.
2
Multiplying by e~* yields
(-2xye' x2 + xe~ x2 )dx + e~ x2 dy =
which is exact. To solve this equation, we compute
g(x, y)
from which we write
Then g(x, y) = ye~
xl
y - ke + |.
4.1 15
Solve
xl
cg/cy = e'
=
x2
— \e~ x2 + c v
(
j
— 2xye~ xl + xe~ x2 )dx = ye~ x2 — \e~ xl + h(y)
+ h\y) = N(x, y) = e~ xl
This yields
.
The solution to the original equation
2
y dx + xy dy = 0.
f Here
M(x, y) = y 2
and
N(x, y) = xy;
1
hence,
fcN
M\dx
cM\_y-2y_
dy J
y
2
1
y
h\y) = 0,
from which
x2
— \e~ x2 = k
is
ye~
h(y)
or
= cv
-
EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS
is
From Problem 4.93, then,
a function of y alone.
= e~ {ll,,)dy = e lny = 1/y. Multiplying the given
ydx + xdy = 0, which has the solution y = c/x.
/(>>)
by /(y) yields the exact equation
differential equation
85
An alternative method would be first to divide the given differential equation by xy 2 and then to note that the
resulting equation is separable.
4.116
Solve
y'
= l/x 2 (l -3)0(3x y — x
f We rewrite this equation in the differential form
dM/dy - dN/dx
3x 2 —
N(x, y) =
and
,
= 3x ,
=
Then
1.
2
dx + 1 dy = 0,
2
)
M(x, y) = 3x 2 y - x 2
„
.
a function only of x, so it follows from
is
ix2dx
with
x
— e \ Multiplying by /(x), we obtain
Problem 4.93 that an integrating factor is I(x) — e !
xi
xi
xi
2
2
which is exact. Its solution (see Problem 4.53) is y = ke~ + 5.
e (3x y — x dx + e dy = 0,
)
4.117
y = l/(2xy).
Solve
I
In differential form this equation is
dN/dx -dM/dy
M
-2y-0
=
= — 2y
is
= e ~ 2ydy = e~ y2
Solve
I
and
N(x, y) =
— 2xy. Then
a function only of y, so it follows from Problem 4.93 that an integrating
Multiplying by it, we obtain
!
I(y)
.
xe~
Integrating this last equation yields
4.118
M(x, y) = 1
with
1
factor is
k = 1/c,
dx — Ixydy = 0,
1
y =
or as
-3yy
+ y3
= c,
x = ce
or
e~
y2
dx — 2xye~
y2
dy =
y2
or
This may be rewritten as
.
y
d(xe~
y2
)
= In |fcx|,
2
= 0.
where
±J\n \kx\.
'
/= x(2m
.
)
3y dx + x(2 + y ) dy — 0,
3
In differential form, this equation is
dN/dx - dM/dy
Then
y2
M
= (2 + y
- 3
3
)
1
1
=-f
3
3y
2
.
M(x, y) = 3y
with
N(x, y) — x(2 + y ).
3
and
,..
a function only of y, so it follows from Problem 4.93 that
'
.
is
3y
an integrating factor is
— e Hy l$-Uiy)dy _ ey3/9-(l/3)ln|>.| _ g(- l/3)ln|y|gy 3 /9 _ ^ln
2
Jty\
3y 2/
Multiplying by it, we obtain
V dx + (2xy~
3/9
Integrating yields the solution in implicit form as
x9y6e
power,
4.119
Solve
(x
y3
—k
k = (c/3)
where
= 2y
and
I
e
V
3
/9
_
-1/3^/9
V
dy = 0, or d(3xy 2/
) = 0.
or after both sides are raised to the ninth
3/9
,3/9
)
= c,
.
= - = /(x),
=
)
the equation is not exact.
However,
dx
xy
dx J
\ dy
3xy
+ xy 8/
2l3 yi/9
ey
9
^— — y,
dy
iV
e
l
+ y 2 + x) dx + xy dy = 0.
2
Because
—
ll3 yi/9
"3
|y
and
e
sf{x)dx
= e )dx,x = e lnx = x
is
an integrating factor. Multiplying
x
by the integrating factor, we obtain
4.120
4
or
)
xy 2 dx + x 2 ydy = d(^x y ),
2
Then, noting that
3x
+ xy 2 + x 2 dx + x 2 y dy =
3
(x
2
(2xy
V + 2xy +
3
= 8xyV
Here
y) dx
-+-
+ (x 2 y
V-x y 2
—
x
or
2xyV + 6xy 2 +
1
3x) dy - 0.
2
and
dM
However,
,
2
, ,
2xe y + 2 - + -t
Now
1
.
dx
- e -*f<iyiy = e -4in>- _ 1/^,4
y
2
(dM
8N „
—
= 8xvV + 8xy + 4 and — —
M\dy
-
dy
V
—- = 2xyV - 2xy -
)
dx + ( x 2 ey -
y J
g(x, y) =
j
(
so the equation is not exact.
dN\
—r =
dxj
— Then
4
.
y
an integrating factor; upon multiplication by /(y) the equation takes the form
dy =
^ - -^
r)
3
)
0,
which is exact.
y
\
f
s
3,
dx
dy
[
1
x4
x3
- +
+ - x2y2 = C
we integrate to obtain
+ 4x 3 + 6x 2 y 2 = c.
Solve
I( y )
x 3 dx + x 2 dx + (xy 2 dx + x 2 y dy) -
X
1
2xe^ + 2 - + -3
\
1
2
dx = x 2 ey +
X
X
—
+ -3 +
h(y),
from which we may write
—
.
CHAPTER 4
86
2
x
do
= xV
v
oy
y
is
xV + — + 4 = c
r
T
-3-r +
x
= xV
/i'(j')
x
— 3—t.
=-
y
y
This yields
h'(y)
= 0,
so that
=c
h(v)
and the primitive
y
-
>'
4.121
3
Solve
+ 4x 2 y + 2xy 2 + xy* + 2y) dx + 2(y 3 + x 2 y + x) dy = 0.
2
(2x y
rM
—
— = 4x y + 4x + 4xv + 4xy + 2
f
3
Here
2
f V
3
^— = 2(2xv +
and
dN\
—
— fcM
— =
1
N \cy
and an integrating factor is
2x.
so the equation is not exact. However,
1):
ex
oy
= e' zxax — ex
I(x)
When l(x) is introduced, the given
.
ex J
3
equation becomes
(2x y
4
x~
+ 4x 2y + 2xy + xy + 2y)e dx + 2(y + x 2y + x)e x dy — 0.
2
2
3
~
which is exact.
Now
g (x, y) = J(2x y
3
xl
+ 4x 2 y + 2xy 2 + xy 4 + 2y)e dx = J*(2xy 2 + 2x 3 y
2
V
2
dx +
Uly + 4x 2 y)e* dx + j xy A ex2 dx
2
= x 2 y 2 exl + 2xyex2 + ^yV + h(y)
2
2
— c,
h(y)
4.122
Solve
2
and the primitive is
Thus
~.
h\y) =
and
'
\
—
=-
v'
(2x y
2
xl
+ 2xex2 + 2y i e x2 + h'(y) = 2(y 3 + x 2 y + x)e x
+ 4xy + A )e x — C.
cg/cy = 2x ye
from which we may write
x
I
Rewriting this equation in differential form yields
_ -1
1
choose
y)
/(.v,
=
r
, t
x[y(l
—x*y— dx
4.123
(xy)
which is exact.
Its
solution is
= —
y
(x In \Cx\)
1
Problem 4.54).
(see
+ 2) dx + v(2 - 2x 2 y 2 )dy = 0.
2
y\x y
yf.(xy)dx + xf2(xy)dy — 0.
The given equation is of the form
factor (see
Multiplying by /(x. y). we obtain
as an integrating factor.
-tj
2
Based on Problem 4.92, we
xy
2
Solve
= 0.
= dy
5
-^
- xy)] - yx
— xy)dx + x(\)dy = 0.
y(l
Problem 4.92).
When
is
it
so
Mx — Ny
=
—
r-y
is
an integrating
3x y
x 2 )- 2 + 2
2 — 2x2y2
=-=
r-=— dx H
dy — 0,
sx y
3X y
—
introduced, the equation becomes
which is exact.
Now
P
**> V) = J
xV + 2,dx
"W"
=
—
dy
5x
.
,
,
+ h'(y)
'
n
3
+ *W
* " 3xV
2
"-
yields
.
h'(y)
— — 2/3y,
and so
3x y
1
.
The primitive is then
=—=
3x~y-
In x
)
—
lnv = lnC,.
x — Cv 2 e
and
l
x ' y '.
3
)
I
integrating factor (see Problem 4.92).
2
1
x 3y2
+ —4r^-3
Now
\
dx +
/
2
1
I
rfv
x y
/
o(x.
y)= f ( -^-^ + —T-.^ dx =
J \x y
x y J
-5-j-
y
= 0,
primitive is then
-In v -
-^
x y
1
Ny
\v
which is exact.
=-y - ,
x y
+ /i'(y) = -5-r
+ -T-T - J
x y
Mx
y
)
-^
+
x y
x
so
When it is introduced, the equation becomes
1
-^—r + —2rr
\x 3 y 4
x y3
— — —— = —r1
yfAxy)dx + xf2 (xy)dy = 0.
The given equation is of the form
Cy
2
,
i
1
1
,
y(2xy + 1 dx + x( 1 + 2xy - x 3 y 3 dy = 0.
Solve
^- =
+
_
3
4.124
\
=
(^ 3xV J *
— 2x y
=—
This
y
1
h(y)=— flny.
2
1
2
2
3fl
from which we may write
r /
= J
x y
+
,
from which we may write
/j( y).
This yields
/j'(
v
)=-lv.
so that
y
^=C
3x y
,
3x y
= Ce- (3x, +1,(3jc3,3)
'
lt
and
v
.
/i(v)=-lnv.
The
!
EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS
4.125
D
87
(2xyV + 2xy 3 + y)dx + (x 2 yV - x 2 y 2 - 3x)dy = 0.
Obtain an integrating factor by inspection for
f When the given equation is written in the form
y*(2xe y dx + x
the leftmost term is the product of y
dy) + 2xy
3
dx - x 2 y 2 dy + y dx - 3x dy =
and an exact differential. This suggests 1/y 4 as a possible integrating
To show that it is an integrating factor, we may verify that it produces an exact equation.
factor.
4.126
4
V
Obtain an integrating factor by inspection for
(x
2
+ 2y)dx + (2x — 2x 3 y 2 )dy = 0.
3
y
dx — 2x i y 2 dy = 0, the term
(ydx + xdy) suggests l/(xy) as a possible integrating factor. An examination of the remaining terms shows
that each will be an exact differential if k = 3.
Thus, l/(xy) 3 is an integrating factor.
# When the given equation is written in the form
2
2(y dx + xdy) + x y
3
fc
4.127
Obtain an integrating factor by inspection for
(2xy
f When the given equation is written in the form
+ y)dx + (x + 2x 2 y — x*y 3 )dy — 0.
2
(x dy
+ y dx) + 2xy(x dy + y dx) — x 4 y 3 dy = 0,
k
two terms suggest l/(xy) as an integrating factor. The third term will be an exact differential if
l/(xy)
4.128
4
is
I
/
+ xy +
dx
y/l
- x 2 y 2 = 0.
x(x dy + y dx) + >/l — x y
2
In differential form, this equation is
1
I
reduces it to
dx
Vl - x y
x
2
dy
y — x>'
dx
x + x y + y
Solve
x dy + y dx
—H
.
xVl - xV
4.129
thus,
an integrating factor.
x2
Solve
the first
k = 4;
2
2
2
dx — 0.
....
whose primitive
= 0,
The integrating factor
.
„
arcsm xy + In x = C.
is
,
— x3
2
3
In differential form, this equation is
(x
3
+ xy 2 — y)dx + (y 3 + x 2 y + x)dy = 0.
When it is rewritten as
xdy — y dx suggest several possible integrating factors.
(x +
)(x dx + y dy) + x dy — ydx = 0,
the terms
2
2
By trial and error we determine that l/(x + y ) reduces the given equation to
2
2
>'
2
xdy -ydx
( xdy - ydx ) /'x
= 0. Its primitive is
5
5— = x dx + y dy H
,
2
x2 + y2
+ (y/x)
2
2
2
2
or x + y + 2 arctan (y/x) = C.
jx + jy + arctan (y/x) = C,
j
a
xdx
+ y dy
H
.
.
1
4.130
Solve
x(4y dx + 2x dy) + y (3y dx + 5x dy) = 0.
3
I Suppose that the effect of multiplying the given equation by x x yp is to produce an equation
(4x*
+i p+l
y
dx + 2x*
+2
y" dy) + (3xV
+4
dx + 5x 3
+
/ + dy) =
3
»
( /)
each of whose two parenthesized terms is an exact differential. Then the first term of (/) is proportional to
d{x
a + 2
/+
= a + 2 x + / +l dx + (p+l )x a + 2 y" dy
*
'
)
'
?L±l = £±i
That is,
(2)
)
(
or
a - 2/? =
(5)
Also, the second term of (7) is proportional to
d{x'
+ l
y
e+4
)
= (a + )*V + 4 dx + (P + 4)x a + / + 3 dy
^±i = ^ti
That is,
a = 2
Solving (J) and (5) simultaneously, we find
equation becomes
4.131
Solve
or
5a - 3/? = 7
(5)
= 1. When these substitutions are made in (/), the
and
3
2 5
(4x y dx + 2x*ydy) + (3x y dx + 5x y*dy) = 0.
3
(4)
1
2
primitive is
Its
x 4 y 2 + x 3 y 5 = C.
(8y rfx + 8x dy) + x y (4y dx + 5x dy) = 0.
2
3
f Suppose that the effect of multiplying the given equation by x"y p is to produce an equation
a
(8x y*
+ '
dx + 8x"
+
y
dy) + (4x*
+
V dx +
+4
5x a
+
V
+3
dy) =
(7)
y
"
CHAPTER 4
88
each of whose two parenthesized parts is an exact differential. The first part is proportional to
d(jc«
+1
/ +1 = (a + l)xV +1 dx + (0
4- l)x»
)
^—
y
-^— =
That is.
8
or
-
y.
+
V
dy
(2)
=
(3)
8
The second part of (/) is proportional to
d(x*
+
V
)
+
+
——
=
4
4- (0 4- 4)x*
or
5a - 4/5
P = 1
4
"ft
3
a
That is,
V dx
+4
= (a 4- 3)x" +
T4
+
V
+3
dy
(4)
(5)
5
Solving (3) and (5) simultaneously, we find
—
y
= 1.
and
1
When these substitutions are made in (7),
Its primitive is
4x 2 y 2 4- x 4 y 5 = C.
4 4
(8xy dx 4- 8x y dy) 4- (4x y dx 4- 5x y dy) = 0.
2
the equation becomes
3
2
5
[Note: In this and the previous problem it was not necessary to write statements (2) and {4) since, after a little
practice, (3) and (5) may be obtained directly from (/).]
4.132
x V(2y dx + x dy) - 5y dx + 7x dy) = 0.
Solve
(
7
f
Multiplying the given equation by x y" yields
(2x
V dx +
+4
a +
+
+1
dx + 7x" +
x' *yt+3 dy) - (5x«/
V
—+—4 = —+—4 and
a
dy) =
x
If the first
- =
of (/) is to be exact, then
we find
-
parenthesized term of (7) is to be exact, then
= —83
y.
= —10/3.
and
Cv
—— and
V dx +
3
1
x
— 50 = —2.
1-x
— 20 = 4.
If the
second part
Solving these two equations simultaneously,
Then (7) becomes
4
dy) - (5x
3
\v
'
4
H
Each of its two terms is exact, and its primitive is iv "V
4
2
v
+ 2x 5/3j " 3 = C or x 3 y 3 + 2 = Cx 5 3 y" 3
y
J
-
(7)
3
V"
2 3
<**
3
?
+ 3x
+ 7x
y
-3 3 y" 10 3
" 3
= Cj.
or
(t dt
dy) =
This may be rewritten as
s
.
Solve
2
— = — +- +-.
:
dy
4.133
v
t
di
I
t
y
This equation has the differential form
Multiplication by the integrating factor
(y
tdt + yi * y
2
+ \-
f
Integrating. we obtain the solution
-2t
(2c-2t)
2c -2t
2
2
t
dy
4.134
Solve
-f-
=
dr
I
=e e
+y = e
3f
= ke
1
2
4-
(r + y
2
t
+ lA=
\ In (r 4-
OT
or
d[i In (f
)
2
+ v 2 )] + d(t) -
2
y + t = c, which we may rewrite as
y = ±{ ke~ 2 - t 2 Z
)
'
as
+ y dy) + (f 2 + y 2 df = 0.
(see Table 4.1) yields
)
1
'
.
)
- 2v
-.
f
This equation has the differential form
(2y - 3f) dt + t dy = 0.
aydt + btdy
equation, the terms in parentheses have the form
integrating factor
f
2 ~
" '
1
'
=
Integrating yields the solution
2
f
or
with
(2y dt 4- f dy) - 3t dt = 0.
a = 2
and
b = 1,
In the latter
which suggests the
Multiplying by it, we get
r.
(2yf dt 4- 1
4.135
+ t) dx 4- y dy = 0.
2
y - f
3
2
dy) - 3f
= c,
=
2
or
2
or
y = (f
3
d(r y)
+ c)/t
- d(r 3 =
)
2
.
Solve
df
I
2yr
This equation has the differential form
2yf dy 4- (f - y
= 0, or y(-y dt 4- 2f dy) +
and
ay dt + bt dy with a = 2
)
dr
= 0.
In the
b = 2,
which
df
1
terms in parentheses have the form
1_1 2_1
= f~ 2 y. Since the expression in parentheses is already multiplied
r"
y
"2
Multiplying by it, we get
7(f, y) = f
latter equation, the
suggests the integrating factor
by y, we try
r
.
(_y 2 r 2 df + 2yr 1 dy) + r
1
dt
=
or
d(ry)4-d(ln|r|) =
—
EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS
1
Integrating yields
n
,
c
Solve
,
dx
=
—
2
3t (t
2
+ In
\t
\
= c,
|r|
- In \k\
where
c=-ln|/c|,
)
.
x
x dt - [3t 2 {t 2 + x 2 ) + t~\ dt = 0,
This equation has the differential form
2
+x
f
= — l/(f + x 2
2
I(t, x)
dx — x dt
t
„ 2,
dt =
,
3t
2
arctan (x/f) - f 3 = c,
Integrating yields
dx
= -In
2
t~ 'y
t In \kt\.
Multiplying the latter equation by
4.137
which we may write as
+ x2 + t
dt
I
2
y — ±yf—
and then as
,
4.136
t~ y
89
„
or
(xdt - t dx) - 3t 2 (t 2 + x 2 ) dt = 0.
we get
),
x
/
d(arctan-) - d{t 3 =
or
)
x = t tan (t 3 + c).
or
x + In t
Solve
dt
#
-tdx + (x + In t)dt = 0,
This equation has the differential form
2
I{t, x) = — 1/f
the latter equation by
,/x\
Int
„
=-dt =
2
,
;
or
__
4.138
„
,
Solve
dx
=
—
3t
2
1
1
t
t
t
Multiplying
— + -lnt + - = c,
lnf
^<ft =
2
d
\t
t
x
(xdf - fdx) + In t dt = 0.
(see Table 4.1), we get
tdx — xdt
Integrating yields
or
t
x = ct - 1 - In t.
or
+ x2
2xt
dt
+ x 2 dt - 2xt dx = 0, or x(x dt - It dx) + 3t 2 dt = 0. The
terms in parentheses in the latter equation have the form axdt + btdx with a =
and b — — 2, which
2 ~
3
~
= x~
suggests the integrating factor
x~
Since the expression in parentheses is already multiplied by x,
-4
Multiplying by
we try I{t, x) = x
we get (x 2 dt — 2x~ 3 dx) + 3t 2 x~*dt — 0, which is not exact even
i
This equation has the differential form
(3t
2
)
1
l
l
!
t
.
it,
.
t
though the first two terms can be expressed as d(x~ 2 t).
In the first differential form above, however, we have
cM/dx-dN/ct
- =
— -= —
2x-(-2x)
2
a function only of t.
,
x replaced by f) that an integrating factor is
in differential
(see
form by /(f) yields
3
+ -y
1
dt
+ x2
and
N(t, x)
= —2xt,
and
,
It
follows from Problem 4.93 (with y replaced by x and
— e _(2/,)d( = e" 21 "
s
— 2 — dx — 0,
'
1
1
= e int
which is exact.
~ 2
= t~ 2
.
Multiplying the equation
Its solution is
x — ±yj3t 2 — kt
—x
Solve
dt
I
I
I(t)
2
Problem 4.64).
dx
4.139
M(t, x) — 3t
2
t{t
x + 1)
xdt + t(t 2 x + 1) dx = 0,
This equation has the differential form
or
{xdt + t dx) + t x dx = 0.
integrating factor for the terms in parentheses in the latter equation is l/(tx)
1
H
x
{xtf
1
Integrating yields
2 2
=c
or
for any n; taking
An
n = 3
gives an
Multiplying by it, we get
integrating factor for the entire equation.
xdt — tdx
3
n
ct
-1
,
7
2
2
dx —
or
2{xt)
x 2 + t 2 x + \ = 0,
2
~\
which may be solved for x explicitly with the
x
quadratic formula.
INITIAL-VALUE PROBLEMS
4.140
Solve
dy
2xy
—
+ ^-^ =
dx
+X
-f- H
0;
y{2)
= 3.
1
I The solution to this differential equation in differential form was found in Problem 4.44 to be
2
2
Applying the initial condition, we get 3 = c 2 /[(2) + 1], from which c 2 = 15.
v = c 2 /(x + 1).
2
solution to the initial-value problem is y = 15/(x + 1).
Thus, the
:
CHAPTER 4
90
4.141
Solve the preceding problem if the initial condition is
y(0)
y = c 2 /(x
f The solution to the differential equation remains
- = c 2 /[(0) 2 + 1],
from which
1
dv
4.142
-f =
Solve
dx
c 2 = - 1.
—
x + sin v
2y — x cos y
2
+ 1). Applying the new initial condition, we get
The solution to the initial-value problem is now y = — l/(x 2
1).
-I-
= n.
y(2)
;
= — 1.
I The solution to this differential equation in differential form was found in Problem 4.45 to be
Applying the initial condition, we get c 2 = i(2) 2 + 2 sin n — n 2 = 2 - n 2
\x 2 + x sin y — y 2 = c 2
.
solution to the initial-value problem is \x 2 + x sin y — y 2 = 2 — n 2
4.143
+ 1) dy/dx + (sin x + x cos x)y = 0;
(x sin x
Solve
.
The
.
= 3.
y(n/2)
The solution to this differential equation in differential form was found in Problem 4.51 to be y = c 2 — xy sin x.
Applying the initial condition, we get 3 = c 2 — (n/2)(3) sin (n/2), from which c 2 = 3 + \n. The solution to
m
y = 3 + § n — xy sin x.
the initial-value problem is
dv
4.144
Solve
t
-f-
+y=
v(-l) = 2.
2
t
:
at
/ The solution to this differential equation in differential form was found in Problem 4.56 to be
y = \t
- k/t.
2
initial-value problem
4.145
Solve
e
3t
2 = \{- l)
Applying the initial condition, we get
y — \(t
is
-/ + 3e 3ty = It;
v(2)
2
- fe/(— 1)
or
fe
= |.
The solution to the
— 5/f).
2
=
1.
dt
I The solution to this differential equation in differential form was found in Problem 4.57 to be
2
M
= (2 2 + k)e~ M2) from which k — e 6 — 4.
Applying the initial condition, we get
y — (t + k)e
2
b
3
solution to the initial-value problem is
y = (f + e — 4)e~
1
.
,
The
'.
y - ty/t
dy
4.146
Solve
-
+-
2
==
r
+y
- yjt + y 2
2
dt
t
><4)
0;
= 3.
I The solution to this differential equation in differential form was found in Problem 4.59 to be
|r
r> - 3ty = k. Applying the initial condition, we get (4 2 + 3 2 3 2 - 3(4)(3) = k, from which
The solution to the initial-value problem is given implicitly by (f 2 + y 2 3 2 — 3ty = 89.
k = 125 — 36 = 89.
1
I
)
'
)
4.147
Solve
(3r
6
x 2 + 5r 4x4)
dx
—
+
5
6t x
3
+ 4t 3 x s = 0;
x(0)
= 0.
<//
I The solution to this differential equation in differential form was found in Problem 4.61 to be
6 3
6 3
+ 4 5 = k, or k — 0. The solution to the
+ 4 x 5 = k. Applying the initial condition, we get
t x
4 5
6 3
initial-value problem is
+ f x = 0, which may be written as f 4 x 3 (r 2 + x 2 = 0. Since is the
r x
f
f
)
independent variable which must take on all values in some interval that includes the initial time, it follows that
4.148
either
x3 =
Solve
dv
=
x—
dx
r + x = 0.
2
or
v
- v
2
;
v(
The last equation is impossible, so
x3 =
or
x(f)
=
is
the only solution.
- 1) = 2.
I The solution to this differential equation in differential form was found in Problem 4.95 to be y = x/(x + c).
Applying the initial condition, we get 2 = (— 1)/( — 1
from which c = \. The solution to the initial-value
c),
problem is v = x/(x + H
-I-
4.149
Solve the preceding problem if the initial condition is
y(0)
= 2.
1 The solution to the differential equation remains as before. Applying the new initial condition, we get
2 = 0/(0 + c),
which cannot be solved for c. This initial-value problem has no solution.
4.150
Solve
y'
= y/(x-l); y(0)=-5.
f The solution to this differential equation was found in Problem 4.98 to be
initial
condition, we get
y = 5(x-l).
— 5 = /c(0 — 1),
from which
k = 5.
y = k(x — 1).
Applying the
The solution to the initial-value problem is
EXACT FIRST-ORDER DIFFERENTIAL EQUATIONS
4.151
Solve
y'
= (x 2 + y + y 2 )/x;
91
= 2.
y(2)
# The solution to this differential equation was found in Problem 4.99 to be arctan y/x) - x = c.
Applying the initial condition, we get arctan (2/2) -2-c or c = \n-2. The solution to the initial-value
(
y — x tan (x + \n — 2).
problem is
4.152
Solve
y'
= -y(l + x 4 y)/x;
y(- 1) = -2.
The solution to this differential equation was found in Problem 4.107 to be
the initial condition, we get
y = l/(ix
4.153
4
———— + (-
-
3
= --.
1)
3
= c.
Applying
The solution to the initial-value problem is
+ fx).
3xy' + y = 0;
Solve
=
c
h - x
y(
- 1) = 2.
xy 3 = c. Applying the initial
The solution to the initial-value problem is y = ( — 8/x) 1/3
I The solution to this differential equation was found in Problem 4.112 to be
condition, we get
c
(
2
2
*i*a
4.154
= — 1)(2) = — 8.
3
*y = y +
+
—
—
*
c
Solve
i
at
<
,
y(
;
—~
2)=—
.
,
1.
y
I The solution to this differential equation was found in Problem 4.133 to be 2 + y 2 = ke~ 2t Applying
— 2) 2 + (— l) 2 = ke~ 2i ~ 2 \ from which k = 5e~ 4 The solution to the
the initial condition, we get
4
2t
2
2
= —(5e~ 2{, + 2) — 2 1/2 where the negative square root is taken
initial-value problem is
y = —(5e~ e~ — t
t
(
.
.
1 '
r
)
)
,
consistent with the initial condition.
4.155
-/= *~ y
Solve
dt
'
y(-5) = 3.
;
t
I The solution to this differential equation was found in Problem 4.134 to be
condition, we get
initial
y = (r
4.156
Solve
3
+ 200)/t 2
f
(
(
or
c
— 200.
2
t
y — t
3
= c.
Applying the
The solution to the initial-value problem is
.
dx
=x+
—
dt
— 5) (3) — — 5) = c
3
2
In f
x(l)
;
= 100.
t
The solution to this differential equation was found in Problem 4.137 to be x = ct — 1 — In t. Applying
100 = c(l) — 1 — In 1, from which c = 101. The solution to the initial-value
the initial condition, we get
x = 101 f - 1 — In t.
problem is
4.157
Solve
dx
=
—
dt
3r
2
4-
x
2
;
x(2)
= -4.
2xt
I The solution to this differential equation was found in Problem 4.138 to be
the initial condition after squaring both sides of this last equation, we get
k =
— 2.
The solution to the initial-value problem is
square root consistent with the initial condition.
x = —y/3t 2 + 2t,
x = ±y/3t 2 — kt.
— 4) = 3(2) 2 — /c(2),
2
(
Applying
from which
where we have chosen the negative
CHAPTER 5
Linear First-Order
Differential Equations
HOMOGENEOUS EQUATIONS
5.1
Show that
I(x, y)
= eSpix)dx
is
an integrating factor for
y'
+ p(x)y — 0,
where p(x) denotes an integrable
function.
I
Multiplying the differential equation by I(x, y) gives
PMdxy + p(x)e; p(x)dx =
y
eS
—
ax
which is exact. In fact, (/) is equivalent to
5.2
(ye
Sp(x)dx
)
(/)
= 0.
Find the general solution to the first-order differential equation
= q{x)
y' 4- p(x)y
if
both p(x) and q(x) are
integrable functions of x.
I We multiply the differential equation by
^SpM^y) =
—
ax
we rewrite it as
C
Finally, setting
c,
(e
S p(x)dx
— —c
es
p(x)dx
Solve
= C e PM"x q{x) dx
y) dx
S
p(x)
— —5
Integration yields
y'
f Here
and
I(x, y)
ye~ 5x = c,
or
p(x)
= 100
Integration yields
Solve
el
Qr
p(x)dx
and
/(x, y)
ye
dy/dt = y/2
100x
= c,
-ip<*)d*
=^
,
5x y'
" 5)dx
= e~ 5x
.
Integration yields
92
P(x)dx
g(x) dx
(/)
(2)
Multiplying the differential equation by /(x, y), we obtain
- 5e' Sx y =
y — ce
or
—
dx
(ye~
5x
)
=
5x
.
= e l00dx = e l00x
s
l00x y'
or
.
Multiplying the differential equation by /(x, y), we obtain
+ 100e 100jc y =
y — ce~
or
^- (ye 100x =
)
dx
100x
.
-p - \y = 0.
Then
p(t)
- -\ and
the differential equation by I{t, y), we obtain
So!ve
eS
+ e -lpi*)d* [ eSplx)dx q(x)dx
dt
.
C
for y(t).
We rewrite the equation as
56
y + Ci =
+ lOOy = 0.
e
5.5
Then, using the results of the previous problem,
and solving (1) for y, we obtain
e'
Solve
.
/ - 5 y = 0.
f Here
5.4
= e lp{x)dx
Integrating both sides with respect to x gives us
q(x).
y = ce
5.3
I(x, y)
ye"" 2 = c,
f+1C=
0.
or
dt
2
y{t)
= ce' 12
dt
.
I(t, y)
= e '^' 2)dt = e~" 2
s
.
Multiplying
LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
I Here
= 1/20 and
p(t)
Q) = ^
I(t,
e"
(I/20)<"
= e" 20
d
20
" 20
e
r+^
20
at
Qe" 20 = c,
Integrating yields
5.7
Solve
rfg/rft
I Here
p(t)
= 0.04 and
Integrating yields
Solve
93
Multiplying the differential equation by l(t, Q), we obtain
.
Q=°
(e^'
t
df
or
/2
°)
-o
.
+ 0.040 = 0.
04 '
£te
dN/dt = kN
Q) = e
I(t,
e
5.8
Q = ce~" 20
or
D
004d
^+
0Mt
= c,
s
= e 004
'
Q = ce~
or
04,
0.04e°
'
Multiplying the differential equation by I(t, Q), we obtain
'.
e=
4
or
(e
OO4,
=
04 *.
for N(t) if fc denotes a constant.
dN/dt — /ciV = 0.
f We rewrite the equation as
Then
p(r)
= —
and
/c
I(t,
N) = es ~ kdt = e~ kt
.
Multiplying
the differential equation by I(t, N), we obtain
^-ke- N =
kt<
e-
Ne' kt — c,
Integrating yields
5.9
Solve
y'
I Here
k,
N(t) — ce
or
kt
.
+ 2xy = 0.
p(x)
= 2x
and
/(x, v)
= e l2xdx = e x \
2
e* y' + 2xe
ye
Integrating yields
5.10
Solve
y'
# Here
x2
or
and
I(x, y)
— 3x
p(x) =
3x2/2
ye~
~
p(x)
= — 3x 2
=
x2
)
x2
.
= e*
~ 3xdx
= e~ 3x2 2
Multiplying the differential equation by I(x, y), we get
'
.
y — ce
or
or
-r-(ye~
3x2/2
)
=
dx
3x2 ' 2
.
I(x,y) = e 5
xi
= c,
x3
y'
~ 3x2dx
=e~ x \
Multiplying the differential equation by I(x, y), we get
— 3x 2 e~ xi —
—
dx
or
(ye~
xi
)
=
y = ce \
x
or
dy/dt + t y = 0.
3
f Here
p(t)
=
3
and
t
/(£, y)
= e 3<" = e'" /4
; '
4
e'
Integrating yields
Solve
{ye
2
= c,
and
ye~
Integrating yields
5.13
—
dx
or
sxWy - 3 xe - ** i2 = o
y
e~
Solve
y =
y'-3x 2 y = 0.
f Here
5.12
x2
Multiplying the differential equation by /(x, y), we get
- 3xy = 0.
Integrating yields
Solve
y — ce~
= c,
e
5.11
— (MT*') =
or
ye'"
14,
— c,
/y + V 4/4 y =
-f {ye'* *) =
1
or
r
y = ce~'
or
Multiplying the differential equation by I{t, y), we find
.
A/4
.
dy/dt + (t - l)y = 0.
/ Here
p(t)
=
t
-
1
and
/(r, y)
2/2
e'
= e i(t
~ l)dt
= e t2/2 ~'.
-'^ + (t- \)e ,2l2 -'y = c
at
Integrating yields
ye'
2/2_,
= c,
Multiplying the differential equation by I(t, y), we find
or
or
^-(ye'
dt
y = ce'~
,2/2
.
212 '')
=
CHAPTER 5
94
5.14
Solve
I
dy/dt + e'y = 0.
Here
= e'
p(t)
and
= e Se
I(t, y)
e
'
= ee
dt
e'
Multiplying the differential equation by I(t, y), we get
'.
4- + e'e e 'v =
— (ye
or
dt
Integration yields
5.15
Solve
e
ye
'
= c,
y = ce
or
e
=
')
dt
e '.
dx/dO - x sin 6 = 0.
f Here
= -sin
p(0)
and
e
cos6
xe
Integration yields
dx
—
+-x =
= e - sin6de = e cos9
— c,
-^ - xe
cosfl
eos0
sin 9
=
= e ]nUl =
|r|.
x = ce~ eose
or
Multiplying the differential equation by I{6, x), we get
s
I(d, x)
.
— (xe
or
cose
)
=
.
1
5.16
Solve
dt
i Here
dx
1/1
1
= 1/f
and
= 0.
When
p(f)
Ifl
h -^- x
df
'
0.
t
/(f,
x) = e
" =
> 0,
t
!il 't)dt
Ifl
f
cix
and this equation becomes
f
dx
dx
—
- + x —
t
is
appropriate for all
¥" 0.
t
t-r-
+ x = 0. When
x = 0,
may be rewritten in differential form as
It
Solve
Here
fx — c,
x = c/t.
or
t
p(t)
7
we obtain
= lit
and
dx
t~
2t\
+
I(t.
x) = c
d
-
Solve
-
I Here
,
'
2
=
2
t
Multiplying the differential equation by /(f, x),
.
7
Integration yields
.v
f
,
= c,
or
0.
p(t)
= 5/l
—+
dN
5
N), we get
and
|f
|
/(t,
5|f
-
—- + 5rN =
for all
N) = ^< 5 ">* = e s ,n
•'•
= eln|
'5
'
=
s
\t
\.
Multiplying the differential equation by
5
- N = 0.
Using the logic of Problem 5.16. we can show that this becomes
|
f
^ 0.
This last equation may be rewritten in the differential form
df
df
5
t
N - c,
p(t)=-5/t
Here
N = ct
or
?
.
and
/(;. /V)
=
t
—
f(
5
=e
51n| '
N=
or
'""
t'
5t~
4
1
dt
5.20
Af
0,
f
by /(f, N) and simplifying field
or
,
N = 0.
Solve
I
—d (rN) =
df
and integration yields
5.19
x = c/t.
f
.dN
f
= e 2inUi = e ln
dt
—+ N=
df
ii2 nd '
(r\) = 0.
or
dt
/(f,
and
-^ + -x = 0.
dt
5.18
—d (fx) = 0,
dt
integration yields
I
|f|
which reduces to the same equation. Thus
dt
5.17
M = -f
< 0,
f
dt
— t—
and the differential equation becomes
Multiplying the differential equation by /(f, x), we get
— rf 5
ln "
5|
= |r 5
— (f" N) =
5
Multiplying the differential equation
|.
Integration then gives
0.
t~
5
N = c.
dt
.
rJV-0.
Solve
df
f Here
=?
r
/>(r)=-5/f 2
and
/(f. A/)
= e S(
5 ,2)<"
= f5
'
1
.
Multiplying the differential equation by /(f, N) and
simplifying, we get
e
5 '"'
-y? 5
df
Integration yields
Ne 5 = c,
'
or
r
N = ce~
_,
'
A/
=
or
-(Ne 5
df
51 '.
'
)
=
LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
5.21
du
2xu
dx
2
Solve
I
x
95
+ 2
„,
We rewrite the equation as
,
du
2x
dx
xl + 2
u = 0,
z
Ji x
from which
2x
=
p(x)
2
x
u \ _ e M-2xHx2
_ e -ln(x2 + 2) _ g ln(x2 + 2)
+ 2)]dx
'
+2
and
_
x2 + 2
Multiplying the rewritten equation by I(x, u), we obtain
du
1
2x
x 2 + 2dx
—1
— c,
or
We rewrite the equation as
du
—
dx
Integration yields
;
x
5.22
Solve
+2
..
"
\
_
x
2
+
- «ctan x du
Integrating yields
e* dT/dx
ue
arc,an *
=c
—
d"7
= e^ 1 '''**** = e* + e_x
e
Integrating yields
f
(e*
and
_d
Qr
(
ue
" "««" *)
=
arclan *.
x+e
"
- 1
e*
H
7 = 0,
Tex+e
~ x
Now
p(x) =
1
dx
Multiplying the rewritten equation by I(x, 7), we get
.
AT
—
+ -e(1
= c,
dT
-— + (1 — e x )7 = 0.
and then as
e
x
)e
x+e
dx
Solve
= — l/(x 2 + 1)
p(x)
dx
u = ce
or
,
^=q
^ - arctan
x^ + 1
dx
5.24
from which
+ (e* - 1)T = 0.
We rewrite this equation as
/(x, 7)
m = 0,
1
1
ax
and
dx \x 2 + 2
Multiplying the rewritten equation by I(x, u), we obtain
'
,
#
or
= c(x 2 + 2).
u
= ef[_1/(*a + 1)1 *' = e arc an *.
e
Solve
u =
+ 2) 2
x2 + 1
.
5.23
2
-^
dx
/(x, u)
(x
A
"T =
-(7e* + e ~*) =
or
dt
T = ce~ (x+e
or
~ x)
.
- \)dT/dx + e x T = 0.
d7
—
We rewrite this equation as -
e*
I
dx
e
x
— 1
7 = 0.
Now
e*
p(x) —
e
x
—
and
1
7) = e J^/(^-l)dx^ e ln|e--l| = |g*_ j|_
I(x,
Multiplying the rewritten equation by I{x, 7) and simplifying, we get
dT
(e*
- 1) —- + e x T =
d
—
dx
or
dx
[T{e
x
- 1)] =
(The equation on the left is the differential equation in its original form. Thus, some work could have been
saved if the original equation had been recognized as being exact.) Integrating yields
or
5.25
x
— 1) = c,
7 = c/{e - 1).
Solve
(sin d) dT/d9
= 7 cos 0.
We rewrite this equation as
—
dU
1(0,
T(e
x
7) = e S
-< cose / sinfl >'' 9
— g-ln|sin8| _ ^ln
:
—- 7 =
Then
0.
p(9)
= - cos 0/sin
and
sin 9
|sin " • fl|
_
1
Multiplying the rewritten equation by 1(0, 7)
sin
-e
CHAPTER 5
96
and simplifying, we get
Integrating yields
5.26
Solve
p(9)
dT
cos
d6
sin
or
T = c sin 0.
=c
7/sin
7=
2
d ( T
—
(^^1 =
d6\sin0
or
= 0.
dT/dO + T sec
# Here
1
sin
= sec
and
T) = e isecede = e i"l»«c«
1(9,
+ «"»l
= sec
+ tan 9\.
Multiplying the differential
|
equation by 1(9, T) and simplifying, we get
(sec 9
+ tan 9)
—+
dT
(sec
+ sec 9 tan 0)7 =
2
— [T(sec0 + tan0)] =
or
d9
Integrating yields
5.27
Solve
dt
+ tan 0) = c
T(sec
T = c/(sec
or
+ tan 0).
dz/dt + z In t = 0.
p(0 = In *
f Here
and (via integration by parts)
I(t, z)
= ?"""" = ?" n '~'.
Multiplying the differential
equation by I(t, z), we obtain
e
Integrating yields
=
—
+ -=
dx
X* + X
Solve
I Here
in '
+
^
dt
= c,
'
or
z
ze'
in,
-'\nt =
n
(ze" '-') =
or
dt
— ce
""'.
1
2z
<fc
5.28
ze'
nn '- ,<
p(x)
I(\ Z)
0.
= 2/(x 2 + x)
and (via partial fractions)
— e 2 ^x2 + x)dx = e llZ
2/(*+ !))<**
*
f
_ g 2ln|x|-21n|x+l| _ glnx -In(x+1) 2 _ ^lnl*- u +
2
l) 2 ]
_
(X +
2
l)
Multiplying the differential equation by I(x, z) and simplifying, we obtain
dz
.x
2
(x+l) dx
zx
= c,
Integrating yields
or
(x + IV
+-
z
2x
ttt z =
d
zx
dx
(x+\)
or
(x+1) 3
=
2
= c(x + l) 2 /x 2 = c(l + 1/x)
NONHOMOGENEOUS EQUATIONS
5.29
Solve
y'
f Here
- 3y = 6.
p(x)
= — 3.
Then
p(x)dx = J -3dx = -3x,
j"
from which
/(x, y)
= e ~ 3x
.
Multiplying the
differential equation by I(x, y), we obtain
~
3x y'
— 3e ixy = 6e 3x
or
—
Av
</x
Integrating both sides of this last equation with respect to x yields
y = ce
5.30
Solve
f
3x
ye
)
ix
= 6e -3x
— J 6e
ix
dx — — 2e
3x
+ c,
— 2.
y'
Here
3x
(ye
+ 6y = 3.
p(x)
=6
and
/(x, y)
= e |6d* = e 6
e
Integration yields
ye
6x
6x y'
= \e bx + c,
*.
Multiplying the differential equation by I(x, y), we get
+ 6e 6x y = 3e 6x
or
y = % + ce
or
d
—
dx
(ye
tx
)
= 3e 6x
or
LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
5.31
Solve
dl/dt + 50/ - 5.
f Here
p(t)
= 50
and
= e 50<" = e 50
!
I(t, 1)
Multiplying the differential equation by this integrating factor,
'.
we get
50 '
e
^ + 50e 7 = 5e
50
50 '
- (Ie 50 = 5e
or
')
dt
Integrating yields
5.32
Solve
Ie
50 '
= io^ 50 + c
'
,
or
I
= i^o + ce -50t
p(t)
= 10
and
= e XOdt = e 10t
s
I(t, q)
10 '
Multiplying the differential equation by I(t, q), we get
.
lo
10r
l0t
'q = 20e
+ ^e 10
e>"><^L+I0e
4dt
Integrating yields
Solve
50t
dt
dq/dt + 10g = 20.
f Here
5.33
qe
— 2e 10 + c,
l0t
i(u
)
= 20e 10
'
dt
q = 2 + ce~
or
'
- (qe
or
10 '
dl/dt + ^l = 6.
I Here
p(t)
= 2g
and
= eSi20l3)dt = e i20/3)t
I(t, I)
Multiplying the differential equation by this integrating
.
factor, we get
e
(2o/3)t "J_
_ / _ 6e
+ e (2o/3>«
dt
Integrating yields
5.34
Solve
{20l3)t
le
<20/3)(
— (Ie
or
(20,3) ')
= 6e (20l3)t
dt
3
= -^e i20,3)t + c,
or
= -fc + ce- (20/3)t
/
.
q + \0q = \.
I Here
p(t)
=10
and
= e 10<" = e 10
s
/(f, /)
e
l0
'q
Multiplying the differential equation by I(t, I), we get
'.
+ We 10 'q = |e 10
~ (qe
or
'
= \e l0t
10t
)
dt
Integrating yields
5.35
Solve
10 '
qe
= 26e 10 + c,
'
p(t)
={
and
= e iill * )dt = e' *.
10 '.
Integrating yields
ve"* = 128e
Multiplying the differential equation by I(t, v), we get
1
I(t,v)
e
Solve
q = jq + ce~
or
dv/dt + \v = 32.
f Here
5.36
dv
1
dt
4
+ c,
or
ti*
,/4
?1 + L e
m v = 32 e"„4
.,„
.
d
= 32e" 4
-f {ve^)
A
or
dt
v
= 128 + ce r/4
dv/dt + 25r; = 9.8.
f Here
p{t)
= 25
and
/(r, y)
e
25 '
= e ;25<" = e 25
^ + 25e
25
't;
Multiplying the differential equation by I(t, v), we get
'.
= 9.8e 25
or
'
^- (ve
dt
Integrating yields
_ __
5.37
_
,
97
I Here
'
k
1
dt
25
—m = ±g
dv
Solve
ve
p(t)
i?
— k/m
= 0.392<? 2S + c,
'
.
and
ve"'""
)
= 9.8e 25
'
or
u
= 0.392 + ce~ 25t
.
,
for k, m, and # constant.
/(r, u)
dt
Integrating yields
25t
dt
= eSWm)dt — e kt,m
Multiplying the differential equation by I(t, v), we get
m
k,/m
= ±^e
+ c,
h
k
.
dt
or
v = ce~
kttm
± ^.
Ir
CHAPTER 5
98
5.38
t + kT = \00k
Solve
I
Here
=k
p(t)
for k constant.
and
T) = e l kdt = e*
I{t,
TV" = J 100/ce*' dt = lOOe*' + c,
Integrating yields
t + kT = ak
Solve
f Here
p(t)
Multiplying the differential equation by I(t, T), we get
.
+ ke k 'T = 100/ce*'
7e*'
5.39
r
= k,
7 = 100 + ce~ kt
so
T) = e 1 *"" = e*'.
/(f,
Multiplying the differential equation by I(t, T), we get
7V" + ke kl T = ak^'
5.40
dv/dt =
Solve
f
Here
TV" = J a/ce*" dr = ae kt + c;
Solve
= 0,
dv/dt = g
I Here
p(t)
so
Solve
I
y'
Here,
=
and
— 1,
l(t, v)
p(x)
= — 2x
as in the previous problem, so we may integrate the differential equation
Doing so, we obtain
and
2
I
ye~"
I
— gt + c.
= e ip{x)dx = e~ x \
I(x, y)
x2
Multiplying the differential equation by I(x, y), we obtain
x2
y = xe
—
dx
or
y = ce
or
x2
(ye~
x2
= xe~ x2
)
— \.
dy/dx + 2xy = Ax.
Here
Solve
v
= J xe~ x2 dx = — \e~ x2 + c,
p(x)
— 2.x
and
and integration then yield
5.44
which indicates that the differential equation can be integrated
— 2xy = x.
Integrating yields
Solve
'.
for g constant.
e~ x \' — 2xe~
5.43
= aJke*'
<*(7V")
T = a + ce~ k
therefore,
— e 0dt — e° — 1,
v = — \t + c.
I(t, v)
directly with respect to time.
5.42
or
-|.
Doing so, we obtain
directly.
5.41
p(f)
.
for a and fc constant.
so
Integrating yields
d{T^) = lOOfce*'
or
y'
Here
J
p(x)dx = J 2xdx = x
ye
x2
= J 4xe
x2
dx — 2e
2
so
,
x2
+ c,
= e xl is an integrating factor. Multiplication
x2
y — 2 + ce~
/(x, y)
or
.
+ y = sin v.
p(x) —
1;
hence
I(x, y)
=e
x
e y
1
'
ix
= ex
.
Multiplying the differential equation by /(x, y), we obtain
+ e x y = ex sin x
d
—
dx
or
x
x
(ye ) = e sin x
To integrate the right side, we use integration by parts twice, and the result of integration is
ye
5.45
= ^e*(sin x — cos x) + c,
x
Solve
y'
/ Here
+ (4/x)y = x 4
p(x)
= 4/x;
y = ce~
or
x
+ \ sin x — \ cos x.
.
hence
= e lp(x)dx = e lnx = x 4
'
/(x, y)
.
Multiplying the differential equation by I(x, y), we
find
x 4 y' + 4x 3 y = x 8
Integrating with respect to x yields
5.46
Solve
yx 4 = ^x 9 + c,
-— (yx 4 = x 8
or
)
dx
y = c/x
or
4
+ ^x 5
.
x dy/dx — 2y = x 3 cos 4x.
m
d
We write the equation as
1
>
dx
u-2/x)dx _ -2in* _ ^inx-2 _ -2
e
e
%
x
x
Then
Multiplying by x
-2 -
dx
Then by integrating we find
2
y = x cos4x.
2x
_3
-2
y = cos4x
x ~ 2 y = \ sin 4x + c,
or
,
p(x)
— — 2/x
and an integrating factor is
we have
—- (x ~ y) = cos 4x
2
or
dx
y = £x sin 4x + ex
2
2
.
—
LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
5.47
Solve
I
x -/ = y + x 3 + 3.x
ax
2
- 2x.
,
.
We rewrite the equation as
ax
—
J =-ln|x|
dy
1
ax
x
so
e
- ,n
y = x
+ 3x — 2.
W=
2y = x
3
d<2
5.48
Solve
-^ +
I Here
— 1/x
and
J
(x + 3 --\dx = -x 2 + 3x - 21nx + c,
+ 6x 2 — 4x In x + ex.
— — 6^ =
3
100
<ft
p(x) =
Then
an integrating factor. Then we have
is
y- = J-(x 2 + 3x - 2)dx =
or
99
2.
t
= 3/(100 - t)
p(r)
and
/(t, Q)
= ^3/(100-r)dr _ e -31n|100-t| _ ^ln
Multiplying the differential equation by I{t, Q), we get
|(100
-
|(
1
00 - f) " ^|
_3
_ I^qq _ (j-3|
"
-j- +
f)
|
at
~
100 —
— Q=
-
|(100
- t)~ 3 \2.
t
By reasoning similar to that in Problem 5.16, we can show that this reduces to
= 2(100 -0" 3
(100- ty 3 -=- + 3(100 -t)- 4
— [(100 -ty Q] = 2(100 -t) -3
3
-
for all
# 100.
t
This last equation may be written as
(100 - r)"
Integrating then yields
3
at
Q = 100 5.49
Solve
t
+ c(100 - f) 3
<W
2
df
10 + 2r
-^ +
Here
=—
p(r)
or
.
Q
* = 4.
——
f
2
2 = (100 - t)~ + c,
2
and
I(t,Q)
= e!2 ll0+2,)dt = g
'
ln
l
10+2f
= [10 + 2t\.
"
Multiplying the differential
equation by I(t, Q) and simplifying, we get
(10 + 2t)-j- + 2Q = 4(10 + It)
Integrating yields
5.50
Solve
(10 + 2t)Q = 40f + 4f
=
^ + ^—g
- t*
dt
+ c,
or
Q=
— [(10 + 2t)Q] = 40 + 8r
——+
4f
40t
2
+c
.
4.
20
I
Here
2
or
p(t)
=
2
and
I(t, Q)
= e i2n2 °-' )dt = e -^m-t\ = e lni2 °- ,r2 = (20 - t)~ 2
.
Multiplying the
differential equation by /(t, Q), we get
(20 - t)~
2
d
Q + 2(20 -t)~ Q = 4(20 -t)~
2
2
^- [(20 - t)~ Q] = 4(20 - t)~
2
or
Integrating yields
5.51
Solve
dl/dt + 201
I Here
p(t)
= 20
(20 - t)~
2
Q = 4(20 - t)~ + c,
l
or
2
dt
dt
Q = 4(20 - t) + c(20 - r)
2
.
= 6 sin It.
and
I(t, I)
= e 520dt = e 20
'.
Multiplying the differential equation by this integrating factor,
we get
d
e
20t J_
dt
+ 20e 20 '/ = 6e 20t sin 2t
or
-f- (7?
dr
20r
= 6e 20 sin 2r
'
)
—
x
x
CHAPTER 5
100
Integrating (and noting that the right side requires integration by parts twice), we obtain
Ie
5.52
20 '
= (-f^-sin 2f - j^rcos 2t)e 20 + c,
= r^-sin2r - -r§rCos2t + c<T 20
/
'.
dq/dt + q = 4 cos 2r.
Solve
I
or
'
Here
p(t)
=
and
1
= esldt = e'.
I(t, q)
dq
—
+
e'
Multiplying the differential equation by I(t, q), we get
— 4e' cos 2r
e'q
—d
or
dt
(qe')
= 4e' cos 2t
dt
Integrating both sides of this equation (with two integrations by parts required for the right side), we obtain
qe'
5.53
— f e' sin It + f e' cos 2f + c,
— + 5/ = —
Solve
dt
/ Here
q = f sin 2f + f cos2r + ce~'.
or
sin 1207cr.
3
p(t)
—5
and
e
= e 5<" = e 5
s
I(t, I)
— + 5e
5'
5,
=
I
e
dt
.
Then
Ie
5'
or
sin 1 207tf
—
110 r
e
,,
5
'
m
.
,
—
110
=
sin 1 20nt dt
J
3
—
(e
s,
I)
=
—
dt
3
=
5'
Multiplying the differential equation by this integrating factor, we get
'.
,,
e
5'
e
5'
sin 1 207rt
3
5sinl207rr- 1207rcosl207rf
—
-n^—
7
25 + 14,4007t 2
3
+c
22 sin 120nr - 24ncos 1207rr
3
5.54
q + lOOq = 10 sin 120t«.
Solve
I
+ 576rc 2
1
Here
p{t)
= 100
qe
and
I(t, q)
= e 100<" = e 100
1
Multiplying the differential equation by /(r, q), we get
'.
+ 100e ,00 'q = 10<? 100 'sin 1207rf
1001
^(qe 100 = 10e loo 'sin 1207tf
or
')
dt
Then
qe
H
100
^
lftn
i00
100
= 10 r e .on,
sin 1207tr dt = \0e
•
'
,
,
'
lOOsin 1207rf - 1207rcosl20rrf
=
'
10,000 + 14,4007r
J
,„„,
10 sin 1207if -
2
+ A
12ncos 1207rf
100 + 144tt 2
or
5.55
I Here
e
o.o4,
p(t)
^+
= 0.04
.04e°
04r
-
1
2n cos 1 207tf
+ Ae~ 100
=
100 + 144tt
dQ/dt + 0.040 = 3.2e
Solve
sin 1 207tf
1
=
q
H
and
2
'
-0.04r
/(f,
Q = 3.2,
Q) = e °
s
04<"
= e° ° 4
04 =
(Qe°
^
at
or
Qe° ° 4
Integrating yields
3.2.
')
at
Multiplying the differential equation by I(t, Q), we get
'.
'
= 3.2* + c,
or
Q = 3.2te-° 04t + ce-° *'.
5.56
dv/dx — xv=—x.
Solve
I
Here
p(x)
= —x
and
e
/(x, y)
-jc 2 /2
= e ~ xdx — e~ x2 2
5
- x- 2L
xe
'
dx
Integrating yields
A
5.57
1
ve~
x2/2
dx
I Here
•
v——xe -xx '*
2 /2
or
v
=
1
+ cexl 2
or
—
dx
"
-
(ve
1 I2\
-xxlz
)=—xe -x*/2
'
.
">
x
p(x)
3
= -2/x
,
2
= - x 3S + ex*.2
9
Z(x, i>)
= e*
dv
2
x dx
i
and
equation by I(x, v), we get
t;
'
.
~--v^-x*.
Solve
or
— e~ x2 2 + c,
Multiplying the differential equation by /(x, v), we get
'
I
—1r
{
~ 2 ' x)dx
2
3
—
/t?\
2
I —= ) = z
dx \x J
3
rf
,
-v = -x 2
x
= e~ 2inM = e ln ° ix2) = 1/x 2
or
.
Multiplying the differential
...023
-^ = +
,
_
*.
Integrating then yields
.
J
x"
9
c,
LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
5.58
Solve
e*
2
2
2/
= e i(xl2)dx = ex2 A
Multiplying the differential equation by it, we obtain
'
/(x, v)
V + \xe
— 6e* 2/4 + c,
ve* '*
Integrating yields
Solve
101
+ \xv = 3x.
v'
I The integrating factor here is
5.59
D
x2
= 3xe x2/4
'*v
or
.
— 6 + ce~ x2/4
v
—
dx
or
(ve
x2 '*)
= 3xe x2 '*
.
v= -2.
v'
x
I The integrating factor here is
-2
2
d ( v\
1
,
xz
5.60
V
-
Solve
—Jr v = —z5x
or
-7- (
v= -5x 2
x
-2
T
r—? = —
x
.
5.61
Solve
= j — 5x
3
dx = fx
— v = —ex
v'
= e i{ ~ 5/x)dx = e" 51n x = e ln|x
I(x, v)
2
+ c,
x
— ve~ x = —
"
= |x" 5
51
l
l
x~V — 5x~ 6 v = — 5x -3
or
u
= 2x + ex 2
.
—
ax
or
Multiplying the differential
|.
(yx~
5
)
= — 5x~ 3
.
Tnen
= fx 3 + or
and
f
I(x, v)
= e ~ ldx = e~ x
.
I The integrating factor here is
v'e~
-^z = - + c,
x
x
.
equation by it and simplifying, we obtain
5
Multiplying by it, we obtain
2
v
fJ
.
Integrating then yields
I
I The integrating factor here is
ux
the same as in Problem 5.57.
,
ax \x /
x
u'
— 1/x 2
I(x, v)
or
1
dv
—
+ - = cos
—
dx
(ve~
x
Multiplying the differential equation by it, we get
s
)=—l. Then
ve~
x
.
= \{—\)dx — —x + c,
and
v
= (c — x)e x
.
1
5.62
Solve
v
dt
t.
t
I The integrating factor here is
simplifying, we get
tv'
I{t, x)
+ v = t cos
= e S{ll,)dt = e "
1
—
or
t
(tv)
=
t
1
=
'1
Multiplying the differential equation by it and
|r|.
Then
cos t.
=J
tv
dt
v
1
c
t
t
cos t dt = r sin t + cos r + c,
t
and
= smt + -cosr + -.
dv
—
+— =
3
5.63
Solve
v
6f.
2t
dr
I The integrating factor here is
I(t, x)
= eSi3l2t)dt = e (3/2,ln|(| = e ta " 3/2 =
l
3/2
|f
|.
Multiplying the differential
equation by it and simplifying, we get
t
3/2
_+_
„ ,*
5.64
3
3/2
'2
5/2
sl2
7l2
— 112
= ji6t
6/
dt = ^-t'
+ c,
Then
ut
vt
r,
dv
2
—
+ -v =
,
Solve
dt
t
we obtain
Solve
(x
v
= 6f 5/2
—
or
(vt
312
)
= 6t 5!2
dt
and
v
d
i? t 2 + cT 3/2
=—
.
4.
2
t
— + 2tv =
At
2
—
or
{vt
2
)
= At 2
.
'
1
1
Then
= e ln
vt
2
'
2
=
2
t
.
Multiplying the differential equation by
= j 4f 2 dt = ff 3 + c,
and
t>
= f t + ct~ 2
dt
- 2) dy/dx = y + 2(x - 2) 3
We rewrite the equation as
= e S(2l )dt = e 21 "
'
l(t, v)
dt
5.65
l >2
t
I The integrating factor here is
it,
t
2
dt
.
-y = 2{x-2) 2
ax
x — 2
.
Then
jp(x)dx=-|
J
-=-ln|x-2|,
x — l
.
,
CHAPTER 5
102
and an integrating factor is
1
y
5.66
r
-= 2
x — 2
e
5.67
e
JCO,xd;c
- 2)dx = (x - 2) 2 + c
= e ln|sinx| = |sinx|,
— — 5e cosx + c.
sin xdx
(x
y ={x- 2)
or
J
3
+ c(x - 2)
.
x 3 dy/dx + (2- 3x 2 )y = x 3
Solve
I
cosx
f
dx = 2
x — 2
eoiX
I An integrating factor is
y sin x = 5 J\ e
1
- 2) 2
(x
Multiplication by it and integration yield
x-2
J
dy/dx + y cot x = 5e
Solve
-ln|x-2| _
y =
Therefore,
_5 e cosi + c
.
:
sin x
.
dy
—
Jx
We rewrite this equation as
and multiplication by it and integration yield
- 3x 2
— =
2
H
y
3
T1
,
r 2
L
Then we have
1.
- 3x 2
—
3
I
—_ m
J
dx =
1
,
^
3
x
-
'
1
an integrating factor is
,
Multiplication by it and integration yield
.
x 2~
x e
>
3
x e
5.68
An integrating factor is
e
y esc 2x =
or
5.69
dx
J
3
x e
1
~ 2e lx + c
1/x2
i
e
2y = x 3 4- cx 3«1/*
or
l
2
dy/dx - 2y cot 2x = 1 - 2x cot 2x - 2 esc 2x.
Solve
I
r
—
IT* 7
' l2col2xdx
j
(esc 2x
= <>-
|n
i
sin2x
l
= |csc2x|.
Then
— 2x cot 2x esc 2x — 2 esc 2 2x) dx = x esc 2x + cot 2x + c
y = x + cos 2x + c sin 2\.
y In y dx + (x - In y) dy = 0.
Solve
With x taken as the dependent variable, this equation may be put in the form
Then
\ In v
<J Jv
=
<
vln >>
In v
Solve
f
y
dv dx + 2v cos x = sin
Here
5.71
v
Solve
f
e
2 S cosxdx
= f2 n *
2x In y = In 2 y + c.
x cos x.
2sinx
is
an integrating factor. Then multiplication by it and integration yield
= {e 2
*'
nx
s\n
2
xcosxdx = ^? 2sin *sin 2 x - ie 2sinA: sinx + ^e 2sinx + c
2sini
1
1
.
dv/dx + v — 4 sin v.
x
—4
I(x. v)
x
e sin x dx
— e ndx = e x
.
Then multiplication by it and integration give
= 2e x (sin x — cos x) + c
or
v
— 2(sin x — cos x) + ce~ x
dv/dx — v — —x.
Using the integrating factor e~ x we obtain
,
dv
v
dx
x
x
=
\
—xe~ x dx — xe~ x + e~ x + c
or
1;
= x + 1 + cex
-1
2
Solve
I
2
s'
ve~
5.73
y
an integrating factor. Multiplication by it and integration yield
s
and the solution is
The integrating factor here is
Solve
m
j
„ —
= j o:„2
sin^x
j sin x + \ + ce
ve
5.72
n y
y In y
2
ve
or
j
— = — In 2 v + K,
J
5.70
= ^iniinyi _
x = -.
1
dy
= —=-.
x*
Using the integrating factor
I(x, v)
= ei(
' 2x)dx
= e~ 2l " M = e inx
' 2
= 1/x 2
we obtain
,
,-3
4-J(-x-«)*c =
1
+c
or
v
x
x
+ ex 2
and
=
LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
D
103
BERNOULLI EQUATIONS
5.74
—
+
dx
Develop a method for obtaining nontrivial solutions to the Bernoulli equation,
p(x)y = q{x)y",
n # 0, 1.
for
y =
# Observe that the trivial solution
y = y
i/(-» + i)
—=- —
and
-
v
"H-" +l
To find others, set
always a solution.
is
>
= y~ n+1
v
Then
.
Substituting these relationships into the Bernoulli equation
.
yields
^H-n+l)J. +
—n +
dx
1
yH-n + l) = xyH-n+l)
q^
p^x
- (» - l)p(x)p = -(« - l)</(x)
Qr
dx
This last equation is linear and may be solved by the method of Problem 5.2.
5.75
Solve
I
y'
+ xy = xy 2
.
This is a Bernoulli equation with
dy
y = l/v
we have
and
~t~
dx
p(x)
— de
— ^y^~2
= q(x) = x
jpr
dy
—
--y=x y
dx
x
Solve
I
1
.
4
—
312
and
=—
dx
3
2
p(x)
= — 3/x,
v
1 '
2
-—.
5.77
dx
.
n dv
dx
-—
4x
reduces it to
3
f
}>'
p(x)
\-
xn
xy' + y = xy
= — \v~ 3l2
2
u
.
and
,
n — 1/3.
dv
...
= x 4 v 1/2
v 3l2
2
or
Setting
2
y
dx
— y~ ll3 + — y 213
1
v
,
.
= -x 4
x
3
— |x 5 + ex 2 (see Problem 5.57). Thus, for the original
5
2 312
y = ±(|x + ex
v
.
)
4v =
q{x)
— 4x,
— x,
n — 5.
The transformation
for which an integrating factor is
e
4
y
4idx =
= v,
e*
x
Then
.
so that, for the original equation,
4
= — x 4+ -„ + ce
-j
—^
4
4
y
3
.
/?
= 3.
= y" 3 + =y
1
Setting
i>
This last equation is linear, and its solution is
x
original equation,
'a
1
or
The original differential equation then becomes
v'.
= —2.
— — 1,
x
= — xe* x + -1a
e
e*
+c
This is a Bernoulli equation with
i/
— x4
x
— — 4 J xe 4jc dx = — xe 4jc + ^6,4x + c,
Solve
q(x)
dx
y *e
5.78
we have
The original differential equation thus becomes
or, explicitly,
4 dx
dx
ve
= |x 5 + ex 2
This is a Bernoulli equation with
y~'-
y = l/v,
y — xy
Solve
I
Problem 5.56). Since
dx
This last equation is linear, and its solution is
213
(see
A
1
-v 112 -
y
+ ce x2/2
-
2
equation,
1
XV — — X
dx
,.,
1/3
A
y = v
or
v
v
v
dv
—
1
x - = X —=
(-
dx
This is a Bernoulli equation with
we have
1
as a set of nontrivial solutions to the original differential equation.
3
5.76
= v~ 2 + = v _1 = 1/y,
The original differential equation then becomes
This equation is linear, and its solution is
+ ce x
v
dx
v
v
1
Setting
1
-\dv
—
y =
n = 2.
and
y
= 2x + ex
or
y =
±
r
/-
i
—j.
v
2
,
we have
y = v~
112
and
— - iT 3/ V + - r
_1/2
= 2x + ex 2
Problem 5.59). Then for the
(see
= y" 3/2
,
or
104
5.79
CHAPTER 5
D
/ + xy = 6x~Jy.
Solve
= y~ 1/2 + = y 112 we have y = v 2 and y' — 2vv'.
The original differential equation then becomes 2vv' + xv 2 = 6xv, or v' + \xv = 3x. The solution to this
xl A
last equation is given in Problem 5.58 as
v = 6 + ce~
Then, for the original equation, y 1/2 = 6 + ce~ x2/A
2/4 2
or y = (6 + ce~* )
I
n = \.
This is a Bernoulli equation with
Setting
1
v
,
'
.
,
.
5.80
Solve
I
+ y = y2
y'
.
= y~ 2+l = y~ l we have y = v' and
The original differential equation then becomes — i?~V + v~ l = v~ 2 or v' — v = — 1.
n — 2.
This is a Bernoulli equation with
— —v~
y'
2
v'.
Setting
V—e
e
x
I(x, v)
—e
5.81
Solve
I
+ y = y- 2
y'
— ^v~
2l3
— 2.
n —
Setting
y =\+ce~
3
3x
,
3x
ve
x
(ve
v
)
—
= —e x
+ cex
1
= y 2 + 1 =y 3
e
jv~
3x
Thus, for the original equation,
.
—
ax
so that
2l3
v'
y = v
we have
,
+v
1 '3
= v~
2li
1'3
or
and
v'
+ 3v — 3.
Multiplying by it, we get
.
or
3x
=
i;
(ve
1
3x
)
= 3e 3x
+ ce~ 3x
Thus, for the original equation,
.
x
or
.
)
xdy + ydx = x 3 y b dx.
equation with
n = 6.
Setting
then becomes
—
V + — v~
= fx 3 + ex 5
Solve
v
i3dx —
+ 3e 3x v = 3e 3x
v'
We rewrite the equation first as
5.83
e
= J 3e 3x dx = e 3x + c,
3x
3
y = (1 + ce~
Then integration yields
v
so that
The original differential equation thus becomes
v'.
e
Solve
—
ax
y
_6/
v
x>''
+ y — x 3 y6
= y~ 6 + =y
= x 2 t;" 6
,
5
or
,
+ — y = x2y6
y'
,
to obtain a Bernoulli
x
5
l
1/5
and then as
we have
y = v~
v
= — 5x 2
v'
115
.
and
y
= — - v~ 6l$
v'.
Our equation
The solution to this last equation is
x
5
v
(see
Problem 5.60); hence
y = v~
= (fx 3 + cx 5 )~ ,/5
1/5
.
x
dy + y dx = y 2 e dx.
+ y = y 2 e x which is a Bernoulli equation with n = 2.
= y~\ we have y = v"
and y' = —v~ 2 v'. The rewritten equation then becomes
x
x
2
— r~ V + iT = v~ e
or v' — v= —e
The solution to this last equation is v — (c — x)e"
x
(see Problem 5.61), so
y — v' — e~ /{c — x).
I
This
Multiplying by it, we obtain
.
.
This last equation is linear with integrating factor
5.82
— e~ x
,
This is a Bernoulli equation with
y'
~ ldx
or
(
1
s
d
— —e x
v
Then integration yields ve~ x = J — e~ x )dx = e~ x + c,
= + ce x and y = (1 + ce x )~ '.
'
,
,
last equation is linear with integrating factor
y~
1
v
This equation may be rewritten as
2 +
v = y~
i
y'
,
Setting
1
1
.
,
1
5.84
Solve
dx
- — x= ( —
—
\
dt
I
I
1
It
2
This is a Bernoulli equation in the dependent variable x and the independent variable t, with
= x~ 3 + = x~ 2
i'
x = r~ 12
we have
1
Setting
,
and
— = —2
iT 3/2
—
The differential equation thus
.
dt
dt
becomes
—
1
v
_,3/z
n dv
2
It
—
\
v
= sin + - cos + -
1
dt
,
.
1;
'
=
/
\
1
cos
2
x = v
1/2
=
1
I
sin t H
- cos t H
cV
1/2
t
or
dv
1
-— + -t; = cost
dt
J
t
c
1
The solution to this last equation is
3/2
My - i,z
f
(see
n = 3.
Problem 5.62). Thus,
3
LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
dx
- — x = -2tx
—
1
5.85
Solve
dt
I
4
.
It
dx
,
—- =
—
— _The
.._
_ .., dt;
1
dt
v
'
.
.
1
dz
- — = -z
—
1
/
_ ± ,,dv
z
1
—= —
dt
'
..*.
t;
'
—
dv
~ 312
(see Problem 5.63).
Therefore,
« = 5.
-/•
+ 2xv
Setting
w
= z _5+1 =z~ 4
we have
,
z
= v~ l>4
and
.
xy 4 = 0,
-I-
. IA
-5/4
or
dv
— tT
dt
2t
1
e~ s
integrating factor
6xdx
3*2
ve~
3
—
3xe~ 3 * dx =
e~
3x2
(cos x
TVi*» transformation
trQncfnrmntirin
The
y" 4
or
+ V = 1(1 ^
dx
3
— 3y ~ 4 — =
-
(2x — l)e
—
x
dx = — 2xe
—
- *
— «•
dy
dv
— y - z^ —
-=—
'
= y;
Using the
y
2
x
= 2x — 1
v
,
for which e
dx
—e x+c
'
1
—^=—\—2x + ce x
or
= cos x — sin x
dv
.
reduces either equation to
-
dx
3x2
2x).
,
it
h ce
2
reduces either equation to
—+y
or
1
6xv — 3x.
3
3
— sin x),
yt»
—3r =
or
y
an integrating factor.
2
+c
2
dx
dx
Then integrating by parts gives
—+y=y
.
dx
y' 3 = v;
I
— v~ 114 = (ft + ct~ 2 )~ 1/4
reduces either equation to
The transformation
=
z
we obtain
3
ve
Problem 5.64), so
(see
—
=—
dx
dx
J
+ I y = I(l-2x)/,
^
dx
t
3
— 3y -4
\
di;
or
dt
2
=
2
=4
— + -p
«,.
= -tT 5/4
/
+ 2xy~ = -x.
dx
-4
= e~ 3x \
...
1/4
= ft + ct' 2
v
y
y~ 3 = v;
The transformation
Solve
6t
The differential equation becomes
•
dx
5.89
v
It
.
The solution to this last equation is
is
3
or
dt
= ^t 2 + ct
: tf
Solve
3
'
5
1
5.88
x = v~
2t
1
Solve
we have
,
dv
—
+— =
_
= ~2tv 4/3
It
v
4
5.87
_,.,
-T--Z-V
'
This is a Bernoulli equation for z(t) with
dz
= x~ 4+1 =x~ 3
.
,
The solution to this last equation is
x = |,-l/3 = (llf 2 + cr 3/2) -l/3
dr
d
dt
3
3
Solve
Setting
differential equation becomes
-t«
5.86
n = 4.
This is a Bernoulli equation for x(t) with
and
105
dx
= sin x — cos x,
t>
for
dx
which e~ x is an integrating factor. Then multiplication and integration give
r
(sin x
1
— cos x)e - * dx = — e x sin x + c
or
J
5.90
Solve
The transformation
which
v
x dy - \y + xy 3 (l + In x)l dx = 0,
I
e
s2dx,x
yx
2
= x2
is
y
- = — sin x + ce x
2
= v;
-2y
or
y
-
-/ - - y~ 2 = 1 + In x.
dx
x
dv
—
= -— reduces
dx
dx
_ 3, dy
dv
—
+ - = -2(1 +
dx
x
2
either equation to
v
an integrating factor. Then multiplication and integration give
4
2
= -2 f(x 2 + x 2 lnx)dx = --x 3 --x 3 lnx + c
or
x2
2
/2
\
-j = -- x 3 - + lnx j + c
I
In x),
for
—
106
CHAPTER 5
Q
MISCELLANEOUS TRANSFORMATIONS
5.91
Develop a method for solving the differential equation
Set
—
+
dx
u
= f{y)
dv
5.92
P(x)i/-
dv
—
=
dx
so that
= Q(x),
f'(y)
dy
—
Then
dx
dy
—
+
ax
/'(v)
f(y)P{x) = Q(x)
for ><x).
the given differential equation may be written as
.
which is linear and may be solved by the method developed in Problem 5.2.
Show that the Bernoulli equation is a special case of the differential equation described in the previous problem.
I
dy
dx
The Bernoulli equation,
\-
p{x)y = y"q(x),
may be written as
{-n + l)y-"^ + (-n + l)p(x)y- n + = {-n + l)q(x)
l
dx
Set
(
<2(x)
= — n + l)q(x)
— n + l)y "
dy
\-
dx
y
n+
P(x) = Q(x),
—
=
dx
dy
Solve
I
sin y
y' n+ l
(cos x)(2 cos y — sin
2
—
dy
/(y) = cosy,
= sin 2 x cos x,
h (cos y)(2 cos x)
dx
P(x) = 2cosx,
Q(x) = sin x cos x.
2
and
which has the form required by
The substitution
r
= cosy
—
2
which is linear. Its solution is
h (2cosx)u = sin xcosx,
dx
(see Problem 5.70). The solution to the original equation is, implicitly,
-
transforms the rewritten equation into
— j sin 2 x — | sin x + { + ce' 2s,nx
~ 2 s,n x
2
cos y = \ sin x — | sin x + \ + ce
v
5.94
.
x).
— sin y -
We rewrite this equation as
Problem 5.91 with
then the Bernoulli equation has the form
which is identical to the differential equation described in the previous
f(y) =
problem for the special case
5.93
P(x) — ( — n + l)p(x);
and
(
Solve
—
dy
-
h
1
.
= 4e y sin x.
dx
I
We rewrite this equation as
ey
dy
1-
e y = 4sinx,
which has the form required by Problem 5.91 with
dx
f(y) = e
y
,
P(x) =
and
1,
Q{x) = 4 sin x.
e y = 2(sin x
equation is, implicitly,
Solve
I
—e
dv
y
transforms it to
—- + v = 4 sin x,
v
— 2(sin x — cosx) + ce~ x
— cos x) + ce~ x
.
Then the solution to the original
y = In [2(sin x — cos x) + ce~
and, explicitly,
x ~\.
dy
x 2 cos y —- = 2x sin v — 1.
dx
We write this equation as
f(y) = sin y,
P(x) =
—
cos y
2
,
and
dy
( 2V -1
—
+ smy
—
x
dx
sinyj \
Q{x) —
—-1y-
whose solution is given in Problem 5.73 as
5.96
v
dx
whose solution is given in Problem 5.71 as
5.95
The substitution
implicitly,
siny = ^x"
Solve
dy
—
=
dx
sin y
1
+ ex 2
(cos y)( 1
which has the form required by Problem 5.91 with
j",
y
The substitution
v
= ^x" + ex 2
and, explicitly,
= sin y
,
transforms it to
dv
2
v
-1
= —^-,
Then the solution to the original equation is,
1
y = arcsin
v
.
—
1
+ kx 3
,
where
k = 3c.
— x cos y).
We rewrite this equation as
= — x,
5
cos y dx
cos y
which has the form required by Problem 5.91 with
LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
f{y) - 1/cosy,
P(x)=-1,
Q(x) = -x.
and
The substitution
1;= 1/cosy
transforms it to
107
--v=-x,
dx
whose solution is given in Problem 5.72 as v = x + 1 + a?*. Then the solution to the original equation is,
x
and, explicitly, y = arcsec(x + 1 + cex
implicitly,
1/cosy = x + 1 + ce
).
dy
5.97
y + 3x y — x
3
x
Solve
dx
Here
(x dy
—
dx
reduced to
3
yield
ue*
y/x = v.
suggests the transformation
3x 2 v = 1,
h
xdy — ydx + 3x 3 y dx — x 2 dx = 0.
or
3
= j e* dx + c
3
for which e*
xl
y — xe~
or
is
\ e
Then
2
Solve
to
(4r s
- 6) dr + r 3 ds = 0,
3
dt
t
6
— dx — dx =
is
x
dx + cxe~
xi
The indefinite integral here cannot be evaluated
.
(rds + s dr) + 3s dr = (6/r ) dr.
2
or
r
r
—
+— =
dr
or
or
. ,_
c
r
«
i
x sin 9 d9 + (x J — Ix 1 cos
I
dy
dy + 2xy dx = x dx,
which reduces the equation
t,
Then r 3 is an integrating factor, and the solution is
r
or
h
x sin 9 d6 + cos 9 dx
«
+ cos 9) dx = 0,
„
rt
,
,
or
5
h
xz
d0 + cos 9 dx
x sin
ay =
xy = cos 0;
The substitution
=
rs
.
= -= + -r.
s
*
3
r
,
-=-.
*
r
= r 4 s = 3r 2 + c,
Solve
6
3
<it
+ 3 - dr = -=• dr,
.
5.99
2
an integrating factor. Multiplication and integration then
xi
I The first term of the second equation suggests the substitution
tr
h 3x
5
x2
terms of elementary functions.
in
5.98
— y dx)
= 0,
2
5
t
,
xL
2 cos 9 dx = xdx.
^
reduces the second equation to
2
2xy = x.
An integrating factor is e x and the solution is
,
dx
ye
xl
=
e
x2
=
I
e
x2
x dx = - e x2 + K
J
x
or
2 cos 9 — x + cxe~
x2
2
INITIAL-VALUE PROBLEMS
5.100
y - 5y = 0;
Solve
= 3.
y(0)
I The solution to the differential equation is given in Problem 5.3 as y — ce 5x
5(0)
— c, so the solution to the initial-value problem is
directly, we have
3 = ce
5.101
y'-5y = 0;
Solve
Applying the initial condition
.
y = 3e
5x
.
= 0.
y(3)
I The solution to the differential equation is the same as in the previous problem. Applying the initial
= ce 3{i \
condition, we get
5.102
y'-5y = 0;
Solve
or
c
= 0.
The solution to the initial-value problem is
y = 0.
= 4.
y(3)
I The solution to the differential equation is the same as in Problem 5.100. Applying the initial condition,
we find that 4 = ce 3(3) = ce 9 or c = 4e~ 9 The solution to the initial-value problem is
~
9 3x
— 4e 3(x 3)
v = 4e~ e
.
,
.
5.103
Solve
y'
- 5y = 0;
y(n)
= 2.
I The solution to the differential equation is the same as in Problem 5.100. Applying the initial condition, we
2 = ce
obtain
5.104
Solve
y'
3{n
or
\
+ 2xy = 0;
c
y(3)
= 2e~ 3n
.
9
y = 4e e~
x2
= 4<?- * - 9)
2
(
.
3n 3x
e
~
= 2e 3(x n)
.
- 4.
f The solution to the differential equation is given in Problem 5.9 as
condition directly, we have
y = 2e~
The solution to the initial-value problem is
4 = ce~
{3)2
= ce~
9
,
or
c
= 4e
9
.
y = ce~
x2
.
Applying the initial
The solution to the initial-value problem is
CHAPTER 5
108
5.105
Solve
>•'
y{-2) = 3.
+ 2xy = 0;
# The solution to the differential equation is the same as in the previous problem. Applying the initial
~ 2)2
= ce~ 4 or c = 3e 4 The solution to the initial-value problem is
condition, we have
3 = ce~
{
,
.
x2
4
= 3e~ (x2 -*\
y = 3e e~
5.106
Solve
dy/dt + (t - l)y = 0;
y(l)
= 5.
2
I The solution to the differential equation is given in Problem 5.13 as y = ce'~' 2 Applying the initial
' 12 2
— ce 112 or c = 5e~ l/2 The solution to the initial-value problem is
condition, we get
5 = ce
-'
2/2
2/2
1/
1/2
=
5e
y =5e- V-'
'
.
1
'
.
,
f
.
5.107
Solve
dy/dt + (t - \)y = 0;
y(- 3) = 0.
f The solution to the differential equation is the same as in the previous problem. Applying the new initial
— ce <_3)_<_3)2/2 = ce~ 1512 or c = 0. The solution to the initial-value problem is
condition, we obtain
y = 0.
,
5.108
N{\) = 1000.
N = 0;
Solve
dt
t
I The solution to the differential equation is given in Problem 5.19 as N = ct 5 Applying the initial
5
condition, we have
1000 = c(l) = c, so the solution to the initial-value problem is N = lOOOf 5
.
.
5.109
N = 0;
Solve
dt
N(2) = 1000.
t
I The solution to the differential equation is the same as in the previous problem. Applying the initial
1000 = c(2) 5 = 32c, or c = 31.25. The solution to the initial-value problem is
condition, we obtain
5
N = 31.25f
.
5.110
Solve
y-3y = 6;
y(0)
= 1.
I The solution to the differential equation is given in Problem 5.29 as y = ce 3x — 2. Applying the initial
= ce 3{0) — 2 — c — 2, or c = 3. The solution to the initial-value problem is
condition directly, we have
ix
- 2.
y = 3e
1
5.111
Solve
>'-3y = 6;
y{l)
= 0.
I The solution to the differential equation is the same as in the previous problem. Applying the initial
= ce Ml) — 2, so that
— 2 — 2e 3{x ~ — 2.
condition directly, we have
problem is
5.1 12
Solve
y — 2e' e
3
y - 3v = 6;
ix
2 = ce
3
or
c
= 2e~ 3
.
The solution to the initial-value
1)
y( -5)
= 4.
f The solution to the differential equation is the same as in Problem 5.110. Applying the initial condition,
~
we find that 4 = ce M 5) — 2, so that 6 = ce~ is or c — 6e 15 The solution to the initial-value problem is
.
15 3x
-2 = 6e 3{x + 5) - 2.
y = 6e e
5.113
Solve
dq/dt + lOq = 20;
q(0)
= 2.
Applying the initial
f The solution to the differential equation is given in Problem 5.32 as q = 2 + ce~ 10
2 = 2 + ce~ 10(0)
The solution to the initial- value problem is q = 2.
condition, we get
or c — 0.
'.
,
5.114
Solve
dq/dt + lOq = 20;
q(0)
- 500.
f The solution to the differential equation is the same as in the previous problem. Applying the initial
10(0)
= 2 + c, or c = 498. The solution to the initial-value
condition to it, we find that 500 = 2 + ce~
- 10t
problem is q = 2 + 498e
.
5.115
Solve
dq/dt + lOq = 20;
q(4)
= 500.
I The solution to the differential equation is the same as in Problem 5.113. Applying the initial condition,
we have
is
500 = 2 + ce~ 10(4)
q = 2 + 498e
40
e-
10 '
,
so that
= 2 + 498e"
498 = ce'
10( '- 4)
.
4
-
or
c
= 498e 40
.
The solution to the initial-value problem
LINEAR FIRST-ORDER DIFFERENTIAL EQUATIONS
5.116
Solve
dv/dt + 25u = 9.8;
u(0)
109
= 5.
/ The solution to the differential equation is given in Problem 5.36 as v = 0.392 + ce~ 25
Applying the
25i0
initial condition, we get
or
5 = 0.392 + ce~
4.608.
cThe
\
solution to the initial-value problem is
-25
v = 0.392 + 4.608e
'.
'.
5.117
Solve
dv/dt + 25v = 9.S;
t<0.1)
= 5.
/ The solution to the differential equation is the same as in the previous problem. Applying the new initial
25 or
25(01)
25 = 56.137. The solution
condition, we get
so that 4.608 = ce~
5 = 0.392 + a?~
c = 4.608e
25
=
0.392 + 56.137e~
to the initial-value problem is
v
,
'.
5.118
Solve
y'
+ y = sin x;
y(n)
= 1.
I From Problem 5.44 the solution to the differential equation is
n
c — \e
= ce~ n + \, or
initial condition directly, we obtain
x
K~x
—
+ sin x — cos x).
y = je"e~ + | sin x
\ cos x = \(e
1
5.119
Solve
x dy/dx — 2y — x 3 cos 4x;
y(n) =
.
y = ce~
problem is
5.120
Solve
>'
1
= \k sin 47r + en = en
= ^x 2 sin 4x 4- (x/7r) 2
x dy/dx — 2y = x 3 cos 4x;
Applying the
1.
2
condition, we obtain
+ jsinx — \cosx.
Thus
2
2
or
,
c
— l/n
Applying the
y = ^x sin 4x + ex
The solution to the initial-value
2
2
I The solution to the differential equation is given in Problem 5.46 as
initial
x
.
2
.
.
y(l)
= n.
I The solution to the differential equation is the same as in the previous problem. Applying the new initial
2
n = |(l 2 )sin4 -(- c(l ),
2
y = \x sin 4x 4- 3.33 lx
condition, we find that
problem is
5.121
Solve
y'
or
c
— n — ^sin4 = 3.331.
The solution to the initial-value
2
+ xy = xy 2
.
y(0)
;
= 1.
The solution to the differential equation is given in Problem 5.75 as
y
=
1
condition, we find that
=
1
1
5.122
Solve
y'
+ xy = xy 2
y(l)
;
+c
or
,
e
= 0.
4- ce
Applving the initial
^h'
y = 1.
The solution to the initial-value problem is
= 0.
Applying this initial condition to the solution found in Problem 5.75, we have
= —1
—
+ ce
which has
rjj,
'
no solution. Thus, there is no value of c that will satisfy the initial condition. However, a Bernoulli equation
also admits the trivial solution
y = 0, and since this solution does satisfy the initial condition, it is the
solution to this initial-value problem.
5.123
Solve
m
y'
+ xy = 6xyjy;
y(0)
= 0.
In Problem 5.79 we found a nontrivial solution to the differential equation to be
y = (6 + ce~
xZ 4 2
)
.
= (6 -(- ce )
Applying the initial condition, we obtain
or c — — 6. One solution to the initial-value
problem is thus y = 36(1 — e~ x2/4 2
The trivial solution to the Bernoulli equation, y = 0, also satisfies the
initial condition, so it is a second solution to the initial-value problem.
2
,
')
.
5.124
Solve
x dy + y dx = x 3 y 6 dx;
y(l)
= 5.
I A nontrivial solution to this Bernoulli equation is given in Problem 5.82 as
= (f + c)~
so that
3
5
5
)"
=
(2.5x
2.49968x
y
the initial condition, we obtain
the initial-value problem is
5.125
Solve
6
x dy + y dx = x 33„6
y dx;
1 '
5
5
,
5~
5
=f + c
or
3
Applying
y = (fx 3 + ex 5
c — —2.49968.
The solution to
'
)
.
l
.
y(l)
= 0.
i A nontrivial solution to the differential equation is given in the previous problem. Applying the initial
condition, we have
y = 0,
= (42 + c) _1/5
or
=
5/2 + c
,
which has no solution. However, the trivial solut
does satisfy the initial condition; hence it is the solution to the initial-value problem.
CHAPTER 6
Applications of First-Order
Differential Equations
POPULATION GROWTH PROBLEMS
6.1
A certain population of bacteria is known to grow at a rate proportional to the amount present in a culture
that provides plentiful food and space.
Initially there are 250 bacteria, and after seven hours
800 bacteria are
observed in the culture. Find an expression for the approximate number of bacteria present in the culture at
any time t.
I The differential equation governing this system was determined in Problem 1.53 to be
N(t) denotes the number of bacteria present and k is a constant of proportionality.
Problem 5.8).
At t — 0, we are given N = 250.
solution becomes
N = 250e'".
Its
dN/dt = kN,
solution is
where
N — ce*'
(see
At
t
= 7,
N — 800. Substituting this condition and solving for k, we get
Now the solution becomes
N = 250?° ,66
so the
800 = 250e* (7)
we are given
k — 4lnf§§ — 0.166.
or
250 = ce* (0) = c,
Applying this initial condition, we get
,
'
(7)
which is an expression for the approximate number of bacteria present at any time t measured in hours.
6.2
Determine the approximate number of bacteria that will be present in the culture described in the previous
problem after 24 h.
= 24.
We require N at
N = 250e° 166,24) = 13,433.
#
6.3
t
= 24
t
into (/) of the previous problem, we obtain
Determine the amount of time it will take for the bacteria described in Problem 6.1 to increase to 2500.
I
We seek a value of
t
N = 2500.
corresponding to
2500 = 250?° lh(",
solving for t, we find
6.4
Substituting
so that
N — 2500
Substituting
10
= e01661
and
t
into (/) of Problem 6.1
A bacteria culture is known to grow at a rate proportional to the amount present.
bacteria are observed in the culture; and after four hours. 3000.
and
= (In 10) 0.166 = 13.9 h.
After one hour, 1000
Find an expression for the number of bacteria
present in the culture at any time t.
I As in Problem 6.1. the differential equation governing this system is
dN/dt = kN,
number of bacteria present and k is a constant of proportionality, and its solution is
N=1000;
hence,
1000 = ce
k
= 4, N = 3000;
hence,
3000 = ce
Ak
At
t=l,
At
t
k = 3 In 3 = 0.366
and
N - 694c0366
'
c
= 1000?
° 366
= 694.
as the number of bacteria present
time t.
In the previous problem, determine the number of bacteria originally in the culture.
I
/V
6.6
.
.
Substituting these values of k and c into the solution yields
6.5
kl
.
Solving these two equations for k and c, we find
at any
where N(t) denotes the
N — ce
= 0.
We require N at
= 694e (0 366)<0) = 694.
t
Substituting
f
—
into the result of the previous problem, we obtain
-
A bacteria culture is known to grow at a rate proportional to the amount present.
Find an expression for
the approximate number of bacteria in such a culture if the initial number is 300 and if it is observed that the
population has increased by 20 percent after 2 h.
f
As in Problem 6.1, the differential equation governing this system is
N = ce ".
proportionality, and its solution is
dN/dt = kN,
where k is a constant of
1
At
f
becomes
At
t
= 0,
we are given
N = 300.
Applying this initial condition, we get
so the solution
N = 300e'".
= 2,
the population has grown by 20 percent or 60 bacteria and stands at
Substituting this condition and solving for k, we get
110
300 = ceki0) = c,
360 = 300<?
i,2)
or
300 + 60 = 360.
k = \ In |§§ = 0.09116.
The number
APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS
111
of bacteria present at any time t is thus
= 300e 0091H"
JV
6.7
fin hours
(/)
Determine the number of bacteria that will be present in the culture of the previous problem after 24 h.
# We require N at
t
= 24.
Substituting this value of t into (/) of the previous problem, we obtain
N = 300e 009116,24) = 2675.
6.8
Determine the number of bacteria present in the culture of Problem 6.6 after
I We require N at
— 7(24) — 168 h.
09116<l68) =
9
=
jV
300e°1.34 x 10
week.
1
Substituting this value of t into (/) of Problem 6.6, we obtain
t
.
6.9
Determine the amount of time it will take the culture described in Problem 6.6 to double its original population.
N = 2(300) = 600.
I We seek the value of f associated with
600 = 300e° 091 16
and then solving for t, we get
6
6.10
or
',
t
=
N = 600
Substituting
—
into (7) of Problem 6.6
= 7.6 h.
0.09116
A certain culture of bacteria grows at a rate that is proportional to the number present. If it is found that the
number doubles in 4 h, how many may be expected at the end of 12 h?
*
—=
dx
—
=
x
dx
Let x denote the number of bacteria present at time t hours.
Then
or
kx,
dt
x = x
— 0, we have c = x
and e Ak = 2. Now when
at time
Ak
=x e
2x
= kt + In c, so that x = ce
Assuming that
= 4. we have x = 2x
then
x = x e
at time
that is, there are eight
x = x e i2k = x (e4*) 3 = x (2 3 = 8x
Integrating the second equation, we have
First Solution:
t
kt
In x
.
k
and
t
k dt.
'\
— 12,
t
;
)
;
times the original number.
Second Solution:
f
= 4,
x = 2x
.
Again we integrate the second equation, this time between the limits
We write
- = k
we integrate between the limits
6.11
t
x = x
= 0,
In
—=
If,
in the previous problem, there are 10
12/c
= 3(4*) = 3 In 2 = In 8.
from which
dt,
\
x = 8x
Then
4
and
t
— In x = 4/c
In 2x
— 12,
x — x,
t
= 0,
- = k
and
4k = In 2.
so that
we get
x — x
from which
dt,
\
Now if
as before.
,
bacteria at the end of 3 h and 4 x 10
4
at the end of 5 h, how many
were there to start?
f
When
First Solution:
x = 10 4
— 3,
t
hence, the equation
;
x = cekt
of the previous problem becomes
hence,
4 x 10 4 = ce 5k
4
10
4
= ce 3k
and so
,
10
c = -^-.
Also, when
t
= 5.
4 x 10 4
—
c =
jt
c
4
10
Second Solution:
^—
from which
e
2k
=4
and so
and
e
k
bacteria.
4x 1
f
= 0,
4
= 3k = 3 In 2 = In 8.
Thus,
8
Integrating the differential equation of the previous problem between the limits
Integrating between the limits
— 2.
,
J10
6.12
4 x 10
;
4
— -^ = —-—
e*
x
4
—^- =
Equating these values of c gives us
10
10
4
10
.
the original number is
In
x = 4 x 10
4
x = x
10
Then
x
4
dX
f5
- — k
dt,
J3
x
from which
\
4
and
t
= 3,
x = 10
4
In 4
/»1 o 4
gives us
|
= 2k,
dx
- = k
t
= 3,
and
f3
I
dt,
= In 2.
k
from which
4
= -—
,
„
as before.
8
In a culture of yeast the amount of active ferment grows at a rate proportional to the amount present.
If the
amount doubles in 1 h, how many times the original amount may be anticipated at the end of 2| h?
I
Let N(t) denote the amount of yeast present at time t. Then
dN/dt = kN,
proportionality. The solution to this equation is given in Problem 5.8 as
where k is a constant of
N = ce
kt
.
If
we designate the initial
CHAPTER 6
112
amount of yeast as N
1
at
t
= 0,
N Q = ce k{0) =
and it follows that
We may then rewrite
c.
N — Nq^.
the solution as
After
N=N
then
,
h, the amount
2N = N eMl \
N = N e 693
N = 2N
present is
so that
e
k
=2
applying this condition and solving for k, we find
;
k = In 2 = 0.693.
and
'.
Thus, the amount of yeast present at any time t is
N = N e° 693
After 2.75 h the amount will be
-
-
<
2
-
75 »
= 6J2N
.
This represents a 6.72-fold
increase over the original amount.
6.13
The rate at which yeast cells multiply is proportional to the number present.
number doubles in
If the original
2 h, in how many hours will it triple?
Let N(t) denote the number of yeast cells present at time f.
m
N=N e
where N
kl
,
N — 2N
we know that
= N ek(2 \
2iV
We seek
N=N e
which
t
=
ln(3iV
/N
)
t
= 2,
e
2k
= 2,
k = \ In 2 = 0.3466.
or
Thus, the number of yeast cells in this culture
0i46(".
N = 3N
for which
t
At
Substituting this condition into the equation and solving for k, we get
.
from which
at any time t is
Then it follows from the previous problem that
designates the initial number present and k is a constant of proportionality.
Substituting for
;
N and solving for
t,
we obtain
3N — N e 03466
from
',
= 3.17 h.
0.3466
6.14
Bacteria are placed in a nutrient solution and allowed to multiply.
Food is plentiful but space is limited, so
competition for space will force the bacteria population to stabilize at some constant level M. Determine an
expression for the population at time t if the growth rate of the bacteria is jointly proportional to the number
of bacteria present and the difference between
I
M and the current population.
Let N(t) denote the number of bacteria present at time t.
— = kN(M —
The differential equation governing this system
dN
was determined in Problem 1.55 to be
N),
where k is a constant of proportionality'.
we
If
at
rewrite this equation in the differential form
N(M - N)
—
M
In
N
M In (M
y
N{M — N)
= ——
N
- N) - kt = c,
or
we see it is separable.
Integrating term
1/M
1/M
1
by term and noting that bv partial fractions
dN — k dt — 0,
-I-
-,
M—N
—In
M
we get
M-N
= c + kt,
from which
N
„cM +A
where C = e cM
M -N
CM
N = -^—
Solving for N, we obtain
becomes
Nn =
CM
and
.
C+
_
If
kMt'
C + e' kM
C =-
.
No
—
M-N
1
we now denote the initial population by N
then at
,
=
t
this
Thus, the solution can be written
.
N_
MiV
N + (M - N )e~ kSU
which is an expression for the bacteria population at any time t. Equation (/) is often referred to as the logistics
equation.
6.15
The population of a certain country is known to increase at a rate proportional to the number of people
presently living in the country.
If after
2 years the population has doubled, and after 3 years the population is
20,000. find the number of people initially living in the country.
f
dN
people initially living in the country. Then.
iV
N
Let /V denote the number of people living in the country at any time f, and let
=N
At
hence, it follows from that
:
t
= 2, N = 2N
.
r
= 3.
N = 7062.
N = 20,000.
N = ce ki0 \
N — NqC 03 *
N = or.
= 0.
kN = 0.
which has the solution
or that
=N
Thus, the solution becomes ,V = N e
2N = N e 2k from which k = | In 2 = 0.347.
Substituting these values, we get
Thus, the solution finally becomes
At
—
denote the number of
c
At
t
.
.
:
1 '.
Substituting these values, we obtain
(0
20,000 = /V e
-
347)(3)
= JV (2.832),
or
kt
.
APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS
6.16
If the population
D
113
of a country doubles in 50 years, in how many years will it treble under the assumption that the
rate of increase is proportional to the number of inhabitants?
#
Let y denote the population at time f years, and y
the population at time
dy
= 0.
f
—=
Then
ky,
or
y = y
at
at
dy
—
= kdt, where
k is a proportionality factor.
y
Integrating the second equation gives us
First Solution:
t
= 0;
then
t
gives
2yo
—=
.
5
and
Integrating this time between the limits
°
dr,
from which
y = y
and
r
=
50 In 3 = 50/cf = f In 2,
and
f
=
/c
f
Jo
y
limits
= 0,
f
1
.
Second Solution:
f
Jyo
y = ce ".
or
y = y e
Then we have 2y = y e 50k or e 50k = 2.
we know y = 2y
Then 3 50 "= e 50kt = {e 50k y = 2', and so t = 79 years.
=y
c
= 50,
kt
3 = e
At
= kt + In c,
In v
Let
time
k '.
y = 3y
50 In 3
,
^
gives us
y = y
= 0,
- In y = 50/c, and so
In 2y
f,
t
50/c
= In 2.
—=k \
f
and
dt,
t
When
y = 3y
= 50,
y = 2y
,
y = y ^'
gives us
Also, integrating between the
from which
In 3
=
/ct.
Then
=m
79 years.
In 2
DECAY PROBLEMS
6.17
A certain radioactive material is known to decay at a rate proportional to the amount present.
is
If initially
there
100 mg of the material present and if after 2 years it is observed that 5 percent of the original mass has
decayed, find an expression for the mass at any time t.
I
Let N(t) denote the amount of material present at time t.
dN/dt — kN,
and its solution is
N — ce*'
= c. Thus, the solution becomes TV = lOOe*'.
— 2,
= 2, 5 percent of the original mass of 100 mg, or 5 mg, has decayed. Hence, at
—
—
=
=
100
5
Substituting
this
condition
in
the
equation
TV
lOOe*"
and
solving
for k, we get
N{2)
95.
this initial condition, we get
At
100 = ce
The differential equation governing this system is
At t — 0, we are given TV — 100. Applying
(see Problem 5.8).
k{0)
t
t
95 = 100e
1
k(2)
,
k = - In
or
—95— = —0.0256.
The amount of radioactive material present at any time t is,
100
2
therefore,
TV
6.18
= lOOe" 00256
'
f
in years
In the previous problem, determine the time necessary for 10 percent of the original mass to decay.
I We require t when JV has decayed to 90 percent of its original mass.
we seek the value of t corresponding to
90= lOOe" 00256
gives us
6.19
(7)
',
so that
N = 90.
Substituting
-0.0256r = In 0.9,
and
Since the original mass was 100 mg,
- 90 into (7) of the previous problem
= -(In 0.9)/0.0256 = 4.12 years.
TV
t
A certain radioactive material is known to decay at a rate proportional to the amount present. If initially there
lost 10 percent of its original
is 50 mg of the material present and after 2 h it is observed that the material has
mass, find an expression for the mass of the material remaining at any time t.
I
Let N denote the amount of material present at time t.
= ce**. At t = 0, we are given N = 50. Therefore,
N
N = 50e*'.
At
t
= 2,
Then
dN/dt - kN =
50 = ce
k(0
\
or
= 50.
Thus, we now have
t
10 percent of the original mass of 50 mg, or 5 mg, has decayed. Hence, at
for
we
get
solving
k,
and
equation
last
the
into
values
these
Substituting
TV = 50 - 5 = 45.
— = -0.053. The amount of mass present
k = - In
50
2
= 2,
45 = 50e
2fc
,
or
at any time t is therefore
N = 50e~ OO53r
6.20
c
and, as in Problem 6.17,
t
in
hours
In the previous problem, determine the mass of the material after 4 h.
I We require /V at t = 4. Substituting t = 4 into (7) of the previous problem and then solving for N, we
N = 50e " ° 053,,4) = 50(0.809) = 40.5 mg.
find that
(
(/)
CHAPTER 6
114
6.21
Determine the time at which the mass described in Problem 6.19 has decayed to one-half its initial mass.
f We require when
0053
find
25 = 50e~
t
',
N - 50/2 = 25.
so that
Substituting
-0.053t = ln|
N = 25
into (7) of Problem 6.19 and solving for t, we
t= 13 h.
(The time required to reduce a decaying
and
material to one-half its original mass is called the half-life of the material. For this material the half-life is 13 h.)
6.22
A certain radioactive material is known to decay at a rate proportional to the amount present. If after
f
Then dN/dt — kt, where k is a constant of
If we designate the initial
The solution to this equation is given in Problem 5.8 as N — ce k
then
N = N Q at t — 0, and we have N Q = ce* (0) = c. Thus, the solution becomes
Let N(t) denote the amount of the material present at time f.
proportionality.
mass as N Q
N=N e
At
t
t
= 1.
,
'.
kt
.
— 1,
The half-life is the time associated with
= N e'
-
105 ',
= N e kn)
0.9A
from which
,
0.9
—e
at
and
,
so that
N — \N
-0.105r = ln£
t
hours
in
(7)
Substituting this value into (7) and solving for t, we obtain
.
and
'
t
= 6.60 h.
Find the half-life of a radioactive substance if three-quarters of it is present after 8 h.
I
Let N(t) denote the amount of material present at time t. Then it follows from the previous problem that
N—N e
kt
,
where N
denotes the initial amount of material and k is a constant of proportionality. If
three-quarters of the initial amount is present after 8 h. it follows that
|/V
= N ek<8)
from which
,
f
A:
t
'
',
f
Sk
—f
r,
we get
(
Radium decomposes at a rate proportional to the amount present.
in
e
00359f".
= g In 4 = —0.03596. Thus the amount of material present at any time is N — N e~
We require when N — \^o- Substituting this value into the previous equation and solving for
"
- (In \) -0.03596) = 19.3 h.
\N = JV <r 003596
from which e 003596 = \ and
and
6.24
k
The amount of radioactive material present at any time t is thus
N = N e' 0105
i/V
N = 0.9N
has decayed, so 90 percent remains. Hence
10 percent of the original mass 7V
Substituting this condition and solving for k, we get
k = In 0.9 = —0.105.
6.23
h it
1
observed that 10 percent of the material has decayed, find the half-life of the material.
is
If half the
original amount disappears
1600 years, find the percentage lost in 100 years.
Let R(t) denote the amount of radium present at time t. It follows from Problem 1.52 that dR/dt = kR.
kt
If we designate the
where k is a constant of proportionality. Solving this equation, we get R — ce
k<0)
— c, so the solution becomes
and apply this condition, we find R Q = ce
initial amount as R
(at
t — 0)
I
.
R = R ekl
.
= 1600. Applying
when
R = |/?
i(,00k
kilb00)
=
which
from
and
R e
%R
\ = e
Since the half-life of radium is 1600 years, we have the condition
this condition to the last
k = (In
equation and then solving for k give
H 1600 = -0.0004332.
r
.
,
The amount of radium present at any time t is thus
R - R e~ 0000 * ii21
t
in years
The amount present after 100 years will be R = £^-0.0004332000) _. o.958i?
R — 9587?
100 = 4.2 percent.
the initial amount R is
(/)
,
so the percent decrease from
—— — —
6.25
A certain radioactive material is known to decay at a rate proportional to the amount present.
If initially \ g
of the material is present and 0.1 percent of the original mass has decayed after 1 week, find an expression
for the mass at any time f.
f
Let N(t) denote the amount of material present at time f. Then
proportionality, and the solution to this equation is
have
\
— cekW = c,
so the solution becomes
N = ce
kI
(see
dN/dt = kt, where k is a constant of
Problem 5.8). Since N = \ at t = 0,
we
N = \d*.
we take the time unit to be 1 week, then 0.1 percent of the initial mass has decayed at t = 1, and
Thus, at t = 1, we have N = 0.999(|) = 0.4995 g. Applying this condition to the last
Hl
equation, we get 0.4995 = \e \ from which e* = 0.999, so that k = In 0.999 = -0.001. The amount
If
99.9 percent remains.
of radioactive material present at any time r is thus
N== y-o.oou
6.26
fin weeks
Determine the half-life of the material described in the previous problem.
(7)
APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS
D
115
f The half-life is the time t associated with the decay of one-half the original mass. Here the original mass is
N = \.
\ g, so we seek the time when
0001 ',
for t, we obtain
\ = |e"
Substituting this value into (/) of the previous problem and solving
-O.OOh =\vl\
from which
and
t
= 693 weeks. The half-life is 693 weeks
or 13.3 years.
6.27
Rework Problem 6.25 using a time unit of 1 day.
N = \e
I Our work through the derivation of the equation
kt
0.4995 = V" 7)
we get
from which
,
e
lk
= 0.999
in
Problem 6.25 remains valid.
N = 0.4995
week, or 7 days, the mass has been reduced to 0.4995 g, so
1
at
t
= 7.
k = 4 In 0.999 = -0.0001429.
and
Now after
Applying this condition,
The amount of
radioactive material present at any time f is thus
N = i e - o.ooo 1429,
6.28
t
i
n days
(/)
Use the result of the previous problem to determine the half-life of the material.
I
Since the original mass is \ g, we require t when
\e~ 00001 * 29 '5 from which
\
N = %.
=
problem, we obtain
Substituting this value into (/) of the previous
-0.0001429t = ln±
and
r
= 4850.
The half-life is
4850 days or 13.3 years (as we found in Problem 6.26).
6.29
A certain radioactive material is known to decay at a rate proportional to the amount present. If initially 500 mg
of the material is present and after 3 years 20 percent of the original mass has decayed, find an expression
for the mass at any time f.
Then dR/dt = kR, where k is a
constant of proportionality. The solution to this equation is given in Problem 5.8 (with R replacing N) as
R — ce kt At f = 0, N = 500; applying this condition yields 500 = ce k(0) — c, so the solution becomes
I
Let R{t) denote the amount of radioactive material present at time f.
.
R = 500e*'.
If
get
— 3, and
= 3, R — 0.8(500) = 400. Applying this condition to the last equation, we
3k
= 0.8 and
= ^ln 0.8 = -0.07438. The amount of radioactive material
e
we take the time unit to be 1 year, then 20 percent of the original mass has decayed at
80 percent remains. Thus, at
400 = SOOe*' 3
so that
',
t
t
fe
present at any time t is then
K = 500e-°- 07438
6.30
t
Substituting
t
— 25
into (/) of the previous problem, we obtain
For the material described in Problem 6.29, determine the amount remaining after 200 weeks.
I
t
6.32
(J)
For the material described in the previous problem, determine the amount remaining after 25 years.
— 25.
I We require R when
K = 500e- OO7438(25) = 77.9mg.
6.31
fin years
'
= 200/52 = 3.846 years.
R = 500<T° 07438(3 846) = 375.6 mg.
Since the time unit in Problem 6.29 is 1 year, we require R when
= 3.846
into (/) of Problem 6.29, we get
t
Substituting
-
Determine the amount of time required for the material of Problem 6.29 to decay to 30 percent of its original
amount.
# We require
150 = 500e~
6.33
f
when
007438 ',
R = 0.3(500) = 150.
from which
Substituting
-0.07438r = ln^g
R - 150 into (/) of Problem 6.29, we obtain
= 16.2 years.
and
t
Determine the amount of time required for the material of Problem 6.29 to decay to 250 mg.
I We require t when
250 = 500e
-007438 ',
R = 250.
from which
R = 250 into (/) of Problem 6.29, we obtain
= 9.3 years. Note that 9.3 years is the half-life
- 0.07438f = In fgg and
Substituting
t
of the material.
6.34
After 2 days, 10 g of a radioactive chemical is present.
Three days later, 5 g is present.
How much of the
chemical was present initially, assuming the rate of disintegration is proportional to the amount present?
f
proportionality, and
the solution to this equation
is
N = ce
dNjdt = kN, where k is a constant of
Measuring time in units of 1 day, we have
Then
Let N(t) denote the amount of chemical present at time t.
k
'.
—
CHAPTER 6
116
N = 10
at
t
= 2;
10 = ce
hence,
2k
N=5
Moreover,
.
at
2 — e~
these last two equations simultaneously for k and c, we find
c
= 10e"
2, "°- 231>
= 15.87.
f
=5
days later); so
(3
3k
N = ce
Substituting these values of c and k into
k
JV
6.35
= 15.87<r
-
231<0)
At
— 0,
t
Solving
'.
= —0.231,
and
N = 15.87<r°- 231r
we obtain
',
as an expression for the amount of radioactive chemical present at any time r.
In 2
k =
so that
,
= ce 5
5
amount is
this
= 15.87 g.
Under certain conditions it is observed that the rate at which a solid substance dissolves varies directly as the
product of the amount of undissolved solid present in the solvent and the difference between the saturation
concentration and the instantaneous concentration of the substance. If 40 kg of solute is dumped into a tank
containing 120 kg of solvent and at the end of 12 min the concentration is observed to be 1 part in 30, find the
amount of solute in solution at any time f. The saturation concentration is 1 part of solute in 3 parts
of solvent.
f
If
Q is the amount of the material in solution at time
t,
40 — Q
then
is
material present at that time, and Q/120 is the corresponding concentration.
the amount of undissolved
Hence, according to the given
information,
dQ
™A
Q\
^2
— = k „«
(40 - QY
*
,, An
= k(40
-Q)[
*'V3
dt
120/
120
Q
——
—
—t = —
^^
-Q)
d
This is a simple separable equation for which we have
k
j
2
(40
Q =
Since
when
t
— 0,
40-4
1
120
40
t
Q
40
120
Also, when
t
— 12,
Q — 3^(120) — 4,
t
-Q
c.
120
so we have
4320
120
1
4320
l
,
from which
——
— = —— + —
40
6.36
k
= —— 12 H
1
Then the solution becomes
c = ^.
we find that
—k +
——— = -^
u
with
solution
;
dt,
dt.
4320
Q = 40
we find that
from which
,
40
t
+ 108
A certain chemical dissolves in water at a rate proportional to the product of the amount undissolved and the
difference between the concentration in a saturated solution and the concentration in the actual solution.
In 100 g of a saturated solution it is known that 50 g of the substance is dissolved. If when 30 g of the
chemical is agitated with 100 g of water, 10 g is dissolved in 2 h, how much will be dissolved in 5 h?
I
Let x denote the number of grams of the chemical undissolved after t hours. At that time the concentration
——- — and
x
30
of the actual solution is
,
dx
30 - x\
( 50
Integrating the latter equation between
nodx_r2o_dx
J30
x
fc„
J30 x + 20
5x
from which
In
3(x + 20)
dissolved after 5 h is
6.37
f
= 0,
x = 30
x = 30
and
t
and
x
= k = -0.46. Then
= 5,
t
Then
= 2,
x = x,
3
e
dx
k
x = 30 - 10 = 20,
we get
.46.
— = - -0
x + 20
.
dx
^| lni= _
fromwhich
f
x + 20
= 0,
5 J°
Integrating now between
—
50
that of a saturated solution is
-
46
we get
= 0.38,
Cx
dx
fX
I
and
x - 12.
dx
— = -k fs
I
dt.
Thus, the amount
5
30 — 12 = 18 g.
Chemical A dissolves in solution at a rate proportional to both the instantaneous amount of undissolved
chemical and the difference in concentration between the actual solution C a and saturated solution Cs A
porous inert solid containing 10 lb of A is agitated with 100 gal of water, and after an hour 4 lb of A is
.
dissolved. If a saturated solution contains 0.2 lb of A per gallon, find the amount of A which is undissolved after 2 h.
I
Let x lb of A be undissolved after t hours.
dx
Then
(
10
-x\
Mx+10)
—
APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS
where
c
— /c/100.
Using the conditions
6.38
117
Separating the variables and integrating, we get
J
x = 3.91 lb
Q
x(x+ io)~ioJVx
— 0,
t
x = 10
and
r
ln
x+ \o)
= 1,
x — 6,
To
*TTo
we find
x =
= c' + c
'
—
5(3/4)'
When
.
t
= 2 h,
of A undissolved.
Find the time required to dissolve 80 percent of the chemical A described in the previous problem.
f
x = 0.8(10) = 2
Substituting
into the last equation of the previous problem and solving for t, we have
5(3/4)'
2=
from which we find
l-(l/2)3/4r
6 $' = 2
or
(|)'
=f
Then
= (In i)/(ln |) = 3.82 h.
t
COMPOUND-INTEREST PROBLEMS
6.39
A depositor places $10,000 in a certificate of deposit account which pays 7 percent interest per annum,
compounded continuously. How much will be in the account after 2 years?
I
Let P(t) denote the amount of money in the account at time t.
the growth of the account was determined in Problem 1.57 to be
dP/dt — 0.07P for an annual interest
— 001
=
P
This equation is linear and separable; its solution is
rate of 7 percent.
P — $10,000.
principal is
The differential equation governing
P = 10,000?° ° 7
We require P when
Applying this condition, we find
ce
° 7(0)
10,000 — ce°
'.
= c,
At
0,
t
the initial
so the solution becomes
'.
P= 10,000?° °
6.40
7,2)
— 2 years.
= $11,502.74.
t
Substituting this value of t into the last equation, we find
How much will the depositor of the previous problem have after 5 years if the interest rate remains constant
over that time?
f
Substituting
P = 10,000?°
6.41
t
° 7<05)
— 5 into the solution derived in the previous problem, we obtain
= $14,190.68.
A woman places $2000 in an account for her child upon his birth. Assuming no additional deposits or
withdrawals, how much will the child have at his eighteenth birthday if the bank pays 5 percent interest per
annum, compounded continuously, for the entire time period?
f
Let P(t) denote the amount of money in the account at time t. The differential equation governing the
growth of the money was determined in Problem 1.57 to be dP/dt — 0.05P for an annual interest rate of
P — ce 005
At t — 0, we have P = $2000, so
5 percent. The solution to this differential equation is
2000 = ce° 05( ° = c, and the solution becomes P = 2000?° ° 5
'.
)
'.
We require the principal at
P = 2000?°
6.42
° 5<18)
How long will
it
t
— 18.
Substituting this value of t into the last equation, we obtain
= $4919.21.
take for the initial deposit to double under the conditions described in the previous problem?
I We seek the value of t corresponding to
in the
6.43
previous problem, we obtain
Solve Problem 6.41
if
P = $4000.
4000 = 2000?"
Substituting this quantity into the solution derived
In (4000/2000)
,
from which
t
=-
the interest rate is 6.5 percent.
f For this new interest rate, the solution derived in Problem 6.41 becomes
we have P = 2000?° 065(18) = $6443.99.
6.44
P = 2000?° ° 65
'.
Then, at
= 18,
t
Solve Problem 6.41 if the interest rate is 9\ percent.
f For this new interest rate, the solution derived in Problem 6.41 becomes
we have P = 2000?° O925(18) = $10,571.35.
6.45
- = 13.86 years.
P = 2000?° ° 925
'.
Then, at
f
= 18,
A man places $700 in an account that accrues interest continuously. Assuming no additional deposits and no
withdrawals, how much will be in the account after 10 years if the interest rate is a constant 1\ percent for
the first 6 years and a constant 8£ percent for the last 4 years?
CHAPTER 6
118
#
For the first 6 years, the differential equation governing the growth is given by Problem 1.57 as
which has as its solution P = ce 0075 (0 < t < 6). At t = 0, P - 700; hence
0015{0)
= c, and the solution becomes P = 700? 0075
700 = ce
At the end of 6 years, the account will have
grown to P = 700e° 075,6) = $1097.82. This amount also represents the beginning balance for the 4-year period.
Over the next 4 years, the growth of the account is governed by the differential equation dP dt = 0.0825P,
which has as its solution P - Ce 00825 (6 < f < 10). At t = 6, P - 1097.82: hence 1097.82 - Ce° 0825 «»,
and C - 1097.82e"° 495 - 669.20. The solution thus becomes P = 669.20*?° ° 825
and at year 10 the account
will have grown to
P = 669.20?° 0825 10 = $1527.03.
dP/dt = 0.075P,
'
'.
'
',
<
6.46
How long will
it
>
take a bank deposit to double if interest is compounded continuously at a constant rate of
4 percent per annum?
I The differential equation governing the growth of the account is dP/dt — 0.04P (see Problem 1.57); this
•°
equation has as its solution P = ce00*'. If we denote the initial deposit as P we have P = ce° 4, ° = c,
°4
and the solution becomes P = P e°
Substituting this value into the last equation and solving for
We seek corresponding to P = 2P
we
)
,
'.
t
obtain
6.47
f,
.
2P - P e° ° 4
How long will
it
from which
',
o)
=
t
= 17.33 years.
°|
take a bank deposit to double if interest is compounded continuously at a constant rate of
8 percent per annum?
I With this new interest rate, the solution derived in the previous problem becomes P = P e° ° 8
We seek
°8
°8
= (In 2)/0.08 = 8.66 years.
2 = e°
corresponding to 2P hence we write 2P = P e°
so that
and
'.
6.48
'
'.
;
t
t
A woman plans to place a single sum in a certificate of deposit account with a guaranteed interest rate of
How much should she deposit if she wants the account to be worth $25,000 at the end
6^ percent for 5 years.
of the 5-year period?
t The differential equation governing the growth of this account was determined in Problem 1.57 and is
dP/dt = 0.0625 P; its solution is
P = ce 00625t Since we want P = 25.000 at t = 5, we have
.
25,000 = a?°°
625(5)
P= 18,290.39e
6.49
from which
,
00625
At
'.
f
= 0,
c
= 25,000e
°- 3125
= 18,290.39.
Thus the solution becomes
P = 18,290.39?° ° 625(0) = $18,290.39.
the initial amount must be
A man currently has $12,000 and plans to invest it in an account that accrues interest continuously. What
interest rate must he receive, if his goal
is
to have $15,000 in 21 years?
I
Let P(f) denote the amount in the account at any time t. and let r represent the interest rate (which is
presumed fixed for the entire period). The differential equation governing the growth of the account is given in
Problem 1.57 as dP/dt = (r/100)P, which has as its solution p = C e (rl00)
At t = 0. P = 12.000; hence
'.
12.000 = ce
{r
100,|0
»
= c,
so the solution becomes
P - 12,000?
,r
10 °".
We require r corresponding to P - 15,000 and t — 2.5. Substituting these values into the last equation
and solving for r, we obtain 15,000 = 12,000? (r ,00,,2 5)
which reduces to 1.25 = e' 40 and yields
r = 40 In 1.25 = 8.926 percent.
.
6.50
What interest rate must the man in the previous problem receive if his goal is $16,000 in 3 years?
I
P = 16.000
Substituting
16,000-= 12.000?
,r
100,
\
and
t
— 3
into the solution derived in the previous problem, we obtain
from which we find that
r
=
— —
In
3
12,000
= 9.589 percent.
y
COOLING AND HEATING PROBLEMS
6.51
Newton's law of cooling states that the time rate of change of the temperature of a body is proportional to the
temperature difference between the body and its surrounding medium. Using Newton's law of cooling, derive a
differential equation for the cooling of a hot body surrounded by a cool medium.
Let T denote the temperature of the body, and let Tm denote the temperature of the surrounding medium.
Then the time rate of change of the temperature of the body is dT/dt, and Newton's law of cooling can be
formulated as dT/dt — —k(T- TJ. or as
I
dT
—
+ kT = kT
m
(1)
°
1
APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS
119
where k is a positive constant of proportionality. Since k is chosen positive, the minus sign in Newton's law is
required to make dT/dt negative for a cooling process. Note that in such a process, 7" is greater than 7m thus
;
7 — Tm
6.52
is
positive.
A metal bar at a temperature of 100F is placed in a room at a constant temperature of OF.
If after
20 min
the temperature of the bar is 50°F, find an expression for the temperature of the bar at any time.
# The surrounding medium is the room, which is being held at a constant temperature of OF, so
Tm =
IT
and (7) of Problem 6.51 becomes
-— + kT — 0.
This equation is linear; it also has the differential form
at
— dT+kdt = 0,
100 = ce' km
temperature of the bar is initially 100°F), it follows that
7 = 100?
we are given that 7 — 50;
= 20,
t
—
-1
k =
50
In
-
=
(
100 = c.
7 = 100
at
(
=
(the
Substituting this value
— 0.693) = 0.035.
50 = lOO?" 20 *,
from which
of the bar at any
The temperature
y time t is then
F
'
(7)
Find the time it will take for the bar in the previous problem to reach a temperature of 25°F.
# We require
25 = 100?'
6.54
or
hence, the last equation becomes
7=100?-° 035
6.53
Since
.
.
20
100
20
kt
~ kt
into the solution, we obtain
At
T = ce~
which is separable. Solving either form, we get
7 = 25.
when
t
035
Substituting
-0.035f = In i
or
'
7 = 25
into (7) of the previous problem, we obtain
Solving, we find that
t
= 39.6 min.
Determine the temperature of the bar described in Problem 6.52 after 10 min.
f We require 7 when
= 10. Substituting
0035 10 =
=
7
lOOe*100(0.705) = 70.5 F.
t
t
= 10
into (7) of Problem 6.52, we find that
**
'
should be noted that since Newton's law is valid only for small temperature differences, the above calculations
It
represent only a first approximation to the physical situation.
6.55
A body at a temperature of 50°F is placed outdoors where the temperature is 100°F. If after 5 min the
temperature of the body is 60°F, find an expression for the temperature of the body at any time.
I
With
Tm = 100
(the surrounding medium is the outside air), (7) of Problem 6.51
dT/dt + kT = 100/c.
7=50
This equation is linear and has as its solution
becomes
T — ce~ kl + 100 (see Problem 5.38).
or c=-50.
Substituting this value into
= 0, it follows that 50 = ce~ k(0) + 100,
T = — 50? "*' + 100.
the solution, we obtain
5k
+ 100. Solving for k,
At t = 5, we are given that T — 60; hence, from the last equation, 60 — — 50?"
5
-50?"
Substituting this value, we obtain
k = — |lnf§ = -|(-0.223) = 0.045.
so that
we obtain -40 =
Since
when
t
*,
the temperature of the body at any time t as
T= -50e 0045 + 100
'
6.56
(7)
Determine how long it will take the body in the previous problem to reach a temperature of 75°F.
f We require t when T = 75. Substituting T = 75 into (/) of the previous problem, we have
0045 + 100 or
we find -0.045r = lni or t=15.4min.
75 = -50?
e~ 0M5 = j. Solving for
'
'
6.57
f,
Determine the temperature of the body described in Problem 6.55 after 20 min.
= 20 into (7) of Problem 6.55 and then solving for T, we find
f We require T when t = 20. Substituting
T = _50e<-° 045 20 + 100 = -50(0.41) + 100 = 79.5°F.
t
><
6.58
>
A body at an unknown temperature is placed in a room which is held at a constant temperature of 30°F.
If
after 10 min the temperature of the body is 0°F and after 20 min the temperature of the body is 15°F, find an
expression for the temperature of the body at time t.
Here the temperature of the surrounding medium, Tm is held constant at 30° F, so (7) of Problem 6.51
becomes dT/dt + kT = 30k. The solution to this differential equation is given by Problem 5.39 (with a = 30)
I
as
,
T=ce~ k + 30.
'
At
t
At
f
= 10,
= 20,
we are given that
we are given that
7 = 0. Hence,
7=15. Hence,
= ce~ 10k + 30 or ce~ l0k = -30.
20k
15 = ce~
+ 30 or ce" 20* = - 15.
120
CHAPTER 6
D
Solving these last two equations for k and c, we find
k = -^ In 2 = 0.069
c =
and
- 30e 10 * = - 30(2) = - 60.
Substituting these values into the solution, we obtain, for the temperature of the body at any time t,
T= -60e-° 069 + 30
'
6.59
(1)
Find the initial temperature of the body described in the previous problem, just as it is placed into the room.
=
f We require 7 at t = 0. Substituting
T = -60e -°- 069,(0) + 30 = -60 + 30 = -30°F.
f
into (/) of the previous problem, we find that
,
6.60
A body at a temperature of
C
D
F is placed in a room whose temperature is kept at 100 F.
10 min the
If after
=
temperature of the body is 25 F, find an expression for the temperature of the body at time t.
I Here the temperature of the surrounding medium is the temperature of the room, which is held constant at
Tm — 100. Thus, (/) of Problem 6.51 becomes dT/dt + kT = 100/c; its solution is given by Problem 5.38 as
= 0, we have 7 = 0; hence
7 = 100 + ce k
At
= 100 + ce* ,0) = 100 + c. Thus c = - 100 and the
'.
f
7 = 100 — lOOe*'.
= 10. we have 7 = 25; hence 25 = 100 - 100e* ,,0)
At
Thus, the last equation becomes
7 = 100 - 100e~° 02877
solution becomes
t
,
or
e
10k
= 0.75,
k = -0.02877.
so that
'.
6.61
Find the time needed for the body described in the previous problem to reach a temperature of 50°F.
We require when 7 = 50. Substituting 7 = 50 into the result of the previous problem and solving for
002877
02877
= 0.5. Then
= (In 0.5)/(- 0.02877) = 24.1 min.
50 = 100 - lOOe'
or e"
f
t,
6.62
r
we find
'
t
Find the temperature of the body described in Problem 6.60 after 20 min.
I
6.63
-
'
Substituting
t
= 20
into the result of Problem 6.60, we have
7 = 100 - 100e"° 02877 2 °> = 43.75°F.
<
A body at a temperature of 50F is placed in an oven whose temperature is kept at 150 F.
C
If after
10 min the
temperature of the body is 75 F. find an expression for the temperature of the body at time f.
I
Here the temperature of the surrounding medium is the temperature of the oven, which is held constant at
Thus (/) of Problem 6.51 becomes dT/dt + kt — 150k; its solution is given by Problem 5.39
(with
a =150) as
T=\50 + ce~ lu At f = 0. we have 7=50. Hence 50 = 150 + ce' k{0 \ so
7m = 150 F.
.
- 100 and the solution becomes 7 = 150 - 100e"*'.
= 10. we have 7 = 75. Hence 75 = 150 - 100^-'I(10)
At
_0 ° 2877
and the last equation becomes 7 = 150 - 100e
c =
t
or
£>- 10 *
= 0.75,
so that
k = 0.02877
'.
6.64
C
Find the time required for the body described in the previous problem to reach a temperature of 100 F.
I
e
6.65
= 050
7 = 100
Then
into the result of the previous problem, we find that
t
=
(i
100 = 150 - lOOe
-0 02877
-
'
or
n o.50)/(- 0.02877) = 24.1 min.
Find the time required for the body described in Problem 6.63 to reach a temperature of 70°F.
I
f
6.66
Substituting
- 0.02877,
Substituting
7=70
into the result of Problem 6.63, we find that
70 = 150 - 100e"
002877
'.
Then
= (In 0.80)/(- 0.02877) = 7.76 min.
Find the time required for the body described in Problem 6.63 to reach a temperature of 200°F.
I Since a body can never reach a temperature higher than that of the surrounding medium, which here is
Tm — 150
the body of Problem 6.63 can never attain a temperature of 200 F.
C
.
6.67
A body whose temperature is initially 100°C is allowed to cool in air whose temperature remains at a constant
20 C. Find a formula which gives the temperature of the body as a function of time t if it is observed that
after 10 min the body has cooled to 40° C.
I
If
we let 7 denote the instantaneous temperature of the body in degrees Celsius and t denote the time in minutes
since the body began to cool, then the rate of cooling is
dT
—
=
k(T — 20).
This equation can be solved either
dt
as a separable equation or as a linear equation.
7-20
= k dt.
Integration yields
In (7
Regarding it as a separable equation, we rearrange it to
— 20) = kt + In
\c\,
from which we write
= e*'
and find that
—
APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS
T = 20 + ce kl
7 = 100
Since
.
when
t
= 0,
it
follows that
c
= 80,
D
121
so that
7 = 20 + 80e*'
(/)
= 10. Under this condition, (/) becomes
7 = 40 when
10
10k
=
We
80e
or
e
=
can
solve
20 +
approximately for k by writing 10/c = -In 4 % - 1.386, so
40
{.
The instantaneous temperature of the body is then given by 7 = 20 + 80e~ 01386
that
k % —0.1386.
To determine the value of k, we use the fact that
t
*,
'.
6.68
Find an alternative expression for the temperature in the previous problem.
#
Rather than solving approximately for k, we can solve explicitly for e\ obtaining
written as
6.69
7 = 20 + 80(/)' = 20 + 80(|)'
e
— (j) 1/10
k
Then (/) can be
.
/10
.
Find an alternative expression for the solution to Problem 6.52.
I Upon applying the second boundary condition in that problem, we obtained
written as
\ = (e~
20
k
e~
to give us
)
k
= (j)
1120
50 = lOOe
-20
which can be
*,
Then the result of Problem 6.52 becomes
.
7 = 100?-*' = I00(e- k = 100(i) ,/2 °.
)'
6.70
Rework Problem 6.53 using the expression obtained in the previous problem.
I We require
25 = 100(i)' /2 °,
t
7 = 25.
when
7 = 25
Substituting
into the result of the previous problem, we obtain
which we rewrite as (f/20) In i = In 0.25. Then t = (20 In 0.25)/(ln \) = 40 min. The difference
between this answer and the one obtained in Problem 6.53 is due to round-off error in computing k in
Problem 6.52.
6.71
Find an alternative expression for the solution to Problem 6.55.
I Upon applying the second boundary condition in that problem, we obtained
written as
0.8
^e
*)
5
e~ k — (0.8) 1/5
to give
— 40 = — 50e
_ 5k
,
which can be
Then the result of Problem 6.55 becomes
.
7 = -50(e- k + 100 = -50(0.8) ,/5 + 100.
)'
6.72
Rework Problem 6.57 using the expression obtained in the previous problem.
= 20.
T = -50(0.8) 4 + 100 = 79.5°F.
f We require T when
6.73
Substituting
t
t
= 20
into the result of the previous problem, we find
According to Newton's law of cooling, the rate at which a substance cools in air is proportional to the difference
between the temperature of the substance and that of the air. If the temperature of the air is 30° and the substance
cools from 100° to 70° in 15 min, find when the temperature will be 40°.
M
at*
Let T be the temperature of the substance at time t minutes.
Then
—- = — k(T — 30)
or
Anr
kdt.
30
at
(Note: The use of — k is optional.
—— = —
T—
—
We shall find that k is positive here; but if we used +k, we would find k to
be equally negative.)
Integrating between the limits
AT
70
f
Jioo
j _ 30
= -k C dt,
Jo
- In 70 = —kt.
= (151n7)/0.56 = 52min.
t
In 10
= 0,
T = 100
In 40 -In
so that
Integrating between the limits
so that
t
5
t
= 0,
t
= 15,
70= -15/c = lnf
'
7=100
Multiplying by
and
and
t
=
t,
T = 70,
and
we obtain
15* = In *
2 = 0.56.
T = 40,
/mo
we obtain
- 15 and rearranging, we obtain
15/cf
— dT™=~k\ dt,
= 15 In 7,
/*t
from which
FLOW PROBLEMS
6.74
A tank initially holds V gal of brine that contains a lb of salt. Another brine solution, containing b lb of salt
per gallon, is poured into the tank at the rate of e gal/min while, simultaneously, the well-stirred solution leaves
the tank at the rate of/ gal/min (see Fig. 6.1).
Find a differential equation for the amount of salt in the tank
at any time t.
I
Let Q denote the amount (in pounds) of salt in the tank at any time. The time rate of change of Q, dQ/dt, equals
Salt enters the tank at the
the rate at which salt enters the tank minus the rate at which salt leaves the tank.
rate of be lb/min.
To determine the rate at which salt leaves the tank, we first calculate the volume of brine in
122
CHAPTER 6
gal/min
e
Fig. 6.1
V plus the volume of brine added et minus the volume of
V + et — ft. The concentration of
the tank at any time t, which is the initial volume
brine removed ft. Thus, the volume of brine in the tank at any time is
Q /(V + et — ft), from which follows that salt leaves the tank at the rate of
dQ/dt = be - f[Q/(V + et - ft)], so that
salt in the tank at any time is then
f[Q/(V
At
6.75
t
+ et- ft)] lb/ min.
Q — a,
= 0,
it
Thus,
dQ
/
dt
V + (e
Q = be
U)
f)t
Q(0) = a.
so we also have the initial condition
A tank initially holds 100 gal of a brine solution containing 20 lb of salt. At
f
= 0,
fresh water is poured into
the tank at the rate of 5 gal/min, while the well-stirred mixture leaves the tank at the same rate.
Find the
amount of salt in the tank at any time t.
V = 100,
Here,
b = 0,
a = 20,
e
= f = 5,
Q — ce'' 20
Substituting these values into the last equation, we find that
Q = 20e~' i2 °.
t -* oo,
Q -»
be rewritten as
Note that as
6.76
At
.
c
= 0,
t
= 20,
0.
The
20
dt
solution to this differential equation is given in Problem 5.6 as
Q — a — 20.
—+—Q =
and (1) of Problem 6.74 becomes
we are given that
so that the solution can
as it should, since only fresh water is being added.
A tank initially holds 100 gal of a brine solution containing
1
lb of salt.
At
t
=
another brine solution
containing 1 lb of salt per gallon is poured into the tank at the rate of 3 gal/min, while the well-stirred mixture
leaves the tank at the same rate.
I Here
dQ
+ 0.032
= 100,
a =
Find the amount of salt in the tank at any time t.
6=1,
1,
and
e
= f = 3;
hence, (1) of Problem 6.74 becomes
The solution to this linear differential equation is
3.
Q = ce~° ° 3
'
100.
-I-
dt
At t = 0, Q = a = 1. Substituting these values into the last equation, we find
-99. Then the solution can be rewritten as Q = -99e~ 003 + 100.
c =
6.77
= ce° + 100,
or
Find the time at which the mixture described in the previous problem contains 2 lb of salt.
I We require
2 =
6.78
1
'
-99<r
-
when
Q — 2.
+ 100
or
t
03 '
e
Substituting
0.03t
98
99'
Q = 2
into the result of the previous problem, we obtain
from which
U = 0.338 min.
i
0.03 In 99
t
A 50-gal tank initially contains 10 gal of fresh water. At
f
= 0,
a brine solution containing 1 lb of salt per
gallon is poured into the tank at the rate of 4 gal/min, while the well-stirred mixture leaves the tank at the rate of
2 gal/min.
Find the amount of time required for overflow to occur.
f Here
a = 0,
any time t is
6.79
6=1,
e
= 4, / = 2,
V + et — ft — 10 + It.
and
V = 10.
We require
t
when
From Problem 6.74, the volume of brine in the tank at
10 + It — 50;
hence,
t
= 20 min.
Find the amount of salt in the tank described in the previous problem at the moment of overflow.
APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS
— + ——
—Q=
+
For this problem, (/) of Problem 6.74 becomes
10
dt
is
given
in Problem 5.49 as
B
At
= 0,
f
Q = a = 0.
Q
* —
40t + 4f
123
whose solution
This is a linear equation
M
4.
It
+c
.
10 + It
Substituting these values into the last equation, we find that
at the moment of overflow, which is
6.80
2
D
t
= 20.
* =
Thus,
10 + 2(20)
c
= 0.
We require Q
— = 48 lb
A tank initially holds 10 gal of fresh water. At t = 0, a brine solution containing \ lb of salt per gallon is
poured into the tank at a rate of 2 gal/min, while the well-stirred mixture leaves the tank at the same rate. Find
the amount of salt in the tank at any time t.
V = 10,
Here
a = 0,
b = \,
and
e
= / = 2.
Hence (/) of Problem 6.74 becomes
— + -0 =
dt
its
1;
5
Q = ce~" + 5 (see Problem 5.39 with k = y, a = 5, and T replaced by 0.
= ce~ 015 + 5 = c + 5, and c=— 5. Thus the last equation becomes
Q = a = 0; hence
15
—
=
5e~' + 5,
which represents the amount of salt in the tank at any time
(2
5
solution is
At
= 0,
t
t.
6.81
Determine the concentration of salt in the tank described in the previous problem at any time t.
I The volume V of liquid in the tank remains constant at 10 gal. The concentration is
6.82
A tank initially holds 80 gal of a brine solution containing \ lb of salt per gallon. At
t
Q/V = — je'" 5 + \.
= 0,
another brine
solution containing 1 lb of salt per gallon is poured into the tank at the rate of 4 gal/min, while the well-stirred
mixture leaves the tank at the rate of 8 gal/min. Find the amount of salt in the tank at any time f.
V = 80,
I Here
a = |(80) - 10,
dQ
—
+
dt
b = 1,
e
= 4,
8
= 1(4)
80 + (4 - 8)r ^
l
dt
,
6.83
Q = 4(20 - f) - ^(20 - t) 2
so that
t
20 - 1*
=4
Q = 4(20 — r) + c(20 — t) 2
so that
c
= —7/40.
.
Applying the initial
Therefore, the amount of salt in
.
V — 0.
From Problem 6.74, we have
V=
= 80 + At — St,
= 20 min.
Determine when the tank described in Problem 6.82 will hold 40 gal of solution.
I We seek
that
6.85
2
Determine when the tank described in the previous problem will be empty.
I We seek t corresponding to a volume
6.84
Then (1) of Problem 6.74 becomes
dQ
+
—
or
;
The solution of this equation is given in Problem 5.50 as
we get 10 = 4(20) + c(20) 2
condition
Q(0) = a = 10,
the tank at time t is
/ = 8.
and
t
t
corresponding to a volume
V = 40.
From Problem 6.74, we have
V = 40 = 80 + 4f — St,
so
= 10 min.
Find the amount of salt in the tank described in Problem 6.82 when the tank contains exactly 40 gal of brine.
f The amount of salt in the tank at any time
f
is
given in Problem 6.82 as
Problem 6.84, the tank will contain 40 gal of solution when
t
= 10.
Q = 4(20 — t) — 4^(20 — t) 2
.
From
At that time,
Q = 4(20 - 10) - ;&(20 - 10) 2 = 22.5 lb.
6.86
Determine when the tank described in Problem 6.82 will contain the most salt,
f The amount of salt in the tank at any time t is given in Problem 6.82 as
Q = 4(20 - t) - 3^(20 - f) 2
(7)
always negative, the maximum value of Q occurs when dQ/dt = 0. Setting the derivative of (7)
— 4 + j^(20 — t) = 0, from which t = 8.57 min. At that time, there will be 22.857 lb of
equal to zero, we get
2
Since d Q/dt
2
is
salt in the tank.
6.87
A tank contains 100 gal of brine made by dissolving 80 lb of salt in water.
Pure water runs into the tank at the
Find the amount of salt
rate of 4 gal/min, and the mixture, kept uniform by stirring, runs out at the same rate.
in the tank at
any time t.
CHAPTER 6
124
V =100, a = 80, b = 0, and e
= 4. Then (7) of Problem 6.74 becomes dQ/dt + 0.04Q = 0,
which has as its solution Q = ce~ 00 *' (see Problem 5.7). Applying the initial condition Q(0) = a - 80, we
obtain 80 = c'° 04(0) = c, so the amount of salt in the tank at time r is Q = 80e"° ° 4
I Here
=f
'.
6.88
Find the concentration of salt in the tank described in the previous problem at any time t.
V = 100.
I Since the outflow equals the inflow of liquid, the volume of liquid in the tank remains a constant
From the result of the previous problem, it follows that the concentration is C = Q/V = 0.8e -0041
.
6.89
Assume that the outflow of the tank described in Problem 6.87 runs into a second tank which contains 100 gal
The mixture in the second tank is kept uniform by constant stirring and is allowed to
run out at the rate of 4 gal/min. Determine the amount of salt in the second tank at any time t.
of pure water initially.
V = 100,
I For the second tank,
(/) of Problem
At
t
= 0,
004
_004
b = 0.8e~
dQ/dt + 0.04Q = 3.2e
6.74 becomes
(see Problem 5.55).
a — 0,
6.90
which has as its solution
= 3.2(0)<T°
Q = a = 0; hence
Q = 3.2te~° ° 4
the second tank at any time t is thus
e — f = 4.
Then
-0
Q = 3.2te~° ° 4 + a? 04
(see the previous problem), and
'
',
04(0)
+ ce~°
The amount of salt in
'.
Determine the amount of salt in each of the two tanks described in Problems 6.87 and 6.89 after 1 h.
— 60 min, we have Q — %0e~° 04(60) = 7.26 lb
I Using the results of the two problems with
004<60)
= 17.42 lb of salt in the second tank.
tank, and
Q = 3.2(60)e"
of salt in the first
t
6.91
Determine when the amounts of salt in the tanks described in Problems 6.87 and 6.89 will be equal.
S0e~° °
I We equate the results of the two problems to obtain
6.92
'
-
'
= c.
04(0)
4
'
= 3.2te~° ° 4
A tank contains 100 gal of brine made by dissolving 60 lb of salt in water.
',
from which
t
= 80/3.2 = 25 mm.
Salt water containing
1
lb of salt per
gallon runs in at the rate of 2 gal/min, and the mixture, kept uniform by stirring, runs out at the rate of 3 gal/min.
Find the amount of salt in the tank at the end of 1 h.
I Here
—+
dt
At
V = 100,
100 - t
f
= 0,
= 2,
b = 1,
a = 60,
e = 2,
and
Q = 100 -
which has as its solution
Q = a = 60;
/ = 3.
60 = 100 + c(100) 3
hence
= 60 min,
At
Q = (100- f)-0.00004(100- r)
3
=
=
37.44
lb.
0.00004(100
60)
Q (100 60)
3
6.93
+ c(100 - f) 3
t
c =
so that
,
(see
Problem 5.48).
-0.00004
and the solution becomes
this equation yields
t
.
Then (7) of Problem 6.74 becomes
A cylindrical tank contains 40 gal of a salt solution containing 2 lb of salt per gallon. A salt solution of
concentration 3 lb/gal flows into the tank at 4 gal/min.
How much salt is in the tank at any time if the
well-stirred mixture flows out at 4 gal/min?
I
Let the tank contain A lb of salt after t minutes.
Then
Rate of change of amount of salt = rate of entrance — rate of exit
Solving the equation
6.94
dA
—
=
dA
lb
dt
min
A
12
subject to
= 3
lb
A
gal
A lb
min
40 gal
x 4
gal
A = 40(2) = 80
at
t
= 0,
t
x 4
gal
min
we find
A = 120 - 40e" rI0
.
A right circular cone (Fig. 6.2) is filled with water. In what time will the water empty through an orifice O of
Assume the velocity of exit is v = kyjlgh, where h is the instantaneous
cross-sectional area a at the vertex?
height (head) of the water level above O, and k is the discharge coefficient.
I
At time t the water level is at h. At time
t
+ dt,
dt > 0,
the water level is at
h + dh,
have
Change in volume of water = amount of water leaving
— nr 2 dh — av dt = akyjlgh dt
where
dh < 0.
We
.
APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS
D
125
Fig. 6.2
From similar triangles OAB and OEF,
Its solution, subject to
r
= Rh/H.
h —
the condition
H
at
dh = ak^flgh dt.
jj^2
Then the above equation becomes
t
= 0,
is
t
2nR
=
H
:
(H 512 - h 512
).
The time required
5akH 2 J2g
for emptying is the time when
h — 0,
%R 2
or
t
—
l2H
= ——-
/
5ak \J g
6.95
A hemispherical tank of radius R is initially filled with water. At the bottom of the tank, there is a hole of radius
r
through which the water drains under the influence of gravity. Find an expression for the depth of the water
in the tank at any time t.
t
Let the origin be chosen at the lowest point of the tank, let v be the instantaneous depth of the water, and let x
be the instantaneous radius of the free surface of the water (Fig. 6.3). Then in an infinitesimal interval dt, the
water level will fall by the amount dy, and the resultant decrease in the volume of water in the tank will be
dV = nx 2 dy.
This, of course, must equal in magnitude the volume of water that leaves the orifice during the
same interval dt. Now by Torricellfs law, the velocity with which a liquid issues from an orifice is v = \flgh,
where g is the acceleration of gravity and h is the instantaneous height, or head, of the liquid above the orifice.
x'+(y-R) 2 ^R 2
Fig. 6.3 Vertical plane section through the
center of a hemispherical tank.
In the interval dt, then, a stream of water of length
v
x dt = \J2gy dt
and of cross-sectional area nr 2 will emerge
2
from the outlet. The volume of this stream of water is dV = area x length = nr yjlgydt.
magnitudes of our two expressions for dV, we obtain the differential equation
nx 2 dy
nr
Now, equating the
Igydt
U)
The minus sign indicates that as t increases, the depth y decreases.
Before this equation can be solved, x must be expressed in terms of y. This is easily done through the use of
x 2 + (y - R) 2 — R 2
or
the equation of the circle which describes a maximal vertical cross section of the tank:
x 2 = 2yR - y 2
.
With this relation, (7) can be written as
n(2yR
2
y )dy = — nr yj2gydt.
2
separable equation that can be solved without difficulty. Separation yields
,
This is a simple
(2Ry 1/2 - y 3l2 )dy = -r 2 Jig dt,
and
—
126
CHAPTER 6
D
%Ry il2 - \y 5 2 = -r 2 sjlg t + c.
integration then gives
and thus
6.96
§
Ry 32 -
'
5 2
'
f>-
= -r jlgt + jf R
Since
y = R
when
t
= 0,
we find
\±R S2 = c,
5/2
.
Determine how long it will take the tank described in the previous problem to empty.
I We require
t
y = 0.
corresponding to
= - r 2 yjlg + jf K 5/2
t
6.97
2
from which
,
t
From the result of the previous problem, we have
=
15 r 2
V2s
A 100-gal tank is filled with brine containing 60 lb of dissolved salt. Water runs into the tank at the rate of
2 gal/min and the mixture, kept uniform by stirring, runs out at the same rate.
after
I
1
How much salt is in the tank
h?
Let s be the number of pounds of salt in the tank after t minutes, so that the concentration then is s/100 lb/gal.
— =—
2s
During the interval dt, 2 dt gal of water flows into the tank, and 2 dt gal of brine containing
s
dt
100
of salt flows out. Thus, the change ds in the amount of salt in the tank is
dt lb
50
5
ds =
Integrating yields
dt.
50
s
= ce -tlso
At t = 0, s = 60; hence, c
6/5
= 60(0.301)= 18 1b.
s = 60^"
6.98
60
and the solution becomes
s
= 60?
'
50
.
When
t
— 60 min.
The air in a certain room with dimensions 150 x 50 x 12 ft tested at 0.2 percent C0 2 Fresh air containing
C0 2 was then admitted by ventilators at the rate of 9000 ft 3 min. Find the percentage of C0 2
.
0.05 percent
after
I
20 min.
Let x denote the number of cubic feet of
x/90,000.
C0
in the
2
room at time t, so that the concentration of C0 2 then is
During the interval dt, the amount of C0 2 entering the room is 9000(0.0005) dt ft 3 and the
,
x
amount leaving6 is 9000
'-
—
dt
3
ft
.
Hence, the change
dx in the interval dt is
fc
90,IXX)
x-45
dx = 9000 0.0005
dt
Integrating yields
dt.
90.000/
10 In (v - 45) =
— + In c,
t
or
10
x = 45 + ce"' "'.
At
t
= 0,
x = 0.002(90.000) = 180.
x = 45 + 135? "' 10
63
When
t
= 20,
Then
c
= 180 - 45 = 135.
x = 45 + 135e
:
= 63.
and the solution becomes
The percentage of CO, is then
= 0.0007 = 0.07 percent.
*
90,000
6.99
.
Under certain conditions the constant quantity Q in calories second of heat flowing through a wall is given by
Q = — kA dT/dx, where k is the conductivity of the material, A (cm 2 is the area of a face of the wall
)
T is the temperature x cm from that face such that T decreases as x
m 2 of a refrigerator room wall 125 cm thick for which
Find the heat flow per hour through
perpendicular to the direction of flow, and
increases.
k = 0.0025,
1
if the
temperature of the inner face is
-5 C and that of the outer face is 75 C. (See Fig. 6.4.)
125 cm
direction of flow
Fig. 6.4
.
APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS
127
—— dx from
dT =
Let x denote the distance of a point within the wall from the outer face. Integrating
D
kA
T=75
x = 0,
to
T = -5,
x=125,
we get
f~
s
dT = --% f
kA J°
J '5
%0kA
Q = -rrr =
6.100
80(0.0025)(100)
125
dx
80 = -^-(125),
or
from which
kA
2
= 16 cal/s.
TXT
3600Q = 57,600 cal.
Thus, the flow of heat per hour is
A steam pipe 20 cm in diameter is protected with a covering 6 cm thick for which
loss per hour through a meter length of the pipe if the surface of the pipe is at
Find the heat
k = 0.0003.
200 C and the outer surface of
the covering is at 30°C. (See Fig. 6.5.)
direction of flow
Fig. 6.5
f At a distance x > 10 cm from the center of the pipe, heat is flowing across a cylindrical shell of surface area
2nx cm 2 per centimeter of length of pipe. From Problem 6.99,
AT
AT
A
Q= -kA — = -2nkx—
dx
T = 30,
Integrating between the limits
2nk
°—
r°°dT= -Q- Jib
f
ji6
J3o
J30
or
,
dx
x - 16
l
I
x > 10 cm
InkdT = —-
=
,„
Check:
170
r*
dx
—
g = ^?^cal/s,
Then
or
x = 10,
,
,.
.
170
x
„
„„
7-30=-In—.
In 1.6
T = 30 +
Q = 245,000 cal.
from the center of the pipe described in the previous problem.
,
,
2
and the
n i_6
x = 16
T — 30,
and
T = T,
x = x,
we get
x
lnl.6Ji6 x
When
100(60)
between the limits
In 1.6
dT
we get
I
340nk dx
Integrating
rr
x = 10,
x
Find the temperature at a distance
J30
T = 200,
= ()(ln 16 - In 10) = Qln 1.6.
340ti/c
—
x
and
heat loss per hour through a meter length of pipe is
6.101
InkdT = -Q
or
——
170
In 1.6
^
Then
T = „„
30 +
„,
16
= 200°C.
170
In 1.6
When
x = 16,
,
16
n—
x
T = 30 +
= 30°C.
In 1.6
6.102
Find the time required for a cylindrical tank of radius 8 ft and height 10 ft to empty through a round hole of radius
1 in at the bottom of the tank, given that water will issue from such a hole with velocity approximately
v = 4.8 yfh ft/s,
where h is the depth of the water in the tank.
f The volume of water that runs out per second may be thought of as the volume a cylinder 1 in in radius and of
height v. Hence, the volume which runs out in dt seconds is
nl
—
(4.S s/h)dt
= —— (4.8^)^-
Denoting by
J
dh the corresponding drop in the water level in the tank, we note that the volume of water which runs out in time
dt is also given by 64rc dh.
Hence,
=
CHAPTER 6
128
n
iao /Z\j
ca j,
—-(4.%yjh)dt=-64ndh
j
or
dt
64(144) dh
-— = -1920 dh
=
144
Integrating between
r
f
4.8
= 0,
h = 10
and
f
=
h = 0,
t,
Jh
—
Jh
d/i
P dt =
we get
Jo
10
,
from which
= -3$40y/h\° = 3840 VlOs = 3 h 22 min.
io
6.103
As a possible model of a diffusion process in the bloodstream in the human body, consider a solution moving
with constant velocity v through a cylindrical tube of length L and radius r. We suppose that as the solution
moves through the tube, some of the solute which it contains diffuses through the wall of the tube into an
ambient solution of the same solute of lower concentration, while some continues to be transported through
the tube. As variables, we let x be a distance coordinate along the tube and y(x) be the concentration of the
solute at any point x, assumed uniform over the cross section of the tube. As boundary conditions, we assume
that
and y(L) = y L(<y ) are known. Find an expression for the concentration y(x) at any point
y(0) = >'
along the tube.
I As a principle to use in formulating this problem, we have Frick's law: The time rate at which a solute diffuses
through a thin membrane in a direction perpendicular to the membrane is proportional to the area of the membrane
and to the difference between the concentrations of the solute on the two sides of the membrane.
We begin by considering conditions in a typical segment of the tube between x and x + Ax (Fig. 6.6). The
concentration of the solution entering the segment is y(x); the concentration of the solution leaving the segment is
y(x + Ax).
In the time Af that it takes the solution to move through the segment, an amount of solute equal to
concentration x volume = y{x)nr
2
Ax
enters the left end of the segment, and the amount
y(x + Ax)rcr
2
Ax
The difference, [y(x) — y(x + Ax)]rcr Ax, must have left the segment by
diffusion through the wall of the tube. The expression for this amount, as given by Frick's law, is
2
leaves the right end of the segment.
Rate of diffusion x time = k(2nr Ax)[y(x + 9 Ax) — c] Af
where
x + 0Ax,
for
< 9 < 1,
is
x + Ax
a typical point between x and
at which to assume an "average"
value of the concentration, and c, assumed constant, is the concentration of the solute in the fluid surrounding the
tube.
Equating the two expressions we have found for the loss of solute by diffusion, we have
[y(x) - y(x + Ax)]7tr
«j/y\
Since
Ax = t>Af,
+J y
2
Ax = k(2nr Ax)[ y(x + 9 Ax) - c] Af
_1_
Av\
Ax
the system were continuous), we get
"JL
-=—
this simplifies to
[y(x + 0Ax) — c]
and, taking limits (as though
rv
dy
—dx
—=—
\y{x)
— c ^ 0,
and we can solve this equation by separating variables to
Ik r
— c\.
rv
y > c;
2k
By hypothesis,
dy
obtain
y — c
we find that
=
hence
dx.
y{x)
Integration then gives
In (y
2k
— c) =
In
B = ln(y — c),
x + In B.
Putting
x =
or
y = c + (y
and
y = y
,
rv
rv
and the solution becomes
\i
r
y
-c
In
=
2k
x,
- c)e~ 2kxlrv
.
rv
Ax = v At
I
y(0) = y«
ii I I
y(x + Ax)
y(x)
y(L)=y L
I
xtttttx Ax
+
x =
X=L
Fig. 6.6 Solve diffusing from a tube through which a solution is flowing.
ELECTRIC CIRCUIT PROBLEMS
6.104
An RL circuit has an emf of 5 V, a resistance of 50 Q, an inductance of 1 H, and no initial current. Find the current
in the circuit at any time f.
.
—
.
APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS
129
I Here £ = 5, R = 50, and L=l, so (7) of Problem 1.87 becomes dl/dt + 50/ = 5. Its solution is
- 50
= 0, 7 = 0; thus,
/ = ce
+ 1 o (see Problem 5.31). At
= aT 50(0) +
The
or c=-&.
50
—
current at any time is then
/ =
+ To3V"
3
'
t
,'„,
'
t
-50
The quantity --foe
in this result is called the transient current, since this quantity goes to zero ("dies
t -* oo.
The quantity jq is called the steady-state current. As t -* oo, the current / approaches
out") as
'
the value of the steady-state current.
6.105
An RL circuit has an emf given (in volts) by 3 sin 2f, a resistance of 10ft, an inductance of 0.5 H, and an initial
Find the current in the circuit at any time t.
current of 6 A.
R = 10,
E = 3>sm2t,
L = 0.5, so (7) of Problem 1.87 becomes dl/dt + 20/ = 6 sin It.
— ce~ 20 + i^sin It — t^j cos It. At
= 0, / = 6; hence,
=
=
=
6
ce~
+ j$[ sin 2(0) t^cos 2(0) or 6 c y^, whence c f§f The current at any time is
20r
then 7 = fgfe+ 1^r sin2r- T^T cos2f.
# Here
Its solution,
20<0)
and
from Problem 5.51, is
'
I
t
t
.
As in Problem 6.104, the current is the sum of a transient current, here fofe
t$i sin It
6.106
-20
and a steady-state current,
',
— y^x cos 2r.
A sin (It — (p).
Rewrite the steady-state current of Problem 6.105 in the form
The angle
<ft
is
called the phase
angle.
I
A sin (2f — (p) = A(s\n 2t cos
Since
^
A cos <p =
2
(tw)
tan <p —
,
and
/I sin
Acos<p
6.107
= ygy.
+ (ttjt) 2 = ^ 2 cos 2
=—
30/101
= /4 2 (cos 2
+ sin 2
(/>)
= A2
Consequently, 7 5S has the required
form if
n
•"
10
cos 2t
..,,., A —
,
.
</>
now follows that
It
+ A 2 sin 2
1
— A sin
sin 2t
/
909
V( 101 r
Viol
Determine the amplitude and frequency of the steady-state current in the previous problem.
I
6.108
/I sin <ft
</>
—
10
= arctan
and
and
3/101
=
— cos 2f sin 0), we require
= j£i sin It — ytn cos 2t = A cos
7S
Thus, we have
</>
r—
A — 3/v 101,
The amplitude is
/=
and the frequency is
—=—
2
1
27t
n
A resistor of 15 Q and an inductance of 3 H are connected in series with a 60-Hz sinusoidal voltage source having
amplitude 110 V. Find an expression for the steady-state current at any time t if initially there is no current in
the system.
R = 15,
I Here
L = 3,
E = Il0sm2n(60)t = UOsmUOnt.
and
dl/dt + 5/ = (110/3) sin 1207tr;
its
/
= 22
- 24n cos
=
2
1 + 576/r
sin 1 207rt
3
When
r
= 0,
7
= 0.
Then
c
=
and
'
,„
+ 576tc 22x
-» oo,
-»•
7
;
3
6.109
1
+ 576tc
/
Since
_ 5,.'
—
,
.
,
.
1
5'
-=
2
.
+ 576/c
,
which is the steady-state current.
Rewrite the steady-state current of the previous problem in the form
i
+ ce
3
,
2
20nt
=—
)
22 sin I20nt - 247rcos 1207rf
t
1
22 sin 1207rt - 24rc cos 1207tt -I- 24ne~
22(24tt)
3(1
As
Thus, (7) of Problem 1.87 becomes
solution, from Problem 5.53, is
Asin(l20nt - </>) = A sin 1207rf cos(p - A cos I20nt sin <p,
A sin (1207tt — <p).
we must have
.4 cos (ft
=
22
^+ ^
3(1
and
/I sin
6 —
(22)(24tt)
=r.
3(1
+ 576tc 2
2
22
3(1 4- 576tt
It
)
now follows that
)'
(22)(24tc)
+
2
g
576rr
3(1
+ 576tc
= A 2 cos 2
2
)
(f>
+ A 2 sin 2 (p = A 2
or
A=
22
3>/l + 576tt
2
2
)
CHAPTER 6
130
.
tan <p =
and
6.110
—
2
A sin $
22(24tt)/3(1 + 576ti
=
= 24n
A cos
22/3(1 + 5767T)
)
= arctan 24;r = 1.56 rad
or
Determine the amplitude and frequency of the steady-state current of the previous problem.
m
A =
The amplitude is
6.111
—
—
")0
3>/l
+ 576tc 2
1
« 0.097,
,
Of\
f =
while the frequency is
= 60.
2tt
Determine the period of the steady-state current in Problem 6.109.
I The period is the reciprocal of the frequency. The results of the previous problem show that the period is 1/60.
6.112
The steady-state current in a circuit is known to be
— yiCOS
py sin t
Rewrite this current in the form
t.
A sin (t — 4>).
I
A sin (f —
Since
follows that
tan cp =
(^)
sin
6.113
= A sin cos
+ (fV) 2 = A 2 cos 2
t
<f>)
A sin
=
cos0
2
— A cos sin 0, we must have A c os = -^
+ A 2 sin 2 - /I 2 so A = y/2/VJ. Also,
t
(f>
= 3/17 = 3
Acoscp
5/17
<f>
3
or
= arctan - = 0.54 rad.
q>
5
A cos (t —
Since
Then
1
-I-
I
—
= A cos
= A 2 cos 2
(f)
+ A 2 sin 2 (p = A 2
—
=
—
sin<p
= Asincf) =
coscf)
Acoscp
5/17
—3/17
sin
= iV
It
then
—
5
cos <p + A sin t sin <f>,
t
J
J
tan (/> =
<f>)
/I
r—
V2/17 sin (f — 0.54).
,
The current is
A cos(r — <p).
Rewrite the steady-state current of the previous problem in the form
I
and
,
5
,
so
(p
3
follows that
it
,
so
-4
A=yJ2/l7,
cos -/> =
as before.
,
„,
— arctan — f)
= —1.03
rad.
,
,
v
— -pj
and
Asincp—j^.
Now, however,
T1
,
The current is
(
72/17 cos (f + 1.03).
6.114
Determine the amplitude and frequency of the steady-state current in the previous three problems.
The amplitude is
6.115
A = yjl/ll,
—
2n
(the numerator of/ is the coefficient of t).
Determine the period of the steady-state current in Problem 6.112.
I From the results of the previous problem, we have
6.116
/=
while the frequency is
period = 1//= 2n.
An RL circuit with no source of emf has an initial current given by /
— —L =
In this case (/) of Problem 1.87 becomes
/
H
0.
Find the current at any time t.
.
Its solution (see
Problem 5.8 with / replacing N and
at
k=-R/L)
1
is
function of time,
6.117
= ce~ {RIL)x At
{R u
/ = I e~
.
t
= 0,
/
= /
;
hence
I
= ce~ {RlL)i0) = c,
and the current is, as a
'.
Determine the steady-state current for the circuit described in the previous problem.
f As
t
-> oo,
/
tends to zero. Thus, when the steady state is reached, there is no current flowing through the
circuit.
6.118
Determine the current in a simple series RL circuit having a resistance of 10 fl, an inductance of 1.5 H, and an emf
of 9 V if initially the current is 6 A.
f
Here
w=
E(t)
9,
R - 10,
and
L=1.5.
Then (7) of Problem 1.87 becomes
'
f.
j
6.119
.
/
dt
= ^ + ar (20/3) (see Problem 5.33). At t = 0,
—
_ 51 ThlK
4- 5i^-(20/3)f
/ —
r — o _
— _9_
— -2inus,
c
10
10 e
10 —
/
dl
20
—
+— =
i
7-6;
hence
6 =
^ + c^-
,
6;
its
.
3
(20/3)(0)
so that
,
-t-
Identify the transient component of the current found in the previous problem.
I
As
t->oo,
7^7, = ^.
The transient component is
= 7 - 7 - ^ + f^e" ,20 3) - & = f^" (2 ° 3 ".
'
7,
'
S
.
solution is
APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS
6.120
131
An RC circuit has an emf (in volts) given by 400 cos It, a resistance of 100 Q, and a capacitance of 10" 2 F.
Initially there is no charge on the capacitor.
Find the current in the circuit at any time t.
f We first find the charge q on the capacitor and then the current using the formula / = dq/dt. Here
E = 400 cos It, /? = 100, and C = 10" 2 Then (7) of Problem 1.90 becomes dq/dt + q = 4 cos It, its
.
q = ce
solution is
At
t
= 0,
4
= ce <0) + - sin 2(0) + - cos 2(0),
hence,
5
= — = - e _, H
dq
.
5
Problem 5.52).
(see
8
q = 0;
— -4 e _, + -8 sin „it + 4- cos 2t
q —
6.121
4
+ - sin It + - cos It
8
_,
,
and
/
4
dt
5
so that
16
—
4
=
.
Thus
8
cos it
5
c
sin It.
5
5
A resistor R = 5Q and a condenser C = 0.02 F are connected in series with a battery E — 100 V.
=
the charge on the condenser is 5 C, find Q and the current /for
> 0.
t
# With £=100, R = 5, and C = 0.02, (7) of Problem 1.90 becomes d<j/df + \0q = 20;
= 2 + ce" 10 (see Problem 5.30). At
= 0,
= 5; hence, 5 = 2 + ce~ 10(0) and c = 3.
i0
10
and / = dq/dt = -30e"
q = 2 + 3e~
'
t
<?
<j
its
solution is
Thus
'.
'
6.122
If at
t
Specify the steady-state and transient components of the current found in Problem 6.120.
f The current is I = fe"' + ^cos 2r — f sin 2t. As r -» oo, the current approaches the steady-state value
The transient component is I, — I — I s = fe~
7 S = ^cos It — f sin 2t.
r
.
6.123
Determine the amplitude, frequency, and period of the steady-state current of the previous problem.
f
A -
The amplitude is
/7l6V
/(
—
V\
6.124
5
—8V = —
7
8
+(
/V /V5
1
p;
I
the frequency is
/=
5
Rewrite the steady-state current in Problem 6.122 in the form
A cos(2r +
—
=
27T7T
2
1
1
—,
and the period is
Since
A cos (2f +
</))
= A cos It cos
<j>).
</>,
</>
71.
/
— /I sin 2t sin
we have A cos = ^ and /I sin $ = f
Asin<j)
r
= 8/5 — -, so
=
previous problem, we know the amplitude A is 8A/5. Also, tan
I
—=
.
From the
1
A cos
6.125
= arctan - = 0.4636 rad.
Thus
h = -7= cos (2f + 0.4636).
capacitor.
Find an expression for the charge on the capacitor at any time t.
I Here
E{t)
= 5,
Jg + ce" 10(
q = ^(l+99e-
R = 10,
C = 0.01,
and
(see Problem 5.34).
10
At
t
so (7) of Problem 1.90 becomes
= 0,
q = 5;
hence
5
q + lOq = £; its solution is
so that c = f§.
Thus,
= ^ + ce - 10(0)
,
').
Determine the current flowing through the circuit described in the previous problem.
#
f
_dq __99
0t
2
dt
6.127
2
An i?C circuit has an emf of 5 V, a resistance of 10 Q, a capacitance of 10" 2 F, and initially a charge of 5 C on the
q =
6.126
6/5
8
1
4>
1
Find the charge on the capacitor in a simple 7?C circuit having a resistance of 10 Q, a capacitance of 0.001 F, and
an emf of 100 sin 1207rf V, if there is no initial charge on the capacitor.
I Here
E{t)
= 100 sin 1207rt,
q
^ + 100(7
^ = 10 sin 1207xr,
A.
q=
I. Q,
<,
= 0;
hence
7?
=
100 + 144tt
C = 0.001.
and
q =
and its solution is
lOsin 1207rf - 127rcos 1207tf
2
= 10,
~
J |^ +
2
3tt
25 + 36n
A,
_ 100t
2
*
Then (7) of Problem 1.90 becomes
— + ——
10 sin 1207rf - J In cos 1207rt
or
2
100
1447T
A =
^^.
+ Ae
Then
100r
(see
Problem 5.54).
—
.
CHAPTER 6
132
6.128
Determine the steady-state current in the circuit described in the previous problem.
#
_,
,
The current is
=
I
dq
—
=
tft£%
10cosl207rt + 127rsinl207rt
1207T
= 0,
E(t)
fc
= - l/RC).
At
q H
RC
=
g
Q
= 0,
t
<?
:
= 0.
Q = ce~
so
q = ce~'
solution is
Its
— e~' RC
RC
=
/
t
-> oo,
/
= /
-»
s
l0) RC
= c,
RC
and
(see
Problem 5.8
q = Q e~'
RC
.
.
.
Find an expression for the charge on the capacitor of a simple series LC circuit (consisting of an inductor and a
and there is no initial current in the circuit.
capacitor only) if the initial charge on the capacitor is Q
Applying Kirchoffs loop law (see Problem 1.81). we have
L
dl
d^Q
—
= L
dt
dt
for the potential
d
drop across C. Then
d 'Q
,
dt
or
Q
LI dl + -dQ = 0.
Since
I
dt
/
=
dl
dl
dQ
dl
dt
dQ dt
dQ
= —- = ±
LC
dt
Q= Q
yjQl — Q
2
,
.
f
= 0,
Q
,
.
,
Q
=
—
—
dQ
C
dl
LI -
n
I
2
Substituting for C, and solving for / yield
which is separable. Integration then yields
dQ
a\L
7
Vel^o
for
u
L
L
so that the last equation
becomes
M
C, = Qq/2C.
we have
:
Q=Q
for the potential drop across L and
{LI 2 + =— = C,
,
f
Since
— = /—
Integration then yields
when
1
2
2
+ =: = 0.
L
—Q- = /, we Lhave —-=- = -- = -
Since
we find
dt
ai
= + C
or
,
sin
JLC
C 2 — rc/2.
Q
-il -—
-
= H
f
,
LC
Qo
Thus, we have
== + rC-,
=-+
sin
Go
or
LC
Q =Q
cos
\LC
Find the amplitude, period, and frequency of the charge in the previous problem.
f The amplitude is Q
,
the period is 27r N LC, and the frequency is the reciprocal of the period,
l/2n\LC.
Determine the current in the circuit described in Problem 6.132.
Since
6.135
x,
Determine the steady-state current in the previous problem.
I As
6.134
->
Find the current in the circuit described in the previous problem.
By differentiating the result of that problem, we get
6.133
t
.
of Problem 1.90 becomes
(/)
with q replacing N and
6.132
As
Find the charge (as a function of time) on the capacitor in a simple RC circuit having no applied electromagnetic
With
6.131
1001
100 + 144ti 2
force if the initial charge is Q Q
6.130
-^
25 + 36;r 2
10 cos \20nt + 127rsin 120;rf
/-/.= 1207T
6.129
3007T
r
100 + 144tt 2
dt
Q— Q
dQ
cos
VLC
dt
Qo
LC
t
sin
.LC
Determine the amplitude, frequency, and period of the current in the previous problem.
# The amplitude is g /VLC, while the frequency and period are identical to those of the charge (see Problem 6.133).
MECHANICS PROBLEMS
6.136
Derive a first-order differential equation governing the motion of a vertically falling body of mass m that is
influenced only by gravity g and air resistance, which is proportional to the velocity of the body.
APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS
133
f Assume that both gravity and mass remain constant and, for convenience, choose the downward direction as
the positive direction.
Then by Newton's second law of motion, the net force acting on a body is equal to the time
rate of change of the momentum of the body; or, for constant mass,
F=m
—
,
where F is the net force on
dt
the body and v is the velocity of the body, both at time t.
For the problem at hand, there are two forces acting on the body: (1) the force due to gravity given by the
weight w of the body, which equals mg, and (2) the force due to air resistance given by
— kv, where
k >
is
a
constant of proportionality. The minus sign is required because this force opposes the velocity; that is, it acts in
The net force F on the body is, therefore,
the upward, or negative, direction (see Fig. 6.7).
F = mg — kv,
so
that we have
mg — kv = m
—
dv
or
dt
dv
k
—
+ -v = g
m
U)
dt
as the equation of motion for the body. If air resistance is negligible, then
|
|
k =
and (7) simplifies to
dv/dt = g.
Falling Body
mm
f Positive x direction
6.137
pj g# ^j
A body of mass 5 slugs is dropped from a height of 100 ft with zero velocity. Assuming no air resistance, find
an expression for the velocity of the body at any time t.
I Choose the coordinate system as in Fig. 6.8. Then, since there is no air resistance, (7) of Problem 6.136
= 0, v =
(initially the body
becomes dv/dt = g; its solution is v = gt + c (see Problem 5.41). When
= g(0) + c, so that c = 0. Thus, v = gt or, for g = 32 ft/s 2 v = 32f.
has zero velocity); hence
t
,
/*k
Body
Fal
Falling
~\0
Ground
x = 100
Positive x direction
6.138
Fig. 6.8
Find an expression for the position of the body in the previous problem at any time t.
f
It then follows from the result of the
Position [as measured by x(t)] and velocity are related by v = dx/dt.
Integrating both sides of this equation with respect to time, we get
dx/dt = 32t.
previous problem that
134
CHAPTER 6
D
x = 16t 2 + c.
x =
But
at
t
=
x = 16r
Thus, the position of the body at any time t is
6.139
= 16(0) 2 + c
(see Figure 6.8), so we have
= 0.
Determine the time required for the body described in Problem 6.137 to reach the ground.
Here
m=2
this differential
With
q — 32,
or
= 128 - 128e"' /4
= — 128.
c
v
f
hence
Find an expression for the velocity
—\v lb.
(/) of Problem
equation is given in Problem 5.35 as
= 128 + ce°,
v
k — j.
2
and
x — 100;
We require when
A body of mass 2 slugs is dropped from a height of 800 ft with zero velocity.
of the body at any time t if the force due to air resistance is
6.141
c
.
f The position of the body is given by the result of the previous problem.
we have 100 = 16f 2 or t = 2.5 s.
6.140
from which
2
—- + - v — 32.
6.136 becomes
— 128 + ce~ xlx
.
At
f
dt
4
= 0,
v
= 0:
The solution to
hence
The velocity at any time t (before the body reaches the ground) is then
.
Find an expression for the position of the body described in the previous problem.
1
Since
v
= dx/dt,
follows from the result of the previous problem that
it
dx
=
—
128 — 128?
"4
Integrating
.
dt
x = 128f + 512e~"
directly with respect to time, we obtain
4
+ c.
Since the positive x direction is downward (see Problem 6.136), we take the origin to be the point at which
the body was released; then the ground is at
c
— —512.
x = 800.
At
t
= 0,
x = 0:
so
= 128(0) + 512e° + c,
or
The position of the body at any time t (before it reaches the ground) is then
x= -512+ 128f + 512* '\
6.142
Find an expression for the limiting (or terminal) velocity of a freely falling body satisfying the conditions of
Problem 6.136.
I The limiting (or terminal) velocity is that velocity for which
(/)
— =
of Problem 6. 136. we find
-\
q
or
v leT
k ^ 0.
If
k
i
dv/dt = 0.
Substituting this requirement into
= mg/k.
in
This equation is valid only when
dv dt —
case, the condition
:
= 0,
then (/) of Problem 6.136 becomes
dv/dt = g.
In that
cannot be satisfied; thus, there is no limiting velocity in the absence of air
resistance.
6.143
Determine the limiting velocity of the body described in Problem 6.140.
I
With
m = 2,
k = \,
and g = 32. we have r [cr = 2(32)/i = 128 ft/s.
Problem 6.140 also tends to v = 128.)
velocity derived in
6.144
(Note that as
t
-» oo,
the
A mass of 2 kg is dropped from a height of 200 m with a velocity of 3 m/s. Find an expression for the velocity
of the object if the force due to air resistance is
I
Here
m=2
k = 50.
and
— 50r N.
a = 9.8,
With
(7) of Problem
6.136 becomes
dv
—
+ 25u =
Its
9.8.
solution
dt
is
i=0.392 + ce" 25
(see
'
Problem 5.36). At
t
= 0,
v
= 3;
6.145
3 = 0.392 + ce°
or
— 0.392 + 2.608e~ 25
hence
velocity at any time t (before the mass reaches the ground) is then
v
c = 2.608.
The
'.
Determine the limiting velocity for the object described in the previous problem.
f
Here we find
r ter
= mg/k — 2(9.8) 50 = 0.392 m/s.
which may also be obtained by letting
t
-* oo
in the
result of Problem 6.144.
6.146
A body weighing 64 lb is dropped from a height of 100 ft with an initial velocity of 10 ft/s. It is known that air
resistance is proportional to the velocity of the body and that the limiting velocity for this body is 128 ft/s.
Find
the constant of proportionality.
I Here
6.147
mg = 64
and
i?
ler
= 128.
It
follows from the result of Problem 6.142 that
128 = 64/fc,
or
k = \.
Find an expression for the velocity of the body described in the previous problem at any time f before that body
reaches the ground.
APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS
mg = 64.
I With
it
m = 2 slugs.
follows that
dv
— + -4 =
Problem 6.136 becomes
We have from the previous problem that
= 0,
6.148
t'
= 10;
v
which has as its solution
32,
10 = 128 + ce°,
hence
so (/) of
1
v
dt
f
k = \.
135
or
c
= 128 + ce~' /4
= — 118. The velocity is then
t;
Problem 5.35). At
(see
= 128 - 118e~" 4
.
A body of mass 10 slugs is dropped from a height of 1000 ft with no initial velocity. The body encounters air
resistance proportional to its velocity.
limiting velocity is known to be 320 ft/s, find the constant of
If the
proportionality.
I With
6.149
m=2
and
vt „
= 320,
320 = 2(32)/fe
the result of Problem 6.142 becomes
fc
= £.
Find an expression for the velocity of the body described in the previous problem.
With
m — 2,
k
— j,
and
g = 32,
(/) of Problem
6.136 becomes
— + — v = 32;
= 320 + ce-' no
v
(see Problem 5.37).
At
t
= 0,
v
= 0;
v
solution is
10
= 320 + a?
= 320(1 — e~" 10
hence
velocity at any time r (before the body reaches the ground) is
its
-
dt
6.150
or
,
so
c
= -320.
The
).
Find an expression for the position of the body described in Problem 6.148.
I
Using the result of the previous problem along with
v
— dx/dt,
we may write
dx
— = 320 —
320c?
r/1 °.
dt
x = 320t + 3200e
Integrating directly with respect to time, we obtain
-,/1 °
+ c.
Since the positive direction is
assumed to be downward (see Problem 6.136), we take the origin to be the point where the body was released
= 320(0) + 3200e° + c so that c=-3200.
and
(and the ground as x = 1000). Then x =
at
f = 0,
~'
The position of the body is then x = 320f - 3200 + 320c- /10
.
6.151
Determine the time required for the body described in Problem 6.148 to attain a speed of 160 ft/s.
f
e
6.152
Substituting
-r/io
=i
Then
v
= 160
f
into the result of Problem 6.149 gives us
160 = 320(1 - e~" °),
l
from which
= _io In | = 6.93 s.
A body of mass m is thrown vertically into the air with an initial velocity v
If the
.
body encounters air
resistance proportional to its velocity, find the equation for its motion in the coordinate system of Fig. 6.9.
ii
Positive x direction
Rising Body
* =
Fig. 6.9
I There are two forces on the body: (1) the force due to gravity given by mg and (2) the force due to air
resistance given by feu, which impedes the motion of the body. Since both of these forces act in the downward or
negative direction, the net force on the body is
-mg -
m — = -mg - kv
feu.
Then from Newton's second law of motion,
dv
we have, as the equation of motion,
or
dv
—
+ -v=-g
m
k
dt
(/)
136
6.153
CHAPTER 6
D
Find an expression for the velocity of the body described in the previous problem.
v
v
=v
;
hence
= \v n +
v
mg
= C e~
(klm)0
— (mg/k),
or
c
=
= ce _( *
— mg/k. At
+ (mg/k). The velocity of the body at any time
I The solution to (J) of the previous problem is given in Problem 5.37 as
t-
v
t
= 0,
t is
then
/ 'n),
mg
(klm)t
'
k
6.154
Find the time at which the body described in Problem 6.152 reaches its maximum height.
# The body reaches its maximum height when
— 0.
v
Substituting this value into the result of the previous
problem, we obtain
!5).-«-._a
;<b+
-(* mil
or
_
1
1
Taking the logarithms of both sides gives us
6.155
m
1
= In
f
+ v k/mg
from which we find
-,
1
+ v k/mg
An object is thrown vertically upward from the ground with initial velocity 1960 cm/s.
t
— — In
(
k
\
+
1
—
mg
Neglecting air resistance,
find its velocity at any time t.
which may be integrated directly to yield
k = 0,
dv/dt — —g,
(7) of Problem 6.152 becomes
= — gt + c. At
= 0, v = 1960; hence we have 1960 = — g(0) + c = c. With this value of c and
I With
v
t
g = 980 cm/s
6.156
2
the velocity becomes
,
v
= 1960 - 980f.
Determine the total time required for the object described in Problem 6.155 to return to the starting point.
I
We require
f
x = 0;
x = 0.
when
dx/dt = 1960 - 980f.
Since
v
— dx/dt,
it
follows from the result of the previous problem that
x = 1960f - 490f 2 + c. At
and the position of the object is given by
Integrating this equation with respect to t, we get
= 1960(0) — 490(0) + c — c,
2
hence we have
x= 1960r-490f
Setting
x =
in (/) and solving for r, we
time needed for the object to return to the ground is
6.157
t
= 0,
2
U)
= 1960f - 490f 2 = 490f(4 - t),
obtain
t
so that
t
=
or 4. The
— 4 s.
Determine the maximum height attained by the body described in Problem 6.155.
I The maximum height occurs when v = 0. Substituting this value into the result of Problem 6.155, we get
= 2 s. For that value of (7) of the previous problem yields
= 1960 — 980f, or
t,
f
x = 1960(2) - 490(2) 2 = 1960 cm.
6.158
A body of mass 2 slugs is dropped with no initial velocity and encounters an air resistance that is proportional
to the square of its velocity.
Find an expression for the velocity of the body at any time r.
The force due to air resistance is —to 2 so that Newton's second law of motion becomes
,
m— = mg — to
dt
or
2
— = 64 — to
2
.
Rewriting this equation in differential form, we have
64 -to 2
dt
separable.
dv — dt = 0,
which is
By partial fractions.
1/8
64 -to 2
so our differential equation can be rewritten as
yjk
8
which can be rewritten as
In
In |8
- v to
8 + yfkv
8
Integration gives
+ y/kv J
- Vto| + 4- 1" |8 + y/kv\ - = c
t
J
y/k
+ Vto =
8Vto + Sy/kc
8 -y/kv
1/8
+
=- dv — dt — 0.
=- H
-(
8 \8
~
8 - y/kv
(8 - Vto)(8 + y/kv)
S
+ y/kv
or as
= c e*^\
i
8 -y/kv
where
cr
= ±e 8>fcc
2
.
=
.
APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS
At
r
= 0,
we are given
= 0.
v
This implies
c,
= 1,
D
-^F^ = e 8v '"
%-yJkv
and the velocity is given by
137
or
o
i>
6.159
= —pr tanh4y/kt.
Without additional information, we cannot obtain a numerical value for the constant k.
A 192-lb object falls from rest at time
=
t
in a medium offering a resistance of 3v 2 (in pounds).
Find an
expression for the velocity of this object.
^
From Newton s second law,
,
.
.
~
m dv/dt = mg — 3v z
.
,
->
192 dv
,
= 192 - 3v 2
so that
,
dv
,
or
2
dt
g
— = 64 -
v
2
dt
Then, separating variables and integrating, we get
dv
y^^ = h
C
Since
6.160
=
v
at
t
we find that
c
1
T6
Then
8^;
8 + v
—
16
1
— 0.
8 + v
,
ln
"'
In
=
t
+C
2
e
t
= -,
8-i?
and
v
2
M - e~ Xt
= 8 —4
e
tt.
'
+ e _4
'
Find an expression for the position of the object described in the previous problem.
I
Since
v
— dx/dt,
x =
we integrate the result of Problem 6.159 subject to
x = 2 In
distance traveled,
6.161
= 0,
c dt
at
t
= 0,
to fird, for the
= 2 In cosh 4f
Determine the limiting velocity for the object described in Problem 6.159.
*t
_ e -*t
lim 8 —r-t
zrr,
#
e
The limiting velocity is
r-oo
+e
t^oo
l- e 8t
1
+e
—r. = 8 ft/s,
which can also be obtained by setting
2
dv
„„
—
= 32 - — =
2
v
e
~
= lim 8
0.
dt
Observe that the result of Problem 6.142 is not valid here, because the resistance of the medium is not
proportional to the velocity of the object.
6.162
It is
A boat of mass m is traveling with velocity v
proportional instead to the square of the velocity.
At
.
r
=
the power is shut off.
Assuming water resistance
proportional to v", where n is a constant and v is the instantaneous velocity, find v as a function of the distance
traveled.
f
Let x be the distance traveled after time
dv
so we have
m — = — kv",
t
> 0.
The only force acting on the boat is the water resistance,
where k is a constant of proportionality. Then we have
dt
dv
m—=m
6.163
dv dx
= mv
dv
—
= —kv. which we
dx
dt
dx dt
Case 1,
n#2:
With
Case 2,
n = 2:
Again with
v
=v
v
at
x = 0,
=v
at
.
write as
mv
,
_„
dv=—kdx.
,
integration gives
x = 0,
v
2 ~"
= vl~"
integration now yields
m
v
(2
— n)x.
= v e~ kx,m
.
A ship weighing 48,000 tons starts from rest under the force of a constant propeller thrust of 200,000 lb. Find
its
velocity as a function of time t, given that the water resistance in pounds is 10,000t>, with v the velocity
measured in feet/second. Also find the terminal velocity in miles per hour.
mass (slugs) x acceleration (ft/s ) = net force (lb) = propeller thrust — resistance,
dv
v
20
48,000(2000) dv
,
= 200,000 - 10,000i;, from which
+ - = -. TIntegrating gives
I
2
Since
32
dt
—
^r/300
—
.
(V /30 ° dt = 20e" 300 + C.
300 J
300
,l30 ° =
20(1 - e-"
v = 20 - 2Qe-
As
t
-* oo,
v -> 20;
Because
,
v
—
—
—
dt
300
300
=
when
.
= 0,
we have
C = -20,
so that
).
the terminal velocity thus is
the differential equation with
t
we have
dv/dt -* 0.
20 ft/s = 13.6 mi/h.
This may also be obtained from
138
6.164
CHAPTER 6
D
A boat is being towed at the rate of 12 mi h.
At the instant
= 0)
(f
that the towing line is cast off, a man in
the boat begins to row in the direction of motion, exerting a force of 20 lb.
If the combined weight of the man
and boat is 480 lb and the resistance (in pounds) is equal to 1.75t\ where v is measured in feet/second, find the
speed of the boat after \ min.
mass (slugs) x acceleration (ft/s 2 ) = net force (lb)
forward force — resistance, we have
480 dv
dv
7
4
It 60 J, _ SOgll 60
71/60 _ 4 f
= 20 - 1.75r from which
l
+
~ 3" Integrating gives
I
Since
Q
,
'
32 dt
When
v
0,
f
=
Now when
= 30,
t
88
12(5280)
(60)
6.165
60
dt
v
216
C=
so that
2
Then
.
g
=^+2
v
e
- ltl60
.
35 = M.6ft/s.
= W + 216
35 e~
A mass is being pulled across ice on a sled, the total weight including the sled being 80 lb. The resistance
offered by the ice to the runners is negligible, and the air offers a resistance in pounds equal to five times the
Find the constant force (in pounds) that must be exerted on the sled to give it a
velocity (v ft/s) of the sled.
terminal velocity of 10 mi/h, and the velocity and distance traveled at the end of 48 s.
I
mass (slugs) x acceleration (ft/s 2
Since
80 dv
dv
= F — 5v,
dt
F
— — ce
v
f 2v == - F,
or
32 dt"
When
F
_
5
"* l,er "
10(5280)
t
= 0,
6.166
we have
where F (in pounds) is the forward force. Integrating then yields
r
=
so that
c
F
=
The required force thus is
.
F
and
44
=—
"(60)
v
= net force (lb) = forward force — resistance,
5
F =
(\-e
v
As
-').
t
»,
220
lb.
3
Substituting this value for F gives us
we have
)
= ^(1 - e~ 9b = ^ ft/s.
)
r
= ^(l— e
So, when
*')•
The distance traveled is
s
t
— 48,
= j^ 8 vdt = ^ $ s (1 - e~ 2, )dt = 697
ft.
A spring of negligible weight hangs vertically. A mass of m slugs is attached to the other end. If the mass is moving
with velocity v
ft/s
when the spring is unstretched, find the velocity v as a function of the stretch x in feet.
I
According to Hooke's law, the spring force (the force opposing the stretch) is proportional to the stretch. Thus,
^ dx
dv
dx
dv
= mg — kx, which we can write as m
— V.
= mv
= mg — kx, since
we have m
dx dt
dx
dt
dt
—
,
,
mv 2 = 2mgx — kx 2 + C.
= 2mgx — kx + mv 2
Integrating then gives
mi'
6.167
2
——
—
=v
when
l"
,
Now
v
.
x =
so that
C = mr 2
and
,.
2
,.
A parachutist is falling with speed 176 ft s when his parachute opens. If the air resistance is Wv 2 256 lb, where
W is the total weight of the man and parachute, find his speed as a function of the time after the parachute opens.
f
-
f
v
2
Since
net force on system = weight of system — air resistance,
dv
dt
- 256
T'
Integrating between the limits
r
= 0.
dv
J 1^6 r
from which we get
v= 16
In
= 176
—
1
dt
-256
v- 16 ~_
v+ 16
r
r.
4t.
or
we have
and
r
- 16
v
+ 16
In
32
f
Exponentiation then gives
Wv 2
g dt
256
from which
t.
V
= V
gives
t
v- 16 5
e
v+ 16 ~6
or
6 + 5e-
6-5e" 4
''
Note that the parachutist quickly attains an approximately constant speed
6.168
Wdv
— the terminal speed of 16
ft/sec.
A body of mass m slugs falls from rest in a medium for which the resistance (in pounds) is proportional to the
square of the velocity (in feet per second).
If the terminal velocity is
150 ft s. find the velocity at the end of 2 s and
the time required for the velocity to become 100 ft/s.
I
Let v denote the velocity of the body at time f.
net force on body = weight of body — resistance,
Then we have
dv
and the equation of motion is
m — — mg — Kxr.
Some
APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS
K = 2mk 2
simplification is possible if we choose to write
— = 2(l6-k
dv
,
dv
,
2
v
2
dt
= 0,
c=—
we find
Then the equation of motion reduces to
.
—
kv — 4
=
Inkv + 4
,
Integrating now gives
f
139
= -2dt.
-J-: —
/rtr
16
or
)
U
from which
+4
fcu
— —+ 44 =—e~
fcu
and
1
kv — 4
— 16/ct + lnc,
16kt
7
Also,
.
i;
=
ibk '.
= ce
Since
= — e -043
when
2
50
1
for
t
-* oo,
so
=-
fc
and our
75
/a?
solution becomes
=
i;
'.
t>+ 150
~ 15 ° = -e' 086 = -0,423 and y = 61 ft/s.
y+150
-0
= 3.7
u=100, e 43 = 0.2 = e -16 so
When
When
6.169
V
= 2,
t
-
'
s.
r
,
A body of mass m falls from rest in a medium for which the resistance (in pounds) is proportional to the velocity (in
feet
per second).
of the medium is one-fourth that of the body and if the terminal velocity
If the specific gravity
24 ft/s, find the velocity at the end of 3 s and the distance traveled in 3 s.
is
I
Let v denote the velocity of the body at time t. In addition to the two forces acting as in Problem 6.136, there is
a third force which results from the difference in specific gravities. This force is equal in magnitude to the
weight of the medium which the body displaces, and it opposes gravity. Thus, we have
net force on body = weight of body — buoyant force
A
1
1
— resistance,
mg — Kv = - mg — Kv.
2
m— = mg
dv
dv
— = 3(8 —
or
kv)
dt
—
In (8
- kv)\" = 3f
I'
|o
|o
k
v
= 24,
so
Since
i;
= 3 dt.
8 — kv
and
fe=l/3
dx
Integrating from
= 24(1 - e~
v
= —- = 24(1 — e
l
Thus, when
).
we integrate between
'),
= 0,
t
K taken as 3mk, the equation becomes
and
v
—
-In (8 - kv) + In 8 = 3kt,
from which
,
g — 32 ft/s
With
and the equation of motion is
t
t
to
= 0,
v
,
—v
v
kv = 8(1 - e~
so that
= 3,
=t
t
gives
3kt
When
).
-* oo,
t
= 24(1 - e~ 3 = 22.8 ft/s.
x =
)
and
t
= 3,
x = x
to find
dt
xfo = 24(r
6.170
-I-
e~')\"
x = 24(2 + e~ 3 ) — 49.2 ft
or
as the distance traveled in 3 s
The gravitational pull on a mass m at a distance s feet from the center of the earth is proportional to m and
2
inversely proportional to s
(a) Find the velocity attained by the mass in falling from rest at a distance 5R from
the center to the earth's surface, where
R — 4000 mi is the radius of the earth, (b) What velocity would
correspond to a fall from an infinite distance; that is, with what velocity must the mass be propelled vertically
upward to escape the earth's gravitational pull? (All other forces, including friction, are to be neglected.)
.
I The gravitational force at a distance s from the earth's center is km/s 2 To determine k, we note that the force is
mg when s = R; thus mg = km/R 2 and k = gR 2 The equation of motion is then
.
.
—
dv
—
dt ds
ds
dsdv
dv
mgR~
=—
5—
2
m — = m — — = mv — =
1'
dt
i
,
,
or
ds
—
= —gR
—gR'-r-,
vdv —
2
the minus sign indicating that v increases as s
,
s
s
decreases.
(a)
Integrating from
- v 2 = gR 2 (--
\R
2
—
=
jRI
)
v
= 0,
- gR,
s
= 5R
so that
to
v
2
v
= v,
s
= R,
we get
= - (32)(4000)(5280).
cv
Then
vdv = —gR 2
v
t*R
ds
from which
-j,
= 2560^165 ft/s
or
5
5
approximately 6mi/s.
(b)
Integrating now from
which
6.171
v
2
= 2gR.
Then
v
v
- 0,
s -» 00
= 6400^33 ft/s
to
v
= v,
s
= R, we get
I
vdv = -gR
I
-^,
from
or approximately 7 mi/s.
A uniform chain of length a is placed on a horizontal frictionless table so that a length b of the chain dangles over
the side. How long will it take for the chain to slide off the table?
f Suppose that at time
t
a length x of the chain is dangling over the side (Fig. 6.10).
per unit length) of the chain is a.
Assume that the density (mass
Then the net force acting on the chain is ogx, and we have
dv
agx = aa—.
,
CHAPTER 6
140
Fig. 6.10
dv
dv dx
dv
—
= -—— = —
dx
dx
Now, since
dt
v
= 0,
we get
=
dx
—
= /v
= 0,
[a
a + Ja
V0
b
2
2
— b2
.
Integrating and using the fact that
Separating the variables, integrating again, and using
x + y/?
we get, finally,
/-In —
T=
yjx
.
x = b
when
x — b
when
a
a
dt
r
v
dt
v
gx
—
=—
dx
dv
this becomes
v
In
t.
Since the chain slides off when
x — a,
the time taken is
- b2
GEOMETRICAL PROBLEMS
6.172
Find the orthogonal trajectories of the family of curves
I
It
y = ex
2
.
follows from Problem 1.95 that the orthogonal trajectories satisfy the differential equation
This equation has the differential form
x dx + J 2y dy = c,
J
or
trajectories are ellipses.
\x 2 + y 2 — c,
xdx + 2ydy = 0,
which is separable.
Its solution is
=
—
dx
2y
.
.
which is the family of orthogonal trajectories. These orthogonal
Some members of this family, along with some members of the original family of
parabolas, are shown in Fig. 6.11. Note that each ellipse intersects each parabola at right angles.
Fig. 6.11
6.173
Find the orthogonal trajectories of the family of curves
m
I
It
x2 + y2 = c2
.
follows from Problem 1.97 that the orthogonal trajectories satisfy the differential equation
solution (see Problem 4.71 or Problem 3.34 with x replacing f) is
y = kx,
dy
—
=—
dx
x
y
.
Its
which is the family of orthogonal
trajectories.
The original family of curves is a set of circles with centers at the origin, while the orthogonal trajectories
Fig. 6.12
APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS
D
141
Some members of each family are shown in Fig. 6.12. Observe that each
are straight lines through the origin.
straight line intersects each circle at right angles.
6.174
xy = C.
Find the orthogonal trajectories of the family of curves
#
follows from Problem 1.99 that the orthogonal trajectories satisfy the differential equation
It
xdx — ydy = 0.
differential form,
or
2
x
- y 2 = C,
where
>•'
This equation is separable; integrating term by term, we get
C = 2k.
= x/y or, in
2
- \\ 2 = k
\x
Both the original family of curves and its orthogonal trajectories are
hyperbolas, as shown in Fig. 6.13.
Fig. 6.13
6.175
Find the orthogonal trajectories of the family of curves
m
I
It
x 2 + y 2 = ex.
Ixy
dy
=—
—
dx
x —
=
follows from Problem 1.96 that the orthogonal trajectories satisfy the differential equation
whose solution is given in Problem 3.127 as
x 2 + y 2 — ky.
2
=-,
2
y
Both the original family of curves and its orthogonal
trajectories are circles.
6.176
p = C{\ + sin 9).
Pind the orthogonal trajectories of the family of cardioids
I
It
follows from Problem 1.100 that the orthogonal trajectories satisfy the differential equation.
dp/p -i- (sec 9 + tan 9) d6 — 0.
This equation is separable; integrating term by term we obtain the equation for
the orthogonal trajectories as
In p
6.177
+ In (sec 9 + tan 9) — In cos 9 = In C
Find the orthogonal trajectories of the family of curves
I
It
p =
or
y = ce
C cos9
sec 9 + tan
= C(1 -sin0)
x
.
follows from Problem 1.98 that the orthogonal trajectories satisfy the differential equation
dy/dx = —l/y
y dy + 1 dx — 0. This equation is separable; integrating term by term, we obtain the
2
equation for the orthogonal trajectories as \y + x = c.
or, in differential form,
6.178
Find the orthogonal trajectories of the family of curves
I
it
2
y
= 4cx.
Differentiating the given equation with respect to x, we obtain
follows that
c
= y 2 /4x.
2yy' = Ac
dy
2c
dx
v
or
Since
dy^
= y_
dx
2x
Substituting this result into the last equation, we obtain
y
2
= 4cx,
as the differential
equation for every member of the given family of curves. The differential equation for its orthogonal trajectories
(see Problem 1.94) is then
dy
=
—
dx
2x
or, in differential form,
ydy + 2xdx = 0.
This equation is separable;
y
2
2
2
where the integration constant has been written as a square
\y + x = k
to emphasize the fact that it cannot be negative since it is equal to the sum of two squares. Typical curves of the
integrating term by term, we find
two families are shown in Fig. 6.14.
,
142
CHAPTER 6
Fig. 6.14
x
—w — - =
—
2
6.179
Show that the family of confocal conies
2
v
H
L
where C is an arbitrary constant, is self-orthogonal.
1,
A,
—
Differentiating the equation of the family with respect to x yields
C
this for C,
C=
we find
Xx
so that
x + yp
C—X=
Xpy
= 0.
-\
V-
where
r-
p —
dy
—
dx
.
Solving
When these replacements are made in the equation
x + yp
of the family, the differential equation of the family is found to be
Since this equation is unchanged when p is replaced by
— 1/p,
it
(x
is
+ yp){px — y) — Xp — 0.
also the differential equation of the
orthogonal trajectories of the given family. The graphs of several members of this family are shown in Fig. 6.15.
If
C > A,
then the graph is an ellipse; if
C < /,
it
is
a hyperbola.
Fig. 6.15
6.180
At each point (x, >') of a curve the intercept of the tangent on the y axis is equal to 2xy
2
(see Fig. 6.16).
Find the
curve.
The differential equation of the curve is
— — x2 + c
or
y
dy
y — 2xy
dx
x
2
x — x v = c\.
2
y — x
dy
—
dx
,
2xjr,
y dx — x dy
or
= 2x dx.
Integrating yields
y
The differential equation may also be obtained directly from the figure as
APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS
143
y
Fig. 6.16
6.181
At each point (x, y) of a curve the subtangent is proportional to the square of the abscissa. Find the curve if it
also passes through the point (1, e).
The differential equation of the curve is
Integrating yields
k In y =
curve has the equation
6.182
/c
\-
C.
y
—
=
dy
/cx
2
x
x = 1,
When
y = e;
x
In y
= — 1/x +
fc
—=
=k
2
or
—
where k is the proportionality factor.
,
y
thus
fc
= — +C
1
C=
or
fc
+ 1,
and the
+ 1.
Find the family of curves for which the length of the part of the tangent between the point of contact (x, y) and
the y axis is equal to the y intercept of the tangent.
We have
x
—
/l+(
—y — x—
\dx I
ax
V
latter equation to
(1
)
x 2 — y 2 — 2xy—
dx
or
,
+ v )dx + Ivxdv = 0,
dx
2v dv
which we write as
h
1
In x
+ In (1
-I-
v
2
)
= In C.
Since
v
= y/x,
we have
xl
1
y — vx
The transformation
.
+ —^ = C
I
+v
— 0.
Integrating then gives
2
x 2 + y 2 = Cx
or
reduces the
as the equation of
the family of curves.
6.183
Determine a curve such that the length of its tangent included between the x and y axes is a constant
a > 0.
Let (x, y) be any point P on the required curve and (X, Y) any point Q on the tangent line AB (Fig. 6.17).
and
The equation of line AB, passing through (x, y) with slope y', is Y — y = y'{X — x). We set X —
OA — y — xy' and O B = x — y/y' = — (y — xy')/y'. Then
Y=
in turn to obtain the y and x in tercepts
y/OA 2 + OB 2 = (y - xy')Vl + y' 2 //- since tms must equal ±a, we have
the length of AB, apart from sign, is
f
on solving for y,
y = xy' ±
ay
ViT7
xp +
ap
ViT?
y =p
where
(i)
*X
Fig. 6.17
To solve (7), we differentiate both sides with respect to x to get
/=p=x
Case 1,
dp/dx = 0:
dp
_ + p± _____
a
dp
In this case
p = c
or
d
±
x +
dx
and the general solution is
(1
+
7f]
y = cx ±
=
ac
"
x/iTT 2
CHAPTER 6
144
Case 2,
dp/dx ^ 0:
In this case, using (/) we find
a 2/3 p 2
,2/3
,2/3
l+p
2
and
'
-\
—
(l+ p
x 213 + y 2 3 - a 2 3
so that
+ p2
1
x =
'
'
.
2)3/2
and
y
'
— +-
«P
J
from which
(1+p2) 3/2'
This is a singular solution and is the equation
of a hypocycloid (Fig. 6.18), which is the envelope of the family of lines found in Case 1 and is the required curve
Fig. 6.18
6.184
Through any point (x, v) of a certain curve which passes through the origin, lines are drawn parallel to the
Find the curve given that it divides the rectangle formed by the two lines and the axes into two
coordinate axes.
areas, one of which is three times the other.
f There are two cases, illustrated in Figs. 6.19 and 6.20.
Fig. 6.19
Case I:
Here
(3)(area
OA P) = area OPB.
Then
3 jo v dx — xy
— j£ y dx,
Ay = y + x
the differential equation, we differentiate with respect to x, obtaining
integration yields the family of curves
Case 2:
Here
y = Cx
area OA P = (3)(area OPB)
the family of curves has the equation
y
3
4feydx = xy.
or
—
or —
dx
dx
dy
To obtain
dy
An
x
3
.
4 j£ y dx = 3xy.
and
dy
_y_
dx
3x'
The differential equation is
and
= Cx.
Since the differential equation in each case was obtained by a differentiation, extraneous solutions may have
been introduced.
It is
necessary therefore to compute the areas as a check. In each of the above cases, the curves
satisfy the conditions.
6.185
The areas bounded by the x axis, a fixed ordinate
x — a,
intercepted by the ordinates is revolved about the x axis.
a variable ordinate, and the part of a certain curve
Find the curve if the volume generated is proportional
to (a) the sum of the two ordinates, and (b) the difference of the two ordinates.
I
(a)
Let A be the length of the fixed ordinate.
n j* y 2 dx = k(y + A)
is
Try
2
= kdy/dx.
The differential equation obtained by differentiating
Integrating then yields
y(C — nx) = k,
from which
Then
n
y
Ja y
2
dx = n
k
\
Ja
2
dx
(C — nx)
C — nx
C — na
= k(y -A)* k(y + A)
Thus, the solution is extraneous and no curve exists having the property (a).
y
—
C — nx
—
1
APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS
145
n ft y 2 dx = k(y - A),
Repeating the above procedure with
we obtain the differential equation
whose solution is v(C - nx) = k. It can be shown (as we tried to do in part a) that this
equation satisfies the condition and thus represents the family of curves with the required property.
(b)
ny
6.186
2
= k dy/dx,
Find the curve such that, at any point on it, the angle between the radius vector and the tangent is equal to
one-third the angle of inclination of the tangent.
f
Let 9 denote the angle of inclination of the radius vector, z the angle of inclination of the tangent, and ip
the angle between the radius vector and the tangent.
tan if/
= tan j9.
\jj
= t/3 =
+ 9)/3,
{\j/
we have
= \9
\jj
and
Now
tan y/ = p
Integrating then yields
6.187
Since
dO
—
=
dp
9
tan 2
= 2 In sin \9 + In C ls
In p
dp
=
—
,
so that
in
cot - d9
2
p
or
p = C x sin \9 = C(l — cos 9).
2
The area of the sector formed by an arc of a curve and the radii vectors to the end points is one-half the length
of the arc.
Find the curve.
I
9 = 9^
Let the radii vectors be given by
and
9 = 9.
Then
- f" p 2 d9 = - f*
2 Je
2 Je
>
'
/(
—
)
\J \d9J
+ p 2 d9.
Differentiating with respect to 9 yields the differential equation
2
P
dp
_
d9
If
p
2
+ p2
= 1,
this latter equation reduces to
of the problem.
p = sec(C + 9).
If
p
2
^ 1,
dp = 0.
and
p = sec (C — 9)
p = 1
satisfies the condition
—-j^=
= ±d9 and
-
we write the equation in the form
p = sec (C + 9)
that
It is easily verified
PVP
p =
Thus, the conditions are satisfied by the circle
Note that the families
6.188
dp = ±pVp 2 ~ 1 d9.
or
y
obtain the solution
2
1
and the family of curves
p = sec(C + 9).
are the same.
Find the curve for which the portion of the tangent between the point of contact and the foot of the perpendicular
through the pole to the tangent is one-third the radius vector to the point of contact.
f
In Fig. 6.21,
p = 3a = 3pcos(7r — i]/) = — 3pcos\j/,
so that
cos i/f =
—|
and
tani/^=— 2V2.
In
JO
Fig. 6.22,
from which
p = 3a = 3p cos ip
—= +
P
—.
and
tan \jj = 2yj2.
tan \\i = p
—
= ±2>J2,
dp
p = Ce
^.
Combining the two cases, we get
The required curves are the families
p = Ce
612
^
and
°i 2
2^2
P
Fig. 6.22
Fig. 6.21
6.189
Find the shape assumed by a flexible chain suspended between two points and hanging under its own weight.
f
Let the y axis pass through the lowest point of the chain (Fig. 6.23), let s be the arc length from this point
We obtain the equation of the curve
to a variable point (x, y), and let w(s) be the linear density of the chain.
from the fact that the portion of the chain between the lowest point and (x, y) is in equilibrium under the action
of three forces: the horizontal tension T at the lowest point; the variable tension T at (x, y), which acts along
the tangent because of the flexibility of the chain; and a downward force equal to the weight of the chain
between these two points. Equating the horizontal component of T to T and the vertical component of T to
T cos 9 = T and T sin 9 = f w(s) ds. It follows from the first of these
the weight of the chain gives
equations that
T sin 9 = T tan 9 = T j-,
so
T y' = f w{s) ds. We eliminate the integral here by
CHAPTER 6
146
Fig. 6.23
differentiating with respect to x:
T °y" =-
h
So
w ^ ds "j
w(s) ds
s Si
1 = w^ + w
1
T y" = w(s)\f\ + (y') 2 is the differential equation of the desired curve, and the curve itself is found by
solving this equation. To proceed further, we must have definit e informa tion about the function w(s).
We shall assume that w(s) is a constant w so that y" = ay/l + (y') 2 where a = w /T
Substituting
Thus
,
y'
=p
and
y" — dp/dx
.
,
dp
then yields
—===== = a dx.
Integration and use of the fact that
p =
when
If
we place the
>/TT?
x =
now give
log (p + Jl
+ p 2 = ax.
)
x axis at the proper height, so that
y = l/a
y =
Solving for p then yields
p =
dy
—
=dx
1
(e
ax
— e ax
).
2
when
(
x = 0,
e °* + e
ax
2a
)
we get
= - cosh ax
a
as the equation of the curve assumed by a uniform flexible chain hanging under its own weight.
is
6.190
This curve
called a catenary, from the Latin word for chain, catena.
A point P is dragged along the xy plane by a string PT of length a.
If
T starts at the origin and moves
along the positive y axis, and if P starts at (a, 0), what is the path of P?
dy
V^ ^
=
—
dx
7
I
It is
easy to see from Fig. 6.24 that the differential equation of the path is
variables and integrating (and using the fact that
y = a In
— yja 2 — x 2
.
y =
when
x — a),
On separating
we find that
This is the equation of a tractrix, from the Latin word tractum, meaning
drag.
Fig. 6.24
.
APPLICATIONS OF FIRST-ORDER DIFFERENTIAL EQUATIONS
6.191
147
A rabbit starts at the origin and runs up the y axis with speed a. At the same time a dog, running with speed b,
starts at the point (c, 0) and pursues the rabbit.
f At time
f,
D = (x, y)
(Fig. 6.25).
t,
ds
=
—
b,
Since the line DR is tangent to the path, we have
dt
Ul
ill
dt
ds
US
11
dx
ds dx
b
—- = ——-= -- V' + (y)
we have
/
;
The substitution
,
xy" = — a
—
dx
where the minus sign appears because s increases as x
k =b
y'
—p
and
v" = dp/dx
dp
reduces this to
the initial condition
1
T/xY
p =
/c x
when
x — c,
we find that
In (p
dx
—
x
=k
,
VI + P
d~x
xy' - y = -at.
r
xv" = Wl+(/) 2
=
or
and the dog at
When these two equations are combined, we obtain the differential equation of the path:
decreases.
dy
—
= -——
dx
x
we begin by differentiating this last equation with respect to x, which gives
dt
P =
R = (0, at)
measured from the instant both start, the rabbit will be at the point
To eliminate
Since
Find a differential equation describing the path of the dog.
2
and on integrating and using
,
+ yjl + p 2 — In —
)
I
Then
.
J
'
2
Fig. 6.25
6.192
x = c are the banks of a river whose current has uniform speed a in the negative
A boat enters the river at the point (c, 0) and heads directly toward the origin with speed b relative
The y axis and the line
v direction.
to the water.
What is the path of the boat?
The components of the boat's velocity (Fig. 6.26) are
— = — cos
b
and
9
dt
dy
dx
-a + bsm6
—b cos
-a + b(-y/y/x 2 + y 2
)
a^Jx
sin 6,
so
+ y 2 + by
bx
-b(x/Jx 2 + y 2 )
+
c\y + \Jx 2 + y 2 ) = x* \
the fate of the boat depends on the relation between a and b.
This equation is homogeneous, and its solution is
2
= —a + b
—
dt
where
k = a/b.
Fig. 6.26
It is
clear that
148
CHAPTER 6
Fig. 6.27
6.193
A boy, standing in corner A of a rectangular pool (Fig. 6.27), has a boat in the adjacent corner B on the end of
a string 20 ft long. He walks along the side of the pool toward C, keeping the string taut. Locate the boy and
boat when the latter is 12 ft from AC.
I We choose the coordinate system so that AC is along the x axis, and AB is along the y axis.
Let (x, y) be
the position of the boat when the boy has reach ed E, and let 6 denote the angle of inclination of the string.
dy
Then
tan 9 = —- =
dx
—
,
-t
V400 - y
x = — V400 — y 2 + 20 In
and
y = 20.
Then
dx — —
or
20 + V400 - y
C = 0,
V400"
dy.
Integrating gives
2
2
\-
and
C
To find C, we note that when the boat is at B, we have
x = -V400 - y 2 + 20 In
20 + v 400 -
r
is
the equation of the boat's
path.
Now AE = x + >/400 — y 2 = 20 In
when
y = 12),
x + 16 = 20 In 3 = 22.
.
x =
Hence, when the boat is 12 ft from AC (that is.
Thus, the boy is 22 ft from A, and the boat is 6 ft from AB.
—
CHAPTER 7
Linear Differential Equations
Theory of Solutions
WRONSKIAN
7.1
Define the Wronskian of the set of functions {z^x), z 2 (x),
function possesses
n
—
1
.
.
The Wronskian is the determinant
W(z u z 2
.
.
,
,
.
Zl
z2
••
A
A
A
••
4
••
z»
A
z„)
1)
Zl
7.2
where each
*r
i)
•
zn
z
(n
~
1)
Find the Wronskian of {sin 2x, cos 2x}.
sin 2x
cos 2x
W^sin 2x, cos 2x) = d(sin 2x)
sin 2x
cos 2x
2 cos 2x
— 2 sin 2x
d(cos 2x)
dx
7.3
a < x < b,
on the interval
z„(x)}
,
.
derivatives.
- -2
dx
Find the Wronskian of {3 sin 2x, 4 sin 2x}.
4 sin 2x
3 sin 2x
W{3 sin 2x, 4 sin 2x) = d(3 sin 2x)
3 sin 2x
4 sin 2x
6 cos 2x
8 cos 2x
d(4 sin 2x)
dx
dx
- (3 sin 2x)(8 cos 2x) — (4 sin 2x)(6 cos 2x) =
7.4
Find the Wronskian of {sin 3x, cos 3x}.
cos 3x
sin 3x
W(sin 3x, cos 3x) — d(sin 3x)
cos 3x
3 cos 3x
— 3 sin 3x
dx
dx
(sin 3x)(
7.5
sin 3x
d(cos 3x)
— 3 sin 3x) — (cos 3x)(3 sin 3x) = — 3 sin 2 3x — 3 cos 2 3x - -3
Find the Wronskian of {1, x}.
X
1
XV
1
W(\, x) = rf(l)
d[x)
1
=
= 1(1) - x(0) = 1
1
dx
dx
7.6
Find the Wronskian of {3x, 5x}.
5x
3x
W(3x, 5x) = d(3x)
Find the Wronskian of {t, t
5x
3
5
= 3x(5) - (5x)(3) =
dx
dx
7.7
3x
d{5x)
2
}.
z
t
t
W{t, t ) -
t
r
l
it
2
r
2
t
dt
dt
7.8
= t(2t) - 2 (l) =
3
Find the Wronskian of {t, t }.
3
t
W(t, t
3
)
= d(t)
dt
t
d(t
3
t
t>
1
3r
= t{3t 2 - t 3 {l) = 2?
)
)
;
dt
149
CHAPTER 7
150
7.9
Find the Wronskian of {t 2
3
t
,
}.
2
W(t 2
7.10
Find the Wronskian of {3f
2
,
3
t
,
)
3
t
r
It
3r
=
W(3t,22
3f
-),3
3
,
2f
2
5f
,
,
2
It
)
3f
=
2
2f
}.
\t
5t
f
)
3t
,
2
15f
3f
,
Find the Wronskian of {e x e
It
3f
= r 2 (3r 5 ) - \t 6 (2t) = It 1
5
It
2
21f
1
= 2f 3 (21r 6 )-3r 7 (6r 2 = 24f 9
)
6
).
e
x
x
e~
= e x (-e~ x )-e- x e x = -2
Find the Wronskian of {5e*, le~ x }.
=
x
,
)
5e
5e
le' x
x
x
-le
~x
= 5e x (-le~ x - le'^Se1 = -70
)
)
Find the Wronskian of {5e 2x le 3x ).
,
W(5e 2x le 3x
,
5e
)
\0e
2x
le
2x
2\e
3x
3x
= 5e 2x(2\e 3x - le 3x(\0e 2x = 35e 5 *
)
)
Find the Wronskian of {le~ 3x 4e~ 3x }.
,
le~ 3x
W(le~ ix Ae~ ix ) =
2\e~
3x
4e~
3x
-12<T 3x
= le' 3x(-\2e- 3x - 4e~ 3x (-2le- 3x =
)
)
Find the Wronskian of {e x xe x ).
,
W(e x xe x =
Find the Wronskian of {x 3
-
f
We have
Then, for
x > 0,
|x
,
,
3
,,
|}
xe
e
x
e (e
)
,
7.20
\t
x
,
W(Se x le
7.19
)
b
t
3
2t
W(ex .e-') =
7.18
)-5f 3 (3t 2 =
}.
6f
7.17
2
7
Find the Wronskian of {2f 3
W(2t\ 3r 7 ) =
7.16
3
= f (15f
2
}.
2
7.15
3
6
W(t 2 ,$t<) =
7.14
= 3r 2 (4r) - 2f 2 (6r) -
4r
3
Find the Wronskian of {t 2
2
3
W(t\ 5t 3
7.13
)
}.
6r
Find the Wronskian of {f 3
= 3f 2 (6r 2 - 2t 3 (6t) = 6r 4
2
6r
2
2r
W(3t 2
7.12
3
2r
)
6r
,
t
)
}.
—
Find the Wronskian of {3f 2
f
3
2f
I
7.11
= 2 (3r 2 - 3 (2r) = r 4
2
x
+ xex - xex (ex = e 2x
)
)
on [— 1, 1].
f
x3
if
x >
d(\x
S°
f
^
— x J3 ifx<0
\)
ax
W(x 3 \x 3 =
,
3x 2
3
\)
3x 2
1„2
3x
-3x 2
=
ifx>0
ifx =
ifx<0
LINEAR DIFFERENTIAL EQUATIONS— THEORY OF SOLUTIONS
For
For
Thus,
7.21
W(x\ \x 3 =
x < 0,
\)
X3
-x 3
2
-3x 2
3x
W(x\ |x 3 =
x = 0,
=
|)
W(x\ |x 3 =
=
on [-1,1].
|)
Find the Wronskian of { 1, x, x
2
}.
I
1
W(l,x,x 2) =
x
x2
1
2x
2x
1
=
2
X
X2
2
1
2x
2x 2
— x.3
-0 X
1
x
+
2
2
=
7.22
Find the Wronskian of {x, 2x 2
W(x,2x 2
,
-x 3
-x 3
,
)
}.
x
2x 2
-x 3
1
4x
-3x 2
-6x
4
-3x 2
-6x
4x
=x
4
x(-12x 2)- l(-8x 3 +
=
7.23
- 2x(0)] - + = 2
l[l(2)
6x
4
2x 2
-x 3
4x
-3x 2
2x
3x 2
2
6x
-4x 3
=
)
+
1
Find the Wronskian of {x 2 x 3 x 4 }
,
2
W(x
,
,
x 3 x 4)
,
X2
X3
2x
3x 2
4x 3
2
6x
12x
3x 2
4x 3
6x
2
12x
4x 3
2x
-x 3
2
12x
2
2
+ x4
x 2 (12x 4 ) - x 3 (16x 3 ) + x 4 (6x 2 ) = 2x 6
7.24
Find the Wronskian of {x 2
— 2x 2 3x 3
X
W{x
2
,
-2x
2
,
3x
3
)
}.
,
,
^
-2x 2
-4x
= 2x
3x 3
-4
2
9x 2
-4.x
-x -4
2
9x 2
-(-2x 2
ISx
2x
9x 2
2
IXx
)
+ 3x 3
?x
-4x
•;
-4
18x
= x 2 (-36x 2 - (-2x 2 M18x 2 + 3x 3 (0) =
)
7.25
Find the Wronskian of {x
2
,
x2
,
)
- 3x}.
2
x" 2
d(x~
H^x
,x
2
,
2-3x) =
d(x
)
dx
2
d (x~
2
)
Find the Wronskian of {e\ e
',
e
- 3x)
x2
2-3x
2x
-3
x' 2
dx
d 2 (x 2 )
d 2 (2-3x)
dx''
dx'
l
die*)
7.27
d(2
)
2x
-3
6x" 4
36x~ 2 -32x
2
}.
e
W(e\ e~\ e
2
dx
dx'
7.26
2-3x
x2
2
2 ')
e~
l
d(e~')
e"
d(e
2
dt
dt
dt
d (e')
2
d (e-')
d (e
dt'
dv
dt'
2
')
2
2t
)
21
e~
l
e~
l
2e
2'
e~'
4e
2'
e
6e
2
Find the Wronskian of {1, sin 2t, cos 2f}.
1
W{l,s'm2t, cos2r) =
sin 2t
cos2r
2 cos 2r
— 2sin2f
— 4cos2f
— 4 sin 2t
8cos 2 2f -8sin 2 2f = -8
-3
151
152
7.28
CHAPTER 7
U
Find the Wronskian of {t, t - 3, It + 5}.
t
W=
7.29
Find the Wronskian of {f 3
3
,
+
t
It
t,
1
3
+
2
3t + 1
t
W = 3f 2
3
,
+
t
2f
t,
6t
- It
-7 =
3
2
6f
3
W = 3f 2
+
2
3t +
6f
6f
12f
cos f
2 sin f
12f
- 7}.
3
3
t
It
t
-1
3
6f
1
= 42f
2
Find the Wronskian of {sin f, cos t, 2 sin t — cos t }.
sin f
W=
Find the Wronskian of {f 3
— sin
cosf
— sin
7.32
It
t
6f
t
7.31
=
2
- It).
3
3
Find the Wronskian of {t 3
+ 5
2f
1
t
7.30
-3
f
f
— cos
f
f
2cosf + sinr
—cosf
— 2 sin + cosf
=
r
2
,
f
t,
,
1
}.
2
r
3f
W=
2
t
t
It
1
1
= 12
2
6f
6
7.33
Find the Wronskian of {e""', e mi
c
w=
',
c
m ",
">><
c
e
mj
'J,
mi
'
m2
'
m 2e
m 22 e m2
m\emi
m\e mit
m e
,mjl
m4
e
t
m >'
2
where m u m 2
mle
mi
'
mle
,
l
m 4e
'
m 3 and m x are constants.
,
t
_ £"1l 'p m2t p mi, p m
m"
*'
m
w3
2
m i e m*<
m\e mit
1
m4t
1
1
m 2 "»3 m 4
m 2 W 3 AM4
m 3 W 3 m4
This last determinant is a Vondermonde determinant and is equal to
(m4 — m 1 )(m4 — m 2 )(m i — m 3 )(m 3 — m,)(w 3 — m 2 )(m 2 — m ). Thus,
(mi+m2+m 3 +m4)t^ _ W] m4 _ m )(m - m )(m
)(m - m )(m
e
{
W-
)(
4
2
3
3
-m
3
l
2
2
- m,).
LINEAR INDEPENDENCE
7.34
Determine whether the set {e x e
,
x
}
is
linearly dependent on (
— oo, oo).
I Consider the equation
c,e
We must determine whether there exist values of c
have
or
c,e
+ c-,e~
=
and c 2 not both zero, that will satisfy (/). Rewriting (7), we
For any nonzero value of c u the left side of this equation is a
-c,e 2x
,
x
constant, whereas the right side is not; hence the equality is not valid.
latter equation,
and therefore to (J), is
(1)
= c 2 — 0.
cx
It
follows that the only solution to this
Thus, the set is not linearly dependent; rather it is linearly
independent.
7.35
Rework the previous problem using differentiation.
#
c,e
We begin again with the equation
— c ,e
x
= 0.
c{e
x
+ c 2 e~ x = 0.
Differentiating it now, we obtain
These two equations are a set of simultaneous linear equations for the unknowns c
Solving them, we find that the only solution is
cl
— c 2 — 0,
x
so the functions are linearly independent.
and c 2
.
LINEAR DIFFERENTIAL EQUATIONS— THEORY OF SOLUTIONS
7.36
153
Rework Problem 7.34 using the Wronskian.
f The Wronskian of {ex e~ x } is found in Problem 7.15 to be -2. Since it is nonzero for at least one point in
,
the interval of interest (it is, in fact, nonzero everywhere), the functions are linearly independent on that interval.
7.37
2
on
x, 1} linearly dependent
Is
the set {x
#
Consider the equation
,
c,x
2
— oo, oo)?
(
+ c 2 x + c 3 = 0.
Since this equation is valid for all x only if
c,
= c 2 = c 3 = 0,
Note that if any of the c's were not zero, then the equation could hold
the given set is linearly independent.
for at most two values of x, the roots of the equation, and not for all x.
7.38
Rework the previous problem using differentiation.
I We begin again with the equation
2cjX + c 2 = 0.
c,x
2
+ c 2 x + c 3 = 0.
linear equations for the unknowns c u c 2 , and c 3
ct
7.39
— c 2 = c 3 = 0,
Differentiating this equation, we get
2c x = 0.
Differentiating once again, we get
These three equations are a set of simultaneous
Solving them, we find that the only solution is
.
which implies that the functions are linearly independent.
Rework Problem 7.37 using the Wronskian.
f The Wronskian of {x 2 x, 1} is found in Problem 7.21 to be 2.
,
Since it is nonzero for at least one point in the
interval of interest (it is, in fact, nonzero everywhere), the functions are linearly independent on that interval.
7.40
Determine whether the set {sin 2x, cos 2x} is linearly independent on ( — oo, oo).
I Consider the equation
c, sin 2x
+ c 2 cos 2x = 0.
2c l cos 2x — 2c 2 sin 2x = 0.
Differentiating, we obtain
These two equations are a set of simultaneous linear equations for the unknowns c x and c 2
The easiest way to
.
solve them is to multiply the first by 2 sin 2x, multiply the second by cos 2x, and then add the resulting equations
together.
Doing so, we find that the only solution is
c x — c2
— 0,
which implies that the functions are linearly
independent.
7.41
Rework the previous problem using the Wronskian.
I The Wronskian of {sin 2x, cos 2x} is found in Problem 7.2 to be —2.
Since it is nonzero for at least one
point in the interval of interest (it is, in fact, nonzero everywhere), the functions are linearly independent on that
interval.
7.42
Determine whether the set {sin 3x, cos 3x} is linearly independent on ( — oo, oo).
I Consider the equation c sin 3x + c 2 cos 3x = 0. Differentiating, we obtain 3c cos 3x — 3c 2 sin 3x = 0.
To solve these two equations simultaneously for the unknowns c and c 2 we multiply the first by 3 sin 3x,
x
x
,
x
The only solution is
multiply the second by cos 3x, and then add.
c{
= c 2 = 0,
which implies that the
functions are linearly independent.
7.43
Rework the previous problem using the Wronskian.
I The Wronskian of {sin 3x, cos 3x} is found in Problem 7.4 to be —3.
Since it is nonzero for at least one
point in the interval of interest (it is, in fact, nonzero everywhere), the functions are linearly independent on that
interval.
7.44
Is
the set {f
2
,
t}
linearly independent on (
f Consider the equation
cxt
2
— oo, oo)?
Since a polynomial equals zero for all values of t in ( — oo, oo) if and
+ c 2 = 0.
t
only if the coefficients are zero, it follows that
c,
= c 2 = 0,
from which we conclude that the original functions
are linearly independent.
7.45
Rework the previous problem using differentiation.
I We begin again with the equation
ctt
2
+ c 2 = 0.
t
Differentiating gives
2c t + c 2 = 0.
These two
x
equations are a system of linear equations which may be solved simultaneously for Cj and c 2
.
Instead, we may
and then from the
It follows from this that
c x — 0,
2c, =0.
differentiate the second equation to obtain
=
=
and the functions are linearly
c2
0,
second equation that c 2 = 0. Thus, the only solution is c
x
independent.
CHAPTER 7
154
7.46
Rework Problem 7.44 using the Wronskian.
f The Wronskian of {t 2 f} is found in Problem 7.7 to be t 2
,
= 2,
we have
linearly independent on (
— oo, oo)?
interval of interest (for example, at
t
Since it is nonzero for at least one point in the
.
W=4#
0),
the functions are linearly independent on
that interval.
7.47
Is the set {t
2
3
,
t
}
I Consider the equation
for all values of fin (
cxt
2
+ c 2 3 = 0.
This is a third-degree polynomial in t. Since a polynomial is zero
t
— x, oo) if and only if all of its coefficients are zero, it follows that
cx
= c 2 = 0,
from
= 0.
These
which we conclude that the functions are linearly independent.
7.48
Rework the previous problem using differentiation.
I We begin again with the equation
cxt
2
+ c 2 3 = 0.
Differentiating, we obtain
t
2c x t + 3c 2 t
2
two equations are a set of linear equations which may be solved simultaneously for c and c 2 Instead, however,
we may differentiate twice more, obtaining successively 2c x + 6c 2 t =
and 6c 2 = 0. It follows from these
equations first that c 2 = 0, and then that c — 0. Thus, the only solution to the first equation is c x = c 2 = 0,
and the functions are linearly independent.
.
x
x
7.49
Rework Problem 7.47 using the Wronskian.
f The Wronskian of {f 2 t 3 } is found in Problem 7.9 to be
,
interval of interest (for example, at
7.50
t
= 1,
W=
^ 0),
I
Determine whether {e x xe x ) is linearly independent on
,
(
4
f
.
Since it is nonzero for at least one point in the
the functions are linearly independent on that interval.
— x, x).
f Consider the equation c e x + c 2 xe x = 0. Differentiating, we obtain c e" + c 2 (e x + xe x = 0. These two
equations may be solved simultaneously for c, and c 2 We begin by subtracting the first from the second, and
x
)
x
.
then recall that e
x
is
never zero. Thus we find that the only solution is
c,
= c 2 = 0.
It
follows that the
functions are linearly independent.
7.51
Rework the previous problem using the Wronskian.
f The Wronskian of {c\ \c
x
is
\
found in Problem 7.19 to be e 2x
.
Since it is nonzero for at least one point in
the interval of interest (it is, in fact, nonzero everywhere), it follows that the functions are linearly independent
on that interval.
7.52
Determine whether
-J
3 sin 2x, 4 sin 2x} is linearly
f Consider the equation
c2
,
independent on (— oo, x
).
= 0. By inspection, we see that there exist constants c and
= -3), that satisfy this equation for all values of x in
c,(3 sin 2x) + c 2 (4 sin 2x)
not both zero (in particular,
c,
=4
and
c2
x
(—x, x); thus, the functions are linearly dependent.
7.53
Can the Wronskian be used to determine whether the functions 3 sin 2x and 4 sin 2x are linearly independent
on ( — x, x)?
f
It is
shown in Problem 7.3 that the Wronskian of these two functions is identically zero, so no conclusions
can be drawn about linear independence.
7.54
Redo the previous problem if in addition it is known that the two functions are solutions of the same linear
homogeneous differential equation.
I
If the Wronskian of a set of functions is zero, and if those functions are all solutions to the same linear
homogeneous differential equation, then the functions are linearly dependent. Thus, it now follows from
Problem 7.3 that 3 sin 2x and 4 sin 2x are linearly dependent.
7.55
Determine whether {3x, 5x} is linearly independent on ( — x, x).
I Consider the equation
both zero (in particular,
c
cx
1
(3x) -I- c 2 (5x) = 0.
= —5
and
c2
By inspection, we see that there exist constants c and c 2 not
that satisfy this equation for all values of x in ( — x, x).
x
= 3)
Thus, the functions are linearly dependent.
7.56
Can the Wronskian be used to determine whether the functions 3x and 5x are linearly independent on — x, x)?
(
LINEAR DIFFERENTIAL EQUATIONS— THEORY OF SOLUTIONS
f
It is
155
shown in Problem 7.6 that the Wronskian of these two functions is identically zero, so no conclusions
can be drawn about linear independence.
7.57
Redo the previous problem knowing that
y
x
= 3x
and
y 2 = 5x
are both solutions of
y" = 0.
I
Since both functions are solutions to the same linear homogeneous differential equation, and since their
Wronskian is identically zero, the two functions are linearly dependent.
7.58
Determine whether {r 3
,
5f
3
(c x
values of t in (
7.59
linearly independent on (
c^t 3 ) + c 2 (5r 3 ) = 0.
f Consider the equation
both zero
is
}
— oo, oo).
By inspection, we see that there exist constants c and c 2 not
l
= — 5, c 2 =
is another) that satisfy this equation for all
is one pair;
c = 10,
c2 = — 2
— oo, oo); therefore the functions are linearly dependent.
1
x
What conclusion can one draw about the linear independence of the functions t 3 and 5t 3 on — oo, oo) by
(
computing their Wronskian?
I
Since their Wronskian is identically zero (see Problem 7.12), no conclusion can be drawn about linear
independence.
7.60
Redo the previous problem knowing that
Vj
= 3
t
and
y 2 = 5t
3
are both solutions of
4
d y/dt* = 0.
I
Since both functions are solutions to the same linear homogeneous differential equation, and since their
Wronskian is identically zero, the two functions are linearly dependent.
7.61
Determine whether {le~ 3x 4e~ 3x } is linearly dependent on ( — oo, oo).
,
I
Consider the equation
not both zero
(c x
—4
+ c 2 {4e~ 3x = 0.
c2 — — 7
is one pair;
c x {le~
and
3x
)
)
By inspection, we see that there exist constants c t and c 2
— 1 and c 2 = —7/4 is another) that satisfy this
c{
equation for all values of x in (— oo, oo); hence the functions are linearly dependent.
7.62
What conclusion can one draw about the linear independence of the functions le~ 3x and 4e~ 3x on (— oo, oo) by
computing their Wronskian?
I
Since their Wronskian is identically zero (see Problem 7.18), no conclusion can be drawn about linear
independence.
7.63
Redo the previous problem knowing that both functions are solutions to
y'
+ 3y — 0.
I
Since both functions are solutions to the same linear homogeneous differential equation, and since their
Wronskian is identically zero, the two functions are linearly dependent.
7.64
What conclusions can one draw about the linear independence of the functions 5e 2x and 7e 3x on — oo, oo) by
(
computing their Wronskian?
I The Wronskian of these two functions is 35e 5 * (see Problem 7.17).
Since it is nonzero for at least one point
in the interval of interest (it is, in fact, nonzero everywhere), the functions are linearly independent on that interval.
7.65
What conclusions can one draw about the linear independence of the functions t 2 and \t 6 on (0, 5) by computing
their Wronskian?
f The Wronskian of these two functions is It 1 (see Problem 7.13).
the interval of interest (for example, at
7.66
t
= 1,
W=2#
0),
Since it is nonzero for at least one point in
the functions are linearly independent on (0, 5).
What conclusions can one draw about the linear independence of the functions 3r 2 and It 2 on (0, 8) by computing
their Wronskian?
f
Since their Wronskian is identically zero (see Problem 7.11), no conclusion can be drawn about linear
independence.
7.67
Redo the previous problem knowing that both functions are solutions to
f
d 3 y/dt 3 = 0.
Since both functions are solutions to the same linear homogeneous differential equation, and since their
Wronskian is identically zero, the two functions are linearly dependent on (0, 8).
CHAPTER 7
156
7.68
What conclusions can one draw about the linear independence of the functions x 3 and |x 3 on (- 1, 1) by
|
computing their Wronskian?
/
Since the Wronskian is identically zero (see Problem 7.20), no conclusion can be drawn about linear
independence.
7.69
Determine whether the set {x 3
f Consider the equation
3
,
cxx
|x |} is linearly
3
+ c 2 \x 3 = 0.
dependent on [—1. 1].
Recall that
\
|x
3
|
= x3
if
x > 0,
and
|x
3
|
= -x 3
if
x < 0.
Thus, our equation becomes
+ c2x3 =
3
3
c x — c2x =
c yx
3
forx >
for x <
t
Solving these two equations simultaneously for c x and c 2
,
we find that the only solution is
ct
= c 2 — 0.
The
given set is, therefore, linearly independent.
7.70
Can both x 3 and |x 3 be solutions of the same linear homogeneous differential equation?
|
I
No, for if they were, then we would have two solutions of the same linear homogeneous differential equation
having an identically zero Wronskian, which would imply that the two functions are linearly dependent. We
know, however, from the previous problem that the two functions are linearly independent on (—1, 1).
7.71
Determine whether x 3 and |x 3 are linearly dependent on [ — 1,0].
|
I
follows from Problem 7.69 that, for linear dependence, we must satisfy
It
thib is the only condition that
is
c,x 3 — c 2 x 3 = 0.
By inspection, we see that there exist constants c t and c 2 not both zero (for example,
variable x.
Observe that
operable here, because we do not include any positive values of the independent
cY
= c 2 = 7)
that satisfy this equation for all values of x in the interval of interest; therefore the functions are linearly
dependent there.
7.72
Determine whether x 3 and |x 3 are linearly dependent on [0, 1J.
|
Problem 7.69 that we must now satisfy c,x 3 + c 2 x 3 = 0. Observe that this is the only
operable condition because we do not include negative values of x. By inspection, we see that there exist
constants c and c 2 not both zero (for example, c, = — c 2 = 3) that satisfy this equation for all values of x in
the interval of interest; hence the functions are linearly dependent there.
I
follows from
It
x
7.73
Determine whether the set {1 — x,
1
+ x,
1
— 3x} is linearly dependent on (— x, oo).
f Consider the equation c,(l — x) + c 2 (l + x) + c 3 {\ — 3x) = 0. which can be rewritten as
— c, + c 2 — 3c 3 )x + (c, + c 2 + c 3 = 0. This linear equation can be satisfied for all x only if both coefficients
are zero. Thus, we require
(
)
— c, + c 2 — 3c 3 =
=
1
c{
+ c2 + c3 =
= — 2c 3 and c 2 = c 3 with c 3 arbitrary. Choosing
as a set of
(any other nonzero number would do), we obtain c = —2, c 2 — \, and c 3 =
Solving these equations simultaneously, we find that
c3
and
r,
,
1
x
constants, not all zero, that satisfy the original equation. Thus, the given set of functions is linearly dependent.
7.74
Determine whether the set {5, 3 — 2x, 2 + x - ^x 2 } is linearly dependent on ( — oo, oo).
2
= 0, which we may rewrite as
c,(5) + c 2 (3 - 2x) + c 3 (2 + x — \x
=
—
The left side is a second-degree polynomial in x. Since a
0.
2c 2 + c 3 )x + (5cj + 3c 2 + 2c 3
(— |c 3 )x +
polynomial is zero for all values of x in — oo, oo) if and only if all of its coefficients are zero, it follows that
f
Consider the equation
)
2
)
(
(
— 2C 3 =
ar, d
— 2c 2 + c 3 =
5c x + 3c 2 + 2c 3 =
and
Solving this set of equations simultaneously, we find that the only solution to it, and therefore to the original
equation, is
7.75
ct
= c 2 = c 3 — 0.
Hence the functions are linearly independent.
Rework the previous problem using differentiation.
— 2 c 3 )x 2 + — 2c 2 + c 3 )x (5c + 3c 2 + 2c 3 = 0.
— c 3 x + — 2c 2 + c 3 =
and — c 3 = 0.
Differentiating this equation twice, we obtain successively
I We begin again with the rewritten equation
(
-I-
(
(
Y
)
)
Solving
LINEAR DIFFERENTIAL EQUATIONS— THEORY OF SOLUTIONS
this set of equations one at a time in reverse order, we find that
c3
= c 2 = c, =
is
157
the only solution;
therefore the functions are linearly independent.
7.76
Rework Problem 7.74 using the Wronskian.
#
Here,
5
3
W=
— 2x
-2
Lv2
2 + x — jx
1-x
= -10
0-1
which is nonzero for at least one point in the interval of interest (it is, in fact, nonzero everywhere); hence the
functions are linearly independent.
7.77
Determine whether {x 2
,
— 2x 2 x 3 } is linearly dependent on (-co, oo).
,
+ c 2 — 2x 2 + c 3 x 3 = 0.
and c 3 not all zero (for example, c -2, c 2 = 1, and
#
Consider the equation
cxx
2
(
)
x
x.
By inspection, we see that there exist constants c
c3
= 0)
x ,
c2
that satisfy this equation for all values of
Therefore, the functions are linearly dependent.
Alternatively, we can rearrange our equation to
{c x
— 2c 2 )x 2 + c 3 x 3 = 0.
The left side is a third-degree
polynomial; it is zero for all values of x in (— oo, oo) if and only if the coefficients are zero, that is, if and only
if
cx
Cy
— 2c 2 =
= 2c 2
,
and
c 3 = 0.
Solving these last two equations simultaneously, we find that
with c 2 arbitrary. Choosing
at the same conclusion as before
7.78
c2
—
1
c3
—
and
(any other nonzero number would do equally well), we arrive
— namely that the functions are
linearly dependent.
What conclusions can one draw about the linear dependence of {x 2 — 2x 2 x 3 } on { — oo, oo} by computing
,
,
their Wronskian?
f The Wronskian of this set is shown in Problem 7.24 to be zero, so no conclusion about linear dependence
may be drawn.
7.79
Redo the previous problem if in addition it is known that all three functions are solutions of
d*y/dx
4
= 0.
I
Since all three functions are solutions of the same linear homogeneous differential equation, and since their
Wronskian is identically zero, the three functions are linearly dependent.
7.80
Determine whether the functions e', e~\ and e 2 are linearly dependent on ( — oo, oo).
'
f
Their Wronskian is
— 6e 2 (see Problem 7.26), which is nonzero for at least one point in the interval of
'
interest (it is, in fact, nonzero everywhere); thus the functions are linearly independent.
7.81
The functions sin t, cos f, and
2 sin t — cos t
are all solutions of the differential equation
y" + y = 0.
Are
these functions linearly independent on ( — oo, oo)?
f The Wronskian of these functions is found in Problem 7.31 to be zero.
Since the functions are all solutions
of the same linear homogeneous differential equation, they are linearly dependent.
7.82
and 2f 3 - 7 are all solutions of the differential equation
functions linearly independent on — oo, oo)?
The functions r 3
3
r
,
+
r,
4
d y/dt* = 0.
Are these
(
f The Wronskian of these functions is 42r (see Problem 7.30).
interval of interest (for example, at
7.83
t
= 1,
W = 42),
Since it is nonzero for at least one point in the
the functions are linearly independent.
- 3,
and 2r + 5 are all solutions of the differential equation
functions linearly dependent on (-co, oo)?
The functions t,
t
d 2 y/dt 2 = 0.
Are these
f The Wronskian of these functions is identically zero (see Problem 7.28). Since the functions are all solutions
of the same linear homogeneous differential equation, they are linearly dependent.
7.84
Determine whether the functions e
2
',
e
3t
,
I Using the result of Problem 7.33 with
e~', and e~
5'
are linearly dependent on (- oo, oo).
m = 2, m 2 = 3, m 3 = -1,
t
and
m 4 = -5, we find that the
Wronskian of these four functions is never zero. Thus, the functions are linearly independent.
,
)
158
CHAPTER 7
D
GENERAL SOLUTIONS OF HOMOGENEOUS EQUATIONS
7.85
d
±
d2y
Show that the equation
2y —
dx
#
If
ax
e
is
v
—
with
wnn ae
dx
dy
,
ax
7.86
a
a solution for some value of a, then the given equation is satisfied when we replace y with
—
ax
2
-=
and --^1 with a 2 e ax Doing so. we obtain
anu
2y = e (a - a - 2) = 0,
dx
dx
dx<
dx
~
a — — 1
or 2. Thus y = e~ x and yv = e 2x are solutions,
,
satisfied when
y = e
has two distinct solutions of the form
.
which is
is a solution of the differential equation of the previous problem for any values
y — c x e * + c2e
of the arbitrary constants c, and c 2
Show that
.
y" — y
— 2y = 0, and since this differential equation is linear and
homogeneous, the result follows immediately from the principle of superposition.
/
7.87
Since both e~
x
and e 2x are solutions to
1
y" — y' — 2y = 0.
Find the general solution of
This is a second-order linear homogeneous differential equation with continuous coefficients on (-co, x
having the property that the coefficient of the highest derivative is nonzero on this interval. This equation
m
possesses two linearly independent solutions.
Two solutions, e~ x and e 2x were produced in Problem 7.85; they
,
e
W—
are linearly independent because their Wronskian is
x
e
2e
is
x
y = c\e
+ c2e
2x
= 3e x 7^ 0.
2x
Hence the general solution
:.
3
7.88
Show that the differential equation
.3
x
d
d)
— 6x—
—?
+ 12y =
)
form
y = x
has three linearly independent solutions of the
dx
dx
r
.
I By making the replacements
dy
y — x
2
= rx r-
d y
1
dx
dx
in the left
3,
or— 2. The corresponding solutions
v
W=
3x
-2x'
;
y = c,x
2
,
2
+ c 2x 3 + c 3 x
x 3 and x
,
3
,
and
y = x
2
which is satisfied when
r
= 2,
are linearly independent because
6x
2
is
a solution of the differential equation of the previous problem for any
,
Since x
y =x
r
= 20 * 0.
values of the arbitrary constants c u c 2 and c 3
M
.
l)(r- 2).v
dx
x
2
6\
Show that
y — x
2
d3y
S-2
^(r 3 - 3r 2 - 4r 4- 12) = 0,
member of the given equation, we obtain
.2
7.89
— r{r —
2
2
are all solutions to
.
x 3 y'" — 6xy' + 12y =
Problem 7.88), and since this
(see
differential equation is linear and homogeneous, the result follows immediately from the principle of superposition.
7.90
Is
f
the solution given in the previous problem the general solution to
Yes.
x 3 /" — 6xy' + 12y =
on (1, 5)?
The differential equation is linear, of order 3, and homogeneous: it has continuous coefficients on (1, 5)
with the property that the coefficient of the highest derivative is not zero on this interval. Thus, this equation
possesses three linearly independent solutions, which we found in Problem 7.88.
The general solution is the
superposition of these three linearly independent solutions.
7.91
Two solutions of
I
y" — 2y' + y =
are e
x
and 5e x
.
Show that
y = c,t
jX
+ 5c 2 e x
is
also a solution.
Since the differential equation is linear and homogeneous, the result is immediate from the principle of
superposition.
7.92
Determine whether
y = c^e
x
+ 5c 2 e x
is
the general solution of the differential equation in the previous problem.
I The differential equation is linear, of order 2. and homogeneous; it has continuous coefficients with the
property that the coefficient of the highest derivative is nonzero everywhere.
is
the superposition of two linearly independent solutions.
However, because
It
follows that the general solution
W(ex 5e x =
,
)
e
x
x
5e
x
=0,
functions are not linearly independent, and their superposition does not comprise the general solution.
the
LINEAR DIFFERENTIAL EQUATIONS— THEORY OF SOLUTIONS
7.93
159
Show that xe x is a solution of the differential equation in Problem 7.91.
= e x + xex and y" = 2e x + xe x into the left side of the differential equation, we
x
x
x
x
obtain y" - 2/ + y = 2e + xe - 2(e + xe + xe = 0. Thus, xe x satisfies the differential equation for all
values of x and is a solution on — oo, oo).
/
y = xe
Substituting
x
y'
,
,
x
)
(
7.94
Determine whether
I
It
of
y" — 2/ + y = 0,
y = cxe
x
+ c 2 xe x
is
the general solution of the differential equation in Problem 7.91.
and xe x are linearly independent. Since both functions are solutions
and since this is a second-order linear homogeneous differential equation with continuous
follows from Problem 7.50 that e
x
coefficients having the property that the coefficient of the highest derivative is nonzero everywhere, the
superposition of these two linearly independent solutions does comprise the general solution.
7.95
Determine whether
I
y = cxe
x
+ c 2 e~ x
is
y" — 2/ + y = 0.
the general solution of
Since y as given is not a solution (that is, it does not satisfy the differential equation when substituted into
the left side), it cannot be the general solution.
7.96
Determine whether
I
y = c e
x
x
+ c 2 e~ x
is
y"'
the general solution of
— y' — 0.
The general solution of a third-order linear homogeneous differential equation with constant
x
x
coefficients must be formed from the superposition of three linearly independent solutions. Although e and e~
are solutions and are linearly independent (see Problem 7.34), they constitute only two functions; they are one
short of the number needed to form the solution of a third-order differential equation.
7.97
It is
not.
Determine whether
it
I
is
y = c sin 2x
a solution of
is
t
y" + 4y =
for any value of the arbitrary constant c,
if
known that sin 2x is a solution.
Since the differential equation is linear and homogeneous, the result follows immediately from the principle
of superposition.
7.98
Determine whether
I
It is
y = C!sin2x
y" + 4y = 0.
the general solution of
is
The general solution of a second-order linear homogeneous differential equation with constant
not.
coefficients must be formed from the superposition of two linearly independent solutions of that equation; here
we have only one such function, namely sin 2x.
7.99
Show that
y,
= sin2x
and
are linearly independent on ( — oo, oo).
y2 = 1
The Wronskian of these two functions is
point in ( — oo, oo). In particular, at
7.100
Determine whether
I Although
is
y,
x = 0,
y = c sin 2x + c 2
x
= sin 2x
and
y2 = 1
is
W
W
sin 2x
1
= — 2 cos 2x,
which is nonzero for at least one
2 cos 2x
--=
— 2 / 0.
Therefore the functions are linearly independent.
the general solution of
y" + 4y = 0.
are linearly independent (see the previous problem), their superposition
not the general solution because one of the functions, namely
y 2 = 1,
is
not a solution to the differential
equation.
7.101
Determine whether y = Cj(3 sin 2x) + c 2 (4 sin 2x)
3 sin 2x and 4 sin 2x are solutions.
#
7.102
It is.
It is
a solution of
y" + 4y =
if it is
known that both
The result follows immediately from the principle of superposition.
Determine whether
#
is
not.
y = c x (3 sin 2x) 4- e 2 {4 sin 2x)
is
the general solution of
y" + 4y = 0.
Although both functions are solutions to the differential equation, they are not linearly independent
Hence the proposed solution is not the superposition of two linearly independent solutions
(see Problem 7.52).
and is not the general solution to the differential equation.
7.103
Determine whether
y = c t sin 2x + c 2 cos 2x
is
a solution of
y" + 4y =
if it is
known that both sin 2x and
cos 2x are solutions.
f
It is.
The result follows immediately from the principle of superposition. See also Problem 2.20.
CHAPTER 7
160
7.104
y = c
Determine whether
I
l
sin 2x
+ c 2 cos 2x
is
the general solution of
y" + Ay = 0.
Since the two solutions are linearly independent (see Problem 7.40), their superposition is the general
It is.
solution of this second-order linear homogeneous differential equation with constant coefficients.
7.105
Determine whether
if it is
I
y — c y + c2x
yt = 1
known that
a solution of
is
y2 — x
and
y" =
for any values of the arbitrary constants c t
and c 2
are solutions.
Since the differential equation is linear and homogeneous, the result follows immediately from the
It is.
principle of superposition.
7.106
Determine whether
I
not.
It is
y — c { + c2x
y'"
the general solution of
is
= 0.
The general solution of a third-order linear homogeneous differential equation with constant
Although y t = 1
y 2 = x are linearly independent (see Problem 7.5). they are only two in number and therefore one short
of the required number of solutions.
coefficients must be formed from the superposition of three linearly independent solutions.
and
7.107
Determine whether
y = c,(l — x) 4- c 2 (l + x) + c 3 (l — 3x)
arbitrary constants c u c 2
,
and c 3 if it is known that
y,
=
is
/" =
a solution of
— x,
1
y 2 = 1 + x,
for any values of the
and
V3
=
— 3x
1
are
solutions.
f
7.108
y — c,(l — x) -I- c 2 (l + x) + c 3 (l — 3x)
Determine whether
f
1
7.109
The result follows immediately from the principle of superposition.
It is.
not.
It is
— x,
-(-
1
is
y'"
the general solution of
— 0.
The general solution is the superposition of three linearly independent solutions, but the functions
and
x,
Determine whether
1
— 3x
are linearly dependent (see Problem 7.73).
y = C] + c 2 x + c 3 x
2
y" =
a solution of
is
if it is
known that 1. x. and x 2 are all
solutions.
f
7.110
It is.
The result follows immediately from the principle of superposition.
Determine whether
I
It is,
y = c, + c 2 x + c 3 x
2
the general solution of
is
y" = 0.
because the three functions 1, x, and x 2 are linearly independent (see Problem 7.37) and their number
(three) is the same as the order of the differential equation.
7.111
Determine whether
f
It is.
y = </,(5) -I- d 2 (3 - 2x) + J 3 (2 + x - \x )
2
The three functions 5,
3
— 2x,
and
2 + x — \x
2
is
y"'
the general solution of
are all solutions of
=
y'"
— 0.
(as may be verified
by direct substitution), and they are linearly independent (see Problems 7.74 through 7.76). Since there are three
such functions, their superposition is the general solution.
7.112
Problems 7.110 and 7.111 identify two general solutions to the same differential equation.
How is this possible
f The two solutions must be algebraically equivalent. We can rewrite the solution given in Problem 7.111 as
2
and
Then with c, = 5c/, + 3d 2 + 2d 3
c 2 = -2d 2 + d 3
y = (5d + 3d 2 + 2d 3 + (-2d 2 + d 3 )x + (-^ 3 )x
=
7.110.
c3
—\d 3 that solution is identical to the one given in Problem
)
l
,
.
,
,
7.113
Show that
d
_y_
^l_fl_
— + 5— — — — 3 fl
5
-r
dx*
y = e
I
is
2y =
-^
-j
dx 2
dx 3
has only two linearly independent solutions of the form
dx
ax
.
Substituting for y and its derivatives in the given equation, we get
satisfied when
a — 1,
1,
1,
—2.
e
x
x
ax
(a*
— a 3 — 3a 2
Since
e
e
e
e'
-2e-
2x
2x
e
#0
but
e
e
the linearly independent solutions are
y — e
x
and
x
e
x
e
x
e
x
y = e
e
2x
x
x
x
x
e*
e~
e
e
e
2x
x
-2e~ 2x
x
4e~
x
-Se~ 2x
2x
=
-I-
5a - 2) = 0,
which
LINEAR DIFFERENTIAL EQUATIONS— THEORY OF SOLUTIONS
7.114
Verify that
y = e
x
y = xe
,
x
y = x e
2
,
x
y = e~
and
,
2x
161
are four linearly independent solutions of the
equation of Problem 7.113, and write the primitive.
y = e
f By Problem 7.113,
x
y = e~
and
2x
By direct substitution in the given equation it is
are solutions.
found that the others are solutions. And, since
e
W=
e
e
e
x
xe
x
xe
x
x 2 ex
+ ex
xe x + 2e x
x
x
xe + 3e
x
x
x 2 e x + 4xe x + 2ex
2
x e
x
Verify that
y = e~ *cos3x
2
+ 6xe
+ 6e
4e
x
2x
1
sin3x
~2x
-Se~
4
2
2x
y = c e
The primitive is
y = e~
and
x
10
110-2 = _ 54^ *
12
13 6-8
-2x
-2e~ 2x
x 2 e x + 2xe x
x
these solutions are linearly independent.
7.115
e
t
x
+ c 2 xe x + c 3 x 2 e x + c 4 e~ 2x
are solutions of
—
^ + 4— + 13y =
dx
dx
z
.
and write the
0,
primitive.
f
Substituting for y and its derivatives, we find that the equation is satisfied by the proposed solutions. Since
W = 3e~ 4x # 0,
7.116
these solutions are linearly independent.
d3y
d 2y
dy
+ 2y =
2-^y
-jj
3
2
-f
Show that the differential equation
dt
dt
2x
y = e~
The primitive is
(c l cos3x
has solutions of the form
+ c 2 sin3x).
y = e
mt
,
where m
dt
denotes a constant.
y —
Substituting
me mt
'
dt~
d2y
,
2
li
' m 2 em
and
',
d3y
—-T3
dt
= m s em
.
,
into the left side of the differential
equation, we get
m 3 e m - 2m 2 e mt - me m + 2e mt - e m '(m 3
m = ±1
This is equal to zero when
or 2. Thus,
2m 2 - m + 2) = e mt (m - \)(m + l)(m - 2)
y l = e\
y 2 = e~\
and
y3 = e
2t
are solutions of the
differential equation.
7.117
Are the three distinct solutions found in the previous problem linearly independent?
I The Wronskian of e', e~', and e 2 was determined in Problem 7.26 to be — 6e 2t
Since it is nonzero for at
'
.
least one point in every interval (it is, in fact, nonzero everywhere), the functions are linearly independent.
7.118
What is the general solution of the differential equation in Problem 7.116?
*
Since the differential equation is linear, homogeneous, and of order 3 with constant coefficients (which implies
continuous coefficients having the property that the coefficient of the highest derivative is nonzero everywhere),
it
follows that the general solution is the superposition of any three linearly independent solutions.
results of the previous two problems, it follows that the general solution is
d y
d y
dy
—
—
2—^-+—
— 2y =
T — 2
Three solutions of
dt
1
+ c 2 e~' + c 3 e 2
From the
'.
2
3
7.119
y = c^e
„
3
dt
2
are known to be sin t, cos t, and
2 sin t — cos t.
Is
the general
dt
y — c t cos t + c 2 sin t + c 3 (2 sin t — cos t)l
solution
I No. The general solution is the superposition of three linearly independent solutions of the differential
equation.
The three solutions given here have a zero Wronskian (see Problem 7.31) and are, therefore, linearly
dependent. Their superposition cannot be the general solution.
3
dy
d
—
T + 4— =
y
7.120
Three solutions of
dt
y — c
f
1
3
are known to be 1, sin 2f, and cos 2f. Is the general solution
dt
+ c 2 sin 2t + c 3 cos 2r?
Yes.
The three functions have a nonzero Wronskian (see Problem 7.27) and are, therefore, linearly
independent. Their superposition is the general solution.
2
7.121
Show that the differential equation
d x
d*x
—
+ 4x =
T dt
a real constant.
A
5 —-=2
dt
has solutions of the form
x = e rt
,
where r denotes
CHAPTER 7
162
#
dx
—
=
x = e",
Substituting
d2x
-—, = rV,
re",
dr
dt
rV - 5rV + 4e
differential equation, we get
when
= ±1
r
or ±2. Thus,
yt
and
dr
n =
4
d x
—
T= V
into the left side of the
r
dt
- 5r + 4) = ^'(r - 4)(r - 1), which is equal to zero
2
=
e
and y4 = e~ 2t are solutions of the differential
y3
e"(r
y 2 = e~\
= e',
4
3
d x
—
T = r V,
2
2
2
',
equation.
7.122
7.123
Is
y = c x e' + c 2 e~' + c 3 e
I
Yes, by the principle of superposition.
Is the solution given in the
i
Yes.
and
/n 4
+ c 3 e~ 2t
2t
a solution of the differential equation of the previous problem?
previous problem the general solution of the differential equation of Problem 7.121?
The Wronskian of the four solutions is nonzero (see Problem 7.33 with m = 1, m 2 = — 1,
= —2). Thus they are linearly independent and their superposition is the general solution.
m 3 = 2,
1
2
7.124
d y
+ 4y =
—
d?
Show that the differential equation
f
y = e"
Substituting
x
IX
xx
x
y'
,
has solutions of the form e
^
— oce",
y" = cc 2 e' x
and
y.
7.125
7.126
e
Is
y — d e
I
Yes.
i2x
a = ±/2,
may be complex.
where
i
= y/—l.
Thus,
y\
— e i2x
a solution of the differential equation in the previous problem?
The result follows immediately from the previous problem and the principle of superposition.
Rewrite the solution given in the previous problem as the superposition of real-valued functions.
I
Using Euler's relations, we have
y = dxe
or
7.127
2
+ d 2 e~ i2x
l
if a
into the left side of the differentia]^ equation, we get
+ 4e* = e (a + 4), which is equal to zero only when
and y 2 = e~' 2x are solutions.
2
xx
i2x
+ d 2 e~ i2x = d,(cos2x + /sin2x) + d 2 (cos2x - z'sin2x) = (d + d 2 cos 2x + (id - id 2 sin2x
y = c, cos 2x + c 2 sin 2x,
where
cx
d 2y
Show that the differential equation
-j-j
2
=d +d
x
-6
dx
f
y = e'
Substituting
x
2
and
—
+ 25y =
dx
dy
= id — id 2
c2
)
y
.
l
has solutions of the form e" if a may be complex.
and its derivatives into the left side of the differential equation, we get
- 6<xe" x + 25e" = e"(a 2 - 6<x + 25), which is equal to zero only when
e
~'* )x
+ i4)x
=
are solutions.
and y 2 = e (3
a
3 + /4.
Thus, y =e (3
2
)
{
ax
<x
a
2
- 6a
-(-
25 =
or
when
l
7.128
Is
y = d e
f
Yes.
(i + iA)x
x
+ d 2 e (3 ~'* )x
a solution of the differential equation in the previous problem?
The result follows immediately from the solution to the previous problem and the principle of
superposition.
7.129
Rewrite the solution given in the previous problem as the superposition of real-valued functions,
f
Using Euler's relations, wc have
y = d ie
{3 + i4)x
= e 3x [(d
or
7.130
y — c1e
3x
l
~ i4)x
= d e 3x e" x + d 2 e 3x e~ i4x = e 3x [d (cos 4x + /sin4x) + d 2 (cos4x + /sin4x)]
+ d 2 cos 4x + (id — id 2 sin 4x]
+ d 2 e {3
{
x
)
)
l
cos4x + c 2 e
3x
s'm4x,
Show that the differential equation
where
cl
d x
dx
—
T + 4
dt
2
—d +d
l
1-
1
lx =
2
and
c2
= id — id 2
l
.
has solutions of the form
x = e",
where a may
dt
be complex.
I
Substituting
V+
+
a=-2±i'V7.
a
7.131
Is
4<xe"
x = dle
{
x — e xt
and its derivatives into the left side of the differential equation, we get
2
which is equal to zero only when a + 4a + 1 1 = 0, or when
" 2 "^ T)
are solutions.
and x 2 = e
= e"'(x 2 + 4a + 11),
~ +
Thus, Xl = e 2 ul)
1 le°"
~ 2 + u 7)t
{
+ d2 e
(
~ 2 ~ ul)t
'
(
'
a solution of the differential equation in the previous problem?
LINEAR DIFFERENTIAL EQUATIONS— THEORY OF SOLUTIONS
f
Yes.
163
The result follows immediately from the solution to the previous problem and the principle of
superposition.
7.132
Rewrite the solution given in the previous problem as the superposition of real-valued functions.
I
Using Euler's relations, we have
^
^
+ d 2 e 2 ^1) = d^- 2 1 + d 2 e- 2, e~ " lt = e^ 2\d x
+ d 2 e~ u 7t
2
= e'^d^cosjlt + isiny/lt) + d 2 {cos>j7t - isinV^t)] - e" '[(^i + d 2 cos Jit + {id - id 2 )s\r\ yftt]
x = <V- 2
-
-
+ ,vl) '
i
(
'
i
'
1
)
)
or
x = c { e~
2t
cosy/lt + e 2 e~
2t
where
sinyflt,
cl
-d +d 2
1
and
c2
x
= id —id 2
l
.
GENERAL SOLUTIONS OF NONHOMOGENEOUS EQUATIONS
7.133
y" — y' — 2y = 2x + 1
Find the general solution
if
I The general solution to the associated homogeneous equation,
x
y — c x e~ + c 2 e
x
2x
— x.
y = c x e~ + c 2 e
to be
7.134
2x
.
y — — x.
one solution is known to be
y" — y' — 2y = 0,
is
found in Problem 7.87
The general solution to the nonhomogeneous differential equation is
Find the general solution of
y" — y' — 2y = cos x + 3 sin x
if
y = —sin x.
one solution is known to be
/ As in the previous problem, the general solution to the associated homogeneous equation is
x
2x —
Therefore, the general solution to the given differential equation is y = c e~ + c 2 e
sinx.
y = Cje
-*
+ c 2 e 2x
.
1
7.135
Find the general solution of
/' — 2y' + y = x 2
if
y = x
one solution is known to be
2
+ 4x + 6.
I The general solution to the associated homogeneous equation, y" — 2/ + y — 0, is found in Problem 7.94
x
x
Therefore, the general solution to the given nonhomogeneous differential equation is
to be
y — c e + c 2 xe
x
x
2
=
c e + c 2 xe + x + 4x + 6.
y
.
{
x
7.136
Find the general solution of
y" —2y' + y = 2e 3x
if
one solution is known to be
y = je
3x
.
i As in the previous problem, the general solution to the associated homogeneous equation is y — c e x + c 2 xe x
x
x
ix
Therefore, the general solution to the given nonhomogeneous equation is y = c e + c 2 xe + \e
x
.
x
7.137
Find the general solution of
x3
•
—
— 6x — + 12_y = 12 In x — 4
^-6x^
-3
dx 3
if
one solution is known to be
dx
y - In x.
I The general solution to the associated homogeneous differential equation is found in Problem 7.89 to be
2
3
2
Therefore, the solution to the given nonhomogeneous differential equation is
y = c t x + c 2 x + c 3 x~
2
3
2
y = c x x + c 2 x + c 3 x~ + lnx.
.
7.138
Find the general solution of
y" + Ay = e 3x
if
one solution is known to be
y = Yje
3x
.
I The general solution to the associated homogeneous equation, /' + Ay = 0, is found in Problem 7.104 to be
y = Cj sin 2x + c 2 cos 2x. Therefore, the general solution to the given nonhomogeneous differential equation is
3x
y = Cj sin 2x + c 2 cos 2x + ^e
.
7.139
Find the general solution of
y" + Ay = 8x
if
one solution is known to be
y = 2x.
I As in the previous problem, the general solution to the associated homogeneous differential equation is
y = Cy sin 2x + c 2 cos 2x, so the general solution to the given differential equation is
y — c sin 2x + c 2 cos 2x + 2x.
{
7.140
Use the information of the previous two problems to ascertain a particular solution of
f
7.141
Since the differential equation is linear, a particular solution is
A particular solution of
y =
f
—J- sin 2x.
>'"'
= sin 2x
is
y = |cos 2x,
Determine a particular solution of
y'"
y = -^e
3x
y" + Ay = e 3x + 8x.
+ 2x.
while a particular solution of
>•'"
= cos 2x
= 3 sin 2x + 5 cos 2x.
Since the differential equation is linear, a particular solution is
y = 3(|cos 2x) + 5(-§sin 2x).
is
.
CHAPTER 7
164
7.142
Find the general solution of
y'"
— 3 sin 2x + 5 cos 2x.
I The general solution of the associated homogeneous differential equation is shown in Problem 7.110 to be
2
Using the result of the previous problem, we conclude that the general solution of the given
y = cy + c2x + c3x
2
equation is y = c + c 2 x + c 3 x + f cos 2x — f sin 2x.
.
x
7.143
A particular solution of y'" - 2y" - y + 2y = x is y = \x + £, while a particular solution of
— 2y" — y' + 2y = sin x is y — j^ sin x + ^ cos x. Determine a solution of
y'" — 2y" — y' + 2y = Ix — 3 sin x.
y'"
I
7.144
y — —3(jx + j) = — fx
2
d x
A particular solution of —-^ — 5 —j + 4x = 20e 3
A
r-
dt
I
'
is
x = \e
if
while a particular solution of
,
dA x
2
d x
d x
— 5 — + 4x =
—
=-
4
dt
-
Since
2
10e
3'
8t
is
x = 2t.
- 20f - ^(20e 3 - f (8f),
y =
a particular solution is
')
d2x
— 5 -r-=- + 4x = 10e 3 — 20f.
—x
4
dt
dt
Determine a F
particular solution of
'
2
^e M - f(2t) = ie
Since
60e
3'
= 3(20e 3
a particular solution of this differential equation is
'),
Find the general solution of
y" + y — x 2
if
,
one solution is
y — x
2
3'
)
— 2,
- 5f.
d 4x
2
y = 3(je
M — \e M
Use the information of the previous problem to determine a particular solution of
I
7.148
/" — 2y" — y' + 2y = |sin x.
y — |(^sinx + ^cosx) = |sinx + ^cosx.
d*x
7.147
4-
Use the information of Problem 7.143 to obtain a solution of
/ The solution is
7.146
y" — 2y" — y' + 2y = — 3x.
Use the information of the previous problem to obtain a solution of
I The solution is
7.145
y = 7(jx + i) — 3(^ sin x + jq cos x).
Since the differential equation is linear, a particular solution is
d x
^ — 5 —y + 4x = 60e
—
~di*~ ~d?
)
3 '.
.
and if two solutions of
y" + y —
are sin x and cos x.
f The Wronskianof {sinx, cosx} is
are linearly independent.
W
sin x
cos x
cosx
—sinx
— sin 2 x — cos 2 x = — # 0,
1
so these two functions
The general solution to the associated homogeneous differential equation, y" + y = 0,
Therefore, the general solution to the nonhomogeneous equation is
y — c sin x + c 2 cos x.
=
sin
c
x + c 2 cos x + x 2 — 2.
y
is
then
x
,
7.149
Find the general solution of
y" + y = x 2
if a
particular solution is known to be
y = x
2
+ 3 sin x — 2.
I As in the previous problem, the general solution of the associated homogeneous equation is
y = C sin x + c 2 cos x; then a general solution to the nonhomogeneous equation is
2
3 sin x — 2.
y = C\ sin x + C 2 cos x + x
x
-!-
7.150
Explain why the results of the previous two problems, in which we have generated two different general solutions
to the same differential equation, are not contradictory.
I The two general solutions must be algebraically equivalent to one another.
2
y = (C + 3) sin x + C 2 cos x + x — 2,
=
solution given in Problem 7.148 if we define
c
Cj + 3 and C 2 = c 2
Problem 7.149 can be rewritten as
1
In particular, the solution given in
which is identical in form to the
.
2
7.151
Find the general solution of
and x
y'"
= 12
if
one solution is
y = 2x
3
and three solutions of
y" =
are
1,
x,
2
.
I The general solution of the associated homogeneous equation, y" — 0, is shown in Problems 7.109 and 7.110
2
Then the general solution to the nonhomogeneous equation is
to be
y = c + c2x + c3x
2
3
=
x
+ 2x
c + c2x + c3
y
.
x
.
x
LINEAR DIFFERENTIAL EQUATIONS— THEORY OF SOLUTIONS
7.152
y = 2x
Rework the previous problem if instead of
3
a particular solution is known to be
y — 3x - 4x
f As before, the solution to the associated homogeneous equation is y = C,
C2x + C3x 2
2
2
solution to the nonhomogeneous equation is y = C, + C 2 x + C 3 x + 3x — 4x + 2x 3
-I-
;
165
2
+ 2x 3
.
thus the general
.
7.153
Explain why the results of the previous two problems, in which we generated two different general solutions
to the same differential equation, are not contradictory.
f The two general solutions must be algebraically equivalent to each other.
In particular, the solution given in
Problem 7.152 can be rewritten as y = C + (C 2 + 3)x + (C 3 - 4)x 2 + 2x 3
which is identical in form to the
and c 3 — C 3 — 4.
solution given in Problem 7.151 if we define c 2 = C 2 + 3
,
{
7.154
Find the general solution of
are 3 sin 2x and 4 sin 2x.
I
y" + Ay = e 2x
if
one solution is
y = \e
lx
and two solutions of
y" + 4y —
Since we do not have enough information to write the general solution of the associated homogeneous equation
(see Problem 7.102), we cannot write the general solution of the nonhomogeneous equation.
CHAPTER 8
Linear Homogeneous Differential
Equations with Constant Coefficients
DISTINCT REAL CHARACTERISTIC ROOTS
8.1
Solve
y" - y' - 2y = 0.
f The characteristic (or auxiliary) equation is
Since the roots
8.2
Solve
y" - ly
A, — —
A2 — 2
and
1
A
2
— A - 2 = 0,
(A + 1)(A — 2) = 0.
+ c 2 e 2x
which can be factored into
y = c x e~
are real and distinct, the solution is
x
.
= 0.
I The characteristic (or auxiliary) equation is A 2 — 7 A = 0, which can be factored into (A — 0)(A — 7) = 0.
Since the roots
A, =
and A 2 = 7 are real and distinct, the solution is y — c e 0x + c 2 e lx — c, 4- c 2 e lx
.
x
8.3
Solve
y" - 5y = 0.
f The characteristic equation is
roots
8.4
= y/5
A,
A
2
A 2 = — V5
and
— 5 = 0,
— >/5)(A + y/5) — 0.
%3x
+ c 2 e,-» 5x
y = c,e
which can be factored into
are real and distinct, the solution is
(A
Since the
Write the solution to the previous problem in terms of hyperbolic functions.
I
In succeeding steps we may write
= c cosh V5* + c sinh N 5.x + c 2 cosh y/Sx — c 2 sinh \/5x
— (Cj + c )cosh >/5x + (c — c 2 )sinh v5x = k cosh y/5x + k 2 sinh v5x
,
,
,
2
8.5
= c, + c 2
and
where
k
Solve
d 2 x/<fr 2 - 16.x = 0.
t
£2
= c, - c 2
t
.
f The characteristic equation is A 2 — 16 = 0,
4
distinct, the solution is
f c 2 e **.
x(t) — c,t'
which has the roots
A = ±4.
Since these are real and
'
8.6
Write the solution to the previous problem in terms of hyperbolic functions.
#
In succeeding steps we may write
\(/)
as
= c [\e*' + \e
4 ')
+ c,(\e*' - \e~*') + c 2 (W 4 +
v(f)
x
'
K 4
')
c 2 (\e*'
- \e~*')
= Cj cosh 4f + c, sinh 4f + c 2 cosh 4f — c sinh 4f
= (Cj + c 2 )cosh4f + (c, - c )sinh4f = k cosh 4f + k 2 sinh4f
2
2
8.7
Solve
d 2 r/dt
2
— to 2 r — 0,
x
where co denotes a positive constant.
I The characteristic equation is A 2 — co 2 — 0, which has the real and distinct roots
where A and B denote arbitrary constants.
is thus
r(t) = Ae
+ Be
10 '
8.8
10
Write the solution to the previous problem in terms of hyperbolic functions.
I
Since
r(t)
= Ae M + Be'
r(f)
C= A + B
Solve
-4-4-£ + y = 0.
2
dt
166
v
2
dv
dt
01
',
we have
- A(k + \e-°") + A{\e - K"") + B$eM + \e-°") - W&" - {e'
= A cosh cot + A sinh cot + B cosh cot — B sinh cot
— (A + B) cosh cot + (A — B) sinh cot = C cosh cot + D sinh cot
where
d
8.9
A = ±co.
',
and
(3 '
D = A - B.
10
'
03
')
The solution
LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS
# The characteristic equation is
,
X
4± V16-4 =
=
y = Cxe
,2 + » 3 >'
2 ± y/3.
3)l
- AX +
1
which, by the quadratic formula, has roots
Since these roots are real and distinct, the solution is
= de 3732 + Cy 02679
'
"
= 0,
om
f-
+ C 2 e {1 '
2
X
167
',
where C, and C 2 denote arbitrary constants.
2
8.10
Solve
dl
d I
—
T + 60 — + 500/ =
dr
0.
dt
# The characteristic equation is
8.11
Solve
— 50
Xx =
the roots
X2 =
and
+ 60A + 500 = 0, which can be factored into (X + 50)(A + 10) = 0.
—10 are real and distinct, the solution is /(f) = c^' 50 + c 2 e~ l0t
2
X
Since
'
.
x + 20.x + 64.x = 0.
I The characteristic (or auxiliary) equation is X 2 + 20/ + 64 = 0, which can be factored into
Since the roots A, = — 4 and X 2 = — 16 are real and distinct, the solution is
(X + 4)(X + 16) = 0.
4
16
=
x
c,e~
+ c 2 e~
'.
'
8.12
Solve
I
x + 128.x + 96.x = 0.
The characteristic equation is
X
2
+ 128A + 96 = 0,
2
X =
-128 ± V(128) —4(96)
X=.Q £<-64 + 20
2
8.13
Solve
v 10)1
d v
dy
dx
dx
= —64 + 20V10.
_~
2_
(D 2 + D — 6)y — {D — 2)(D + 3)y — 0.
since they are real and distinct, the primitive is
Solve
Q g -127.2t
_j_
-4 + -i-6>' = 0.
I We write the equation as
8.14
Since these roots are real and distinct, the solution is
q e (-64-20v 10)f _ Q e -0.7544f
_|_
which by the quadratic formula has roots
r-
2
d3v
d v
dv
dx
dx
dx
y = C e
x
3
2
d v
d
ax
dx
(D 3 + 2D 2 - 5D - 6)y = (D - 2)(D + l)(D + 3)y = 0.
They are real and distinct, so the primitive is
2m 2 — 5m + 2 =
d x
dx
—
+
dt
dt
-4+9
2
m = 1/2
14.x
—2
—7
roots
Xt —
Solve
q + 1000<j + 50,000^ = 0.
and
X2 =
f The characteristic equation is
—
=—
X =
and
X
2
q = c x e-*
219t
Solve
+ c 2 e- 9 *n2t
d Q
n dQ
—
y + 1000
1000^
-ty
4r +
dt
2
dt
or
(2m — l)(m — 2) = 0,
Then the general solution is
+ 9X + 14 = 0,
X
2
2x
Then the characteristic
+ C 2 e~ x + C 3 e _3x
.
so the real and distinct
y = c^e
xl2
+ c 2 e 2x
.
which can be factored into (X + 2)(X + 7) = 0.
2
7
x = c { e~ + c 2 e~
+ 1000A + 50,000 = 0,
= —52.79
.
2
8.19
m — 2.
are real and distinct, the solution is
2
is
y — C{e
= 0.
f The characteristic equation is
8.18
;
.
2y" - 5/ + 2y = 0.
characteristic roots are
Solve
+ C 3 e~ 3x
dx
f The auxiliary equation is
8.17
Then the characteristic roots are 0, 4
4x
dv
v
roots are 2,-1, and —3.
Solve
y — C, + C 2 e
-4J + 2-4-5-^- 6y = 0.
I We write the equation as
8.16
Then the characteristic roots are 2 and — 3;
.
(D 3 — D 2 — l2D)y = D(D — 4)(D + 3)y = 0.
and —3. They are real and distinct, and the primitive is
Solve
+ C 2 e~ 3x
-4 --4 -12-^ = 0.
I We write the equation as
8.15
2x
160,000<2 = 0.
'
Since the
'.
which, by the quadratic formula, has roots
and —947.2. Since these roots are real and distinct, the solution
—
168
CHAPTER 8
I The characteristic equation is X + 1000/ + 160,000 = 0, or (/. + 200)(A + 800) = 0. Since the roots
-200
+ c 2 e -800
and k 2 = -800 are real and distinct, the solution is Q = c^
zlj = -200
2
'.
'
2
dx
d x
n
—
+ k—=
,
8 20
Solve
=-
dt
,
,
,
where k denotes a real constant.
0,
2
dt
m 2 + km = 0,
I The characteristic equation is
m, =
m 2 ——k
and
2
8.21
Solve
d x
g
dt
10
—5
mim + k) = 0.
x = C t + C 2 e~ kt
x = 0,
The roots
which can be factored into
are real and distinct, so the solution is
.
where g denotes a positive constant.
m 2 - g/10 = 0,
I The characteristic equation is
*
x = C 1 e-/°rs + C 2 e^^^'.
which has the roots
m = ±-Jg/\0.
The solution is then
k = ±yjk/m.
The solution is thus
t
2
8.22
Solve
d x
—
dr
k
^
m
where both k and m denote positive constants.
x = 0,
X2 - k/m = 0,
I The characteristic equation is
8.23
Solve
y" - \y
which has as its roots
+ 2y = 0.
m 2 — §m + 2 — 0, which may be written as 2m 2 — 9m + 4 = 0. This last
equation can be factored into (2m — \)(m — 4) = 0. The roots are m = | and m 2 = 4; the solution is
x 2
+ Be* x
y = Ae
I The auxiliary equation is
x
'
.
8.24
Solve
y" - {y' - |y = 0.
— \k — \ = 0, or
The roots are k = \ and
f The auxiliary equation is
(2A — l)(4x + 1) = 0.
8.25
Solve
X
2
x
8/ 2 — 2/. — 1
X 2 = — i;
= 0.
This last equation can be factored into
y = c^e*'
the solution is
2
+ c 2 e _Jt/4
.
y - 2y + \y = 0.
m 2 — 2m + \ — 0, or 2m 2 — 4m +1=0. Using the quadratic formula, we
2
-4(2)(l)
/4±V(-4)
,,
u*
•
.
—
= ± V2/2.
The solution is
m —
obtain
the roots
to .v
this last equation as
f The characteristic equation is
•
1
+ Be{1 -^' 2)t
{1+Vil2)t
Solve
y
Jy + 2 T- = n
-A-l
1
dx
dx
dx
d2
.
1
.
rfv
°-
2
L
i
—
ri
y = Ae
d3
8.26
.
I The characteristic equation is k 3 - 3/. + 2X = 0, which may be factored into k(k - 1)(/. - 2) = 0.
= 2, are real and distinct. The solution is y = c + c 2 ex + c 3 e 2x
roots,
X x =0, k 2 = 1, and
3
2
3
8.27
Solve
d y
dy
dx
dx
The
.
/.
x
—^ - t~ = 0.
I The characteristic equation is X 3 - k = 0, which may be factored into
x
x
and ± 1, are real and distinct, so the solution is y = c t + c 2 e + c 3 e~
k{k
- 1)(A + 1) = 0. The roots,
.
8.28
Solve
- 6y" + lly' - 6y = 0.
y'"
f The characteristic equation is
The roots are
8.29
Solve
y
(4)
X x = 1,
X 2 = 2,
k
3
- 6X2 + \IX — 6 = 0,
and
A 3 = 3;
which can be factored into
hence the solution is
y = c1e
x
{k
— 1){X — 2)(k - 3) = 0.
+ c 2 e 2x + c 3 e 3x
.
- 9y" + 20y = 0.
I The characteristic equation is A 4 - 9/ 2 + 20 = 0, which can be factored into
The roots are ^=2, A 2 = -2, A 3 =V5,
(k - 2)(A + 2)(A - V5)(A + v^) = 0-
and
/.
4
= -y/5;
the solution is
y
c1e
2x
+ c 2 e" 2x + c 3 e v5A: + c 4 e~* 5x = /^cosl^x +
/c
2
sinh2x + k 3 coshyJlx + /c 4 sinh>/5x
hence
LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS
D
169
+ 36 = 0, which may be factored first into (A 2 - 4)(A 2 - 9) =
=
3)(A + 3)
The roots, ±2 and ±3, are real and distinct. The solution is
(A
2)(A + 2)(A
0.
and
d 2x
d*x
8.30
—dt*x - 13 —dt
Solve
-=-
2
„,
+ 36x = 0.
X* - 13A
I The auxiliary equation is
then into
x = cxe
2t
+ c 2 e~ 2t + c 3 e 3 + c 4 e' 3 = /cjCosh2f +
'
'
d*x
dx
d x
d x
- 10 —x + 35 — - 50 — + 24x =
—r
dt*
dt
dt
dt
Solve
fe
2 sinh2t
+
/c
3
cosh3f + /c 4 sinh3r
2
3
8.31
2
-=-
3
0.
2
I The characteristic equation is m* — 10m 3 4- 35m 2 — 50m 4- 24 = 0, which can be factored into
(m — l)(m — 2)(m — 3)(m — 4) = 0. The roots are 1, 2, 3, and 4, so the solution is
x = C e + C 2 e 2t + C 3 e 3 + C> 4
l
'.
'
x
5
8.32
Solve
2
3
d*x
d x
d x
d x
dx
- — r + 35 —^ - 50—T + 24 — =
—
dr
dr
„
=-
^
„
10
dr
3
dt
2
0.
dt
m 5 — 10m 4 + 35m 3 — 50m 2 + 24m = 0, which may be factored into
m(m — l)(m — 2)(m — 3)(m — 4) = 0. The roots are m =0, m 2 = 1, m 3 = 2, m 4 = 3, and m = 4;
solution is
x = c + c 2 e' + c 3 e 2 + c 4 e 3 + c 5 e 4t
I The characteristic equation is
5
x
'
.
t
8.33
Solve
the
'
x (5) - 3x <4) - 3x (3) + 9x + 2x - 6x = 0.
— 3A 4 — 3A 3 + 9A 2 + IX — 6 = 0, which we factor into
or (A - 1)(A + 1)(A - Jl){X + ^2)(X - 3) = 0. The roots are ±1, ±^2, and 3,
(a - \){k - 2)(X - 3) =
21
where A, B, C, D,
which are real and distinct; thus the solution is x = Ae' + Be~' + Ce^2 + De^^ + Ee 3
I The characteristic equation is
X
5
2
2
'
',
and E are arbitrary constants.
8.34
Solve
y'
- 5 y = 0.
I The characteristic equation is
8.35
Solve
m = — 1/2,
Solve
which has the single root
A x = 5.
The solution is then
y = c^ *.
5
(2D 3 - D 2 - 5D - 2)y = 0.
f The auxiliary equation is
8.36
X — 5 = 0,
2m 3 — m 2 — 5m — 2 =
— 1, and 2. The general solution is then
or
(2m + l)(m + l)(m — 2) = 0.
y = c l e~ xl2 + c 2 e~
x
+ c 3 e 2x
The roots are
.
(D 3 + D 2 - 2D)y = 0.
I The auxiliary equation is m 3 + m 2 — 2m =
x
2x
solution is then
y = C^ + C 2 e + C 3 e~
or
m(m — l)(m + 2) = 0,
so that
m = 0,
1,
and — 2. The
.
8.37
Solve
(D + 6)y = 0.
I The characteristic equation is
y = Ae~
8.38
Solve
m + 6 = 0,
which has the single root
m = — 6. The general solution is
2X — 5 = 0,
which has the single root
X — 5/2.
6x
.
(2D - 5)y = 0.
I The characteristic equation is
5xl2
y = Ae
The general solution is
.
8.39
Solve
(D 3 - 2D
2
- D + 2)y = 0.
# The characteristic equation is X 3 - 2X 2 — X + 2 =
2x
x
x
the solution is y = c±e + c 2 e~ 4- c 3 e
or
(X
2
— l)(X — 2) = 0. The roots are ±
1
and 2, and
.
8.40
Solve
(D
4
- 8D 2 + 15)x = 0.
2
2
or (A - 3)(A - 5) = 0.
f The characteristic equation is A 4 - 8 X 2 + 15 =
_y5
5
3
3t
and the solution is x = C e^ + C 2 e~^ + C 3 e^ + C 4 e
'
'
The roots are ±^3 and ±V5,
'.
l
8.41
A second-order linear homogeneous differential equation for y(x) with constant coefficients has
A 2 = 4 as the roots of its characteristic equation. What is the differential equation?
Xx = 2
and
170
CHAPTER 8
I The characteristic equation is
y" — 6/ + 8 v = 0.
is
8.42
— 2)(A — 4) =
(A
# The characteristic equation is
equation is y" — 17 y = 0.
— 6/ + 8 = 0.
The associated differential equation
k = ±\l\l
What is the differential equation?
or
(
A
— 17 = 0.
2
The associated differential
Find a second-order linear homogeneous differential equation in x(t) with constant coefficients that has, for the
2 + ^5-
I The characteristic equation is
differential equation is
[A
— (2 + \/5)][A — (2 — V5)] =
or
A
— 4/ — 1 = 0. The associated
2
x — 4x — x = 0.
Find a third-order linear homogeneous differential equation in x(t) with constant coefficients that has, for the
roots of its characteristic equation,
—1 and ±>/3.
f The characteristic equation is
[m — (— l)](m — V3)[m — ( — V3)] =
m 3 + m 2 — 3m — 3 = 0.
or
The
—
2
3
d x
dx
—^T + ~ji2~~ 3 17 - 3x = 0.
~dt
~dt
~di
d x
associated differential equation is
8.45
2
— \JTl)[?. — — >/l7)] =
(A
roots of its characteristic equation,
8.44
A
A second-order linear homogeneous differential equation for y(x) with constant coefficients has
as the roots of its characteristic equation.
8.43
or
A fourth-order linear homogeneous differential equation in y{t with constant coefficients has, as the roots of its
— ± >/8. What is the differential equation?
and
\ + y/5
)
characteristic equation,
1
I The characteristic equation is [m - (§ + \f5)][m — ft — >/5)][m - (-1 + V8)][m - (- - V8)] = 0, which
or
m* + m 3 — "m 2 — \m + *£* = 0. The associated
we can simplify to (m 2 — m — ^)(m 2 + 2m — 1) =
1
differential equation
4
</
v
dt*
8.46
Show that
d
is
3
dt
v
55d 2 v
5dv
133
3
2
2dt
4
4 dt
(D - a)(D - b)(D - c)y = (D- b)(D - c)(D - a)y.
I We expand both sides of this equation and show they are equal:
~
d 2y
A%- cy
3
d y
d
dx
dx
dy_
_
(b + c)^-+
bey
c)
-(b
dx 2
2
dx
dy
y
—^-(a
+ b + c)—^ + (ab + bc + ac)-~
3
dx
__
2
d y
(D - b)(D - c)(D - a)y = (D - b)(D - c)(j^- ay
3
d2
dy
dx
dx
dx
d
'
8.47
Verify that
y = C^e
ax
= t-4y3 - {a + b + c) —4yz + (ab + ac + be)-
+ C 2 e bx + C 3 ecx
I We need to show that
satisfies the differential equation
(D - a)(D - b){D - c){C l e!
,x
+ c)
dy
_
+ acy
abcy
(D - a)(D - b)(D - c)y = 0.
+ C 2 e bx + C 3 ecx = 0.
)
(a
For the first term on the left, we
have
(D - a)(D - b)(D - c)C x e
ax
= (D - b)(D - c)(D - a)C^ x =(D- b)(D - c)0 =
and similarly for the other two terms.
DISTINCT COMPLEX CHARACTERISTIC ROOTS
8.48
Show that if the characteristic equation of a second-order linear homogeneous differential equation with real
constant coefficients has complex roots, then these roots must be complex conjugates.
f Denote the two roots as a + ib
[X - (a + ib)~\\_k — (c + id)] =
or
and
c
+ id,
where
i
= V — 1.
Then the characteristic equation is
The associated
)} - X[(a + ib) + (c + id)] + [(a + ib)(c + id)] = 0.
differential equation (with y as the dependent variable and x as the independent variable) is
^-
[(a
+ ib) + (c + id)] £+[(« + ib)(c + id)]y = 0.
LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS
171
+ ib) + (c + id) = {a + c) + i(b + d) must be real, which
2
(a + ib)(c - ib) = (ac + b
+ ib(c - a). This
b =
or c = a. We discard b =
coefficient is real only if
as a possibility, because if
is so then the
roots are not complex as hypothesized. Thus, for the differential equation to have complex roots a + ib
and
we must have d=-b and c = a, which implies that the roots are complex conjugates.
c + id,
If the coefficients
requires that
of this equation are real, then
d = -b.
(a
Then the coefficient of y becomes
)
it
8.49
Derive a real-valued solution for a second-order linear differential equation with real constant coefficients if
the roots of its characteristic equation are complex.
I Assume the unknown function is y(x). By the previous problem, the roots of the differential equation must
be complex conjugates; denote them as A = a + ib and X 2 — a — ib. Then two linearly independent
~
~
+ lb)x
solutions are e
and e (a lb)x and the general complex solution is y(x) = d e + ib)x + d 2 e ib)x Using
lbx
,bx
— cos bx + sin bx and e~ — cos bx — sin bx, we can rewrite this solution as
Euler's relations,
e
t
(a
la
,
(a
.
l
i
i
+ d 2 e ax e~ ibx = eax{d e ibx + d 2 e' ibx
= eax [d (cos bx + sin bx) + d 2 (cos bx — sin bx)~]
= eax [{di + d 2 cos bx + i{d^ — d 2 sin bx]
y = dxe
ax ibx
e
)
x
i
i
1
)
If
we define
solution as
)
— d + d 2 and c 2 = i(d — d 2 as two new arbitrary constants, we can write the general
ax
ax
=
c e
cos bx + c 2 e sin bx. This equation is real if and only if c and c 2 are both real, which
y
cl
y
)
x
x
x
occurs if and only if d l and d 2 are complex conjugates. Since we are interested in the general real solution, we
must restrict d and d 2 to be a conjugate pair.
x
2
8.50
d y
Solve
-j-2
dx
2
dy
~ 6 ~t + 25y = 0.
dx
I The characteristic equation is
X2 — 6A + 25 = 0.
a —
,
-(-6)±V(-6) -4(25) = 6±V^64 =
3 ± i4.
a — 3
(from the result of Problem 8.49 with
8.51
-f +
%-"%
Solve
-j-4
10
2
8.52
and
-(-10) ± V(-10) -4(29)
A
2
b — 4)
— 10A + 29 = 0.
10±y^T6
=
/.
=
is
(from the result of Problem 8.49 with
y — e
3x
(c { cos 4x
+ c 2 sin 4x).
a = 5
Using the quadratic formula, we find its roots to be
= 5 + /2.
Since these roots are complex conjugates, the solution
6 = 2)
and
y = e
5x
(c l cos 2x -f c 2 sin 2x).
d2y
I The characteristic equation is
X
2
+ 9 = 0,
which has as its roots
are complex conjugates, the solution is (from Problem 8.49 with
8.53
.
.
,
+ 9y = 0.
73
d?
Solve
y = e
.
,
Since these roots are complex conjugates, the solution is
29y = 0.
f The characteristic equation is
,
Using the quadratic formula, we find its roots to be
„.„„.,
2
0x
Solve
(c 1 cos 3x
-I-
r-
2
8
A = ± i3 =
and
± ;3.
Since these roots
b — 3)
c 2 sin 3x) = c t cos 3x + c 2 sin 3x.
dx
d x
—
+ — + 25x =
dt
a =
0.
dt
I The characteristic equation is
A
2
+ 8A + 25 = 0.
Using the quadratic formula, we find its roots to be
-8± V(8) -4(25) = -8 ±7^36 = — ,.„„..
Since these roots are complex
2
A =
,
x = e
8.54
Solve
,
4 ± i3.
-4
'(Ci cos 3t
,
.
.
conjugates, the solution is
+ c 2 sin 3r).
-^2 + 4 — + 8x = 0.
d x
dx
dt
dt
I The char acteristic e quation is A 2 4A + 8 = 0.
_4 + V(4) 2 -4(8)
.„
=—
= -4±V^6 = — 2„ i2.
=
-I-
±
,
A
x = e~ 2 \c^ cos2r + c 2 sin2r).
Using the quadratic formula, we find its roots to be
.
.
,
.
,
Since these roots are complex conjugates, the solution is
172
CHAPTER 8
D
2
8.55
d Q
Solve
--,-f
2~
„dQ
M 8S -^ + 52Q = 0.
+
^
~dl
A 2 + 8A + 52 = 0.
f The characteristic equation is
-8±V(8)
A =
2
-4(52)
=
-8 + V-144
Using the quadratic formula, we find its roots to be
= — 4 ± i6.
Since these roots are complex conjugates, the solution is
4
Q = e~ '(ci cos 6f + c 2 sin 6t).
8.56
d2 l
dl
dt
dt
tnn —
+ 50,000/ = 0.
-^2 + 100
Solve
m 2 + 100m + 50,000 = 0,
f The characteristic equation is
- 100 ± V(100) - 4(50,000)
2
m—
/
8.57
r—
en
rn
= — 50
± i50V19.
m 2 + 16 = 0,
which has as its roots
i'4.
Since these roots
m= +
1'8
=
± i8.
Since these roots
and
).
X x = 2\
which has roots
= —2i.
Since these roots
2
y = c x cos 2x + c 2 sin 2x.
y + 50y = 0.
Solve
I The characteristic equation is
y = c, cos \/50f + c 2 sin V50r.
X
2
+ 50 = 0,
which has roots
A = ±i'v50-
The solution is
x + 96x = 0.
Solve
2
and A 2 = — iy96.
a = i>J96
A + 96 = 0, which has roots
=
x
C 2 cos y/96t + C sin \f96t, where C 2 and C denote arbitrary constants.
The characteristic equation is
solution is
x
x
x
x + 12.8x + 64x = 0.
Solve
f The characteristic equation is
m=—
—
Solve
x + ^x + 96x - 0.
m 2 + 12.8m + 64 = 0,
= -6.4 ± i'4.8.
f The characteristic equation is
=
+
x — c x cos 8f + c 2 sin 8f.
are complex conjugates with real part equal to zero, the solution is
8.63
=
x
)? + 4 = 0,
f The characteristic equation is
8.62
i"4
y" + 4y - 0.
Solve
f
m= +
which has as its roots
x = c cos At + c 2 sin4r.
m 2 + 64 = 0,
are complex conjugates, the solution is
8.61
-(1/32) ±7(1/32)'- 4(96)
The solution is
m 2 + y^m + 96 = 0,
_ _
which has roots
+
x = e'
6A \c cos4.8t + c sin4.8r).
2
l
which has roots
The so|u
2
8.64
x = e-°
015625
Solve
-T2 + 20 — + 200/ = 0.
'(C 1 cos9.7979f
d2/
d/
dt
dt
+ C 2 sin 9.7979r).
# The characteristic equation is
A2 =
8.65
,
x + 64x = 0.
Solve
I The characteristic equation is
8.60
.
,
Since these roots are complex conjugates, the solution is
x + 16x = 0.
Solve
are complex conjugates, the solution is
8.59
„.
= ^- 50 '(c, cos 50Vl9t + c 2 sin 50Vl9f)-
I The characteristic equation is
8.58
which has as its roots
Solve
- 10 - iTO. The solution is
y" - 3y' + 4y = 0.
+ 20A + 200 = 0, which has roots
10,
/ = e~
(ci cos lOt + c 2 sin lOf).
A
2
kx =
- 10 + HO
and
The
~
.
LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS
/
The characteristic equation is
The solution is
8.66
y = c x e°
l2)x
cos
A
2
- 3A + 4 = 0,
—- x +
V7
c2e
i3l2)x
which has roots
—
A,1
173
3/73
——
—
=-+
and
i
2
2
i
-v/7
=
A,
1
2
2
Jl
sin
x.
y" + 4/ + 5y = 0.
Solve
# The characteristic equation is A 2 + AX + 5 = 0,
2x
2x
solution is y = c e~
cos x + c 2 e
sin x.
which has roots
Xl —
—2 +
which has roots
m=
—
A2 =
and
—2 —
1.
The
x
8.67
(D 2 + D + 2)y = 0.
Solve
The characteristic equation is
m 2 + m + 2 = 0,
=
2
/
solution is
8.68
y = e~
xl2
\.A cos
Jjx
—
—
— B —\flx\
sin
h
2
where A and B denote arbitrary constants.
This solution
y
)
d2q
).
^ nn dq
^
+ 200
+ 400,000g = 0.
-yf
2
dt
dt
Solve
/ The characteristic equation is
-200 +
~ V40,000 -1,600,000
A =
,
'
2
~ 100 '(C! cos
X
2
+ 200A + 400,000 = 0,
/—
= - 100 + ilOO V39.
The solution is
dq
q
Mnn
400 -T + 200,000^ = 0.
-^
+ 400-^
TT
dt
dt
Solve
2
I The characteristic equation is X 2 + 400 X + 200,000 = 0,
~ 200t (A
is
cos 400f + B sin 400r).
q = e
— + 40— +
d2 l
8.71
Solve
dt
8.72
2
dl
A =
which has roots
- 200 + *'400. The solution
800/ = 0.
dt
2
I The characteristic equation is
A
- 20 - i20. The solution is
/
X2 =
which has roots
100V39r + c 2 sin 100^390-
d2
8.70
+ 40A + 800 = 0, which has the roots
= e~ 20f (c cos 20f + c 2 sin 20t).
Xx =
— 20 + i20 and
1
x + 25x = 0.
Solve
m 2 + 25 = 0,
which has roots
m = /5
I The char acte ristic equation is m 2 + 128 = 0,
x = C cos y/l2St + C 2 sin y/l2&t.
which has roots
m — ±i-Jl2S. The solution is
f
.
2
(D 2 - 2D + I0)y = 0.
Solve
q = e
— The
J,
# The characteristic roots are 1 + 3i, so that the primitive is y = ex{C cos 3x + C 2 sin 3x).
may also be written as C 3 e x sin (3x + C 4 or C 3 e x cos (3x + C 5
8.69
_hi
The characteristic equation is
t
and
m 2 ——i5.
The solution is
x = Cj cos 5f + C 2 sin 5r.
8.73
x + 128x = 0.
Solve
x
8.74
x + 3^x =
Solve
where g denotes a positive constant.
m 2 + 3g — 0,
which has roots
m=±i^j3g.
# The characteristic equation is m 2 + 488 = 0,
y=C cos 22.09x + C 2 sin 22.09x.
which has roots
m = +1^488 = +i22.09.
I The characteristic equation is
The solution is
x = C, cos v^^ + C 2 sin v^^8.75
{D
Solve
x
2
+ 488)y = 0.
The solution is
174
8.76
CHAPTER 8
D
Determine the characteristic equation of a second-order linear homogeneous differential equation with real
coefficients if one solution is e"'cos 5f.
8.77
8.78
This particular solution corresponds to the roots
[A
- (-1 + /5)][A - (- 1
= 0,
or
)
2
2'
This particular solution corresponds to the roots
[A
- (2 + i'3)][/ - (2 - i3)] = 0,
or
k
2
Thus, the characteristic equation is
sin 3r.
f
2 + i3.
Thus, the characteristic equation is
- Ak + 13.
Solve Problem 8.76 if one solution is cosv3f.
+ iyJ3.
This particular solution corresponds to the roots
[A - /\/3][;.
- - V3)] (
or
1
k
The characteristic equation is
+ 3 = 0.
2
Find the general solution to a second-order linear homogeneous differential equation for x(t) with real coefficients
if
8.80
i"5)]
Solve the previous problem if one solution is e
f
8.79
— 1 + i5.
+ U + 26 = 0.
I
one root of the characteristic equation is
3
+ il.
I
Since the roots of the characteristic equation must be a conjugate pair (see Problem 8.48), the second root
is
3
— il.
x = e 3, (ci cos It + c 2 sin It).
The general solution is then
Solve the previous problem if, instead, one root of the characteristic equation is
f
It
follows from
— 18.
Problem 8.48 that the second root is +i'8. The general solution is then
x — f, cos 8f + c 2 sin 8f.
8.81
Find the general solution to a second-order linear homogeneous differential equation for x{t) with real coefficients
if a
2
particular solution is e 'cos 5t.
A second linearly independent solution is e 2 sin 5f, so the general solution is
=
(The given particular solution is obtained by taking
and c 2 — 0.)
I
'
<
8.82
x = e
2,
(c x cos 5f
+ c 2 sin 5/).
1
,
Solve the previous problem if, instead, a particular solution is 3e 'sin4r.
I Such a particular solution can occur only if the roots of the characteristic equation are — ± /4,
which implies that the general solution is x = e~'(c cos4r + c 2 s'm4t). The given particular solution is the
special case
c, =0,
c 2 — 3.
1
{
8.83
Solve Problem 8.81
f
if,
instead, a particular solution is 8cos3f.
Such a particular solution can occur only if the roots of the characteristic equation are
+ /3, which
x = c, cos 3f + c 2 sin 3t. The given particular solution is the special case
implies that the general solution is
c,
c 2 = 0.
= 8,
4
r
</
8.84
~4 + 10
Solve
ax
~+
d2 v
9v = 0.
ax
I The characteristic equation is m 4 + 10m 2 + 9 = 0, or
has roots m, =
m 2 = —i, m 3 — /3, and w 4 = —
ix
ix
=
d^e + d 2 e~
v
+ d 3 e i3x + d A e~ i3x
it
;'3.
i,
(m 2 + \){m 2 + 9) = 0;
Since the roots are distinct, the solution is
.
Using Euler's relations (see Problem 8.49), we can combine the first two terms and then the last two terms,
rewriting this solution as
y = c, cos x + c 2 sin x + c 3 cos 3x -(- c4 sin 3x.
d4 v
8.85
d3 v
d2 v
+ t4 + ^4 + 2 = °t4
dx*
dx
ax
Solve
>'
J
z
w 4 + m 3 + m 2 + 2 = 0. which we factor into (m 2 - m + l)(m 2 + 2m + 2) = 0;
Since the roots are distinct, the solution is
its roots are
and — ±
1/2 + iyJ3/2
~^ 3/2)x
W2+i 3 2)x + d e
+ d 3 e<- +i)x + d 4 e^ l,) \
y = d e
2
f The characteristic equation is
1
{i
-
2
/.
'
l
Using Euler's relations (see Problem 8.49). we can combine the first two terms and then the last two
x
x;2
y = e {c cos V3x/2 + c 2 sin \f3x/2) + e~ (c 3 cos x + c 4 sin x).
terms, rewriting this solution as
8.86
Solve
y
w + Ay" -
y'
+ 6>- = 0.
x
LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS
m 4 + Am 2 - m + 6 = 0,
f The auxiliary equation is
its
^ e
roots are
— 1/2 ± i>/TT/2
(l/2 + iV7)x
(l/2-iV7)x
and
1/2
± ijl.
which we factor into
y = d e
The solution is
{
(m 2 + m + 3)(m 2 - m + 2) = 0:
u 1)x + d e 12 '-" 2 " +
~ X2> '^
{
2
x
+^ g
Using Euler's relations, we can combine the first two terms and then the last two terms, rewriting this
8.87
Solve
y
(4)
x,1
y = e~
solution as
(c l cos y/llx/2
- 4y'" + 7y" - Ay' + 6y - 0.
and
4
— 4A 3 + IX 2 - AX + 6 = 0, has roots A t = 2 + iv'2,
+
~^ 2)x + <*>'* + d e _,x
The solution is y = c/,e <2 vl) * + d 2 e i2
4
f The charateristic equation,
A 3 = i,
+ c 2 sin VTTx/2) + e* /2 (c 3 cos V"7x + c4 sin >/7x).
XA = -i.
A
''
.
/.
2
= 2 — i>/2,
Using Euler's
relations, we can combine the first two terms and then the last two terms, to rewrite the solution as
2x
2x
cos y]2x
c2e
sin V2x
c 4 sin x.
cxe
c 3 cos x
y
=
8.88
+
+
+
Find the solution of a fourth-order linear homogeneous differential equation for x(r) with real constant
2 + i3
coefficients if the roots of its characteristic equation are
x
1
Solve Problem 8.88 if the roots are
\ ± i\
|
± i\.
l
-7 ±
Solve Problem 8.88 if the roots are
1
Solve Problem 8.88 if the roots are
and ± i3.
+ c 4 sin3f = {c^' 11 + c 3 )cos3f + (c 2 e~ 7 + c 4 )sin 3f
c 3 cos3f
-I-
±
2
i'3
'
and ±
i
i.
x = e 2, (c cos t + c 2 sin r) + c 3 cos t + c 4 sin f.
f The solution is
8.93
and
x = e" 2 (c cos \t + c 2 sin ^r + c 3 cos ^r + c 4 sin ^f).
/ The solution is
1
x = e~ '{c cos 3r + c 2 sin3f)
8.92
-7 ± i83.
and
1
x = e~ 15t (c cos 1 If + c 4 sin lit) + e~ 7 '(c3Cos83r + c 4 sin83f).
I The solution is
8.91
-15 + i\
Solve Problem 8.88 if the roots are
# The solution is
8.90
— 5 + 18.
x = e 2 '(c cos 3f + c 2 sin 3r) + e~ 5, (c 3 cos 8t + c 4 sin 8f).
# The solution is
8.89
and
l
Determine a differential equation associated with the previous problem.
I The characteristic equation is
= [X - (2 + i)][A - (2 - i)!P - CP -(-')] - (A 2 - 4A + 5)(X 2 + 1)
= A 4 - AX 3 + 6X 2 - AX + 5
x
d x
dx
d x
—
r — 4 —r + 6 — — 4 -— + 5x =
dt*
jj
The associated differential equation is
dt
8.94
-^
2
3
dt
0.
dt
Determine a differential equation associated with Problem 8.90.
i The characteristic equation is
=
<H
29
,
1
\->\
2
2
,
13
n
-2
+
1
k2 - k +
J
4
\){"
5
An associated differential equation is
d*x
d3x
29d 2 x
13 dx
which may be written as
8.95
(32D 4 - 64D 3 + 58£>
,d x co d x
dx
d4 x
c
64 7T+ 58 TT- 26 T + 5x = °
7J4
df
dt
3
5
2 ^T
+^x=
-7X2
3 + T7-TT-T7-r
16 dt
32
16 dt
dt
dt*
2
n
or
32
2
3
<ff
2
(if
- 26D + 5)x = 0.
Find the solution of a sixth-order linear homogeneous differential equation for x(r) with real coefficients if the
— 3 ± j'5, and — 1 ± \2.
2 ± i'4,
roots of its characteristic equation are
I
Since the roots are distinct, the solution is
x(t)
= e 2 \c cos At + c 2 sin 4f) + e~ 3 \c 3 cos 5r
x
-I-
c 4 sin 5t)
+ e~'(c s cos 2f + c 6 sin 2t)
175
x
CHAPTER 8
176
8.96
I
-\ ± il,
2 ± i\,
and +i4.
Since the roots are distinct, the solution is
x(f)
= e -,/2 (c, cos 2f + c 2 sin 2t) + e 2 '(c 3 cos \t + c 4 sin ^f) + c 5 cos 4f + c 6 sin 4f
Find the solution of an eighth-order linear homogeneous differential equation for y(x) with real coefficients
if the
I
roots of its characteristic equation are
3
+
—3 ±
/',
1
±
and
i'3,
— + i3.
1
= e 3x (c cos x + c 2 sin x) + e~ 3x(c 3 cos x + e 4 sin x) + e x(c 5 cos 3x + c 6 sin 3x) + e~ x{c 7 cos 3x + c 8 sin 3x)
{
7 + /8.
Solve the previous problem if the roots are, instead,
I
/',
Since the roots are distinct, the general solution is
y(x)
8.99
-2 ± i64.
and
= e~ 16 Vi cos 108f + c 2 sin 108f) + e~ 14,(c 3 cos 53f + c 4 sin 53f) + <?~ 2 '(c 5 cos 64f + c 6 sin 64r)
Solve Problem 8.95 if the roots are, instead,
I
8.98
- 14 ± z'53,
Since the roots are distinct, the solution is
x(f)
8.97
- 16 ± i'108,
Solve the previous problem if the roots are, instead,
8 ± i'9,
\ ± /4,
and
— j + i\.
Since the roots are distinct, the general solution is
= e lx (c cos 8x + c 2 sm 8x) 4- e Sx (c 3 cos 9x + c 4 sin 9x) + e x 2 (c 5 cos 4x + c 6 sin 4x)
y( \)
x
+ ^ _x/2 (c 7 cos {x + c 8 sin \x)
8.100
Find the general solution to a fourth-order linear homogeneous differential equation for x(f) with real
coefficients if two roots of the characteristic equation are
I
4
d
d
—4r-6—
t+15 —T - 18 -- + lOx =
dt
x
x
3
2
dt
dx
- 6/ + 15/ - 18/. + 10 = 0.
2
if a
particular solution is 5e
x =
:
t'
'((
i
The general
is
2'
cos t.
of the characteristic equation
2 ±
2
[A
(2
(2 + i)][/.
0] = k - 4A + 5 is a factor of the
Thus,
i
Dividing by this factor, we find that
characteristic polynomial (the left side of the characteristic equation).
}} - 2/ + 2
j'4.
dt
This particular solution corresponds to the complex roots
3
—2 +
x
dt
I
and
x = e 2 \c cos 3f + c 2 sin 3f) + e~ 2 '{c 3 cos 4f + c 4 sin At).
d x
Solve
A
2 — /'3
Since the roots must be in conjugate pairs, the other two roots are
solution is then
8.101
—2 — ;4.
and
2 -(- /3
also a factor.
-
Thus, two additional roots are
cos I + c 2 sint) + e'iCiCOSt + c 4 sinf)-
1
±
i,
and the general solution is
The given particular solution is the special case
cx
= 5,
ci = ?i = c 4 = 0.
2
3
8.102
Solve
dv
d*v
d v
d v
—
+ 20v =
44 + 4—4 + 9^?+ 16
dx~
dx
dx
if a
particular solution is sin 2x.
dx-
i
This particular solution corresponds to the roots ±/2 of the characteristic equation
2
3
Thus [A - i2][A - ( - 1"2)] = /. 2 + 4 is a factor of the characteristic
X* + 4/. + 9/. + 16/. + 20 = 0.
polynomial (the left side of the characteristic equation). Dividing by this factor, we find that
-2 ±
The general solution is then
)} + 4/. + 5
is also a factor. Thus, two additional roots are
i.
2x
2x
sin x.
cos x + c A e~
y = c, cos 2x + c sin 2x + c 3 e~
2
6
8.103
Solve
s
3
2
dx 3
dx 2
d
dv
d v
d*v
d v
d v
-^-4—
4+ 16-4- 12-4 + 41-4-8-4 + 26v =
v
dx b
dx-
dx*
if
two solutions are sin
dx
and e 2x sin 3x.
I The particular solution sin x corresponds to the characteristic roots + while e 2x sin 3x corresponds to
2
the characteristic roots
2 +
and
so both
[/ — /][/. — — 0] = / + 1
'/?
— 4/. + 13 are factors of the characteristic polynomial. Dividing the
[/. — (2 + i'3)][/. — (2 — i'3)] =
i,
i'3,
characteristic polynomial
X
2
+2
is
also a factor.
6
/.
(
— 4/. 5 + 16A 4 — 12/ 3 + 41/. 2 — 8/ + 26
by both factors successively, we find that
Thus, two additional roots are ±i'V2. The general solution is
y — c, cos x + c 2 sin x + c 3 e
2x
cos 3x + c x e
lx
sin 3x
+ c 5 cos \'2x + c 6 sin s/2x
LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS
177
DISTINCT REAL AND COMPLEX CHARACTERISTIC ROOTS
8.104
y" + 2y" + 5/ - 26y = 0.
Solve
I The characteristic equation is
its
k x = 2,
roots are
2x
~ 2 + i3)x
3
X
+ 2X 2 + 5X - 26 = 0,
-2 + i3, and
X2 =
~
-
which we factor into
X 3 = -2 - 0.
- 2)(/ 2 + 4A + 13) = 0;
Since they are all distinct, the solution is
+ d2e
+ d 3 e 2 i3)x
y = c,e
Using the result of Problem 8.49, we can rewrite this solution as
i
(X
{
.
y = Cl e
2x
+ e~ 2x (c 2 cos 3x + c 3 sin 3.x),
which is real.
8.105
/" + 5/' + 7/ + 13y = 0.
Solve
f The characteristic equation is X 3 + 5X 2 + IX - 13 = 0, which can be factored into (X - \){X 2 + 6X + 13) = 0.
Its roots are
X 2 — — 3 4- i2,
and A 3 — —3 — /2. Since they are all distinct, the solution is
/ = 1,
~ 3 + i2)x
- 3 ~ i2)x
x
=
+ d3e
y
Cl e + d 2 e
Using the result of Problem 8.49, we can rewrite this solution as y = c e x + e~ 3x(c 2 cos 2x + c 3 sin 2x),
x
(
{
.
x
which is real.
8.106
y"'
Solve
+ 8y" + 37/ + 50y = 0.
f The characteristic equation is X 3 + SX 2 + 37/1 + 50 = 0, which we factor into (X + 2)(X 2 + 6X + 25) = 0.
Its roots are
X - — 2,
X2 = 3 +
and X 3 = 3 — /4. Since they are all distinct, the solution is
~
2x
+ d 2 e {3 + ,4)x + d 3 e i3 i4)x Using the result of Problem 8.49, we can rewrite this solution as
y - c e~
2x
+ e 3x(c 2 cos 4x + c 3 sin 4x).
y = c e~
i'4,
x
.
{
y
d3
8.107
Solve
d2
dy
-4y - 7 -4y - -^- + 87^ = 0.
dx
3
dx
2
dx
which has the roots X x = — 3,
~
3x
Since they are all distinct, the solution is y — c y e~
+ d 2 e (5+i2)x + d 3 e l5 i2)x
3x
rewritten, using Euler's relations (Problem 8.49), as
+ e 5x (c 2 cos 2x + c 3 sin 2x).
y — c x e~
f The characteristic equation is
A
3
— 7A 2 — A + 87 = 0,
X 3 — 5 — x2.
8.108
Solve
A 2 = 5 + z'2,
and
This can be
.
/" + y' = 0.
m 3 + m = 0, with roots m = 0, m 2 =
and m 3 = —
Since the roots
0x
x
x
are all distinct, the solution is
By the result of Problem 8.49, this can be
y = c^e + d 2 e' + d 3 e~'
rewritten as
y = c + c 2 cos x + c 3 sin x.
f The characteristic equation is
;',
i.
t
.
{
8.109
Solve
>'"
+ Ay' = 0.
f The characteristic equation is
m 3 = —\2.
m 3 + Am =
Solve
y'"
m(m 2 + 4) = 0,
y = c^e
Since these roots are all distinct, the solution is
which has roots
0x
wij
+ d 2 e i2x + d 3 e' i2x
.
=0, m 2 = i'2, and
By the result of
y = c t + c 2 cos 2x + c 3 sin 2x.
Problem 8.49, this can be rewritten as
8.110
or
- 6y" + 2/ + 36y - 0.
— 6X 2 + 2X + 36 = 0, with roots X — — 2, X 2 = 4 + i'V2, and
+
This can be rewritten, using Euler's
The solution is y = c e~ 2x + d 2 e l4 "^2)x + d 3 e (4_,v^
A 3 = 4 — j-v/2.
2x
4
+ c 2 e * cos \/2x + c 3 e 4x sin ^2x.
relations (Problem 8.49), as
y = c e~
# The characteristic equation is
X
3
x
)Jc
'
.
l
x
d3x
8.111
Solve
d2x
.„dx
-TT--Tt-12— -40x = 0.
- X 2 - \2X - 40 = 0, whose roots are
~ ~
~2+
St
+ d3e 2
and A 3 =-2-i2. The solution is thus x = c e + d 2 e
2t
5t
Problem 8.49) as x = c e + e~ {c 2 cos It + c 3 sin 2t).
# The characteristic equation is
X
3
(
x
i2)t
(
X l = 5,
i2)t
.
A2 =
-2 + i2,
This can be rewritten (see
x
8.112
Solve
^ + 5^ + 26^-150x^0.
dt
3
dt
2
dt
m 3 + 5m 2 + 26m - 150 = 0, which has roots m, = 3, m 2 = -4 + iy/34,
,_4
3t
'(c 2 cos V34^ + c 3 sin >/34t).
The solution is x = c e +
I The characteristic equation is
and
m 3 = -4 - iy/34.
x
£
CHAPTER 8
178
8.113
Solve
f
+ 25^- 125Q = 0.
-7T--5—
2
dt
dt
dt
3
I The characteristic equation is m 3 - 5m 2 + 25m - 125 - 0, which we factor into (m - 5)(m 2 + 25) = 0;
has as roots m, =5, m 2 =
and m 3 = -/5. The solution is then Q = c^e 5 + c 2 cos 5t + c 3 sin 5r.
d
— + — -27 =
—3
8.114
Solve
dr
I
d2I
3
z
dr
dl
2
0.
dr
I The characteristic equation is
m 3 = —iyjl.
d3r
8.U5
^
Sol*
The solution is
d2r
/
m 3 - m 2 + 2m - 2 = 0; its roots are m^ = 1, m 2 = i\/2,
= c^ + c 2 cos >/2r + c 3 sin V2r.
and
dr
+ .18^ + 64-.
f The characteristic equation, m 3 + 12.8m 2 + 64m = 0, can be factored into m(m 2 + 12.8m + 64) =
has as its roots m, = 0, m 2 = -6.4 + t'4.8, and m 3 = -6.4 - /4.8. The solution is thus
y + 64y = 0.
Solve
3
A + 64A = 0, may be factored into k(X 2 + 64) =
—
A3 =
i%.
The solution is y = c + c 2 cos 8t + c 3 sin 8f.
f The characteristic equation,
k 2 = i'8,
Xi =0,
8.117
and
= Cl e oe + e- 6A \c 2 cos 4.80 + c 3 sin 4.80) = c % + c 2 e~ 6Ae cos 4.80 + c^ 64 " sin 4.80
r
8.116
it
'
i"5,
and
and has as roots
x
^-81y = 0.
T
dx
Solve
4
The characteristic equation is A 4 — 81 = 0, which we factor into (A 2 — 9)(A 2 + 9) = 0;
A, =3,
A 2 = — 3, A 3 = /3, and
A 4 = — i3.
Since they are all distinct, the solution is
f
i3x
its
roots are
i3x
3
3
+ d 2 e~
y = c^ * + c 2 e~ * + d e
Using the result of Problem 8.49 on the last two terms, we may rewrite this solution as
.
x
y = c e
3x
Solve
= 0.
-T7-y
clx
t
+ c 2 e~ 3x + c 3 cos 3x + c 4 sin 3.x.
d*y
8.118
A
I The characteristic equation, A 4 — = 0,
k = 1,
and A 4 = —
X 2 = — 1, x 3 =
x
x
ix
ix
y = c e + c 2 e' + d e + d 2 e~
1
i.
/,
l
can be factored into
(A
2
— 1)(A 2 + 1) = 0;
its
roots are
Since they are all distinct, the solution is
.
x
x
Using the result of Problem 8.49 on the last two terms, we may rewrite this solution as
y = cxe
8.119
Solve
x
+ c 2 e~ x + c 3 cos x + c 4 sin x.
2
d*y
d
dx*
dx*
v
-4 + -4 - 20y = 0.
I The characteristic equation is
Its
roots are
X x — 2,
A2 =
X* + A
— 2,
2
— 20 = 0,
A 3 = iy/5,
and
which we factor into
A 4 = —iy/5.
(A
2
— 4)(A 2 + 5) = 0.
Thus, the solution is
+ c 2 e~ + d l e'~"* + d 2 e~' w$x which may be rewritten as
y = cye
2x
+ c 2 e~ 2x + c 3 cos v5x + c 4 sin >/5x.
y = c^e
2x
2x
x
,
8.120
Solve
^ + 2^4 + 5^-26^ =
ax
ax J
dx
0.
dx
- 2)(A 2 + 4/. + 13) = 0.
I The characteristic equation is A 4 + 2A 3 + 5A 2 - 26/ = 0, which we factor into
Its roots are
and A 4 = —2 — B. Since they are all distinct, the
k = 0,
A 2 = 2,
A 3 = — 2 + /3,
~ 2 ~ i3)x
2x
2 * 13 ^
=
solution is
d
e
c
c
e
+
+
+
y
2
2
Using the result of Problem 8.49 on the last two terms, we may rewrite this solution as
2x
+ e~ 2x (c 3 cos 3x + c 4 sin 3x).
y = Ci + c 2 e
/.(/.
1
x
8.121
Solve
y
(4)
d^
(
.
+ 5y <3) + 7y" + 13/ = 0.
f The characteristic equation is m 4 + 5m 3 + 7m 2 + 13m = 0, which we can factor into
and m 4 =-3-i2.
m(m - l)(m 2 + 6m + 13); its roots are m = 0, m 2 = 1, m 3 = -3 +
x
i'2,
The solution
LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS
'
x
+
3
i2)x
+ d2 e
y = c, + c 2 e + d e
x
3x
—
cos
2x
sin
c
e
+
e~
(c
c
2x).
c\
+
+
y
2
3
4
is
8.122
thus
y (4) + y
Solve
(
{
~ 3 ~ i2)x
which may be rewritten (see Problem 8.49) as
,
x
179
- / - y = 0.
{3)
m 4 + m 3 - m - = 0, which we can factor into (m 2 - l)(m 2 + m + 1) = 0;
m, = 1, m 2 = — 1, m 3 = -1/2 +_iyf3/2, and m 4 = -1/2 - iyfl/2. The solution is
its roots are
- 2 + r3/2)x
~
~
x
x
=
c,e~
c
e
+
+d e
+ d 2 e 1/2 ,v3/2)x This may be rewritten (see Problem 8.49) as
y
I The characteristic equation is
l '
{
^
(
.
l
1
y = c^* + c 2 e
8.123
i
Solve
y
(4)
+e
c 3 cos
'
(
— +
w 4 = — 1/2 — VI 5/2.
y
<4)
c 4 sin
—
x
m 4 + m 3 — Am — 16 = 0, which we factor into
Its roots are
m = 2, m 2 = — 2, w 3 = - 1/2 + />/l5/2,
1
The solution is
/
Solve
x
+ y <3) - 4/ - 16y = 0.
# The characteristic equation is
2
(m - 2)(m + 2)(m + m + 4) = 0.
8.124
1
>'
— c^ 2 * + c 2 e
2x
+ ^ "^l c 3 cos—— x + c 4 sin
and
—— x
- 6y (3) + 16/' + 54/ - 225y = 0.
I The characteristic equation is m* — 6m 3 + 16m 2 + 54m — 225 = 0, which we factor into
(m — 3)(m + 3)(m 2 — 6m + 25) = 0. Its roots are m = 3, m 2 = — 3, m 3 = 3 +
and m 4 = 3 — i4.
The solution is then y — c e 3x + c 2 e~ 3x + e 3x (c 3 cos4x + c 4 sin4x).
i'4,
x
x
8.125
/ 4) + / 3) - 2/' - 6/ - 4y = 0.
Solve
m 4 + m 3 — 2m 2 — 6m — 4, with the roots m = — 1,
The solution is y — c e~ x + c 2 e 2x + e~ x (c 3 cos x + c 4 sin x).
I The characteristic equation is
and
8.126
m4 = — 1 —
t
m 2 — 2, m 3 = — 1 +
r
(D 3 + 4D)y = 0.
Solve
I
i.
This equation may be factored into
D(D 2 + 4)y = 0,
with characteristic roots
and ±2i. The solution is
y — C + C 2 cos 2x + C 3 sin 2x.
x
8.127
(D
Solve
I
4
+ 5D 2 - 36)y = 0.
This equation may be factored into
y = Ae
(D 2 — 4)(£> 2 + 9)y = 0.
2x
2x
+ Be~ + C 3 cos 3x + C 4 sin 3x.
y — C cosh 2x + C 2 sinh 2x + C 3 cos 3x + C 4 sin 3x, since
sinh2x = i(e -e" 2x
the primitive is
x
The characteristic roots are ± 2 and ± 3i, and
This may be written as
cosh 2x = \{e
2x
+ e' 2x
)
and
2jc
).
8.128
Solve
(D
4
- 16)y = 0.
2
2
or (m + 4)(m - 4) = 0,
m 4 - 16 =
2x
+ c 4 e~ 2x
y = c cos 2x + c 2 sin 2x + c 3 e
# The auxiliary equation is
general solution is
8.129
Then the
x
Find the solution of a fourth-order linear homogeneous differential equation for x(t) with real coefficients
and + 1.
2 ± i
if the roots of its characteristic equation are
f The solution is
8.130
with roots ±2i and ±2.
.
x = e
2t
(ci cos t
+ c 2 sin t) + c 3 e' + c 4 e~'.
Determine a differential equation associated with the previous problem.
t The characteristic equation is
= [A - (2+ i)][A - (2 - i)JA ~ IP - (-1)] = (^ 2 - 41 + 5)(P - 1)
= A 4 - AX 3 + AX 2 + AX - 5
An associated differential equation is
8.131
dx
d x
d*x
d x
—
^ - 4 —j + A —^ + 4 — — 5x =
0.
Find the solution of a sixth-order linear homogeneous differential equation for x(t) with real coefficients if
3 ± in,
± i2n, and ± 5.
the roots of its characteristic equation are
i,
180
CHAPTER 8
D
t
Since the roots are distinct, the solution is
x — e 3 \c x cos nt + c 2 sin nt) + c 3 cos 2nt + c 4 sin 2nt + c 5 e 5
8.132
-3 ±
Solve Problem 8.131 if the roots are, instead,
4, 5, and
i'3,
'
+ c 6 e~ 5
'.
±6.
I Since the roots are distinct, the solution is
x = e
8.133
'^ cos 3f + c 2 sin 3f) + c 3 e M + c 4 e 5 + c 5 e 6t + c 6 e~ 6
'
'.
Solve Problem 8.131 if the roots are ±i\, 1. 2, 3, and 0.
x — c cos ^f + c 2 sin|t + c 3 e' + c A e 2
# The solution is
8.134
-3
+ c5e3 + c6
'
'
x
.
Determine a differential equation associated with the previous problem.
I The auxiliary equation is
= (m - i\)(m + i\)(m - l)(m - 2)(m - 3)(m - 0) = m 6 - 6m 5 + *fm 4 - fm 3 + ^m 2 - fm
4m 6 — 24m 5 + 45m 4 — 30m 3 + 1 lm 2 — 6m = 0.
or
(4D
8.135
6
Find the solution of a twelfth-order linear homogeneous differential equation for x(t) with real coefficients
if
I
the roots of its auxiliary equation are
— 2 + /3,
2 + i'3,
2t
(c l cos 3f
+ c9e
±/19, ±13, 3, and 0.
13 '
+ c 2 sin 3f) + e 2 '(c 3 cos 3f + c 4 sin 3f)
+ c l0 e'
13 '
+ c lx e 3t + c x2
-I-
c 5 cos 5f + c 6 sin 5f + c 7 cos 19f -I- c s sin 19f
.
Find the general solution to a fifth-order linear homogeneous differential equation for x(t) with real coefficients
if
I
three solutions are cos 2f, e~' sin 3f, and e
If cos 2f
2 '.
and e~' sin 3f are solutions, then so too are sin 2f and e~' cos 3f. Since these are the remaining two
linearly independent solutions, the general solution is
8.137
±i'5,
Since all the roots are distinct, the solution is
x = e~
8.136
An associated differential equation is
- 24D 5 + 45D 4 - 30D 3 + 11D 2 - 6D)x = 0.
x = c, cos 2x + c 2 sin 2f -I- e"'(c 3 cos 3f + c 4 sin 3f) + t\e 2
'.
Determine the characteristic equation of a third-order linear homogeneous differential equation with real
coefficients if two solutions are cos 3r and e
3 '.
I To generate cos 3f, two roots of the characteristic equation must be +i'3. To generate e 3 another root
= (/. — /3)(/ + i'3)(/. — 3) = 3 — 3/. 2 + 9/. — 27.
must be 3. Thus, the characteristic equation is
',
/.
8.138
Solve
x'+ 7x + x + Ix =
I To generate sin
is
particular solution is sin f.
two roots of the characteristic equation must be ±
Solve
Therefore, the roots are
3
—
dt*
—
—
dt
dt
dt
I The particular solution 8 corresponds to the root
characteristic polynomial,
equation by
3
/.
+ 25/,
/.
The general solution is then
8.140
Solve
-I-
4A
we obtain
x = c
t
if
/.
t
is
f.
two solutions are 8 and | sin 5f.
=
of the characteristic equation, while the solution
/.
(/.
if
two solutions are -3e' sin \t and e~'.
I The particular solution — 3e's\r\jt corresponds to the roots
the root - 1. Thus,
[/. - (- 1)][/ - (1 + /^)][/. - (1 - i\)] =
± ij. while the solution e~' corresponds to
- 2 - |/. + | is a factor of the characteristic
3
- 4/. 2 - 3/. + 5. Dividing by this last factor
4/.
1
3
/.
/.
- 40/ 4 - 20/ 3 + 50/ 2 -IX- 5, as is
2
— 2/ — = 0, which implies that
= \ and X = — \
yields
8/.
4
solution is then
x = e\c cos \t + c 2 sin \t) + c 3 e~' + c A e ,a + c 5 e~'
32A
— /)(/. + i) = X2 + 1
- 0)(/ - i5)[X - (-j'5)] = 3 + 25/. is a factor of the
+ 33A 3 + 100/ 2 + 200A. Dividing both sides of the characteristic
2
— 2±/2 are two other roots.
which implies that
A + 4/. + 8 = 0,
2
cos
2r
c
+ 5 sin It).
+ c 2 cos 5f + c 3 sin 5f + e~ \c A
Thus,
(32D 5 - 40D 4 - 20D 3 + 50D 2 - ID - 5)x =
polynomial,
(/.
'
—
j sin 5t corresponds to the roots ±i'5.
4
5
which implies that
/.
5
d x
d4x
dx
d3x
d2x
j- + 33
=- + 200
=
—r4
+
r + 100
5
3
2
dt
i,
1
+ I/ + X + 7. Dividing by this factor, we find that X + 7
—7 and +/, and the general solution is x = t,e~ 7 + c 2 cos + c 3 sin
a factor of the characteristic polynomial,
also a factor.
8.139
f,
if a
5
1
are two additional roots.
/.
x
.
The general
LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS
181
REPEATED CHARACTERISTIC ROOTS
8.141
y" + 4/ + 4y = 0.
Solve
# The characteristic equation is
2x
+ c 2 xe~ 2x
y — c e~
2
A
+ 4/ + 4 = 0,
which has the roots
+ 6A + 9 =
or
(A
+ 3) 2 = 0,
which has the roots
A,
= X 2 = -3.
= A2 =
A,
= X 2 = — 2.
The solution is
.
x
8.142
Solve
/' + 6/ + 9>- = 0.
f The characteristic equation is
y — c e~
The solution is
8.143
Solve
3x
X
_y
X
— 2A +
1
=
or
(A
— l) 2 = 0,
which has the roots
A,
+ 2/ +
1
=
or
(A
+ l) 2 = 0,
which has the roots
X l — X2 =
— 4) 2 = 0,
which has the roots
y — C e~
x
l
A
2
+ C 2 xe~
— 1.
.
x - 8x + 16.v = 0.
'
x
Solve
The
x
I The characteristic equation is A 2 — 8A + 16 =
The solution is x - C> 4 + C 2 te*' = e M {C + C 2
8.146
1.
y" + 2y' + y = 0.
The solution is
Solve
2
= Cje* + c 2 xe*.
f The characteristic equation is
8.145
.
/' - 2y' + y = 0.
solution is
Solve
3x
+ c 2 xe~
t
f The characteristic equation is
8.144
1
or
(A
A,
= A 2 = 4.
t).
x + lOi + 25x = 0.
I The characteristic equation is m 2 + 10m + 25 =
The solution is x = C e~ il + C 2 te~ 5
(m + 5) 2 = 0,
or
which has the roots
m, — m 2 — —5.
'.
x
2
8.147
Solve
dQ
d Q
-4+
dt
-^- + 250,000Q = 0.
1000-^
+ 1000
-pf2
dt
or (m + 500) 2 = 0,
m 2 + 1000m + 250,000 =
500t
The solution is Q = c e'
+ c 2 te~ 500t
f The characteristic equation is
8.148
=m = -500.
roots
m,
Solve
x + 16x + 64 = 0.
which has the repeated
.
2
x
or
I The characteristic equation is m 2 + 16m + 64 =
8t
8t
—
=
=
=
solution
is
x
c
c
te~
m2
8.
The
e~
+ 2
Wj
(m + 8) 2 = 0,
which has the repeated roots
.
{
2
8.149
Solve
dl
d
- 60 — + 900/ =
—
l
0.
=-
dt
2
dt
i The characteristic equation is
root. The solution is
8.150
Solve
/
= Ae
30t
- 60A + 900 =
or
+ Bte 30t = (A + Bt)e i0t
A
4A 2 - 4A + 1 =
x 2
+ Bxex 2
y — Ae
2
= 0,
which has
A =\
as a double root.
.
16m 2 + 8m + 1 =
x
xl
y = Ae~ '* + Bxe~ *.
or
(4m + l) 2 = 0,
or
(m + 15)
which has
m — —%
which has
m = — 15
as a
(D 2 + 30D + 225)y = 0.
double root. The solution is
Solve
as a double
(16D 2 + 8D + l)y = 0.
f The characteristic equation is
8.153
A = 30
which has
'
'
double root. The solution is
Solve
(2A - l)
or
f The characteristic equation is
8.152
- 30) 2 = 0,
(4D 2 - 4D + l)y = 0.
The solution is
Solve
(A
.
f The characteristic equation is
8.151
2
y" = 0.
m 2 + 30m + 225 =
y = c x e~
lSx
4-
c 2 xe~
l5x
.
2
= 0,
as a
182
CHAPTER 8
I The characteristic equation is
0x
+ c 2 xe 0x — c + c 2 x.
y — c e
which has the roots
/,
= X 2 = 0.
The solution is then
x
x
8.154
= 0,
2
a
= 0.
y'"
Solve
I The characteristic equation is X 3 — 0, which has zero as a triple root. The solution is then
0x
+ c 2 xe 0x + c 3 x 2 e 0x = c, + c 2 x + c 3 x 2
y = c e
.
x
4
d y
8.155
—£ = 0.
Solve
ax
I The characteristic equation is
2
3
y = c + c 2 x + c 3 x + c4x
A
4
= 0,
which has zero as a quadruple root. The solution is then
.
x
8.156
3
2
dx
dx
d v
d v
dv
-±
+ 6 -\ + 12 -f + 8v = 0.
Solve
dx
X = —2
3
I The characteristic equation is X3 + 6X 2 + 12/ + 8 =
or (/ + 2) = 0. which has
2x
2x
2x
2x
2
=
root. The solution is
c e~
+ c 2 xe~ + c 3 x e~
e~ (c + c 2 x + c 3 x 2
y
x
8.157
Solve
y
+ 8>'" + 24 v" + 32/ + 16>- = 0.
,4)
i The characteristic equation is
4
+ 24/ 2 + 32/ -4-16 =
or (/ + 2) = 0.
2x
2x
2x
+ c 2 xe~ + c 3 x 2 e~ + c^x 3 e~ 2x
y = c e~
X* + 8/
root of multiplicity four, the solution is
3
2
Solve
d
2
3
3
X x = —1
Since
a
is
.
i
dQ
Q
Q
—£
+ 3-^ + 3-^ + 6 =
dt
dt
dt
d
8.158
as a triple
).
x
0.
I The characteristic equation is A 3 + 3/ 2 + 3X + 1 =
or
triple root. The solution is
Q = C e + C 2 te~' + C 3 2 e~'.
'
(/
+ l) 3 = 0,
which has
X — —1
as a
t
y
8.159
Solve
Q
i4)
+ 4Q i3) + 6Q + 4Q + Q = 0.
or
+ 4/. 3 + 6/ 2 + 4/ + =
+ l) 4 = 0, which has
•"'
or
The solution is Q = C e + C 2 te
e
+ C 3 t 2 e + C 4 t 33„-i
4
f The characteristic equation is
as a root of multiplicity four.
Q = (C
8.160
l
Solve
+C
2t
/.
-
'
1
+ C 3 2 + C 4 3 )e "'.
t
t
Q ,5) + 5C ,4) + 10g (3) + 10^ + SQ + Q = 0.
X = —1
A
5
+ 5X4 + 1(U 3 + 10A 2 + 5X + 1 =
The solution is
as a root of multiplicity five.
d 4r
Solve
/ =
'
x
I The characteristic equation is
8.161
(/.
1
d 3r
^- *
,2
d 2r
^-
+ 54
or
(/+1) 5 = 0,
which has
4„-r
Q = C e~' + C 2 te~' + C 3 2 e~' + C 4 3 ^"' + C s t*e
t
t
x
dr
s + 8,r = a
108
+ 54/ 2 - 108/ + 81=0 or (/ - 3) 4 = 0,
3e
r = Ae
as a root of multiplicity four. The solution is
+ B0e 3e + C0 2 e 3e + D6 3 e 30
f The characteristic equation is
X* - 12/
3
/ = 3
which has
.
8.162
Find the solution of a third-order linear homogeneous differential equation for x(t) with real coefficients if its
characteristic equation has
f The solution is
8.163
x = c e'
2
as a triple root.
+ c 2 te' 2 + c 3 2 e' 2
t
x
.
Find the solution of a fourth-order linear homogeneous differential equation for x(f) with real coefficients if its
characteristic equation has
# The solution is
8.164
X =\
X =\
x = e' 2 (c
x
as a quadruple root.
+ c 2 + c 3 2 + c4 3
t
t
f
).
Determine a differential equation associated with the previous problem.
I The characteristic equation is
=
(/.
- i) 5 = A 5 - |A 4 + f;. 3 - f/ 2 +&X-&
An associated differential equation is
(32D 5 - 80D
4
or
32/
5
- 80/ 4 + 80;. 3 - 40A 2 + 10/ -
+ 80D 3 - 40D 2 + 10D - l)x = 0.
1
=
LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS
8.165
183
Find the solution of a sixth-order linear homogeneous differential equation for x(/) with real coefficients if its
characteristic equation has
—8
/ =
x = e~ s '(c
f The solution is
8.166
D
1
as a root of multiplicity six.
+ c 2 + c 3 2 + c 4 3 + cy 4 + c 6 t 5
t
t
t
).
Solve the previous problem if the differential equation has order
10.
f A tenth-order differential equation must have 10 roots to its characteristic equation, and since we are given
We can, however, say that the solution to the previous problem
only six of them we cannot solve the problem.
will
8.167
be part of the solution to this problem.
Determine a differential equation associated with Problem 8.165.
I The characteristic equation is
= (/ + 8) 6 = X 6 + 48/ 5 + 960/. 4 + 10,240/ 3 + 61,440A 2 + 196,608/ + 262,144
An associated differential equation is
x (6) + 48x (5) + 960x ,4) + 10,240x ,3) + 61,440x + 196,608x + 262,144x =
8.168
Find the solution of a ninth-order linear homogeneous differential equation for x(r) with real coefficients if
its
characteristic equation has a single root of multiplicity nine.
x = e x \c x + c 2 t + c 3 r 2 + c 4 r 3 -I- c 5 t 4 + c 6 f s + c 7 r 6 + c 8 r 7 + <V 8 )-
f Denote the root by X. The solution is
CHARACTERISTIC ROOTS OF VARIOUS TYPES
2
d 3y
8.169
Solve
d y
—^-3—^
+ 4y =
d?
dx^~
0.
/ The characteristic equation is m 3 — 3m 2 + 4 = 0; its roots are m, = m 2 — 2 and m 3 = — 1. Since the
2x
first two are repeated,
c e
+ c 2 xe 2x is part of the solution. The last root, which is distinct from the others,
l
gives c 3 e'
x
y — cxe
The complete solution is then
as part of the solution.
d2 y
dy
d
—
—
4 4-3-^--18y
=
-4—+ 4—
lx
+ c 2 xe 2x + c 3 e~ x
.
*y
8.170
Solve
f
,
|-
A
-t
3x
cxe
2x
m 3 + 4m 2 — 3m — 18 = 0,
(4£>
3
+ c 2 xe~
3x
with roots —3, —3, and 2. Since the first two
The last root, which is distinct from the
The complete solution is then y — c e~ 3x + c 2 xe 3x + c^e 2 *.
part of the general solution.
is
as part of the solution.
others, gives c 3 e
Solve
0.
dx
The characteristic equation is
are repeated,
8.171
3
dx 2
dx 3
x
-28D 2 -31D-8).v = 0.
/ The characteristic equation is 4m 3 — 28m 2 — 31m — 8 = 0,
xi2
solution is
+ c 2 xe' x 2 + c 3 e 8x
y — c e"
with roots —j, —j, and 8. The general
'
.
x
8.172
Find the general solution to a fourth-order linear homogeneous differential equation for y(x) with real coefficients
the roots of its characteristic equation are — 1, — 1, — 1, and 2.
if
I
which is distinct from the others, adds c A e
y = e'
8.173
e~
Since the first three roots are repeated,
x
(Ci
+ c 2 x + c 3 x 2 + c 4 e 2x
)
2x
x
{c l
+ c2x + c3x2
)
is
to the general solution.
part of the general solution.
The last root,
The primitive is then
.
- 1, - 1, 2, and 2.
Solve Problem 8.172 if the roots are
x
The first two roots are repeated and contribute e~ (c + c 2 x) to the general solution. The last two roots,
2x
which are different from the first two and are also repeated, contribute e (c 3 + c 4 x) to the general solution.
2x
x
The complete solution is y = e~ (c + c 2 x) + e (c 3 + c 4 x).
I
}
x
8.174
Solve Problem 8.172 if the roots are |, \, 3, and 3.
f The general solution is
8.175
y = e
xt2
{cA
+ c 2 x) + e 3x (c 3 + c 4 x).
Solve Problem 8.172 if the roots are ±/2 and ±/2.
CHAPTER 8
184
# The root \2 has multiplicity two, so it contributes
double root —
d 3 e~' 2x + d 4.xe~ l2x
contributes
1'2
y = dxe
i2x
dxe
i2x
+ d 2 xe i2x
to the general solution.
Similarly, the
The complete solution is
to the general solution.
+ d 2 xe i2x + d 3 e~ i2x + d^xe~ i2x = {d e i2x + d 3 e~ i2x + x(d 2 e i2x + d 3 e' i2x
)
x
)
Using Euler's relations (see Problem 8.49) on each set of terms in parentheses, we can rewrite the latter solution
y = (c-j cos 2x + c 2 sin 2x) + x(c 3 cos 2x + c 4 sin 2x).
as
8.176
Solve Problem 8.172 if the roots are
f The root
+
3
i'5
± \5
3
and
3
± i5.
has multiplicity two, so it contributes
Similarly, the double root
3
— 5
d3e
contributes
i
(3 ~' 5)x
d 1 e (3
+ d 4 xe
+ ,5)x
+ d 2 xe i3+,5)x
<3_,5) *
to the general solution.
to the general solution.
The
complete solution is
~
{3 + i5)x
- ,5,JC
+ d 2 xe ti + i5)x + d 3 e (3 i5)x + d 4 xe (3
= e 3x \{d e i5x + d 3 e~ i5x + x(d 2 e i$x + d 4.e~ i5x
= e 3x [c cos 5x + c 2 sin 5x + x(c 3 cos 5x + c 4 sin 5x)]
y = dxe
)']
)
y
x
8.177
y = e
I The general solution is
8.178
1
y — e
7 + i23
lx
(c x cos 23x
7
± i'23.
+ c 2 sin 23x) + xe lx (c 3 cos 23x + c 4 sin 23x).
±4 and ±4.
y — e
There are two real roots of multiplicity two, so the general solution is
Ax
(c l
+ c 2 x) + e' Ax {c 3 + xc 4
).
Solve Problem 8.172 if the roots are -6, -6, and 2 ± /4.
e
2x
(c 3 cos 4x
+ c 4 sin 4x).
e"
6jc
(c 1
+ c 2 x)
to the general solution; the distinct complex roots
The complete solution is
y = e~
bx
(c x
+ c 2 x) + e 2x(c 3 cos 4x + c 4 sin 4x).
Solve Problem 8.172 if the roots are 2, 2, and ±/2.
Solve
(D
4
y = ^
#
(D
4
Solve
(D
4
y — C e
2x
y
+ C 2 e~
2x
+ C 3 e~
-D -9D -\\D - A)y =
3
2
3x
+ C A xe~
y
Solve
(D
4
which has characteristic roots 2, -2, -3, and -3.
3x
.
0.
(D + l) 3 (D — 4)v = 0,
Ax
x
2
=
e~ (C + C 2 x + C 3 x + C A e
y
)
which has characteristic roots — 1, — 1, - 1, and 4. The
.
+ 4D 2 )y = 0.
This may be rewritten as
solution is
#
)
This may be rewritten as
primitive is
I
+ xc 2 + c 3 cos 2x + c 4 sin 2x.
(Cj
(D - 2)(D + 2)(D + 3) 2 y = 0,
This may be rewritten as
Solve
2jc
+ 6D 3 + 5D 2 - 24D - 36)y = 0.
The primitive is
8.187
and
,
Solve Problem 8.172 if the roots are
I
8.186
1
y = c cos 4 x + c 2 sin Jx + x(c 3 cos ^x + c 4 sin |x).
I The general solution is
8.185
- ±I
and
i
y
The general solution is
contribute
8.184
i'3.
y = e~ \c cos x + c 2 sin x + x(c 3 cos x + c 4 sin x)].
/ The double real root —6 contributes
8.183
+
Solve Problem 8.172 if the roots are ±i\ and ±i'i
f
8.182
\
cos 3x + c 2 sin 3x + x(c 3 cos 3x + c 4 sin 3x)].
,
- ±
Solve Problem 8.172 if the roots are
f
8.181
[c
and
x
f The general solution is
8.180
x 2
Solve Problem 8.172 if the roots are
I The general solution is
8.179
\ ± i'3
Solve Problem 8.172 if the roots are
D 2 (D 2 + 4)y = 0,
which has the characteristic roots 0, 0, and + 1'2. The general
y = c + c 2 x -f c 3 cos 2x + c 4 sin 2x.
t
- 6D 3 + 13D 2 - 12D + 4)y = 0.
This differential equation may be rewritten as
1, 1. 2. and 2. The primitive is
(D — 1) (D — 2) 2 y = 0,
2
x
2x
y — e (c + c 2 x) + e {c 3 + c 4 x).
l
which has the characteristic roots
LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS
8.188
Solve
y
(4)
X t — 2 + il
v
8.189
Solve
185
- 8v'" + 32/' - 64/ + 64y = 0.
m 4 - 8m 3 + 32m 2 - 64m + 64 =
has roots 2 + il and
A2 = 2 —
are both roots of multiplicity two. The solution is
f The characteristic equation
hence
D
and
2 ± il;
j'2
= e 2x (c! cos 2x + c 3 sin 2x) + xe 2x (c 2 cos 2x + c 4 sin 2x) = (c t + c 2 x)e 2x cos 2x + (c 3 + c 4 x)e 2x sin 2x
—
— + lOOy = 0.
^-12^
- 12-4 + 56-4
120-^
-120
-4
56
4
dx 3
dx 3
dx
dx
# The characteristic equation has roots
multiplicity two.
The solution is
y = (c
and 3 ±
3 ±
so both
3 +
+ c 2 x)e 3x cosx + (c 3 + c 4 x)e 3x sin x.
i,
i
{
i
—
and
3
—1 +
i'2
j"
are roots of
4
8.190
Solve
3
^ y
d* y
d 2y
dy
4-4+14-4
^- + 25y = 0.
-4
43 + 14 -4 + 20
^-4 + 4
20/
ax
ax
ax
dx
—
— 1 + i2
I The characteristic equation has roots
8.191
Solve
y
(4)
— + il,
and
so both
1
x
y = (c t + c 2 x)e~ cos2x + (c 3 + c 4 x)e
The solution is
roots of multiplicity two.
Solve
y
,4)
— — il
1
are
y = (cj + c 2 x)cos2x + (c 3 -'- c 4 x)sin2x.
+ 14y" + 49y = 0.
m — il
I The characteristic equation has roots
y = (cj + c 2 x) cos 7x + (c 3 + c 4 x) sin 7x.
8.193
and
sin 2x.
+ 8y" + 16y = 0.
f The characteristic equation has roots ±il and ±/2. The solution is
8.192
_x
and
t
m 2 = — f7,
both of multiplicity two. The solution is
Find the general solution of a sixth-order linear homogeneous differential equation for y(x) with real coefficients
if its
characteristic equation has roots i5 and
— i5, each with multiplicity three.
I The solution is
+ d 2 x + ^x 2 + e" i5x (d 4 + d 5 x + d 6 x 2
= {d e i5x + d^e~ i5x + x(J 2 5x + d 5 e _i5x) + x 2 (d 3 e i5x + d 6 e-'' 5x
= (c cos 5x + c 4 sin 5x) + x(c 2 cos 5x + c 5 sin 5x) + x 2 (c 3 cos 5x + c 6 sin 5x)
y = e
i5x
{d x
x
)
)
e''
)
)
t
= (c + c 2 x + c 3 x 2 cos 5x + (c 4 + c 5 x + c 6 x 2 sin 5x
x
8.194
8.197
1'3,
each of multiplicity three.
2
2
—3 +
y — (c x + c 2 x + c 3 x )e
2
Solve Problem 8.193 if the roots are
f The solution is
—
y = (c x + c 2 x + c 3 x ) cos 3x + (c 4 + c 5 x + c 6 x ) sin 3x.
Solve Problem 8.193 if the roots are
I The solution is
8.196
)
Solve the previous problem if the roots are z'3 and
I The solution is
8.195
)
0.2
_3x
2
02x
each of multiplicity three.
2
cos 5x + (c 4 + c 5 x + c 6 x )e
+ i0.7
y = (c t + c 2 x + c 3 x )e
—3-/5,
and
i"5
and
0.2-/0.7,
_3x
sin 5x.
each of multiplicity three,
2
cos 0.7x + (c 4 + c 5 x + c 6 x )e
02x
sin 0.7x.
Find the general solution of an eighth-order linear homogeneous differential equation for y(x) with real coefficients
and \ — /3, each with multiplicity four.
characteristic equation has roots
\ + i'3
if its
f The solution is
8.198
+ c 4 x 3 )e x/2 cos 3x + (c 5 + c 6 x + c 7 x 2 + c 8 x 3 )e xl2 sin 3x.
y = (c x + c 2 x + c 3 x
2
— 3 + i\
y = (cj + c 2 x + c 3 x
Solve Problem 8.197 if the roots are
2
and
—3 — i%,
each of multiplicity four.
+ c 4 x 3 )e~ 3x cos ^x + (c s + c 6 x + c 7 x 2 + c 8 x 3 )e~ 3x sinjx.
Solve Problem 8.197 if the roots are id and
f The solution is
8.200
2
Solve the previous problem if the roots are
# The solution is
8.199
y — (c x + c 2 x + c 3 x
— i6, each of multiplicity four.
+ c 4 x 3 cos 6x + (c s + c 6 x + c 7 x 2 + c 8 x 3 sin 6x.
2 + il
)
)
and
-2 ± il,
each of multiplicity two.
CHAPTER 8
186
and 2 — \2 are both roots of multiplicity two, they contribute
Since 2 + i2
+ c 2 x)e 2x cos 2x + (c 3 + c 4 x)e 2x sin 2x to the general solution (see Problem 8.188). Similarly,
—2 +
and — 2 — il are roots of multiplicity two, they contribute
since both
2x
2x
sin 2x
cos 2x + (c 7 + c 8 x)e~
to the general solution. The complete solution is the sum
(c 5 + c 6 x)e~
I
(c t
i'2
of these two contributions, namely
y = (Cj + c 2 x)e
8.201
cos 2x + (c 3 + c 4 x)e 2x sin 2x + (c 5 + c 6 x)e _2x cos 2x + (c 7 + c s x)e~ 2x sin 2x
3±i2
Solve Problem 8.197 if the roots are
f The solution is
8.202
2x
y = (a + c 2 x)e
3x
4 ± i'5,
and
cos 2x + (c 3 + c 4 x)e
—3 +
Solve Problem 8.197 if the roots are
each of multiplicity two.
3x
sin 2x
+ (c 5 + c 6 x)e* x cos 5x + (c 7 + c 8 x)e 4x sin 5x.
—5 ±
each of multiplicity three, and
i"5,
j'6,
each of
multiplicity one.
—3 +
Since
and — 3 —
are both roots of multiplicity three, they contribute
+ c 2 x + c 3 x 2 )e~ 3x cos 5x + (c 4 + c 5 x + c 6 x 2 )e~ 3x sin 5x to the general solution (see Problem 8.195).
5x
—5 +
In contrast,
are both simple roots, so they contribute
c 7 e~
cos6x + c 8 e~ 5x sin6x to the
I
i'5
i'5
(c x
i*6
general solution.
The complete solution is the sum of these two contributions, namely
y = (Ci + c 2 x + c 3 x )e~
2
8.203
2
— 16 ± (25,
y — (c
Solve
y
<5)
+ c 2 x + c 3 x 2 )e~ 16x cos 25x + (c 4 + c 5 x + c 6 x 2 )e _16x sin 25x + (c 1 + c 8 x)e~ x 2
— 16 ± (25,
y = (c,
each of multiplicity two, and
x
8.206
x
— \ of multiplicity four,
.
- y (4) - If + 2y" + v' - y - 0.
a root of multiplicity three and
y = c e
.
+ c 2 x)e~ 16x cos25x + (c 3 + c A x)e~ 16x sin 25x + (c 5 + c 6 x + c 7 x 2 + c 8 x 3 )e" x/2
I The characteristic equation can be factored into
is
— j of multiplicity two.
each of multiplicity three, and
'
x
Solve Problem 8.197 if the roots are
f The solution is
8.205
cos 5x + (c 4 + c 5 x + c 6 x )e~ 3 *sin 5x + c 7 e" 5 *cos6x + c 8 e _5jc sin6x
Solve Problem 8.197 if the roots are
I The solution is
8.204
3x
X2
= —
is
1
+ c 2 xe* + c 3 x 2 ex + c x e x + c 5 xe~ x
— 1) 3 (A + l) 2 = 0;
(/.
a root of multiplicity two.
hence,
/.,
= 1
The solution is
.
Find the general solution of a fifth-order linear homogeneous differential equation for y(x) with real
if its characteristic equation has roots 2, 2, —3, —3, and 4.
coefficients
f The solution is
8.207
2x
i
+ c 2 e* x + (c 3
y = (c,
ix
2
-(-
c 4 x -I- c s x )e~
+ c 2 x + c 3 x 2 + c 4 x 3 )e 2x + c 5 e Ax
y = (c i + c 2 x + c 3 x )e
2
y = c e
1
2x
2x
.
3
± /4.
and
-(-
c 5 sin 4x).
± i4.
3
+ (c 2 + c 3 x)e 3x cos4x + (c 4 + c 5 x)e 3x sin 4x.
Solve Problem 8.206 if the roots are 2, 2, 2, 2, and
I
.
+ e ix (c 4 cos 4x
3 ± i4,
Solve Problem 8.206 if the roots are 2,
I The solution is
8.211
y = c e
Solve Problem 8.206 if the roots are 2, 2, 2, and
f The solution is
8.210
.
Solve Problem 8.206 if the roots are 2, 2, 2, 2, and 4.
I The solution is
8.209
+ c 2 x)e 2x + (c 3 + c A x)e~ ix + c 5 e 4x
Solve the previous problem if the roots are 2, —3, —3, —3, and 4.
f The solution is
8.208
y ={c i
3
+
i'4.
This cannot be. Since the coefficients of the differential equation are real, the complex roots must occur in
conjugate pairs. Thus, if
3
+ i4
is
a root, then
3
— i4
is
also a root.
We now have six characteristic
roots for a fifth-order differential equation, which is impossible.
8.212
Find the general solution of a sixth-order linear homogeneous differential equation in x(t) with real coefficients
if its
characteristic equation has roots
I The solution is
1,
2, 2, 3, 3,
x = c e' + (c 2 + c 3 t)e 2
r
'
and 3.
+ (c 4 + c 5 + c 6 2 )e 3
t
r
'.
LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS
8.213
Solve Problem 8.212 if the roots are 2, 2, 3, 3, 3, and 3.
/ The solution is
8.214
t
x = (c t + c 2 t)e~ 3
x = c x e 2t + c 2 e 3
'.
'
'
Solve
-3 ± in.
— 1 ± 2i.
+ e~ 3,(c 5 cosnt + c 6 sinn*).
'
— 3 + in,
— 3 + in.
and
+ (c 3 4- c 4 r)e~ 3 'cos7rf + (c 5 + c 6 t)e~ 3 sinnt.
'
— 3 ± in,
and
— 3 ± in.
+ (c 3 + c^t)e~ 3 cos nt + (c 5 + c 6 t)e~ 3 sinnt.
'
— 3 + in,
—2±in,
'
and
'
— 3 + in.
'
— 3 + in,
— 3 ± m,
and
— 3 + in.
x = (c, + c 2 t + c 3 t 2 )e~ 3t cosnt + (c 4 + c 5 t + c 6 t 2 )e~ 3 sinnt.
'
4
(D + 2) (D - 3) (D
3
.
x = c x e~ 2 cos nt + c 2 e~ 2t sinnt + (c 3 + c^t)e~ 3 cos nt + (c 5 + c 6 t)e~ 3t sinnt.
I The auxiliary equation
2
+ 2D + 5)y = 0.
4
(m + 2) 3 (m — 3) (m
2
+ 2m + 5) =
has roots —2, —2, —2, 3, 3, 3, 3, and
The general solution is
y = (<?! + c 2 x + c 3 x )e~
2
8.221
t
t
x = (c x + c 2 t + c 3 t 2 + c 4 r 3 )e~ 3
Solve Problem 8.212 if the roots are
I The solution is
8.220
t
+ (c 2 + c 3 + c 4 2 + c 5 t 3 + c 6 t*)e 3t
'
x
Solve Problem 8.212 if the roots are
I The solution is
8.219
x = c e2
Solve Problem 8.212 if the roots are 2, 3,
I The solution is
8.218
t
Solve Problem 8.212 if the roots are —3, —3,
I The solution is
8.217
+ c 2 t)e 2 + (c 3 + c 4 + c s 2 + c 6 3 )e 3
'
x
Solve Problem 8.212 if the roots are -3, -3, -3, -3, and
I The solution is
8.216
x = (c
Solve Problem 8.212 if the roots are 2, 3, 3, 3, 3, and 3.
# The solution is
8.215
187
2x
+ (c 4 + c 5 x + c 6 x 2 + c 7 x 3 )e 3x + e~ x (c 8 cos 2x 4- c 9 sin 2x)
Find the general solution to a fourth-order linear homogeneous differential equation for x(t) with real coefficients
if
two particular solutions are 3e 2 and 6t 2 e~'.
'
must be a characteristic root. To have 6t 2 e~' as a solution, m = — 1
must be a root of multiplicity three. Thus, we know four characteristic roots, which is the complete set for this
differential equation. The general solution is
x = c e 2t + {c 2 + c 3 t + c A t 2 )e~'.
m=2
I To have 3e 2r as a solution,
x
8.222
Solve the previous problem if two particular solutions are \te~
x
and — Ste 2
'.
i To generate these solutions both m, = - 1 and m 2 = 2 must be roots of multiplicity two. Thus,
we know four characteristic roots, which is the complete set here. The general solution is
x = (c l + c 2 t)e~' + (c 3 + c±t)e
8.223
2t
.
Determine the differential equation associated with the previous problem.
I The characteristic equation is [m — — l)] 2 (m — 2) 2 = 0, so the differential equation is
3
2
This may be expanded to (D* - 2D - 3D + 4D + 4)x = 0.
(
8.224
(D + \) 2 {D — 2) 2 x — 0.
Determine the form of the general solution to a fifth-order linear homogeneous differential equation for x(r)
3
with real coefficients if a particular solution is 1 e
i To have t 3 e 3t as a solution,
m=3
characteristic equation are 3, 3, 3, and 3.
real root.
must be a root of at least multiplicity four. Thus, four roots of the
Since the differential equation is of order 5, there must be one additional
then 3 is a root of multiplicity five, and the general solution is
then the general solution is
k ± 3,
If
f
t
2
+ c 4 r 3 + c 5 4 )e 3
y = Cl 4- c 2 + c 3
2
+ C4.t 3 )e 3 + c 5 e
y = (Ci + c 2 + c 3
(
.
Denote it as k.
k = 3,
If
3t
t
'.
kt
t
8.225
t
'
.
Find the general solution to a sixth-order linear homogeneous differential equation for x(r) with real coefficients
2
if one solution is t sin t.
188
D
CHAPTER 8
f The t 2 portion of the given particular solution implies that the associated characteristic root has multiplicity
three.
+
i
+
Since sin t can be generated only from roots
are roots of multiplicity three.
which must occur in conjugate pairs, it follows that both
i,
Since this yields six roots, we have the complete set, and the general solution is
x = (c l + c 2 t + c 3 t ) cos t + (c 4 + c 5 t + c 6 t ) sin t.
2
8.226
2
Solve the previous problem if two particular solutions are sin t and Ate~ 3t cos 2t.
f The particular solution sin t can be generated only from the characteristic roots ±i.
— 3 + \2.
be generated only from the characteristic roots
— 3 ± /2
—3 +
Similarly, e~
— 3 ± il.
and
cos It can
Since this function is multiplied by t, it follows that
are both roots of at least multiplicity two. Thus, we have identified as characteristic roots
j'2,
3t
±
i,
These six roots form a complete set of characteristic roots for a sixth-order differential
equation, so the general solution is
x = c cost + c 2 sinr + (c 3 + c 4 r)e
_3
x
8.227
Determine the differential equation associated with Problem 8.225.
f The characteristic equation is
equation is
8.228
'cos2r + (c 5 + c 6 f)e~ 3 'sin2f
(D
2
+ l) x = 0,
3
(m — i) 3 [m — ( — i')] 3 = 0,
which may be expanded to
(m 2 + l) 3 = 0.
or
(D
6
The corresponding differential
+ 3D 4 + 3D 2 + l)x = 0.
Find the general solution of a twelfth-order linear homogeneous differential equation for x(t) with real coefficients
characteristic equation has roots
if its
1, 2, 2, 3, 3, 3,
2 + i'3.
2 ± i3,
±i,
I The solution is
x = c,e
8.229
(
+ (c 2 + c 3 t)e 2 + (c 4 + c 5 t + c 6 f 2 )e 3 + c 7 cos + c 8 sin + (c 9 + c l0 t)e 2t cos3f + (c n + c 12 r)e 2 'sin 3f
'
'
t
t
Solve the previous problem if the roots are 0, 0,
±
i,
±
2 + i'3,
2 ± i'3,
/',
2 + i'3.
f The solution is
x = c, + c 2 t + (c 3 + c 4 f) cos r + (c 5 + c 6 t) sin t + (c-, + c s t + c 9 t 2 )e 2 cos 3r + (c,
'
+ c u t + c 12 2 )e 2 sin 3r
t
'
EULER'S EQUATION
8.230
Develop a method for obtaining nontrivial solutions to Euler's equation,
b n x"y
where bi
(j
— 0, 1, ...
,
n)
is
{n)
+ b n . x"- y-
»>
x
+
•
•
+ b 2 x 2 y" + b xy' + b y =
x
a constant.
I An Euler equation can always be transformed into a linear differential equation with constant coefficients through
the change of variables
z
= In x
x = ez
or
D=
With the notation
.
d
—
dz
,
it
follows from this equation and
the chain rule that
dy
dy dz
dy \
1
dx
dz dx
dz x
x
Similarly,
d3y
1
ax
x
—^ = -^ D(D - 1)(D - 2)y,
and in general
pL
L D{D _ 1)(D _ 2)(D -
dx
x
3)
•
•
•
(D - n + l)y
By substituting these derivatives into an Euler equation, we obtain a linear differential equation without variable
coefficients, which
8.231
may be solved like the other problems in this chapter.
2x 2 /' + Uxy' + 4y = 0.
Solve
This is an Euler equation.
If
we set
x = ez
,
it
follows from Problem 8.230 that
y'
— — Dy
and
x
y" = —= D(D — l)v,
and the given differential equation becomes
2D{D — l)y + HZ)y + 4y =
or
x
(2D 2 + 9D + 4)y = 0.
Now all derivatives are taken with respect to z. From the result of Problem 8.23 (with
LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS
D
189
z replacing x), the solution is
y = Ae
8.232
Solve
z' 2
z
+ Be* = A(e z l/2 + B(e z f = Ax 112 + fix 4
)
x 2 y" - 2y = 0.
This is an Euler equation with
6,
= 0.
If we set
x = ez
follows from Problem 8.230 that
it
,
y'
= - Dy
x
y" = -j D(D - l)y,
and
- D - 2)y = 0.
\2
2
(D
D(D - l)y - 2y =
and the given differential equation becomes
x
or
Now all derivatives are taken with respect to z. From the result of Problem 8.1 (with z
replacing x), the solution is
y = c^
8.233
Solve
-2
+ c 2 e 2z = c (e z y + c 2 (e z 2 = c^x -1 + c 2 x 2
l
)
{
x 2 y" - 6xy' = 0.
This is an Euler equation with
= 0.
b
If we set
x = ez
follows from Problem 8.230 that
it
,
y'
= - Dy
x
y" = -j D(D — l)y,
and
D{D — l)y — 6Dy =
so that the given differential equation becomes
or
(D 2 — lD)y = 0.
y = Ci + c 2 e
8.234
Solve
7z
Now all derivatives are taken with respect to z. By Problem 8.2, the solution is
= Ci + c 1 (e z 1 = c t + c 2 x 7
.
)
x 2 y" + xy' - 5y = 0.
This is an Euler equation.
If
we set
x = ez
,
it
follows that
y'
= -Dy
y" = —, D(D — l)y,
and
D(D — l)y + Dy — 5y =
that the given differential equation becomes
so
1
x
x"
(D 2 — 5)y = 0.
or
The independent
variable in this last equation is z, and by Problem 8.3 the solution is
y = c x e^
8.235
Solve
I
z
+ c 2 e~^ z = c^Y1 + c 2 {e z )~^ = c x v/5 + c 2 x~^
x
x 2 y" + 5xy' + 4y = 0.
This is an Euler equation. Using the substitutions suggested in Problem 8.230, we obtain
D(D — l)y + SDy + 4y =
or
(D
2
+ 4D + 4)y = 0.
This equation is similar in form to that given in Problem
8.141, except now the independent variable is z; its solution is
8.236
Since
x = e
Solve
x 2 y" - xy' + y = 0.
I
z
and
z
= In x,
it
follows that
y = c^x'
2
y = c x e~
+ c 2 (\n x)x~
2
+ c 2 ze~ 2z = c (e z )" 2 + c 2 z{e z )~ 2
= {c + c 2 In x)/x 2
2z
x
.
.
l
This is an Euler equation. Using the substitutions suggested in Problem 8.230, we obtain
D(D — l)y — Dy + y =
or
(D 2 — 2D + l)y = 0.
This equation is similar in form to that given in Problem
8.143, except now the independent variable is z; its solution is
so this solution may be rewritten as y = c x + c 2 x In x.
y = cxe
z
+ c 2 ze z
.
But
x = ez
and
z
= In x,
x
8.237
Solve
I
x 2 y" + xy' + 4y = 0.
This is an Euler equation. Using the substitutions suggested in Problem 8.230, we rewrite the differential
2
or (D + 4)y = 0. This equation is similar in form to that given in
D(D — l)y + Dy + 4y =
Problem 8.59, except now the independent variable is z; its solution is
equation as
y = c x cos 2z + c 2 sin 2z = c^ cos (2 In x) + c 2 sin (2 In x) = c l cos (In x ) + c 2 sin (In x )
2
8.238
Solve
f
2
x 2 y" + xy' + 50y = 0.
This is an Euler equation. Using the substitutions suggested in Problem 8.230, we rewrite the differential
2
or (D + 50)y = 0. This equation is similar in form to that given
D(D — \)y + Dy + 50y =
equation as
in Problem 8.60, except now the independent variable is z; its solution is
y = c x cos >/50z + c 2 sin \/50x = c x cos (^50 In x) + c 2 sin (V50 In x).
8.239
Solve
x 2 y" - 5xy' + 25y = 0.
This is an Euler equation. Using the substitutions suggested in Problem 8.230, we rewrite the differential
2
or (D — 6D + 25)y = 0. This equation is similar in form to
equation as D{D — l)y — SDy + 25y =
I
1
190
CHAPTER 8
D
that given in Problem 8.50, except now the independent variable is z; its solution is
= e 3z (c cos4z + c 2 sin4z) = (e z 3 (c cos4z + c 2 sin4z). But since x = e z
3
4
3
4
y = x [c cos (4 In x) + c 2 sin (4 In x)] = x [c, cos (In x + c 2 sin (In x )].
v
)
x
x
8.240
z = lnx,
we have
x 3 y'" - 3x y" + 6xy' - 6y = 0.
2
Solve
f
and
)
l
This is an Euler equation. Using the substitutions suggested in Problem 8.230, we rewrite the differential
D(D - 1)(D - 2)y - 3D(D - l)y + 6Dy - 6y = 0, or (D 3 - 6D 2 + UD - 6)y = 0. This equation
is similar in form to that given in Problem 8.28, except now the independent variable is z; its solution is
equation as
y = cxe
8.241
+ c 2 e 2z + c 3 e 3z = c e z + c 2 (e 2 2 + c 3 (r) 3 = c x + c 2 x 2 + c 3 x 3
)
x
x
x 3 y'" - 2xy' + 4y = 0.
Solve
I
z
2
This is an Euler equation with the coefficient of x y" equal to zero. Using the substitutions suggested in
Problem 8.230, we rewrite the differential equation as D(D — 1)(D — 2)y — 2Dy + 4y =
or
This equation is similar in form to that given in Problem 8.169, except now the
(D 3 — 3D 2 + 4)y = 0.
independent variable is z; its solution is
y = c e
2z
x
8.242
x
)
)
l
)
— c x 2 + c 2 x 2 In x + c 3 x x
4x 3 y'" - 16x 2 y" - 55xy' - 8y = 0.
Solve
f
+ c 2 ze 2z + c 3 e z = c {e 2 2 + c 2 z(e z 2 + c 3 (e z
This is an Euler equation. Using the substitutions suggested in Problem 8.230, we rewrite the differential
AD(D - 1)(D - 2)y - 16D(D - 1) - 55Dy - 8y = 0,
or
+ c 3 e H:
z
(4D 3 - 28D 2 - 31D - 8)y = 0. This
equation is identical in form to that given in Problem 8.171. except now the independent variable is z; its
equation as
y = c e~
solutionis
y = c,x"
8.243
+ c 2 x~
In x
+ c3x
.
Since
x = ez
and
= lnx,
this solution becomes
8
.
D(D - \){D - 2)y + 2D(D - \)y + 4Dy - Ay = 0,
equation is
/
y = c e
x
z
3
— A + 4/ — 4 = 0,
2
which has as its roots
1
or
(D
3
- D 2 + AD - A)y = 0.
Its
characteristic
and ±/2. The solution is
+ c 2 cos 2z + c } sin 2z = c x + c 2 cos (2 In x) + c 3 sin (2 In x) = c x x + c 2 cos (In x 2 + c 3 sin (In x 2
)
x
)
x 3 y " - 3x 2 y" - 3xy' + 36y = 0.
Solve
f
This is an Euler equation. Using the substitutions suggested in Problem 8.230, we rewrite the differential equation
as
D(D - \)(D - 2)y- 3D(D - l)y- 3Dy + 36y = 0.
similar in form to that given in
y = c e'
x
8.245
z 2
+ c 2 ze~
This is an Euler equation. Using the substitutions suggested in Problem 8.230, we rewrite the differential
equation as
8.244
1/2
x 3 y" + 2x 2 y" + 4xy' - 4y = 0.
Solve
I
zl2
x
,/2
(D 3 - 6D 2 + 2D + 36)y = 0.
This equation is
Problem 8.1 10. except now the independent variable is z; its solution is
+ c 2 e 4z cos\f2z + c 3 e* sin \J2z = c,x^ 2 4- c 2 x 4 cos( x 2 In x) + e 3 x*sin(\ 21nx)
z
x 4 y (4) + 14x 3 y (3) + 55x 2 y" + 65xy' + 16y = 0.
Solve
m
2z
or
This is an Euler equation. Using the substitutions suggested in Problem 8.230, we rewrite the differential
equation as
D(D - 1)(D - 2)(D - 3)y + IAD(D - 1)(D - 2) + 55D(D - 1) + 65Dy + 16y =
or
(D
4
+ 8D 3 + 24D 2 + 32D + 16)y = 0.
This equation is identical in form to that given in Problem 8.157,
except now the independent variable is z; its solution is
y = c x e~
8.246
+ c 2 ze' 2z + c 3 z 2 e~ 2z + c^z 3 e~ 2z = c x' 2 + c 2 (lnx)x~ 2 + c 3 (lnx) 2 x" 2 + c 4 (lnx) 3 x" 2
x
x 4 y (4) + 6x 3 y (3) - 2x 2 y" - 8xy' + 20y = 0.
Solve
f
2z
This is an Euler equation. Using the substitutions suggested in Problem 8.230, we can rewrite the equation as
D(D - 1)(D - 2)(D - 3)y + 6D(D - l)(D - 2)y - 2D(D - l)y - 8£>y + 20y =
or
(D
4
— 9D 2 + 20)y = 0.
This equation is similar in form to that given in Problem 8.29, except now the
independent variable is z; its solution is
y = c e
x
2z
+ c 2 e~ 2z + c 3 e^ lz + c 4 e~ %3r = c x 2 + c 2 x~ 2 + c 3 x >3 + c 4 x"
x
v
s
CHAPTER 9
The Method of
Undetermined Coefficients
EQUATIONS WITH EXPONENTIAL RIGHT SIDE
9.1
Solve
- 5 v = e 2x
y'
.
I We assume a particular solution of the form y p — A e 2x The general solution to the associated homogeneous
5x
equation is shown in Problem 8.34 to be y h — c e
Since y p is not a linear combination of y h there is no
.
.
,
x
need to modify it.
2A e 2x — 5A e 2x — e 2x
Substituting y p into the given nonhomogeneous differential equation, we obtain
— 3A e
2x
—e
2x
,
y = yh + yp = c e
nonhomogeneous equation is
9.2
y + 6y = e 3x
Solve
A = — \.
from which we find
y p = — \e
Then
Sx
x
— \e 2x
2x
or
and the general solution to the
,
.
.
I We assume a particular solution of the form
yp — A e
3x
where A
,
denotes an unknown cons' ant which must
be determined. The general solution to the associated homogeneous equation is found in Problem 8.37 to be
yh = AeSince y p and y h have no terms in common except perhaps for a multiplicative constant, there is no need to
modify y p Substituting it into the given differential equation, we obtain 3A e 3x + 6A e 3x = e 3x
or
9A e 3x = e 3x from which A = %. Then y p = %e 3x and the general solution to the nonhomogeneous equation
6x
.
.
,
,
is
9.3
At>~
y = y h + y p = Ae
Solve
I
,
y'
6x 4- lo 3x
+ \e
+ 6y = 18e 3a
Both y p and y h of the previous problem are valid here. Substituting y p into this differential equation, we obtain
3A Q e 3x + 6A e 3x = 18<? 3x or 9A e 3x = \Se 3x Thus, A = 2 and y p = 2e 3x
nonhomogeneous differential equation is then y — y h + y p — Ae~ 6x + 2e 3x
.
,
.
The general solution to the
.
9.4
Solve
y'
+ 6y = Ae
~ 5x
.
I We assume a particular solution of the form y p — A e~ 5x with y h — Ae~ 6x as in Problem 9.2.
Since y p and y h have no terms in common except perhaps for a multiplicative constant, there is no need to
,
modify y p
.
Substituting it into the nonhomogeneous differential equation, we get
— 5A e~ 5x + 6A e~ 5x — 4e~ 5x
or
,
A e' 5x — 4e~ 5x
to the nonhomogeneous equation is thus
9.5
Solve
y'
+ 6y = 6e 6x
y p — 4e~
Then A = 4, and
6x
+ 4e~ 5x
y = y h + y p — Ae
.
5x
.
The general solution
.
.
6x
1 We assume a particular solution of the form
with y h as in the previous problem.
yp = A e
Since y p and y h are linearly independent, there is no need to modify y p Substituting it into the given
,
.
6A e 6x + 6A e 6x = 6e 6x or l2A e 6x = 6e 6x Thus A = \, and
6x
bx
The general solution to the nonhomogeneous equation is then y = y h + y p — Ae~ + \e
differential equation, we obtain
6x
y p — \e
9.6
.
2y'-5y = 6e 6x
Solve
.
,
.
.
I We assume y p as in the previous problem, but now the general solution to the associated homogeneous differential
y h = Ae
equation, as found in Problem 8.38, is
yp
is
5x ' 2
.
Since y p and y h are linearly independent, no modification of
necessary.
Substituting y p into this differential equation, we obtain
A =f
Thus
y = yk + y, =
9.7
and
^ +
,
x' 2
y" - 7y' = 6e
Solve
yp =
^e
b
e
6x
.
2{6A e
,
— 5A e 6x = 6e bx
,
or
lA e 6x = 6e 6x
.
6x
.
6x
.
6x
yp = A e
8.2
to
be
Problem
in
found
yh - c
differential equation is
and e
)
The general solution of the given differential equation is then
I We assume a particular solution of the form
lx
bx
there is no need to modify y p
.
,
The general solution of the associated homogeneous
+ c 2 e lx
.
Since e
6x
is
not a linear combination of 1
.
191
CHAPTER 9
192
Substituting y p into the given nonhomogeneous differential equation, we obtain
-6A e
or
6x
= 6e
= e Sx
y" - ly
Solve
A =-l,
Thus
.
y = yh + yp = c
equation is then
9.8
6x
and
+ c 2 e lx — e 6x
x
y p = -e
6x
36A e 6x - l(6A e 6x
)
= 6e 6s
The general solution to the given differential
.
.
.
f We assume a particular solution of the form y p = A e Sx with y h as in the previous problem. Since y h has no
terms in common with y p except perhaps for a multiplicative constant, there is no need to modify y p
,
.
64A e Sx — 7(8/l e 8x ) = e 8x or &A e Sx = e 8x
The general solution to the nonhomogeneous differential equation is then
Substituting y p into the given differential equation, we get
A = 5, and y p = ^e Sx
lx
+ & 8x
y = yn + P = c, + c 2 e
Thus
>'
9.9
.
I We assume a particular solution of the form
8.142 to be
y — c e~
c
x
3x
+ c 2 xe~
3x
)
y = yc + y
2x
The complementary solution is found in Problem
.
= c^e
+ c 2 xe
3x
.
into the given differential equation, we obtain
25A e 2x = \00e 2x
or
,
3x
= yp
)•
Thus
.
A = 4,
y. = 4e
and
2x
The general
.
+ 4e
y" — y' — 2y — e ix
Solve
We assume a particular solution of the form
I
e
,
4A e 2x + 6(2A e 2x + 9A e 2x = \00e 2x
solution is then
yp — A
Since y p is not part of yc there is no need to modify y p
.
Differentiating y p twice and substituting
9.10
.
.
y" + 6/ + 9y = 100e 2x
Solve
.
,
x
y h — c e~ + c 2 e
equation is found in Problem 8.1 to be
3x
yp — A e
The general solution to the associated homogeneous
and since y p is not part of y h no modification
.
2x
,
x
,
of y p is required.
9A e ix — 3A e 3x — 2A e 3x = e 3x
Substituting y p into the differential equation, we obtain
Thus
9.11
A — 5,
and
fy
dy
y p = {e
-4-4-f
" dt + y = 3e
dt
Solve
3x
y = c e~
The general solution then is
.
x
+ c2e
x
2x
+ \e
4A e 3x = e 3x
or
,
.
3x
.
2 '.
2
I We assume a particular solution of the form
yp — A e
2
The general solution of the associated homogeneous
2
Problem 8.9 to be y h — C e 3 732r + C 2 e° 2679
since e
cannot be written as
2679
3 132t
a linear combination of e
and e°
there is no need to modify y p
'.
differential equation is shown in
-
'
';
x
-
',
.
Substituting y p into the given nonhomogeneous differential equation, we obtain
4A e 2 - 4(2A e 2 + A e 2 = 3e 2
or
-3A e 2 = 3e 2 Thus A Q = -1, and y p = -e 2
3 732
solution to the nonhomogeneous equation is then
y = y h + y = C,e
+ C 2 e° 2679 — e 2
'
')
'
'
',
'.
'
2
Solve
,
dt
2
'.
The general
'.
dy
-4y - 4 -f + y = 2e 3
<^
9.12
'
'.
dt
3
y p = A e ', and y of the previous problem is valid here as well.
Since y p and y h have no terms in common except perhaps for a multiplicative constant, there is no need to modify y p
We assume a particular solution of the form
I
fc
.
9A Q e 3 — 4{3A e 3 ') + A e 3 = 2e 3
Substituting y p into the given differential equation yields
Thus A =-l, and v p =-e 3
3132
=
+ C 2 e 0261 9t - e 3
y
y h + y p = C.e
— 2A e 3 = 2e 3
'
is
'.
then
'.
'
'
'
'
',
or
The general solution to the given differential equation
'.
2
9.13
Solve
d x
dx
—
-y + 4— + %x = e~
dt
2
2 '.
dt
I We assume a particular solution of the form
x p — A e~ 2
'.
The general solution to the associated homogeneous
x h = c t e~ 2t cos 2f + c 2 e~
problem is found in Problem 8.54 to be
2'
sin 2t.
Since the functions e~
2
',
e'
2'
sin 2t, and
2
e~ 'cos2f are linearly independent, there is no need to modify .v p
.
Substituting x p into the given nonhomogeneous differential equation, we get
-2
2
2
= <'~ 2 ', or 4A e' 2 = e~ 2t Thus A = ±
4/V- + 4(-2/V- ') +
8/V
'
'
'
.
genera! solution to the nonhomogeneous equation is then
9.14
Solve
+ 900/ = 5e 10
^T-60—
dt
dt
and
x p = \e' 2
'.
The
'(Ci cos 2r + c 2 sin 2r + |).
'.
2
I We assume a solution of the form
is
x — x h + x p — <T
2
found in Problem 8.149 to be
Ih
I
p
= Ae
= A e 10
30 '
'.
+ Bte 30
The general solution of the associated homogeneous problem
Since / and I h have no terms in common, except perhaps
'.
.
THE METHOD OF UNDETERMINED COEFFICIENTS
for a multiplicative constant, there is no need to modify I
p
193
Substituting it into the given nonhomogeneous
.
lO0A o e lOt - 60(l0A o e 10 ') + 900A e 10
differential equation, we obtain
U
= 5e i0
'
or
',
10 '.
4O0A o e 10 = 5e 10
'
Thus
'.
A = 0.0125, and I p = 0.0125e
The general solution (o the nonhomogeneous equation is then
i0
i0
10
Bte
+
0.0125<?
I = I h + I = Ae
+
p
9.15
Solve
'
'
<Px
d2x
dx
dr
dt
dt
'.
-^ + 5^2 + 26 — - 150x = 20<T'.
f We assume a particular solution of the form
x p = A e~'. The general solution to the associated homogeneous
Problem 8.112 to be x h = c x e Zi + e" 4 '(cj cos y/34t + c 3 sin>/34r). Since x p
cannot be obtained from x h by any choice of the arbitrary constants c ls c 2 and c 3 there is no need to modify it.
Substituting x p into the given nonhomogeneous equation, we get
-A e~' + 5A e-' + 26(-A e~')- l50A o e~' = 20e~\ or - 172^ e _t = 20e~'. Thus A = -20/172, and
x = — Yi2.e~'. The general solution to the nonhomogeneous equation is then
differential equation is found in
,
,
x h + X P = c i g3t + e M ( c i cos v34f + c 3 sm V340 —
9.16
Solve
1250 =
—J -5— + 25-^A
-60<?'
f-
dt
Ae
2
3
</r
f We assume a particular solution of the form Q p = A e 7
The general solution to the associated homogeneous
equation is shown in Problem 8.113 to be Q h = c^ 5 + c 2 cos 5f + c 3 sin 5t. Since Q p cannot be obtained from
Q h by any choice of the constants c lt c 2 and c 3 there is no need to modify Q p
Substituting Q p into the given differential equation, we get
or
\4SA e 7t = -60e 7
343A e 7 - 5(49 A e 7 + 25{lA e 7 - \25A e 7t = -60e 7
Therefore A = -£f,
7
—
—
and Q p
j^e
The general solution to the nonhomogeneous differential equation is then
7
Q = Qh + Q P = c e5 + c 2 cos 5f + c i s n 5f _ 37^
'.
'
,
'
')
.
,
')
'.
',
'.
'
9.17
Solve
y
<4)
'-
J
\
- 6y (3) + 16/' + 54/ - 225>> = 100e~
i We assume a particular solution of the form
2Ar
.
yp = A e
~ 2x
The general solution to the associated homogeneous
+ c 2 e~ + c 3 e 3x cos4x + c 4 e 3x sin 4x. Since y p cannot
be obtained from y h by any choice of the constants c through c 4 there is no need to modify it.
Substituting y p into the given nonhomogeneous differential equation, we obtain
l6A e~ 2x - 6(-SA e- 2x + \6(4A e~ 2x + 54(-2A Q e~ 2x ) - 225A e~ 2x = 100e 2x
or
-205A o e~ 2x = 100e~ 2x Thus A = - 100/205 = -|f and y p = -^e' 2x The general solution to the
nonhomogeneous equation is then
equation is shown in Problem 8.124 to be
yh = c e
.
3x
3x
{
,
x
)
)
,
.
.
3x
y = yh + y P — c \ e3x + c 2 e~
+ c 3 e 3x cos4x + c 4 e 3x sin4x — j^e~ 2x
.
EQUATIONS WITH CONSTANT RIGHT SIDE
9.18
Solve
/ - 5v = 8.
I We assume a particular solution of the form
yp = A
where A
is
a constant to be determined. The general
yh = c 1 e
solution of the associated homogeneous equation is found in Problem 8.34 to be
nonzero constant A
is
linearly independent of e
5x
,
there is no need to modify y p
Substituting y p into the nonhomogeneous equation and noting that
A = — f. Then y p = — f
5x
-f
y = yh + yP = c e
y'
p
— 0,
5x
.
Since any
.
we get
— 5A = 8,
or
and the general solution to the nonhomogeneous equation is
x
9.19
Solve
/' -
/ - 2y = 7.
The general solution to the associated homogeneous
yp — A
2x
Since y p cannot be obtained from y h by
Problem 8.1 to be yh = c x e~ x + c 2 e
any choice of c and c 2 there is no need to modify it.
Substituting y p and its derivatives (all of which are zero) into the nonhomogeneous differential equation, we get
— - 2A = 1 or A = —\. Then y p = — \ and the general solution to the nonhomogeneous
I We assume a particular solution of the form
differential equation is shown in
,
x
equation is
2
9.20
Solve
y = c { e~
x
+ c 2 e 2x — \.
d Q
dO
—^
+ 8-^ + 520 = 26.
.
.
194
CHAPTER 9
Q
The general solution to the associated homogeneous
I We assume a particular solution of the form Q p — A
-4
-4
cos 6t + c 2 e
sin 6f.
Since Q p is not part of Q h it
equation is shown in Problem 8.55 to be Q h — c^
requires no modification. Substituting Q p into the nonhomogeneous differential equation, we obtain
+ 8(0) + 52A = 26 or A — \. Thus Q p = \ and the general solution to the nonhomogeneous equation
4t
is
Q = c e~* cos6t + c 2 e~ sin6t + \.
.
'
'
,
,
l
9.21
+ 100 -r + 50,000*? = 2200.
4r
dt
dt
Solve
2
1 We assume a particular solution of the form
qp — A
The general solution to the associated homogeneous
.
q h = c 1 e"
equation is found in Problem 8.56 (with q replacing /) to be
50
'cos 50vT9f + c 2 e~
50,
sin 50>/l9f.
Since q p is not part of q h no modification to q p is necessary.
,
+ 100(0) + 50,000.4 o = 2200,
Substituting q p into the nonhomogeneous differential equation, we get
^o = 1 P — 250-
or
The general solution to the nonhomogeneous equation is then
= Qh + Q P = c^' 50 cos 50^19? + c 2 e _50, sin 50vT9t + ^.
'
<7
9.22
q + 20? + 200q = 24.
Solve
We assume a particular solution of the form
I
qp = A
equation is, from Problem 8.64 with q replacing /,
constant A
modify q p
The general solution to the associated homogeneous
.
10
10
cos lOt + c 2 e~
sin lOf.
q h — c e~
Since an arbitrary
cannot be expressed as a linear combination of e~ 10( cos 10t and e~ 10 'sin lOf, there is no need to
'
'
x
.
= 24, or A = q p — ^.
I0
q = c^'^'cos lOf + c 2 e~ 'sin lOf + 23.
Substituting q p into the nonhomogeneous differential equation, we obtain
The general solution to the nonhomogeneous equation is then
9.23
q + 400q + 200,000? = 2000.
Solve
We assume a particular solution
f
200.4
q h = e~
thus, q p needs no modification.
found in Problem 8.70 to be
qp = A
200
.
The general solution to the associated homogeneous equation is
+ fisin400f), which has no terms in common with q p
'{A cos400f
;
Substituting q p into the given nonhomogeneous differential equation, we get
= 2000,
200,000/1
or
A = q p — 0.01. The general solution to the nonhomogeneous equation is then
200
'(/4cos400r + Bsin400f) + 0.01.
q = <?"
9.24
q + 1000? + 50,000? = 2200.
Solve
qp — A
5219
q h = c e~
We assume a particular solution of the form
f
equation is shown in Problem 8.18 to be
there is no need to modify q p
The general solution to the associated homogeneous
.
'
y
+ c 2 e~ 9A1 2
'.
Since q p and q h have no common terms,
.
Substituting q p into the given nonhomogeneous differential equation, we obtain
50,000>1
= 2200,
or
A = q p = 0.044. The general solution to the nonhomogeneous equation is then
52 79
+ c 2 e- 9 * 12 + 0.044.
q = c e~
'
'
x
9.25
Solve
—
1000-^
-^ + 250,OO0Q =
y++ 1000
-yf
dr
I
24.
dt
e
The general solution to the associated homogeneous
Qp = A
500
500
=
Since Q p is not a linear combination of
c
c^e'
+
2 te'
Qh
We assume a particular solution of the form
equation is shown in Problem 8.147 to be
-soor
.
'
'.
ancj ^-500^ no modification of Q p is required.
+ 1000(0) + 250,000/1 = 24,
A = Q p = 9.6 x 10~ 5 The general solution to the nonhomogeneous equation is then
-5
Q = Cl e- 500 + c 2 te- 500 + 9.6 x 10
Substituting Q p into the given nonhomogeneous differential equation, we get
or
.
'
'
.
9.26
Solve
d 2x
—^
q
dt~
10
q
x = -,
where g denotes a positive constant.
5
I The right side of this equation is a constant, so we assume a particular solution of the form
xp = A
.
The
general solution to the associated homogeneous equation is found in Problem 8.21 to be
T
x h = (7^^°' + C 2 e" % ^ °'.
need to modify x p
Since x p cannot be obtained from x h by suitably choosing C\ and C 2
,
we do not
.
Substituting x p into the nonhomogeneous differential equation, we get
The general solution to the nonhomogeneous equation is then
9
- -- A = -,
x — C x e" 9 10
'
or
+ C 2 e~ %9 10 — 2.
'
A = x p = — 2.
THE METHOD OF UNDETERMINED COEFFICIENTS
195
2
d x
g
——x =—
—
20
2>g
9.27
Solve
=-
where g denotes a positive constant.
,
10
at
I Both x p and x h are as in the previous problem.
°~m A
=
°
%
or
3
9.28
Solve
A ° = x p = -\
p
into this differential equation, we get
x = xh + xp = C 1 <r®T*' + C 2
'
e-^ -l
t
2
d x
d x
dx
—
T + 5 —T + 26 — - 150x =
dt
Thus
Substituting x
3
dt
2
30.
dt
I We assume a particular solution of the form
xp = A
The general solution to the associated homogeneous
differential equation is found in Problem 8.1 12 to be
x h = c x e 3t + e~ A \c 2 cos y/34t + c 3 sin >/34t). Since x
p
cannot be obtained from x h no matter how the arbitrary constants c through c 3 are chosen, there is no need to
modify x p
Substituting x p into the given differential equation, we obtain
150A = 30, or A = x p =—\. The
3
M
general solution is then x = c^ + e~ (c 2 cos yf$4t + c 3 sin \/34f) — j.
.
x
.
'
d
9.29
Solve
3
Q
d2Q
3
2
dQ
—f-5—^ + 25-^- 1250 = 1000.
dt
dt
dt
I We assume a particular solution of the form
The complementary function is showr. in Problem
Qp = A
5
5
—
c
e
cos
5f
sin
Since
is
not
+
c
c
5r.
a linear combination of e
cos 5r, and sin 5t,
+
Q
2
Qp
3
there is no need to modify Q p
— 5(0) + 25(0) — 125A = 1000. Thus
Substituting Q p into the given differential equation, we obtain
~
—
—
and
the
general
solution
to
the
nonhomogeneous
differential
equation is
8,
QP
^o
Q = Qc + Q P = c e$ + c 2 cos 5f + c 3 sin 5t - 8
8.1 13 to
be
.
'
c
',
{
.
'
-
i
9.30
Solve
y
(4)
- 6y (3) + 16/' + 54/ - 225y = -75.
I We assume a particular solution of the form
yp = A
The complementary function is found in Problem 8.124
cos4x + c 4 e sin 4x. Since y p cannot be obtained from ye by any choice of
the constants c through c 4 there is no need to modify y p
— 225A — — 75. Therefore, A — y p — \, and
Substituting p into the given differential equation, we find
the general solution to the nonhomogeneous equation is
3x
+ c 2 e~ 3x + c 3 e 3x cos 4x + c 4 e 3x sin 4x + ^.
y — y c + v'p — c e
to be
yc — c e
3x
+ c 2 e~
1
3x
+ c3e
.
3x
3jc
,
x
.
v'
x
EQUATIONS WITH POLYNOMIAL RIGHT SIDE
9.31
Solve
f
y'-5y = 3x+l.
Since the right side of the differential equation is a first-degree polynomial, we try a general first-degree
We assume y p = A x x + A
where the coefficients A y and A must be
determined. The solution to the associated homogeneous equation is shown in Problem 8.34 to be y h = c x e 5x
Since no part of y p solves the homogeneous equation, there is no need to modify y p
polynomial as a particular solution.
,
.
.
y' — A
u we get
p
Equating the coefficients of like powers
Substituting y p into the given nonhomogeneous equation and noting that
A — 5{A x + A
x
x
)
= 3x + 1,
or
(
— 5/4 )x + {A — 5A
x
{
)
= 3x + 1.
of x, we obtain
=3
or
A -5/4 = 1
or
5^ x
y
9.32
— |x — ^,
Then
yp =
Solve
/ - 5y = 8x.
and the general solution is
- - 15
—
^0 —
25
A
/ij
y = yh + yp = c e
x
5x
- |x — ^.
The right side of this differential equation is a first-degree polynomial, so both y p and y h of the previous problem
A - 5{A x + A Q = 8x, or
Substituting y p into the given differential equation, we get
(- 5/4 Jx + (A - 5A ) = 8x + 0. Equating the coefficients of like powers of x, we obtain
f
are valid here.
x
x
x
-5A
t
=8
A - 5A =
x
or
or
Ax =
A =
_§_
25
y p = — fx — ^, and the general solution to the given nonhomogeneous equation is
5x
=
- fx - &.
+
y
y P = c,e
}'h
Thus
)
CHAPTER 9
196
9.33
/ - Sy = 2x 2 - 5.
Solve
f
Since the right side is a second-order polynomial, we assume a general second-order polynomial for y
yp = A 2 x
2
+A x+A
namely
The complementary solution remains as it was in the previous two problems:
Since y p and y h have no terms in common except perhaps for a multiplicative constant, there is no
x
5x
yh = c x e
need to modify y p
.
.
.
Substituting y p along with
y'
p
2A 2 x + Ay- 5{A 2 x 2 + A x x + A
= 2A 2 x + A
= 2x 2 - 5,
into the given differential equation, we find that
x
(-5A 2 )x 2 + (2A 2 - 5A )x + (A - 5A
or
)
x
x
)
= 2x 2 + Ox - 5.
Equating coefficients of like powers of x, we obtain
=
5A 2
2A 2 -5Ay
or
Ay- -5A = -5
y p = — 0.4x — 0.1 6x + 0.968,
5x
- 0.4x 2 - 0.16x + 0.968.
=
c
e
y
2
Then
A 2 = -0.4
A = -0.16
A = 0.968
or
2
—
-
x
or
and the general solution to the given differential equation is
x
9.34
2/ - 5 y = 2x 2 - 5.
Solve
f The particular solution y p of the previous problem is valid here, but the complementary solution is now
(see Problem 8.38). Substituting y into the differential equation, we get
yc = Ae
p
2
2(2A 2 x + A ) - 5(A 2 x + A x + A ) = 2x 2 - 5, or (-5A 2 )x 2 + (4A 2 - 5A )x + (2A
5xl2
x
x
x
X
- 5A
= 2x 2 + Ox - 5.
)
Equating coefficients of like powers of x, we obtain
=
=
-5/1 2
4A 2 -5A
X
2/1,
y = c e
The general solution is then
9.35
y" - y' - 2y = 4x 2
Solve
5xl2
x
or
A 2 = -0.4
A = -0.32
or
A =
or
2
-5A = -5
x
0.872
— 0.4x 2 — 0.32x + 0.872.
.
yp = A 2 x
We assume a particular solution of the form
2
The general solution to the associated
+ A x + A
x
2x
=
homogeneous differential equation is found in Problem 8.1 to be yh
Since y p and y h have
c e' + c 2 e
no terms in common except perhaps for a multiplicative constant, there is no need to modify y p
Substituting y p into the given differential equation, we get
2A 2 — {2A 2 x + A — 2(A 2 x 2 + A x + A = 4x 2
2
—
—
or, equivalently,
2A 2 )x +
2A 2 — 2A )x + (2A 2 — A — 2A = 4x 2 + Ox + 0. Equating the coefficients
I
.
x
.
x
.
x )
(
(
x
)
x
)
x
of like powers of x, we obtain
—4
=0
— 2A 2
-2A 2 -2A
2A 2 — A — 2A =
X
x
yp =
Then
—2x + 2x — 3,
2
2
9.36
C e
-
'
x
+ C2e
- 4A + A
X
x
-
or
AQ = — 3
2
x
y = yh + y p — c e~
x
x
+ c 2 e 2x — 2x 2
-I-
2x — 3.
The complementary solution is found in Problem 8.9 to be
yp — A x t + A
Since y p and yc have no terms in common, there is no need to modify y p
into the given differential equation, we get
and y" =
t
y'
,
.
.
=A
p
+ A = 3t - 4,
Substituting y p
x
,
p
A t + (-4A + A
or
x
X
yields
= 3f + 8,
Then
y
Solve
- 4 -f + y = 2 - 2t + 3.
-f
2
dt
dt
2
f
A =
Since the right side is a first-order polynomial in t, we assume a general first-order polynomial in f as the form
We try
9.37
or
+ y = 3f-4.
-4-4-f2
dt
dt
of a particular solution.
2619 '.
3 732
=
yc
A2 — — 2
dy
d y
Solve
I
and the general solution is
or
d y
.
)
A
x
=3
-4A
x
+A =-4
and the general solution is
= 3f - 4.
y = ye + y
Equating coefficients of like powers of t
or
A = 8
=C e
31i2
x
'
+ C 2 e°- 2619t + 3r + 8.
dy
t
Since the right side of this differential equation is a second-degree polynomial, we try a second-degree
polynomial (with undetermined coefficients) as the form of y p namely
,
yp — A 2 t
2
+A
x
t
+A
.
The
THE METHOD OF UNDETERMINED COEFFICIENTS
D
197
complementary solution of the previous problem is valid here, and since y and yc have no terms in common,
p
there is no need to modify y p
.
Substituting y p into the given differential equation, we obtain
2A 2 - 4(2A 2 t + A
A 2t
2
+ A 2 2 + A t + A = 2 - It + 3, or
+ (-8/l 2 + A )t + (2A 2 - 4/1, + A = 2 - It + 3. Equating coefficients of like powers of
t
x )
t
x
t
)
x
=
A2
-%A 2 + A
we get
1
= -2
y
t,
2A 2 -4A + A =
3
X
or
A = 6
or
A = 25
x
2
+ 6t + 25, and the general solution to the nonhomogeneous equation is
yp =
3132t
=
C e
+ C 2 e 2619 + 2 + 6t + 25.
y
Then
t
-
'
t
x
9.38
—£ - 4 -^ + y =
Solve
df
I
2
2t
3
+ 3f 2 - 1.
dr
Since the right side of the differential equation is a third-degree polynomial, we assume a general third-degree
We try
polynomial as the form of a particular solution.
yp — A 3 t
3
+ A2 2 + A
t
x
+A
t
.
The complementary
solution is again that of Problem 9.36.
Substituting y p along with
y'
p
= 3A 3 t 2 + 2A 2 + A
t
y" = 6A
and
x
p
+ 2A 2
3t
into the nonhomogeneous
differential equation, we get
+ A 3 3 + A 2 t 2 + A t + A = 2t 3 + 3r 2 A 3 3 + (-12A 3 + A 2 )t 2 + (6A 3 - SA 2 + A )t + {2A 2 - AA + A = 2t 3 + 3r 2 + Ot - 1
6A 3 t + 2A 2 - 4(3A 3 t 2 + 2A 2 t + A
or
t
x )
t
1
x
x
)
X
Equating coefficients of like powers of t, we obtain
A3
=
2
=
3
or
=0
A 2 = 27
or
2/4 2 -4/1, + A = -1
or
A = 204
= 761
/1
-12A 3 + A 2
6A 3 - SA 2 + A
x
x
yp = 2t + 21t + 204r + 761, and the general solution to the nonhomogeneous equation is
=
C e 3132 + C 2 2679 + 2t 3 + 21t 2 + 204f + 761.
y
3
Then
2
-
'
'
<?
x
2
9.39
dx
d x
—
+4—+
Solve
=-
dt
2
8.x
= -3t + 1.
dt
The complementary solution is shown in
I We try a particular solution of the form x p — A + A
Problem 8.54 to be x = c e~ 2 cos 2t + c 2 e~ 2 sin 2t. Substituting x p along with its derivatives into the given
+ 4A + S(A + A = -3t + 1, or (SA )t + {4A + SA = -3f + 1.
differential equation, we obtain
x
c
t
.
'
'
x
X
x
t
)
x
X
)
Equating coefficients of like powers of t, we find that
= —3
8/4,
4/1,
Thus
y p = — |r + ^,
2t
cos2f + c 2 e
x = c e
x
9.40
—£ + 4 -^ +
Solve
dt
I
2
8.x
+ 8/l =
1
A =
A =
or
i
x
or
16
and the general solution to the nonhomogeneous equation is
2<
— ft + _5_
sin 2f
= 8t 2 + 8r + 18.
dt
Since the right side is a second-degree polynomial, we try
.x
p
= A2 2 + A t + A
t
x
.
The complementary
solution of the previous problem is valid here. Substituting x p along with its derivatives into the given differential
+ 4{2A 2 + A + S(A 2 2 + A + A = St 2 + St + 18, or
2
{SA 2 )t 2 + (SA 2 + SA )t + (2A 2 + 4A + SA = 8r + 8f + 18. Equating coefficients of like powers of
equation, we get
2/4 2
x
t
t
x )
X
x
t
)
)
we obtain
=8
=8
or
A2 =
or
A =0
2A 2 + 4A +%A = 18
or
/1
SA 2
SA 2 + 8/1,
X
1
x
=2
Thus x p = t 2 + 2, and the general solution to the nonhomogeneous equation is
2
2t
x = c e~ 2 'cos2t + c 2 e~ s\r\2t + t + 2.
x
t,
198
CHAPTER 9
D
,dx„
— 8x = -
d 2x
9.41
—5+4
2
Solve
+
,
2
t
.
dt
dt
I
Since the right side of the differential equation is a second-degree polynomial, y of the previous problem is
p
appropriate here. The complementary solution is given by y c of Problem 9.39, and since it has no terms in
common with the particular solution, no modifications are necessary.
Substituting y p and its derivatives into the given differential equation, we get
2A 2 + 4(2A 2 t + A ) + S(A 2 t 2 + A,t + A = -t 2 or
2
(SA 2 )t + (8A 2 + iAi)t + (2A 2 + 4A + SA = -t 2 + Of + 0.
we obtain
)
x
,
M
2
+ 8/4,
8/4 2
2A 2 + 4A
x p — — gf 2 + gf — y^,
Then
x = c e~
2t
x
cos 2t + c 2 e~
Solve
dx
—
+ 8x =
dt
dt
2
= -1
or
=0
A 2 = -g
or
At =
I
A = -^
+$A =
1
or
and the general solution to the nonhomogeneous equation is
sin 2t
—=- + 4
d x
9.42
2t
Equating coefficients of like powers of t,
)
X
16f
3
- gf 2 +& — £%.
- 40f 2 - 60r + 4.
I The complementary solution is that of Problem 9.39. Since the right side of the given nonhomogeneous equation
is
a third-degree polynomial, we try a third-degree polynomial as the form of a particular solution.
xp = A 3t 3 + A 2 t 2 + A
t
x
+A
xp = 3A 3 t 2 + 2A 2 t + A x
Then
.
x"p = 6A 3 t + 2A 2
and
.
We assume
Substituting these
quantities and x p into the given differential equation, we obtain
= 16f 3 - 40t 2 - 60f + 4
3
2
(8-4 3 )t + (12,4.3 + 8/l 2 )t + (6.4 3 + 8/4 2 + SA ,)r + (2A 2 + 4A
+ SA = 16f 3 - 40r 2 - 60f + 4
6A 3 t + 2A 2 + 4(3/4 3 f 2 + 2A 2 t + A ) + 8(/l 3 f 3
x
or
-I-
A 2t2 + A t + A
)
x
)
X
Equating coefficients of like powers of f, we get
= 16
12/4 3 + 8/4 2
= -40
6.4 3 + 8/4 2 -1-8.4,
= -60
4
2A 2 + 4A +%A =
8/4 3
X
x p — 2f
Then
3
— 8f —
2
x = c e~ 2 'cos2t + c 2 e
x
t
2
+ 3,
'sin It
d 2x
9.43
dx
—T + 4 —
+ 8x =
Solve
dt
2
8f
4
=
or
<4 3
or
A 2 = -8
or
At = -1
or
A =
2
3
and the general solution to the nonhomogeneous equation is
+ 2f 3 - 8r 2 - t + 3.
+
1
6f
3
-
1
2f
2
- 24r - 6.
dt
I The complementary solution is. again. xe of Problem 9.39.
differential equation
is
Since the right side of the given nonhomogeneous
a fourth-degree polynomial, we assume a general fourth-degree polynomial as the form of
That is, we let xp = A 4 t 4 + A 3 t 3 + A 2 t 2 + A t -I- A
Since x p has no term in common
with the complementary solution except perhaps for a multiplicative constant, there is no need to modify x p
2
3
and x"p = '2/l 4 f 2 + 6A 3 t + 2A 2 into
Substituting x p and its derivatives
x' = 4/4 4 f + 3A 3 t + 2A 2 t + A
p
a particular solution.
.
x
.
x
the nonhomogeneous differential equation, we get
12.4 4 r
2
+ 6/4 3 f + 2A 2 + 4(4Aj 3 + 3A 3 2 + 2A 2 + A J + 8(/l 4 f 4 + A 3 3 + A 2 2 + A
t
t
4
t
t
x
+ A
t
)
= 8f + 16f - 12f - 24f - 6
2
4
3
(8/4 4 )t + (16/4 4 + 8-4 3 )r + (12/4 4 + !2/4 3 + SA 2 )t + (6/4 3 + SA 2 + %A )t + (2A 2 + 4A + 8/4
= 8t 4 + 16r 3 - 12f 2 - 24f - 6
or
3
2
x
X
Equating coefficients of like powers of t yields
8/4 4
+ SA 3
=
16
or
12/4 4
+ 12/4 3 + 8/l 2
- -12
= -24
= -6
or
+ 8^1 2 + %A
2A 2 + 4A + 8^1
l
X
x p = f 4 — 3r 2
,
or
or
AA =
A3 =
A 2 = -3
At =
A =
and the general solution to the nonhomogeneous equation is
2
4
2
2,
x = c e~ 'cos2f + c 2 e" sin2f + f - 3f
1
or
16.4 4
6/4 3
Then
8
.
1
)
A
A
A
THE METHOD OF UNDETERMINED COEFFICIENTS
9.44
Solve
d 2I
dl
dr
dt
199
D
—y - 60 — + 900/ = 1800r - 300f.
3
I We assume a particular solution of the form
= A3 3 + A2 2 + A + A
The general solution of the
30
associated homogeneous equation is found in Problem 8.149 to be
I = Ae
+ Bte 30
Since/ and I have
I
t
p
t
t
t
.
'
'
.
c
c
no terms in common, there is no need to modify I p
Substituting I p and its derivatives into the given nonhomogeneous differential equation, we obtain
.
6A 3 t + 2A 2 - 60(3A 3 t 2 + 2A 2 t + A J + 900(A 3 t 3 + A 2 t 2 + A t + A
x
(900^ 3 )r
3
)
= 1800r 3 - 300t,
or
+ (— I8O/I3 + 900A 2 )t + (6A 3 - 120A 2 + 900/1, )t + (2A 2 - 60/1, + 900
2
)
= 1800r 3 - 300t
Equating coefficients of like powers of t yields
=
or
A,=
2
or
A2 =
0.4
i
or
1
or
A l = -0.2933
A = 0.0204
900 A 3
1800
-180/4 3 + 900A 2
= -300
60A +900A =
00 A
6A 3
2/1-
Then I p = 2t 3 + 0.4r 2 — 0.2933f + 0.0204, and the general solution to the nonhomogeneous equation is
30
/ = Ae
+ Bte 30 + 2t 3 + OAt 2 - 0.2933r + 0.0204.
'
9.45
Solve
'
— - 60 — + 900/ = 900r + 1800t - 3600f
4
2r
dt
3
2
.
dt
= A 4 t4 + A 3 3 + A 2 2 + A + A
The complementary
solution is I of the previous problem, and since it has no term in common with I p there is no need to modify I
I We assume a particular solution of the form
I
t
t
p
x
t
.
,
c
Substituting I p and its first two derivatives into the given nonhomogeneous equation, we get
\2A 4 t 2 + 6/l 3 f + 2A 2 - 60(4/l 4 r 3 + 3A 3 t 2 + 2A 2 t + A ) + 900(/l 4 t 4 + A 3 t 3 + A 2 t 2 + A x t + A )
x
= 900f 4 + 1800r 3 - 3600f 2
4
+ (-240/4 4 + 900A 3 )t 3 + (12/1 4 - iS0A 3 + 900^1 2 )r 2 + (6A 3 - \20A 2 + 900A )t
- 900f 4 + 1800r 3 - 3600r 2 + Of +
+ (2/4 2 - 60/4 + 900
or
(900/l 4 )r
t
)
!
Equating coefficients of like powers of t, we obtain
900A A
-240/l 4 + 900/l3
12/1 4
=
900
or
At =
=
1800
or
-3600
or
A 3 = 2.267
A 2 = -3.560
- I8O/I3 + 900A
«M 2
0/4 2
+ 900/4!
2A 2 -
60
y
A x = -0.490
A = 0.025
or
+900^1
=
1
or
Then I p = t 4 + 2.267r 3 - 3.56r 2 - 0.490f + 0.025, and the general solution is
30
/ = Ae
+ Bte 30 + t 4 + 2.261t 3 - 3.56r 2 - 0.490f + 0.025.
'
9.46
Solve
'
—T - 60 — + 900/ = 4500f
dt
2
5
.
dt
I We assume a particular solution of the form
I
p
— A 5 5 + /l 4 4 + A 3 3 + A 2 2 + A^t + A
t
t
f
t
,
which is a
The complementary solution is I c of Problem 9.44. Substituting I p and its
2
3
and I'p = 5/l 5 t 4 + 4^ 4 t 3 + 3^ 3 r 2 + 2A 2 t + A
first two derivatives,
I = 20A 5 t + l2A 4 t + 6A 3 t + 2A 2
p
into the given differential equation and rearranging yield
general fifth-degree polynomial in t.
1
,
4
+ (20A 5 - 240A 4 + 900A 3 )t 3 + (12/1 4 - IS0A 3 + 900^ 2 K 2
5
4
+ (6/4 3 - 120A 2 + 900/4j)r + (2A 2 - 60/1, + 900A = 4500f + Or + Of 3 + Or 2 + Of +
(900A 5 )t 5 + (-300/1 5 + 900A 4 )t
)
Equating coefficients of like powers of f, we get
900/4 5
-300/4 5 + 900/l 4
20/4 5
- 240 A 4 + 900 A 3
12/4 4 - I8O/I3
-I-
900A 2
6/13-120/12+900/1,
2A 2 -
60/1,
= 4500
=
=
=
=
+900A =
f
CHAPTER 9
200
A 5 = 5, A x = 1.6667, A 3 = 0.3333, A 2 = 0.0444, A = 0.0037,
A = 0.0001. Thus, the general solution to the nonhomogeneous equation is
30t
+ Bte 30 + 5f 5 + 1.667r 4 + 0.3333* 3 + 0.0444t 2 + 0.0037r + 0.0001.
I = I + I p = Ae
The solution of this system is
l
and
'
C
3
9.47
Solve
2
d x
dx
d x
—
+ — + 26 — - 150x =
dv
5
r-
dt
3
-=-
600f.
dt
I We try a particular solution of the form
xp = A
equation is found in Problem 8.112 to be
x h = c,e 3
to modify x p
+
,
x
t
'
The general solution to the associated homogeneous
+A
_4
+ c 2 e 'cos V34f + c 3 e _4r sin V34t. Since there is no need
.
we substitute it and its derivatives into the given differential equation, obtaining
= 600f, or (- 150/1 ,)r + (26/1, - 150A = 600f. It follows that
+ 26,4, - 150M,r + A
)
)
-150/1!
- 150/1
26/1,
-600
or
A = -4
=0
or
A = -0.693
l
Then x p — — At — 0.693^ and the genend_solution is
x = c,e 3 + c 2 e~*' cos V34r + c 3 e'*' sin ^34f -At- 0.693.
'
d x
dx
—T + 5 —
T + 26 — - 50.x = 600f
d x
9.48
Solve
dt
3
dt
2
1
2
.
dt
xp = A 2t2 + A
1 We assume a particular solution of the form
x
+ A
t
while the general solution x h of the
,
previous problem is valid here. Substituting x p into the given differential equation yields
+ 5(2A 2 + 26(2A 2 + A - \50(A 2 2 + A + A = 600f 2 or
(- 150 A 2 )t 2 + (52/1 2 - 150/1,)! + (10 A 2 + 26/1, - \50A = 600t 2 + Or + 0.
t
)
t
x )
t
t
)
,
Equating coefficients of like powers
)
of t, we find
=600
-150A 2
- 150/1,
=
10A 2 + 26/1,
\50A =
52/1 2
9.49
Then
x — xh + x
Solve
—p + 5 —
dt
3
—= c x e
3'
+ c2e
,
-~
dt
2
'4
'
cos >/34r + c 3 e
+ 26-^- 150x = 600f 3
A 2 = -4
or
A x = -1.387
A = -0.507
or
sin -J},4t - -4f
'
2
- 0.507.
- - 1.387r
.
dt
I We assume a particular solution of the form
valid here.
4
or
xp = A 3 t 3 + A 2 t 2 + A
{
t
+ A
while x h of Problem 9.47 is
,
Substituting x p and its derivatives into the given differential equation and equating coefficients of like
powers of t in the resulting equation, we obtain
=600
=
-150/13
1SA 3 -\50A 2
=0
3O/I3+ 52/4 2 - 150/1,
6A 3 + 10/1 2 + 26/1, - 150A =
9.50
Then
x = x h + x p = c,r" + c 2 e~*' cos >{3At + c 3 e~
Solve
+ 5
*^-r
3
dt
~
dt
2
+ 26 -? -
1 50.x
= 600f 4
valid here.
sin V34'
A 3 =-4
or
A 2 = -2.08
or
A = -1.521
or
A = -0.562
x
- 4t 3 - 2.08t 2 - 1.52k - 0.562.
.
dt
xp = A4t
I We assume a particular solution of the form
is
At
or
4
+ A 3 3 + A 2t2 + A
t
Y
t
+ A
,
while x,, of Problem 9.47
Substituting x p and its derivatives into the given differential equation and equating coefficients of
like powers of t in the result, we get
150^4
= 600
or
104^4 - 150^3
=
or
=
or
6O/I4 +
24/l 4
+
78/l 3
—
30/13 +
6A 3 +
1
50 A 2
52A 2
150 A,
=
or
10A 2 +
26/1, -
=
or
Then the general solution is
3t
4t
cos v34r + c 3 e
x = xh + xp = c y e + c 2e
4I
sin v '34t
- 4f
.773r
3 --
3.042r
A A = -4
A 3 = -2.773
A 2 = -3.042
A = -2.249
A = -0.704
x
2
- 2.249t - 0.704
THE METHOD OF UNDETERMINED COEFFICIENTS
d2
d3
9.51
Solve
201
dQ
Q
Q
+ 25-^-125(2= -625r 4 + 250t -150f + 60r+ 137.
-^-5-^
3
2
I We try a particular solution of the form Q p = A+t* + A 3 3 + A 2 t 2 + A + A
The complementary solution
5
is shown in Problem 8.1 13 to be
Q = c e + c 2 cos 5t + c 3 sin 5t. Since Q p and Q have no terms in common,
there is no need to modify Q p Substituting Q and its derivatives into the given differential equation and equating
p
t
t
t
.
'
c
{
t
.
coefficients of like powers of t in the result, we obtain
= -625
= 250
= -150
=
60
—
137
o
-125A 4
\00A 4 - 125/4 3
-60A 4 + 75 A 3 - 125 A 2
24A 4 - 30A 3 + 50/4 2 - 125/4,
6/4 3 10A 2 + 25/4,
Solve
dt
2
or
A3 =
2
or
A2 =
or
A =
or
A = -1
t
'
5
5
3
5
c
—= - —% + 25 ~ - 125Q* = 5000f - 3000f
dt
At =
Q - Q + Q p = c^ 5 + c 2 cos 5t + c 3 sin 5t + 5r 4 + 2t 3 - 1.
Then the general solution is
9.52
or
3
.
dt
i We assume a particular solution of the form
5
3
2
while Q
Qp = A 5 + A4f + A 3 + A 2 + A + A
remains as in the previous problem. Substituting Q p and its derivatives into the given differential equation and
t
t
t
y
t
,
c
equating coefficients of like powers of t in the result, we get
-125^ s
125/4 5 - 125X4.
-100/4 5 + 100/4 4 - 125X3
60A 5 - 60X 4 +
24X 4 -
This system may be solved to yield
75A i -l25A 2
30X 3 +
50/4 2
- 125/4,
6X3 -
10/4 2
+ 25A
A 5 = A 4 = —40,
A 3 = 24,
-1
125/4
A 2 = 14.4,
= 5000
=
= -3000
=
=
=
A = —7.68,
and
x
A = — 1.536.
Then the general solution is
Q = Q P + Qc = c^ 5 + c 2 cos 5t + c 3 sin 5f - 40r 5 - 40t* + 24t 3 + 14.4f 2 - 7.68t - 1.536.
'
9.53
Solve
(D 4 - 16)y = 80x 2
.
I We assume a particular solution of the form
yp — A 2 x
2
+A x+A
associated homogeneous equation is shown in Problem 8.128 to be
The general solution to the
.
y
2x
+ c 4 e~ 2x
y h = c, cos 2x + c 2 sin 2x + c 3 e
Since y p and y h have no terms in common, there is no need to modify y p
.
.
D 4 v p = 0, we find
= 80x 2 + Ox + 0. Equating
- 16(/4 2 x + A x + A = 80x
or (-l6A 2 )x + (-\6A )x + (-16A
— 16/1 2 = 80, so that A 2 = —5, and /4, = A = 0. Thus
coefficients of like powers of x then yields
2
—
= c, cos 2x + c 2 sin 2x + c 3 e 2x + c 4 e~ 2x — 5x 2
=
solution
is
the
general
5x
and
y
yp
Substituting y p into the nonhomogeneous equation and noting that
2
2
2
)
x
,
t
)
.
,
9.54
Solve
(D 4 - 16)y = 80x 5 - 16.
while y h of
+ X 4x4 + A 3 x 3 + A 2 x 2 + A x + A
4
=
120X
x
24X
into
the
given
Substituting
and
D
well.
+
valid
here
as
the previous problem is
yp
yp
5
4
f We assume a particular solution of the form
yp = A 5 x
5
,
y
differential equation, equating coefficients of like powers of x in the result, and solving for the coefficients of
A s = — 5,
y p we find that
,
and the general solution is
A 4 = A 3 = A 2 = 0, A = -37.5, and A =l. Then yp = — 5x s — 37.5x + 1,
2x
+ c 4 e~ 2x - 5x 5 - 37.5x + 1.
y = c, cos 2x + c 2 sin 2x + c 3 e
x
EQUATIONS WHOSE RIGHT SIDE IS THE PRODUCT OF
A POLYNOMIAL AND AN EXPONENTIAL
9.55
Solve
y'
- 5y = xe 2x
1
.
Since the right side of this equation is the product of a first-degree polynomial and an exponential, we
assume a particular solution of the same form a general first-degree polynomial times an exponential. We try
2x
The solution to the associated homogeneous differential equation is found in Problem 8.34
y p = (/4,x + A Q )e
5x
=
Since
to be
c
e
y p and y h have no terms in common, there is no need to modify y p
yh
l
—
.
.
.
^
CHAPTER 9
202
Substituting y p into the given differential equation, while noting that
obtain
(
p
x
x
x
x
we
,
,
x
)
x
x
2x
2x
— 3A
=1
l
Then
y p = (— ^x — ^)e
y = yh + y P =
Solve
c i e5x
and the general solution to the nonhomogeneous differential equation is
,
W -¥
lx
-
x
or
x
2x
A — —\
A — —
or
A — 3A =
9.56
= A e 2x + 2(A x + A Q )e 2x
y'
A e + 2(A x + A )e — 5(A x + A )e = xe
which may be simplified to
— 3A )x + (A — 3A = x. Equating coefficients of like powers of x, we obtain
2x
2x
/ - 5y = (2x - \)e 2x
lx
-
.
I The right side of this equation is again a first-degree polynomial times an exponential, and both y p and y h
of the previous problem are valid here. Substituting y p into the given differential equation and simplifying, we
obtain
{
— 3A )x + (A — 3A
)
x
x
= 2x — 1.
Equating coefficients of like powers of x yields
— 3A
X
— 3/l
/!,
Then
y p — (— fx -I- %)e
y = yh + yP = c^
9.57
Solve
1
y'
5x
2x
or
=—
or
1
At = —\
A = ^
and the general solution to the nonhomogeneous differential equation is
,
+ (-£* + hV
- 5y = 3x 2 e 2x
=2
x
-
.
Since the right side of this equation is a second-degree polynomial times an exponential, we assume a
particular solution in the form of a general second-degree polynomial times an exponential:
>'
p
= {A 2 x 2 + A x + A )e 2x
From Problem 8.34 we have, as the general solution of the associated
.
x
5x
Because y p and y h have no terms in common except perhaps
yh — c e
for a multiplicative constant, there is no need to modify y p
homogeneous differential equation,
.
x
.
Substituting y p and
after simplification,
= [2A 2 x 2 + (2A 2 + 2A )x + Ay + 2A Q ~\e 2x into the given differential equation yields,
p
— 3A 2 )x 2 + (2A 2 — 3A )x + (A — 3A = 3x 2 + Ox + 0. Equating coefficients of like
y'
r
(
t
)
x
powers of x, we obtain
3A 2
2A 2
--
3/1,
A, — 3/4
Then
y p = (— fx
y = y„ + y P = c e
2
Solve
y'
or
A2 =
or
A =
"3
or
A =
'9
1
2
x
2
— fx — \)e 2x and the general solution to the nonhomogeneous differential equation is
+ (-|x 2 - fx - \)e 2x
,
ix
.
{
9.58
=3
=
=
- 5y = (-9x 2 + 6x)e 2x
.
I The right side of this equation is again a second-degree polynomial times an exponential, and y p y'p and
,
,
y h are all as in the previous problem.
Substituting the first two into the given differential equation yields
\2A 2 x 2 + (2/1 2 + 2A )x + A x + 2A ]e 2x - 5(A 2 x 2 + /l,x + A )e
2x
x
= (-9x 2 + 6x)e 2 *
After simplifying and equating the coefficients of like powers of x, we have
-3A 2
9.59
Then
y p = 3x e
Solve
>•'
I
2x
2
,
2A 2 -3A
X
A
l
= -9
or
=6
A2 = 3
or
A
or
Ao =
-3A =
t
=0
and the general solution to the given differential equation is
- 5y = (2x 3 - 5)e 2x
y = yh + yp = c x e
5x
+ 3x 2 e 2x
.
Since the right side of this equation is the product of a third-degree polynomial and an exponential, we
assume a particular solution in the form of a general third-degree polynomial times an exponential:
5x
2x
3
2
The complementary solution is found in Problem 8.34 to be y h = c e
y p = (/l 3 x + A 2 x + A x x + A )e
Since y p and y h have no terms in common, there is no need to modify y p
2x
2
+ 2(/l 3 x 3 -I- A 2 x 2 + A x + A )e 2x into the given
Substituting y and y'p = (3A 3 x + 2A 2 x + A )e
.
x
.
p
x
x
differential equation yields, after simplification,
(-3/4 3 )x 3 + (3^3 - 3A 2 )x 2 + (2A 2 - 3A x )x + (A X - 3A 2 ) = 2x
3
+ Ox 2 + Ox - 5
.
.
THE METHOD OF UNDETERMINED COEFFICIENTS
203
Equating coefficients of like powers of x and solving the resulting system of equations, we obtain
A = - 9> an d A Q = ty. The general solution is then
5x
=
=
c,e
+ (-fx 3 - fx 2 -%x + ^)e 2x
y
yh + yP
^3 - ^2 = ~h
\
.
9.60
Solve
y'
+ 6y= 18xe" 3 *.
3x
y p - {A x + A Q )e~
I We assume a particular solution of the form
which is the product of a first-degree
polynomial and an exponential. The complementary function is shown in Problem 8.37 to be y c = Ae~ 6x
Since
y
,
.
y p and yc have no terms in common, there is no need to modify y p
Substituting y p into the given differential equation yields, after simplification,
.
(3
Equating coefficients of like powers of x then yields A —d and A = — 2, so
6x
general solution to the nonhomogeneous equation is y = Ae~
+ (6x — 2)e~ 3x
x
A )x + (A + 3A — 18x + 0.
y p — dx — 2 and the
t
l
)
.
9.61
Solve
y'
+ 6y = 9xV - 12xV.
f The right side of this equation is (9x 3 — \2x 2 )e x which is the product of a third-degree polynomial and
x
x
2
3
Also, y of the previous problem is valid here; since it
e so we assume
y p = (A 3 x + A 2 x + A x + A )e
has no term in common with y p there is no need to modify y p
,
,
.
t
c
.
,
Substituting y p into the given differential equation yields
(3A 3 x 2 + 2A 2 x + A^)e
x
+ (A 3 x 3 + A 2 x 2 + A x + A )e x + 6(A 3 x 3 + A 2 x 2 + A x + A )e x - (9x 3 - 12xV
t
x
After this equation is simplified and the coefficients of like powers of x are equated, we have
=
9
= -12
7/4 3
3A 3 + 1A 2
2A 2 + 1A
=0
X
A + 1A =
y = y h + y p = Ae~
9.62
Solve
y" - ly' = (3 - 36x)e* x
I We try
y p = {A l x + A
6x
A3 =
or
A 2 = -2.27
or
A - 0.65
A = -0.09
1.29
x
or
x
The general solution is then
or
+ (1.29x 3 - 2.27x 2 + 0.65x - 0.09 )e x
.
.
4x
a first-degree polynomial times an exponential, as a particular solution. The
lx
Since y p and yc have no term in
complementary solution is shown in Problem 8.2 to be y c = c + c 2 e
common, there is no need to modify y p
4x
and y'p = {\6A x + %A + 16/l )e 4x into the given differential
Substituting y'p = (AA x + A + 4A )e
— 12^ )x + {A - \2A ) = -36x + 3. Equating coefficients of like
equation and simplifying, we obtain
powers of x yields a system of two equations from which we find A = 3 and A = 0. The general solution
'
)e
,
.
x
.
l
,
x
(
x
t
X
x
x
is
9.63
then
Solve
y = yc + y p = c x + c 2 e
lx
+ 3xe* x
y" - ly' = ( - 80x 2 - 108x + 3S)e
.
2x
.
i We try as a particular solution y p = (A 2 x 2 + A^x + A )e 2x while y of the previous problem is valid here
as well. Since y has no term in common with y p there is no need to modify y p
2x
2
and
Substituting y'p = {2A 2 x + 2A 2 x + 2A x + A + 2A Q )e
2x
y" = (4A 2 x 2 + 8/4 2 x + 4A x + 2A 2 + 4A + 4A )e
into the given differential equation, simplifying the result,
,
c
.
,
c
x
x
X
x
and then equating coefficients of like powers of x, we get
=
-80
or
A2 =
=-108
-6A 2 -lOA
38
2A 2 - 3A - \0A o =
or
At = 6
A - -4
-\0A 2
l
or
X
The general solution to the given nonhomogeneous equation is then
9.64
Solve
y" - y' + 2y = (6x
2
+ 8x + l)ex
I We try as a particular solution
Problem 8.1 to be
to modify y p
.
y c = c x e~
x
+ c2e
.
{
+ c 2 e lx + (8x 2 + 6x - 4)e 2x
.
.
y p = (A 2 x
2x
y = c
8
2
+ A x + A )e x
The complementary solution is found in
have no term in common, there is no need
solutions
two
Since these
x
.
204
D
CHAPTER 9
= (A 2 x 2 + 2A 2 x + A x + A + A )e*, and
2
y'p — (A 2 x
+ 4A 2 x + A x + 2A 2 + 2A + A )e x into the given differential equation, simplifying, and then equating
Substituting y p
y'
p
,
x
x
X
l
coefficients of like powers of x, we get
=6
=8
or
A2 = 3
A = \
A +2A = 1
or
Ao =
2A 2
2A 2 + 2A
2Ai +
y — c x ee~
Thus, the general solution is
9.65
y" - y' + 2y = (x 2 - x + 4)e x
Solve
X
or
X
x
+ c 2 e 2x + (3x 2 + x)e x
x
.
I The expressions for y p y'p y'p and y of the previous problem are all valid here.
,
,
,
Substituting the first three
c
into the given differential equation and simplifying, we get
(2A 2 )x
2
+ (2A 2 + 2A )x + {2A 2 + A + 2/l
x
x
= x 2 — x + 4.
)
y = yf + y p — c e~
solution is then
9.66
Solve
By equating coefficients of like powers of x and
A Y = — 1, and A = 2. The general
A 2 =\,
solving the resulting system of equations, we find that
+ c 2 e 2x + (\x 2 — x + 2)ex
x
x
y" - y' + 2y = (x 2 - x + 4)e 4x
.
.
I The complementary solution is that of Problem 9.64: yc — c e~ x + c 2 e 2x and we try a particular solution
Ax
2
Since y and y p have no terms in common, we need not modify y p
of the form
y p — (A 2 x + A^x + A Q )e
2
y'
=
{4A
4/l,x
Substituting y p
x
+
2A
x
+
+ A + 4A )e 4x and
2
2
p
4x
y' = (16/l x 2 + \6A x + 16/4,x + 2A
into the nonhomogeneous differential equation and
2
2 + 8/1, + \6A )e
2
p
,
x
.
.
c
,
,
x
simplifying yield
(\4A 2 )x 2 + (\4A 2 + 14/4 ,)x + (2A 2 + 1A
X
+ 14/4
- x2 - x + 4
)
A 2 = 0.071,
Equating the coefficients of like powers of x, we obtain a system of three equations that yields
A = —0.143, and A — 0.347. The general solution is then
x
2x
+ (0.071x 2 - 0.143x + 0.347)e 4x
y = yc + y p = c e~ + c 2 e
x
.
x
9.67
Solve
-^ + 4 — + 8x = (20f + 16f - 78)e
d x
dx
dt
dt
2
2 '.
I The complementary solution is shown in Problem 8.54 to be x = c,e _2 'cos2t + c 2 e~ 2 'sin2r. We try a
particular solution of the form
x p = (A 2 t 2 + A
+ A )e 2
and since this function has no term in common
c
',
t
x
with xc there is no need to modify it.
Substituting x p
Xp = (4A 2 t
2
+ SA 2
= (2A 2 2 + 2A 2 + 2A + A + 2A )e 2
and
+ 4A + 2A 2 + 4A + 4A )e 2 into the given differential equation, we get, after simplifying,
x'
,
t
p
t
x
t
x
',
t
x
'
t
X
(20A 2 )t 2 + (\6A 2 + 20A
)t
x
+ (2A 2 + SA + 20A
)
X
= 20f 2 + 16f - 78
Then equating the coefficients of like powers of t yields
=20
=16
or
+ 8A +20A = -78
or
20/4 2
16A 2 + 20A
2/4 2
d 2x
Solve
X
x
x = c e~ 2 cos2t + c 2 e~ 2 'sin2r + (t 2 —4)e 2
'
Thus the general solution is
9.68
X
A2 = 1
A =
A = -4
or
'.
x
+ 8x = (5r - \4t + U)e
-^
2++ 4 —
~dl
~dl
dx
2
I The complementary solution x c of the previous problem is valid here, but now we assume a particular solution
3
2
Substituting x p
x' = (-3A 2 t + 2A 2 - 3A t + A - 3A )e~
of the form
x p = (A 2 t 2 + A t + A )e~ 3
p
3
2
into the given differential equation yields, after
and x'p = {9A 2 t - \2A 2 t + 9A t + 2A 2 - 6A + 9A )e~
'.
',
t
,
x
x
x
'
x
X
simplification,
(5A 2 )t
2
+ (-4A 2 + 5A )t + (2A 2 - 2A + 5A
x
X
)
= 5t 2 - Ut + 11
By equating coefficients of like powers of t and solving the resulting system of equations, we find that A 2 = 1,
2
3
2
2
and A = 1. The general solution is then x = c e" 'cos2f + c 2 e' sin 2t + (t - 2r + l)e~
A t = -2,
'
1
'.
A
A
THE METHOD OF UNDETERMINED COEFFICIENTS
d 2x
9.69
dx
—
+ 8x =
yj- + 4
Solve
(29f
+ 30r 2 - 52f - 20)e 3
3
205
'.
# The complementary solution of the preceding two problems is valid here, and we try a particular solution of
x p = (A 3 3 + A 2 2 + A + A )e 3
This trial solution needs no modification.
the form
3
2
2
Substituting x p
x' = {3A 3 t + 3A 3
3A
+
+ 2A 2 + 3A + A + 3A )e 3 \ and
2
p
3
2
2
x p = (9A 3 t + \SA 3 + 9A 2 t + 6A 3 + l2A 2 + 9A + 2A 2 + 6A + 9A Q )e il into the given differential
t
t
'.
t
x
t
,
t
t
t
t
t
x
x
t
x
t
l
equation yields, after simplification,
(29 A 3 )t
3
2
+ (30A 3 + 29
+ (6A 3 + 20A 2 + 29 A )t + (2A 2 + 1(M, + 29
2 )t
= 29t 3 + 30f 2 - 52* - 20
)
x
Then equating the coefficients of like powers of t and solving the resulting system of equations, we find that
A 3 = 1, A 2 = 0, A = — 2, and A — 0. The general solution is thus
l
x = x c + x p = c e~ 2t cos2t + c 2 e~ 2, sin2t + (t 3 - 2t)e 3t
l
9.70
Solve
x + 9x + 14x = (I2t
2
.
+ 22t + 21)e~'.
I The complementary solution is shown in Problem 8.17 to be xc — c e~ 2 + c 2 e~ lx We try a particular
solution of the form
x p = (A 2 t 2 + A t + A )e~\ which needs no modification.
Substituting x p
x p = — A 2 2 + 2A 2 — Aj + A — A )e ~', and
2
— 4A 2 + A it + 2A 2 — 2Ai + A )e~' into the given differential equation and simplifying the result,
x" = (A 2
'
.
l
Y
t
(
,
t
t
x
t
we get
(6A 2 )t 2 + (14A 2 + 6Ai)t + (2A 2 + 1A X + 6A ) = \2t 2 + 22t + 27
Then equating the coefficients of like powers of t, we obtain
= 12
=22
UA 2 + 6Ai
2A 2 + lAi + 6A Q = 27
6A 2
c x e~
Thus, the general solution is
9.71
Solve
2x
A2 = 2
A x = -1
A = 5
or
or
or
+ c 2 e~ 11 + {2t 2 - + 5)e _l
\
t
x + 9x + 14x = (144r 3 + 156t 2 + 24t)e
2t
.
I The complementary solution x of the previous problem is valid here as well. We try a particular solution of
2
which needs no modification.
the form
x p = (A 3 3 + A 2 2 + A + A )e
2
2
3
2A
3A
t
+
+ 2A 2 + 2A^ + A + 2A )e 2 and
Substituting x p
x' = (2A 3
+
2
3
p
3
+ \2A 3 t 2 + 4A 2 2 + 6A 3 + SA 2 + 4A + 2A 2 + 4A + 4A )e 2t into the given differential
x'p = (4A 3 t
c
t
t
x
',
t
t
t
,
t
t
',
t
t
x
t
t
X
equation and simplifying yield
(36^ 3 )f
3
+ (39A 3 + 36A 2 )t 2 + {6A 3 + 26A 2 + 36,4^ + (2A 2 + l3A + 36A
)
t
= I44t 3 + 156f 2 + 24f
Equating the coefficients of like powers of t, we then have
or
A3 = 4
39A 3 + 36A 2
= 144
=156
or
A2 =
6A 3 + 26A 2 + 36Ai
=24
or
A
=0
or
A =
36A 3
2A 2 + l3Ai + 36A
The general solution is thus
9.11
Solve
x = x f + x p = c^
x + lOx + 25x = (2t - 10)e
-2
'
+ c 2 e~ nt 4- 4t 3 e 2t
X
=Q
.
3 '.
-5
+ C 2 te' 5 We assume a
f The complementary solution is found in Problem 8.146 to be x = C^
+ A )e 3 and since it has no terms in common with x it requires
x p = {A
particular solution of the form
'
'.
c
x
',
t
f,
no modification.
Substituting x p
,
x p = (3/4 x r + A
x
+ 3A )e 3
equation, we have, after simplification,
',
x p = (9/1, t + 6A
and
(64Ai)t + (16/lj + 64v4
)
X
+ 9A )e 3
= 2x - 10.
Ax = —
we then form the general solution
x = (C\ •+ C 2 t)e~
5'
into the given differential
Equating the coefficients of like
-21
1
powers of t, we get a pair of equations whose solution is
'
and
52
+ (At - 21)e 3 '/128.
A 2 = -rrr-.
1
Zo
Combining x c and x p
,
A
206
9.73
CHAPTER 9
D
Solve
x + lOx + 25.x = (320r
+ 48f 2 - 66r + 122)e 3
3
'
f The complementary solution x c of the previous problem is valid here, but we now assume a particular
x p = (A 3 t 3 + A 2 t 2 + A t + A )e 3
solution of the form
Substituting x p and its first two derivatives (see
'.
x
Problem 9.69) into the given differential equation and simplifying, we get
(64/l 3 )f
3
+ (48^3 + (AA 2 )t 2 + (6A 3 + 32A 2 + 64/4,)f + {2A 2 + 16/1, + 64A
)
= 320f 3 + 48r 2 - 66t + 122
Then equating coefficients of like powers of t, we obtain
=
=
320
or
A3 =
48
or
= -66
=
2A 2 + 16/1, + 64A
122
or
A 2 = -3
*i =
A = 2
64A 3
48/i 3 + 64A 2
6A 3 + 32A 2 + 64/1,
The general solution is then
9.74
Solve
- 6y" +
y'"
1 1 y'
x = (C, + C 2 t)e~
- 6y = 2xe~ x
or
+ (5f - 3f + 2)e
3
5'
2
5
3t
.
.
I The complementary solution is found in Problem 8.28 to be yc = c e x + c 2 e 2x + c 3 e 3x We try a particular
x
solution of the form
which has no term in common with y and therefore requires no
y p — (A x + A )e~
.
x
,
l
modification.
c
Substituting
yp =
-A xe' x + A e' x - A e' x
l
x
= A xe' — 2A e~ x + A e~ x
x
y'p = — A xe~
+ 3A e' x — A e~ x
y'
x
'
p
and
x
x
x
x
into the given differential equation and simplifying, we obtain
(
— 24/1 Jx 4- (26/1, — 24
= 2x + 0. Equating
A = — 2 and A = —j4*.
= c e* + c 2 e 2x + c 3 e 3x — j2xe~ x — &e~ x
)
coefficients of like powers of x and solving the resulting system yield
Then
yp -
d
9.75
Solve
3
-j^xe'" — fee'*,
Q
-,-£ - 5
dt
3
d2Q
and the general solution is
dQ
_v
—dff + 25 -f - \25Q = -522r + 465f - 387)
2
(
]
x
,
.
x
-'.
t
dt
+ c 2 cos 5f + c 3 sin 5f. We try a
which requires no modification because Q p has no
+ A Q )e
term (except perhaps for a multiplicative constant) in common with Q
Substituting Q p
I
The complementary solution is found in Problem 8.1 13 to be
2
Q p — {A 2 + A
particular solution of the form
t
2
t
t
Q = c,e
5'
c
'.
c.
,
= [2A 2 t 2 + 2A 2 + 2/1, + .4, + 2A )e 2
Q'
'
r
t
p
2t
2
Q p = (4A 2 + SA 2 + 4/1, + 2A 2 + 4/1, + 4A Q )e
Q' = (8/l 2 + 24A
+ 8/4,f + \2A 2 + 12/1, + %A )e 2
2
2
p
t
and
t
t
'
'
f
i
into the given differential equation and simplifying, we get
(-S7A 2 )t 2 + (34A 2 - 87/1, )f + (2A 2 + 17/1, - 87/l
)
= -522r 2 + 465f - 387
Equating the coefficients of like terms, we obtain
-87/1,
34/12-87/1,
2/4 2
The general solution is then
9.76
Solve
(D
4
+ 17/1, - 87/l
= -522
- 465
= -387
or
A2 =
or
A = -3
or
A =
x
2
2
Q = c,? 5 + c 2 cos 5f + c 3 sin 5t + (6r — 3f + 4)e
+ 4D 2 )y = (3.x - \)e* x
6
'
4
'.
.
I The complementary solution is found in Problem 8.186 to be y c = c, 4- c 2 x + c 3 cos 2x + c 4 sin 2x. We
x
which needs no modification because y p has no
assume a particular solution of the form y p = (.4,x + A )e*
.
terms in common with \\.
= (256A x + 256/1, + 256A )e Ax into the given
= 3x - 1. Equating coefficients of like
320/1 ,)x + (288/1, + 320/1
differential equation and simplifying yield
terms and solving the resulting system, we obtain /l, = 0.00938 and A = -0.01156. The general solution
4
is then
y = y + yp = c x + c 2x + c 3 cos 2x + c4 sin 2.x + (0.00938.x - 0.01 156)e *.
y'
p
Substituting
= (16/1,-x + 8.4, + 16.4 )e 4x
(
c
and
y' "
p
l
)
A
THE METHOD OF UNDETERMINED COEFFICIENTS
D
207
EQUATIONS WHOSE RIGHT SIDE CONTAINS SINES AND COSINES
Solve / — 5y = sin x.
9.77
f The complementary solution is found in Problem 8.34 to be y = c,e 5 *. We assume a particular solution
of the form
y p = A sin x + B cos x, which needs no modification because y p and y have no terms in
common except perhaps for a multiplicative constant.
c
c
Substituting y p and its derivative into the given differential equation yields
cos x — B sin x — 5(A
A
sin x
(
+ B cos x) = sin x,
— 5A — B
sin x
)
which we rearrange as
+ (A — 5B
)
cos x = 1 sin x +
cos x
Equating the coefficients of like terms, we obtain the system
-B =1
-5j4
A = — j§ and B = — jg.
5x
=
=
^^
y
y + yP
js sin x - ^ cos x.
from which we find
A -5B =
Then the general solution is
c
9.78
Solve
/ + 6y = -2 cos 3x.
I The complementary solution is shown in Problem 8.37 to be y = Ae~ 6x We assume a particular solution
of the form
y p = A sin 3x + B cos 3x, which need not be modified.
Substituting y p and its derivative y'p — 3A cos 3x — 3B sin 3x into the differential equation and rearranging,
.
f
we obtain
— 36
(6A
)
sin 3X
+ (3A + 6B
)
cos 3x =
sin 3x
+ — 2) cos 3x
(
A = — y$
Equating coefficients of like terms and solving the resulting system, we find
9.79
Solve
2y'
and
B = —-fe.
The
6x —
13 sin 3x — 13 cos 3x.
y — Ae~
general solution is then
— 5y = sin 2x — 7 cos 2x.
f The complementary solution is shown in Problem 8.38 to be yc — Ae 5xl2 We assume a particular solution
of the form
y p = A sin 2x + B cos 2x, which needs no modification. Substituting y p and its derivative into
.
the differential equation and rearranging yield
(
— 5A — 4B
)
sin 2x
+ {4A — 5B Q cos 2x =
)
1
sin 2x
9.80
Solve
y = y c 4- y p — Ae
5xl2
(
A — — ff
Equating coefficients of like terms and solving the resulting system, we find
general solution is then
+ — 7) cos 2x
and
B Q = ||.
The
— ffsin 2x + ||cos 2x.
y" - ly' = 48 sin 4x + 84 cos 4x.
f The complementary solution is found in Problem 8.2 to be yc = c l + c 2 e lx We assume a particular
solution of the form
y p = A sin 4x + B cos 4x, which needs no modification.
y'
=
AA
cos 4x — 4B sin 4x and y"p = — 16/1 sin4x — 16B cos4x into the given differential
Substituting
p
.
equation and rearranging give
(-16/4
+ 28B )sin4x + (-28/l - 16B )cos4x = 48sin4x + 84cos4x
Equating coefficients of like terms and solving the resulting system, we get
y = c x + c2e
solution is then
9.81
Solve
7x
A = —3
B = 0.
and
The general
— 3 sin 4x.
y" - 6y' + 25y = 48 sin 4x + 84 cos 4x.
3x
yc = c 1 e cos4x + c 2 e sin4x, while y p and
its derivatives are as in the previous problem. There is no need to modify y p because it is not a solution of the
f The complementary solution is shown in Problem 8.50 to be
associated homogeneous differential equation for any choice of the arbitrary constants A
e
3x
3j:
and B
.
(Observe that
sin 4x is linearly independent of sin 4x.)
Substituting y p
,
y' ,
p
and y'p into the given differential equation and rearranging yield
{9A
+ 24B
)
sin 4x
+ - 24
(
+ 9B
)
cos 4x = 48 sin 4x + 84 cos 4x
Equating coefficients of like terms and solving the resulting system, we obtain A = -2.411
3x
cos 4x + c 2 e sin 4x - 2.41 1 sin 4x + 2.904 cos 4x.
y =
The general solution is then
c^
and
B = 2.904.
.
208
9.82
CHAPTER 9
y" + 4y
Solve
+ 5y = 2 cos .x — 2 sin x.
I The complementary solution is shown in Problem 8.66 to be ye = c e~ 2x cosx + c 2 e~ 2 *sinx. We assume a
particular solution of the form
y p = A sin .x + B cos x, which needs no modification because y p cannot be
made equal to y by any nonzero choice of the arbitrary constants.
l
c
Substituting y p and its derivatives into the differential equation and rearranging, we obtain
- 4B
{4A
)
sin x
+ (4A + 4B
)
— 2 sin x + 2 cos x
cos x =
A =
Equating coefficients of like terms and solving the resulting system yield
9.83
_2
y = Cie
solution is then
*cosx + c 2 e
_2
and
B = §.
The general
*sinx + ^cosx.
x + 4x + 8x = 16 cos 4t.
Solve
I The complementary solution is shown in Problem 8.54 to be x = c,e~ 2 cos2f + c 2 e~ 2 'sm 2f.
particular solution of the form
x p — A sin 4f + B cos 4f, which needs no modification.
'
c
We try a
Substituting x p into the differential equation and rearranging, we get
(-8^ - 16S )sin4r + {\6A - 8B )cos4r = 0sin4f + 16cos4f
Equating coefficients of like terms yields the system
Solving, we find that
x = c e
2'
x
9.84
Solve
- 16B =
-8/l
,
cos 2\ + c 2 e
A —f
~ 2
'
B — — §.
and
16/l
-8Bo=16
The general solution is then
+ f sin 4f — \ cos 4f
sin 2(
x + ^x + 96.x = 96 cos 4?.
I The complementary solution is found in Problem 8.63 to be xc = e 0015 « 5 '(C, cos9.7979r + C 2 sin9.7979t).
We assume a particular solution of the form xp = A s'm4t + B cos4t which, because it has no term in
common with xc needs no modification.
,
Substituting x
p
and its first two derivatives into the differential equation and rearranging, we obtain
- |5
(80/l o
)
sin 4t
+ (|/4 + 80S
)
cos 4f = 96 cos 4f +
Equating coefficients of like terms and solving the resulting system, we find that
The general solution is then
sin 4f
A = 0.0019
and
B = 1.2000.
x = e -o.oiS6as^ Cj cos 9.7979? + C 2 sin 9.79790 + 0.0019sin4f + 1.2000 cos 4f.
2
9.85
Solve
d x
—
+ 25x = 2 sin
-r-
dt
2
2f.
I The complementary solution is shown in Problem 8.72 to be x c = C, cos 5f + C 2 sin 5f.
particular solution of the form
x p = A sin2t + B cos2t. which requires no modification.
x p and its second derivative into the differential equation and rearranging, we get
and
(2\A )s\n2t + (2\B )cos2t = 2sm2t + 0cos2t. It follows that /4
=^
solution is then
9.86
Solve
B = 0.
We assume a
Substituting
The general
x = xt + x p — C, cos 5f + C 2 sin 5f + ^ sin 2t.
x + 128x = 512(sin ?t + cos It).
I The complementary solution is
x c = C, sin >/128l + C 2 cos \\2St (see Problem 8.73). We assume a
which requires no modification. Substituting x p
sin 2t + B Q cos 2f,
xp = A
particular solution of the form
and its second derivative into the differential equation and rearranging, we get
+ (124B )cos2f = 512 sin 2r + 512cos2t. It follows that
x = C, sin >/l28t + C 2 cos v^128f + 4.129(sin 2t + cos It).
(124/l )sin2<
solution is
9.87
Solve
A = B = &£ = 4.129.
The general
Q + SQ + 520 = 32 cos It.
I The complementary solution is Q = c e 4 'cos6f + c 2 e~*'s'm6t (see Problem 8.55). We assume a particular
solution of the form
Q p = A sin 2f + B cos 2t, which requires no modification. Substituting Q p and its first
c
x
two derivatives into the differential equation and rearranging, we obtain
(48/l
- 16B )sin2r + (16 A + 48B )cos 2f = Osin 2t + 32cos2r
By equating the coefficients of like terms and solving the resulting system, we find
general solution is then
Q = c e" 4 'cos6f + e 2 e~ 4 sir\6t + |sin2f + f cos2f.
1
'
A =j
and
B = |.
The
-
A
.
THE METHOD OF UNDETERMINED COEFFICIENTS
9.88
Solve
209
Q + SQ + 25Q = 50 sin 3t.
f The complementary solution is shown in Problem 8.53 (with Q replacing x) to be
A
We assume a particular solution of the form Q p = A sin 3t + B cos 3f.
y
This trial solution requires no modification because no part of it can be obtained from Q c by any choice of the
Q = e' \c cos 3f + c 2 sin 3r).
c
arbitrary constants c t
and c 2
(Note the additional e" 4 term in Q c .)
'
.
Substituting Q p and its derivatives into the differential equation and rearranging, we get
(16/4„
- 24B
)
sin 3f
+ (24
+ 16B
)
cos 3i = 50 sin 3r +
Equating the coefficients of like terms and solving the resulting system, we obtain
cos 3f
A = f-°
and
fl
= — |f.
Q = e " 4( (c cos 3t + c 2 sin 3f + f§ sin 3t — |f cos 3r.
The general solution is
)
,
w
9.89
Solve
g + 9q + I4q = \ sin t.
I The complementary solution is found in Problem 8.17 (with q replacing x) to be q = c e~ 21 + c 2 e~ 7
assume a particular solution of the form q p — A sin t + B cos
which requires no modification.
c
'.
x
We
t,
Substituting q p and its derivatives into the differential equation and rearranging, we get
(13/4
— 96
)
+ (9A + 13fi
sin t
)
cos t = \ sin t +
Equating coefficients of like terms yields the system
cos t.
13,4o-9flo = 2
9.90
Thus
A = j&o
Solve
I
B = — joo-
and
+13B =
9,4
q = c^e~
The general solution is then
2t
+ c 1 e~ lt + y^sin — Jjq cos
t
t.
+ 100/ + 50,000/ = -400,000 sin lOOf.
I The complementary solution is shown in Problem 8.56 to be I c = e -50 '(ci cos 50\/l9r + c 2 sin 50-J\9t).
We assume a particular solution of the form I p — A sin 100r + B cos lOOf, which needs no modification
because it has no terms in common with I c
(40,000,4o
.
Substituting l p into the differential equation yields, after rearranging,
- 10,000B o )sin lOOf + (1 0,000/1,, + 40,000B )cos lOOr = -400,000 sin lOOf + Ocos lOOr
By equating coefficients of like terms, and solving the resulting system, we find
40,000/1
Thus
A = — ^f-
9.91
Solve
#
B = j^,
and
/
- 10,000B o = -400,000
+ 40,000B o =
10,000/4
and the general solution is
= e~ 50, (ci cos 50yi9f + c 2 sin 50>/i90 -
W
sin lOOf
+ f^cos lOOf
q + lOOq + 50,000q = 4000 cos lOOt.
This problem is very similar to the previous one. The associated homogeneous differential equations are
identical, except that here q replaces /.
Thus, q c is identical in form to I c
.
We assume the same form for the
particular solution (with q replacing /), so most of the analysis of the previous problem remains valid.
Only the
right sides of the differential equations are different.
Substituting the particular solution into this differential equation, rearranging, and equating coefficients of like
terms yield the system
- 10,000B o -
40,000/4
Now A = yfo
and
B = yj^,
q = e
9.92
Solve
y" — y
~ 50,
+ 40,0005 = 4000
10,000,4
and the general solution is
(ci cos 50 y/l9t
+ c 2 sin 50 Vl9f) +
&
sin lOOr
+ $fc cos lOOf
— 2y = sin 2x.
x
2x
We assume a particular
I The complementary solution is shown in Problem 8.1 to be y = c e~ + c 2 e
solution of the form
y p = A sin 2x + B cos 2x, which needs no modification.
and y p = -4A Q sin2x - 4B cos2x into the differential
y' = 2A cos 2x - 2B sin 2x,
Substituting y p
p
c
i
.
,
equation and rearranging yield
(
6A + 2B
)
sin 2x
+ (-6B - 2A
)
cos 2x = 1 sin 2x +
cos 2x
Equating coefficients of like terms and solving the resulting system, we find that A =
x
2x
cos 2x, and the general solution is y = c l e~ + c 2 e
Then y = - sin 2x +
p
9.93
Solve
^
^
q + 400q + 200,000^ = 2000 cos 200f
-^ and B = ^.
& sin 2x + ^ cos 2x.
CHAPTER 9
210
q c = e~
f The complementary solution is
qp — A
particular solution of the form
200
+ B sin 400r) (see Problem 8.70). We assume a
+ B cos 200r, which has no term in common with q and
'(A cos 400r
sin 200t
c
therefore needs no modification.
Substituting q c into the given differential equation and rearranging yield
(160,000^l
- 80,0005
)
sin 200f
+ (80,000/4 + 16Q,0O0B o cos 200f =
)
sin 200r
A — 0.005
+ B sin 400f) + 0.005 sin 200f + 0.01 cos 200t.
Equating coefficients of like terms and solving the resulting system, we get
9.94
200
q = e~
general solution is
'(A cos 400t
+ 2000 cos 200t
and
B = 0.01.
The
q + \000q + 50,000q = 4000 cos lOOf.
Solve
f The complementary solution is q = c e~ 52 19 + c 2 e~ 9A1 2 (see Problem 8.18). We try a particular solution
of the form
q p = A sin lOOf + B cos lOOf, which needs no modification.
'
c
'
y
Substituting q p into the given differential equation and rearranging, we get
(40,000/J
- 100,0O0B o sin lOOr + (100,000,4 + 40,000B )cos lOOr = 4000 cos lOOr
)
Equating coefficients of like terms yields
- 100,000B o =
40,000,4
100,000/4
A = jg and B = yfj.
9A12t
5219t
=
+ c 2 e'
c x e~
+ ^sin lOOt + yfjcos lOOf.
q
Solving this system, we find
2
9.95
dl
d
—
T + 40 — + 800/ = 8 cos
Solve
dt
+ 40,000B o = 4000
The general solution is
I
2
t.
dt
f
The complementary solution is found in Problem 8.71 to be
assume a particular solution of the form I p = A sin + B cos t,
/,.
t
= c e~ 20t cos20f + c 2 e" 20 'sin 20t. We
which has no terms in common with I and
x
c
so needs no modification.
Substituting I p into the given differential equation and rearranging, we have
(799v4
— 40B
)
I
20
= c,e
'
1
— 799B
sin t 4- (40/l
cos 20t + c -,e
cos t = 8 cos t.
)
A = 320/640,001
resulting system, we find
20
'
sin 20r
+
—— —
and
-
—
+ — —- cos
The general solution is then
6392
320
640.001
Equating coefficients of like terms and solving the
B = 6392/640.001.
sin t
t.
640.001
d 2r
9.96
Solve
—^ — (.or — — g sin cot
when both g and (o are positive constants.
/ The complementary solution is r — Ae at + Be'™' (see Problem 8.7). We assume a particular solution
of the form
r — A sin cot + B cos cot,
which requires no modification.
p
c
Substituting r p into the differential equation and rearranging lead to
(
— 2co 2 /4
)
sin a»f
A = g/{2co 2
)
+ — 2cD 2 B )cosa)t = —g sin cot + Ocosojf.
By equating coefficients of like terms, we find that
(
and
B = 0.
The general solution is then
r
= Ae m + Be~°" + -— sin cor.
'
r-
2a>
d
9.97
Solve
i
—
dt
—
dQ
Q - d2Q
- 1250 = 504 cos 2f - 651 sin It.
f
5
+ 25
f
2
3
—
dt
dt
I The complementary solution is Q = c e St + c 2 cos 5t + c 3 sin 3f (see Problem 8.1 13). We try a particular
solution of the form
Q p — A sin It + B cos It, which needs no modification.
c
x
Substituting Q p into the given differential equation and rearranging, we obtain
(-l05A o -42B o )sm2t + (42A - 105B )cos2f - -651 sin 2r + 504 cos 2r
Equating coefficients of like terms and solving the resulting system, we find
general solution is
9.98
Solve
y
(4)
A = 1
and
B — —2,
so the
Q = c 1 e 5 + c 2 cos 5r + c 3 sin 5t + 7 sin 2r — 2 cos 2t.
'
- 6y (3) + 16/' + 54/ - 225y = 1152 cos 3x - 3924 sin 3x
f The complementary solution is. from Problem 8.124, yc = c^ 3 * + c 2 e~ 3x + c 3 e 3:t cos4.x + c 4 e 3x sin4.v. We
assume a particular solution of the form yp = A sin 3x + B cos 3x, which has no terms in common with y
c
and needs no modification.
THE METHOD OF UNDETERMINED COEFFICIENTS
211
Substituting y p into the given differential equation and rearranging yield
(
- 288,4 - 324B
) sin
- 288fl
3x + (324A
)
cos 3x =
- 3924 sin 3x +
1
1
52 cos 3x
By equating coefficients of like terms, we obtain the system
- 288^ - 324B = - 3924
324/4
- 288B =1152
A = 8 and B — 5. Then the general solution is
3x
=
c e
+ c 2 e~ 3x + c 3 e cos4x + c 4 e ix s'm 3x + 8 sin 3x + 5 cos 3x.
y
and find that
3jc
x
EQUATIONS WHOSE RIGHT SIDE CONTAINS A PRODUCT
INVOLVING SINES AND COSINES
9.99
/ + 6y = 3e 2x sin 3x.
Solve
6x
(see Problem 8.37). Since the right side of the nonhomogeneous
yc = Ae'
differential equation is the product of an exponential and a sine, we try a particular solution of the form
I The complementary solution is
yp = A e
2x
sin 3x
+ B Q e 2x cos 3x.
Since y p has no term in common with ye (except perhaps for a multiplicative
constant), there is no need to modify y
Substituting y p and
y'
p
p
.
= 2A e 2x sin 3x + 3A e 2x cos 3x + 2B e 2x cos 3x - 3B Q e 2x sin 3x
into the differential
equation yields
2A e 2x sin 3x 4- 3A e 2x cos 3x + 2B e 2x cos 3x — 3B e 2x sin 3x + 6(A e 2x sin 3x + B e 2x cos 3x) = 3e 2x sin 3x
which may be rearranged and simplified to
— 3B
(8A
)
sin 3x
+ {3A + 8B
)
cos 3x = 3 sin 3x +
cos 3x.
Equating coefficients of like terms, we obtain the system
- 3B = 3
8A
A = ff and
6x
2x
2x
—
=
Ae~
sin
3x
+
^%e cos 3x.
y
7ie
from which we find that
9.100
B — —-$%.
3A + 8B =
The general solution is then
/ + 6>' = 2e~ x cos 2x.
Solve
I The complementary solution of the previous problem is valid here, and we assume a particular solution of the
x
form y p = A e~ x sin 2x + B e~ cos 2x. Because it has no term in common with y y p needs no modification.
c,
Substituting y p into the given differential equation and rearranging, we obtain
(5y4
— 2B
)
sin 2x
+ (2A + 5B
cos 2x =
)
sin 2x
y = Ae~
9.101
Solve
6x
y'
+ 2 cos 2x.
A = j§
the resulting system of equations, we find
and
Equating coefficients of like terms and solving
B = ^§.
The general solution is then
+ ^e~*sin2x + ^§e~*cos2x.
+ 6y = lle x sin x + 23e* cos x.
f The complementary solution again is that of Problem 9.99. We try a particular solution of the form
x
x
y p = A e sin x + B e cos x, which has no term in common with y c and therefore requires no modification.
Substituting y p into the differential and rearranging lead to
(1A
—B
)
sin x
+ (A + 1B
)
cos x = 1 1 sin x + 23 cos x.
A =2
resulting equations, we find that
x
x
6x
2e sin x + 3e cos x.
y = Ae~
9.102
B = 3.
The general solution is then
+
2/ - 5>> = 41<T x cosx - lle"
Solve
and
Equating coefficients of like terms and solving the
x
sinx.
f The complementary solution is y c - Ae 5x 2 (see Problem 8.38). We try a particular solution of the form
x
x
which needs no modification.
y. = A e~ sin x + B e~ cos x,
'
Substituting y p into the given differential equation and simplifying yield
(
- 1A - 2B
)
sin x
+ (2A - 7B
the resulting system, we obtain
9.103
)
cos x =
A = 3
-
1 1
and
sin x
+ 41 cos x.
B = -5.
Equating coefficients of like terms and solving
The general solution is
+ 3e~ x sin x - 5e~ x cos x.
y = Ae
5xl2
Solve
y" + 6/ + 9y - 16?
-x
cos 2x.
3x
(see Problem 8.142). We try a particular solution of
I The complementary solution is y c = c e~ 3x + c 2 xe'
x
x
modification because it has no term in common
no
which
needs
=
A e~ sin 2x + B e~ cos 2x,
the form
y
t
with yc
.
A
212
CHAPTER 9
= — A Q — 2B )e~ x sin2x + (2A - B )e~ x cos2x, and
— 3A + 4B )e~ sin 2x + — 4A — 3B )e~ x cos 2x into the given differential equation and rearranging
y'p —
yield
— 8B sin 2x + (SA cos 2x = 16 cos 2x. By equating coefficients of like terms, we find A — 2 and
B = 0. The general solution is then y = c e~ 3x + c 2 xe~ 3x + 2e~*sin2x.
Substituting y p
y'
(
p
,
x
(
(
(
)
)
x
9.104
d2y
Solve
*|-
sinx.
5 >'
I The complementary solution is ye = e 3x (Cj cos4x + c 2 sin4x) (see Problem 8.50). We try a particular
3x
3x
solution of the form
y p = A e sin x + B e cos x. Since no part of y p can be generated from y by any
c
choices of the arbitrary constants, there is no need to modify y p
.
= (3A — B )e ix sin x + (A + 3B )e 3x cos x. and
y' — (&A
— 6B )e sin x + (6A + SB )e 3x cos x into the given differential equation and simplifying, we get
p
(\5A sin x + (15B cos x = 30 sin x. Therefore, A = 2 and B = 0, and the general solution is
3x
3x
y — e (c cos 4x 4- c 2 sin 4x) + 2e sin x.
Substituting y p
y'
.
p
3x
)
)
9.105
(D 2 - 2D)y = e x sin x.
Solve
y = C, + C 2 e
f The complementary solution is
2x
As a particular solution we try
.
y = A^sxnx + Be cosx,
x
which requires no modification.
Substituting y,
equation yields
B = 0.
and
9.106
(D 4 -
Solve
Dy = {A — B)ex sin x + (A + B)e x cos x, and D 2 y = - 2Be x sin x + 2Aex cos x into the given
(-2/1) sinx + (-2B)cosx = sinx.
Equating coefficients of like terms, we find that A = — \
1
Hence, the primitive is
y = Cx + C 2 e
2x
— \e* sin x.
1% = 60?* sin 3x.
I The complementary solution is, from Problem 8.128, yc = c cos 2x + c 2 sin 2x + c 3 e + c x e
We try a
x
x
=
which
requires
no
modification.
particular solution of the form
A e sin 3x + B e cos 3x,
Since
yp
x
— (A — 3B )e x sin 3x + (3A + B )e x cos 3x
y' = - SA
- 6B )e x sin 3x + (6A - 8B )^ cos 3x
p
y'p
(
3)
yj,
41
and
\;
= (-26y4 + 18B )f x sin3x + (- 18/4 - 26S Kcos 3x
= (2%A + 96B )e x sin 3x + -96
(
+ 2SB )e x cos 3x
the given differential equation becomes, after substitution and simplification.
+ 96B )sin 3x + — 96.4 + 12B )cos3x = 60 sin 3x. Equating coefficients of like terms and solving
A — -^ and B = y^. The general solution is then
2x
=
=
cos
2x
sin
2x
c3e
+ c x e~ 2x 4- Y*e x sin 3x + fie" cos 3x.
+
c
+
+
y
y
y
d
2
(\2A
(
the resulting system, we find that
c
3
9.107
Solve
d Q
dt
2
_
d Q
,
l
di
dQ
+25^1250
v = 5000e"'cos2f.
dt
I The complementary solution is Q c = c,e 5 + c 2 cos 5f + c 3 sin 5f (see Problem 8.1 13). We try a particular
solution of the form
Q p — A e~' sin 2r + B e~' cos 2f, which requires no modification. Since
'
Q p = (-A -2B )e~'s'm2t + (2A - B )e~'cos2t
Q' = (-3A + 4B )e~'sin2t + (-4A - 3B )e~'cos2t
P
"
Q p =(1M + 2B )e~' sir\2t + (-2A + B)e cos 2f
'
and
1 1
'
the given differential equation becomes, after substitution and simplification,
— 68B
+ (68A — 124B cos 2r = 5000 cos 2f. By equating coefficients of like terms and
A — 17 and B = —31, and the general solution is
Q — c^ 5 + c 2 cos5f + c 3 sin5f + 17e"'sin2r — 31e"'cos2f.
(— 124/4
)
sin 2r
)
solving the resulting system, we find
'
9.108
Solve
y'
+ 6y = (20x + 3) sin 2x.
f The complementary solution is shown in Problem 8.37 to be
y c = Ae~
6x
.
Since the right side of the differential
equation is the product of a first-degree polynomial and sin 2x, we try a particular solution of the form
y p = (/l t x + 4
modification.
)
sin 2x
+ (BjX + B
)
cos 2x.
This trial solution has no terms in common with y c so it needs no
,
THE METHOD OF UNDETERMINED COEFFICIENTS
y' = (/!,- 2B x - 2fl )sin2x + (2/1 ,x
Substituting y p and
+ 2A
p
equation and simplifying, we get
x
-2B )xsin2x + (A + 6A - 2B
(6/1!
x
x
)
sin 2x + (2/1 ,
= 20x sin 2x + 3 sin 2x + Ox cos 2x +
+ fl,)cos2x
213
into the given differential
+ 6B,)xcos2x + (2A + B +6fi )cos2x
x
cos 2x
Equating coefficients of like terms yields the system
= 20
A + 6A
-2B = 3
=
2A
+ 6B
2A + B + 6B =
-2B,
6/1,
x
y
X
x
- 1,
= 0.05, B —
= —
A x = 3, A o —
6x
=
Ae~
(3x
sin
2x
0.15)cos2x.
+
+
0.05)
+(-x
+
y
which we solve to find that
9.109
x
Determine the form of a particular solution to
and
B = 0.15.
The general solution is then
+ 6y = (x 2 — 1) sin 2x.
y'
6x
(see Problem 8.37). Since the right side of the given differential
yc = Ae~
equation is the product of a second-degree polynomial and sin 2x, we assume a particular solution of the more
# The complementary solution is
general form
y p — {A 2 x
2
+ A x + A )sin 2x + (B 2 x 2 + B x + B )cos 2x.
x
This form of y p has no terms in
x
common with the complementary function, and therefore it requires no modification.
9.110
Determine the form of a particular solution to
+ 6y = (20x 2 — lOx + 7) cos 2x.
y
I The complementary solution is that of the previous problem. The right side of the given differential equation is
the product of a second-degree polynomial and cos 2x, so y p of the previous problem is again an appropriate trial
form for the particular solution. Since it has no terms in common with y c
9.111
Determine the form of a particular solution to
,
it
needs no modification.
+ 6y = 18x cos 5x.
y'
6x
(see Problem 8.37). Since the right side of the given differential
yc — Ae~
equation is the product of a first-degree polynomial and cos 5x, we try a particular solution having the more
# The complementary solution is
general form
y p — {A x + A
)
x
sin 5x
+ (B x + B
)
x
cos 5x.
This form of y p has no terms in common with yc and
so requires no modification.
9.112
Determine the form of a particular solution to
2/ — 5y = — 29x 2 cos x.
5xl2
(see Problem 8.38). Since the right side of the differential equation
yc — Ae
is the product of a second-degree polynomial and cos x, we assume a particular solution having the more general
I The complementary solution is
2
2
y p = (A 2 x + A x x + A )sin x + {B 2 x + B x + B )cos x.
there is no need to modify y p
form
Since yc and y p have no terms in common,
x
.
9.113
Solve the differential equation of the previous problem.
f
Substituting yp of that problem and
y p = (2A 2 x + Ai- B 2 x
2
- B x - B )sinx + (A 2 x 2 + A x + A + 2B 2 x + B,)cosx
x
x
into the given differential equation and simplifying, we obtain
(-5A 2 -2B 2 )x 2 sinx + (4/l 2 - 5A - 2B )xs'mx + (2A - 5/l - 2B )sinx + (2/1 2 - 5B 2 )x 2 cosx
+ (2/1, 4- 4B 2 - 5B,)x cos x + (2A + 2B, - 5B cos x = -29x 2 cos x
t
X
x
)
By equating coefficients of like terms and solving the resulting system of six simultaneous equations, we find that
B 2 = 5, fi,=f|, and B = l$. Substituting these results into y p
A 2 =-2, /*! = -f§,
and yc of the previous problem, we generate the general solution
2
5x 2 2
(2x + §§x + J8^) sin x + (5x + ffx + Hf cos X.
y = y c + y p = Ae
40=-^'
'
9.114
)
Determine the form of a particular solution for
y" — ly' — (2x - 1) sin 2x.
lx
(see Problem 8.2). Since the right side of the given differential
yc = c + c 2 e
equation is the product of a first-degree polynomial and sin 2x, we assume
f The complementary solution is
y = (A x x + A
modify y p
.
)
sin 2x
x
+ (B x + B )cos 2x.
x
Since y p and yc have no terms in common, there is no need to
x
CHAPTER 9
214
9.115
Solve the differential equation of the previous problem.
I
Substituting y p
,
/P = {A
and
y'p
-2fi x-2fl )sin2x + (2,4 x + 2A + 5,) cos 2x
1
1
1
= {-4A x - 4A -4B!)sin2x 4- (44, - 4B^x -4B )cos2x
x
into the given differential equation and simplifying yield
{-4A + UBi)xsinx 4 {-!A - 4A - 4B + 14B )sinx + (-14A -4B )xcosx
t
X
t
1
l
+ (4A - 14A - 7B, -4B )cosx = 2x sin 2x + (-l)sin2x
1
Equating coefficients of like terms leads to a system of four equations whose solution is
^o = —0.0571,
A — —0.0377,
1
Bj = 0.1321,
and B = -0.0689. Then the general solution is
lx
v = yc + yp = c i + c 2 e
+ (-0.0377x - 0.0571)sin 2x + (0.1321x - 0.0689) cos 2x.
9.116
y" — 7y' = Ax cos 2x.
Determine the form of a particular solution for
i The form is identical to that of y p in Problem 9.114.
9.117
y" — ly = 3x cos 3x.
Determine the form of a particular solution for
yc = c, 4 c 2 e
f As in Problem 9.1 14, the complementary solution is
lx
Since the right side of the differential
.
equation is the product of a first-degree polynomial and cos 3x, we try
y p = (yi,x 4 /4 )sin 3x 4 (B,x 4 B )cos 3x.
be modified.
9.118
This expression has no terms in common with ye , so it need not
y" — ly' - (3x 2 — 2x + 1 l)cos 27tx.
Determine the form of a particular solution for
f
We try a particular solution of the form
yp = (A 2 x
2
4 A x 4 A )sin2nx 4 {B 2 x 2 4 B,x
-1-
x
B )cos27rx,
which is a generalization of the form of the right side of the given equation. Since y p has no terms in common with
lx
the complementary function
it needs no modification.
yc = c, + c 2 e
.
9.119
y" — 7/ = 6x 2 sin 2nx.
Determine the form of a particular solution for
I The form is identical to that of yp in the previous problem.
9.120
yc = c e
I The complementary solution is
Since y
x
x
y p = (A 3 x* + A 2 x
9.121
y" — y' — 2y — x i sin Ix.
Determine the form of a particular solution for
2
+ c 2 e 2x
(see Problem 8.1).
+ A x + A Q )sin 7x + (B 3 x
3
4-
x
We try a particular solution of the form
2
B 2 x + B,x + B )cos7x
and yc have no terms in common, y need not be modified.
d 3y
d
dx
dx
dy
2
dx
2x
2x
(see Problem 8.156). We try a particular
+ c 2 xe 2x + c i x 2 e
yc — c e
yp = (A x + A ) sin 3x + (BjX + B cos 3x. Since y p and y c have no terms in
I The complementary solution is
solution of the generalized form
2
—^3 + 6 —^y + 12 -f- + 8y = 2x sin 3x.
Determine the form of a particular solution for
{
)
x
common, y p needs no modification.
9.122
Determine the form of a particular solution for
y
(4)
+ y (3) — 2y" — 6y' — 4y = (x 2 — 5x)cos x.
f The general solution to the associated homogeneous differential equation is shown in Problem 8.125 to be
x
2x
+ c 3 e~ x cos x 4- c A e~ x sin x. We try a particular solution of the form
y h = c e~ + c 2 e
2
2
y p = (A 2 x + A x + A ) sin x + (B 2 x + B x + B cos x. Since no part of y p can be obtained from yh by any
x
x
)
t
choice of the constants c 1 through c 4 there is no need to modify yp ; it has the proper form as written.
,
9.123
y" — fy' + 2y = (3x — 4)ex sin 2x.
Determine the form of a particular solution for
I The complementary solution is
yc — Ae
x' 2
+ Be* x
(see
Problem 8.23). Since the right side of the differential
equation is the product of a sine term, an exponential, and a first-degree polynomial, we try a particular solution
of the same but more generalized form, noting that wherever a sine term occurs, the particular solution may
have an identical component with a cosine term.
We try, therefore,
x
x
y p - (A^ + A )e sin 2x + (B t x 4- B )e cos
Since yc and y p have no terms in common, there is no need to modify y p
.
THE METHOD OF UNDETERMINED COEFFICIENTS
9.124
y" — 10/ + 29y — x 2 e
Determine the form of a particular solution for
I The complementary solution is, from Problem 8.51,
yc - c\e
$x
2x
215
cos 3x.
cos 2x + c 2 e
5x
sin 2x.
Since the right side of the
differential equation is the product of a cosine term, an exponential, and a second-degree polynomial, we try a
particular solution of the same but more generalized form, noting that wherever a cosine term occurs, a particular
We try
solution may have an identical component with a sine term.
y p = (A 2 x
2
+ A x + A )e~
2x
{
Since yc and y p have no terms in common, there is no need to modify y
9.125
2x
sm 3.x + (B 2 x 2 + B,x + B )e
p
cos 3x
.
y" — 10/ + 29y = (103x 2 — 27x + ll)e" 2x cos 3x.
Determine the form of a particular solution for
I The form is identical to that of y p of the previous problem.
2
9.126
I The complementary solution is
Ic
d I
, n dl
dr
dt
yy + 60 — + 500/ = (2f - 50)e
Determine the form of a particular solution for
— c^' 50 + c 2 e~ 10
'
Problem 8.10).
(see
'
10(
sin 50f.
We try a particular solution
having the same general form as the right side of the given differential equation, coupled with an associated term
involving cos 50r:
I
= (A l t+ A )e- 10 sin 50r + {Bj + B ]e~ 10 cos 50f
'
p
'
Since no part of I p can be obtained from I c by any choice of the constants c, and c 2
follows that I p does not need
it
,
modification.
9.127
Determine the form of a particular solution for
d*x
2
dt
dt
d x
—-r — 13 —
-y + 36x = f(5 — I2t )e
I The complementary solution is, from Problem 8.30,
2t
2
x c = c^e 2
'
cos St.
+ c 2 e~ 2 + c 3 e 3 + c 4 e~ 3
'
'
The right side of
'.
the differential equation is actually the product of a third-degree polynomial, an exponential, and a cosine term.
We therefore try
Xp = (A 3 t
i
+ A2 2 + A
t
x
t
+ /l )e" 2 'sin8r + (B 3 3 + B 2 2 + B
t
t
x
t
+ B )e~ 2 'cosSt
Since no part of x p can be formed from x c by any choice of c x through c 4 there is no need to modify x p
,
.
MODIFICATIONS OF TRIAL PARTICULAR SOLUTIONS
9.128
Determine the form of a particular solution to
/ — 5y = 2e 5x
.
5x
(see Problem 8.34). Since the right side of the given
yc = c i e
nonhomogeneous equation is an exponential, we try a particular solution of the same form, namely A e 5x
I The complementary solution is
.
However, this is exactly the same as y c except for a multiplicative constant (both are a constant times e 5x
5x
which is distinct from yc
Therefore, we must modify it. We do so by multiplying it by x to get y p — A xe
).
,
and is an appropriate candidate for a particular solution.
9.129
Solve the differential equation of the previous problem.
I
Substituting y p of the previous problem and its derivative
y'
p
5x
— A e 5x + 5/l xe
5x
- 5(A xe = 2e
A e = 2e
so that
or
x
5x
5*
=
=
=
2x)e*
2xe
(c\
solution
is
c^
+
+
+
and the general
y
y
yp
we get
(A e
5x
+ 5A xe
5x
5x
5x
)
)
,
,
Determine the form of a particular solution to
I The complementary solution is
A = 2.
into the differential equation,
yc = Ae~
6x
(see Problem 8.37).
9.131
,
,
Since the right side of the given differential
this is the same as yc except for a multiplicative constant, it must be modified.
= A xe~ 6x
5x
/ + 6y = 3e 6x
equation is an exponential, we try a particular solution having the same form, namely A e~
y
y pX = 2x^
Then
.
c
9.130
Sj:
6x
.
However, since
Multiplying by x, we get
which is different from y c and the appropriate candidate for a particular solution.
Solve the differential equation of the previous problem.
f
Substituting y p of the previous problem and
simplifying, we obtain
y = Ae~
6x
+ 3xe~
6x
.
A e~ 6x = 3e~ 6x
,
y'
p
so that
= A e ~ 6x - 6A xe ~ 6x into the given differential equation and
A = 3. Then y p = 3xe~ 6x and the general solution is
,
216
9.132
CHAPTER 9
D
2/ — 5y = — 4e 5x 2
Determine the form of a particular solution to
'
# The complementary solution is
5x ' 2
by multiplying by x. The new candidate is
5x 2
y p = A xe
.
(see Problem 8.38). We try a particular solution having the same
yc = Ae
form as the right side of the given differential equation, namely A e 5x 2 Since this is identical to yc we modify it
'
.
,
which is distinct from yc and therefore an
'
,
appropriate trial solution.
9.133
Solve the differential equation of the previous problem.
I By substituting y p of the previous problem and y'p = A e 5xl2 + jA xe 5x/2 into the given differential equation
and simplifying, we find that A = — 2. Then y p = —2xe 5x!2
and the general solution is
,
5xl2
- 2xe 5x 2
y = y + y p = Ae
'
e
9.134
.
y" — y' — 2y = e 2x
Determine the form of a particular solution to
+ c 2 e 2x
.
Problem 8.1). We try the particular solution A e 2x
which is similar in form to the right side of the given equation. However, it also has the form of part of the
complementary function (let c 2 = A ), so it must be modified. Multiplying by x, we get y p = A xe 2x which
has no term in common with the complementary function and is, therefore, in proper form for a particular solution.
y = c e~
I The complementary solution is
x
x
(see
,
,
9.135
Solve the differential equation of the previous problem.
I
Substituting y p of the previous problem with its derivatives
2x
= 4A e
2x
=
\xe
yp
y'
p
9.136
,
+ 4A xe
2x
y'
p
— A e 2x + 2A xe 2x and
A = \. Then
into the given differential equation and simplifying we find that
x
2x
+ \xe 2x
y — c e~ + c 2 e
and the general solution is
x
Determine the form of a particular solution to
.
y" — y' — 2y = 2e~ x
x
.
I The complementary solution is ye = c^e~ + c 2 e
(see Problem 8.1). We try a particular solution of the form
A e~ x but since this has the same form as part of y (let c = A
we must modify it. Multiplying the right side
x
=
we
get y
A xe~
since this is distinct from ye it is in proper form for a particular solution.
by x,
p
,
2x
c
9.137
),
x
,
;
Solve the differential equation of the previous problem.
I
Substituting yp of the previous problem and its derivatives into the differential equation, we obtain
-2A e~ x + A xe~ x — (A e~ x — A xe~ x — 2(A xe~ x = 2e' x from which we find that A = -§. Then a
~x
particular solution is
and the general solution is y = c e~ x + c 2 e 2x — \xe~ x
yp = — %xe
)
)
,
.
,
9.138
x
Determine the form of a particular solution to
y" + 4y' + 4y = e
~ 2x
.
I The complementary solution is found in Problem 8.141 to be y = c e~ 2x + c 2 xe~ 2x We try a particular
2x
solution of the form A e~
similar to the right side of the given differential equation. However, this is also part of
2x
=
which is also part of y
A
so we modify it by multiplying by x. This gives us A xe~
y c (let c,
2x
=
Since this result
(let
A
and must be modified. Multiplying again by x, we generate y p = A x 2 e'
c2
has no terms in common with y
is the proper form for a particular solution.
.
c
x
,
),
,
c
.
)
c,
9.139
it
Solve the differential equation of the previous problem.
I
Substituting \ p of the previous problem and its derivatives
2x
2x
y'
p
— —2A x 2 e~ 2x + 2A xe~ 2x
and
2x
— 4A x e~ — %A xe~ + 2A e~
into the differential equation and simplifying, we find that
p
Then yp — \x 2 e~ 2x and the general solution is y — y + y p = c e~ 2x + c 2 xe~ 2x + \x 2 e~ 2x
y'
2
'
,
9.140
c
Determine the form of a particular solution to
y" + 6y' + 9y = \2e~*
A = j.
.
x
x
.
I The complementary solution is shown in problem 8.142 to be yc — c e~ 3x + c 2 xe~ 3x We try the particular
3x
solution A e~
which is a general form of the right side of the given equation. However, it has a term in common
3x
But this also is part
with yc (take c = A
so it must be modified. To do so, we multiply by x, to get A xe~
3x
2
which
of) (let c 2 = 4
and must be modified. Multiplying by x once more, we obtain y p = A x e~
has no term in common with yc and so is the proper form for a particular solution.
x
.
,
c
9.141
.
),
x
,
)
Solve the differential equation of the previous problem.
f
Substituting y p of the previous problem and its derivatives into the given differential equation yields
(9x
from which we find that
2
- 12x + 2)A e~ 3x -F 6(-3x 2 + 2x)A e~ 3x + 9AoX 2 e~ 3x =\2e~ 3x
3x
2
A = 6. Then y p = 6x 2 e~ 3x and the general solution is y = (c + c 2 x + 6x )e~
,
x
.
4
THE METHOD OF UNDETERMINED COEFFICIENTS
9.142
-4
x + 20x + 64x = 60e
Determine the form of a particular solution to
217
D
'.
I The complementary solution is x c = c e~ M + c 2 e~ 16 (see Problem 8.1 1). We try a particular solution of the
form A e~*', but because it is part of x c (let c = A
it must be modified.
We multiply it by the independent
variable, and obtain
x p - A te~ A
Since this has no term in common with x c it is the proper form for a
'
x
t,
),
x
'.
,
particular solution.
9.143
Solve the differential equation of the previous problem.
I
Substituting x p of the previous problem and its derivatives into the given differential equation, we get
-4
'
16/V*?
A — 5.
from which
x = c,e
9.144
*'
16(
+c e
- 8/V~ 4 + 20(- A te~ 4t + A e~ M + 64A te~*' = 60e" 4
'
'
)
x p = 5te~*',
Then a particular solution is
and the general solution is
+ 5te
x + lOx + 25x = 20e
Determine the form of a particular solution to
5 '.
f The complementary solution is x = C e~ 5t + C 2 te~ 5 (see Problem 8.146). We try the particular solution
A e~ 5t which is similar in form to the right side of the given differential equation. But because it is part of
we modify it by multiplying by
x c (let C = A Q
The result, A te~ $t is also part of x c (let C 2 = A
so it
'
c
x
,
t.
),
x
),
,
too must be modified.
Multiplying again by the independent variable t, we obtain
with x c
9.145
,
it is
'.
Since this has no tern: in common
the proper form for a particular solution.
Solve the differential equation of the previous problem.
1
Substituting x p of the previous problem and its derivatives into the given differential equation, we obtain
25A t 2 e' 5
9.146
x p = A t 2 e~ 5
'
- 20A o te~ 5t + 2A e'
which can be simplified to
A — 10.
x = C,<r
2
5'
+ C 2 te-
5'
+ iot e
5t
Then
- 5t
+ 10(-5/4 ' 2 e~ 5 + 2A te~ 5t + 25A t 2 e~ s = 20e -5
'
'
'
)
x p = 10r 2 e
-5
',
and the general solution is
.
Determine the form of a particular solution to
n
—
\
-3
3
dx
—
+ 6 —-j2 + 12 it
y
**-y
dx
I-
,-2x
8y = \2e
dx
I The complementary solution is y — {c + c 2 x + c 3 x 2 )e~ 2x (see Problem 8.145). We try the particular
2x
solution A e~
which is similar in form to the right side of the differential equation. But it is part of yc (let
2x
is also part of y (let
and must
c2 — A
so we modify it by multiplying by x. The result, A xe~
Cj = A
c
x
,
),
,
)
c
also be modified.
2x
3
y p = A x e~
by x, we obtain
,
2x
which again is part of yc (let c 3 = A ). Multiplying once more
which has no terms in common with yc and thus is the proper form for a
2
Multiplying again by x, we get A x e~
,
particular solution.
9.147
Solve the differential equation of the previous problem.
I
of the previous problem and its derivatives
Substituting y
-2A x 3 e~ 2x + 3A x 2 e 2x
2x
t A v„~2x
3„-2x
2x _ n
2
3
- UA x^2„-lx
e
xe
+ 6A
y'p = 4A x e'
2x
2x
2
3
y';' = -SA x e~
+ 36A x e~ - 36A xe' 2x + 6A e~ 2x
yp =
a
and
,
A = 2.
into the given differential equation and simplifying, we find that
solution is
9.148
y = (c + c 2 x + c 3 x )e~
2
t
2x
3
+ 2x e~
Determine the form of a particular solution to
y p = 2x e~
3
Then
2x
and the general
2x
.
d3Q
2
—
dQ
^d Q
n
+ Q = 5e
-jT + 3 ~TT + 3
.
Q = C e~ + C 2 te~' + C 3 t e~'. We try a particular
equation, namely A e~'. Since this is part of the
differential
solution similar in form to the right side of the
2
or
the result will be of the
If we multiply by
it must be modified.
complementary solution (let C, = A
3
2
we
form A te~' or A
e~' and will also be part of the complementary function. However, if we multiply by
3
e~\ which is not part of the complementary function and is, therefore, the proper form for a
obtain Q p = A
I The complementary solution is, from Problem 8.158,
),
t
t
particular solution.
2
l
c
x
t
f
,
t
,
218
9.149
CHAPTER 9
U
Determine the form of a particular solution to
Q
w + 4Q
I The complementary solution is, from Problem 8.159,
+ 6Q + 4Q + Q = -23e~'.
(3)
Q = C x e~ + C 2 te~' + C 3
2
l
c
r
e~' +
Cj 3 e~'. We try a
particular solution having the same form as the right side of the differential equation, namely A e~'.
Since
C, = A ), it must be modified. To do so, we multiply by the smallest positive integral
power of t that eliminates any duplication between it and Q c This is the fourth power, and the result is the
this
is
part of Q c (for
.
Qp — A
proper form for a particular solution.
9.150
Determine the form of a particular solution to
4
t
e~'.
Q <5) + 5<2 ,4) + 10Q {3) + 10Q + 5Q + Q = -3e~'.
I The complementary solution is, from Problem 8.160, Q p = (C\ + C 2 + C 3 2 + C 4 3 + C 5 t*)e~'. We first try
a particular solution of the form A e~'. Since this is part of Q we modify it by multiplying by the smallest
positive integral power of that eliminates any duplication of Q
This is the fifth power, and the proper form for
5
=
a particular solution is
A
e~'.
Qp
t
c
r
t
,
t
c.
t
9.151
Determine the form of a particular solution to
y
(4)
+ 8y (3) + 24/' + 32/ + 16y = le~ 2x
.
I The complementary solution is found in Problem 8.157 to be yc — c e' 2x + c 2 xe~ 2x + c 3 x 2 e~ 2x + c^x 3 e~ 2x
We first try a particular solution of the form A e' 2x Then, since this is part of y we modify it by multiplying
x
.
c
by the smallest positive integral power of x that eliminates any duplication of yc
proper form for a particular solution is y p — A Q x x e~ 2x
.
.
,
This is the fourth power, and the
.
9.152
Determine the form of a particular solution to
(D 4 - 6D 3 + 13D 2 - 12D + 4)v = 6e*.
x
yc — e {c
I The complementary solution is. from Problem 8.187,
l
+ c 2 x) + e 2x {c 3 + c 4 x).
solution having the same form as the right side of the differential equation, namely A e x
we modify it by multiplying by x 2
x
Thus, the proper form is y p = A x 2 e
of yc
,
,
.
We try a particular
Then, since this is part
the smallest positive integral power of x that eliminates any duplication.
.
9.153
Rework the previous problem if. instead, the right side of the differential equation is — 7>e 2x
.
I The complementary solution y of that problem remains valid, and a first try for the particular solution is
A e 2x which has the same form as the new right side. It must be modified, however, because it is part of
We multiply by x 2 the smallest positive integral power of x that eliminates any duplication of
yc (for c 3 = A
2 2 '.
yc The result is yp = A x e
c
,
).
,
.
9.154
Determine the form of a particular solution to
/' — 7/ =
— 3.
f The complementary solution is y — c +c 2 e lx (see Problem 8.2). We try the particular solution A but since
must be modified. Multiplying by x, we get y p — A x, which is not part
this is part of y (for
c, = A
c
c
,
x
it
),
of yc and is, therefore, the proper form for a particular solution.
9.155
Solve the differential equation of the previous problem.
I Substituting y p y p — A
and y'p =
into the given differential equation yields
A Q = 3 Then y p = 3 x, and the general solution is y = c'i + c 2 e lx + 3 x.
,
,
— 1A
=—
3,
or
.
9.156
Determine the form of a particular solution to
/' — 7/ —
— 3x.
I The complementary solution y of Problem 9.154 remains valid. Since the right side of the differential
which is a general first-degree polynomial. But
equation is a first-degree polynomial, we try A x + A
part of this trial solution, namely A
is also part of y
so it must be modified. To do so, we multiply it by the
c
,
x
,
c
,
smallest positive integral power of x that will eliminate duplication.
y p = x{A x + A
x
9.157
)
x
Solve the differential equation of the previous problem.
into the given differential
and y"p = 2A
= — 3x + 0. Equating coefficients of like
2
y p = ^x + ^x, and the general solution is
y' = 24,x + A
Substituting y p of the previous problem,
p
equation yields, after rearrangement, (— 14/lJx + (2/1! — 1A
#
powers of x, we find
A
1
=Yi
and
A = ^.
Then
,
X
)
elX
+ T4* 2 + 49*y = C + C2
l
9.158
This is the first power, which gives us
= A x 2 + A x.
Determine the form of a particular solution to
y" — 7/ =
— 3x 2
.
THE METHOD OF UNDETERMINED COEFFICIENTS
219
f As in Problem 9.154, the complementary solution is y = c, + c 2 e lx We try as a particular solution
A 2x 2 + A x + A
which has the same degree as the right side of the given differential equation. But part of
this duplicates part of y (let
c, = A
so it needs to be modified. We multiply it by x, which is the smallest
3
2
positive integral power of x that eliminates duplication, obtaining
y p = A 2 x + A x + A x. Since there is
now no duplication, y p is in proper form.
.
c
,
x
),
c
x
9.159
Determine the form of a particular solution to
(Z)
4
+ 4D 2 )y = 6.
f The complementary function is, from Problem 8.186, y = c, + c 2 x + c 3 cos 2x + c 4 sin 2x. We first try the
particular solution A
Because this is part of y (for c = A
we modify it by multiplying by the smallest
c
.
c
positive integral power of x that eliminates duplication.
yc (for
9.160
c2
=A
),
x
The first power will not work, because A x is part of
yp = A x
The second power does work, however, so a particular solution is
).
2
.
Redo the previous problem if, instead, the right side of the differential equation is 6x.
I yc remains valid, but we now use a general first-degree polynomial as our initial try for y p A±x + A
However, this is part of y (for Cj = A
c2 = A
so we modify it by multiplying by x. The result is
A{x 2 + A Q x. But part of this (namely A x, a constant times x) is also part of y so we must modify it by
multiplying again by x. The result, y p = /4 x 3 + A 2 x 2
does not duplicate any part of y and is therefore in
:
.
v ),
,
c
c,
,
x
c
proper form.
9.161
6x 2 — 2x + 5.
Redo Problem 9.159 if the right side of the differential equation is
I The complementary solution yc of Problem 9.159 remains valid, but we now use a general second-degree
polynomial as our initial try for y p
A 2 x 2 + A^x + A
Because parts of this expression (namely A x and A
are also part of the complementary function, we must modify it. To do so, we multiply by x 2 the smallest
3
2
positive integral power of x that eliminates duplication of yc The proper form is then
y p = A 2 x* + A x + A x
:
.
)
t
,
.
9.162
Determine the form of a particular solution to
x
.
y" — 9x 2 + 2x + 1.
y c — c l + c 2 x (see Problem 8.153). Since the right side of the differential
equation is a second-degree polynomial, we try a general second-degree polynomial as the form of y p
1 The complementary function is
:
A 2x2 + A x + A
However, this trial solution has a first-power term and a constant term in common with
We
modify
it by multiplying by the smallest positive integral power of x that eliminates duplication between
must
yc
A
3
2
2
y p and y c This is x which gives y p = A 2 x + A x + A x
.
x
.
9.163
.
,
.
x
Solve the differential equation of the previous problem.
f
Differentiating y p of the previous problem twice and then substituting into the differential equation, we obtain
\2A 2 x 2 + 6A x + 2A
4
2
3
y p = fx + ^x — \x
= 9x 2 + 2x — 1,
x
A 2 = f, ^1=3, and
4
2
3
y = c x + c + fx + ^x — ^x
A = — \.
from which we find
Then
and the general solution is
The particular solution also can be obtained by twice integrating both sides of the differential equation with
,
.
x
respect to x.
9.164
Determine the form of a particular solution to
y" — 3x 2
.
yc = c y + c 2 x + c 3 x
differential equation is a second-degree polynomial, we try
2
f The complementary function is
Problem 8.154). Since the right side of the
However, since this is identical
3
the smallest positive integral power of x that
in form to yc it must be modified. To do so, we multiply by x
A
5
3
eliminates any duplication of y c The result is
y p = A 2 x + A x + /l x
(see
A — A x + A 2x2
x
.
9.165
.
,
,
.
x
Determine the form of a particular solution to
y'"
= — 2x 2 + 9x + 18.
f The particular solution is identical to that of the previous problem.
14
9.166
Determine the form of a particular solution to
—^ = 12x —
2
60.
3
y c = c, + c 2 x + c 3 x + c 4 x
Since the right side of the given differential equation is a second-degree polynomial, we try, as a particular
2
f The complementary solution is, from Problem 8.155,
solution, the general second-degree polynomial
.
A + A x + A 2x2
x
.
But this is part of y c for suitable choices of
4
and so must be modified. To do so, we multiply by x the smallest positive integral power of x
<?! through c 4
4
5
6
that eliminates any duplication of y c The result is a proper particular solution:
y p = /l x + A x + A 2 x
,
,
.
y
.
CHAPTER 9
220
9.167
Solve the differential equation of the previous problem.
I
Differentiating y p four times successively and substituting the result into the given differential equation, we
= 12x 2 — 60. Then, by equating the coefficients of like powers of x, we
A = — f, A = 0, and A 2 = yg. Thus y p = jqx 6 — fx 4 and the general solution is
+ c 2 x + c 3 x 2 + c 4 x 3 + ^x 6 — fx 4
24 A Q + 120,4 jx + 3604 2 x
get
conclude that
y = ct
9.168
2
,
x
.
d*y
—
= 30x — 2x +
ax
2
5
Determine the form of a particular solution to
z
4
yc = c v + c 2 x + c 3 x
5.
+ c 4 x 3 The right side of the differential
equation is a fifth-degree polynomial, so we try the particular solution A 5 x 5 + A A x* + A 3 x 3 + A 2 x 2 + A x + A
f The complementary solution is, again,
2
.
x
But because this trial solution and yc have terms in common, it must be modified. Again we multiply by x 4
the smallest positive integral power of x that eliminates any duplication of y c
J
9
4
8
6
s
in proper form,
yp — A 5 x + A 4x + A 3 x + A 2 x + A { x + A x
.
.
,
The result is a particular solution
.
d 2x
9.169
dx
+ k— - g,
—^
dr
dt
Solve
,
where k and g are positive constants.
x c = C, + C 2 e~ kt
I The complementary solution is
(see
Problem 8.20). Since the right side of the
differential equation is a constant, we first try the particular solution A
Cj = A ), so we modify it by multiplying by t. The result is
comprising x c
xp = A
.
t,
But this term is part of x,. (for
which is distinct from the terms
.
+ kA = — g,
Substituting x p into the given differential equation, we get
x = xc + x p = C
general solution is then
9.170
differential equation
is
y" -I- Ay — cos 2x.
y r = c, cos 2x + c 2 sin 2x
a cosine term, we try
y p — A x sin 2x + fi x cos 2x,
A Q sin 2x
Problem 8.59). Since the right side of the given
(see
B Q cos 2x as a particular solution. But this trial
c2 — A
so we must modify it. Multiplying by x, we
-I-
= B
and
),
which is distinct from yc and therefore in proper form.
solution is identical in form to yf (with
get
The
k
Determine the form of a particular solution to
I The complementary solution is
9.171
x
A — —g/k.
from which
9
+ C 2 e kt — - t.
c,
Redo the previous problem if, instead, the right side of the differential equation is —3 sin 2x.
I The particular solution is identical to y p of the previous problem.
9.172
x + 16x = 2 sin 4t.
Determine the form of a particular solution to
I The complementary solution is
differential equation
is
x r = c, cos 4r + c 2 sin 4f
A
a sine term, we try
sir\4t
9.173
t
sin 4f
+ B Q cos 4f.
t
Since the right side of the
as a particular solution.
We do so by multiplying by
solution are of identical form, so we must modify it.
xp = A
(see Problem 8.57).
+ B cos4t
f,
Since all the terms of x p are distinct from those of x c
,
But y c and this trial
getting
it
is
in
proper form.
Solve the differential equation of the previous problem,
f By substituting x p and
x" = ( - \6A
t
- 8B
)
sin At
+ (8.4 - 16B f) cos At
into the given differential equation
and simplifying, we get
(
- 85
)
sin At
+ (8A
Then equating the coefficients of like terms yields
)
cos At = 2 sin At +
A =
and
cos At
B = — £.
The general solution is thus
x = x c -\- x p = c, cos At + c 2 sin At — \t cos At.
9.174
x + 64x = 64 cos St.
Determine the form of a particular solution to
I The complementary solution is
x f = x x cos 8f + c 2 sin 8r
A
c2 = A
sin 8f -(- B
differential equation is a cosine term, we try
is
identical to x c when
ct
=B
and
by t obtaining x p = A t sin 8f -I- B t cos St.
x c by a suitable choice of Cj or c 2
,
,
Since the right side of the
as the form of a particular solution.
so we must modify the trial solution.
But it
We do so by multiplying
This is in proper form, because no part of it can be formed from
.
9.175
(see Problem 8.58).
cos 8f
Solve the differential equation of the previous problem.
A
THE METHOD OF UNDETERMINED COEFFICIENTS
D
221
f By substituting x p and its second derivative into the given differential equation, we obtain
(- 64
t
- 16B
)
sin St
+ (16 A - 64B t) cos 8t + 64(A
t
sin St
+ B
t
which can be simplified to (-16B )sin8t + (\6A )cosSt = 0sin8f + 64cos8f.
A = 4 and B = 0. Then the general solution is
terms, we conclude that
cos 8f) = 64 cos 8f
Equating coefficients of like
x = Cj cos 8f + c 2 sin St + 4r sin St.
9.176
Solve
3c
+ 3gx = 3g sin -J3gt,
where # is a positive constant.
§ The complementary solution is, from Problem 8.74, x = C cos y/lgt + C 2 sin -Jlgt. We try
A sin >/3gf + B cos v3# as a particular solution, and then modify it to x p — A sin y/3gt + B
c
x
f
t
t
cos \J3gt
by multiplying by t.
Substituting x p into the given differential equation and simplifying, we get
— 2yJ3gB sin y/3gt + (2yJ3gA cos \J3gt = 3g sin \j3gt. By equating coefficients of like terms, we find that
B = — jy/3g and A = 0. The general solution is then
x = x + x p = Cj cos yJ3gt + c 2 sin \J3gt — \\l3gt cos yJ3gt.
(
)
)
c
9.177
Determine the form of a particular solution to
f The complementary solution is
(D 2 + 4)v = x 2 sin 2x.
yc = C x cos 2x + C 2 sin 2x.
Since the right side of the differential equation is a
2
+ Ex + G) sin 2x + (Ax 2 + Cx + F) cos 2x as a
particular solution. But this trial solution has terms in common with yc namely a constant times sin 2x and a
second-degree polynomial times a sine term, we try
(Bx
,
We do so by multiplying by the smallest positive integral
constant times cos 2x, so we must modify it.
power of x that eliminates all commonality between ye and y p
3
y p = Ax cos 2x + Bx
9.178
3
.
This is the first power. As a result, we find
2
+ Cx cos 2x + Ex 2 sin 2x + Fx cos 2x + Gx sin 2x
sin 2x
Solve the differential equation of the previous problem.
f
Differentiating y p twice yields
D 2 y p = -4 Ax 3 cos 2x - 4Bx 3 sin 2x + (12B - 4C)x 2 cos 2x + (- 12,4 - 4£)x 2 sin 2x
+ (6/1 + 8£ - 4F)x cos 2x + (65 - 8C - 4G)x sin 2x + (2C + 4G) cos 2x + (IE - 4F)sin 2x
Then substituting this and y p into the given equation yields
12Bx 2 cos 2x - \2Ax 2 sin 2x + (6A + 8£)x cos 2x + (6B - SC)x sin 2x + (2C + 4G) cos 2x + (2£ - 4F) sin 2x
= x 2 sin 2x
By equating coefficients of like terms, we find that A = — -fa, B = C = G = 0,
Then y p = — y^x 3 cos 2x + j^x 2 sin 2x + j^x cos 2x, and the primitive is
£ = yg,
and
£ = ^.
2
3
y = C t cos 2x + C 2 sin 2x — ^x cos 2x + ^x sin 2x + ^x cos 2x.
9.179
Solve
(D 2 + 4)y = 8 sin 2x.
I The complementary solution is yc = c cos 2x + c 2 sin 2x. For a particular solution we would normally try
A cos 2x + B sin 2x. However, since its terms appear in the complementary solution, we multiply by x to obtain
the particular solution
y p = x(A cos 2x + B sin 2x). Then, substituting in the given equation yields
and B = — 2. Then the required general solution is
—4A cos 2x — 4B sin 2x = 8 sin 2x, so that A =
t
y = c t cos 2x 4- c 2 sin 2x — 2x sin 2x.
9.180
Determine the form of a particular solution to
y" + 4y = 8x sin 2x.
I As in the previous problem, the complementary solution is yc = c, cos 2x + c 2 sin 2x. We try a particular
because the right side of the given differential
solution of the form (A x + A sin 2x + (B x + B cos 2x
equation is the product of a first-degree polynomial and sin 2x. But two of the summands in this trial solution
are identical in form with the summands in yc so it must be modified. Multiplying by x, we get
2
2
y p = (A x + Aqx) sin 2x + (B x + B x) cos 2x, which has no terms in common with yc and is therefore in
)
x
)
x
,
x
1
proper form.
9.181
Solve the differential equation of the previous problem.
f
Differentiating y p of the previous problem twice in succession yields
y" = (-4A x x 2 - 4A x + 2A l - SB^ - 4B )sin 2x + (SA t x + 4A
- 4B x 2 - 4B x + 2Bj)cos 2x
x
222
CHAPTER 9
D
Substituting this and y p into the given differential equation and simplifying, we obtain
[(-8B,)x + (2A 1 - 4B )] sin 2x + [(8/t,)x + (4A
+ 25,)] cos 2x = 8x sin 2x
By equating the coefficients of like terms and solving the resulting system of equations, we find that
A = j, B, — — 1, and B — 0. Combining these results with ye and yp of the previous problem, we
A = 0,
x
form the general solution
9.182
2
y" + Ay = (8 - 16x) cos 2x.
Solve
I
y = c, cos 2x + c 2 sin 2x + jx sin 2x — x cos 2x.
y c and y p of Problem 9.180 are valid here, as is y"p of the previous problem. Substituting y p and y"p into the
given differential equation and simplifying, we get
[(-8B,)x + (2/1, -4B )]sin2x + [(8A
x
)x
+ (4A + 2B,)]cos2x = -16xcos2x + 8cos2x
By equating coefficients of like terms and solving the resulting system of equations, we find that A = -2,
B, = 0, and B = - 1. Combining these results with y and y of Problem 9.180, we form the
p
x
A = 2,
c
y = (-2x
general solution
9.183
2
+ 2x + c 2 )sin2x + (-x + c,)cos2x.
Determine the form of a particular solution of
x + 16x = (80r — 16) sin 4r.
f The complementary solution is xc = c cos4f + c 2 sin 4t (see Problem 8.57). We try a particular solution
of the form
(A
+ A )s'm4t H (B,f + B )cos4t, but since it contains summands which are identical in
form to the summands of x it must be modified. Multiplying by f, we get
x p = (A^ 2 + A t)sm4t + (B,r 2 + B f)cos4f.
Since none of the summands of x p is identical to a summand of x,. except perhaps for a multiplicative constant,
x
t
x
f,
x p is the proper form for a particular solution.
9.184
Solve the differential equation of the previous problem.
f
Differentiating x
.\
r
p
of the previous problem twice yields
= (-16.4,r 2 - 16/V + 2/1, - 16S,f -8B )sin4f + (l6A
x
t
+ %A - 16B,f 2 - 16B
f
+ 2B,)cos4f
Then by substituting .v p and x p into the given differential equation and simplifying, we get
[(-16B,)f + (2/1, -8B )]sin4f + [(164 ,)f + (8.4
+ 2B,)] cos4f = (80f - 16)sin4f + (Or + 0)cos4r
Equating coefficients of like terms, we get a system of equations whose solution is A = 0, A — 1.25,
By— —5, and B = 2. Combining these results with x p and xe of the previous problem, we form the general
x
9.185
x = c, cos 4f + c 2 sin 4f +
.25f sin 4f
+ - 5f 2 + 2t) cos 4t.
Determine the form of a particular solution to
x + 64x = 8f 2 cos 8f.
solution
1
(
xe = c, cos 8f + c 2 sin 8f
I The complementary solution is
(see
Problem 8.58). Since the right side of the
given differential equation is the product of a second-degree polynomial and a cosine term, we try a
(A 2 t 2 + A
2
+ B,f + B )cos8r.
Two of the summands in this trial solution also appear in x for suitable choices of c, and c 2 so it must be
particular solution of the form
+ /1
t
x
)
sin 8f + (B 2 f
,
c
modified.
Multiplying by t, we get
Xp = (Ait* + A
2
t
x
+ A t)smSt + (B 2 3 + B t 2 + B r)cos8r
t
x
Since the summands of x p are linearly independent of the summands of x c
,
we have the proper form for a
particular solution.
9.186
Determine the form of a particular solution to
x + 96x = (r
3
+ 3) sin yj%t.
I The complementary solution is, from Problem 8.61, x = C, sin ^96' + C 2 cos V96f. We try a particular
2
solution of the form
(A 3 t 3 + A 2
+ A t + A )s\n v96f + (B^ + B 2 2 + B,f + BoJcosv^r. But this trial
solution has summands in common with x so it must be modified. We multiply it by the smallest positive
c
t
t
y
c
,
integral power of t that eliminates any duplication of terms of x f
X p = (A 3 t 4 + A 2 t 3 + A t
x
9.187
2
—the
first.
Thus, we have
+ A f) sin V96f + (B 3 t* + B 2 3 + B
f
Determine the form of a particular solution to
(D
4
2
x
t
+ B f)cos v96r
+ 4D 2 )y = x cos 2x.
f The complementary solution is, from Problem 8.186, x, = c, + c 2 x + c 3 cos 2x c 4 sin 2x. We try a
(A x + A sin 2x + (B,x + B cos 2x. But this trial solution and x have
particular solution of the form
t-
x
)
)
c
THE METHOD OF UNDETERMINED COEFFICIENTS
D
223
summands in common, so the trial solution must be modified. Multiplying it by x, we get
2
2
y p = (AiX + A x) sin 2x + (B x + B x) cos 2x.
Since each summand of y p is linearly independent of the summands of
ye it is the proper form for a particular
t
,
solution.
2
9.188
d*y
d y
—
—
+ 16y =
z +
ax*
dx
Determine the form of a particular solution to
8
\
1
(1
— 8x) sin 2x
I The characteristic equation of the associated homogeneous equation can be factored into
the roots are +i'2, each of multiplicity two.
('/}
+ 4) 2 = 0,
so
follows from Problem 8.175 that the complementary function is
It
y c = c, cos 2x + c 2 sin 2x + c 3 x cos 2x + c 4 x sin 2x.
Since the right side of the given differential equation is a first-degree polynomial times sin 2x, we try the
(A { x + A )sin 2x + (S,x + B )cos 2x.
particular solution
c3
=B
x
,
and
c4
= Av
But this is identical to y c when
We must modify the trial solution, and we do so by multiplying
positive integral power of x that eliminates any duplication.
Cj
it
= B
,
c2
=A
,
by x 2 the smallest
,
The result is
3
2
2
3
y p = {A x + A x )sin2x + (BjX + B x )cos2x.
t
The summands of y p are distinct from those of yc so it is the proper form for a particular solution.
,
9.189
(D 2 + 9) 3 y = (2x
Determine the form of a particular solution to
2
— 3x + 5) cos 3x.
I The roots of the characteristic equation of the associated homogeneous equation are + i3, each of multiplicity
three.
The complementary function is
yc = (c : + c 2 x + <r 3 x ) cos 3x + (c 4 + c 5 x + c 6 x ) sin 3x
2
We try a particular solution of the form
is
(A 2 x
identical in form to yc and must be modified.
2
2
+A x+A
Y
sin 3x
)
+ (B 2 x 2 + B,x + fi
)
cos 3x.
However, this
To do so, we multiply by x 3 the smallest positive integral
,
power of x that results in summands distinct from those of y c The result is
5
A
3
4
3
y p = (A 2 x + A x + 4 x )sin 3x + (B 2 x5 + B]X + B x )cos 3x, which is the proper form for a particular
.
x
solution.
2
9.190
Determine the form of a particular solution for
d y
dy
—
— 6 -—h 25y — 6e
-r
dx
cos 4x.
dx
ix
y c = e (c cos4x + c 2 sin4x) (see Problem 8.50). Since the right side of the
differential equation is an exponential times a cosine term, we try the particular solution
I The complementary solution is
x
A e 3x sin 4x + B e 3x cos 4x. Because this has the same form as y we modify it by multiplying by x. The result,
3x
ix
y p — A xe sin 4x + B xe cos 4x, consists of terms that are different from those of y so it needs no further
c,
c,
modification.
9.191
Determine the form of a particular solution to
I The complementary solution is
yc = e
—-^ — 10 — + 29 y = —8e Sx sin 2x
dx 2
5x
(Cj cos 2x
dx
+ c 2 sin 2x)
(see
Problem 8.51). The right side of the
given differential equation is an exponential times a sine, so we try the particular solution
A e 5x sin 2x + B e 5x cos 2x. Since this is identical in form to y we modify it by multiplying by x. The result,
5x
B xe Sx cos 2x, consists of terms that cannot be obtained from yc by any choice of the
y — A Q xe sin 2x
c,
-I-
constants c, and c 2 and so needs no further modification.
9.192
Determine the form of a particular solution to
y" + 4/ + 5y = 60e~ 2x sinx.
I The complementary solution is yc — c e~ 2x cosx + c 2 e~ 2x s'mx (see Problem 8.66). We try, as a particular
solution,
4 e -2jc sinx B e" cosx; but because it is identical in form to y it must be modified. We
_2
2x
sinx + B xe *cosx, which has no terms in common with yc and so is
multiply it by x, getting y p = ,4 xe~
x
2Ar
-I-
c,
in proper form.
9.193
Solve the differential equation of the previous problem.
I
Differentiating y p of the previous problem yields
= (-2A x + A - B x)e~ 2x sinx + (A x - 2B x + B )e~ 2x cosx
2x
2x
y" = (3^ ^ - 4/4 + 4B x - 2B )e~
s\nx + (-4A x + 2A + 3B x - 4B )ecosx
y'
p
and
x
CHAPTER 9
224
Substituting y p and its derivatives into the differential equation and simplifying, we get
- 2B e ~ 2x sin x + 2A e ~ 2 cos x = 60e ~ 2x sin x;
~
and B = — 30. Then y p = — 30xe 2x cos x,
2x
cosx + c 2 e~ sinx — 30xe~ 2 *cosx.
y — c e~
A =
by equating coefficients of like terms, we find that
and the general solution is
2jc
1
9.194
Determine the form of a particular solution to
(D 2 - 2D + 10)y = 18e* cos 3x.
f The complementary solution is, from Problem 8.68, y = C e x cos 3x + C2e" sin 3x. We try the particular
A ex sin 3x + B e x cos 3x, but since this is identical in form to yc it must be modified. We therefore
solution
x
multiply by x, obtaining y p = A xe sin 3x + fl xe*cos 3x.
Since there is no duplication between y and y
p
c
t
,
c
,
the latter is in proper form.
9.195
Solve the differential equation of the previous problem.
I By differentiating y p of the previous problem twice, we get
= Mo x + ^o — 3B x)e sin 3x + (3/l x + B x + B )ex sin 3x
y' = - 84 x + 2A
- 6B x - 6B )e x sin 3x + (6A x + 6A Q - SB Q x + 2B Q )ex cos 3x
p
Ar
y'
p
and
'
(
Substituting y p and its derivatives into the differential equation and simplifying then yield
— 6B )e x sin 3x + (6A )ex cos 3x = 18^ cos 3x. By equating coefficients of like terms, we find that A = 3
and B = 0. Combining these results with yc and y p of the previous problem, we form the general solution
x
x
x
y = C e cos 3x + C 2 e sin 3x + 3xe sin 3x.
(
x
9.196
Determine the form of a particular solution to
y
<4)
— 8y (3) + 32/' — 64/ + 64y = 30e 2x sin 2x.
I The complementary solution is, from Problem 8.188, yc — (Cj + c 2 x)e 2x cos 2x + (c 3 + c A x)e 2x sin 2x. We
try as a particular solution
A e 2x sin 2x + B e 2x cos 2x, but since this is part of the complementary function
(for
it must be modified.
c3 = AQ
c, = B
Multiplying by x will also duplicate terms of y c so we multiply
2x
2x
2
by x to get y p = A x 2 e sin 2x + B x 2 e cos 2x. Since the summands of y p are distinct from those of yc yp
),
,
,
,
is
the proper form for a particular solution.
d3
d*\
9.197
Determine the form of a particular solution to
2
dy
ax
ax
d
—7 + 4 —-^y + 8 —-jy + 8
ax
ax
J
\-
4y — 2e'
x
cos x.
I The characteristic equation of the associated homogeneous differential equation can be factored into
— ±
are both roots of multiplicity two, and the complementary solution is
(m 2 + 2m + 2) 2 = 0, so
x
' x
sin x.
yc = (c, + c 2 x)e~ cosx + (c 3 + c A x)e
We try as a particular solution A Q e x sinx + B e' x cosx, but because this trial solution has terms in
We thus multiply by x 2 the smallest positive integral power of x that
common with ye we must modify
x
2 -Jc
2
sin x + B x e" cosx.
eliminates any duplication, obtaining y p — v4 x e
1
/
it.
,
9.198
,
Determine the form of a particular solution to
(D 2 + 2D + 2) 3 y = 3e~*sinx.
I The characteristic equation of the associated homogeneous differential equation is
—1 ±
so
are both roots of multiplicity three, and the complementary solution is
i
+ c i x + c 3 x 2 )e"*cosx + (c 4 + c 5 x + c 6 x 2 )e _Jc sinx.
We try as a particular solution A e~ x s\nx + B e~ x cosx,
yc = ( c
(m 2 + 2m + 2) 3 = 0,
i
but because this trial solution is part of yc
,
it
must be modified. We multiply by x 3 the smallest positive integral power of x that eliminates any duplication
ofy c The result is y . = A x 3 e~ x sin x + B x 3 e" x cosx, which is the proper form of a particular solution.
,
-
d*y_
9.199
Determine the form of a particular solution to
-j-j
2
—
— 66-*y h 25y = (2x — l)e ix cos 4x.
dx
dx
I The complementary solution is, from Problem 8.50, y c = c,e 3x cos4x + c 2 e 3x sin4x. We try as a particular
3x
ix
Two of the summands of this trial solution are identical
solution {A x + A )e
cos 4x.
sin 4x + (B x + B )e
in form to the summands of y c so it must be modified. Multiplying by x yields
ix
3x
2
2
y p = (A x + A x)e sin 4x + {B^x + B x)e cos 4x, which has no summands in common with yc
t
t
,
.
x
9.200
Determine the form of a particular solution to
d2y
-r-=
—
dx
+ 10 -
dx'
i The complementary solution is, from Problem 8.51,
5x
solution
(A^x + A )e 5x sin 2x + (BjX + B Q )e cos 2x.
29y — xe
h
s
5x
sin 2x.
dx
yc = c e
i
5x
cos 2x + c 2 e
5x
sin 2x.
We try as a particular
Since two of the summands of this trial solution are
THE METHOD OF UNDETERMINED COEFFICIENTS
225
identical in form to the summands in yc we must modify it. We do so
by multiplying by x, which results in
2
5
2
5x
y p = {A x x + ,4 x)e *sin 2x + (B lX + B x)e cos 2x. Because y p and yt have no terms in common, this is of
the proper form for a particular solution.
,
9.201
y" + 4/ + 5y = (x 2 + 5)e~ 2 *sin x.
Determine the form of a particular solution to
# The complementary solution is, from Problem 8.66, y = c e ~ 2x cos x + c 2 e ' 2x sin x. We try as a particular
2
solution
(,4 2 x
+ A x + ,4 )<T 2x sinx + (fl 2 x 2 + B x + B )e~ 2x cosx. But since this trial solution has
c
x
x
y
summands that are identical in form to those of y c it must be modified.
,
y p = (A 2 x
3
2
+ A x + ,4 x)e~
2x
x
sinx + (fl 2 x 3 + B x 2 + B x)e~ 2x cosx
x
Since y p does not duplicate any of the summands of yc
9.202
Determine the form of a particular solution to
,
of the proper form for a particular solution.
it is
w - 8y
y
Multiplying by x, we get
+ 32/' - 64/ + 64y = x 2 e 2x sin 2x.
<3)
I The complementary solution is, from Problem 8.177, y = (c + c 2 x)e 2x cos 2x + (c 3 + c 4 x)e 2x sin 2x. We
2
try as a particular solution
(,4 2 x
+ A x + A )e 2x sin 2x + (B 2 x 2 + B x + B )e 2x cos 2x. Because this trial
solution has terms in common with y it must be modified. If we multiply by x, it will still have terms in
common with yc so we multiply by x 2 getting
c
l
x
x
c,
,
,
y p = (A 2 x* + A x
+ A x 2 )e 2x sin 2x + (B 2 x 4 + B x 3 + B x 2 )e 2x cos 2x
3
x
x
Because each of its summands is different from those in yc y is in proper form.
p
,
A
9.203
2
d y
d y
dy
^y
—
£ + 4 —^ + —V + 8 — + 4v =
dx*
dx
dx
dx
Determine the form of a particular solution to
8
3
Wv-J
J-v-^
2
A-v*-
J--
(x
- 4)e~ x cos x.
I The complementary solution is, from Problem 9.197, y = (c + c 2 x)e~ x cosx + (c 3 + c 4 x)e 2x sinx. We try
x
as a particular solution
{A x + A )e' sinx + (B x + B )e~ x cosx, but because it is identical in form to yc
it must be modified.
To eliminate any duplication of y we multiply by x 2 obtaining
c
x
x
x
,
c,
y p = (A x x
3
,
+ ,4 x 2 )<?"*sinx + (B^ 3 + B x 2 )e~ x cosx
as the proper form for a particular solution.
9.204
(D 2 + 2D + 2) 3 = xe~ x sin x.
Determine the form of a particular solution to
f The complementary solution is, from Problem 9.198,
x
2
2
yc = (Ci + c 2 x + c 3 x )e" cosx + (c 4 + c 5 x + c 6 x )e~*sinx
We try as a particular solution
(A x + A )e~
x
x
sin x
+ (B x + B )e~ x cos x,
x
because the right side of the
x
and a sine term. Since each summand in
the trial solution also appears in yt except for the arbitrary multiplicative constants, it must be modified. We
3
multiply by x the smallest positive integral power of x that eliminates any commonality with yc The result is
differential equation is the product of a first-degree polynomial, e~
,
,
.
3
y p = {A^x* + /4 x )e~*sinx + (B l x
A
+ B x 3 )e~ x cosx
which is the proper form for a particular solution.
9.205
Redo the previous problem if the right side is replaced with
(5
— 3x)e" x cosx.
I The particular solution here is identical to y p of the previous problem: Since the right side of the new
differential equation is the product of a first-degree polynomial, e~
the same as that in the previous problem.
solution y p
x
,
and a cosine term, the trial solution here is
And it must be modified in the same way to the same particular
.
EQUATIONS WHOSE RIGHT SIDE CONTAINS A COMBINATION OF TERMS
9.206
/ - 5y = e 2x + 8x.
Solve
I A particular solution corresponding to a right side of e 2x is found in Problem 9.1 to be
a particular solution corresponding to a right side of 8x is found in Problem 9.32 to be
y x = — ^e
y 2 = — fx — £.
2x
,
and
A particular solution to the given differential equation is then y + y 2 = —\e 2x - fx - 2\. When combined
5x
(see Problem 8.34), it yields the general solution
with the complementary solution y = c e
X
C
e
— e
255
3
x
c
Jf
\
x
e
226
9.207
CHAPTER 9
D
Solve
/ - 5y = e
2
I A particular solution corresponding to a right side of e 2x remains y, of the previous problem, while a
particular solution corresponding to sin x is found in Problem 9.77 to be
y 2 = —^ sin x — ^cos x.
A particular solution to the given differential equation is then y + y 2 — —\e 2x — ^sinx — ^gcosx; when
combined with the complementary solution, it yields the general solution y = c,e 5x — \e 2x — ^sin x - y^cos x.
t
9.208
Solve
y
- 5y = 8x + sin x.
I A particular solution corresponding to 8x is found in Problem 9.32 to be y t = — |x — j$, and a particular
solution corresponding to sinx is found in Problem 9.77 to be
y 2 — — ^sinx — ygcosx.
A particular solution to the given differential equation is then y x + y 2 — — |x — ^ — ^sinx — t^cosx,
which, when combined with the complementary solution (see Problem 8.34) yields the general solution
9.209
y = cxe
5x
Solve
y'
- fx - £ - 2% sin x - ^ cos x.
— 5y = 8x — sin x.
f The expressions for y and y 2 of the previous problem are valid here, but now a particular solution is the
difference of those two solutions, namely
y, — y 2 = — fx — ^ — (— ^sinx — ygcosx).
x
Combining this with the general solution to the associated homogeneous problem (see Problem 8.34), we
obtain the general solution to the nonhomogeneous equation as
9.210
Solve
y'
- 5y = 8 - 2e 5x
x
5x
— fx — ^ + jg sin x + j§ cos x.
.
A particular solution corresponding to a right side of 8 is found in Problem 9.17 as
#
particular solution corresponding to 2e
5x
is
found in Problems 9.128 and 9.129 to be
complementary solution (see Problem 8.34), it
Solve
y
= — f,
=
2xe 5x
y2
and a
y,
.
— y 2 = — f — 2xe 5x Combined with the
5x
— 2xe 5x — f.
yields the general solution
y - c e
A particular solution to the given differential equation is then
9.211
y = c e
y,
.
x
- 5y = xe 2x + 2x 2 - 5.
A particular solution corresponding to xe 2x is found in Problem 9.55 to be y = (—\x — %)e 2x and a
particular solution corresponding to
2x 2 — 5 is found in Problem 9.33 to be y 2 = — 0.4x 2 — 0.16x + 0.968.
A particular solution to the given differential equation is then y, + y 2 = (— 5X — ^)e 2x — 0.4x 2 — 0.16x + 0.968.
When combined with the complementary solution (see Problem 8.34), it yields the general solution
2x
5x
2
y = Cl e + {-\x - \)e - 0.4x - 0.16x + 0.968.
I
9.212
Solve
I
,
x
y'
— 5y = xe 2x + 8x + sin x.
y, of the previous problem is valid here.
Problem 9.32,
y 2 = — fx — 2%;
y 3 = —2% sinx — ^cosx.
,v,
In addition, a particular solution corresponding to 8x is, from
a particular solution corresponding to sin x is found in
Problem 9.77 to be
A particular solution to the given differential equation is then
+ y2 +
>'
3
=(-i* - 9)e
2x
+(-!* - A) + (-^sinx -
'
26
c°sx)
We combine this result with the solution to the associated homogeneous equation (see Problem 8.34) to obtain
Sx
— \xe 2x — \e 2x — fx — 2% — je sm x ~ 26 cos x
y = c e
the general solution
9.213
Solve
y'
-
x
+ 6y = —2 cos 3x 4- 3e 2x sin 3x.
A particular solution corresponding to —2 cos 3x is found in Problem 9.78 to be
— y$ sin 3x — ^cos 3x, and a particular solution corresponding to 3e 2x sin 3x is found in Problem 9.99
=
y
2x
2
to be
y 2 = ffe * sin 3x — ^%e cos 3x. A particular solution to the given differential equation is then
f
t
y t + y 2 = — y$ sin 3x — 33 cos 3x + yfe
2x
sin 3x
— r%e 2x cos 3x
When combined with the complementary solution (see Problem 8.37), this yields the general solution
y = Ae ~
9.214
6x
— ys sin 3x — y§ cos 3x + ff 2 * sin 3x — ^e 2x cos 3x
Rework the previous problem if the term e 3x is added to the right side of the differential equation.
3x
is found in
y x and y 2 of the previous problem are valid here. A particular solution corresponding to e
Problem 9.2 as y 3 = <je 3x
A particular solution corresponding to the new right side is then y x + y 2 + y 3
when combined with the complementary solution (see Problem 8.37), it yields the general solution
I
.
y = Ae
~ 6x
- tV sin 3x - -& cos 3x + ^e 2x sin 3x - %e 2x cos 3x + %e 3x
:
THE METHOD OF UNDETERMINED COEFFICIENTS
9.215
Solve
y'
+ 6y = 4e' 5x - 6e 6x + 3e~ 6x
227
.
I A particular solution corresponding to a right side of 4e~ 5x is found in Problem 9.4 to be y, = 4e~ 5x a
6x
tx
particular solution corresponding to 6e
is found in Problem 9.5 to be
and a particular solution
y 2 = \e
;
\
corresponding to 3e~ 6x is found in Problems 9.130 and 9.131 to be
6x
y 3 = 3xe~
—
y
When combined with the
y2 + y3
complementary solution (see Problem 8.37), it yields the general solution y = Ae~ 6x + 4e~ 5x — \e ex + 3xe~ 6x
A particular solution to the given differential equation is
9.216
Solve
y" - 7/ = 6e 6x + e Sx
.
.
t
.
I A particular solution corresponding to 6e 6x is found in Problem 9.7 to be
8x
is
Solve
y" - ly' =
6x
and a particular
,
.
When combined with the
+ c 2 e lx — e 6x + %e Sx
.
y = c
complementary solution (see Problem 8.2), it yields the general solution
9.217
y{ = — e
8x
found in Problem 9.8 to be y 2 = \e
A particular solution to the given differential equation is then y\ + y 2
solution corresponding to e
.
.
x
- 3x + 48 sin 4x + 84 cos 4x.
I A particular solution corresponding to — 3x is found in Problems 9.156 and 9.157 to be y = -^x 2 + ^x.
A particular solution corresponding to 48 sin 4x + 84 cos 4x is found in Problem 9.80 to be y 2 = — 3 sin 4x.
x
A particular solution to the given differential equation is then y + y 2 when combined with the
complementary solution (see Problem 8.2), it yields the general solution y = c x + c 2 e lx + ^x 2 + ^x — 3 sin4x.
;
i
9.218
Solve
y" - / - 2y = 7 + e 3x
.
I A particular solution corresponding to a right side of 7 is found in Problem 9.19 to be
and a
y = —f
3x
=
\e
y2
A particular solution to the given differential equation is y + y 2 When combined with the complementary
x
2x
— J + \e ix
solution (see Problem 8.1),
yields the general solution
y = c e~ + c 2 e
particular solution corresponding to e
2x
is
found in Problem 9.10 to be
.
x
it
9.219
.
1
Use the results of the previous problem to solve
y" — y' — 2y = 14 — 3e 3x
t The right side of the given differential equation may be written as
particular solution to this differential equation is
2yj — 3y 2
2(7)
.
— 3(e 3x
Solve
y" - y' - 2y = e 2x + 2e~ x
Then we conclude that a
).
where y x and y 2 are as in the previous problem.
,
y = c l e~
When combined with the complementary solution, this yields the general solution
9.220
x
+ c 2 e 2x — 7 — fe 3 *.
.
I A particular solution corresponding to e 2x is found in Problems 9.134 and 9.135 to be
particular solution corresponding to 2e~
x
is
y t = \xe
y t + y2
complementary solution (see Problem 8.1), it yields the general solution
Use the results of the previous problem to solve
;
2x
.
A
x
y 2 = —\xe~
when combined with the
found in Problems 9.136 and 9.137 to be
A particular solution to the given differential equation then is
9.221
,
x
.
y = c x e~
y" — y' — 2y = 3e 2x - \%e~
x
.
+ c 2 e 2x + ^xe 2x — \xe~ x
.
x
.
— 9(2e~*). It then follows from the
When combined with
previous problem that a particular solution to this differential equation is 3y — 9y 2
I The right side of this differential equation may be written as
3(e
2x
)
{
y = c l e'
the complementary solution, this yields the general solution
9.222
Solve
y" - y' -2y = 7 + e
+ e 2x + 2e~ x
.
+ c 2 e 2x + xe 2x + 6xe~ x
.
.
Combining y and y 2 of both Problem 9.218 and Problem 9.220 with the complementary function, we obtain
I
t
the general solution
9.223
ix
x
Solve
y" - y
y = c l e~
x
+ c 2 e 2x - \ + \e 3x + \xe 2x - \xe~ x
.
- 2y = 4x 2 - sin 2x.
f A particular solution corresponding to a right side of 4x
2
is
found in Problem 9.35 to be
— 3. A particular solution corresponding to sin 2x is found in Problem 9.92 to be
y = —2x + 2x
—
—
sin
2x
+
jq cos 2x.
jq
y2
when combined with the
to the given differential equation is then
solution
particular
A
y - y2
2
x
x
complementary solution (see Problem 8.1), it yields the general solution
y = c x e'
9.224
Solve
x
+ c 2 e 2x - 2x 2 + 2x - 3 + ^ sin 2x - ^ cos 2x.
^ - 4^
dt
2
dt
+ y = 3e 2 + 3t - 4.
'
;
228
CHAPTER 9
D
I A particular solution corresponding to 3e 2t is found in Problem 9.11 to be y t = —e 2t and a particular
3f — 4
= 3t + 8.
is found in Problem 9.35 to be
solution corresponding to
2
A particular solution to the given differential equation then is y^ + y 2 When combined with the
complementary solution (see Problem 8.9), it yields the general solution y = C e 3 132t + C 2 e°- 2679t — e 2t + 3f +
,
_y
.
t
d 2x
9.225
Solve
+ 4 — + 8x = <T
^r
It *
It
dx
2'
+ 20 ' 2 + l6t ~ 78 )* 2
'-
(
2
I A particular solution corresponding to e~ 2 is found in Problem 9.13 to be x =\e~ 2t A particular
2
2t
solution corresponding to
(20r + 16r — l%)e
is found in Problem 9.67 to be
x 2 = (t 2 - A)e 2t
A particular
solution to the given differential equation then is
When combined with the complementary solution
Xj + x 2
'
.
l
.
.
(see Problem 8.54), it yields the general solution
x = c 1 e" 2, cos2r + c 2 e~ 2 'sin2r + \e~ 2t + (r 2 - 4)e 2
9.226
Solve
d x
dx
—
r + 4 — + 8x = -t +
2
dt
2
5t
2
e~ 3
'
- I4te~ 3 + lie" 3
'
'
'.
dt
«
I A particular solution corresponding to — is found in Problem 9.41 to be x, = — £r 2 + %t — 322
3
particular solution corresponding to
(5f — 14f + ll)e~
is found in Problem 9.68 to be
2
3
Then a particular solution to the given differential equation is x, + x 2
x 2 = (t — It + l)e~
when
2
i
t
'
'.
;
combined with the complementary solution (see Problem 8.54), it yields the general solution
x = c e
2t
x
9.227
Solve
f
cos It + c 2 e
q + 400q + 200,000c/ = 2000(1
2'
sin 2f
- £f 2 + %t - j^ + {t 2 - 2t + l)e -3c
+ cos 200f).
A particular solution corresponding to 2000 is found in Problem 9.23 to be q — 0.01. A particular solution
q 2 = 0.005 sin 200f + 0.01 cos 200t\ Then a
{
corresponding to 2000 cos 200t is found in Problem 9.93 to be
particular solution to the given differential equation is
g,
+ q2
when combined with the complementary
\
solution (see Problem 8.70), it yields the general solution
q = e~
9.228
Solve
200
x + lOx + 25x = 20e
'(A cos 400f
-5
'
+ B sin 400r) + 0.01 + 0.005 sin 200f + 0.01 cos 200t
+ 320fV + 48fV - 66re 3 + 122e 3
'
'.
A particular solution corresponding to 20e~ 5 is found in Problems 9.144 and 9.145 to be x, = \0t 2 e~ s
2
3
3
a particular solution corresponding to (320f + 48f — 66f + 122)e
is found in Problem 9.73 to be
it
3
2
x 2 = (5f - 3f + 2)e
A particular solution to the given differential equation then is x, + x 2 When combined with the
I
'
';
'
.
.
complementary solution (see Problem 8.146), it yields the general solution
$l
x = (d + C 2 t)e- + 10rV 5 + (5f 3 - 3t 2 + 2)e 3
'.
'
3
9.229
Solve
2
d Q
dQ
d
- —Q
- 1250 =
—
f + 25
=-
dt
-=-
5
3
dt
2
1000(1
+ 5^"'cos2f).
dt
I A particular solution corresponding to a right side of 1000 is found in Problem 9.29 to be
particular solution corresponding to 5000e
Q 2 = 17^"'sin2t — 31e"'cos2f.
~5
d=—
A
8.
cos It is found in Problem 9.107 to be
A particular solution to the given differential equation then is
Q + Q2
x
;
when combined with the complementary solution (see Problem 8.113), it yields the general solution
Q = Cl e 5 + c 2 cos 5t + c 3 sin 5f - 8 + lie'' sin 2f - 31e"' cos It
'
9.230
Rework the previous problem if the term — 60e 7
'
is
added to the right side of the differential equation.
an d Q 2 of the previous problem remain valid. In addition, a particular solution corresponding to — 60e
7
is found in Problem 9.16 to be
Then a particular solution to the new differential equation is
Q 3 = —^e
f
(2i
'.
Qi + 0.2 + 63^
an d the general solution is
Q = Cl e 5 + c 2 cos5t + c 3 sin5t - 8 + 17e" sin2f - 31e"'cos2r - yfe
f
'
9.231
Solve
v
(4)
- 6y (3) + 16y" + 54/ - 225y = 100e~ 2x +
1
152 cos 3x - 3924 sin 3x.
7'
7'
"
THE METHOD OF UNDETERMINED COEFFICIENTS
f A particular solution corresponding to lOOe" 2x is found in Problem 9.17 to be
D
229
= -£?e~ 2x a particular
solution corresponding to
152 cos 3x - 3924 sin 3x is found in Problem 9.98 to be y = 8 sin 3x + 5 cos 3x.
2
Then a particular solution to the given differential equation is y, + y 2
When combined with the
y,
;
1
.
complementary solution (see Problem 8.124), it yields the general solution
3x
y = Cl e
9.232
+ c 2 e' 3x + c 3 e ix cos4x + c 4 e 3x sin 4x - f?e
2x
+ 8sin3x + 5cos3x
(D 4 - I6)y = 80x 2 + 60e* sin 3x.
Solve
I A particular solution corresponding to 80x 2 is found in Problem 9.53 to be y = — 5x 2 A particular
solution corresponding to 60e* sin x is found in Problem 9.106 to be y 2 — Y^e x sin 3x + ^ cos 3x. Then a
particular solution to the given differential equation is
when combined with the complementary
y + y2
.
x
;
x
solution (see Problem 8.128), it yields the general solution
y = c cos 2x + c 2 sin 2x + c 3 e
x
9.233
(D 2 + 2D + 4)y = 8x 2 + 12e _x
Solve
2x
+ c x e~ 2x — 5x 2 + Yie x sin 3x + Y$ex cos 3x
.
I The complementary solution is e~ x (c cos V3x + c 2 sin V3x). To obtain a particular solution, we may
assume the trial solutions ax 2 + bx + c and de~ x corresponding to 8x 2 and \2e~ x respectively, since none of
2
x
in the given
these terms is present in the complementary solution. Then substituting y — ax + bx + c + de~
x
,
equation, we find
4ax 2 + (4a + 4b)x + (2a + 2b + 4c) + 3de~ x = 8x 2 + \2e~ x
Equating corresponding coefficients on both sides of the equation and solving the resulting system yield
b =
— 2,
solution is
9.234
c
= 0, and d — 4. The particular solution is then 2x 2 — 2x + 4e~ x
x
2
x
y = e~ (c cos V3x + c 2 sin V3x) + 2x — 2x + 4e~
.
a = 2,
Thus the required general
.
x
Solve Problem 9.233 if the term 10 sin 3x is added to the right side.
f Corresponding to the additional term 10 sin 3x we assume the additional trial solution
h cos 3x
+ k sin 3x,
whose terms do not appear in the complementary solution. Substituting this into the equation
2
(D + 2D + 4)y = 10 sin 3x, we get (6k — 5h) cos 3x — (5k + 6h) sin 3x = 10 sin 3x, from which we find that
Then the required general solution is
k — — £?.
h = — gf and
y = e~
9.235
(D 5 - 3D
Solve
4
x
(c, cos
V3x + c 2 sin \J3x) + 2x 2 — 2x + 4e
+ 3D 3 - D 2 )y = x 2 + 2x + 3e x
— ^fcos3x — £ysin3x
*
.
m - 3m 4 + 3m 3 — m 2 =
m 2 (m — l) 3 = 0.
Thus m — 0, 0, 1, 1, and 1,
x
and the complementary solution is c, + c 2 x + (c 3 + c 4 x + c 5 x )e
2
2
Corresponding to the polynomial x + 2x, we would normally assume the trial solution ax + bx + c.
2
4
3
2
However, some of its terms appear in the complementary solution. Multiplying by x yields ax + bx + ex
which has no term that is in the complementary solution and so is the proper trial solution.
x
x
But since this term as well
Similarly, corresponding to 3e we would normally assume the trial solution de
3
x
2 x
x
our
assumed
trial solution is
we
must
use
dx
Thus
e
as dxe and dx e are in the complementary solution,
2
3 x
4
3
Substituting this in the given differential equation, we get
ax + bx + ex + dx e
5
I The auxiliary equation is
or
2
.
,
.
.
.
- 12ax 2 + (72a - 6b)x + [lib - 72a - 2c) + 6de x = x 2 + 2x + 3ex
from which we find a= —&, b — — f, c = — 9, and d = £.
2
4
3
x
3 x
2
y = Cl + c 2 x + (c 3 + c 4 x + c 5 x )e + \x e - ^x - fx - 9x
The general solution is then
.
9.236
Find a complete solution to the equation
I The characteristic equation is
2x
y" + 5y' + 6y = 3e' 2x + e 3x
m 2 + 5m + 6 = 0,
+ c 2 e~ 3x
and its roots are
.
m, = -2
m 2 = -3.
and
Hence the
complementary solution is c e~
2x
we would normally use Ae~ 2x However, e~ 2x is a part
For a trial solution corresponding to the term 3e~
3x
the normal choice for a trial
of the complementary solution, so we must multiply it by x. For the term e
2x
3x
+ Be 3x
is satisfactory as it stands. Hence we assume
solution, namely, Be
y p = Axe
~ 2x
Ae
+ 30Be 3x = 3e~ 2x + e 3x from
Substituting this into the differential equation and simplifying yield
2x
3
+ ^e *, and a complete solution is
which we find A = 3 and B = ^. Hence y p = 3xe~
.
x
.
.
,
,
v
= c e~ 2x + c 2 e 3x + 3xe~ 2x +
x
^
3x
-
230
9.237
CHAPTER 9
D
y" + 3/ + 2y = \0e 3x + Ax 2
Find a particular solution to the equation
.
3x
If we wished, we could find y p by beginning with the expression
Ae + Bx 2 + Cx + D, which means that
we would handle the various terms all at the same time. On the other hand, we can also find y p by first
/
3
finding a particular integral corresponding to 10e *, and then finding a particular integral corresponding to 4x 2
,
and finally taking y p to be their sum.
y" + 3y' + 2y = 10e 3x
Using the second method, we assume y t = Ae 3x
substitute into the equation
and
3x
2
—
A
—
find that
\e
Then we assume y 2 = Bx + Cx + D, substitute into the equation
\ and
y
y" + 3y' + 2y — Ax 2
and find after equating coefficients of like terms that B — 2, C — — 6, and D — 1.
2
3x
—
Hence y 2 — 2x
6x + 7 and, finally y p — y, + y 2 — \e + 2x 2 — 6x + 7.
,
,
.
{
,
9.238
(D 2 - 2D + 3)y = x 3 + sin x.
Solve
/ The complementary solution is y = e x (C cos -Jlx + C 2 sin V2x). As a particular solution try
3
2
y p — Ax + Bx + Cx + E + F sin x + G cos x. Substituting y p and its derivatives into the given equation then
c
x
yields
3 Ax
3
+ 3{B - 2A)x 2 + (3C - AB + 6A)x + (3£ - 2C + 2B) + 2(F + G) sin x + 2(G - F) cos x = x 3 sin x.
A = \,
Equating coefficients of like terms yields
C = % E = —j^, and F — G = j. Thus, a
—
\x 3 + f x 2 + |x — ^ + i(sin x + cos x), and the
yp
B — §,
particular solution of the given differential equation
is
,
primitive is
x
y = e {Ci cos V2x + C 2 sin >/2x) + 2J7(9x
9.239
(D 3 + 2D 2 - D - 2)y = e x + x 2
Solve
+ 18x 2 + 6x - 8) + |(sin x + cos x)
3
.
I The complementary solution is y — C e x + C 2 e~ x + C 3 e~ 2x We take as a particular solution
2
-2^x 2 - 2(B + A)x + (4A - B - 2C) + 6Ee x = ex + x 2
yp = Ax + Bx + C + Exe*. Substitution then gives
By equating coefficients of like terms and solving, we find that A = — \, B = j, C — — |, and E — £.
Hence y p — — 2 x 2 + 2 x — I + <,xr\ and the general solution is
x
x
2
2x \x + {x - %-\ \xe*.
y = C e + C 2 e~ + C 3 e~
.
c
l
x
.
l
x
9.240
Solve
y>
— 5y = x 2 e x — xe Sx
.
yc = c^e
I The complementary solution is
51
and the right side of the differential equation is the difference
x
2
two
each
in
manageable
form.
For
of
terms,
x e* we assume a solution of the form e (A 2 x 2 + A x + A
For xe 5x we would try the solution
,
).
t
e
5x
(fi,x
+ B
)
= B xe 5x + B e 5x
x
But this trial solution would have, disregarding multiplicative constants, the term e
therefore multiply by x to obtain
yp = e*(A 2 x
2
+ A x+ A
x
)
e
5x
(B l x
2
5x
in
common with yc We
.
Now we take y p to be the difference
+ B x).
- e 5x(B x 2 + B x).
l
Substituting into the differential equation and simplifying, we get
e
x
[(-4A 2 )x 2 + (2A 2 - A A ,)v + (/!,- 4/l )] + e 5x [(-2B )x - B Q ] - e x (x 2 + Ox + 0) + e 5x [(-l)x + 0]
l
A 2 = — |, A — — |, A
32'
ix
=
=
c
e
+ e x — jx 2 — |x — j^) — {x 2 e 5x
+
y
y
yp
Equating coefficients of like terms and solving the resulting system yields
B = \,
x
9.241
and
6 = 0.
Then the general solution is
Determine the form of a particular solution for
y'
c
x
{
x
.
— 5y = (x — 1) sin x + (x + 1) cos x.
f The solution to the associated homogeneous equation is shown in Problem 8.34 to be y h — c e 5x An
assumed solution corresponding to (x — l)sinx is (A x + A )s'mx + {B x + B )cosx, and no modification
is
(C,x + C )sinx + (D x + D )cosx.
(x + l)cosx
is required. An assumed solution corresponding to
(Note that we have used constants C and D in the last expression, since the constants A and B already have
.
l
x
x
x
been used.)
We therefore take
y p = (A x + A )sinx + (B { x + B )cosx + (CjX + C )sinx + (D,x + D )cosx
x
Combining like terms we arrive at
9.242
yp = (£,x + £
)
sin x
Solve the differential equation of the previous problem.
+ (f ,x + £
)
cos x
as the form of the particular solution.
THE METHOD OF UNDETERMINED COEFFICIENTS
/
231
Substituting y p of the previous problem and its derivative into the differential equation and simplifying, we obtain
(-5£, - £,)xsinx + (-5£ + £, - £ )sinx +(-5F, + E,)xcosx + (-5F + £ + £,)cosx
= x sin x —
1
sin x
+ lx cos x +
1
cos x
Equating coefficients of like terms and solving the resulting system of equations lead to
£i = ~T3,
and
£ = -^.
Solve
I
— fc,
71
E = 33S-
By combining these results with ye and yp we obtain the general solution
,
y = Cl e
9.243
£, =
5x
+ (-& x + 3&) sin x - (^ x + 36398 cos x
)
/ - 5>- - 3e x - 2x + 1.
y h of Problem 9.241 is valid here. The right side of the given differential equation is the sum of two manageable
x
For 3e x we assume a particular solution of the form A e ix for — 2x + 1 we
x
assume a solution of the form B t x + B
Thus, we try y p = A e + B,x + B
Substituting y p into the differential equation and simplifying, we obtain
— 4^0)^ + ( — 5B,)x + (i?! - 5B ) = 3e x + ( — 2)x + 1. Then by equating coefficients of like terms, we find that
(
A = — |, B, = |, and B = — ^5. Hence the general solution is y = c e 5x — \e x + fx — ^5.
functions: 3e
and
— 2x + 1.
;
.
.
x
v
CHAPTER 10
Variation of Parameters
FORMULAS
10.1
Discuss the solution of the nth-order linear differential equation
n
d y
d?
,
+ a "~ l(x)
d
n~ l
y
+ " + ai{x)
[
d^
d2y
'
+ fllW
d~x>
dy
^+a
° {x)y
= ^ {X)
U)
by the method of variation of parameters.
I
Let yi(x), y 2 (x),
equation
— that
is,
.
.
.
,
y„(x) be a set of n linearly independent solutions to the associated homogeneous differential
(/) with
= 0.
</>(x)
The complementary function is
yc = c y (x) + c 2 y 2 (x) +
1
+ c n y„(x)
l
(2)
A particular solution to the nonhomogeneous differential
where c u c 2 •••.<„ denote arbitrary constants.
,
equation is
y P = ^i>'i + v 2 y 2 +
where
y\
= y,(x)
for
i
= 1, 2,
.
.
.
,
and
n,
r, (i
+ v„y„
= 1,2,..., n)
(3)
an unknown function of x which still must
is
be determined.
To find the r,, we first solve the following linear equations simultaneously for the v\:
'"'a.V
i
"i/i
+ v'2 y 2
+
+
+
i
':>':
•
+
InXn
=
•
+
I'nV'n
=
•
+ i/By;- = o
•
•'
(4)
2)
+ v':^:
»>
+ r',
V'itf
c\y<?
1
,"
-2.
+
"+
.
•
•
2)
+ r;ylnl, = 0(x)
We then integrate each »J to obtain c,, disregarding all constants of integration. (This is permissible because
we are seeking only one particular solution.)
10.2
Finally, we find the general solution as
y = yc + y p
.
Specialize (4) of the previous problem to a third-order differential equation,
f
Equations (4) become
'•'i.Vi
+ ''2.V; + i-3.V 3 =
+ v'2 y'2 + v'3 y'3 v\y'[ + 14/2 + ^3/3 = <P( X
v\y\
10.3
Specialize (4) of Problem 10.1 to a second-order differential equation,
f
Equations (4) become
v\y\
10.4
+ v'2 y 2 =
v\y\
+ v'2 y'2 -
</>(.x)
Specialize (^) of Problem 10.1 to a first-order differential equation.
I
10.5
)
Equations (4) become
v\\\
= $(x).
Compare the method of variation of parameters to the method of undetermined coefficients.
f The method of variation of parameters can be applied to all linear differential equations. It is therefore
more powerful than the method of undetermined coefficients, which generally is restricted to linear differential
equations with constant coefficients and to particular forms of 0(x). Nonetheless, in those cases where both
methods are applicable, the method of undetermined coefficients is often the more efficient and, hence, the
preferable method, because no integration is involved. As a practical matter, the integration of v[{x) may be
impossible to perform.
232
VARIATION OF PARAMETERS
233
FIRST-ORDER DIFFERENTIAL EQUATIONS
10.6
x dy/dx + v = In x, for
x dy/dx + y = 0.
Solve
x > 0,
y = 1/x
if
is
one solution of the associated homogeneous differential
equation
—
+-y=dx
x
x
We first divide the nonhomogeneous differential equation by x, obtaining
In x,
which has the
y c = c,(l/x),
form of (7) in Problem 10.1; the lead coefficient is now unity. The complementary function is
we assume a particular solution of the form y p — v {l/x).
so
x
= - In x,
it follows from Problem 10.4 that
v\ - = - In x,
so that
v\ = In x.
Then
x
in which we have disregarded all constants of integration. Now we have
v x = j In x dx = x In x — x,
—
—
=
=
(x
In
x)(l/x)
In
x
and the general solution to the given nonhomogeneous differential equation
x
1,
yp
=
=
c^l/x)
is
In
X
1.
+
y
yc + y p
With
10.7
Solve
4>(x)
xx
+ 2xy = 4x.
y'
2
I The solution to the associated homogeneous equation is found in Problem 5.9 to be
y p = v t e~
assume a particular solution of the form
Since
Then
and (j)(x) — 4x,
y x — e~
2
x
=
—
v\
dx
§4xe* dx — 2e \
vx
J
x2
x2
y p - 2e e~*
2
= 2,
it
follows from Problem 10.4 that
in
v\e~
xl
— Ax,
which we disregard constants of integration.
and the general solution is
y c — ce~* ,
so we
so that
= 4xe*
x *.
y = yc + y p = ce~
x2
+ 2.
v\
2
.
Now we have
(Compare this problem with
Problem 5.43.)
10.8
Solve
- 5y = e 2x
y'
.
yc = c x e
I The complementary solution is found in Problem 8.34 to be
Here
and
is
10.9
yx
=e
5x
0(x) = e
and
2x
v\e
5x
= J v\ dx = j e~ 3x dx — — \e~ ix Now we have y p = — ±e~ ix e 5x = —\e lx
5x
— \e 2x (Compare with Problem 9.1.)
y = y + yp = c e
.
c
—e
2x
.
Then
v\
t
5x
.
= e~ 3x
and the general solution
,
.
x
y'-5y = 8.
Solve
v\e
5x
5x
we assume
,
.
c
yx = e
and
,
5x
.
_5x
= 8.
,
— J v\ dx = j &e~ 5x dx — — fe~ 5x
5x —
(Compare with Problem 9.18.)
y = y + yp = c e
f.
vl
yp = v t e
5x
Here (/>(x) = 8, and it
=
and
integration
gives
Thus, v\
8e
Then we have y p — — je~ 5x )e 5x = — f, and the general solution is
yc = c x e
follows from Problem 10.4 that
Solve
yp = v x e
so we assume
,
and it follows from Problem 10.4 that
,
vx
f As in the previous problem,
10.10
5x
(
x
y'
- 5y = 3x + 1.
yc — c e
§ As in Problem 10.8,
5x
follows from Problem 10.4 that
v\(e
5x
5
)
= 3x + 1.
Then
v\
and
y\
= e 5x
= (3x + l)e~ 5x
,
.
Here 0(x) = 3x + 1,
and integration gives
so it
= \v\ dx = J(3x 4- \)e- 5x dx = {-lx-%)e- ix
Vl
y p = (-fx - f^)e~
Now we have
y p = t^e *,
we assume
,
x
5x 5x
e
= -|x - ^,
and the general solution is
y = cYe
5x
- fx - ^.
(Compare with Problem 9.31.)
10.11
Solve
y'
— 5y = sin x.
5
Problem 10.4 that
v\e
5x
5x
yp = v e
from which we obtain
yc = c^e *,
f As in Problem 10.8,
= sinx,
we let
= \e
t>!
5x
,
x
and
v\
yx = e
_5
=e
(
Solve
y'-5y = 2e 5x
Here 0(x) = sinx. It follows from
and (using integration by parts twice)
.
*sinx
sinx^x = (— ^sinx — 26COSx)e"
Now y p = -^ sin x - ^ cos x)e $x e 5x = -^ sin x - ^ cos x,
5x y = Cl e
j§ sin x - j§ cos x. (Compare with Problem 9.77.)
10.12
5x
5x
so that the general solution is
.
I As in Problem 10.8,
5x
yc = c x e
= 2e 5x
Problem 10.4 that v\e
and the general solution is
,
5x
5x
yp = v x e
from which we obtain
,
we let
y = yc + y P = c x e
,
5x
+ 2xe 5x
.
and
= e 5x
Here
= 2<? 5 *.
It follows from
and
=|2dx = 2x. Then y p = 2xe 5x
(Compare with Problems 9.128 and 9.129.)
v\
y,
=2
.
i;
1
4>{x)
,
~
CHAPTER 10
234
10.13
Solve
- 3x 2 y - 12x 2
y'
.
I The complementary solution is found in Problem 5.11 to be yc — ce x \ so we assume y p — v e xi
xi
x
and (p(x) = 12.x 2
= 12x 2 or v\ = \2x 2 e' x \
it follows from Problem 10.4 that
With y, = e
v\e
— J \2x 2 e~ xi dx = — 4e~ x \ Then y p = — 4e~ x *exi = —4, and the general solution is
v
Integration gives
x3
y = yc + y P = <^ - 4.
.
x
*
,
,
x
10.14
Solve
- 3x 2 y - - 12x 3 + 4.
y'
x
yc = ce \
f As in the previous problem,
It
follows from Problem 10.4 that
v\e
x
and y = e x \ Here 0(x) = - 12x 3 + 4.
=
(- 12x 3 + 4)e ~*\ Then
v\
yp = v x e \
we assume
= - 12x 3 + 4
xi
so that
x
= ji-Ux 3 + 4)e~ xi dx= ( -I2x 3 x~ xi dx+ Ue' x dx
'
t;,
— 3x 2 e~ x3 dx) gives us
— \2x e~ dx = 4xe~ - 4e~ dx, so that v = 4xe~ x3 — J 4e _x3 dx + J 4e x3 dx = 4xe~ xi
J
xi
xi
xi
Then > p = (4xe~ )e = 4x, and the general solution is y = y + y p = ce + 4x.
u = 4x
Integration by parts (with
x3
3
dv =
and
xi
x3
j"
.
x
c
10.15
Solve
m
- 3xy = e 3x2 2
'
y'
yp = v e
'
.
l
y,
= x,
Solve
vt
'
it
v\
= 1.
Thus
'
'
.
.
— 3xy = — 6x.
y'
yc = ce
I As in the previous problem,
It
so we assume a particular
.
jx2 2
= e 3x2/2 — </>(x), and follows from Problem 10.4 that v\e 3x2 2 = e 3x2 2 or
2
and y p = xe 3x2 2
The general solution is then y = yc + y p — ce 3 * 2 + xe 3x2 2
Here
10.16
3xl 2
yc — ce
The complementary solution is found in Problem 5.10 to be
solution of the form
vt
.
follows from Problem 10.4 that
= j — 6xe
Vv
:
3x
dx = 2c
2
'
.
3 *1
we assume
,
v\e
Then
— e 3xl 2
Here 0(x) = — 6x.
— — 6x, from which we conclude that v\ = — 6xe 3x2
and
3x2 2
3x2,2 3xi 2
=
—
=
solution
is
ce
the
general
2.
and
+ 2.
2e~
e
y
yp
2
yp = v e
3x2 2
and
,
x
y,
.
3x2/2
2
2
10.17
y - 3xy = -dxe' 3 " 2
'
Solve
\\=ce 3x22
I As in Problem 10.15.
It
"'
y
Solve
= — 3xV,A
2
— 6xe 3x2/2
Then i, = j — 6xdx — — 3x 2
v\ = — 6x.
3x2
— 3x 2 e 3l2/2 = (c — 3x 2 )e~ 3x2/2
y = y + y = ce
3x2 ' 2
yp = v l e
—6xe 3x2 ' 2 or
we assume
,
follows from Problem 10.4 that
and
10.18
.
v\c
ix22 =
6x
y, = e~
.
,
.
and
= 18e
</>(x)
3jc
It
.
yc = Ae~
follows from Problem
bx
solution is
v\
y p — v t e~
so we assume
,
10.4 that
9x
= We9x and so r, = J lSe 9x dx = 2e 9*. Then y p = 2e
6x
=
+ 2e 3x (Compare with Problem 9.3.)
y
yc + y p = Ae~
conclude that
Solve
0(x) =
Here
.
c
I The complementary solution is found in Problem 8.37 to be
10.19
= e3* 2
y,
+ 6y=18e lx
y'
Here
2
and
2
The general solution is then
.
,
i/,e"
bx
6*
= 18e *,
= 2e 3x
e
3
6x
.
from which we
and the general
,
.
y + 6y - -2 cos 3x.
6x
Here
and y x =e~ 6x
As in the previous problem yc = Ae~ 6x we assume yp = v 1 e~
<p{x) — — 2cos3x.
6x
= -2cos3x or v\ = -2e 6x cos3x. Applying integration by
v\e
It follows from Problem 10.4 that
tx
— 2e cos 3x dx — — ^ sin 3x — ^ cos 3x)e bx Then
r =
parts twice, we find that
J
f
(
,
10.20
Solve
y'
+ 6y = 3e~ 6x
.
tx
~ 6x
- ^ sin 3x - fs cos 3x)e e
bx y = Ae
y$ sin 3x - fs cos 3x.
yp =
and the general solution is
.
,
,
(
= - ^ sin 3x - y§ cos 3x
(Compare with Problem 9.78.)
.
6x
6x
Here #x) = 3e 6x
and y,-f"
yp = v l e~
6x
6x =
from which we conclude that v\ = 3 and
3e~
It follows from Problem 10.4 that
v\e~
" 6Ar
6x
=
=
+ 3xe~ 6x (Compare with
and the general solution is y = Ae
3 dx
Then y = 3xe~
t!
3x.
I As in Problem 10.18,
y c = Ae~
6x
we let
.
.
.
,
,
J
p
.
,
Problems 9.130 and 9.131.)
10.21
Solve
iy - 5y = 2x 2 - 5.
I The complementary solution is found in Problem 8.38 to be
yx = e
5xi2
,
yc = Ae
Sx' 2
,
so we assume
yp = v t e
5x' 2
.
Here
but before we can determine 0(x). we must write the differential equation in the form of (/) of Problem
.
VARIATION OF PARAMETERS
10.1; that
get
is,
the coefficient of the highest derivative must be unity.
— \y =
y'
- f,
2
-x
so that
2
twice, we find
= J (x - \)e~
»,
5x/2
10.22
(3D - \)y = 6e 3x
Solve
M
5x ' 2
= x 2 - § or v\ = (x 2 - \)e~ 5x a
dx = -|x - £x + \^)e~ 5x!2
Then
v\e
Using integration by parts
.
2
(
.
V = (
Jp
\
— lx
4- i09\ p ~ 5x12 e 5x12 _ — 2 V 2 — 8 „
X 2 — -&-X
— 5*
25-* ^ 125/ e
X + 109
5
,
125
25
5xl2
- §x 2 - ^x + |§|
y = Ae
The general solution is thus
Dividing the differential equation by 2, we
= x 2 -f.
</>(x)
now follows from Problem 10.4 that
It
235
(Compare with Problem 9.33.)
.
.
(3D — l)y — 0,
The associated homogeneous equation is
which has as its characteristic equation
and as its characteristic value m = 1/3. The complementary function is yc = c^' 3
so we
assume a particular solution of the form y p — v^ " 3
Thus, y = ex 3
To apply variation of parameters, we must have the coefficient of the highest derivative in the differential equation
3x
equal to unity. Dividing the given differential equation by 3, we get (D — j)y — 2e 3x
so that
4>(x) = 2e
x3
3x
8x
3
Sx
3
8x/3
- 2e
It follows from Problem 10.4 that
Then v t = J 2e
v\e
or v\ = 2e
dx = |e
and
x/i
3x
±e 8xl3 e x 3 — %e 3x
—
=
=
The
general
>
solution
is
c
e
+
\e
yp
yc + yp
3m —
=
1
,
3
'
.
.
x
.
,
''
.
,
'
.
.
i
10.23
(2D - \)y = t 3 e" 2
Solve
.
dy/dt — \y — jt e
I We first divide the differential equation by 2, obtaining
3
tl2
which is in the form of (1) of
Problem 10.1 with t as the independent variable. The complementary function is found in Problem b.5 to be
2
hence we assume y p = v r e tl2
yc — ce"
2
3 12
from which we
Here y\ = e' 12 and (f)(t) = \t 3 e ia
It follows from Problem 10.4 that
v\e" = \t e'
4
3
2
conclude that v\ = \t
and v —^t
Then y p — ^rV
and the general solution is
2
2
y = y c + >' P = ce" + y*e"
,
.
;
.
,
.
{
.
10.24
dy/dt + e'y = e'.
Solve
I The complementary solution is found in Problem 5.14 to be yc — ce~ e \ so we assume y p — i\e~ et
It follows from Problem 10.4 that
v\e~ = e', from which we conclude
Here y\ — e~ e
and 0(f) = e'.
.
e>
'
that
=e'e
i',
y = ce
~e
'
+
e
'
and that
dx
—
+-x=
Solve
= J e'e e dt = e e
'
'.
y p = e 'e~
e
Then
= 1,
e'
and the general solution is
1
1
10.25
vj
dt
2
t
.
t
I The complementary solution is found in Problem 5.16 to be x = c/t, so we assume a particular solution
x p = vjt.
of the form
2
2
from which we conclude
v\(\/t) =
It follows from Problem 10.4 that
Here x, = \it and 4>{t) =
3
4
3
4
3
= /4. Then x p = (f / 4 )/t = if
and the general solution is x = c/r +
and
that
v\ =
f
t
f
t>,
dx
—
+-x =
f
.
f
,
4-f
,
.
1
10.26
Solve
dt
sin It.
t
I As in the previous problem.
It
v,1
x c = c/t,
then follows from Problem 10.4 that
=
f
J
tsinltdt =
— sin7f - - cos
49
c
1
t
49f
x = - H
„
sin 7f
we assume
v\(\/t)
so that
7f,
x p = vjt,
= sin It
x. =
or
x t = 1/f.
and
=
t
Here
</»(?)
= sin It.
Integration gives
sin It.
—- sin It - - cos It.
The general solution is
7
49/
7
v\
1
— - cos 7f.
7
SECOND-ORDER DIFFERENTIAL EQUATIONS
10.27
y" - 2/ + y = e x /x.
Solve
I The complementary solution is found in Problem 8.143 to be
x
yc =
c^ + c xex
2
;
x
y p = v e + v 2 xe
x
Since y\ = e
.
x
,
y 2 = xe
x
,
and
4>{x)
v\e
x
= e x /x,
+ v'2 xe x =
it
follows from Problem 10.3 that
v\e
x
e
x
+ v'2 (e x + xe x = —
)
hence we assume
j
CHAPTER 10
236
Solving this set of equations simultaneously, we obtain
— —1
v\
yp — —xe
and
10.28
x
+ xe In |x|.
y" - 2/ + y = ex /x 2
Solve
— 1/x.
v'2
Thus,
1
A = f-ldx=-x
v
x
and
= j-dx = ln\x\
v2
x
y = yc + y p = c x e
The general solution is therefore
x
+ (c 2 — l)xe x + xe x In |x|.
.
I The general solution of the associated homogeneous differential equation is found in Problem 8.143 to be
x
x
hence we assume y p = v e x + v 2 xex
y = c e + c 2 xe
x
x
Since y = e
and $(x) = ex/x 2
it follows from Problem 10.3 that
y 2 — xe
c
;
x
,
x
.
x
,
,
v\e
+ v'2 xex —
x
v\e
Solving this set of equations simultaneously, we obtain
vi
W+
y = yc + y P = fci y" - 2y
Solve
m
It
\\.
\
,, and y 2 of the
and
— 1/x 2
v'2
Then
.
= JV2 dx = f —2 dx
v2
x
The general solution is then
.
.
y p = r,e* + v 2 xe
previous problem remain valid, and we assume
x
Here
.
</>(x)
— ex/x 3
follows from Problem 10.3 that
"i** + v'2 xe
x
=
v\e
x
Solving this set of equations simultaneously, we find that
v.
and
- e x in |x|.
K
c 2 xe
+ y = e x /x 3
+ v'2 (e x + xe*) = —
— —1/x
v\
— dx=-ln|x|
xe x = — e x In |x| — e x
y p — —In |x|e*
so that
10.29
=£v\dx=[
x
\
r
=
dx — 1/x
r
xz
J
and
=
v2
r
J
1
—Jr ax = —
x
—
2x
)
Thus,
r.
= -
v\
1
z
+ v2 (e* + xex — —3
x2
I
and
/
1
y. =
1
— —^z
h I
xe
v'2
\
— 1/x 3
xe* =
—- ? x
1
.
and the general
2x
2x y
\
Then
.
solution is
= y + y P =c l e* + c 2 xe x +
v
10.30
c
e
x
.
2
yc = c e
f The complementary solution is found in Problem 8.143 to be
x
y p = v e + v 2 xe
y
= ex
x
,
y 2 = xe
x
,
and
= e 2x
<j>{x)
x
=
j"
e
dx = e
x
.
y = yc + y P = c i e* + c 2 xe
x
y" + 6y' + 9y = e~
Solve
so we assume
follows from Problem 10.3 that
+ v'2 xe =
y p = { — xe
Thus,
It
.
x
Solving this set of equations, we find that
v2
+ c 2 xe*,
.
v\e
x
x
x
x
l
Here
10.31
1
y" - 2y' + y = e *.
Solve
and
—
v\e
x
+ v'2 {e x + xex = e 2x
)
= —xe x and v'2 = e x Then v x = J -xe x dx = —xex + ex
+ e")ex + ex xex = e 2x and the general solution is
v\
x
.
,
+ e 2 x.
3x
/x
5
.
yc = c e~
I The complementary solution is found in Problem 8.142 to be
3x
+ v 2 xe' 3x
y p = v e~
x
3x
+ c 2 xe~ 3x
;
hence we assume
.
x
Here
y,
= e~ 3x
,
y 2 = xe'
v\e~
3x
3x
,
5
v\
= —x
-4
and
= f-x~ 4 dx = 5X" 3
~~
3x
3x
3
= j^x
+ ( — ^x " *)xe
y p = jx " e ~
3x
3
3x
=
c
xe~
C\e~
+ 2
+ Y2X~ e~ 3x
y
yc + y P
Then
3x
It
.
follows from Problem 10.3 that
v\{-le~ 3x ) + v'2 (e-
+ v'2 xe~ 3x =
Solving this set of equations, we get
t>,
0(x) = x~ e~
and
.
~3
e~
v'2
and
3x
,
=x
-5
t?
,
2
3x
- 3xe~ 3x = x^e" 3 *
)
from which
x" 5 (fx = -5X" 4
=
J
and the general solution is
.
VARIATION OF PARAMETERS
10.32
237
y" + 6/ + 9 y = lOOe2x
Solve
f The complementary solution is found in Problem 8.142 to be y = c e 3x + c 2 xe~ 3x so we assume
3x
+ v 2 xe~ 3x
y p = v e~
3
2
Here yi = e~ 3x
y 2 = xe~ *, and 0(x) = lOOe *, so the results of Problem 10.3 become
c
,
x
.
t
,
t/,<?-
^(-Se -3 *) + v 2 (e~ 3x - 3xe' 3x = 100e
+ v'2 xe- ix =
3*
Solving this set of equations simultaneously, we obtain
vt
2jc
)
= — lOOxe 5 *
v\
= f - lOOxe 5 * dx = - 20xe 5 * + 4e 5 *
and
u2
and
= lOOe 5 *,
«/
2
from which
= f 100e 5 * dx = 20e 5 *
-3
5
Sjc
5x
3x
= 4e 2x The general solution is
y p = ( — 20xe * + 4e )e * + 20e xe'
3x
ix
2x
(Compare this with the result of Problem 9.9.)
+ c 2 xe~ + 4e
y = yc + y p — c t e~
Then
.
.
1033
y" + 6y' + 9y = 12e~ 3 *.
Solve
y c = c t e~
I The complementary solution is found in Problem 8.142 to be
3x
+ v 2 xe~ 3x
y p = v t e~
3x
+ c 2 xe~ ix
so we assume
,
.
y t = e~
Here
3x
,
y 2 = xe~
3x
v\e~
3jc
and
,
</>(x)
= 12e~
3jc
follows from Problem 10.3 that
It
.
3x
v\(-3e~ 3x ) + v'2 (e~ 3x - 3xe'
+ v'2 xe~ 3x =
= \2e~ 3x
)
= 12. Then
and
v\ = — 12x
2
3x
3x
2
2
=
=
=
= 6x 2 e~ 3x and the
Thus,
-6x
e~
\2x 2 e~
and
12
dx
12x.
+
-12xdx=
-6x
v
yp
On=J
2
J
3x
3x
3x
2
(Compare with Problems 9.140 and 9.141.)
general solution is y = y + y p — c e~
+ c 2 xe~ + 6x e~
Solving this set of equations simultaneously, we obtain
v'
,
.
c
10.34
x
(D 2 - 6D + 9)y = e
Solve
3x
/x
2
.
m The complementary solution is y c = c t e 3x + c 2 xe 3x so we assume
3x
and 0(x) = e 3x/x 2 ] that
Problem 10.3 [with y = e 3x y 2 = xe
yp = v e
,
3
i/,e *
+ i/,xe
Solving this set of equations, we obtain
3jc
=
v\(3e
is
1035
v2
=\—2dx=
—
v\
—
—
y = yc + y p = (c x - l)e
y" - ly' = 6e
Solve
3x
3x
)
1
y p = ( — In \x\)e
Then
.
)
1
and
3x
+
v'2
—
I
——
J
x^
\xe
,
x
v'2
+ v'2 e 7x =
= %e~ x
and
= ji/l dx = j-%e 6x dx = -$e6x
x
Thus y p = -4e6 * + (-je~ )e
(Compare with Problem 9.7.)
= — e 3x \n \x\ — e 3x
3x
—
1
dx — —In Ixl
X
and the general solution
,
lx
= -e 6x
,
v\(0)
v\
yc = c x + c 2 e
.
;
hence we assume
follows from Problem 10.3 that
+ v'2 {le lx = 6e 6x
= -je
and
It
lx
)
6x
Then
.
v2
= §v'2 dx =
x
fe~
dx = -fe"*
y = ye + y p = c x + c 2 e
and the general solution is
lx
- e 6x
.
y"-7y'=-3.
Solve
yp = v x + v 2 e
and again we assume
y c y x and y 2 are as in the previous problem,
(f>(x)
/•
=
.
Solving these equations, we obtain
f
u,
+ c 2 xe 3x - e 3x In |x|.
v\(\)
1036
so that
=•,
6x
.
Vl
follows from
+ v'2 (e 3x + 3xe 3x = -T
I The complementary solution is found in Problem 8.2 to be
lx
lx
and (/>(x) = 6e 6x
Here y = 1, y 2 = e
y = Vl + v 2 e
p
It
.
x
X
and
+ v 2 xe 3x
,
,
t
3x
x
,
,
= -3.
It
lx
.
Here, however,
follows from Problem 10.3 that
v\(l)
+ v'2 e 7x =
^(0) + v'2 {le
lx
)
= -3
lx
3
3
so that
dx = ix and
v1 =
and v'2 = - e~
The solution to this set of equations is v\ =
ff
x
7
lx
lx
general
solution
is then
The
=
=
+
^e"
=
^.
fx
Thus y p
fx +
^e~
dx
v2 =
J -fe"
7A:
and
9.154
9.155.)
Problems
with
+ fx. (Compare
y = yc + y = c + ^ + c 2 e
.
p
x
V
,
CHAPTER 10
238
10.37
y" -ly' = -3x.
Solve
f
It
+ v 2 e lx
y p = t-,
y c y u and y 2 are as in Problem 1035, and again we assume
,
0(x) =
Here, however,
.
-3x.
follows from Problem 10.3 that
v\(l)
+ v'2 e 7x =
The solution to this set of equations is v\ = fx
" lx
~
~
dx = ^xe 7x + ^e lx
v 2 = J -fxe
Thus
r',(0)
and
v'2
+ v'2 (le lx = -3x
)
= -^xe~ lx
Then
.
=
y,
\
}x dx = -^.x
2
and
.
yp = A-x
and the general solution is
v
+ (^xe~ lx + jhe- lx )e lx = ±x 2 + £x + 343
2
=y + y, = c +-£s + c e lx + ^x 2 + ^x.
e
(Compare with Problems 9.156 and
2
1
9.157.)
10.38
y" - y
Solve
- 2y = e 3x
.
>'p= V e ~*
+ V 2 e2X
Here
= e~ x
l
y,
y2 = e
,
2x
,
0(x) = e
and
t>,
r',
I
''
\\,
so we assume that
,
follows from Problem 10.3 that
It
Di(-e~*) + v'2 (2e
= — \e* x
v2
— \e x
and
v2
and
2x
)
= e 3a
so that
,
= f \e x dx = \e x
y = y,
and the general solution is
>
y" - y' - 2y = 4.x 2
Solve
.
= f - \e* x dx = - iV 4 *
Then y p = — 2 4v x + V'^'' x = J^ 3x
(Compare with Problem 9.10.)
t'
3x
+ v'2 e 2x =
x
The solution to this set of equations is
10.39
+ c 2 e 2x
-
i\e
,'
x
yc = c x e~
I The complementary solution is found in Problem 8.1 to be
+ y p = c x e~* + c 2 e 2x + \e ix
.
.
y,, and y, are as in the
yp = v t e
previous problem, and we let
+ v 2 e 2x
x
Here
.
$(x) — 4x
2
It
.
follows
from Problem 10.3 thai
x
r\c
+ i'2 e 2x =
The solution to this set of equations is v\ — — fxV
twice on each successive integral, we calculate
r,
Thus
}'
10.40
—
=
y,,
)',
=
fxV dx -
|'
<-'\
- 2x 4- 2]e*
2
= %x 2 e~ 2x
/',
and
)
)
2x
- \\x 2 c
r2
Then, using integration by parts
.
2
2
:
'
2x
dx = -\(2x 2 + 2.x + \)e
— 2x + 2)eY
',(2v
+ 2.x + )c V = —2.x 2 + 2x — 3,
2x
2
— 2x f 2x — 3. (Compare with Problem 9.35.)
e ~* + c 2 e
}(.x
+ y,, —
f(x
and
+ v\j2e lx = 4x 2
x
r,|-i'
2jt
and the general solution is
1
y" — y' — 2y = sin 2.x.
Solve
yp = r,e
f y. y,, and y, are as in Problem 10.38, and we let
Problem
+ p2 e 2x
v
Here
.
0(.x)
= sin 2.x.
It
follows from
10.3 that
v \c
*
+ v'2 e 2x =
Solving this set of equations simultaneously yields
v'ii-e'
1
+ v'2 (2e 2x = sin 2.x
)
)
= — \e x sin 2.x
v\
and
r'
2
= jt'^ 2 * sin 2.x.
Then, using
integration by parts twice on each successive integral, we obtain
r,
=
— %ex sin2xdx = — ^^(sin 2x — 2cos2x)
yp =
Thus
solution is
10.41
v2
=
je"
2x
sin2xdx = — ^2 e~ 2x {s\n 2x + cos2x)
— ^€x(sin2x — 2cos2.x)e
x
— j^e" 2x(sin2x + cos 2x)e 2x = — 2%sin2x + 2ocos2x, and the general
x
2x —
=
2%sin2x + 20 cos 2.x. (Compare with Problem 9.92.)
y
yc + y p — c e' + c 2 e
y" — y' — 2y = e 2x
Solve
and
l
.
I yc y,, and y 2 are as in Problem 10.38, and we assume
,
yp = v e
1
x
+ v 2e 2x
.
Here
0(x) = e
2x
.
It
follows
from Problem 10.3 that
v\e'
x
+ v'2 e 2x -
v\(
-e~ x + v'2 {2e 2x = e 2x
)
)
The solution to this set of equations is v\ = —^e ix and v'2 = \: integrating directly gives
~
Then y p = —$e 3xe x + \xe 2x = (j.x — \)e 2x and the general solution is
v 2 = j.x.
x
+ (c 2 - k)e 2x + \xe 2x (Compare with Problems 9.134 and 9.135.)
y = yc + y p = t,f
,
.
i\
= —^e ix
and
VARIATION OF PARAMETERS
10.42
d2y
dy
- 4 -f + y = 3e
-f
2 "
dt
dt
Solve
D
239
2 '.
# The complementary solution is found in Problem 8.9 to be
p^-73* + 2 e 268
yp =
= c e 3132 + C 2 e° 268
'
ye
so we assume that
'
i
"
'.
yt = e
Here
3,7321
y2 = e
,
Die
0MSt
and
t
3 7321
'
<£(«)
= 3e 2
+ B'2e 268 =
-
u'1 (3.732e
'
The solution to this set of equations is
v\
= 0.866e _1 li2
Dj= J0.866e _1 732, <it= -0.5e -1 732
-
-
_1732
so that
y = de
10.43
y p = -0.5e
3132
026S
'
+ c2e
d2y
'
'e
3 732f
-e
2
'.
follows from Problem 10.3 that
It
'.
+ i'2 (0.268e 0268 = 3e 2
')
and
'
= -0.866e' 732
i/2
'.
Then
= J-0.866e' 132t dt = -0.5e 1132
v2
'
'
')
'
V° 268 - -«? 2
The general solution is then
(Compare with Problem 9.11.)
- 0.5f
732
3 - 732
J
'
'.
dy
f-4-p + y = 3t-4.
—
dt
dt
Solve
.
2
I yc y
x ,
0^3.7321 +
^ eo.268« = Q
The solution to this set of equations is
Then
and
v\
p
3/T32t
+
)
3732
and
'
+ v 2 e°- 2b8t
268 ') =
2 (0.268e°
i''
= (0.866r - 1.155)e -3,732
u'2
3f
= J(0.866f- I.155)e
= (-0.232r + 0.248 )e
3.732(
v2
= J(-0.866f + 1.155)6--° 268 dt = (3.231t + 7.748)?
0.268f
'</f
Here
.
-4
= (-0.866f + 1.155)e"° 268
t>,
'
'.
= (-0.232f + 0.248)e~ 3 732 'e 3 7321 + (3.2311 + 7.748)?'° 268, e° 268 = 2.999f + 7.996
-
>'
'
v 1 (3J32e
3132t
yp = v t e
and y 2 are as in the previous problem, and we assume
It follows from Problem 10.3 that
0(f) = 3t — 4.
,
-
y = ye + yp = C,c
Thus, the general solution is
'
3
-
732 '
+ C 2 ?°- 268 + 2.999f + 7.996.
'
(Compare this result
with that of Problem 9.36; the differences are due solely to roundoff.)
+ 4 — + 8.x = e
—V
dt
dt
dx
d x
10.44
Solve
~2
'.
2
I The complementary solution is found in Problem 8.54 to be xc = e ? _2 'cos2f + c 2 e 2 'sin 2t, so we
assume a particular solution of the form x p = v x e~ 2t cos 2f + v 2 e 2l sin 2t.
_2
It follows from Problem 10.3 (with x replacing
Here Xj = e~ 2 cos2t, x 2 — f 'sin2f, and <p(t) = e~ 2
1
'
'.
>')
that
2t
cos2t + v'2 e~ 2 'sm 2f =
2'
s'm2t + 2e~
2t
— ^cos 2f,
and integration yields
v\e~
y'
1
(-2e'- 'cos2f - 2e~ 'sin2f) + v'2 (-2e
2
2
The solution to this set of equations is
and 1 2 = | sin 2f. Then
xp = (i cos 2t)e "
The general solution is then
2'
v\
= — \ sin 2f
cos It + (i sin 2f )(«? ~
x = c x e~ 2t cos 2f + r 2 e
_2
and
2'
v'2
sin It) =\e~
'sin2f + If
2,
-2
'.
(cos
2
cos2t) = e~
2t
vt
— |cos 2f
+ sin 2 It) = %e~ 2t
2f
(Compare with Problem 9.13.)
dx
d
—
+ 4— + 8x = 16cos4f.
x
10.45
Solve
^-
dt
I x
4>(t)
c,
2
dt
x,, and x 2 are as in the previous problem, and again we let
= 16cos4f.
It
x
2
'cos2f + r 2 t'' 2 sin 2f.
'
Here
follows from Problem 10.3 (with y replaced by x) that
2'
v\ e
2
(-2e>- 'cos2r - 2e~ 'sin2f) + v'2 (-2e
2
t/1
The solution to this set of equations is
i?!
xp — v e
v\
= - 8?
2'
cos 4f sin 2?
= f 2, (sin 2f - cos 2f - ^ sin 6f + | cos 6f)
v2
sin It
and
+ 2c
v'2
=
'cos2f) = 16cos4f
2I
cos 2r + v'2 e
~ 2'
2
sin It
= 8f 2 cos At cos 2f.
'
Integrating yields
= e 2 '(sin 2f + cos 2f + J sin 6f + 5 cos 6f)
)
.
CHAPTER 10
240
Then
.x
p
= (sin 2f - cos It — \ sin 6f + § cos 6t) cos It + (sin 2f + cos 2f + § sin 6f + \ cos 6f) sin It
= 2 sin 2f cos 2f - (cos 2 It - sin 2 2t) - ^(sin 6f cos 2f - cos 6f sin It) + § (cos 6f cos It + sin 6f sin It)
= sin 2(2f) - cos 2(2f) - ^ sin (6f - 2f) + 1 cos (6f - 2f)
= f sin 4f — § cos At
x = c e~ 2t cos It + c 2 e~ 2t sin2t + f sin4f — f cos4r.
and the general solution is
10.46
(Compare with Problem 9.83.)
x
x + 25.x = 5.
Solve
I The complementary solution is found in Problem 8.72 to be xc = C, cos 5t + C 2 sin 5f, so we assume a
particular solution of the form
xp = r, cos 5f + v 2 sin 5t.
Here x, = cos 5f. x 2 = sin 5f. and 0(f) = 5. It follows from Problem 10.3 [with .x(r) replacing y(.x)] that
v\ cos 5t
+ d'2 sin 5f =
The solution to this set of equations is
»i = 5 cos 5f
and
= 5 sin 5f,
v2
i/x
= —sin 5f
r',
(— 5 sin 5f) + v'2 (5 cos 5f) = 5
and
v'2
= cos 5t.
Then integration yields
so that
x p — 5 cos 5f cos 5t + \ sin 5f sin 5f = |(cos 2 5f + sin 2 5r) = j
x = xc + x
The general solution is then
10.47
Solve
.x
+ 25.x = 2 sin 2f
I
\,.
0(f)
= 2sin2f.
v,.
and .x 2 of the previous problem are valid here, and again we assume
It follows from Problem 10.3 (with y replaced by x) that
(',
cos5f + r'2 sin 5r =
The solution to this set of equations is
=
f,
= C, cos 5f + C 2 sin 5f + \.
p
I
I
sin 2/ sin 5f dt
t',
5
,'
s
and
and
sin 7f
v2
=
v2
= § sin 2f cos 5f.
2
sin 2r cos 5f dt
Then
= 75 cos 3f — ^ cos 7f
x p = ( - v5 sin 3f + 35 sin It) cos 5f + (,'5 cos 3f - 3 5 cos It) sin 5f
and
!
=
,'
5
(sin 5f cos 3t
— sin 3f cos 5f) + ^(sin It cos 5f — cos It sin 5r
= ^5 sin (5f - 3f) + 35 sin {It - 5f) =
The general solution is
10.48
Also,
-5 sin 5r) + i'2 (5cos 5f) = 2 sin 2r
(',(
= - 2 sin 2f sin 5t
= — / sin 3l +
x p = r, cos 5r + o 2 sin 5f.
x = C, cos 5t + C 2 sin 5f
2
42 ,
sin 2t.
2
2 ,
sin It
(Compare with Problem 9.85.)
x + 16.x = 80.
Solve
= c, cos4r + c 2 sin4f, so we assume a
I The complementary solution is found in Problem 8.57 to be
=
particular solution of the form
c, cos4f + v 2 sin 4f.
p
Here x, = cos4f, x 2 = sin4f, and 0(f) = 80. It then follows from Problem 10.3 (with x replacing y) that
.x
c
,x
[',
cos 4f + v 2 sin 4f =
t',(
-
-
4 sin 4f
)
+ i'2 (4 cos 4f) = 80
The solution to this set of equations is r', = — 20sin4f and r 2 = 20cos4f. Integration yields r, - 5cos4f
= 5sin4t, so that .x p = (5 cos4f)cos4f + (5 sin4f)sin4f = 5(cos 2 4f + sin 2 4f) = 5. The general
and
2
solution is then
x = xc + x p = c, cos 4f + c 2 sin 4f + 5.
i'
10.49
x + 16x
Solve
= 2sin4f.
x p = i\ cos 4f + t- 2 sin 4f.
I xc x,. and x 2 of the previous problem are valid here, and again we let
It follows from Problem 10.3 (with x replacing y) that
0(f) = 2 sin4f.
,
c\ cos 4f
+ i/2 sin 4f =
The solution to this set of equations is
V{
=
-
i
t'2
v\(— 4 sin 4f) + r 2 (4 cos 4f) = 2 sin 4f
=-|sin 2 4f
Uin : 4f dt = -jf + yjsinSf
and
and
r2
= { sin 4f cos 4r.
r2
=
J
Then
2sin4f cos4f dt = ^ sin
But sin 8f = sin 2(4f = 2 sin 4r cos 4f,
4f sin4f.
= - {f cos 4f + yg sin 4r. The general solution is then
x = c, cos4f + (c 2 + 6 )sin4f - |f cos4t. (Compare with Problems 172 and 173.)
and
x p = ( -\t + y2 sin 8f) cos 4f +
simplification,
xp
,'
1
1
6 sin
Also,
2
)
2
4f
so that, after
-
VARIATION OF PARAMETERS
10.50
D
241
y" + y = secx.
Solve
I The characteristic equation of the associated homogeneous differential equation is A 2 + = 0, which
admits the roots A = ±i. The complementary function is y = c, cos x + c sin x, and we assume a
2
particular solution of the form
y p = v cos x + v 2 sin x.
Here y\ = cosx, y 2 = sinx, and cj)(x) = secx. Then it follows from Problem 10.3 that
1
c
t
+ v'2 sin x =
v[ cos x
v\ (
— sin x) + v'2 cos x = sec x
— — tanx and v'2 — 1. Then v 2 = x and
y p = (In |cosx|)cosx + xsinx. The general solution is then
y = c cos x + c 2 sin x + (In |cos x|) cos x + x sin x.
The solution to this set of equations is
v = j — tanxdx = In |cosx|,
so that
v\
x
x
10.51
Solve
y" + Ay = sin 2 2x.
yc = c t cos 2x + c 2 sin 2x,
I The complementary solution is found in Problem 8.59 to be
y p = Tj cos 2x + v 2 sin 2x.
With y = cos 2x, y 2 — sm 2x,
and
1
v\ cos 2x
= sin 2 2x,
+ v 2 sin 2x =
The solution to this set of equations is
»!
</>(x)
v\
y' (
x
follows from Problem 10.3 that
— 2 sin 2x) + v'2 {2 cos 2x) = sin 2 2x
= — ^sin 3 2x
and
— |sin 3 2xdx = |cos2x — t^cos 3 2x
=
it
so we assume
— \ sin 2 2x cos 2x.
v'2
and
v2
=
)
i sin
2
Then
2x cos 2x dx =
^ sin 2x
3
= ^cos 2 2x — j2(cos 4 2x — sin 4 2x). But cos 4 2x — sin 4 2x = cos 2 2x — sin 2 2x, so
2
y p = \ cos 2x + Y2 sm2 2x. Then the general solution is y — c cos 2x + c 2 sin 2x + ^cos 2x + ji sm2 2x.
so that
_v p
2
{
10.52
Solve
y" + 4y = csc2x.
I
y c and the form of y p assumed in the previous problem are valid here. Also,
It then follows from Problem 10.3 that
0(x) = csc 2x.
v\ cos 2x
+ v'2 sin 2x —
v'2 (
_y,
= cos 2x,
y 2 — sin 2x,
— 2 sin 2x) + v'2 (2 cos 2x) = csc 2x
and v'2 =|cot2x. Then v = — \x and
v\ = —\
= J j cot 2x dx = \ In jsin 2x|, so that y p — —\x cos 2x + \ In |sin 2x| sin 2x. The general solution is thus
y = yc + V'p = Cj cos 2x + c 2 sin 2x — \x cos 2x + | In |sin 2x| sin 2x.
The solution to this set of equations is
x
r2
10.53
(D 2 + A)y = 4 sec 2 2x.
Solve
I yc y u and y 2 of Problem 10.51 are valid here, and again we let
,
0(x) = sec
2
2x.
y p — v cos 2x + v 2 sin 2x.
x
Also,
Then it follows from Problem 10.3 that
v\ cos 2x
+ v'2 sin 2x =
v\(
— 2 sin 2x) + i/2 (2 cos 2x) = 4 sec 2 2x
Solving this set of equations, we find
v\
= —2
——
sin 2x
^
cos
and
v2
2x
= 2 sec 2x
Then integration yields
,
sin 2x
JC
=
I
—2
dx — — cos
","T
2
l
2x = — sec 2x
cos 2x
2 sec 2x rfx =
J
The general solution is then
Solve
I
sec 2x + tan 2x
dx = In sec 2x + tan 2x
,
.
,
— 1 + In |sec 2x + tan 2x| sin 2x.
y p = —sec 2x cos 2x + In |sec 2x + tan 2x| sin 2x =
and
10.54
mn zxi
2x)
2x -tsec zxisec
2x(sec ax
+ tan
2
y = c, cos 2x + c 2 sin 2x - 1 + In |sec 2x + tan 2x| sin 2x.
x + 64x = sec 8f.
The complementary solution is found in Problem 8.58 to be
x p = r, cos8t +
i;
2 sin8t.
x f = c cos St + c 2 sin 8f,
x
so we assume
and
242
CHAPTER 10
U
x,
Here, also,
= cos St,
x 2 = sin 8f,
and
<j>(t)
= sec 8t.
It
follows from Problem 10.3 (with x replacing y)
that
v\
cos 8f + v'2 sin 8f =
v\(
— 8 sin St) + v'2 (S cos St) = sec 8f
v\ = — |sec St sin St = — gtan 8f
and v'2 — g. Then
= J — gtan8fdr = ^ln|cos8r| and w 2 = 8 f T nus x p = ^ In |cos 8r| cos 8r + gf sin 8r, and the general
solution is
x = x + x p = c, cos St + c 2 sin St + ^ In |cos 8f| cos 8f + gf sin St.
The solution to this set of equations is
vi
-
c
10.55
x + 64x = 64 cos 8f.
Solve
f
x c x 1; and x 2 of the previous problem are valid here, and we assume x p as in that problem. With
= 64 cos 8f, we have
,
0(f)
v\ cos St
+ v'2 sin St =
The solution to this set of equations is
vx
10.56
t?'
t(
- 8 sin 8f) + v'2 (S cos 8f) = 64 cos 8f
= — 8sin8fcos8f
v\
= f - 8 sin St cos St dt = \ cos 2 St
v2
and
v'
2
= 8cos 2 8f.
Integrating yields
= \S cos 2 St dt = At + \ sin 16f = 4f + \ sin 8f cos St
Then
x p = ^ cos 2 8f cos St + {At + \ sin 8f cos 8f
and the general solution is
x = (c, 4- ^)cos St + c 2 sin 8f + 4f sin 8f.
)
sin 8f
= 4f sin St + \ cos St
(Compare with Problems 9.174 and 9.175.)
Rework the previous problem, taking a different antiderivative for v v
I
Integrating v\ differently, we obtain
= j — 8 sin St cos St dt = — 2 sin 2 8f.
t>,
With v 2 in its original form, we
then have
x p = -\ sin 2 8f cos 8f + (4f + \ sin St cos 8f) sin St = At sin St
x = c, cos 8f + c 2 sin 8f + At sin 8f,
The general solution now is
which is identical in form to the previous answer
because c x denotes an arbitrary real number.
10.57
y - Ay + 3y = (1
Solve
+ e~V
'.
I The associated homogeneous differential equation is y — Ay + 3y = 0, which has as its general solution
3
We assume as a particular solution to the nonhomogeneous equation y p = v^e' + v 2 e il
yc — c e + c 2 e
follows from Problem 10.3 that
Now, with y = e', y 2 = e 3t and 0(f) = (1 + e~')~
f
'.
.
x
',
,
x
+ v'2 e 3t =
v\e'
The solution to this set of equations
is
M
v\'
v\e'
2
-
1
-1
+ v'2 (3e 3 = (1+
')
— and
+e~'
3t
=~
—
21 +e
e''
1
=
it
e
1
v'2
.
Setting
u =
+e
1
'
yields
'
rl
--—e — dt = - \-du
= -lnu = -]n(l+e
'
1
1
2\ + e
u = e
And setting
'
y p = [\\n{l
2
Solve
d y/dt
2
2
2
l«j-* + I e -«-iln(l +
-u 2 - u + In (1 + u)
+ £?"')>' + [-^" 2 + ^"' - |ln(l + e"')> 3 = \(e - e il)\n(\ + e<) - \e + \e 2x
and the general solution is
10.58
2 J u
yields
-4/("- 1+ rb)*-Then
'
l
f
'
y = {c l - {)e
- y = (1 + e~T 2
l
'
+ c 2 e 3 + j(e' - e 3
'
')
ln(l
+ e~') + }e 2
'.
-
I The associated homogeneous differential equation, d 2 y/dt 2 — y — 0. has as its characteristic equation
m 2 — = 0, which we may factor into (m — l)(m + 1) = 0. The characteristic roots are ±1. so the
complementary function is y — c e + c 2 e~'. We assume a particular solution of the form y p = r^' + v 2 e~'.
1
,
c
It
l
follows from Problem 10.3 that
v\e
l
+ v'2 e'' =
v\e'
+ v'2 (-e
')
= (1 + e -t\-2
')
—
=
'
'
VARIATION OF PARAMETERS
The solution to this set of equations is
—
v\
and
2
2(1 +e~')
—
=
v'-,
2(1
+e
1
1
-j.
Setting
u
D
= 1 +e
243
'
')
yields
-J
u = e
Setting
noting that
',
e
X
C
X
1
-e' + 1
,
-—
1/2
1
+ e _t
'
1
d+")
2
-J
du
2
1/2
— - e — In e
v\e'
2t
+
+ w) 2
1
+ u
du
he
In (1
+e
)
— \e — + e
+ e-'ln(l + e"*).
x
vp
1
+e
'
1
'
+ v'2 e
~
'
so that the general solution is
'),
'.
—
The solution to this set of equations is
we obtain
Setting
u = e~',
j(e~
(1
7 + 111(1 +<?-')
1
y c and the form of yp assumed in the previous problem are valid here.
—
u
e
1
d 2 y/dt 2 — y = e~' s'me~' + cos e
vi
+
1/2
2
c
f
:
— e'+l-—+^7 + ln(l+0
y = y + y P = (c, + iK + (c 2 + l)e~' Solve
2 J u
2
The latter equation may be simplified to
10.59
and using partial fractions, we obtain
,
dt
+ 21n(l +u)
+ u
1
+
e'
21 + e _l
2
')
1
+o
- r + ln(l
-
1+e
2
..
2\+e
l
1
1
-U„-t\-2
(1
u
rdu
1
.
dt
e-'(e-')
C
iiIn u
2
1
+ e"«)<>-
r\2
1
y„
2(1
'
2(1
and
e~*
= e 'e 2 = e \e
e'
a
1
+ v'2 — e
v\ e'
v\
~
')
(
f
(
follows from Problem 10.3 that
= e ' sin e ~ + cos e ~
'
= \{e~ 2t sin e~' + e~' cose
sme~ + e~'cose~')df = \
It
'
-
and
')
v'2
— usin u — cosu)du — —\
— — i(sin e~' + e'eose
usin udu — \
I
').
cosudu
J
= — j(sinu — ucos u) — \ sin u = —sin u + ^ucosu = —sine"' + je~'cose~'
For
t,-
2>
integration yields
v2
=
I
— 2-(sine _r + e' cos e~')dt = —\
j
d(e'cose~') =
— — sin e " + \e ~ cos e " ')e' + — \e cos e " ')e " = — e' sin e "
e sin e
>V + y P = c e + c 2 e"
Then
l
y
(
'
(
'
'
— |e'cose~'
and the general solution is
'
x
10.60
y" + - y
— =m
f
> 0>
homogeneous differential equation are
y,
Solve
2
?
f'
^or
=
known that two linearly independent solutions to the associated
y 2 = 1/r.
and
t
y c = c,f + c 2 (l/f),
I The complementary solution is
follows from
s
if it
y p = v^ + v 2 (\/t).
so we assume
With
0(f)
= In
t,
it
Problem 10.3 that
+—=
v\t
1
= \nt
+ V2
v\
t
The solution to this set of equations is
v\
= ^ln
and
t
v'2
= -\t 2 In
t.
Then integration yields
3
3
lnf + iVt
-j-^m rrfr
and
18'
-Jin. (/f = In - ir
2
3
3
- ft 2 The general solution is then
)(l/t) = jt In
so that
y p = (^r In - ^f)f + (-gf In +
2
2
y = yc + y P = Cit + c 2 (i/o + ^ in - ft
1
'
f
f
^
f
f
*
10.61
Solve
2
t
y — 2ty + 2y = t In f,
f
f
f
f
.
.
for
t
> 0,
homogeneous differential equation are
y
x
if it is
=
known that two linearly independent solutions of the associated
and
t
y2 — t
2
.
2
1
l
y — 2t~ y + 2t~ y = t~ In f, which has the form of
The complementary solution is the same for either form of the
In t.
(pit) = t~
(7) in Problem 10.1. Now
2
and we assume yp = v x t + v 2 t 2
associated homogeneous differential equation, so
yc = c t + c 2 f
2
I We first divide the differential equation by
t
,
obtaining
l
.
x
It follows from Problem 10.3 that
v \t
+ v'2 2 =
t
v\
+ i/2 (2t) =
"
f
'
In t
—
~
CHAPTER 10
244
The solution to this set of equations is
= \-t~
vt
l
2
\nt - t~
1
v'2
and
v2
=
2
)t
2
y = y + yP = (c, - i)t + c 2 t - \-tdn t) - tin
Then
In t.
t
= ft -2 In
= — ^f(ln t) 2 - t\nt -
2
l
and
\r\t
\ntdt = -\{\nt) 2
l
y p = -|(lnt) t + (-f"
so that
= -t~
v\
dt =
t
-f"
1
- f"
In t
1
The general solution is
t.
2
t.
c
10.62
V
if it is known that two linearly independent solutions to the associated homogeneous
y — ty — r
2
differential equation are
y — 1 and y 2 = t
Solve
2
t
.
x
We divide the differential equation by 2 obtaining y
t
y — te',
,
so that the coefficient of the highest
yc = c x + c 2 t
The complementary solution of the differential equation in either form is
2
—
v + v2 t
Since
0(f) = te',
it follows from Problem 10.3 that
yp
derivative is unity.
so we assume
x
+ v2 2 =
'
The solution to this set of equations is
vt
t
and
+ v'2 (2t) = te'
v'
2
= f -\t e' dt = -\t e' + te' - e'
y p = —\t e' + te' — e' + \e't
Solve
(x
2
2
= te' — e',
2
— l)z" — 2xz' + 2z = (x 2 — l) 2
homogeneous differential equation are
= \e'.
Therefore,
and
f& dt = W
=
v2
y = y c + yp = c
and the general solution is
and
x2 — 1
z2
= x2 +
z'
z
xz +
+ -=z
so that the coefficient of the highest derivative is unity,
2
x
1
t
1.
2
The complementary solution remains
z — x — 1.
+
2
we assume z p — t,x + r (x + 1).
2
— 1,
Since
follows from Problem 10.3 (with z replacing y) that
(p(x) = x
obtaining z"
+ c 2 2 + te' — e'.
x
known that two linearly independent solutions to the associated
if it is
=x
z,
I We divide the differential equation by
2x
t-'i(O)
— — \t 2 e'
v\
2
Then
2
,
.
v\
10.63
2
zc
— c,x + c 2 (x 2 + 1),
so
1
2
it
v\x + v'2 (x
The solution to this set of equations is
= jx
so that
z p = (-3X - x)(x) +
2
4
2
z = z + z p = c x + c 2 (x + 1) + ^x - \x
v2
,
10.64
Solve
c e
x
x
2 \
(
c
v\
2
)(v
+ v'2 (2x) = x 2 -
1
and t'2 = x. Integration yields r, = — jx 3 — x
+ 1) = £x4 - \x 2 Then the general solution is
= —x 2 —
v\
3
2
+ 1) =
2
2
and
1
.
.
x
+ x)z" + (2 — x 2 )z' — (2 + x)z = x(x + l) 2
2
(x
+ c 2 x~
complementary function is known to be
if the
'.
We divide the differential equation by x + x so that the coefficient of the highest derivative is unity,
2
2
2 - x
— z 2 - x z = x + The complementary solution remains z = c e x + c x so we
obtaining z" H
2
I
'-
1.
5
f
+ X
X + X
assume z p — v e x + v 2 x~\ It follows from Problem 10.3 (with y replaced by z) that
X
2
x
,
x
v\e
x
+ v'2 x~ =0
'
v',e
x
+ v'2 (-x~ 2 = x +
)
1
x
- e'
Integration yields
r, = —xe~
The solution to this set of equations is i', = xe~ x and v'2 = —x 2
_1
x
3
2
x
x
3
=
general
Then
the
-5X
x
1.
(-3X
)(x
and v 2 = -3X
so that
z = {-xe~ - e~ )(e
+
)
p
x
2
solution is
z = z c + z = c e + c 2 x~ — 5X — x — 1.
p
x
.
)
,
'
x
10.65
Solve
^4-60— + 900/ = 5^ 10
dt
2
'.
dt
I The complementary solution is found in Problem 8.149 to be
solution of the form
Here
Ix
= e i0t
,
=v e
p
30
I 2 = re
I
i0t
x
',
v\e
30 '
+ v 2 te
and
0l
so that
l
= 5e 10
It
= -5te~ 20
= (it + io)e- 20 'e 30 -ie- 20 'te 30 = ioe 10
(Compare with Problem 9.14.)
'
so we assume a particular
',
follows from Problem 10.3 (with / replacing y) that
v\(30e
'
v'l
'.
= §-5te- 20 'dt = (it + ^)e~ 20
'
p
4>(t)
+ v'2 te 30 =
The solution to this set of equations is
= Ae 30 + Bte 30
Ic
30 '.
'
'.
'
'
30 ')
+ v'2 (e 30 + 30te 30 = 5e 10
and
and
'
&2 = 5e~ 20t
v2
'
')
.
Integration then gives
= ^5e~ 20 dt = -ie~ 20
'
'
The general solution is thus
= Ae 30 + Bte 30 + g ^ 10
1
/
'
'
'-
Q
—
~
VARIATION OF PARAMETERS
d2I
10.66
I
Ic
dl
- 60 — + 900/ = 4500r
—i
2
dt
dt
Solve
5
.
and the assumed form of /„ in the previous problem are valid here. Now, with
v\e
30 '
245
+ v'2 te
30
'
=
The solution to this set of equations is
v\(30e
30
= -4500f 6 e
v\
+ i/2 (e
')
~
30
'
30
+ 30?e
'
and
v'
2
30 ')
= 4500? 5
0(f)
we have
,
= 4500? 5
= 4500? 5 e
30 '.
Integration yields
= (150? 6 + 30? 5 + 5? 4 + §? 3 + V 2 + 2ht + efso)^ 30
5
4
3
- £t - j^)e " 30
v 2 = - 50? - 25? - ^t 6
4
5
Ip = (150 - 150)? + (30 - 25)? + (5 - ^)? + (| - i)? 3 + (A - £)? 2 +
225 - r&o)' + <rfe>
'
t;,
and
t
(
so that
^
1
(
Then the general solution is
with Problem 9.46.)
10.67
Solve
'
/
= I + I p = Ae 30 + Bte 30 + 5? 5 + ft 4 + ^? 3 + 425 2 + 24
'
/
+ 40/ + 800/ = 8 cos
'
?
c
assume
I
= r,e
p
'cos20? + v 2 e~
20
y',(-20e-
'
so we
Then it follows from Problem 10.3 (with / replacing y) that
sin 20?.
cos 20? - 20e~
20 '
and
+ i/2 (-20e~
sin 20f)
The solution to this set of two equations is
which we find
c
(Compare
— c e' 20t cos 20? + c 2 e~ 20 sin 20?,
Ic
v\e~
20t
+ ^.
t.
I The complementary solution is found in Problem 8.71 to be
20
?
v\
20t
20 '
'
x
20 '
cos 20? + v'2 e
20
sin 20? + 20e
— — fe 20r sin 20? cos ?
and
'
-
sin 20?
cos 20?) - 8 cos ?
— \e 20t cos 20? cos
v'2
from
?,
,=(-A inl9, + i^co .9 -A sin21 + J^ cos21 ,)^..
S
S
(
4
,
v2
= ^rz Sinl9
sin 19? 4Sin21,
21? + ^4r cos 21?
+ ^r COSl9
cos 19?' + -7^ sin
I
=
p
(
'
3§55
——
4
42y5
e
)
20
'
8?i
- sin 19? cos 20?) + —
-^- (cos 20? cos 19? + sin 20? sin 19?)
19
(sin 20? cos 19?
3805
761
4
(sin 21? cos 20?
—•sin (20? - 19?) +
/ol
4 \
4
761
Si
84lJ
— sin 20? cos 21?) +
cos
t—
JoUj
——
21
(cos 21? cos 20? + sin 21? sin 20?)
- 19?) - -—sin (21? - 19?) + -—cos (21? - 20?)
(20?
hZOj
841
/ 19
320
6392
21 \
°S ' "
S '"
C °S '
+
" ' + (m05 + i205j C
64*001
'
640loT
6392
320
The general solution is then
/
= I + Ip = c e 20 cos 20? + c 2 e 20t sin 20? + tt^t sin +
cos
fi4nn01
'
e
f
x
?.
(Compare with Problem 9.95.)
10.68
d 2 Q ^,dQ
-=- + 8S
+ 52Q = 26.
J1.2
2~
li
lt
~
Solve
I
assume
Here
c
'cos 6? + c 2 e
4
'
sin 6?.
so we
4'
4t
Q p = t e" cos6f + c 2 e~ sin 6?.
4
Q — e~*'cos6t, Q 2 = e" 'sin6?,
,
4
Q = Cj«
The complementary solution is found in Problem 8.55 to be
1
and
t
= 26.
(p(t)
It
follows from Problem 10.3 (with Q replacing y)
that
v\e~
v\
(-Ae'
At
cos 6? - 6e
4'
M cos 6? + v' e
At
2
4
sin 6?) + v'2 - 4e~ ** sin 6? + be
(
The solution to this set of equations is v\ - -^e* sin 6? and v'2 = ^e
4
Xt
M
Xl
cos 6? - \e sin 6? and v 2 - \e cos 6? + |e sin 6?, so that
», = \ e
1
4
'
sin 6?
cos 6?) - 26
cos 6?.
'
=
Integration yields
'
Q p = (i e
4'
cos 6? - \e
M sin 6?)e" 4 cos 6? + {\e M cos 6? + \e M sin 6?)e
The general solution is then
'
Q=Q +
e
= c^' 4 cos 6? + c 2 e
*
P
4'
sin 6?
*'
sin 6?
+ 2
.
- | cos 2 6? + \ sin 2 6? - 2
(Compare with Problem 9.20.)
"
246
10.69
)
.
CHAPTER 10
Solve
d2 Q
—— +
dQ
at
dt
+ 52Q = 32cos2f.
# Q and the form assumed for Q p in the previous problem are valid here. With
v\e'
v\(
cos 6f - 6e
" 4'
sin 6f )
+ v'2 - Ae
4'
(
4
cos 6f + v'2 e
'
~ 4'
+ 6e
sin 6f
=
cos 60 = 32 cos 2r
'
sin 6f
— — ^e 4 sin 6f cos 2f and i/2 = ^ cos 6f cos It. Integration yields
= - 3 sin 4f + ^ cos 4f - -^ sin 8r + ^ cos 8f )e 4 and w 2 = 3 sin 4f + j cos At + fs sin 81 + ^ cos 8t)e 4
with solution
v,
- Ae
~ 4r
A
= 32 cos 2r, we have
4>(t)
c
v\
'
'
(
',
(
so that
Q p = j (cos 6r cos 4f + sin bt sin 4f) + 3 (sin dt cos 4/ — sin At cos 6f)
— ^(sin 8f cos 6f — sin 6r cos 80 + j^(cos 8f cos 6f + sin St sin 6f)
= 3 cos (6f - At) + 3 sin (6? - 4t) - ^ sin (8? - 6f) + j^ cos (8f - 6f) = § cos 2f 4- y sin 2f.
Q — c,e 4 cos 6f + e 4 sin 6f + 5 cos 2r + 5 sin 2r.
The general solution is
10.70
'
(Compare with Problem 9.87.)
'
Q + 8Q + 250 = 50 sin 3f.
Solve
f The complementary solution (from Problem 8.53 with Q replacing x) is Q — c,e 4 cos 3f + c 2c -4* sin 3t,
4
so we assume
Q p — v (e 'cos 3f) + v 2 {e *' sin 3/). Then with 0(f) = 50sin 3f, we have
4
4
i/,^ 'cos 3f + v'2 e
sin 3r =
~
4
4
4,
4
r',(-4f
sin 30 + is(-4c
sin3f + 3?
'cos3f - 3e
cos 30 = 50sin3f
'
c
t
'
'
'
from which
v\
= - ^"t' 4 sin
and is = 53V*'sin 3f cos 3f. Integration yields
2
4
and
, = (^ sin 6f so that
q cos 6f)e
§| cos 6f )e
'
2^
+ r,£sin6?
y, = (-
f2
p
\
3f
4'
— -\ 52 cos 3? + 2<,(sin bt cos 3t - sin 3f cos 6f + ff(cos 6r cos 3f + sin 6f sin 3f)
= - 2 |cos3f + |f sin(6f - 30 + ^cos(6f - 30 = -^|cos3f + || sin 3r
)
Q = Q + Qp = c e
e
4
Q + 8Q + 250 = 90c
Solve
',
1
The general solution is then
with Problem 9.88.)
10.71
2
'
4
'
x
cos 3t + c 2 c
4
'
- l 2 cos 3f + \l sin 3f.
sin 3t
cos It.
I Q and the form of Q p assumed in the previous problem are valid here.
In addition, we have
c
4
v\e
r\(-Ae
1
5f
+ 2
'
+ r'2 -4f
sin 30
'
4
(
'
1
1
2
4
3fl>
4
cos 3/ + v'2 e~
4
+ 3e
sin 3f
'
sin 3f
=
4
'cos 30 = 90e~ 'cos 3f
Integration yields
"it.
r',
Q p - -5sin
t,
= — 5 sin 2 3r
and
Then
4
'cos30 + (15r + 5 sin 3t cos 30(«
'sin 30 - I5te
4
'sin 3f
Q = c e 4 cos 3/ + c 2 e 4 sin 3t + I5te 4 sin 3f
and the general solution is
10.72
4
'cos3/ - 3e
= — 30cos 3f sin 3/ and v'2 = 30 cos 2
sin 6f = 5f + 2 sin 2(3f) =
5/ + 5 sin 3f cos 3/.
with solution
v2 =
4
(Compare
'
'
'
x
Rework the previous problem, integrating v\ differently.
I With v 2 as in the previous problem, but with
vt
= J —30 cos 3f sin 3r
^if
= 5 cos 2 3r.
we have
4
4
2
sin 3f
cos 30 +
5f + 5 sin 3f cos 3f)(«
Q p - (5 cos 30(e
4
= 5e 4 'cos3f(cos 2 3r + sin 2 30 + 15fe" 'sin3f = 5e~ 4 'cos3f + 15f^" 4 'sin3f
'
'
( 1
The general solution is then
Q = Qc + Q P = (ci + 5)e
2
10.73
Solve
d Q
4
'cos3f + c 2 e
4r
+ 15fc
sin 3f
4
'sin3f
dQ
+ 1000 -=- + 250,000 = 24.
—f
df
dt
i The complementary solution is found in Problem 8.147 to be
Q = y,e
500 '
+ v 2 te'
500 '.
Q =c e
c
500
'
x
Then we have
v\e-
v\(-500e- 5W + v'2 (e
')
5OO,
+ v'2 te- soo =
5OO =
24
500fe"
'
')
+ c 2 te
500 ',
so we assume
VARIATION OF PARAMETERS
from which we find
u,
= -24te S00t
v\
and
= 24e 500t
'
t
J
(500)
24
3
(500)
31,250
Then integration yields
.
-—
+ -^t
500
= f -24/ t 500 dt=
>
v'2
e
2
500
Q p = ——J2 = tt^^t. The cgeneral solution is then
so that
247
fj
'
24 gSOOl
= f24^ 500 J/
v?2
'
J
500
5001
Q
e~ 5001 + cl22 te"
"
1
+
"
* = Cl
"
1
*
l
'
31,250'
(Compare with Problem 9.25.)
10.74
d2 Q
dQ
1000
1000-^
f++ tnnn
-^ + 250,0000 = 24ft 500r
-fdr
dt
—
Solve
I Q and the form of Q p assumed in the previous problem are valid here. Then we have
c
+ v'2 te~ 5oo =
+ v'2 (e~ 500t - 500/e~ 500 = 24te~ SOOt
5OO '
v\e'
^(-SOOe" 500
')
')
Solving this set of equations simultaneously yields
v\
500
— —24t 2 and v'2 = 24t,
+ I2t 3 e 500f = 4/ 3 e 500
= -8/ and v 2 = 12/
Then Q p = -8/ e
q = Cl e- 500 + c 2 te- 500 + 4r 3 t- 500 = (c, + c 2 + 4/ 3 )e- 500
3
r,
2
10.75
3
.
'
'
d2x
g
— 77^ xX -= g-,
-jy
2
5'
Solve
~di
~W
'
'
'
/
',
from which we find
and the general solution is
'.
where # denotes a positive constant.
f The general solution to the associated homogeneous equation is found in Problem 8.21 to be
9 10
x,. = C eso we assume a particular solution of the form
+ C 2 f v9/10
x p = u,^^15 + v 2 e~^r6
~
'
',
'
'.
x
Then
we have
v'l
JL eV57To
10
<
e^IWt + v'2 e-^}TGt =
- l± e -smot\ = l
+ v >(
2
Solving this set of equations yields v\ = jg/lQe~" 9n0t
- v°IWt
¥n>t
=
= -
-e
vl
solution is
10.76
and v 2 =-e^
Thus x
l0t 2.
x = x f + x p = C,^ 9/10 + C 2 e '^'
.
x
yp = v + v 2 e
t
.
and
2x
yc = C t + C 2 e
It follows from Problem 10.3 that
= — 2 e x sinx
i
v\
,
t
.
(Compare with Problem 9.26.)
,
+ v2 e 2x =
so we assume a particular solution of the form
'
v\
Then
t
v2
sin x.
I The complementary solution is
2x
5
)
— —^g/\0e y/BlT^
from which we find
^7T5t -Vg7Tg = _
The general
e -^""c^"»' - e
g
2
'
(D 2 - 2D)y = e
Solve
VlO
V
and
t/2
t>',(0)
= |e~*sinx,
+ v'2 (2e 2x = e x sin x
)
and integration yields
u,
= — ^(sinx — cosx)
= — £e"*(sinx + cosx). Thus y p = — ^(sinx — cosx) — |e~*(sinx + cosx)e 2x = — ^''sinx,
2x (Compare with Problem 9.105.)
|e* sin x.
general solution is y = y + yp = C + C e
v2
c
10.77
and
and the
2
x
q + 20q + 200q = 24.
Solve
f The complementary solution is found in Problem 8.64 (with q here replacing /) to be
10
10
10
i0,
'cos 10/ + c 2 e" 'sin 10/.
'sin 10/,
so we assume q p = v e
cos lOf + c 2 e
q c = c e~
x
1
Here we have
q { = e~
10
'cos 10/,
q2 = e
_10
and
'sin 10/,
= 24.
4>(t)
It
follows from Problem 10.3 (with q
replacing y) that
v\e~
v\(-\0e
10
'cos 10/ - 10t>
,0
'sin 100 + v 2 (- 10e'
10,
10,
cos 10/ + v'2 e~
sin 10/ +
10c"
10
cos 10/) = 24
10,
10
'
sin 10/
The solution to this set of equations is v\ = -f^ 'sin Of and v2 = f^'°'cos lOf.
10
10
so that
and v 2 = (^cos 10? + 2\ sin 10f)t
v = (^cos lOf - ^ sin 10f)?
1
Integration then yields
',
'
t
q p - ^(cos lOf
- sin Wt)e l0, (e
Thus the general solution is
10
3
'cos lOf) + 2 5 (cos lOf + sin I0t)e
q = qe + q p = c,e~
10 '
cos 10/ + c 2 e
w '(e
10 '
10,
sin 10/
sin 10?)
- 235 (cos 2 10/ + sin 2 10/) - £
+ 23
(Compare with Problem 9.22
5 .
)
—
CHAPTER 10
248
10.78
q + 20q + 200q = 2Ae~
Solve
I
10t
sin lOf.
q c and the form assumed for q p in the previous problem are valid here.
v\e~
10
t/^-lOe
'cos lOr - 10<r
10,
+ i/2(-10e
sin lOt)
10r
_10f
Thus
cos lOf + v'2 e~
sin lOr
10t
sin lOt
=
+ 10<>- 10 'cos lOr) = 24<?- 10, sin lOt
from which we find v\ = —2.4 sin 2 lOt and v'2 = 2.4 sin lOf cos lOt.
and v 2 = -0.12 cos 2 lOr, so that
v = -1.2* + 0.12 sin lOt cos lOf
Then integration gives
t
qp = (- 1.2t + 0.12 sin lOf cos 10r)<?~
q — c^"
and the general solution is
10.79
10
'
10 '
cos lOr + ( -0.12 cos
cos lOr + c 2 e~
I0t)e~
10 '
sin lOt
= - 1.2te~ 10r cos lOf
— 1.2te _10 'cos lOf.
'sin lOr
Redo the previous problem, integrating v'2 differently.
I With b, as in the previous problem, but with
qp = (-l.2t + 0.12 sin lOf cos 10r)<?"
= -1.2re-
10
'cosl0f + 0.12f-
10
10 '
= J 2.4 sin lOtcos lOf dt — 0.12 sin 2 lOr,
v2
cos lOf + (0.12 sin
'sin 10r(cos
q — c 1 e~
and the general solution becomes
10.80
10
2
l0
2
lOr
+ sin
2
10 '
I0t)e~
2
we have
sin lOr
= -1.2t^ 10 'cos 10f + 0.12e" 10 'sin lOt
lOr)
'cos lOf + (c 2 + 0.12)e"
10,
— 1.2fe~ 10 'cos lOf.
sin lOf
q + 400q + 200,000q = 2000.
Solve
q c — e'
f The complementary solution is found in Problem 8.70 to be
assume a particular solution of the form q p — f e"
Here q = e~ 200, cos400f, q 2 = e~ 200 sin 400r,
200
1
'
'cos400r + t'2e~
and
x
200
200
= 2000.
0(f)
+ Bsin400f),
'(A cos400f
so we
'sin400r.
It
follows from Problem 10.3 (with q
replacing y) that
v\e~
'cos400f - 400e"
2OO
The solution to this set of equations is
v\
r,(
:oo
200e
r,
'sin400f) + i/2 (-20O<r
— — 5e 200 sin 400f
'
q p = 0.01 cos 400f + 0.01 sin 400f = 0.01.
200
'(A cos400f + Bsin 400f) + 0.01.
q = q c + qp = e~
2
so that
10.81
Solve
f
2
v2
2OO
'
cos 400f + v'2 e-
'sin400f + 400e"
and
'
= 0.01e 2O0 'cos400f - 0.005e 200 sin 400f
200
200 '
200
=
'cos400f) = 2000
— 5e 200 'cos400r.
v'2
sin 400t
Integration gives
= 0.005? 200 cos 400f + 0.01e 200 'sin400t
'
and the general solution is
(Compare with Problem 9.23.)
q + 400q + 200,000^ = 2000 cos 200f.
qc and the form assumed for q p in the previous problem are valid here. Thus, it follows that
v\e~
v\(
-
200<?" 200
'
cos 400r - 400?
" 200
'
sin 400f)
The solution to this set of equations is
and integration yields
'
(
sin 400f
= - 5e 200 cos 200f sin 400f
'
cos 400f + v'2 e
+ 400<T
and
v'
2
200
~ 200
'
=
(
^
After some simplification, these values for v t
q p = rlo sin (400f - 200f ) +
^
=
cos 400r) = 2000 cos 200r
= 5e 200 cos 200f cos 400f,
'
^
^
i^o sin 200r
sin 400*
'
200
sin 600r +
cos 600f)<?
+ j^ cos 200r 200
cos 200f + gtb sin 600t + goo cos 600r)e
v 2 = (j|o sin 200r +
r,
and
v\
+ v'2 -2(Xk"
200
200t
'
'
and v 2 give
cos (400t - 200f) -
^
sin (600t
- 400f) + goo cos (600f - 400t)
= 2^0 sin 200f + j^o cos 200r
The general solution is then
Problem 9.93.)
q = e
~ 200 ' (A cos 400f
+ B sin 400f + 205 sin 200r +
)
^
cos 200t
.
(Compare with
VARIATION OF PARAMETERS
10.82
249
D
q + 9q + I4q = {- sin t.
Solve
q c = c e'
# The complementary solution is found in Problem 8.17 (with q here replacing x) to be
q p = v l e~
so we assume
2'
+ v 2 e~
v\e-
= j^e 2 sm r ancl
v 2 — — 500 sm
+ soo c osf)e 7
so that
v\
v 'i
'
f
(
<?
2t
',
P
The general solution is then
7
x
2'
+ c 2 e~ 1
',
Then
'.
+ v'2 e~ 11 =
v\(-2e- 2 ') + i/2 (-7e" 7 ') = ^sin t
= -lo^ 7 sin 5t.
Integration yields
'
t>,
= (^ sin - ^ cos f)e 2
t
and
'
and thus
= (A - soo) sin + ~ so + soo) cos = 5^ sin - 5§o cos
(
*
q = q c + q p = c { e'
t
2'
+ c 2 e~
7t
t
+ j^s'mt — j^cost.
t
(Compare with
Problem 9.89.)
HIGHER-ORDER DIFFERENTIAL EQUATIONS
10.83
x 3 y'" + x 2 y" — 2xy' + 2y = x In x,
Solve
yc = c l x~
+ c2 x + c3x
1
x > 0,
for
if the
complementary solution is
2
.
I We first divide the differential equation by x 3 so that the ccoefficient of the highest derivative is unity, as
2
3
2
y'" + x~ y" — 2x~ 2
in (/) of Problem 10.1. The result is
y + 2x~ y = x~ In x, for which 4>(x) = x~ In x.
2
We assume a particular solution of the form y p = Vix" + v 2 x + v 3 x
It follows from Problem 10.2 that
l
1
.
v\x~ 1 + v'2 x + v'3 x 2 =
v\(-x~ 2 ) + v'2 + t/3 (2x) =
u;(2x "
3
)
+ v'2 (0) + v'3 (2) = x " 2 In x
The solution to this set of simultaneous equations is
v\
= £x In x,
v'2
~ 1
= — \x
In x,
and
v'3
= |x " 2 In x.
Integration yields
vy
= T2~x 2 \n x — J4X 2
— — j(lnx) 2
v2
yp —
and, after substitution and simplification, we have
v3
— — \x~
x
lnx — ^x" 1
— |x[(lnx) 2 + lnx] — fx.
Then the general solution is
y = y + y P = c ,x" + (c 2 - |)x + c 3 x - ix[(ln x) + In x].
2
2
'
c
10.84
Solve
>•'"
+ / = sec x.
I The complementary solution is found in Problem 8.108 to be
particular solution of the form
y p — v + v 2 cos x + v 3 sin x.
y c = c, + c 2 cosx + c 3 sinx,
so we assume a
{
Here
y,
= 1,
y 2 — cos x,
y 3 = sin x,
0(x) = secx.
and
It
follows from Problem 10.2 that
+ v'2 cos x + sin x =
— sin x) + v'3 cos x =
y'i(0) + v'2
u',(0) + v'2 — cosx) + v'3 — sinx) = secx
v\
1/3
(
(
(
and v'3 = — tanx. Thus,
v\ = secx,
v'2 = — 1,
= Jsecx^x = ln|secx + tanx|, v 2 = — \dx=— x, and v 3 = j -tanxdx = In |cosx|. Substitution then
yields
y p = In |sec x + tan x| - x cos x + (sin x) In |cos x|. The general solution is therefore
Solving this set of equations simultaneously, we obtain
i?!
y = yc + y p = Cj + c 2 cos x + c 3 sin x + In |sec x + tan x| — x cos x + (sin x) In |cos x|
10.85
Solve
(D 3 + D)y = esc x.
y c and the form assumed for y p in the previous problem are valid here.
follows from Problem 10.2 that
I
Now, however,
4>(x)
- esc x,
and it
+ v'2 cos x + v'3 sin x =
v\(0) + v'2 — sin x) + v'3 cos x =
t/^O) + u'2 (-cos x) + v'3 — sin x) = esc x
v\
(
(
The solution to this set of equations is v\ = esc x, v'2 = -(cos x)/sin x, and v'3 = - 1, from which we find
and v 3 =-x. Then substitution gives
v 2 = -ln|sinx|,
that
p, = -ln|cscx + cotx|,
x sin x. The general solution is thus
(In |sin x|) cos x
y = - In |csc x + cot x|
p
y = Cl + c 2 cos x + c 3 sin x
- In |csc x + cot x| - (In |sin x
|)
cos x - x sin x
250
10.86
CHAPTER 10
+ 4y' = 4 cot 2x.
y'"
Solve
# The complementary solution is found in Problem 8.109 to be ye = c 4 c 2 cos 2x 4 c 3 sin 2x,
y p — v t 4 v 2 cos 2x + v3 sin 2x. It follows from Problem 10.2 that
so we assume
r
+ v'3 sin 2x =
— 2 sin 2x) + v'3 (2 cos 2x) =
!>i(0) + t/2
—
v\(0) 4 r'2
4 cos 2x) + v'3 — 4 sin 2x) = 4 cot 2x
v\ 4- v'2 cos 2x
(
(
(
The solution to this set of equations is
v\( cos 2x) sin 2x.
v'2
— — (cos 2 2x)/sin2x,
and
=— cos2x.
r'
3
Integration then gives
vt
= | In |sin 2x|
— — 4 In |csc2x — cot 2x| — |cos 2x
v2
=— ^sin2x
r3
Substitution into the expression for y p and combination with yc finally lead to
y — c
4 c 2 cos 2x + c 3 sin 2x + \ In |sin 2x| — | In |csc 2x — cot 2x| cos 2x — ^(cos 2 2x 4 sin 2 2x)
x
— c — 2 4 c 2 cos 2x 4 c 3 sin 2x + \ In sin 2x| — \ In |csc 2x — cot 2x| cos 2x
l
10.87
— 3y" + 3y' — y = e*/x.
y'"
Solve
|
I The complementary solution is
follows from
yc = c e* 4 c 2 xe* 4 c 3 x e
2
x
JC
y p — r,^ 4- i^xe* 4 t^x ^.
2
so we assume
,
It
Problem 10.2 that
4 v' xe x + r'jx 2 ?* =
x
Die" + v'2 (e 4 xe*) 4 i'3 (2xr 4 xV) =
v\e
x
2
t
i\e
x
e
x
4 i/2(2e* 4 xe") 4 v 3 (2e x 4 4xe 4 xV) = —
JC
x
= 2 x. r 2 = — 1, and r 3 = 2 x
The solution to this set of equations is
2
and r 3 = |ln |x|. Thus y p = §x e*ln |x|, and the general solution is
i
y = yc + )'p = e*(Cj 4 c 2 x 4 c 3 x
10.88
2
d*y
d
Y
dy
</v
Ja-
dx
2
',
l
i\
= x2
v2
,
= — x,
.
f The complementary solution is found in Problem 8.156 to be
2x
4 .\c 2a + v 3 x 2e 2v Then we have
y p = v e~
;
so that
,
4 ix 2 In |x|).
-4 + 6-r44- 12/ + iy=l2e~ 2x
Solve
_1
2x
yc — (c t 4 c 2 x 4 c 3 x )e
2
,
so we assume
.
:
,-Zx*
,2„-2x
-* 4
r,?
4 v 2 xe,-2x
D'aX-'e" =
'
=
4 i/2(e~ - 2xe" *) 4 v'3 (2xe- 2x - 2xV
-2
2x
2x
2x
2
i\(4e~
4 v'2 (-4e- 4 4xe *) 4 3 (2<T - Sxe~ 4 4xV *) = \2e~ 2x
i\(-2e'
2x
2*
2
2jc
)
)
2jt
t<
)
v'2
,
,
10.89
y = y f 4 y_ = (c, 4 c 2 x 4 c 3 x
Solve
y" — 6y" 4 1 1/ — 6y = 2xe
2
?
,
2x
3
i?
thus
= — 12x,
3
and v'3 = 6
so that
p, = 2x
2x
2x
3
3
—
The
general
solution is
6x e
4 6x e
2x e
(Compare with Problems 9.146 and 9.147.)
The solution to this set of equations is v\ — 6x 2
2
and
Then yp = 2x 3 e~ 2x
» 2 = — 6x
3 = 6x.
.
4 2x 3 )e~ 2
x
.
yc =
I The complementary solution is found in Problem 8.28 to be
x
2x
4 v 3 e 3x Then we have
y p — i\e 4 v 2 e
c^ 4 c e 2x 4 c e 3x
2
3
r3
= xe~* x
,
so we assume that
.
4 v'2 e 2x 4 r'3 e 3 * =
x
2x
v\e 4 r'2 (2<?
4 3 (3<? 3 *) =
3x
x
2
= 2xe~ x
v\e + r 2 (4e *) 4 v 3 (9e
v\e
x
t<
)
)
The solution to this set of equations is
= xe~ 2x
v\
,
v'
2
= -2xe' 3x
and
,
,
so that integration
yields
= e~ 2x(-\x - i)
Vi
Substitution and simplification then give
x
Solve
y'"
x
,21
10.90
- 3y" 4 2y' =
+e
f
1
'
= e~ 3 *(fx 4 §)
v3
- e" 4jc (-ix -
^
y p — e'^—j^x — y^), and the general solution is
(Compare with Problem 9.74.)
4 c 2 e 2x 4 c 3 e 3x 4 ^ _x(— i^x — j^).
y = c e
v2
VARIATION OF PARAMETERS
I The complementary solution is found in Problem 8.26 (with
yp = v
Thus, we assume
+ v 2 e' + v 3 e
{
2
yt = c, + c 2 J + c 3 e
here replacing x) to be
t
251
2 '.
and it follows from Problem 10.2 that
',
+ v'2 e' + v'3 e 2 =
2
v\(0) + v'2 e' + v'3 (2e
=
v\
'
')
v\(0)
+ v'2 e' + v 3 (4e 2 = -±L+e
')
1
The solution to this set of equations is
e
= 1
v\
2'
v'7
:,
2 1 + e'
u =
substitution
-e'
—
and
,
+ e'
1
v\3
1
1
=
2 1
+ er
Usine the
we find
4- e\
1
=-
I
e'
r
—
r w
1
1
1
1
^ = 2lrT7^^2/-^ d " = 2 (1+e) -2 ln(1+e,)
y2
,
,
= f 7^-7 e dr= -ln(l +e')
J
+e
f
'
1
1
and
U
1
/•
1
1
y = c, + c 2 f' + c 3 e
2'
•>
1
2'
+ 1).
In (e~'
Then the general polution is
- (| + e') In (1 + e') - \e 2t In (e~' + 1).
e
/" — 3y" + 2/
Find a particular solution to
1
I
1
yp = \{\ + e') - \ In (1 + e') - e' In (1 + e') - |e
and it follows that
10.91
1
/•
dt =
r« "'A=-zln(e-*+D
f^T—
^
3=^f
2 J TT-T
2
e
+
2
+r
3'
+e'
y c and the form assumed for y p in the previous problem are valid here.
2t
=
2
=
+ v'2 e' + v'3 e
v\
v\(0)
+ v'2 e' + v'3 (2e
v\(0)
+ v'2 e' + v'3 (4e 2t
')
e
)
v\
= --
=
2=
v'2
-,,
2TT7'
"
3'
+e'
1
The solution to this set of equations is
follows from Problem 10.2 that
It
— — and ^21+e*
=v'3
,
TT7'
p.
Then integration
yields
„!
y„ = i(l +e )
r
so that
10.92
Solve
=i(i +e') 2 -(l +<?') + ±ln(l +e')
t
3 y"'
+ 3t 2 y" — 1,
differential equation are
y
= -(1 + e*) + ln(l + e')
t>
= ± In (1 •+ J)
3
-(l +e') + iln(l +et) + [-(l + eO + ln(l + e')]e' + {-e 2 In (\ + e')
2
'
for
f
=
~
x
y2
t
> 0,
if three
y 2 = 1,
1
,
linearly independent solutions to the associated homogeneous
and
y 3 = f.
so we assume a particular solution of the form
t The complementary solution is yc = c x t~^ + c 2 + c 3
=
differential
equation by 3 so that the coefficient of the
We divide the nonhomogeneous
+ v2 + v3
v t~
yp
y"'
+ 3t~ y" = t~ 3 therefore, (j)(t) = t~ 3
highest derivative is unity, as in (/) of Problem 10.1. The result is
t,
l
t
t.
x
1
.
;
It
follows from Problem 10.2 that
l
v\t~
Dj^-^f"
1
y = yt + y P = ci*
10.93
Solve
v\=j,
so that
,
-1
)
= —
Then integration yields 1?!=^,
and v'3 =jt 2
—
—
—
=
=
The general solution is then
Int.
t
%t~H \nt \t~
yp
l
v'2
t
3
v2
=-\nt,
x
+ c 2 + c 3 - in
t
2
.
,
t.
+ 25^-1252=
^-5^
^
dr
v\{2C 3 ) + v'2 {0) + v'3 (0) = f 3
v\{-r 2 + w'2 (0) + v 3 =
t
from which we find
and
+ v'2 + v'3 =
-60^ 7
'.
^r
f The complementary solution is found in Problem 8.1 13 to be
assume Q p = u,e 5 + f 2 cos ^^ + ^3 sin 5t. Then we have
Q =c e
c
{
5'
+ c 2 cos 5t + c 3 sin 5t,
'
+ v'2 cos 5t + v'3 sin 5t =
+ «/2 (-5 sin 5f) + y'3 (5 cos 5f) =
v\e
5
y'i(5e
v\(25e
5t
)
')
5'
+ i/2 (-25 cos 5f) + u'3 (-25 sin 5t) = -60e 7
'
so we
-
252
CHAPTER 10
The solution to this set of equations is
= -fe 2
v\
',
= fe 7 '(-sin 5f + cos 50,
v'2
and
v'3
= fe 7 '(sin 5f + cos 50,
and integration gives
v,
3„2r
= -ie
_
Then
= i85 e7 6 cos 5t - sin 5t)
v2
3-2(_5l
2
5t
le
'e
= jfje^cos 5r + 6 sin 5r)
15„7t
+ ^'(6 cos 2 5t + 6 sin 2 5t) = (-f + t%)<? 7 = -$e
'
Q = Q + Q p = c^ 5 + c 2 cos 5f + c 3 sin 5t - ^e 1
and the general solution is
'
c
d Q
— —d Q
—
3
10.94
v3
'(
Find a particular solution to
2
dt
5
5
3
dt
dQ
=2
25
(Compare with Problem 9.16.)
'.
1250, = 1000.
dt
I Q and the form assumed for Q p in the previous problem are valid here. Then we have
c
+ v'2 cos 5r + v'3 sin 5r =
+ v'2 (-5 sin 50 + v'3 (5 cos 50 =
v\e
v\(5e
5
v\(25e
Sl
)
5'
+ i/2(-25cos5t) + v'3
')
25 sin 5t) = 1000
(
5
v\ — 20e~
v'2 = 20sin5f — 20cos5f,
and
5(
—
—
—
—
=
v'
20sin5t
20cos5f; integration yields v
4e
v 2 — — 4 cos 5r — 4 sin 5f
3
v 3 = 4 cos 5f — 4 sin 5f
Thus
The solution to this set of equations is
',
,
x
,
and
.
Q p = (-4e
5,
)e
+ (-4cos5f - 4 sin 5f)cos 5f + (4cos5f - 4sin5f)sin5f = -8
5'
(Compare with Problem 9.29.)
10.95
Find a particular solution to
d3 Q
2
— —Q +
dr
d
dQ
25 —
-=-
5
-r-=
dt
2
1250 = 5000e
cos 2t.
dt
I Q and the form assumed for Q p in Problem 10.93 are valid here. Then we have
c
v\e
v\(25e
')
5
')
(
'
The solution to this set of equations is
v'3
+ v'2 cos 5f + v'3 sin 5f =
+ v'2 - 5 sin 5r) + v'3 (5 cos 5f) =
+ p'2 (-25cos5t)+ r'3 (-25sin 5f) = 5000<? " cos 2r
v\(5e
5
5'
= 100?~ 6 'cos 2t;
v\
= 100e~'( — sin 5f cos2f — cos5f cos20=e
v2 = e
v'2
= 100e~'( — cos 5rcos 2t + sin 5rcos2f);
and
Integration yields
'(-15cos2f + 5 sin 2t)
r,
- 20 sin 3f - 10 cos 3f - 8 sin It - 6 cos It)
v 3 = e"'(-10sin3f + 20cos3f - 6 sin It + 8 cos It)
'(
Substitution and simplification then give the particular solution
Q p = e~'[— 15 cos 2f + 5 sin 2t + 20(sin 5f cos 3f — sin 3f cos 5f) - 10(cos 5f cos3r + sin 5rsin 3f)
- 8(sin It cos 5/ - sin 5f cos It) - 6(cos It cos 5f + sin It sin 50]
= e~'[-15cos2t + 5sin2f + 20sin(5f - 3f) - 10cos(5f - 3r) - 8sin(7r- 5f)-6cos(7f - 50]
= e"'(-31 cos2r + 17 sin 20
(Compare with Problem 9.107.)
d*y
10.96
Solve
ax
9^
= 54x
dx
2
2
'
I The complementary solution is
y p = t>! + v 2 x + v 3 e
J*
+ v Ae
ix
It
yc — c
x
+ c 2 x + c 3 e ix + c x e
follows from Problem
ix
,
10.1 (with
3x
so we assume
n = 4)
that
3x
—
=
v\(0) + V2 + v'3 (3e
+ v'A (-3e3x
3x
v\(0) + v'2 (0) + v'3 (9e
+ v't(9e3x
3x
= 54x 2
v\(0) + v'2 (0) + v'3 (21e
+ v' {-21ev\
+ v'2 x + v'3 e
3x
3x
)
)
)
)
The solution to this set of equations is
v\
— 6x 3
,
v'2
+ v'^e
)
)
At
= —6x 2
,
and
i4
= — x 2 e ix
,
and
VARIATION OF PARAMETERS
253
integration yields
p,
= fx 4
v2
= -2x 3
v3
= e- 3x (-±x 2 - lx - A)
t;
= e 3x (- 3 x 2 + §x - &)
4
By substituting these results into the expression for y p and simplifying, we obtain y = ~ix 4 - §x 2 - =&.
p
general solution is then y = yc + y p = c t - yj + c 2 x + c 3 e 3x + c 4 e~ 3x - \x x - fx 2
The
.
d*y
10.97
= 5x.
t4
ax
Solve
I The complementary solution is found in Problem 8.155 to be yc = c, + c 2 x + c 3 x 2 + c 4 x 3 so we assume a
2
3
3
particular solution of the form
Here y, = 1, y 2 = x, y 3 = x 2
y p = v + v 2 x + v 3 x + v4 x
y4 = x
and 0(x) = 5x. It follows from Problem 10.1 (with n = 4) that
,
.
x
,
,
+ v'2 x + t/3 x 2 + u 4 x 3 =
2
=
»'i(0) + V2 + v'3 (2x) + u 4 (3x
i/,(0) + i/2 (0) + t/3 (2) +
4 (6x) =
v\
)
i;
+ v'2 (0) + v'3 (0) + v'M = 5x
v\(0)
Solving this set of equations simultaneously, we obtain
— — fx 4
v\
v'2
,
= fx 3
v'3
,
= — |x 2
,
and
u4
= fx.
Then
y = c
— £x 5 + |x 4 (x) — |x 3 (x 2 + ^x 2 (x 3 = ^x
Thus, the general solution is
+ c 2 x + c 3 x 2 + c 4 x 3 + j^x 5 This solution also can be obtained simply by integrating both sides of the
yp =
and
5
.
)
)
.
t
differential equation four times with respect to x.
10.98
Solve
y
(4)
+ 8y (3 + 24y" + 32/ + 16y = 120<T 2 7x 2
>
.
I The complementary solution is found in Problem 8.157 to be y = c e~ 2x + c 2 xe~ 2x + c 3 x 2 e~ 2x + c 4 x 3 e~ 2x
2x
so we assume y p — v e~
+ v 2 xe~ 2x + v 3 x 2 e~ 2x + v^x 3 e~ 2x It then follows from Problem 10.1 (with n = 4)
l
,
.
x
that
+ v'2 xe~ 2x + v'3 x 2 e~ 2x + v\x 3 e~ 2x —
v\(-2e- 2x + v'2 (e' 2x - 2xe~ 2x) + v 3 (2xe~ 2x - 2x 2 e~ 2x + 4 (3x 2 e" 2x - 2x 3 e" 2x =
2x
v\(4e~
+ v'2 (-4e- 2x + 4xe~ 2x + v'3 (2e' 2x - Sxe~ 2x + 4x 2 e~ 2x
+ 4 (6x<T 2 * - \2x 2 e' 2x + 4x 3 e~ 2x =
v\e~
2x
)
t;
)
)
)
)
)
i>
v\(-8e~
2x
)
+ v'2 (\2e~
2x
- Sxe~
2x
)
+ v'3 (-\2e-
The solution to this set of equations is
Integration then yields
yp =
2
-\0x e~
2x
vl
= —\0x 2
+ (60x)(xe~
The general solution is then
10.99
)
120e"
2x
x2
= — 20x, v'2 — 60, v'3 = — 60x
and v\ — 20x 2
_1
v 2 = 60x,
v 3 = — 601n |x|,
and v\ = — 20x
so that
1
v\
,
2
)
-60(ln \x\)x e~
Find an expression for a particular solution to
x
2x
)
2x
y = [c t + c 2 x + (c 3 + 30)x
y p = v + v 2 x + y 3 cosx + y 4 sinx.
- 8x <T
2
.
,
,
2x
f The complementary solution is
+ 24xe"
2x
- 36xe" 2x + 36xV 2x - 8x 3 e~ 2x =
2x
4- v' (6e~
4
)
2x
(D
4
3
+ (-20x" )(x e1
= 30x e"
2
)
+ c 4 x 3 — 60x 2 In |x|]e" 2x
2
2x
- 60x 2 e _2x ln |x|
.
+ D 2 )y = f{x).
y c = c x + c 2 x + c 3 cos x + c 4 sin x,
It follows from Problem 10.1 with
v\
2x
so we assume
n = 4
and
0(x) =/(x)
that
+ v'2 x + 3 cos x + v\ sin x =
v'2 + v'3 — sin x) + v 4 cos x =
— cos x) + f 4 — sin x) =
3
— cos x) = /(x)
v'3 sin x +
4
i^'
(
t;' (
(
i>
The solution to this set of equations is
vx
and
= - (*x/(x) dx
yp =
v\
v2
= - x/(x),
= §f(x) dx
v'2
v3
(
= /(x),
t/3
= /(x) sin x,
= \f(x) sin xdx
- fx/(x) dx + x f/(x) dx + cos x
J
vA
and
v\ = -f(x) cos x,
= -\f(x) cos x dx
f(x) sin x dx - sin x \f(x) cos x dx
so that
254
10.100
CHAPTER 10
Solve
(D
5
- 4D 3 )y = 32e
2-x
I The complementary function is y = c + c 2 x + c 3 .x 2 + c x e 2x + c 5 e~ 2x
Zx
2
2
o^e ' + v 5 e~
It follows from Problem 10.1 (with
y p = D] + 2 v + d3x
f
-i-
t
,
1
.
[',
+ r 2 (.\) + r 3
2
.x
+ i\e
2x
+ v e'
2x
5
so we assume that
n = 5)
that
=
lx
:t
+ rf£-2e
) =
x
2x
=
^(O) + r 2 (0) + r 3 |2) + r 4 |4 t
+ r 5 (4<r
^(0) + r : + vT3(2x)+ rj2t
i
)
)
1
2x
=
+ r 3 (0l + v'Ji&e *) + i'5 -8<T
2x
2x
= 32e 2x
- r 4 (16e + r'5 (16<r
r,(0) +
: l0l + 1,10)
r,(0) + r : (0)
(
t
)
)
)
The solution to this set of equations is
D,
= — 4.x.2-e 2x
p,
= (-2.x H-2x-2)e :t
2? 2*
r,
_
= 8v„2*
Sxe'
p
= (4.x-2)* 2x
..'
= -4e 2j
1
i\
= e^
so that
:
Substituting these quantities gives, after simplification.
y = yt + y, =Cj + c 2 .x + c 3 .x
2
+ (c 4 - l)e
2x
+ c5e
-2*
r3
= -2e :t
y p = [x — A.)e-
+ xe
2jt
.
X
.
and so
!„•**
CHAPTER 11
Applications of Second-Order
Linear Differential Equations
SPRING PROBLEMS
11.1
A steel ball weighing 128 lb is suspended from a spring, whereupon the spring stretches 2
ft
from its natural
The ball is started in motion with no initial velocity by displacing it 6 in above the equilibrium position.
Assuming no air resistance, find the position of the ball at t = n 12 s.
length.
This is free, undamped motion. The differential equation governing the vibrations of the system is shown in
i
Problem 1.75 to be x + 16x = 0. Its solution (see Problem 8.57) is x — c, cos4f + c 2 sin4r, from which
we find v — x ~ — 4c, sin 4r + 4c 2 cos At.
The initial conditions for this motion are x(0) = — £ft (the minus sign is required since the ball is initially
displaced above the equilibrium position, which is the negative direction) and
r(0) = 0.
Applying these
— 2 = x(0) = c, and
= i(0) = 4c 2 respectively.
conditions to the equations for x and r, we obtain
Therefore C, = — \, c 2 = 0, and x = - 2 cos4f. At t = n/12, we have
x(7r 12) = -£cos(4tt/12)= —
ft.
,
i.
;
11.2
k — 48 lb ft
hangs vertically with its upper end fixed. A mass weighing 16 lb is attached to
From rest, the mass is pulled down 2 in and released. Find the equation for the resulting motion
A spring for which
the lower end.
of the mass, neglecting air resistance.
x + 96x = 0,
(7) of Problem 1.69 becomes
Problem 8.61).
Differentiating with respect to t yields
v = dx dt — x 96 (C, cos x 96f — C 2 sin x 96/). _When
the initial conditions
x = £ and r = 0. Then C 2 — £, C 1 =0, and x = ^cos N 96f.
11.3
k = 48 lb ft.
f With
m = 16 32 =J).5 slug
solution
x = C, sin N 96? + C 2 cos N 96f
and
which has as its
(see
=
t
we have
A 20-lb weight suspended from the end of a vertical spring stretches it 6 in. Assuming no external forces and no
air resistance, find the position of the weight at any
time if initially the weight is pulled down 2 in from its rest
position and released.
I
or
From Hooke's law (see Problems 1.69 through 1.74). the spring constant k satisfies the equation k{\) — 20
With m = 20 32 slug. (/) of Problem 1.69 becomes x + 64x = 0. which has as its solution
k = 40 lb ft.
x = C] cos 8f + c 2 sin St
and
Since x(0) = I
11.4
(see
Problem 8.58).
x(0)
= 0.
we have
c,
=^
and
c2
— 0.
Thus we have
x = £cos8r.
Solve the previous problem if the weight is initially pulled down 3 in and given an initial velocity of 2 ft s
downward.
I
As in the previous problem
v
= c
;
cos 8f + c 2 sin 8?.
Now, however,
x(0)
— | ft
and
x(0)
= 2 ft/s.
Applying these initial conditions, we find that
i
Thus
11.5
= x(0) = Cj cos
c,=c 2 =
4;
and
+ c 2 sin
= c,
and
2 = x(0) =
—8c, sin
= 8c 2
+ 8c 2 cos
x = |cos 8f + |sin 8f.
Determine the motion of a mass m attached to a spring suspended from a fixed mounting if the vibrations are
free
I
and undamped.
— =
k
The differential equation governing such a system is found in Problem 1.70 to be
= N —k m
x H
x
0,
where k is
= - N -k m, or. since
both k and m are positive, A, = x k m and X2 = —iy/k m. Its solution is x = c, cos N k mt + c 2 sin N k mt.
we obtain c, = x
and c 2 = oN m k. Thus
and r(0) = r
Applying the initial conditions x(0) = x
the spring constant.
The roots of its characteristic equation are
/.,
and
/.,
i
t
.
the solution becomes
x = x cos N k mt + i 0N m/k sin x k mt
11.6
Rewrite the displacement x found in the previous problem in the form
(I)
x = A cos (cot - <f>).
255
256
CHAPTER 11
D
/
Since
A cos (cot — cp) = A cos cot cos
A cos cot cos
</>
2
— yjk/tn,
co
= ^2(1) = A 2 (cos 2
A — -n/x 2 + t>o(m//c)
we have
A cos cp = x
given implicitly by
is
= v jm/k.
A sin
and
,
+ sin 2 0) = (/I cos 0) 2 + (A sin 0) 2 = (x
and the phase angle
,
we require
cf>,
+ A sin cot sin cp = x cos ^/k/rnt + v >Jm/k sin yfk/mt
For this equality to hold, we must have
^4
+ A sin cot sin
c\>
2
)
+
yjkfm)
(i?
cos c\> = -
Now, since
2
and
/I
Im/k
To find
sin
we write
explicitly,
sin
tan cf> =
v
= arctan
so that
— Im/k
x
cos
11.7
\jm/k
Determine the circular frequency, natural frequency, and period for the motion described in Problem 11.5.
I The motion described by (/) of Problem 11.5 is called simple harmonic motion. The circular frequency of such
motion is given by
co
= yfk/m.
f = co/2n — (l/2n)yjk/m.
The period of the motion, or the time required to complete one oscillation, is T = \jf = 2nyjm/k.
The natural frequency, or number of complete oscillations per second, is
11.8
Determine the circular frequency, natural frequency, and period for the vibrations described in Problem 11.1.
I
In that problem,
Circular frequency:
Natural frequency:
Period:
11.9
In that problem
Using the formulas of the previous problem, we have
= yJ64/4 — 4 cycles per second = 4 Hz
f — 4/2n = 2 1 Hz
T = l/(2/n) = n/2 s
co
it
k — 48 lb/ft
m — 0.5 slug.
and
Using the formulas of Problem
1
1.7,
we have
= ^48/0.5 = 9.80 Hz
Circular frequency:
co
Natural frequency:
f = 9.80/271 = 1.56 Hz
T = 1/1.56 = 0.64 s
Period:
Determine the circular frequency, natural frequency, and period for the vibrations described in Problem 11.3.
f
k = 40 lb/ft
T = 1/(4/tt) = tc/4 s.
In that problem
and
11.11
m = 4 slugs.
and
Determine the circular frequency, natural frequency, and period for the vibrations described in Problem 11.2.
I
11.10
k — 64 lb/ft
m = 0.625 slug. Then
and
Write the displacement found in Problem
I We have
k
- 40 lb/ft,
1
1.4 in the
m = 0.625 slug,
\
form
- J ft.
c
cp
= arctan
2^0.625/40
= arcian
1
- n/4.
= V40/0.625 = 8 Hz; / = 8/2tt = 4/tt Hz;
x = A cos (cot — cp).
= 2 ft/s,
Substituting these values into the formulas derived in Problem
and
co
1
and
1.6 yields
o)
- 8 Hz
(from Problem
11.10).
A = V(i) 2 + (2) 2 (0.625/40) = 0.35 ft
x — 0.35 cos (8f — it/4).
Thus
1/4
11.12
Write the displacement found in Problem
I
Since
must have
1
1.4 in the form
A sin (cot + cp) = A sin cot cos cp + A sin cp cos cor
co
— 8,
/I
cos c/> = |,
2
(i)
^ sin
2
+(i)
</>
tan
Also, because
and
A %'\x\ ^> — \.
= /l 2 cos 2
=
—=
1/4
1,
tf>
x = A sin (cot + cp).
+ /4 2 sin 2 c6 = A 2
we have
x — \ cos 8r + \ sin 8r,
and the displacement is
we
Squaring and adding give
cp
or
— arctan = n/4.
1
A=
^=
Thus
0.35
x = 0.35 sin (8r + n/4).
-4cosc6
11.13
A 20-g mass suspended from the end of a vertical spring stretches the spring 4 cm from its natural length.
Assuming no external forces on the mass and no air resistance, find the position of the mass at any time t if it is
pulled 1 cm below its equilibrium position and set into motion with an initial velocity of 0.5 cm/s in the upward
direction.
APPLICATIONS OF SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS
I By Hooke's law, the spring constant k satisfies the equation
addition,
m = 20 g,
x
=
1
and
cm,
= -0.5 cm/s
v
20(980) = /c(4),
D
257
k = 4900 dynes/cm.
so that
In
(the minus sign is required because the initial velocity
is in the upward or negative direction). Then from Problem 1.70 we have
mass at any time t is given by (J) of Problem 11.5 as
x + 245x = 0.
The position of the
x = cos y/l45t - 0.5^20/4900 sin ^245* = cos V245r - 0.03 94 sin J245t.
1
11.14
Determine the circular frequency, natural frequency, and period for the vibrations described in the previous
problem.
I Using the formulas of Problem
T = 1/2.49 = 0.40 s.
11.15
1
we have
1.7,
w = 74900/20 = 15.65 Hz; / = 15.65/2tt = 2.49 Hz;
and
A 10-kg mass attached to a spring stretches it 0.7 m from its natural length. Assuming no external forces on the
mass and no air resistance, find the position of the mass as a function of time if it is pushed up 0.05 m from its
equilibrium position and set into motion with an initial velocity of 0.1 m/s in the upward direction.
f
It
v
— —0.1 m/s,
follows from Problem 1.74 that
k = 140 N/m.
Then with
m = 10 kg,
= -0.05 m,
x
and
the motion of the mass is given by (/) of Problem 11.5 as
x = - 0.05 cos Vl40/10r - 0. 1 VlO/140 sin Vl40/10r = - 0.05 cos yfl4t - 0.0267 sin yf\At
11.16
Find the amplitude and period of the vibrations described in the previous problem.
I It follows from Problem 1.6 that the amplitude is
A = V*o + vl(m/k) = V(-0.05) 2 + (-0.1 2 10/140) = 0.0567 m.
1
)
It
11.17
(
T = In^mjk = 2^^10/140 = 1.68 s.
follows from Problem 11.7 that the period is
A 10-kg mass attached to a spring stretches it 0.7 m from its natural length. The mass is started in motion from
the equilibrium position with an initial velocity of 1 m/s in the upward direction.
if the force due to air resistance is
Find the subsequent motion,
— 90x N.
I The differential equation governing the vibrations of this system is given in Problem 1.78 as
and we have the initial conditions x(0) =
and x(0) = — 1. The solution to the
2t
—
differential equation is found in Problem 8.17 to be
x(t)
c e~
+ c 2 e~ 7t and differentiation yields
x + 9x -f 14x = 0,
,
x
x{t)
= -2c e~ 2t -lc 2 e- lt
.
x
Applying the initial conditions to these last two equations, we get
= x(0) - c, + c 2
and
Solving this set of equations simultaneously yields
11.18
c}
-
= —^
- x(0) = - 2c, - 7c 2
1
and
c2
= \,
so that
x(f)
= |(e~ 7 — e~ 2
'
').
Classify the motion described in the previous problem.
I The vibrations are free and damped. The roots of the characteristic equation are real (see Problem 8.17), so the
system is overdamped. Since
11.19
x ->
as
t
-> oo,
the motion is transient.
A mass of 1/4 slug is attached to a spring, whereupon the spring is stretched 1.28 ft from its natural length. The
mass is started in motion from the equilibrium position with an initial velocity of 4 ft/s in the downward
Find the subsequent motion of the mass if the force due to air resistance is -2x lb.
direction.
f The differential equation governing the vibrations of this system is given in Problem 1.77 as
x + 8x + 25x = 0,
and x(0) = 4. The solution to the differential
x(0) =
= c^" 4 cos 3f + c 2 e~ At sin 3f.
4
4(
4
4
sin 3r + 3<?~ 'cos3r).
x(f) - c (-4e~ 'cos 3t - 3e" 'sin3f) + c 2 (-4e~
and we have the initial conditions
equation is given in Problem 8.53 as
Differentiation of x(t) yields
x(f)
'
1
Applying the initial conditions, we obtain
= x(0) = cj
from which we find
11.20
c,
=
and
c2
- f.
and
Then
x(t)
4 = x(0) = -4c, + 3c 2
= %e 4 sin It.
'
Classify the motion described in the previous problem.
f The vibrations are free and damped.
Since the roots of the characteristic equation are complex conjugates
(see Problem 8.53), the system is underdamped.
Furthermore,
x -
as
t
-> oo,
so the motion is all transient.
258
11.21
CHAPTER 11
D
A mass of | slug is attached to a spring having a spring constant of 1 lb/ft. The mass is started in motion by
displacing it 2 ft in the downward direction and giving it an initial velocity of 2 ft/s in the upward direction.
Find the subsequent motion of the mass if the force due to air resistance is — lx lb.
f Here m = \,
x + 4.x + 4x = 0.
a = 1, k = 1, and the external force F(t) = 0, so (/) of Problem 1.69 becomes
The solution to this differential equation is x(f) = c e~ 2 + c 2 te~ 2
[see Problem 8.141 with
'
'
x
replacing y(x)].
x(t)
Differentiation yields
x(t)
= — 2c e~ 2 + c 2 (e~ 2t - 2te~ 2
'
l
').
Application of the initial conditions to the last two
equations gives
2 = x(0) = Cj
Thus,
11.22
ct
=2
and
c2
= 2,
so that
x(t)
-2 = x(0) = 2c + c 2
and
j
= 2(1 + t)e~ 2t
.
Classify the motion described in the previous problem.
# The vibrations are free and damped.
Since the roots of the characteristic equation are real and equal (see
Problem 8.141), the system is critically damped. Furthermore,
11.23
x ->
as
Show that free damped motion is completely determined by the quantity
a
t
-* oo,
2
— A km,
so the motion is all transient.
where a is the constant
of proportionality for the air resistance (which is assumed proportional to the velocity of the mass), k is the spring
constant, and m is the mass.
I
For free damped motion,
F(t)
=
.,,
,
of the associated characteristic equation are then
If
a
2
— 4km > 0,
.
/.,
=
—a + yja 2 — 4km
2m
2
a — 4km — 0,
x +—x H
m
and
/.
k
=
2
x
The roots
0.
—a — >Ja 2 — 4km
.
2m
— 4km < 0.
The corresponding motions are, respectively, overdamped. critically damped,
the roots are real and distinct; if
the roots are complex conjugates.
—m =
a
and (/) of Problem 1.69 becomes
the roots are equal; if
a
2
and oscillatory damped. Since the real parts of both roots are always negative, the resulting motion in all three
(That the real parts are always negative follows from the fact that for overdamped motion
cases is transient.
we must have
11.24
\Ja
— 4km < a,
2
whereas in the other two cases the real parts are both
—a 2m.)
A 20-lb weight suspended from the end of a vertical spring stretches the spring 6 in from its natural length.
Assume that the only external force is a damping force given in pounds by at. where v is the instantaneous
Find an expression for the vibrations of the system if
motion by displacing it 2 in below its equilibrium position.
velocity in feet per second.
is
set into
a = 8 slugs/s
and the mass
I With mg = 20, it follows that m = 20/32 = 0.625 slug; then from Hooke's law 20 = k(k) or
k = 40 lb/ft.
With a = 8 and the external force F(t) = 0, (/) of Problem 1.69 becomes
64 sin4.8f.
6 4,
cos4.8f + c 2 e~
x + 12.8x + 64x = 0. Its solution is found in Problem 8.62 to be x(f) = c,e"
3
Applying the initial conditions x(0) = g ft and x(0) = r(0) = 0, we find that c, = 8 and c 2 = ,%.
' 6 4 '(3
Thus, x(t) =
cos 4.8f + 4 sin 4.8f
'
&
11.25
,
-
).
Classify the motion described in the previous problem.
f The vibrations are free and damped.
(see
11.26
Since the roots of the characteristic equation are complex conjugates
Problem 8.62), the system is underdamped.
Solve Problem 11.24 if
a = 12.5.
a/m = 12.5/0.625 - 20, so (7) of Problem 1.69 becomes x + 20x + 64x = 0. Its solution is found
Problem 8.11 to be x(t) - c,e" 4 + c 2 e~ l(".
Applying the initial conditions x(0) = £ and x(0) = 0, we find c t = 3*5 and c 2 =~ys- Thus,
I Here
in
x(t)
11.27
'
= is(4e-*'-e- 16
').
Classify the motion described in the previous problem.
f The vibrations are free and damped.
(see
11.28
Since the roots of the characteristic equation are real and unequal
Problem 8.11), the system is overdamped.
Solve Problem 11.24 if
a = 10.
APPLICATIONS OF SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS
I Here
a/m = 10/0.625 = 16,
Problem 8.148 to be
x(t)
so (/) of Problem 1.69 becomes
= c\e~ s + c 2 te~ St
11.29
= £(l + 8t)e~ 8
Its
solution is found in
'
Applying the initial conditions
x(t)
x + 16x + 64x = 0.
259
x(0)
=£
.
and
= 0,
x(0)
we find that
ct
=$
and
c,=f.
Thus,
'.
Classify the motion described in the previous problem.
f The vibrations are free and damped.
Since the roots of the characteristic equation are real and equal (see
Problem 8.148), the system is critically damped: a smaller value of a would result in an underdamped system; a
larger value, in an overdamped system.
11.30
Solve Problem 11.2 if the system is surrounded by a medium offering a resistance in pounds equal to t>/64, where
the velocity v is measured in feet per second.
m = 0.5 slug,
I As in Problem 11.2,
a = -£4 slug/s,
k = 48 lb/ft,
x(0)
= \,
and
x + j^x + 96x = 0.
so (J) of Problem 1.69 becomes
x(0)
= 0.
In addition, we now have
Its solution is found in
Problem 8.63 to be
= C e-° 015625 'cos9.7979t + C 2 e-° 015625 'sin9.7979t.
x(r)
1
Differentiating yields
x( t) = *>-°
015625
'[(9.7979C 2 - 0.015625CJ cos 9.7979* - (9.7979Q + 0.01 5625C 2 ) sin 9.7979r]
Applying the initial conditions, we obtain
i = x(0) = C,
- x(0) = -0.015625^ + 9.7979C 2
and
from which we find C, = I and C 2 = 0.0002657. Then
X (t) = f>-°- 015625 '(i C os9.7979r + 0.0002657 sin 9.7979r).
11.31
Find the amplitude and frequency of the motion of the previous problem.
I The natural frequency is
the amplitude is
A = V(^
/'
= 9.7979/27T = 1.56 Hz;
_0015625( 2
+ (0.0002657£T
)
it
remains constant throughout the motion. In contrast,
0015625f 2
)
= 0.167<T° ° 15625
decreases with each oscillation, owing to the effect of the damping factor e
At
t
— 0,
11.32
-0.015625r = i
5
orwhen
f
w
= 70s.
Find the amplitude and frequency of the motion described in Problem 11.24.
I The natural frequency is / = 4.8/27T = 0.76 Hz.
11.33
_
there is no damping, and the amplitude is at its maximum of 0.167. The amplitude reaches
-°-°i5625r _ 2^
or h en t — 26 s. It is one-third of its maximum when
e
two-thirds of its maximum when
e
Thus the amplitude
'.
-° 0l5625t
The amplitude is
A = ^ 6 4 'V(ns) 2 + (ye) 2 = A e~ 6A
-
'-
Solve Problem 11.2 if the system is surrounded by a medium offering a resistance in pounds equal to 64u, where
the velocity v is measured in feet per second.
I With
m = 0.5,
k = 48,
= 0,
= \,
= 0,
a = 64,
and now
we have, from (7) of Problem
7544
+ C 2 e~ I27 2
x + 128x + 96x = 0. Its solution is given in Problem 8.12 as x = C^e'
_ 127 2
7544r
- 127.2C 2 e
v = -0.7544C,e~°
Differentiating once with respect to t yields
When t = 0, we have x = £ and v = 0. Thus
F(t)
x(0)
x(0)
'.
'
1.69,
'.
+C = i
and
C 2 = -0.0001.
Then
C
so that
11.34
Ci= 0.1677
and
l
2
x = 0.1677^"°
7544
'
- O.OOOle" 127 2
-
'.
Describe the motion of the system of the previous problem.
I The motion is not vibratory but overdamped.
the position of equilibrium as t increases.
11.35
-0.7544C, - 127.2C 2 =
After the initial displacement, the mass moves slowly toward
The motion is completely transient.
Assume the system described in Problem 11.3 is surrounded by a medium that offers a resistance in pounds
equal to ax, where a is a constant. Determine the value of a that generates critically damped motion.
f
Here
when
m = 0.625 and k = 40. It follows from Problem 11.23 that critically damped motion will occur
= a 2 - 4km = a 2 - 4(40)(0.625), or when a = 10 slugs/s.
,
CHAPTER 11
260
11.36
Assume the system described in Problem 11.1 is surrounded by a medium that offers a resistance in pounds equal
Determine the value of a that generates critically damped motion.
to ax, where a is a constant.
/ Here
m = 128/32 = 4 slugs,
critically damped
11.37
and k = 64 lb/ft (see Problem 1.71).
motion will occur when
= a 2 — 4km = a 2 — 4(64)(4)
It
follows from Problem 11.23 that
or when
a = 32 slugs/s.
Assume the system described in Problem 11.13 is surrounded by a medium that offers a resistance in pounds
equal to ax, where a is a constant. Determine the value of a that generates critically damped motion.
f Here
m = 20 g and k = 4900 dynes/cm. It follows from Problem^ 11.23 that critically damped motion
= a 2 - 4km = a 2 - 4(4900)(20) or when a = 280 v 5 g/s.
will occur when
11.38
A 10-kg mass is attached to a spring having a spring constant of 140 N/m. The mass is started in motion from
the equilibrium position with an initial velocity of 1 m/s in the upward direction and with an applied external force
F(t)
= 5 sin
# Here
becomes
t.
Find the subsequent motion of the mass if the force due to air resistance is — 90x N.
m = 10, k = 140, a = 90, and F(t) = 5 sin
The equation of motion, (1) of Problem 1.69,
x + 9x + 14x = ^sin
Its solution is found in Problem 9.89 (with q replaced by x) to be
t.
t.
9
= c e~ 2 + c 2 e~ 11 + s^sinf - § 6o
cost.
Applying the initial conditions x(0) =
and
x = ^o(-90e" 2 + 99e~ 7 + 13 sin -9cosf).
x(t)
'
l
'
11.39
'
x(0)
= — 1, we obtain
t
Identify the transient and steady-state portions of the motion of the previous problem.
I The exponential terms that comprise the homogeneous (or complementary) solution represent an associated
free overdamped
motion. These terms quickly die out. and they represent the transient part of the motion. The
terms that are part of the particular solution (see Problem 9.89) do not die out as
steady-state portion of the motion.
t -* oo,
so they comprise the
Observe that the steady-state portion has the same frequency as the forcing
function.
11.40
A 1-slug mass is attached to a spring having a spring constant of 8 lb/ft. The mass is set into motion from the
Find the subsequent
F(f) = 16cos4f.
equilibrium position with no initial velocity by applying an external force
motion of the mass, if the force due to air resistance is —4.x lb.
I
Here
m—
1
k = 8 lb/ft,
slug,
x + 4.x + 8.x = 16cos4f.
2
cos 2f + c 2 e
\( i) —
e
'
c
Its
2|
solution
sin 2f
•
x(t)
a = 4 slugs/s,
is
and
F{t)
— 16cos4f.
Then (7) of Problem 1.69 becomes
given in Problem 9.83 as
+ * sin 4r — § cos 4t
and differentiation yields
,
=(-2c, + 2c 2 )e~ 2, cos2f + (-2c, - 2c 2 )e" 2 'sin2f + ^cos4f + ^sin4?
Applying the initial conditions, we obtain
= .x(0) = c x - \
so that
11.41
= I
c,
and
c2
= -|.
Then
= .x(0) = - 2c, + lc 2 + ^
and
x = e" 2 '( 2 cos 2f - f sin It) + f sin4f - |cos4r.
Describe the motion of the system of the previous problem.
f
The motion consists of overdamped transient vibrations which are due to the homogeneous (or
complementary) function, namely e~ 2 '(f cos2f — f sin 2f), along with a harmonic component that does not
—
tend to zero as
The latter, namely f sin4f
|cos4f, is the steady-state part of the solution. The
t -+ oo.
steady-state oscillations have a period and frequency equal to those of the forcing function.
F(t) = 16cos4f.
The natural frequency is
/ = 4/2tt = 0.637 Hz,
while the amplitude of the steady-state vibrations is
A = V(4/5) + (-2/5) = V20/5.
2
11.42
2
Derive the differential equation governing the motion of the mass in Problem 11.30 if, in addition, the "fixed" end
of the spring undergoes a motion
y = cos 4f ft.
# Take the origin as the equilibrium position of the spring with the mounting fixed
the distance of the mass from the origin (see Fig. 11.1).
(y = 0).
and let x denote
The restoring force on the spring is
-k{x-y)= -48(x - cos4r).
The force due to air resistance is — ^x lb, so by Newton's second law of motion we have
— 48(x — cos 4r) — ^x = mx. Since m — 0.5 slug, this equation may be written as
while x(0) = 0.
x(0) = y(0) + £ = 1 + £ = |,
t = 0,
Note that at
x + j^x + 96x = 96 cos 4t.
APPLICATIONS OF SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS
261
'////////////////////,
T
X
L_
11.43
Fig. 11.1
Find an expression for the motion of the mass in the previous problem.
f The solution to the differential equation of the previous problem is found in Problem 9.84 to be
x — e 00,56, (C cos9.8f + C 2 sin9.8r) + 0.0019sin4f + 1.2 cos At.
1
0.0156r
With
11.44
=
i*0)
0.01561
[(9.8C 2 - 0.0156C!)cos9.8t - (9.8C,
Differentiating once with respect to t yields
+ 0.01 56C 2 sin 9.8f] + 0.0076 cos At - 4.8 sin At
)
and x(0) - 7/6, we find that C x = - 1/30, C 2 = -0.0008,
(-0.0333 cos 9.8r - 0.0008 sin 9.8f) + 0.0019 sin At + 1.2 cos At.
and
Describe the motion of the system of the previous problem.
# The motion consists of a damped harmonic component which gradually dies away (a transient component)
and a harmonic (steady-state) component which remains. The steady-state oscillations have a period and a
frequency equal to those of the forcing function y — cos4f, namely, a period of 27r/4 = 1.57 s and a
frequency of
11.45
4/2rc
= 0.637 Hz.
The steady-state amplitude is
v^O.0019)
2
+ (1.2) 2 = 1.2
ft.
A mass of 20 lb is suspended from a spring which is thereby stretched 3 in. The upper end of the spring is then
y = 4(sin 2f + cos 2t) ft. Find the equation of the motion, neglecting air resistance.
given a motion
I Take the origin at the center of the mass when it is at rest. Let x represent the change in position of the
mass at time
The change in the length of the spring is x — v, the spring constant is 20/0.25 = 80 lb/ft,
cos 2r).
and the net spring force is — 80(x — v). Thus d 2 x/dt 2 + 128x = 512(sin 2t
t.
-I-
Assuming that the spring starts from rest without any additional displacement, we have the initial conditions
x(0) =
+ 4(sin + cos 0) = 4 and x(0) = 0.
11.46
Find an expression for the motion of the mass in the previous problem.
I The solution to the differential equation of the previous problem is found in Problem 9.86 to be
Differentiating once with respect to
yields
4.129(sin 2f + cos 2f).
x = Cj cos >/l28f 4- C 2 sin ^JYlSt
=
v
%fmc sin yfl2St + y/l2&C 2 cos Jmt + 8.258(-sin It + cos It).
we have
i(0) = 0,
Since x(0) = 4 and
t
-I-
l
4 =
C = 0.129
d + 4.129
and
y/l2&C 2 + 8.258 =
C 2 = 0.730. Then
x = -0.13cos>/l28f-0.73sin^l28r + 4.13(sin2f + cos2f).
from which
11.47
l
and
k = 50 lb/ft
A mass of 64 lb is attached to a spring for which
at time r if a force equal to 4 sin 2r
is
and brought to rest. Find the position of the mass
applied to it.
f Take the origin at the center of the mass when it is at rest. The equation of motion is then
d2x
.
64d 2 x
m
A
=
or
-—r
It
+ 25x 2 sin 2f. Its solution is found in Problem 9.84 to be
4
sin
50x
+
2
2
dt
32 dt
x = C, cos 5f + C 2 sin 5r + yr sin It.
Differentiating once with respect to t yields
conditions
x =
and
v =
x = -0.038 sin 5r + 0.095 sin 2f.
different periods.
v
= -5Cj sin 5t + 5C, cos 5f + 24 cos 2t.
we find C, = 0, C 2 = -j^j, and
,
Then, from the initial
when t = 0,
The displacement here is the algebraic sum of two harmonic diplacements with
CHAPTER 11
262
11.48
A 128-lb weight is attached to a spring having a spring constant of 64 lb/ft. The weight is started in motion with
no initial velocity of displacing it 6 in above the equilibrium position and by simultaneously applying to the
weight an external force F{t) — 8 sin 4f. Assuming no air resistance, find the equation of motion of the weight.
m = 4,
f Here
k = 64,
a = 0,
and
= 8 sin 4r; hence, (1) of Problem 1.69 becomes x + 16x = 2 sin At.
x = c, cos4t + c 2 sin4t - ^f cos4r. Applying the initial conditions
F(t)
solution is found in Problem 9.173 to be
Its
= —\
we obtain, finally, x = — \ cos At + yg sin At — £ t cos At.
? -> x.
This phenomenon is called pure resonance. It is due to the forcing function
F(t) having the same frequency as the circular frequency (see Problem
1.7) of the associated free undamped system.
x(0)
and
x(0)
= 0,
|x|
-» oo
as
Note that
1
11.49
Solve Problem 11.3 if, in addition, the mass is subjected to an externally applied force
m — 0.625,
I With
is
k = 40,
a — 0, (/) of Problem 1.69 becomes
= c, cos 8f + c 2 sin 8f + 4f sin 8f.
ft
and x(0) = 0, we find that
x(0) —
11.50
Its solution
x(f)
Applying the initial conditions
x(f)
= 40 cos 8r.
x + 64x = 64 cos 8f.
and
found in Problem 9.175 to be
F(f)
},
ct
={
and
c2
— 0.
Then
= gCos8f + 4f sin8f.
Describe (physically) the motion of the previous problem as t increases.
I
As t increases, the term At sin 8f increases numerically without bound so that the amplitude of the motion
The spring will ultimately break. This illustrates the phenomenon of resonance and
shows what can happen when the frequency of the applied force is equal to the natural frequency of the system.
increases without bound.
11.51
A mass of 16 lb is attached to a spring for which k — 48 lb ft.
y — sin N 3gt ft.
Find the motion of the mass if. from rest, the
support of the spring is given a motion
We take the origin at the center of the mass when it is at rest, and let x represent the change in position of the
mass at time t. The stretch in the spring is then x - y, and the spring force is — 48(.x — y). Thus, the
2
r16<* X
d2X
- = — 48(x
—
—-=T + 3#x = 3g sin y/3gt. Its solution is found
equation of motion is
48(.v
sin
\3at)
or
-j-j
3</M
N
T
I
g
in
Problem 9.176 to be
x = (', cos v Jgt
Differentiation gives
v
= -
using the initial conditions
\
dt-
til-
=
'
(
',
\
C
2 sin \ 3gt
\ 3</ sin x 3gt
x =
and
\
\ lai cosy/Jot.
+ C 2 \ 3y cos N 3gt —
r-0
when
— 0.
t
\
\ 3<y cos N 3gt +
we find that
— sin y/3gt.
t
C, =
and
Then.
C 2 = ':
thus,
\\ ?>ut cos v 3gt.
sin N 3gt
The first term of this solution represents a simple harmonic motion, while the second represents a vibratory
motion with increasing amplitude (resonance). As
is
mechanical breakdown.
t
increases, the amplitude of the oscillation increases until there
;i
MECHANICS PROBLEMS
11.52
A particle P of mass 2 g moves on the v axis toward the origin O acted upon by a force numerically equal
Determine the differential equation governing the motion of the particle.
to 8x.
f Choose the positive direction to the right (Fig.
and so is — 8x.
x < 0,
When
d2X
:
— = - 8x or —r +
</
law.
2
.Y
r-
</r
1
When
1.2).
the net force is to the left (i.e., negative)
4.x
— 8x. Thus by Newton's
= 0.
in-
O
11.53
> 0.
\
the net force is to the right (i.e., positive) and so is also
P
Fig. 11.2
Find an expression for the position of particle P of Problem 11.52 as a function of time if the particle is
initially at rest at
x — 10 cm.
f The solution to the differential equation of the previous problem is found in Problem 8.59 [with x(t) here
replacing v(.x)] to be
x — c, cos 2/ + c 2 sin It.
Applying the initial conditions, we get
becomes
x=10cos2f.
Then
v
10 = x(0) = c,
=
—= —
2cj sin It + 2c 2 cos 2f.
at
and
= t(0) = 2c 2
or
c2
= 0.
Then the solution
~
"
APPLICATIONS OF SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS
11.54
263
Describe the motion of the particle of the previous problem.
# The graph of the motion is shown in Fig. 11.3.
It is simple harmonic motion with an amplitude of 10 cm, a
period of n s, and a natural frequency of \/n Hz. The particle starts out at x = 10 cm at time zero and begins
moving toward the origin, picking up speed as it moves. Its velocity is greatest in absolute value (i.e., its speed
is greatest) at time
t = n/4,
when the particle reaches the origin. The velocity is negative at that time, so
the particle continues through the origin; it begins slowing as its acceleration changes sign.
zero at time
(x
= n/2,
x = - 10.
when
(Thus, at time
t
= n/2,
The velocity reaches
the particle is at rest 10 cm to the left of
The particle then begins picking up velocity as it is accelerated toward the origin. It reaches the
at
t = 3n/4
with maximum speed. Once it is through the origin, its velocity decreases until it
the origin.)
origin
t
= 0)
again comes to rest at time
t
= n,
x = 10.
now at
This completes one cycle of the motion; the particle will
continue to repeat that cycle in the absence of other forces.
Fig. 11.3
11.55
Solve Problem 11.53 if the particle is also subject to a damping force (or resistance) that is numerically equal to
eight times the instantaneous velocity.
The damping force is given by —8 dx/dt, regardless of where the particle is. Thus, for example, if x <
and
then the particle is to the left of O and moving to the right, so the damping force must be acting to
f
dx/dt > 0,
2
the left (i.e., negative).
Thus by Newton's law,
2
d x
„ dx
—
—= — 8x — 8 —
T =-8x-8
-r-
dt
2
d 2x
—- + 4x =
—
— + 4 dx
or
=-r-y
2
dt
dt
The solution to this equation is found in Problem 8.141 [with x(f) here replacing j(x)] to be
2
when t = 0, we have c = 10 and
'(c
+ c 2 t). Since x = 10 and dx/dt =
x = e
:,
x= 10f
11.56
t
2,
0.
dt
c2
— 20.
Then
+2r).
(1
Describe the motion of the particle in the previous problem.
f
For all
t
> 0,
x — 10?
2
'(1
+ It)
positive; furthermore, x tends to zero as
is
t
-> oo.
Thus particle P
approaches the origin but never reaches it. In addition, the velocity of the particle, v — dx/dt = — 40te~ 2
indicating that the particle is always heading in the same direction (the negative x
negative for all
t > 0,
'
is
direction); thus the motion is nonoscillatory.
11.57
A particle of mass m is repelled from the origin O along a straight line with a force equal to
distance from O.
k >
times the
Determine the differential equation governing the motion of the mass.
I
Denote the line on which the particle moves as the x axis, taking the positive direction to be to the right of the
When x > 0, the net repellent force is to the right (i.e., positive) and is kx. When x < 0, the net
repellent force is to the left (i.e., negative) and is also kx. Since the repellent force is kx in both cases, by Newton's
2
k
d2x
d x
x = 0.
second law of motion we have m -—2r = kx or —-=2
origin.
dt
dt
11.58
m
Find an expression for the position of the particle of the previous problem if it starts from rest at some initial
position x
.
# The solution to the differential equation of the previous problem is found in Problem 8.22 to be
x(f)
= c e^ kmt + c 2 e~ y,TI7"
t
Differentiating this equation yields
.
x
x'(f)
= c sfkjme- kJmt - c 2 >Jkjme~
-
* "".
,
Applying the initial conditions, we obtain
x
= x(0) = c, + c 2
Solving these two equations simultaneously, we find that
Wm + e~ a7^") = x cosh sfk/mt.
x(t) = \x {e^
= x'(0) = c, yjk/m — c 2 yjk/m
and
c
{
—c 2 — \x
,
so that
264
11.59
CHAPTER 11
D
k =
Find the position and velocity of the mass described in Problem 1 1.58 after 2 s, if numerically
m
and
initially the particle starts from rest 12 ft to the right of the origin.
k —
m and x = 12, the result of the previous problem becomes x(r) = 6(e' + e~'). Therefore, the
= dx/dt — 6(e' — e~'). At
= 2, these equations become x(2) = 6(e 2 + e~ 2 = 45.15 ft and
velocity
2
2
=
=
43.5ft/s.
6(e -ei;(2)
I With
v(x)
is
t
)
)
11.60
Determine when the particle described in the previous problem will be 18 ft from the origin, and find its velocity
at that time.
I
Setting
= 6{e' + e~') — 18
x(t)
formula. Thus, we find
= - '-——
e'
discard the negative choice and take
r(0.962) = 6(e
11.61
1
and solving for t, we obtain
equation by e' and rearranging then yield
2
(e')
— 3e' +
and
— 0.962
t
1
= 0,
+ e~' — 3 = 0.
Multiplying this last
which may be solved for e' with the quadratic
—
= In —=-
f
e
.
Since time is positive in this problem, we
At that time, the velocity is
s.
— e 0.986 )= 13.4 ft/s.
0.962
Determine the equation of motion for a mass m that is projected vertically upward from a point on the ground
if the
resistance of the air is proportional to its velocity.
We designate the point on the ground from which the flight began as the origin O. We take upward as positive,
and let x denote the distance of the mass from O at time t (Fig. 1 1.4). The mass is acted upon by two forces, a
dx
gravitational force of magnitude mg and a resistance of magnitude
Kv = K
both directed downward.
I
dx
Hence,
d2 x
dx
dx
~d~X
m —= = —mg — K
— + k— =
2
dx
d x
or
dx
2
g,
K = mk.
where
dt
Fig. 11.4
11.62
Find the maximum height attained by the mass in the previous problem if it is projected with initial velocity t
.
I The solution to the differential equation of the previous problem is found in Problem 9.169 to be
x = C, + C 2 e~ k
'
—-
X
.
Differentiating yields
r
=
— = —kC e~ —
kt
2
dx
k
When
x
= 0,
x =
and
v
=v
.
C = —C 2
Then
1
replacements, we get
x — —r (g + kv ){l — e
*')
—
k
occurs when
e~
kt
11.63
T
+
F
Making these
The maximum height is reached when
x.
:
g (1
or
f
—
= T In -
v
= 0.
This
Then the maximum height is
.
9
g + kv Q \
if
g
g + kv
kv {
A perfectly flexible cable hangs over a frictionless peg with 8
other side (see Fig.
C> =
k
= - ——
= r^r^
2
k C2
g + kv
x = p- (g + kv )[
and
x
-.
k
11.5).
ft
of cable on one side of the peg and 12 ft on the
Find the differential equation for the motion of the sliding cable.
f We denote the total mass of the cable by m, and the length (in feet) of cable that has moved over the peg at time
by x. At time x there are 8 - x ft of cable on one side and 12 + x ft on the other. The excess of
x
)
APPLICATIONS OF SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS
265
Hi
Fig. 11.5
2
4 + 2x ft
on one side produces an unbalanced force of
(4
+ 2x)mg/20 lb.
m —2 = (4 + 2x)
Thus,
=-
c/f
d2x
-r-y
2
— _0_
—x = -.
or
20
Observe that the motion is not influenced by the mass of the cable.
10
dt
11.64
—
Find the time required for the cable of the previous problem to slide off the peg, starting from rest.
I The solution to the differential equation of the previous problem is found in Pro blem 9.26 to be
x = C e" 9/101 + C 7 e v g! Oi - 2. Differentiating once with respect to t yields v = y/g/lOiC^ 9 10 - C 2 e _vWT7>
1
'
'
x
When
Hence
peg,
11.65
t
= 0, x =
and
= yflO/gcosh'
t
v = 0.
C, = C, =
Then
\{x + 2) = y/l0/g\n
1
1
and
x + 2 + V* 2 + 4x
')-
x = e v9/10 + e' /97n5 -2 = 2 coshyfg/Wt - 2.
'
'
When
x = 8 ft
of cable has moved over the
= y/10/g In (5 + 2 V6) s.
t
Determine the differential equation for the motion of the cable described in Problem 1 1.63 if, in addition, the force
of friction over the peg is equal to the weight of 1 ft of the cable.
I The force of friction retards the motion of the cable (so it is negative) and in absolute value is equal to mg/20.
According to the analysis developed in Problem 11.63, it follows that the net force on the cable is
rng
d2x
mg
mg
= (3 + 2x)-. Then Newton's second law of motion gives m —^ — (3 -I- 2x)
(4 + 2x)
_-_
2
d x
dt
11.66
g_
2
X =
10
df
—
or
20
3g
20'
Find the time required for the cable in the previous problem to slide off the peg, starting from rest.
f The solution to the differential equation of the previous problem is found in Problem 9.27 to be
v?7T °
and v{0) = 0, we find
x = de v?7T "' + C 2 e"
-f. Applying the initial conditions x(0) =
-v9/10r
v9/l0r
I
_
=
1).
e
+ e
so that x
C,=C 2 =
I
| (cQsh
=
=
y/g/l0t - 1),
or cosh -Jg/lOt = ^f.
write
(cosh
Thus,
we
8
which
8.
for
x
We seek the value of t
f
2
= 710/^(2.53268) = 1.42 s, where
Then JgJlOt = cosh" ^ = In (^ + V(19/3) - 1) = 2.53268, and
we have taken g — 32 ft/s 2
r
J^
t
x
t
.
11.67
Show directly that for
v
> 0,
cosh
l
v
= \n{v ± yjv 2 - 1).
1
y
y
y
y
y = cosh" v, so that v = cosh y = %(e + e~ ); then we rewrite this as e + e~ - 2v = 0.
y
y 2 y
+1=0,
and
the
quadratic
formula
2ve
gives
then
yield
(e
Multiplying by e and rearranging
)
f We set
e>
11.68
=
—=^-
= v± y/v 2 — 1.
Show that if cosh"
for
v
1
v is
y = In (v ± yjv
Thus
— 1).
Since
y = cosh
known to be positive (from physical considerations), then
'
the identity follows.
v,
cosh
'
v
= \n(v + yjv 2 - 1
> 0.
f We have, from the previous problem, that
cosh
"l
minus sign in front of the square root as a possibility.
that
2
v
> 1.
It also follows that
2t;
> 2
and that
v = In (i> ± s]v
-1>
-
1 ),
y = cosh
If we set
2v
2
1.
so all that remains is to eliminate the
v,
it
follows that
v
= cosh y
and
266
CHAPTER 11
D
Then multiplying by — 1 reverses the sense of the inequality, and adding v2 gives us
Factoring yields
(v
then it must be that
- l) < V'
2
In (v
or
1
— y/v 2
v
is
1 )
— 1 < V^
S* - 1,
which we write as
o
2v + 1 < v 2 - 1.
„2
— yjv —
2
1
< 1
If this is so,
But this is a contradiction if cosh ~ * v is known to be positive
negative.
so the minus sign is impossible.
HORIZONTAL-BEAM PROBLEMS
11.69
Derive the differential equation for the deflection (bending) of a uniform (in material and shape) beam under
specified loadings.
I
It is
Fig.
convenient to think of the beam as consisting of fibers running lengthwise. In the bent beam shown in
1 1.6,
the fibers of the upper half are compressed and those of the lower half are stretched, the two halves
being separated by a neutral surface whose fibers are neither compressed nor stretched. The fiber which originally
coincided with the horizontal axis of the beam now lies in the neutral surface along a curve called the elastic
curve (or curve of deflection).
We seek the equation of this curve.
P(x.v)
Fig. 11.6
Consider a cross section of the beam at a distance x from one end. Let AB be its intersection with the neutral
surface, and
to
P its intersection with the elastic curve.
It
is
shown in mechanics that the moment M with respect
AB of all external forces acting on either of the two segments into which the beam is separated by the cross
section is independent of the segment considered and is given by
elasticity of the
EI
M=—
R
,
where E is the modulus of
beam, / is the moment of inertia of the cross section with respect to AB, and R is the radius of
curvative of the elastic curve at P.
P.
For convenience, we think of the beam as being replaced by its elastic curve, and the cross section by point
We take the origin at the left end of the beam, with the x axis horizontal, and let P have coordinates (x, y).
Since the slope dy/dx of the elastic curve is numerically small at all points, we have, approximately,
R =
[1
+ (dy/dx) 2 ] 312
2
d y/dx 2
1
~ d 2 y/dx~'
so that
M = El d^y_
dx 2
The bending moment M at the cross section (or at point P of the elastic curve) is the algebraic sum of the
moments of the external forces acting on the segment of the beam (or segment of the elastic curve) about line
AB in the cross section (or about point P of the elastic curve). We assume that upward forces give positive
moments and downward forces give negative moments.
11.70
Find the bending moment at a distance x from the left end of a 30-ft beam resting on two vertical supports (see
Fig. 1 1.7) if the beam carries a uniform load of 200 lb per foot of length and a load of 2000 lb at its middle.
30-x
±*-
±(30 -x)
±*
15 -x
i
2000
4000
6000 -200x
200x
Fig. 11.7
4000
APPLICATIONS OF SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS
267
I The external forces acting on OP in Fig.
1 1.7 are an upward force at
0, x feet from P, equal to one-half the
}[2000 + 30(200)] = 4000 lb, and a downward force of 200x lb, which we assume is
concentrated at the middle of OP and thus ^x ft from P. The bending moment at P is then
= 4000x - 200x(|x) = 4000x - 100x 2
total load, or
M
11.71
.
Show that the bending moment in the previous problem is independent of the segment used to compute it.
I Consider the forces acting on segment PR. They are (1) an upward force of 4000 lb at R, 30 - x ft from P; (2)
the load of 2000 lb acting downward at the middle of the beam,
15 - x ft
from P; and (3) a force of
200(30 — x) lb
|(30 — x) ft
downward, assumed to be concentrated at the middle of PR,
from P. Then the
total moment is, again,
M = 4000(30 - x) - 2000(15 - x) - 200(30 - x) |(30 - x) = 4000x - 100x
•
11.72
2
A horizontal beam of length 2/ ft is freely supported at both ends. Find the equation of its elastic curve and its
maximum deflection when the load is w lb per foot of length.
I We take the origin at the left end of the beam, with the x axis horizontal as in Fig.
1
1.8.
Let P be any point
on the elastic curve, with coordinates (x, y), and consider segment OP of the beam. There is an upward force
of w/ lb at O, x ft from P; there is also a load of wx lb at the midpoint of OP, \x ft from P. Then, since
2
EI d 2 y/dx 2 — M,
we have
El
= w/x —
—
dx
d y
r
2
wx(jx) = w/x
jWX 2
1
ix—Np<*.y) V x '
.
l
v'=o
Fig. 11.8
El dy/dx = \wlx 2 — £wx 3 + C v
and
At the middle of the beam, x =
2
3
3
—
—
— yw/ 3
=
—
=
jwlx
£wx
El
dy/dx
Applying
these
conditions
then
yields
C
^w/
so
that
dy/dx
0.
A second integration now gives Ely — ^w/x 3 — 2iwx* — ^wl 3 x + C 2 At point O, x = y — 0.
Integrating once yields
/
,
t
.
.
Thus,
11.73
C2 =
and
y=
w
——
(4/x
3
- x 4 - 8/ 3 x).
Determine the maximum deflection of the beam in the previous problem.
f The deflection of the beam at any distance x from O is given by — y. The maximum deflection occurs at the
4
w
5vv/
-y max = -^7777 (4/ 4 - 4 - 8/ 4
middle of the beam (x = /) and is
I
24£/
11.74
Solve Problem 1 1.72 if there is, in addition, a load of
)
24£/
W lb at the middle of the beam.
I We choose the same coordinate system as in Problem
1
1. 72.
1
wx
I
(b)
L<x<2L
P(*Ty)
v>U$W
Since the forces acting on a segment OP of the
beam differ according to whether P lies to the left or right of the midpoint, two cases must be considered.
268
CHAPTER 11
D
When
< x < I [Fig. 11.9(a)], the forces acting on OP are an upward force of wl + {W lb at 0, x ft
from P, and the load wx acting downward at the midpoint of OP, {x ft from P. The bending moment is then
M = (wl + {W)x - wx({x) = wlx + {Wx - {wx
[Fig. 11.9(6)], there
an additional force: the load of W lb at the midpoint of the beam,
2
When
x — / ft
(1)
/ < x < 2/
is
from P. The bending moment is then
M = (wl + {W)x - wx({x) - W(x -l) = wlx + {Wx - {wx - W(x yield the bending moment
M = {wl + {Wl when x = The two cases may be treated at
2
I)
Both (/) and (2)
(2)
2
/.
the same time by noting that
wlx + {Wx - {wx 2 = wlx - {wx 2 - {W{1 - x) + {Wl
wlx + {Wx - {wx 2 - W(x - I) = wlx - {wx 2 + {W(l - x) + {Wl
and
EI d 2 y/dx 2 = wlx - {wx 2 + {W(l — x) + {Wl with the understanding that the upper
sign holds for
< x < /, and the lower for / < x < 21.
Integrating this last equation twice yields
Ely = ^wlx 3 — ji wx * + YiW(l — x) 3 + {Wlx 2 + C x + C 2
Using
the boundary conditions
at
and x = 2/ and y =
x=y=
at R, we obtain
C 2 = Wl 3 and
2
Then
C, = -{wl 3 -{Wl
Then we may write
.
1
^
.
Ely = |w/x 3 - ^wx 4 - ]n/ 3 x + ? 2 W{1 - x) 3 + {Wlx 2 - \Wl 2 x + &W1 3
3
- ±wlv
—
6 wjx
L,.,v4
24
-
W - &W\l 3
*
x|
3
+ {Wlx 2 - {Wl 2 x + ^Wl 3
W
u
4
2
3
4/x3 - * " 8/3 *) +
< 3/ *
" I' " x " 6/2x + /3 )
y = ^TF7
7^F7
24£/
2£7
and
<
!
11.75
Determine the maximum deflection of the beam in the previous problem.
I
11.76
x — I,
The maximum deflection occurs at the middle of the beam where
A horizontal beam of length
/
ft
is
fixed at one end but otherwise unsupported.
— y_„
}m3X —
and is
5vv/
4
Wl 3
1
2AE1
.
6EI
Find the equation of its elastic
curve when it carries a uniform load of w lb per foot of length.
Fig. 11.10
f
We take the origin at the fixed end and let P have coordinates (x, y). Consider the segment PR in Fig. 11.10.
n(/ - v) lb
from P. Then
,(/ - x)
at the midpoint of PR.
2
—
—
—
—
—
=
=
w(l
x)]
{w(l
and integrating once yields Eldy/dx — ^w(l — x) 3 + C,.
El d y/dx
x)[£(l
x)
3
and dy/dx = 0; thus C, = -£w/ 3 and we have EI dy/dx = gu(/ - x)
At 0, x =
iw/
Applying the conditions x = y =
Integrating once again gives us
Ely = —ji w(l — x) — h w x + C 2
The only force is the weight
2
ft
2
,
3
.
'
at
11.77
O then yields
C 2 = 2J4 VV 4
'
-
Substitution and simpification finally give
y
(4/x
3
- 6/ 2 x 2 - x 4
).
24EI
Determine the maximum deflection of the beam in the previous problems.
The maximum deflection, occurring at point R (where
11.78
.
x = /),
is
— y,
1
w/
4
8£T
A horizontal beam of length 3/ ft is fixed at one end but otherwise unsupported. It carries a uniform load of
w lb/ft and two loads of W lb each at distances and 2/ ft from the fixed end (see Fig. 11.11). Find the equation of
/
its elastic
curve.
APPLICATIONS OF SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS
D
269
y
<~(Zl-x)w
Fig. 11.11
I We take the origin to be at the fixed end, and let point P have coordinates (x, y). There are three cases to be
<x <
< x < 21, or 2/ < x < 3/. In each case,
considered, according to whether P is on the interval
I,
I
use will be made of the right-hand segment of the beam in computing the three bending moments.
<x <
When
(P = P l
/
in Fig. 11.11), there are three forces acting on
assumed to act at the midpoint of P R, \(3l — x) ft from P,; the load
lb,
from P
The bending moment about P, is then
load
(2/ — x) ft
l
W
W
PjP: the weight
lb,
(/
— x) ft
(3/
— x)w lb
from P,; and the
.
x
d
2
v
1
M, = £7-4z = _(3/-x)w
W(l - x) - W(2l - x) = -- w(3/ - x) 2 - W(l - x) - W(2l - x)
(3/
dx
EI dy/dx = >(3/ - x) 3 + \W(l - x) 2 + \W{2\ - x) 2 + C v
At point O, x =
and
Wl 2 so that
C =
A second integration yields
EI dy/dx = >(3/ - x) 3 + {W{1 - x) 2 + \W{21 - x) 2 - |w/ 3 - f Wl 2
at 0, we have
EIy= -^w(3/-x) 4 -^(/-x) 3 -^(2/-x) 3 -fw/ 3 x-|VK/ 2 x + C 2 Since x = y =
C 2 = ^w/ 4 + fW/ 3 and
Integrating yields
dy/dx = 0;
these conditions give
,
x
.
.
£/ y = -^w(3/ - x) 4 - ±W(l - x) 3 - iW{2l - x) 3 - fw/ 3 x - \Wl 2 x + ^w/ 4 + \Wl 3
When
/
(P — P 2
< x < 2/
in Fig. 11.11), the bending moment about
M — EI d y/dx = — |w(3/ — x) — W(2/ —
2
2
2
2
U)
P 2 is
Integrating twice, we obtain
x).
Now we note that when x =
Ely — ~2iw(3l — x) 4 — ^W(2l — x) 3 -I- C 3 x + C 4
C 3 = Cj and C 4 = C 2 Thus, we have
agree in deflection and slope, so that
.
this equation and (I) must
/
-
Ely = -2^w(3/ - x) 4 - iW{2l - x) 3 - fw/ 3 x - \W\H + ^w/
When
2/
< x < 3/
{P = P 3
M = EI d y/dx = — \w(3l — x)
2
2
in Fig. 11.11), the bending moment about P 3
2
3
.
4
+ f W/ 3
(2)
is
Then, integrating twice and noting that the result must agree with (2) in
deflection and slope, we obtain
-2^w(3/ - x)
EIy= -£w(3!-x)4 + C 5 x + C 6
4
- fw/ 3 x - \Wl 2 x - ^Wl* + § Wl 3
Finally, we combine (7), (2), and (5) as follows:
w
(12/x
24£7
3
W
- 54/ 2 x 2 - x 4 + i— (2x 3 - 9/x 2
)
v
<x <
)
W
w
2
2
3
4
2
2
3
-— 3
^=S 24£7 (12/x - 54/ x - x + 6£7 (x - 6/x - 3/ x +
w
(12/x
/
)
v
3
-54/ 2 x 2 -x 4 +
)
1
6£7
—W
(3/
3
/
)
-5/ 2 x)
< x < 2/
2/
< x < 3/
;/
11.79
Determine the maximum deflection of the beam in the previous problem.
f
The maximum deflection, occurring at point R (where
x = 3/),
—
1
is
-y max =
,
(81w/
+ 48 Wl-
(3)
V
x
CHAPTER 11
270
11.80
A horizontal beam of length
ft is
/
fixed at both ends.
Find the equation of its elastic curve if it carries a
uniform load of vv lb/ft.
Fig. 11.12
f
We take the origin at the left end of the beam and let P have coordinates (x, y), as in Fig. 11.12. The external
OP are a couple of unknown moment K exerted by the wall to keep the beam horizontal
forces acting on segment
upward force of |m7 lb at 0, x ft from P; and the load wx lb acting downward at the midpoint of OP,
2
lux
EI d 2 y/dx 2 = K + kwlx
Integrating once and using the conditions
x —
and
3
dy/dx =
at 0, we obtain
El dy dx = Kx + \wlx 2 - 'u\y
At point R, x —
and dy/dx =
(since the beam is fixed there). Substituting these values into the last
= Kl + £w/ 3 — ^w/ 3 from which K — — 2 vv/ 2 Substitution for K, integration, and the
equation yields
at O; an
\x ft
from P. Thus.
.
.
(
l
,
x = y =
use of
at
.
O finally yield
Ely =
2
wl x
+ — w/x 3
2
24
11.81
4
or
v
=
12
MX"
(2/x
-
2
I
24£/
Determine the maximum deflection of the beam in the previous problem.
I
11.82
n.v
24
4
The maximum deflection, occurring at the middle of the beam (where
Solve Problem
I
I
ISO if, in addition, there is a weight of
We use the coordinate system of Problem
1
1.80.
x = ^/),
— ym
,
is
vv/
384£/
W lb at the middle of the beam.
Figure 11.13 shows that two cases must be considered: x
When
< x < [I. the external forces on the segment to the left
of Pi(x,y) are a couple of unknown moment K at 0: an upward force of \[w\ + W)]b
at O, xft from P,:
and the load wx lb. Iv
from /',. Thus. El d 2 y dx 2 - K + \{wl + W)x
wx - K + \wlx - {wx 2 + [Wx.
\ =
and dy/dx —
at O. we obtain
Integrating once and using the conditions
between
and '/. and x between '/ and
/.
"
It
= Kx + iwlx 2
/:/
,',»
\
'
\
\Wx 2
Integrating once again and using
.
dx
Ely = \K
:
+ r^w/x
3
-
,'
4
x = j
at O,
we get
hv 4 + feWx 2
(/I
±(wl + W)
Fig. 11.13
When
<x <
'/
/.
there is in addition the weight
W lb at the middle of the beam,
Thus. El d y dx =K+ \wlx - 2Wx + \Wx - W(x - \l). Integrating twice yields
\W{x - I/) 3 + f j.v + C 2 When x = \l,
Ely = \Kx 2 + iW-x 3 - 24WX4 + &Wx 3
dy dx here must agree with those for (/). Thus. C.=Ct = O. and
2
x - \l ft
from P 2
.
2
2
.
Ely
\Kx 2 + ,U7x 3 - 24 1VX + hWx
2
U (A - \l)
the values of y and
2
(2)
)
APPLICATIONS OF SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS
x = ,/
To determine K, we let
K = - fowl 2 - IWI,
dy/dx =
and
in the equation for EI
y =
W
<>
—w—
(2/x
o<x<
;/
< x <
/
w
- 2 x 2 - x 4 + -—(/*_ 6/ 2 x + 9/x 2 - 4x 3
3
l
)
48£/
24£/
|/
)
Find the maximum deflection of the beam in the previous problem.
—y
The maximum deflection, occurring at the middle of the beam, is
11.84
dy/dx above. 1 his yields
so that (/) and (2) become
vv
11.83
271
A horizontal beam of length
/
-=
(w/
384£r
fixed at one end and freely supported at the other end.
ft is
the elastic curve if the beam carries a uniform load of vv lb/ft and a weight
I We take the origin at the fixed end (Fig.
1
1.14) and let
4
+ 2 WP
j
'
Find the equation of
W lb at the middle.
P have coordinates (x, y). There are two cases to be
considered.
Fig. 11.14
R,
< x < jl, the external forces acting on the segment P,P are an unknown upward force of S lb at
When
- x ft from P,; the load w(/ - x) lb at the midpoint of PjP, 2 (l - x) ft from P,; and W lb,
jl
—x
I
from P,. Thus, we have
ft
EI d 2 y/dx 2 = S(l - x) - u(/ - x)[\(l - x)] - W{\1 - x) - S(l - x) - \w(l - x) - W(\l - x). Integrating once and
at 0, we obtain
and dy/dx =
using the conditions x =
2
3
2
x
=
\W({1
x)
x)
S(l
x)
+
£w(/
+
+ {SI 2 - £w/ 3 - ^Wl 2 Integrating again
EI dy/dx
2
2
.
x = y —
and using the conditions
Ely = iS(l - x)
3
at
O
yield
- ML - x) - hW{# - x) 3 + (iSl 2 - ^w/ 3 - iWl 2 )x - {SI* + ^w/4 + ^Wl 3
4
(1)
from P 2 and
(/ — x) ft
the forces acting on P 2 R are the unknown upward thrust S at R,
- x) lb, \{l - x) ft from P 2 Thus, EI d 2 y/dx 2 = S{1 - x) - \\v(l - x) 2 from which we get
When x = \l, the values of Ely and EI dy/dx here and in (/)
Ely = lS(l - x) 3 - 24H'(/ - xf + CjX + C 2
the
values of the constants of integration in (/). so that
must
have
must agree. Hence, C, and C 2
When
\l
the load
<x <
I,
,
u(/
.
,
.
Ely = ISO - x)
3
y =
To determine S, we note that
w
(5/x
3
- iMl ~
4
-v)
+ (iS/ 2 - £w/ 3 - iWI 2 )x - kSI" + &w/4 + 2Ls Wl>
x = /.
when
- 3/ 2 x 2 - 2x 4 +
48£/
y =i
vv
(5/x
3
- 3/ 2 x 2 - 2x 4 +
)
48£/
11.85
W (11.x -9/x
3
)
S = |wi + j^W.
so from (2),
It is
clear from the result of Problem
the beam.
y =
vv
48£/
Thus we substitute
/
= 10
1
This yields
< x<U
2
)
96£/
W
(2/
3
96EI
- 12/ 2 x+ 15/x 2 - 5x 3 )
Locate the point of maximum deflection of the beam in the previous problem when
#
(2)
M<x<
/
= 10
/
and
W - lOw.
1.84 that the maximum deflection occurs to the right of the midpoint of
and
W = lOw
2
(_2x 4 + 25x 3 + 450x - 6000x + 10,000).
in the second part of the solution, obtaining
Since
dy/dx =
at the point of
maximum deflection,
272
CHAPTER 11
D
we solve
8.x
3
— 75x 2 — 900x + 6000 = 0,
x — 5.6
finding that the real root is
(approximately). Thus, the
maximum deflection occurs approximately 5.6 ft from the fixed end.
BUOYANCY PROBLEMS
11.86
Determine a differential equation describing the vertical motion of a cylinder partially submerged in a liquid of
density p, under the assumption that the motion is not damped.
I Denote the radius of the cylinder as R, its height as H, and its weight as W.
Archimedes' principle states that
an object that is submerged (either partially or totally) in a liquid is acted on by an upward force equal to the
weight of the liquid displaced.
Equilibrium occurs when the buoyant force upward on the cylinder is precisely equal to the weight of the
Assume that the axis of the cylinder is vertical, denote it by y, and take the upward direction to be the
cylinder.
nR 2 hp = W
At equilibrium, with the cylinder partially submerged,
positive direction.
h —
or
W/nR 2 p,
where h is the number of units of the cylinder's height that are submerged at equilibrium (see Fig. 11.15). Let
the distance from the surface of the water to the equilibrium position as if it were marked on the
y(t) denote
We adopt the convention that
cylinder.
surface, and
<
y\t)
if it is
y{t)
>
if the
equilibrium position on the cylinder is above the
below the surface.
Z>
Equilibrium
r</)
position
y(t)
L_
Equilibrium/^
position
I
I
l_t
I
I
I
(b)
(a)
(c)
Fig. 11.15
According to Newton's second law, the total force F acting on the cylinder can be written
W
F(t)
= — y (t)
y
W
+ buoyant force.
where g is the gravitational constant. The force F{t) can also be described as F(t) = nR\h - y(?)], we have by Archimedes' principle
Since the submerged volume of the cylinder can be written
= nR 2 hp. so it
that
+ nR 2 [h - y(i)]p = - + nR 2 hp - nR 2 y(t)p. But h was chosen so that
Fit) = -
W
W
follows that
F(t)
= -nR 2 vU)p.
and hence
W
W
-
v"(f)
'
= -nR 2 v(t)p.
9
Therefore, the initial-value problem describing the motion of the cylinder is
y
where y
11.87
is
H
w
v
-
=
the initial position of the cylinder, and v
v(0)
=y
its initial
If the
weight
W of the cylinder
is
=v
c
velocity.
In the notation of the previous problem, discuss what happens when
i
y'(0)
W > nR Hp.
2
greater than the buoyant force generated when the entire cylinder is
submerged, then the cylinder must sink to the bottom of the liquid.
11.88
A cylinder with radius 3 in and weight 5rc(~ 15.71) lb is floating with its axis vertical in a pool of water (density
p = 62.5 lb/ft
raised
1
in
3
Determine a differential equation that describes its position y(t) relative to equilibrium if it is
above equilibrium and pushed downward with an initial velocity of 4 in s.
).
APPLICATIONS OF SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS
R = { ft,
# Here
p = 62.5 lb/ft
3
and
,
W = 5\b,
so the result of Problem
1
D
273
1.86 becomes
2
7t(l/4) (62.5)(32)
y
+~
- v =
—2
or
+ 25y = 0.
y
y — c, cos 5t + c 2 sin 5/.
With v(0) = T2 ft and >-'(0) = -\ ft/s
Then y = ^(5 cos 5f — 4 sin 5f).
solution is found in Problem 8.72 (with y here replacing x)
Its
to be
11.89
as initial conditions, it follows that
c,
= 1/12
and
c2
= -1/15.
Solve the previous problem if, instead of water, the cylinder is floating in a liquid having density 125 lb/ft 3
With the data as given, the result of Problem
1
1.86 becomes
-y =
y" H
\"
or
.
+ 50 v = 0.
571
Its
previous problem, we find
11.90
y = c, cosV^Of + c 2 sin y50t.
solution is found in Problem 8.60 to be
cl
— 5/60
and
Applying the initial conditions of the
y = ^q(5 cos y/50t — 2 y/2 sin yj^t).
so that
c 2 ——2yJ2/60,
Find the period and the amplitude of the motion described in Problem 11.86.
# The motion is defined by the initial- value problem that is the result of Problem 11.86. The differential
equation is of the second order and linear, with constant coefficients, having as the roots of its characteristic
— ± iR s/npg/yJW. Hence the general solution to the differential equation is
y(t) — c cos (R\Jnpg/Wt) + c 2 sin (Ryjnpg/Wt).
The initial conditions imply c = y
and
equation
/.
t
c2
t
— \'Wv /RyJnpg,
so that
fnpg
\
sfWv
Therefore, the motion of the cylinder is periodic, and the period
(
.
[npg
2jnW
T=
T and amplitude A are given by
Ry/pg
A =
and
j y%
+ —^2——
R npg
The amplitude depends on the initial position and velocity, but the period is
.
independent of the initial conditions. Notice also that the motion is independent of the height H (as long as the
amplitude of the motion is smaller than H/2).
11.91
A cylindrical can, partially submerged in water (density 62.5 lb/ft 3
)
with its axis vertical, oscillates up and down
with a period of \ s. Determine the weight of the can if its radius is 2 in.
I Taking
g = 32 ft/s
,
and
T = \ Hz,
UjnW
24*W
1
R = £ ft,
2
nW = 2000/(24) * 3.4722
2
Therefore,
2
11.92
(
and
W * 1.105
W < nR Hp\
2
Since the can does not sink, it follows from Problem 11.87 that
H>
W
-~
nR 2 p
1
105
^F
2
= 0-203 ft = 2.43 in.
A cylindrical buoy 2 ft in diameter floats in water (density 62.5 lb/ft 3
f
R = \,
With
thus
tt(1/6) (62.5)
slightly and released, it vibrates with a period of 2 s.
flf
= 32,
and
T = 2,
it
)
with its axis vertical.
Therefore,
W = 2000/tt = 636.6
When depressed
Find the weight of the cylinder.
follows from Problem
1
1.90 that
2 =
—2JnW
1
11.94
lb.
^2000
1/6) v (62.5)(32)
What is the minimum height of the can in the previous problem?
f
11.93
we have, from the previous problem,
=2
V(62.5)(32)
lb.
Solve the previous problem by beginning with the differential equation of motion for the cylinder,
f With the numerical values given in the previous problem, the result of Problem 11.86 becomes
7r( l)
*
2
(62.5)(32)
W
^
+
2WOn
W_
y = c, cos y/2000n/Wt + c 2 sin yf2W0n/Wt.
wh ich has as its solution
nW
/—
/
-.
V2000
CHAPTER 11
274
2tt
The period of these oscillations is
2n
2 =
from which we find that
V'2000jW
11.95
W = 2000/71 = 636.6 lb
What is the minimum height of the buoy in Problem
#
=s
nR 2 p
as before.
1.93?
1
Since the buoy does not sink, it follows from Problem
W
H>- -=
11.96
Since the period is known to be 2 Hz, it follows that
1
W < nR Hp:
2
1.87 that
thus
636.6
= 3.24 ft.
—7-^-r^r
2
7t(l)
(62.5)
Suppose a cylinder oscillates with its axis vertical in a liquid of density p v What is p l if the period of the
oscillation is twice the period of oscillation in water?
f
Let p denote the density of water (62.5 lb/ft 3 ). Then it follows from the result of Problem 11.90 that the period
Tw = 2-JnW /Ryfpg,
in water is
If
11.97
T = 2Tw
t
,
whereas the period in the liquid of unknown density is
2
-— = -—.
1
= = 2-
then
ir
T, = 2yJnW/R yfp 1~g.
Pl = p/4 = 62.5/4 - 15.625 lb/ft
Thus
3
.
Solve the previous problem if the period of oscillation in the liquid of density p y is three times the period of
oscillation in water.
T, = 3T H
Using the notation of the previous problem, we have
hence
;
—
=3
3
Thus
or
= p/9 = 62.5/9 - 6.944 lb ft 3
==,
.
y/p'
\fp\
11.98
Px
—
R\'pg
RyJPiG
Suppose a rectangular box of width S,. length S 2 and height H is floating in a liquid of density p, as
indicated in Fig. 11.16. As for the cylinder of Problem
1.86. let \it) denote the position of the box relative to
How large should // be so that the box will oscillate?
is the weight of the box.
equilibrium, and suppose that
,
1
W
/
lum
position
y
T
s
/
v(f)
/
i
i
//
/
/
/
/
J
VX&
f
The volume of the box is SiSjH, and it will displace S S 2 Hp lb of liquid when the box is completely submerged.
i
If the
11.99
Fig. 11.16
box is to oscillate and not sink, then it must be that
W < S S Hp.
x
Thus.
2
H > W/S S p1
2
Find a differential equation for the position y(t) of the box described in the previous problem.
W (see Fig. 11.16). (We adopt
The buoyant force acting on the box when y(t) units of
S S 2 p[/i — y(0]- The net force on the box then is
I At equilibrium, h units of height of the box are submerged, such that
the sign conventions and notations of Problem 11.86.)
height are displaced from the equilibrium position is
— W + S S 2 p[h — y{i)\
x
S S 2 hp —
1
1
d2 v
so by Newton's second law of motion, we have
m —r = — W + S S 2 p\h — yit)].
x
at
Setting
y
+
m — Wjg
S S 2 pg
x
W
y = 0.
and noting that
W — S S ph =
l
2
0,
we simplify this differential equation to
—
—
APPLICATIONS OF SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS
11.100
Suppose the box described in Problem 1 1.98 is displaced from the equilibrium position by
y
initial velocity v
Find an equation that describes its position y as a function of time.
275
units and given an
.
I
Set
= y/S^iPg/W.
co
y" + to 2 y = 0,
Then the result of the previous problem becomes
roots of its characteristic equation
= ico
/,
and
/.
= -ico.
2
which has as the
Since the differential equation is of the second
y = c, cos on + c 2 sin on.
= y(0) = c, and v Q = y'(0) = c 2 co\
order, linear, and homogeneous with constant coefficients, its general solution is
= - c t co sin cot + c 2 co cos on, the initial conditions imply
c, — >'o
and c 2 = v /co. Then
Since
y'
that is,
y = y cos com
—
sin on
= y cos
°>
11.101
——w
t
•
-11
Tt,
u oscillations
The ffrequency offtthe
is
w
f — —— =
r
•
VS^^g
—
2?r
A =
The amplitude is
/(y
)
= sm
.
/
t
W
V
slS,S 2 pg
^
,
How does the period of the oscillations change if S
doubled, then the period becomes
If Sj is
2n4w
doubled in the previous problem?
is
x
so the period is
,
In-Jw
T=
= — — litsfw
^=. Thus the period
1
reduced to
1
of its original value.
How does the period of the oscillations in Problem
f „t
is
N/2vS S 2 p0
V(2S,)S 2 pg
about 71°
2njw
1
T=—=
f y/s^S iP9
.
—
y = /yj +
vh/Uo
+
f
11.103
+
Determine the period and amplitude of the oscillations described in the previous problem.
1
11.102
/
V
y
„
Si
,_
If both
,
o
•
,
,
•
,
,
1
change if both Si and S 2 are tripled?
1.101
T=
,
and S 2 are tripled, the period becomes
271^
=
.
—2nyfw
1
Thus the period is
,
reduced by a factor of three.
11.104
A prism whose cross section is an equilateral triangle with sides of length L is floating in a liquid of density p,
with its height
H parallel to the vertical axis. How large should H be so that the prism will oscillate?
f The volume of the prism is -j3L 2 H/4, so it will displace yj3pL 2 H/4 lb of liquid when it is completely
submerged. If the prism is to oscillate and not sink, then it must be that W < y/3pL 2 H/4 or H > 4W/y/3pL 2
where
11.105
W
is
the weight of the prism.
Find a differential equation for the position y(t) of the prism described in the previous problem,
f We adopt the conventions and notations of Problem
1
1.86.
At equilibrium, h units of height of the prism
y/3pL 2 h/4 = W.
The buoyant force acting on the prism when y(t) units of height
2
are displaced from the equilibrium position is
sJ3pL [h — y(t)]/4. The net force on the box then is
are submerged, such that
r 2 [h — y(f)]/4,
— W + yj3pL
Setting
11.106
m = Wjg
so by Newton's second law of motion we have
and noting that
The prism described in Problem
velocity v
/
Set
co
.
2
1
W — >/3pL h/4 =
2
we simplify this equation to
y" +
vFx
4W
i 2
—y=
0.
units from its equilibrium position and given an initial
1.104 is displaced y
Find its position y as a function of time.
= yf3pgL 2 /4W.
y
= y cos on H
=y
and
sin wt
= y cos
y(0)
—
v
.
co
y'(0)
=v
=
(see
2yfWv Q
tfisfpgLt
2yfw
h
1
2n
—
4tiV^
/
0)
tfeJpgL
T=-=
Solving this
Problem 11.100), we obtain
—
tfijpgL
Determine the period of the oscillations obtained in the previous problem.
The period is
y" + co 2 y = 0.
Then the result of the previous problem becomes
equation with the initial conditions
11.107
d2
r 2 \)i — y(f)]/4.
m —yy = — W + \j7>pL
.
sin
=
tfJJp~gLt
2JW
,
276
11.108
CHAPTER 11
D
How does the period of oscillations change if L is doubled in the previous problem?
#
follows from result of the previous problem that if L is doubled, then the period is halved.
It
ELECTRIC CIRCUIT PROBLEMS
11.109
Describe how to obtain two initial conditions for the current in a simple series RCL circuit with known emf, if
conditions for the current and the charge on the capacitor are given at
initial
t
— 0.
# Denote the current in the circuit and the charge on the capacitor at time t by 7(r) and qUi respectively. We
are given 7(0), which is one initial condition for the current. From Kirchoffs loop law [see (7) of Problem 1.81].
dl
—
+—q—
1
Rl + L
we have
dl
R
1
= — £(0)
E(t)
C
dt
= 0.
Solving this equation for dl/dt and then setting
t
— 0,
we obtain
1
— q(0)
7(0)
as the second initial condition.
It
11.110
Find two initial conditions for the charge on the capacitor in the circuit of the previous problem.
I We are given q(0) and 7(0); the first of these quantities provides one initial condition for q(t).
dq/dt = 7,
it
dq
—
follows that
dt
11.111
= 7(0)
1
the second initial condition.
is
=
R = 10 Q,
A series RCL circuit has
C=10"
no initial current and no initial charge at
L = \ H, and an applied voltage E = 12 V. Assuming
when the voltage is first applied, find the subsequent current
2
F,
=
/
the system.
in
I
d2I
_
Kirchoffs loop law (see Problem 1.85) gives
/
= e
I
""(<
,
cos 10f + c 2 sin lOf)
he initial conditions are
1(0)
(see
-j-j
dr
dl
+ 20 — + 2007 = 0,
which has as its solution
dt
Problem 8.64).
These conditions yield
=
c,
u^
JO
1/2
1/2
dl
—
and, from Problem
1
—
1.109,
~
dt
11.112
Since
and
— y;
c2
thus.
/
=
'
5
-c~
10
1
(0)(
2 , (0)
(1/2)(10
= 24.
'sin lOf.
Solve the previous problem by first finding the charge on the capacitor.
I The differential equation for the charge on the capacitor is given in Problem 1.84 as q + 2Qq + 200q = 24.
10
Its solution is found in Problem 9.22 to be
q = e~
Vi cos lOf + c 2 sin lOf) + ,\
and q{0) = 0; applying them, we obtain c, = c 2 = —3/25.
Initial conditions for the charge are
q(0) —
q =
Therefore,
,(
-e
" (
h cos lOr + A sin 10r) + 2\,
and
7
= -^ =
dt
11.113
R = 6 Q,
A series RCL circuit with
C = 0.02 F,
Using Kirchoffs loop law (see Problem
1.86).
= c e~ 50 + c 2 e~ l0
'
(see
'
x
— —50c, — 10c 2
di
1
.
n
as before.
C and the initial current is zero.
which has as its solution
-^
at
We are given the initial condition
Problem 8.10).
— = — 50c,^ -50 — 10c e"
Differentiation of our expression for 7 yields
dl
,'
— + 60— + 5007 = 0.
we get
dt
I
sin lOf
has no applied voltage. Find the
subsequent current in the circuit if the initial charge on the capacitor is
f
10 '
5
L = 0.1 H
and
— e~
'
2
= 0,
7(0)
10 ',
so that
c
,
+ c 2 = 0.
from which we conclude that
Moreover, from the result of Problem 11.109 we have
16
=
dl
It ,_
=
—
0.1
(0)
0.1
(0)
'
11 =
-50
(0.1)(0.02) 10
— 50 = —50c, — 10c 2 This is a second equation in c, and c 2
simultaneously, we obtain
and c 2 = — |, so that / = |(c 50t — c" 10
c, = f
which implies that
.
.
Solving the two
').
11.114
Solve the previous problem by first finding the charge on the capacitor.
I
With
R = 6,
C - 0.02,
L = 0.1,
and
E - 0,
(7) of Problem 1.82
becomes
—
% + 60
dt
which is the same differential equation as in the previous problem (with 7 replaced by q).
q = c el
50 '
+ c 2 c"
10 '.
-j-
+ 500q = 0,
dt
Its solution is
APPLICATIONS OF SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS
We are given the initial condition
q **-&-*» + fa-
that
so we have
q(0) = ,'
,'„ = c, + c
Also, wc have
2
Solving these two equations simultaneously yields c, =
4',,
-50c, - 10c 2 = 0.
so that
101
,
Then
.
/
=
^
= 5«-*ot _ *
4
lit
11.115
/(())
.
-to.
as before
= #0) = 0,
c2 -
and
1
.
.
4
A series 7?C£ circuit consists of an inductance of 0.05 H, a resistance of 5 fi, a capacitance of 4x10
a constant emf of
277
4
I
.
and
Find the current flowing through the circuit as a function of time if initially there is
10 V.
no current in the circuit and no charge on the capacitor.
L = 0.05,
C = 4 x 10 4
R = S,
and £=110, (2) of Problem 1.81 becomes
+ 100/ + 50,000/ =0. The solution to this equation was found in Problem 8.56 to be
50
50
/ - c,c
cos 50Vl9f + c 2 e~
sin 50>/l9r.
Applying the initial condition /(0) = 0, we find that c, = 0, so this last equation becomes
50(
sin 50V19t,
which has as its derivative
/ — c 2 e~
I With
,
/
'
'
—=
With
= 1(0) = 0,
q(0)
it
/=
11.116
——
5O,
sin50 > /i9t + 50Vl9g- 50 'cos50 > /r90
follows from the result of Problem
2200 = 7(0) = c 2 (50vT9);
we make use of (7) to find that
—
44\/19 e- 5Or
c 2 (-50e-
1.109 that
7(0)
c 2 = 44/VT9
thus,
-
£(0)
=
—
= 2200.
Then
and
,
sin50 x/l9f.
Solve the previous problem by first finding the capacitance.
q + \00q + 50,000*/ = 2200. The solution to this
_50
'sin 50y/\9t + 2yo
Since q =
I We make use o. (I) of Problem 1.82, which becomes
q = c,e"
equation is found in Problem 9.21 to be
t
— 0,
5O
ct —
or
50
-
'
(
—
50
'cos 50yfl9t + c 2 e
.
at
—Ho-
Since differentiation yields
'cos50>/79j - 50>/i9e~
= q(0) = -50c, + 50Vl9c 2
we have
"
= q(0) — 0.
7(0)
q = c,(-50e
q = e
= c, + jio,
this gives us
We also have
11.117
1
(7)
50
'
or
sin 50>/19f)
c2
= -
1 1
j— sin 50^19*) +
cos 50y/\9t
+ c 2 (-50e
5O
'sin50Vl9r + 50^\9c
^19/4750.
Then
and
- -j =
7
— — c~
-
50
'
50
'
sin 50/l9f
cos 50>/l9f)
as before.
Solve Problem 11.115 if instead of the constant emf there is an alternating emf of 200 cos lOOf.
I Now we have L = 0.05, R = 5, C = 4(10)~ 4 and £ = 200 cos OOt, so that (2) of Problem 1.81 becomes
Its solution is found in Problem 9.90 to be
7 + 1007 + 50,0007 = -400,000 sin lOOf.
1
,
7
- c^-so'cosSOyfWt + c 2 e' 50 'sin 50jl9t + f^cos lOOf - ^Q sin lOOf
Applying the initial condition
With
7(0)
7(0)
= q(0) — 0,
= - £(0) =
7
7(0)
=
x
or
c,
'sin 50 v/l9t + 50^19 cos 50^19/)
cos lOOf
/—
/
/
50
17
4000 - 7(0) =s -50c, + 50>/19f 2
and c 2 yields
1.109:
16,000
nn
sin lOOf
17
so we have
1
Then differentiation gives
- c,(-50e- 5O 'cos50Vl9r - 50,/l9 sin 50>/l90 + c 2 (-50e
.
= — ff.
we obtain the second initial condition by using the result of Problem
(200) = 4000.
4000
11.118
= c + 4°
yields
t=—,
1640x/l9
r—
40
16,000
= e' 5Q 'i --cos50 N/l9f + -
^
or
c2
=
1640^19
—
——
o L
Substitution of the values of c,
.
.
.
40
r— \
(cos lOOf - 4 sin 100f).
sin50V19H +
—
Solve the previous problem by first finding the charge on the condenser.
f With L = 0.05, R = 5, C - 4 x 10 4 and £ - 200 cos lOOf, (7) of Problem 1.82 becomes
q + 1004 + 50,000r/ - 4000 cos 100*. Its solution is found in Problem 9.91 to be
,
q = cxe
~ 50
'
cos 50/l9r + c 2 e"
50 '
sin 50 Vl9r
+ t% cos 100f + TTo sin lOOf.
278
CHAPTER 11
D
Then differentiation yields
q = c^-SOe-^'cosSOs/Wt - 50n/19V
50 '
sin
SOv^O + c (-50e- 50 sin 50 % T9f + 50^19^ 50 cos 50Vl9t)
'
2
'
-^sinlOOf + f^cosl00f
Applying the initial conditions, we obtain
= q(0) = C, + p^,
or
=./(0) = q(0) = -50c, + 50 sf\9c 2 + #,
from which we find
c2
q = e"
give
same expression for
11.119
——
—— cos 50 v 19f
50 '(
= dq/dt
/
+
sin 50yf\9t
—
C, — — p^,
and
= - 12^19/1615. These values for c, and c 2
(4 cos lOOf
+ sin lOOr).
Differentiation yields the
J
as before.
An RCL circuit has R = 180Q, C = 1/280 F, L = 20 H, and an applied voltage £(r) = 10 sin t. Assuming
no initial charge on the capacitor, but an initial current of 1 A at t =
when the voltage is first applied, find
the subsequent charge on the capacitor.
q + 9q + \4q — |sin t,
1 The differential equation governing this system is formulated in Problem 1.82 as
where q denotes the charge on the capacitor. The solution to this equation is found in Problem 9.89 to be
q = c e
21
+c 2 e~
lt
+ 5Viosinf- jffecosr.
Applying the initial conditions q{0) —
x
Hence,
11.120
q - ^o(l \0e~
- lOle
$0) = 1,
and
we obtain
c,
= 110/500
and
c2
= — 101,500.
-*
x.
+ 13 sin f - 9 cos r).
7
'
Determine the transient and steady-state components of the charge in the previous problem.
I
Since the homogeneous (complementarj
transient component.
g,
11.121
2'
)
$$e
function.
2l
- ^e
lt
tends to zero as
,
t
it
is
the
The steady-state component is the remaining part of the charge, namely
= 5ou(13sint -9 cost).
Determine the amplitude, period, and frequency of the steady-state charge of the previous problem.
I The amplitude is
A = M M >/(13) 2 + — 9) 2 — 0.0316.
'
(
,
The natural frequency is
f — 2n.
so the period is
T = l/f=l/2n.
11.122
A series circuit contains the components
the circuit is completely passive (thai
is
suddenly switched into the circuit.
is.
L =
1
R =1000 0,
H,
Q — I = 0),
while
C = 4 x 10
and
6
F.
At
t
= 0,
a battery supplying a constant voltage of
while
E — 24 V
Find the charge on the capacitor as a function of time. (Here Q denotes
the charge on the capacitor.)
f
Substituting the numerical values for R. I.. C. and £ into (/) of Problem 1.82 yields
2
—
dQ
d Q
dt
=2
-I-
—
1000
Q
r = 24.
h-r
dt
The solution to this equation is found in Problem 9.25 to be
"
4 x 10
Q = Cl e- 500 + c 2 te- 500 + 9.6 x 10" 5
'
'
Differentiation now yields
.
^ = /= -500c,f-
500
'
+ c 2 (e~ 500 - 500f<T 500
'
').
Substituting
Q=
at
dt
= 0, we find
= 0, we find c, = -9.6 x 10 5 Substituting / =
= -500c, + c 2
at
500 2
500
5
2
10~
10~
x
=
=
-9.6
x
-4.8
x
9.6
x
4.8
10
fc"
Hence
lO'V
+
c2
Q
t
t
.
'
'
.
11.123
,
so that
.
Determine the current as a function of time in the circuit of the previous problem.
/
= —- = 24rc
500 ',
in amperes.
dt
11.124
A series RLC circuit has
R = 4 Q,
C = ye F,
L = | H,
and a constant emf of 13 V. Find the charge on the
capacitor as a function of time if initially the circuit is completely passive.
f Substituting the given values for R. L. C, and the emf into (/) of Problem 1.82, we obtain
Q + 8Q + 52Q = 26, where Q denotes the charge on the capacitor. The solution to this equation is found in
4
The current / is then
sin 6r + \.
Problem 9.20 to be Q = c,c" 4 cos 6r + c 2 e
'
/
=
—=
dt
'
c,(-4c" 4 'cos6f - 6c" 4r sin6f) + c,(-4^
_4
4
'sin 6r + 6c" 'cos6r)
APPLICATIONS OF SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS
= £(()) = c, + 2
Applying the initial conditions, we obtain
from which
= — i-
c2
or
,
Substitution of these values then yields
= -\,
c,
= -c
0(f)
= /(()) - -4c, + 6c 2
and
4
- e
'
'
4!
=
/
—=—
dt
zero as
-> oo,
t
2
e'
4
'
Since this tends to
sin 6t
3
the steady-state current is zero.
Find the steady-state charge on the capacitor in the circuit of Problem
The steady-state charge may be computed as
lim Q = lim
»->oo
11.127
+
3
Find the steady-state current in the circuit of the previous problem.
Differentiation of the result of the previous problem yields
11.126
,
S
2
11.125
279
D
— e~ 4t
I
r-»a>
1.124.
1
e~
4'
2
\
h - J
= -.
2/
3
2
E = 16cos2f.
Solve Problem 11.124 if, instead of being constant, the emf is
I The differential equation becomes Q + 8() + 52Q — 32 cos 2f, which has as its solution
Q — c e~ 4 cos6r + c 2 e 4 'sin6f + f cos2f ^sin 2f (see Problem 9.87). Thus the current is now
'
-I-
}
= -f- = c (-4e~ 4t cos6t -6e~ 4 'sin6f) +
/
c-
l
(-4e~ 4, sin6r + 6e-~ 4 'cos6r) - | sin 2f + fco:>2r
2
at
= Q(0) = c x + f,
Applying the initial conditions, we obtain
from which we find
c2 =
—
iV-
11.128
= 1(0) — —4c'!
and
6c 2 + f,
-I-
— -J- —
s
Since the homogeneous (complementary) function,
— 35 e' 4 cos6t — ,^~ 4 'sin6r,
the particular solution is the steady-state component.
The steady-state charge is thus
A sin (2f +
Since
</>
+ A sin 2t cos
cj)
t
-> oo,
Q = I cos 2r + ^ sin 2f.
s
A sin (2? + <£).
— § cos 2r + \ sin 2t,
we require
A sin
</>
=|
and
= 5.
/I cos
2
now follows that
It
— A cos 2f sin
<f>)
tends to zero as
'
Express the steady-state charge of the previous problem in the form
/
tan </> -
11.130
— — f,
Find the steady-state charge on the capacitor in the previous problem.
I
11.129
c{
.,cos6f
__3.-«_
_- 7.-.« sin6f + 3 cos 2r + sin 2f
.
fl
or
Substitution of these values yields
/4Sm<
(f)
= -J- = 3,
^
4cos0
+ A 2 cos 2
= <4 2
= arctan 3 * 4tt/10.
Thus
+ (j) 2 = A 2 sin 2
so
</>
so that
,
A = VlO/5.
& ^ (VTo/5) sin
(2f
Moreover,
+ 4tc/10).
1/5
An inductance of 2 H, a resistance of 16 Q, and a capacitance of 0.02 F are connected in series with an emf
£ = 100 sin 3f. At t =
the charge on the capacitor and the current in the circuit are zero. Find the charge
> 0.
at
t
f
Letting Q and / be the instantaneous charge and current at time f, we find by KirchhofTs laws (see Problem
-yy + 8 -=- + 25Q = 50 sin 3f.
1.82) that
The solution to this equation is found in Problem 9.88 to be
Q = c e~ 4 cos 3f + c 2 e~ 4t sin 3f + f§ sin 3f - 5! cos 3f.
'
The current is then
x
= Cl (-4e~ 4 cos 3r - 3e~ 4, sin 3f) + c 2 (-4e -4 sin 3f + 7>e~ 4 cos 3f) + ^cos 3r '
/
'
'
= £(0) = c, - £f,
Applying the initial conditions, we obtain
= /(0) = -Ac + 3c 2 + it,
l
so that
Q = If (2 sin 3t - 3 cos It) + ff e~
11.131
4,
c 2 = |f.
(3 cos 3f
c,
+ 2 sin 3f
(2
K
dt
3f
'
3 sin 3f)
4,
and
).
f
>
The
Th«
/=^
6cos3/).
cos
(17sin3f + 6
e
(17
= z~ = ~ (2cos3f
cos
+ 3sin3f)-^52
j
= H,
sin 3f
These values yield
Determine the current in the circuit of the previous problem at
*
or
^
sin 3f 4-
3f).
first
term is the steady-state current and
52
the second, which becomes negligible as time increases, is the transient current
.
280
11.132
CHAPTER 11
D
An electric circuit consists of an inductance of 0.1 H, a resistance of 20 Q, and a capacitance of
6
25 [iF = 25 x 10" F.
and
I
i
— dq/dt =
L = 0.1,
Since
2
= 0.
f
R = 20,
c
^ nn dq
+ 200 ^T + 400,000q = 0.
-jy
2
dt
dt
d q
—
TT
q = c ,e~
100
= 25xl0 -6
,
and
= 0,
E(t)
of Problem 1.82 reduces to
(7)
The solution to this differential equation is found in Problem 8.69 to be
'cos 100^39/ + c 2 (?"
100t
lOOV^-
sin
Differentiation yields
q = c^-lOOe-^'cos 100V39r - 100V39V
+ c 2 (-100<T
Applying the initial conditions, we obtain
which
11.133
= 0.05/ ^39 = 0.008.
c2
q = 0.05 C
Find the charge q and the current i at time t, given the initial conditions
when
100,
sin 100V39r
'cos 100 V^r)
= q(0) = - \00c + 100V39c 2
and
q = e~
'sin l(X)V39r)
,00
+ 100V39V
0.05 = q(0) = c x
Substitution then gives
100
l
100,
,
from
(0.05 cos 624.5r + 0.008 sin 624.5r).
Find the steady-state current for the circuit of the previous problem.
f
dq
Differentiating the result of the previous problem, we find
/
= -j- = -lie
100
'
Since this
sin 624.5f.
dt
quantity tends to zero as
11.134
-* oo,
the current is all transient and there is no steady-state current.
Solve Problem 11.132 if there is an initial current of —0.2 A in the circuit.
I
This change affects only the initial conditionsjbr the problem, from which we now obtain
as before, and
becomes
11.135
t
-0.2 = q(0) = - 100c, + 100>/39c 2
q = e'
100
'(0.05 cos 624.5f
A circuit consists of an inductance of 0.05 H, a resistance of 20 CI, a capacitance of 100 /iF, and an emf
=
= 0.
E = 100 V. Find and q. given the initial conditions q —
and
when
- -~
400 -f
+ 200,000^/ = 2000.
TT
2 +
T
dt
dt
1.82 becomes
i
q = e~
- 200
200
A = —0.01
-0.01 cos 400f- 0.005 sin 400/) + 0.01
'(
'(A cos400f
+ Bsin400f) + 0.01.
and
and
B = —0.005.
i
= 5e
200
'
Then substitution yields
sin 400f.
Solve the previous problem if the constant emf is replaced with a variable emf
The differential equation now becomes
d2q
200
i
= e~
'(A cos400f
200,
[(
E(t)
= 100 cos 200f.
.„„ dq
-^1 + 400 — + 200,000^ = 2000 cos 200f,
dt
q = e'
Differentiating with
= -^ = 200e" 2OO '[(->J + 2B)cos400f + (-B- 24)sin400r].
Use of the initial conditions yields
q = e
The solution to this differential
-I-
equation is found in Problem 9.23 to be
respect to t yields
t
/'
Here (/) of Problem
which has as its solution
dt
+ B sin 400f) + 0.01 cos 200f + 0.005 sin 200f
(see
Problem 9.93). Therefore,
- 200 A + 400B) cos 400f + - 200fi - 400 A) sin 400f ] - 2 sin 200r + cos 200f
(
A = -0.01
Use of the initial conditions yields
q = e"
and
11.137
x
so the result
+ 0.0077 sin 624.5f).
i
11.136
0.05 = q(0) = c
c 2 = 0.0077,
This latter equation yields
.
i
= e~
2CO '(-0.01
200,
(
and
B = -0.0075.
Then
cos400f - 0.0075 sin 400f) -I- 0.01 cos200f + 0.005 sin 200r
- cos 400f + 5.5 sin 400r) - 2 sin 200f + cos 200r
A series circuit contains an inductance L = H, a resistance R = 1000 Q, and a capacitance
C = 6.25 x 10~ 6 F. At t = 0, with the capacitor bearing a charge of 1.5 x 10" 3 C, a switch is closed
so that the capacitor discharges through the (now) closed circuit. Find Q and as functions of
1
/'
I
With
E(t)
= 0,
(/)
of Problem 1.82 becomes
2
dQ
d Q
-^~
1000-^
+ 1000
f
-j£ + 160,0000 - 0.
2
—
Q = c.e' 200 + c^
-800 ',
and differentiating yields
Substitution of the initial data into the equations for Q and i yields
'
1.5
x 10" 3 = c, +c 2
Q
'
i
solution is found in
= -lOOc^ 200 - 800c 2 e" 800
'
0- -200c! - 800c
2
=2x 10"
c2
Solving these two equations simultaneously, we obtain c,
= 2 x lO" 3 ?" 200 - 5 x 10- 4 e- 800 and i = -0.4e- 200
'
Its
dt
dt
Problem 8.19 to be
t.
'
+ 0.4e-
3
and
800 '.
— — 5 x 10 -4
.
Hence
'.
5
APPLICATIONS OF SECOND-ORDER LINEAR DIFFERENTIAL EQUATIONS
11.138
When is the absolute value of the current a maximum for the circuit of the previous problem?
# The extreme value of the current occurs when
^ = 80e-
2OO '-320<?- 8OO
600r = In 4
which
11.139
281
'
= 0.
and
t
= 0.
di/dt
Hence we must solve the equation
Multiplying by e 200 and dividing by 320, we obtain
'
= 0.00231
e"
600
'^,
from
s.
What is the extreme value of the current in the circuit of Problem 11.137?
I We use the result of the previous problem, noting that the extreme value of the current occurs at
= -0.4e" 462 + 0.4e" 1848 = -0.189 A.
Then
f
= 0.00231.
i
11.140
A series RCL circuit with
C=10 _2 F,
R = 5 Q,
L = |H
and
has an applied voltage
— sin
E(t)
t.
Find
the steady-state current in the circuit.
I
2
dl
d I
._.._..
....
We use (2) of Problem 1.81, which becomes -^2 + 40n — + 800/ = 8 cos
dt
to be
I
—e
20,
(c,cos20t + c 27 sm20t)-\
>
cost
1
Its solution is found in Problem 9.95
t.
dt
The complementary function, composed
v
sin t.
-I
640,001
640?0 oi
of the first term of this equation, tends to zero rapidly, leaving as the steady-state solution
/s
11.141
—
=
(6392 cos t + 320 sin t).
640,001
Solve the previous problem if there is no applied voltage.
# With no external emf, the current in the circuit must tend to zero.
current is always zero. In any event,
/ = 0.
If,
initially,
the system is passive, then the
s
11.142
Find the amplitude and frequency of the steady-state current in Problem 11.140.
I
17
A =
The amplitude is
6392 Y^
7
+
/(
I
Y^
320
A
J
.
nnm
The period of both sin t and cos r is 2n, so the
=0.001.
)
frequency is 1/27T.
11.143
For a series circuit consisting of an inductance L, a resistance R, a capacitance C, and an emf
derive the formula for the steady-state current
i
=
X
E o( R
— —
- — cos cot \ =
sin cot
I
E
—
,
sin (cor
E{t)
= E sin cot,
mi.
- 0),
where
J
X = Leo - 1/Cco, Z = yjx 2 + R 2
2
I
da
d q
q
L —-f + /?— + - = £ sm cor
By differentiating
— + R — + — = LD + RD —
2
-
dt
2
dt
C
-\
(
C
\
,
= X/Z
sin 6
) i
cos 9 = R/Z.
and
dq
.
and using
i
= •—
C
dt
dt
L
and 9 is determined from
,
we obtain
,
at
= co£ cos cot.
The required steady-state solution is the particular
integral of this equation:
co£
LD + RD+IC
i
2
coE
cos cot =
r-r
-,
—
cos cot
Dn — ( Leo — —— \ co
RD
C ")
\
1
I
coE (RD + Xco)
2
R D
=
11.144
2
-X
E fR
— sm
—
z\z
2
co
.
sin cot
2
cos cot =
X
E
—
.
j (R sin cot
R + X2
2
— X cos cot)
\
- — cos cot = -y sin (cot - 9)
z
)
)
A series circuit consisting of an inductance L, a capacitance C, and an emf E is known as a harmonic oscillator.
= 0. Assume
and
=
when
Find q and when E = E cos cof and the initial conditions are q = q
that
co # lVCL.
i
i
—T + — = -j- cos
•
With
R = 0,
(J) of Problem 1.82
becomes
2
i
t
I?
cot.
If
co
# lA/CL,
the solution to
^
282
CHAPTER 11
D
this differential equation is
„
1
q = A cos
f
/
=
/
1
.
VCL
JCL \
f
—
- CL
^—
— CL
£ Cw
\
1
+ B cos
=
f
.
JCL
JCL J
\
2
/
-
B = yfCLi
and
1
11.145
E C
2
d q
—
+
dr
=-
co
2
q —
E (\
= co( — A sin cot + B cos cot) + -— — sin cot +
co
Jcl
.
j— sin cot
-
t
- co CLJ
1
Solve the previous problem under the condition that
Here (/) of Problem 1.82 becomes
£ Cw
1
.
z
JCL \
I
- co 2 CL
1
j—\ sin -== -
1
t
JCL
Then
.
j—- cos cot
t
-== - -== / q 1
cos cor
E C
1
.
t
cos
- co 2 CL
.
f—r
cos -== + VCLi sin -== +
Uo - 1 - co 2iTrr
CLJ
Jcl
Jcl
\
(
j
=
1
sin cot
1
and
-1
co
1
E C
<o
1
A =q
Use of the initial conditions then yields
4 =
E C
1
t
1
— A sin
1
1
+ —- — 2 —
cos cot — A cos —== + B sin
r
^ ^ + 1/CL
V^Z
V^Z
;
VCL
and
E
1
.
+ B sin
t
.
1
- co z CL
= l/JCL.
E
—
cos
L
Then
cot.
q — A cos cot + B sin cot H
E—
2Lco
.
and
i
I
t
cos cot
2L \co
Use of the initial conditions now yields
—
q = q cos cot H
'o
o>
sin cot
+-
—
Eo
2Lco
A = q
.
t sin
a>f,
and
.
/
=
and
5=
i
/aj.
Then
E o (\
i
cos cot — q co sm cot + ——
I
— sin cot +
2L \co
t
cos cot
\
.
J
Note that here the frequency of the impressed emf is the natural frequency of the oscillator, that is, the
frequency when there is no impressed emf. The circuit is in resonance, since the reactance X = Leo — \/Cco
zero when co—\/JCL. The presence of the term (£ f/2L)cosa>f, whose amplitude increases with f,
indicates that eventually such a circuit will destroy itself.
is
CHAPTER 12
Laplace Transforms
TRANSFORMS OF ELEMENTARY FUNCTIONS
12.1
Find the Laplace transform of
I We have
= 1.
f(x)
= P° e~ sx(i)dx.
J§?{1}
f°°
°°
sx
e
s
= 0,
^"rfx = I*" e- (0)W dx = lim
Hence the integral diverges. For
I
For
f* I dx = lim xl* = lim
# 0,
s
dx = lim
sx
e
I
dx — lim
—
(
e
s
— si? > 0;
< 0,
I
12.2
Find 2>{e
^{l} = -,
Thus,
When
s
s
— sR < 0'
> 0,
hence, the
> 0.
ax
).
,(a-s)x\x = R
I
00
ax
&{e ) = f
e~
sx ax
e
dx = lim
,(a-s)R
= lim
Note that when
< a,
s
-
f* e
la
~ s)x
dx = lim
1
for
a — s
\
r-><d
12.3
for
1
S
R-*oo V
hence the limit is oo and the integral diverges.
limit is 1/s and the integral converges.
1
- e - SsRR + -
= lim
s*
R->oo
When
= oo
K =
>a
s
the improper integral diverges.
Find the Laplace transform of
=
f(t)
&(t) = f
x
e
t.
-s
'(t)dt
T te
P-»™ J
= lim
J
— lim
<fr
e
= lim ( t—S
P-»r^ V
P—x
for
s
~»\ P
S
\
JV
e
1
g 'I
_s,
>
where we have used integration bv parts.
12.4
Find the Laplace transform of
f(x)
= x2
.
# Using integration by parts twice, we find that
i?{x
1
2
}
'
rx <T"dx= lim
2
= r°°e-"x 2 rfx= lim
R-x\
J°
R-ooJ
Hm
K e-"--3-«—
-—
-3
2
2/?
,
e
vK
e
- S*
— sx
x=R
2x
s
x=
+
R-»ao
;
f!
For
s
< 0,
lim
I
e~
sK
]
= oo,
and the improper integral diverges. For
s
> 0,
it
follows from repeated
R->ac
use of L'Hopital's rule that
R2
-sr\
lim
R->x
S
/
lim
R-»x
5
/
~ R2
~ 2R
V
R->ooSe
/{-.oc^f
se
R-j» f
k- x
,
V
V
"2
A
R->x J e
283
.
CHAPTER 12
284
Also,
—re
lim
sR
directly; hence the integral converges, and
= 2/s 3
F(s)
For the special case
.
R->oo
s
12.5
= 0, we have
e~" x 2 d x=l
combining all cases, we obtain
if {x 2 } = 2/s 3
Find the Laplace transform of
/(f)
= te"',
for
,
s
> 0.
(a-s)l
S
te"
r°o
— s) Jo
(a
t-x ( a — s
a - s
(a
- s)
2
(a-s)T
lim
r^ x a — s
a — s
r-x
— s)
(a
for
s
2
>a
Find if {sin at}, where a denotes a constant.
r
Using integration by parts twice and the formulas
e™ sin /Jr dt =
e
xt
(tx sin
Bt
z
C
o
e" cos fit dt =
*,
j
—
—
— we
e *'(* cos P* + P sin P*)
•J
ex
'
1°°
st
e
Jo
= lim
sin at dt
e
p—w
s
2
+a
f
s
'sin at dt
_ Jo
sF
aP + a cos aP)~\
(s sin
e
= lir
e
P
— um
2
- B cos Bt)
and
+ B2
obtain
,
j + B-J
if {sin af} =
— s sin at — a cos at)
st
{
p^
s
2
+ a2
-I-
2
s
2
for
+ a2
s
>
Find if {cos at}, where a denotes a constant.
I
Using integration by parts twice along with the formulas given in the preceding problem, we obtain
if {cos af} =
" cos at dt — lim
e
\
e
I
" cos at dt - lim
e
w
(
-scosaf + asinaf)
s
e
= lim
s
12.8
- s)
(a
2
1
- s)
+ hm
2
f
12.7
'
1
1
,(fl-s)T
12.6
/r e'
(a -sir
1
(a
,{a - s)t
"V
c (a-s)(
a — s
T-x
Finally,
oo.
where a denotes a constant.
roc
= lim
—=
x 2 dx = lim
e" s(0) x 2 dx = lim
,
c
I
2
+a
(s cos
2
s
2
2
+ a2
aP — a sin aP)
+ a2
s
2
+ a2
for
s
>
Solve the previous two problems by utilizing complex numbers.
I Assuming that the result of Problem 12.2 holds for complex numbers (which can be proved), we have
<?{e""}
=s
- = -^1
— ia
s
=.
+ a1
Using Euler's formula, we also have
if {*?'<"} = f* e'
=
r
e~
\
sl
(cos at + i sin at)
s'
cos at dt + i f
x
e'
s'
sin at dt
e""
— cos at + isinaf,
so
= ¥ {cos at} + i'if{sinar}
Equating our two expressions for J?{e""} and then equating real and imaginary parts, we conclude that
£f {cos at} =
s
12.9
2
and
+ a2
if {sin at} —
s
Find the Laplace transform of
Find the Laplace transform
f(t)
— e 4(
+ a2
'
ix
f(x) = e
.
Using the result of Problem 12.2 with
12.10
2
a = 3,
we have
¥{e 3x } —
s-3
.
f The Laplace transform of f(t) is the same as that of f(x), so the result of Problem 12.2 (with
2>{x e -*'} =
'
l
-
s-(-4)
=
—
s
+4
a —
— 4)
yields
—
LAPLACE TRANSFORMS
12.11
Find the Laplace transform of sin 7if.
#
a — n,
Using the result of Problem 12.6 with
n
if {sin nt\ — —2
we have
T
.
+ n2
s
12.12
Find the Laplace transform of cos 2x.
a = 2)
i The Laplace transform of f(x) is the same as that of /(f), so the result of Problem 12.7 (with
if {cos 2f } =
s
12.13
2
+ 22
s
2
Find the Laplace transform of
f(x) = sin( — 3x).
Find the Laplace transform of
/(f)
a =
a— —5,
Using the result of Problem 12.7 with
S
if{cos( — 5f)}
we obtain
s
Find the Laplace transform of
/(f)
= te 4
Find the Laplace transform of
a = —4,
f(x) = xe
Find if {/(x)} if
X
f(x) = \~]
[
+(-5)
s
2
+ 25
we have
<£\te~
X
M =
\
=
^
-.
(-4-s) 2
(s
=
=
T
e-* {-\)dx + £" e~
sx
'
+ 4Y
.
a = 1,
y\xe x \ =
we have
.
~].
x > 4
1
se{f(x)} =
I
S
2
x
Using the result of Problem 12.5 with
12.17
2
'.
Using
B the result of Problem 12.5 with
12.16
— 3)
= cos( — 5f).
m
12.15
°°
e-°*f{x)dx =
Jo
x
J*
(l)dx
x =-4A
+ lim
S
1
,.
+ hm
S
5
sx
e
J*
dx
4s
2<T
l-\
„
A \
Rs
=
- -e+ -e~ 4s
1
1
R-oo V s
S
for
S
I
s>0
S
0<
:
!>3'
\l
^if(t)} =
J
" e-«f(t)dt =
5j
12.19
gives
+4
f The Laplace transform of f(x) is the same as that of /(f), so the result of Problem 12.6 (with
12.14
285
Find the Laplace transform of
V
*{A*)} = jo
e~
'(5)^f
s
+
f"
3s
5(1 -e~
3
s
Vf = 5-
2
e" 2
2
sx
A*)dx =
«J 2
(— sx — 1)
,
5
s
e" '(0) di
)
<x<2
[x
/(x) = <
s
£^
Jo
e~**xdx + £" e~*\2)dx
+ hm 2 C
i-
->
m
M
dx
e
J2
\(-2s- l) + -r+ hm /
2e"
M->oo V
s
<s
1
S
,.
Ms
+
2e~
s
2s
\
/
I
- e~ 2s
s
yields
286
D
CHAPTER 12
<x <
[1
12.20
Find the Laplace transform of
f(x)
= <e x
1
<x<4
1
x>4
r
sx x
= P e sx (\)dx + j* e~
Jo e'^fWdx
<f{f{x)} -
£
dx +
e
"(0)dx
e
p—s
= P e' sx dx + P e~ - 1)x dx =
is
s-
1
-e
1
s-
1
< x < 2
12.21
Find the Laplace transform of
f(x)
x>2'
£ e""(0)ix + £ e"»(4)dx
#{/(*)} = /J e "/( vj^/x 4(?
_sx
4e
dx
-s>-
+
lim
s2>
4g
2s
1
=4
P- x J "
12.22
Find the Laplace transform for the (Heaviside) unit step function about the point c, defined by
u(x-c)
[0
x < c
Hi
^c
x >
1
•
-
2>(x - c)} = P e' sx u(x - c)dx = Pe "(0)dx + f " e
"Ox = lim
e
R-x J'
12.23
Find the Laplace transform of
/(x)
H
= sin 2 ox,
(l)dx = f '
=-e
—S
•
tx
for
i
>
sin
^
ax =
c<
>
where a denotes a constant.
axdx=
sin
e
J°
'
2s
2a
2
:
M.v
axdx= lim <e
/.
:
r
,•
.
sl
-,
—
_ Vl I
2a sin 2ax — s cos 2axT) L
—— —— —
1
z
L 2s
I
2(s
2
=
+ 4,r)
s(cos2aL)e
+4<r)
2(s
2
+ 4<r)
2(s
2
+ 4a 2
)
I
2
for
4a 2 )
:
f 4a )
2(.s
"sin
J"
2|s
1
2s
>
2a(sin2aL)e
1
a
e
-,
/
= lim
ci-
.
lim
.
>
S
-
„<
s
e~"dx
t
s
>
<\
12.24
Find the Laplace transform of
/(.v)
^{smfax + b)} = Km
'\<
o
PAfnr^noiu^
U, -
^(o-K+fc)
Aii:0,Ol>(d4rtJ»)
~S*.
c(y =
^:C
/r-H
— se
,
vP
Sl
i
y{cos(ax + ft)} = lim
P-.
+ a cos
+ a2
f
J°
r
e
for
Find <f{ft [t)}, where
/€ (f)
s
2
+
sin ft
s
2
Ci
+ a cos 6]
+
>
— se sx cos (ax +
2
+ a
ft
for
< <e
f
f
>6
s
>
2
s
sin (aP
+
ft)
2
+ ae
sx
sin (ax + ft)
2
+ a
— a sin ftH
s cos
P^X
— a sin
+ a2
2
+
2
cos (aP + b) + ae
s
1/e
12.26
+ ft) — ae " cos (a;
ax + ft) T
where both a and ft denote constants.
P- X
s
+a
s
lim
2
2
"cos (ax + ft) dx — lim
— se
s cos ft
sin (ax
ft
cos (ax + ft),
Find the Laplace transform of
sv
sin(aP + ft) — ae "cos(aP + ft)
s
a-V-jV^
se
s
im
2
where both a and b denote constants.
ft),
x + ft)</.v = lim
J
s sin ft
12.25
= sin (ax +
ft)
ft
+
s
2
+ ^~ "J
>
Jj
"
LAPLACE TRANSFORMS
&{/#)} = J7 e-*fJLt)dt =
12.27
Show that
lim Sf{ft (t)} =
in
1
-
re
I
1
.
Problem 12.26.
p e~"-dt + J/ e
Is this limit
€-0
I
'
the same as if{lim/e (f)}? Explain.
f
-o
The required result follows at once, since
se
c^o
It
«(0) dl
o
-e~ s
287
se
€ -oo
€-o \
2!
/
also follows by use of L'Hospital's rule.
Mathematically speaking, lim f€(t) does not exist, so that £f {\im fe {t)} is not defined. Nevertheless it proves
useful to consider
S(t)
— lim/e (f)
to be such that
We call d(t) the Dirac delta function or impulse
= 1.
J&?{<5(f)}
€-0
function.
TRANSFORMS INVOLVING GAMMA FUNCTIONS
12.28
T(p + 1) = pT(p)
Define the gamma function T(p), and then show that
p > 0.
for
T{p) —
I The gamma function is defined, for any positive real number p, as
xp
~ 1
x
e~ dx.
Using integration by parts, we have
T(p+ 1)= f°° xlp+1)
-l
e' x dx= lim
P x pe~ x dx = lim
r-»oo "
"
= lim (-r p e~ + 0) + p f°° x
r
12.29
lim r p e~
Prove that
T(l)
I
r
—
is
e
_Jc
1-1
)
J
y
r
as r p/e
r
and then using L'Hopital's rule.
e-^x= lim fV x </x = lim (-<?"*)' = lim (-e' + 1) r
**
r->oo
Prove that if n is a positive integer, then
r->oo
First we consider
n = 1.
=
=
=1
=
we
have
T(l
1T(1)
1!.
12.29,
+ 1)
1(1)
Next we assume that T(n + 1) = n\ holds for n = k
From Problem 12.28 with
p — k + 1
T[(fc
1
r-»oo
T(h + 1) — n\.
I The proof is by induction.
Using Problem 12.28 with
p = 1
and then Problem
and then try to prove its validity for
+ 1) + 1] = (k + l)r(it + 1) = (k + l)(k!) =
(/c
+ 1)!
r(rc
Prove that
T(p + k + 1) = (p + k)(p + k - 1)
•
•
(p
0!
= T(0 + 1) = T(l) = 1.
+ 2)(p + l)T(p + 1).
I Using Problem 12.28 repeatedly, where p is replaced first by
p + k,
then by
p + k — 1,
obtain
r(p + k + 1) = r[(p + /c) + i] = (p + fcjr(p + *)
=(p + /c)r[(p + /c - i) + i] = (p + k)(p + k- i)r(p + * - i) =
= (p + fc)(p +
12.32
fc
- 1)
•
•
(p
+ 2)(p + l)T(p + 1)
Evaluate r(6)/2r(3).
I
n = k + 1.
and using the induction hypothesis, we have
+ 1) = n! is true by induction.
Note that we can now use this equality to define 0!; that is,
12.31
\
I
= 1.
T(l)= f°°x
Thus,
-x p <H + ['px" -1 ?"* dx
dx = pY{p)
easily obtained by first writing r p e
"
12.30
p_1
r
(
**
r-»oo
The result
r-*co
T(6)
5!
2T(3)
2(2!)
=
(5)(4)(3)(2)
(2)(2)
=
•
•
and so on, we
'
~
.
CHAPTER 12
288
12.33
~
~
Evaluate
1X5/2)
ni/2)-
H5/2)
(3/2)r(3/2)
T(l/2)
r(l/2)
(3/2)(l/2)r(l/2)
3
Hl/2)
4
r(3)r(2.5)
12.34
Evaluate
r(5.5)
12.35
Evaluate
r(3)r(2.5)
2!(1.5)(0.5)r(0.5)
16
H5.5)
(4.5)(3.5)(2.5)(1.5)(0.5)r(0.5)
315
6H8/3)
5H2/3)"
6T(8/3)
6(5/3)(2/3)r(2/3)
=
5T(2/3)
12.36
2.37
4
3
5H2/3)
x
Evaluate P° x 3 e' dx.
f° x 3 e~ x dx = f(4)
I
1
=
Evaluate
Let
J
" x6e
2x
2x — y.
= 3! = 6
dx.
Then the integral becomes
l\\-,lL-±
— = —=
e
6
"
1
(
J°
\2/
2
2
7
I
y
v e
dv =
J°
—
2
=7
= -"r7 = —
2
8
2
12.38
e~* dx as a gamma function.
Express
I
Let
z
= x2
x = z 1/2
dx — \z
to oo so does z, we have
then
;
that as x goes from
ll2
and
Substituting these values into the integral and noting
dz.
,
12.39
Prove that
f
If
We have
I
12.41
If
*
we let
Evaluate
I
r(^)
4 1"*
Evaluate
= f°° x
Jo
T
yjye~
3
y
3
y3
e
l
2
x
e
2
dx — 2 \* e~" du,
where we have substituted
p:
u — pcos(f)
pdpdd> = 4\
= x,
the integral becomes
d(p
\
= n,
= p sin
Hi) = yjn.
and
and so
r
.
follows that
It
the last integral
(/>,
~
Jx^e x (\x 2 3 dx) = \ J*o x Xile~ x dx = \Y{\) = ^/3.
*
Jo
*
'
_4z2
d"z.
x
We write the integral as
dx
Evaluate
Jo'
-{e' p
x — u2
dy.
and the integral becomes
12.42
6)
= Vtt.
we change now to polar coordinates (p, <f>), substituting
becomes
12.40
T(|)
fw - --*-5 r
i
xr-"-*«-jr--6--" )*-ijr-
y^hT
Jq
f- e
Jo
3
~ 4"2
dz =
£
(e
* l2
ln 3
)
dz -
£
e
,4 ln
M "' rfz.
Now we let
(4 In 3)z
m \ —
Hl/2)
sfc
*
xJ
f" x -1/2 c -*A
= —==.
=
ax = —
d\
Jo
4 v lriT
2 V 41n3
V4hO/ 2 N/4ln^
l
.
I
'
,
/
2
- x,
u
LAPLACE TRANSFORMS
#
Let
-lnx = u.
x = e~".
Then
x = 1,
When
u = 0;
and when
x = 0,
u= x.
289
Thus, the given
a-
—— du =
°°
integral becomes
°°
)
J
12.43
T(p + 1) = pr(p)
Using the relationship
(a)
I
u~ 1/2 e~" du = T(|) = yfn.
of Problem 12.28 as the definition of T(p) for nonpositive p, find
T(-i); (6) r(-|); (c) r(-|); (d) r(0); (e) T(- 1); (/) T(-2).
For
For
For
For
For
For
(a)
(b)
p=-i
p=-l
Then r(-£) = -2TQ) = -2<Jn.
T(i) = -|r(-i).
r(-£) = -fr(-f). Then r(-|) = -fr(-|) = W«.
we have
we have
we have
r(-f) = -fr(-f). Then r(-f) = -|r(-f)= -&>/£.
=
=
T(l)
0r(0).
It follows that T(0) must be infinite, since
0,
r(l) = 1.
p
—
—
=
=
(e)
T(0)
T(
and it follows that T(- 1) must be infinite.
p
1,
p = - 2, r( - 1) = - 2T( - 2), and it follows that T{ - 2) must be infinite.
(/)
In general, if p is a positive integer or zero, T{ — p) is infinite and
(c)
(d)
p=-f,
1
1 ),
r(-p-{) = (-ir
12.44
Prove that
<£{t
/ We have
12.45
Let
)
n
n
^
y e~
5£{t \=
for
n> -1,
dt.
Letting
n
sl
t
i?{r 1/2 } = Vn/s,
Prove that
I
=
n
»=-l/2
where
s
/2\ /2\ (2
st
> 0.
s
= u
and assuming
Find the Laplace transform of
J^{r 1/2 = *-^ =
}
(i/2)r(i/2)
5
12.47
> 0,
we obtain
'
^=
s
'
/-.
\] s
f(x) = vx-
I Using the result of Problem 12.44 with x replacing
ro/2)
s
> 0.
s
12.46
\2p+l
1/VVW
where
Problem 12.44. Then
in
l
n = \,
we have
r_
i
3/2
and
t
2
Find the Laplace transform of
f{x) = x"~
1/2
= 1, 2,
(n
.
.
.).
n — 1/2
replacing n, we have
Then repeated use of the formula of Problem 12.28 yields
f Using the result of Problem 12.44 with x replacing t and
i?{x"- 1/2 } = r(n + |)/s
°^i x
12.48
n + 1/2
.
r(n + 1/2)
„
/
s"
(w
- l/2)(n - 3/2)
+ 1/2
Find the Laplace transform of
•
s
f(x) = x"
•
•
(5/2)(3/2)(l/2)r(l/2)
(2w - l)(2n - 3)
n + 1/2
2V
•
•
•
(5)(3)(l)y^
+1/2
for n a positive integer.
I The Laplace transform of f(x) is identical to the Laplace transform of f(t), so it follows from Problem 12.44
that
J?{x"} =
———
t
12.49
,
which, as a result of Problem 12.30, may be written as
Find the Laplace transform of x 4
.
I
It
12.50
follows from the previous problem that
Find the Laplace transform of x 14
4!
4
}
24
= -^y = -j-.
.
I
It
if {x
follows from Problem 12.48 with
n = 14
14'
that
J^{x 14 } = -yj.
if{x"} = —j-r.
290
CHAPTER 12
D
LINEARITY
12.51
Prove the linearity property of Laplace transforms: If both f(x) and g(x) have Laplace transforms, then for
any two constants c t and c 2
^{cjix) ± c 2 g(x)} — c x if{/(x)} + c 2 ££{g(x)}.
,
2>{cJ{x) ± c 2 g(x)} =
I
Jj"
e-"[ Cl f(x) ± c 2 a(x)] dx = Cl " e~ sxf(x) dx ± c 2 j* e~"g{x) dx
J"
= Cl JSf{/(x)} ± c 2 2>{g(x)}
12.52
Find the Laplace transform of sinh ax, where a denotes a constant.
f Using the linearity property and the result of Problem 12.2, we have
(V>*
^{sinhax} = Se{
_ e -«)
1 (s
2
12.53
1111
i
i
= -<e{eax --^{e'"} =-
\
J2
2
+ a) — (s — a)
(s — a)(s + a)
}
2s — a
2
2 s + a
a
s
2
— a„i
Find the Laplace transform of cosh ax, where a denotes a constant.
I Using the linearity property and the result of Problem 12.2, we have
{e
+ e ax )
\
2
[
12.54
1111
=
+ --
~
ax
if{coshax} = if^
1
1
= -&{?*}+ -&{ e ->*}=-l
2
J
J
l
2
2a
1
s
Find
Find the Laplace transform of
f(x) = x + x
JS?{20x + 4x
I
2
}
= i'{x}+i'{x 2 }=i +
,
if {20.x + 4.x 2 } - 20i"{x} + 4i^{.x
2
+ 4a 2
s(s
)
)
A
J
r
1
2
J
2!
20
s
s
8
=20^ + 4-^ = ^ + -^
z
J
S
j^{-15x 2 + 3x}.
,
2!
,
1
if{-15x 2 + 3x} = -15j^{x 2 } + 3i"{x} = -15-^ + 3^ =
f(x)
f
4
if {15x
Find the Laplace transform of
= 15x 4 — x 2
s
- x 2 = 15^{x 4 - J^{x 2
f(x)
}
30
^-^
3
s
s
.
}
}
4'
2'
360
2
s
s
s
s
=15^-^ = ^
,
= 2x 2 — 3x + 4.
if{2x 2 -3x + 4} = 2if{x 2 } - 3^{x} + 4i?{l} =
§
- 2a 2
}.
Find the Laplace transform of
Find
+ 4a 2
2
.
s
12.61
s
=-+^
2^
s
s
s
s
12.60
2
2
I
12.59
-a j
2
i?{x + x
Find
2
2
1
")
s
Find
s(s
1
2>{3 + 2x 2 } = 3JSf{l} + 2if {x 2 } = 3- +
12.58
5
^{3 + 2x 2 }.
m
12.57
2 s + a
2
Using the linearity property and the results of Problems 12.1 and 12.23, we have
if {cos 2 ax} = if { 1 - sin 2 ax} = ¥{ 1 } - if {sin 2 ax} =
12.56
-=
Find the Laplace transform of cos 2 ax, where a denotes a constant.
I
12.55
s
-
-
2s-o
J
2^-3- + 4- = T -^ + -
if{-7x + 4x 2 + 1}.
1
2'
1
j^{-7x + 4x 2 + 1} = -7if{x} + 4if{x 2 } + if{l} = -7^ + 4^ + - =
s
s
s
8
7
1
s
s
s
^-^ + -
LAPLACE TRANSFORMS
12.62
/(x) = 79x
Find the Laplace transform of
I
14
- 8x 2 + 32.
14'
13
1
S
5
Find
32
16
_ 79(14!)
J^{9x 4 - 16 + 6x 2 }.
J^{9x
4
Find
^- — + ^
- 16 + 6x 2 = 9j^{x 4 - 16j^{1} + 6j^{x 2 }=9^16 - + 6^ =
3
J
}
}
s
12.64
2'
J^{79x 14 - 8x 2 + 32} = 79j^{x 14 } - 8^{x 2 } + 32if{l} = 79 -^ - 8 -^ + 32 S
12.63
s
s
s
5
S£{ - 14x
5
s
8!
6
26
80
s
s
s
s
5
+ 6x 4 - 100x} = - 14i^{x 5 } + 6i^{x 4 } - lOOif {x} = - 14 % + 6 -^ - 100 \
144
1680
s
Find
+
6
s
s
2
Find
jSf{17>/x - 10/>/x + 25x
2
}
j^jnTx" - -^L + 25x 2 j = 17JSf{Vx"} - lOSe \^=\ + 25if{x 2 } = 17 i V^s" 3/2 - lOV^s"
17
= — y/ns
Find
if {14x
3 2
'
'
s
+ 13x - 10x I/2 }
'
14-^+ 13-^-10-^
>
4s
21 Jn
2s
Find the Laplace transform of
512
13
+ ^r2
s
s
2s
+
—+ = - + + -—+
+ 3 -2
1
s
1
2
s
-
4
s
3s
2
s
1
2
4
65 sin 7x — 8 cos ( — 3x).
s
Find the Laplace transform of
2
2 sin x + 3 cos 2x.
—+ — -
j^{65sin7x-8cos(-3x)} = 65if {sin 7x} - 8i^{cos(-3x)} = 65^2
12.71
s
3/2
Se{2 sin x + 3 cos 2x} = 2S£ {sin x} + 3 if {cos 2x} = 2 -2
Find the Laplace transform of
5/2
5Jn
s
12.70
+ 25 ^
50
2
3!2
li2
ll2
\ = \4Sf{x
\\9>\x v
\
\lS?lx\ - 10i*{x
\OS?lx 1/2
\ = 14
J^{14x 3 2 + 13x- 10x 1/2
}
}
} + US*{x}
f
1 '2
^+^
Vs
IOVtt
r- _,.,
3 2
2
12.69
^
^{19x 3 -40>/x~} = 19j^{x 3 } -40&{yfx} = 19 -^ - 40- Vrcs _3/2 =-i
I
12.68
100
5
^{19x 3 - 40 Vx"}
I
12.67
s
+ 6x 4 - 100x}.
S£{ - 14x
12.66
J
+ ^- - 26 - + 40^
^
s
s
Find
s
s
S?{x 8 + x 3 - 26 + 40x 2 }.
S?{x 8 + x 3 - 26 + 40x 2 } = if{x 8 } + ^{x 3 } - 26^(1} + 40if{x 2 } =
12.65
291
8
49
s
2
+9
s
2
+ 49
s
2
+9
6 cos 4x - 3 sin ( - 5x).
S
^{6cos4x - 3 sin (- 5.x)} = 6if {cos4x} - 3if{sin(-5x)} = 6
s
2
+16
—+ — = +16 + +
- 3 -2
s
S
25
s
2
s
2
25
5
1
292
12.72
CHAPTER 12
U
Find
if {5 sin x + 10 cos x}.
if {5 sin x + 10 cos x} = 5if {sin
x}J + 10i"{cos
x} = 5 -,
1
l
'
12.73
5
s
2
10 cos lOx — sin(— lOx).
Find the Laplace transform of
if {lOcos lOx - sin(-lOx)} = 10JSf{cos 10x} - if {sin(-lOx)} - 10
s
12.74
3
Find
if {9 sin 4x + 20 cos ( - 5x) + \0e
10
^{9sin4x + 20cos(-5x) + 10e
I
s
-10
10s +10
+ 100
s
2
+ 100
Find
#
if{ 103e"
6x
*}
= 9if{sin4x} + 20if {cos(-5x)} + 10if{e 10x
——
4
2
14
s
+ 25
s
10
s
2
10
20s
+ 16
— + 2- = +
14!
s
9/2
+ 16
— + 10 -
}
36
1
+
s
2
+
+ 25 s - 10
+ 2} = 103if{e- 6x } - 18i^{cos 5x} - 9if {x 14 } + 2if{l}
= 103
if {2x
+ 20 -z2
s
- 18 cos 5x - 9x 14 + 2}.
J^{103<r 6jc - 18cos5x-9x
Find
—^ + 7-7
}.
s
12.77
+ 100
2
10x
=9
12.76
2
Find if {/(x)} if f(x) = sin 3x + x 3 - 25x.
&{f(x)} = if {sin 3x} + 2>{x } - 25J^{x} -
12.75
= IJL^l
+ 10 -^—
2
2
s + 1
s + 1
+ 1
18
+ 6
s
2
9
+ 25
.1
„
103
1
s
s
18s
6
s
9(14!)
+ 25
1
s
+ 16e*}.
I Using the results of Problems 12.2 and 12.47, we have
if{2x 9 2 + 16e*} = 2^{x 9/2 } + 16^{^} = 2
'
12.78
Find if {/(f)} where
/(f)
11 = 4<f{e
<e{f{t)\
4
s-5
12.79
s
{W
^} Vnr + s-17=
(
* ]
1
16
945V"
+
-. 1,,„
1/2
32s
16
s-1
= 4e 5 + 6f 3 - 3 sin 4f + 2 cos 2f.
'\
'
+ 6^{/ 3 - 3^{sin 4f} + 2if {cos 2f} = 4
}
36
+ -r-^2
s*
Find the Laplace transform of
s
12
/(f)
2s
—+
+16
=
s
2
where
+4
1
if
2
if
4
if
<
<
3 <
-2
if
4<f
<
1
f
f
f
s
s- 5
+ 6^-3
-^—j
+ 2 -^—
A
4
2
2
s
s + 16
s +4
> 5.
<
< 3
< 4'
1
/
4 (-1
3 r
2
1
I
I
i
1
1
2
3
I
(J
4
5
6
1
2
Fig. 12.1
+
2
.
LAPLACE TRANSFORMS
D
293
I The graph of this function is shown in Fig. 12.1. Using the unit step function (see Problem 12.22). we can write
= + u(t - 1) + 2u(t - 3) - 6u(t - 4). It then follows from linearity that
/(f)
1
+ &{u{t - 1)} + 2if{u(f - 3)} - 6if{u(r - 4)}
Se{f{t)} = if {1}
1
=-+
—+
e~>
2
if
12.80
Find the Laplace transform of
I
it
a(f)
=<
1
e
~4s
for
s
>
s
s
[0
is
„ e
2
<
f
<a
if
a<t<b
if
b < t
< a < b.
for
This function is known as a square pulse; its graph is given in Fig. 12.2. Since
= u(t — a) — u(f — b),
g(t)
follows from linearity and Problem 12.22 that
if {a(f)> =
&W
„- as
bs
e
~ a)} - y{u(t - b)} =
— e ba
s>0
*
--
f
Fig. 12.2
12.81
Find
i^{}(sinh at — sin at)},
where a denotes a constant.
if {^(sinh at - sin af)} = fSf {sinh at} - \S£ {sin af} =
la
2 s
12.82
Find
if {^(cosh at — cos at)},
1
2 s
Find
if {^(sinh af + sin af)},
+ a2
2 s
s
4
— a^4
4
- a4
where a denotes a constant.
if {|(cosh af - cos at)} = |if {cosh af} - |if {cos af}
12.83
la
—a
2
s
1
-a 2
2 s
s
2
+ a2
s
where a denotes a constant.
as
—
i^{|(sinh af + sin af)} = \$£ {sinh af } + \S£ {sin af} = - 2
T +
2
2
2
2 s - a
2 s + a
12.84
Find
if {^(coshaf + cosaf)},
-a
4
- a4
where a denotes a constant.
if {^(cosh af + cos af)} = |if {cosh af} + i^{cos af} = - —2
- a 27
2 s
12.85
s*
Use the linearity property to find if {sin
2
+
2 s
2
+ a2
s
ax}, where a denotes a contant.
J^{sin ax} = if {^(cos 2ax - 1)} = ^{cos 2a.x} - \tf {1}
1
1
2a
1
2
(Compare this with Problem 12.23.)
12.86
Find
^£
a-b
,
where a and b are constants.
2s
2
+ (2a)
2
2 s
s(s
2
2
+ 4a 2
)
'
CHAPTER 12
294
,{-L^l
—
'
a
b
—
r x-b - be-""}
'
12.87
Find
5£'{
(ae~
,
»
a-fc
M
Find
¥ l-^(eM —
1
xb -
1111
be~ xl"\
1
— at)},
— a)(s — b)
1
_
at
at
1
2
s
a(a-b)s-(-l/a)
{I
+ as){l + bs)
where a denotes a constant.
l\ (e - - at)] = \ J?{e
a
(s
J
b(a-b)s-(-\/b)
12.88
a - b s — b
a - b s - a
where a and o denote constants.
>.
abya — b)
* {0
«
a - fr
J
a
j
}
2
- \2 ^{1} - - ^{t}
a
a
a
2
s
—a
a
2
s
a s
2
2
s (s
— a)
FUNCTIONS MULTIPLIED BY A POWER OF THE INDEPENDENT VARIABLE
12.89
& if(t)} =
Prove that if
F(s),
¥{t nf(t)} = (- 1)"
then
—
d"
F{s)
= (- l) n F
<n,
(s),
where
= 1,2,3,....
n
as"
I We have
1^
F(s)
— f* e~*'f(t)dt.
Then by Leibnitz's rule for differentiating under the integral sign,
r
-
^ J e- /(0* - £ | «~*/M* - J -te-«f{t)dt - -J" e-*{t/(t)} A = ~^{t/W}
= F(s) =
"
n
t/F
Thus
= — —— = — F'(s),
y|f/(f)|
n =
which proves the theorem for
1.
ds
To establish the theorem in general, we use mathematical induction. We assume the theorem is true for
that is, we assume
k
s
(^ e
'{t
= (-l)*F(k)(s).
J(t)\ dt
Then
d
—
e
Is J
'
_s,
{f*/(0}
q
A = (- l)*F +1,
Leibnitz's rule.
-P" e-° {tk+if{t)}dt = {-lf'Fik+1 \s)
t
That is,
12.90
Find <f{te
Since
n
f" e
l
{r*
'
Since
Since
bince
'
Thus the theorem is true by induction.
"(x).
\.
'J\c
=-
2
'}
—2
,
ds\s — 21
^ '}~, n< '}=^(^)
2
2
i^'^
^e
2 '^
=
e
2
(s
— 2)
(s
- 2)-
=
i?h 3 e 2 x =
x\t
'
J
s
_2
,
]
&3 ^ _ 2 J
I
1
=
(s
- 2) 4
Find ^{xe 4 *}.
f
Since
12.94
Y |k- : '!
(*
Find y{tV'}.
I
12.93
—
l
Find ^{fV}
I
12.92
= (-l)* H F
2t
s
12.91
/(t)) dt
*
4
if {e *} =
,
,
1
f
*
4x ,
-,
if {xe
-,
if{xV*}
4
s —
}
=
d /
\
1
ds V s — 4 /
1
(s - 4)
,2
Find if {xV*}.
I
,
Since
,
,
<£ {e Ax }
=
1
,
s-4
l
,
._.
d3 /
1
dsMs-4/
(s-4)
(
"
(s)
or, by
n — k;
LAPLACE TRANSFORMS
12.95
Find &{x 6 e*
I „
x
}.
...
1
}
1
„..
<£
!
'
'
'
12.96
Find y{x 5 e~
Since
= -i-,
3x
]
5
f(x) — cos ax,
)
'
'
we have
—
S
ds\s 2 + a 2 )
(s
—
—ds^ f\s + ) = -^.
+
5
3/
F(s)
6
(s
—
2
3)
Then
.
J
fl
+ a2 2
2
—
— if {/(x)} — -^—
A
s
+ a'
'
)
d
s
if {cos ax} = -=2
2
(
i£\x 2 cos ax} = —^2
=,
+ a2
s
(
s
r
+ a 2 /)
2
oV Is
ds
\i'
J
— 6sa 2
2
(s
+ a2 3
2s
\
-=
3
)
Find if {f sin af }, where a denotes a constant.
f
,
d /
a
.
Y'{sinai}=^2
Since
+a
s
'
\
a
if {? sin af}
'
j,
2
las
'
2
'
0"sV5
2
+a /
(s
2
+ a2 2
)
Find ¥{t 2 sinaf}, where a denotes a constant.
Since
JSf {sin at)
= —.2
s
/M v
Find if{x
f
I
Define
-
<*
,
+a
if
2? ,
d
r sin at = —T2
i
,
.
ds
'
a
/
\s
2
2
3
das - 2a
~
2
2 3
I
(s
+a
\
-=
7
2
+a )
'
)
2
= y/x.
f(x)
x 7/2 = x 3 Vx = x 3/(x)
Then
12
y{/(x)J = Se{yJx~}={Jiis-*
and, from Problem 12.46,
d
Therefore,
.
3
¥{x 3 Jx~} = (- I) 3 -piles' 3 2 = f^J-" 2
'
)
(Compare this with the result of Problem 12.47 for
12.102
- 4) 7
(s
Find i?{x 2 cos ax}, where a is a constant.
Since
12.101
J^{x 5 <r 3 *} -
+3
*
12.100
720
\
'
Find if{xcosax}, where a is a constant.
i?{xcosax} =
12.99
- 4/
\s
\s-Aj
}.
jSf{e-
Taking
12.98
1
ds
3x
'
12.97
d6
(
....
1
= - - 6hk (
(xV
&„,{eAx = -s-4-,
Since
n = 4.)
Find ^{x 4 /^}.
I
From Problem 12.45, we have
&{l/yfx} = s/ns^
1 '
2
.
-Wr-<^MmmV"-"
12.103
We know that
s
¥ (cosh 3x] = -= — 9
.
Therefore,
if x cosh 3x} = —
™Jts- 9 2
'
s
—-25—
d /
5
Since
if{sinh5t}=^2
s
,
,
if }rsinh5f}
'
l
2
Find J^{r sinh4f}.
I
,,.,„,
2
J^{f
d2 /
sinh4f}
ds
2
\s
\
24s
- 16 I
2
4
2
(s
2
+ 128
- 16)
5
\
ds\s -25 1
2
2
10s
2
(s
s
-5
Find^{tsinh5t}.
I
\
d /
—
«v \s 9
—9/
d.v \.s
.sr
12.105
=
.
Find if {x cosh 3x}.
#
12.104
.
-25) 2
/
(s
2
2
+9
- Q\2
9)
-
295
'
CHAPTER 12
296
TRANSLATIONS
12.106
Prove the first translat ion or shifting property: If
&{f(t)} = f" e~ f{t)dt = F(s).
I We have
st
^{e*/(0} =
12.107
2
2'
We have
sekt
1
1
}
J
= -3 = —3
then
j§f{e"'/(t)}
= F(s - a).
Then
x
V/(0} dt = Jo e^-^f(t)dt = F(5 - a)
2
=
if {tV}
l
J
Then
.
s
s
s
- 3 T3
.
Find ^{e" 2 sin At).
'
I
We have i*{sin4f}=^2
s
12.109
«~
= F(s),
Find &{t 2 e 3t }.
I
12.108
J7
JSf {/(t)}
4
+
—
16
J^{e~ 2 'sin4r} =
Then
.
(s
—4 — = -— 4
+
+ +
+
2
16
2)
s
2
4s
20'
Find if{<? 41 cosh 5r}.
Since
if {cosh 5t} =
-2
s
2
—
-25
if{<?
,
—4
2
- 25
(s - 4)
s
4
'
cosh 5f] -
—4
- 8s - 9
s
s
2
Alternative Method:
i^{<?
4
'
cosh 5f } = i^e 4
e
Jt
'
+e
<£{e
+ e„-»\ _
<"
9x
111
.
s-9
2
12.110
s+
J^{cos It } = -=2
s
s
+49
&{e 2t cos 7f } = F(s - 2) =
= F(s),
-2
2
+ 49
(s - 2)
'
With F(s) as defined in Problem 12.110, we have
s-3
^{e 3 'cos7t} = F(s -3) =
,
,
'
With F(s) as defined in Problem 12.110, we have
Find ^{e'
I
5t
+ 3
-j2
(s + 3)
+ 49
^{e~ 3 coslt} = F[s - (-3)] = F{s + 3)='
s
cos It}
With F(s) as defined in Problem 12.1 10, we have
if{e" 5 cos 7r} = F[s - (-5)] = F(s + 5) '
+5
+ 49
(s + 5)
s
-j—
2
Find if {e" 5 cos 6t}.
'
Setting
F{s)
*
= i^{cos 6f } =
s
2
+ 36
2
^g
+ 25
,
we have
J^{<?~
5
'
cos 6t} = F[s - (-5)] = F(s + 5) = 7777^—
2
(s
+ 5) + 36
Find se{e~ ix cos 5t}.
Setting
F(s)
5
= if{cos5f} =
s
12.116
- 3) 2 + 49
Find ^{e~ 3 cos It}.
|
12.115
-8s -9
s
(s
12.114
2
Find 5£ {e 3 cos It}.
I
12.113
s
'
Since
12.112
1
Find if {<? 2 cos It}.
#
12.111
s-4
1
+
,
we have
J5f{e"
5'
cos 5t} = F[s - (-5)] = F(s + 5) = 777-^72—
2
(s
Find ¥{e' cos 5r}.
With F(s) as defined in Problem 12.1 15, we have
<£{f cos 5f} = F(s - 1) =
(s
-5—
- l) 2 + 25
+ 5) + 25
—
LAPLACE TRANSFORMS
12.117
Find
F(s)
= if {sin 5t} = -2
s
Find if {<?~
5'
—+ —
if {e' sin 5t} = F(s - 1) =
we have
,
J
25
'
(S
- l) 2 + 25
sin 5f}.
Using F(s) as defined in the previous problem, we have
5r
<£{e
= F[s - — 5)] = F(s + 5)
sin 5t}
(
(s
12.119
F(s)
—
= ^{sin6f} = -=2
1
;
Find
if{e
-2
'
(3 cos 6r
2
&{e
Find ^{e~
2x
Setting
'(3 cos 6f
- 5 sin 6f =
}
Find if {e
.
Let
F(s)
_x
+ 2) - 30
2
(s + 2)
+ 36
3s
3(s
s
2
6
s
3s
30
2
- 24
+ 4s + 40
—+ —
= ^{sin 5x} = -2
,
^{e~ 2x sin 5x} = F(s + 2) =
we have
v
l
25
'
+ 2) 2 + 25
(s
x cos 2x}
/(x) = xcos2x.
From Problem 12.97 with
a — 2,
s
we obtain
F(s)
2
= —^2
(s
-4
—2
Then
.
+ 4)
(s+ l) 2 -4
j^{e" xcos2x} = F(s+ l) =
x
[(s
+ l) 2 + 4]
Find J^{xe 4 *}.
Setting
F(s)
= ¥{x) = -T
we have
,
if{e
4jc
x} = F(s - 4) =
(Compare this with Problem 12.93.)
— 4)
=-.
Then
2>{x 3 e
4x
Then
y{x 6 e* x } = F(s - 4) =
s
12.124
.
+ 5) 2 + 36
sin 5x}.
s
12.123
(s
- 5 sin 6f)}.
j^{3 cos 6r — 5 sin 6t} = 3if {cos 6t} - 5^{sin 6r} = 3-2
so that
I
=
'
—— - 5^—— = ——-
f
12.122
5t
- (-5)1
^{e'
sin 6t\ = F[s
l
L
n =
we have
,
+ 36'
s
12.121
+ 5) 2 + 25
Find Sf {e~ St sin 6t}.
Setting
12.120
297
<£{e' sin 5t}.
Setting
12.118
D
(s
Find if{x 3 e 4jc }.
f
Setting
f(x) = x
,
we have
F(s)
= i*{x 3 = 3!/s 4
,
we have
F(s)
= ^{x 6 } = 6!/s 7
3
}
.
}
3'
~.
= F(s - 4) =
(Compare
with Problem 12.94.)
12.125
Find 2>{x 6 e* x }.
#
Setting
f{x) = x
6
.
6<
_'
with Problem 12.95.)
12.126
Find &{e 3x y/x}.
I From Problem 12.46 we have
12.127
_3/2
,
so
&{e 3x Jx} = ^(s - 3)" 3/2
Find i^e" 4 *^}-
I
12.128
if{Vx~} = iV^s
Since
£f{4x} - |V^s" 3/2
,
we have
j^{*T
4jc
V*} = W«(s + 4)"
3/2
.
Find if{e 2 '/Vf}.
I From Problem 12.45 we have
^{l/yft} = yfn/s.
Then
if{e /Vt} = V*/(s - 2).
2,
(Compare
4)
(S
.
298
12.129
CHAPTER 12
D
Find J?{e-
15
2t
t
}.
2>{ e
-2
T(8.5)
=
7 5
-
't
}
+ 2)
(s
12.130
n = 7.5
follows from Problem 12.44 with
It
^{f 75 } =
that
H8.5)
Then
,8.5
(7.5)(6.5)(5.5)(4.5)(3.5)(2.5)(1.5)(0.5)r(0.5)
8.5
+ 2)
(s
(15)(13)(11)(9)(7)(5)(3)(1)^
8
8.5
2 8/.
(S
-)\8.5
+ 2)
F(s)
= if {f sin t} =
.
Find &{te 2t sin t}.
I
If
we set
/(f)
=
f
a =
then it follows from Problem 12.99 with
sin f,
that
1
—
(s
12.131
Therefore,
Se{te
Find ^{f 2 ?
-
f
If
we set
'
2t
12.132
/"(f)
—
2
t
sin 3f,
= i"{f 2 sin 3f} = —-2
- 2)
2
[(5 - 2) + \y
Setting
—
-3-.
2
Therefore,
i?{t e
/(x) = sinax,
we have
'
= if {sin ax} = -^z
F(s)
a
j.
+ a'
Then, using the principle of linearity, we obtain
= \<£{e ax sin ax} - ^{e"" sin ax} = \F(s - a) - |F(s + a)
i^{sin ax sinh ax} = i <(sin ax)
7
a
1
2
2
2 (s - a) + a
2(s + a)
a
2
+a
2
+ las + 2a 2 - (s 2 - las + 2a 2
2
- 2as + 2a 2 )(s 2 + 2as + 2a 2
(s
(s
2
2
)
)
2a
)
s
4
2
s
+ 4a
Find i^{sin ax cosh ax}, where a denotes a constant.
f
With F(s) as defined in the previous problem, we have
~ajc
^ — ax
1
= £if {e" sin ax] + \!£{e~ ax sin ax} = ^F(s - a) + |F(s + a)
i^sin ax cosh ax} = Sf < (sin ax)
2
2 (s - a) + a
+
2
2 (s + a)
a(s
2
+a
2
+ 2a 2
4
4
s + 4a
a
I
2
)
Find if {cos ax cosh ax}, where a denotes a constant.
I
Setting
/'(x)
= cosax,
we have
F(s)
= i?{cos ax} = —2
s
+e„ - ax
.ax
1
.s
\
1
\
(s
2
Then
+ a)(s 2 - 2as + 2a 2 + (s - a)(s 2 + las + 2a 2
2
2
2
2
(s + 2as + 2a )(s - 2as + 2a
)
)
if {/(f)} = F(s)
Prove the second translation or shifting property: If
I
t.
+ a2
= \<£{eax cos ax} + \Se{e~ ax cos ax} = ^F(s - a) + |F(s + a)
s-a
2(s-a) 2 + a 2
+a
2(s + a) 2 + a 2
i"{a(f)}
s
1
*
i?{cos ax cosh ax} = SB \ (cos ax)
12.135
that
+ l) 2 - 54
sin 3f} = F(s + 1) =
2
[(s + l) + 9]18(s
-
s
12.134
a — 3
then it follows from Problem 12.100 with
- 54
2
Find i^sin ax sinh ax}, where a denotes a constant.
I
12.133
.
2
sin 3f}.
18s
F(s)
-
+ l)T
2(s
= F(s - 2) =
sin f}
2s
2
and
a(f)
=<
= e~ as F(s).
se{g(t)}
=
=
e~
J7
st
g{t) dt
a
|o
<?-*(0) 0"f
+
= |; e" 'a(f) df + £" *-«#) A
s
X
Ja
e" /(f - a) rft s,
= e~ as f" e~ suf(u)du = e~ as F(s)
where we have used the substitution
t
— u + a.
X
JQ
e'
s{u + a)
f(u) du
s
)
s
4
3
+ 4a 4
then
LAPLACE TRANSFORMS
299
Observe that y(t) may be written compactly as git) = f(t - a)u(t - a), where u(t - a) denotes the unit
Thus the second shifting property may be stated as follows: If
step function (see Problem 12.22).
J?{f(t)} = F(s),
12.136
£f{f(t-aMt-a)}=e- aif(s).
then
Find the Laplace transform of
=
git)
(t-2) 3
f
[0
i
12.137
^{t } = 3!/s
3
Since
Find if {/(f)} if
f(t)
4
it
,
a = 2)
follows from the previous problem (with
cos (f - 2n/3)
=
> 2
t<2
,
t
t
&{g(t)}
that
= 6e~
> 2?r/3
< 271/3'
2n,v 3
J^{cosf} =
Since
12.138
i^{e'}
=
s
12.139
Find &{f(t)} if
f
If we
-,
2
follows from Problem 12.135 (with
it
/(0 = {°_ 3
Find X{f(t)} if
Since
-
/(f)
write
e'
—
follows from Problem 12.135 (with
= j°
f
<3
|
>3"
r
= e 3 e'~ 3
,
then
Find the Laplace transform of
write
e
4'
Sf{f(t)\ = -
that
s
3
If we
a = 3)
l
&{f(t)} = ^{e e'~
12.140
'—
&{f{t)} = 2
that
>y
|
it
,
a = 2tc/3)
/(f)
3
u(t
= eV~ 3 u(f - 3),
f
>3
f
<3'
= e l2 e M '~ 3 \
then
/(f)
= e l2 e M,
' 3)
u(t
- 3).
Since
S-
-e
-
S-
¥{e M =
\
1
—4
it
,
follows that
3(s
12.141
l2
eM
'- 3)
u(t
.
\
3(s-l)
3s
1
s
&{f{t)} = 2>{e
-e~ 3s
and
- 3)} = e 3 ^{e'~ 3 u(t - 3)} = e 3
/(f)
—
- 3)} = e 12 lf{e M,i) u(t - 3)} = e
-e~ 3s = -
s-4
4)
s-4
Discuss the graphical relationship between an arbitrary function /(f) defined for all nonnegative x and the function
u(x
— c)f(x — c),
where c is a positive constant.
With fix) defined for
x > 0,
the function
u{x — c)f(x — c) = <
[f(x - c)
translation, of/(x) by c units in the positive x direction.
u(x — c)f(x
— c)
is
given graphically by Fig.
1
x > c
represents a shift, or
For example, if fix) is given graphically by Fig. 12.3, then
2.4.
u(x — c) f{x — c)
x
Fig. 12.3
Fig. 12.4
300
12.142
CHAPTER 12
D
f{x) — u{x — n)cos 2(x — n).
Graph the function
I
f(x) is sketched in Fig. 12.5.
Fig. 12.5
12.143
f(x) = \{x - \) u(x - 1).
2
Graph the function
I
f(x) is sketched in Fig. 12.6.
1
>/(*)
4
3
2
/i
/
f
1
1
1
1
1
.5
X
1
12.144
Ms
if {cos 2x) = -=2
'
12.145
'
\(x — \) u(x — 1).
f
1
i?{ix
2
}
x < 4
[(x-4) 2
x>4'
f(x) = x
define
Noting that
2
9 (x) = |° 2
x
it
follows that
3
,
+4
e~**.
we conclude that
X
^*
x > 4
1
if{|(x - l) u(x - 1)} = -, e
2
then g(x) can be given compactly as
,
if{/(x)} = 2/s
Find if {g(x)} if
1
("0
f
If we
/2'\
I
s(x)_
s
s
2
= }if{x 2 } = -I -j = -y,
Find if {g(x)} if
Fig. 12.6
5
Sf{u(x - 7r)cos 2(x — n)} = -=2
follows that
Find the Laplace transform of
Since
12.147
it
.
+4
s
4
u(x — rc)cos 2(x — n).
Find the Laplace transform of
Since
i
i
3
2
1
^{^(x)} = e
g(x)
4s
^.
= u(x — 4)/(x — 4) = u(x — 4)(x — 4) 2
.
LAPLACE TRANSFORMS
301
f We first determine a function f(x) such that f(x - 4) = x 2 Once this has been done, g(x) can be written
Now, f(x - 4) = x 2 only if /(x) =/(x - 4 + 4) = (x + 4) 2 = x 2 + 8x + 16.
as
g(x) = u(x - 4)/(x - 4).
.
Then
~ + -I + -
&{f(x)} = if {x } + 8^{x} + 16if {1} =
2
&{g{x)} = 5£{u{x - 4)/(x - 4)} = <?- 4s (— +
and it follows that
\S
J
4+—
s
2
s
).
J
TRANSFORMS OF PERIODIC FUNCTIONS
fie-*f(t)dt
12.148
T > 0,
Prove that if f(t) has period
then
Se{j\t)}
1
I We have
Sf{f(t)} =
s
f* ?
In the first integral let
= u;
t
which we write as
'f(t)dt,
^{/(f)} =
-e sT
J V"/W
* + J" e~*/X0 A + £* ^'/'(f) A +
in the second integral let
t
= u + T;
in the third integral let
= u + IT;
t
and so on. Then
&{f(t)} =
=
su
e~ f(u)du +
/J"
T
f*
e-« u
e~ f(u) du + e~ sT
su
|o
+ T)
f(u +T)du +
<r s<u
^ e~
e" Y(M) </u + e~ 2sT
s
JJ
+ 27
y(u + 2T)du +
JJ"
su
f(u) du +
•
•
•
•
•
V-/(«)Ai
J"o
1
where we have used the periodicity of /(f) to write
that
1
+ r + r2 + r3 +
•
•
•
=
—r
1
12.149
Graph the function
j\t)
for
,
\r\
f(u + T) = f(u),
f(u + IT) = f{u),
.
.
.
n < t < 2n
along with the fact
extended periodically with period 2n, and find if {f(t)\.
2
"
f
e~
s,
f(t) dt
Lr- f* e~
=
s'
T = 2n,
we have
=
U=-
sin t dt
e
s
\ — ssin t — cost)
s
2
+ 1
I
1
1
1
,
"
= \
—;-
s'
< 1.
I The graph appears in Fig. 12.7. By Problem 12.148, since
&{f(t)} =
-e"
-e" 2 "
s
2
+
1
(1
-c _,B)(s 2 + 1)
The graph of the function /(f) is often called a half-wave-rectified sine curve.
F(t)
-»
f
2tt
Fig. 12.7
12.150
Find the Laplace transform of the function graphed in Fig. 12.8.
T = 2n, and in the interval
< x < 2n it can be defined
<x <n
Thus, using the formula of Problem 12.148 with x replacing
n < x <2n
f Note that /(x) is periodic with period
analytically by
/(x) =
x
2n — x
t,
302
CHAPTER 12
D
2
we obtain
i?{/(x)} =
"
e~ sxf(x)dx
^—-e
f
_,..
2 its
.
1
2
*
J*o
g
_
_sx
/(x)<ix =
Jj
e
"xdx + £" e _sx(2n - x)dx = ^(e~ 2K5 - 2e~ ns + 1) = -^(e'" - l)
2
follows that
it
„.
,, u =
if{/(x)}
(l/s
- e"->
){e-" —)(e-"-l)
= — (\— —
:—
3^___
^r
-e
-e'
+ e~
\l+e
2
2
(l/s
2
2 ns
(1
1
l)
2
— tanh —
1
KS
ns
)(\
s
)
2
I
1
I
f(x)
i
12.151
3it
2jt
77
5tt
47T
-^ x
6tt
Fig. 12.8
77T
Find y{/(x)} for the square wave shown in Fig. 12.9.
fix)
2
1
3
|
4
I
I
I
6
5
I
I
i
I
7
|
I
I
I
I
-HI
Fig. 12.9
f
by
f{x) = <
nn*)} =
<x<
1
f
-
1
1
it
can be defined analytical!)
Thus, from the formula of Problem 12.148 with x replacing r. we have
< x < 2
-.
-e 2s
2
f
sx
Since
e~ f(x)dx =
2
P e sx (\)dx + f e _SJC(-l)<fcc = - (f 2s - 2e~ +
5
1)
follows that
(eF(s)
s
-D
2s
!
(l-O
s»2
2
s(l-e" )
_g-s
^/2
s(l
=
—
e* s(l+0
gi/2
_ e -*/2
p:
s{e*
2
)(l
+
)
-e
s(l+<T s
1
i=r
+ e s2
Find the Laplace transform for the function shown in Fig. 12.10.
1
O
-e
_s
r
a
12.152
<x < 2
and in the interval
1
JV»/<x) dx
1
it
7=2,
Note that f(x) is periodic with period
-
= - tanh s
2
)
= - (e
w
LAPLACE TRANSFORMS
I
We note that f(x) is periodic with period
fix) =
1
1
< x<
< x <2
T = 2,
and that in the interval (0, 2) it is defined analytically as
follows from Problem 12.148 that
It
(
J'\f(x)\
uy n
= J"
V»/(x) dx
- e 2v
But
1
1
2
Jo
Therefore,
Sf{f(x)} =
e-»/(x)dx -
e-"(l)dx +
JJ
— — ^27— = —-e
7;—
e
1
J'
1
g-"(0)ix = -- e
= -d ~e *)
s
1
-s
(1
)(l
+^"
s
s(l+e" s )
)
fix)
1
.
14
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
2
3
4
5
6
7
Fig. 12.10
12.153
Find the Laplace transform for the function shown in Fig. 12.11.
I
This function is periodic with period
T — \,
and it is defined as
/(f)
=
e
H
on the interval (0, 1). Using
f
the formula of Problem 12.148, we have
*{fm) =
1
—
.V
•- tdt
/.'
e
e
s
-e
1
-e
s
1
2
-e
s
2
1
-e
Find the Laplace transform for the function shown in Fig. 12.12.
fit)
i
1
2
3
4
5
6
s
(s
+ 1)
s {\ - e~
2
Fig. 12.11
12.154
303
7
Fig. 12.12
s
)
"
1
CHAPTER 12
D
304
f
0<
t
t
<
^e-"f(t)dt
Therefore.
e --/T[t)A = /;e--tA +
SC{f(t)}
=
e
2
s (\
l
-e
2s
se
2s
— se
- e~
But
J7«-1(l)&-(-£e---ie--Y+^
2s
)
Find the Laplace transform for the function shown in Fig.
I
i?{/(r)} =
Using the formula of Problem 12.148, we have
^
55
In the interval (0, 2) it is defined analytically as
l
fit)
1 2.
7 = 2.
This function is periodic with period
7=2.
This function is periodic with period
12. 13.
and it is defined on the interval (0. 2) as
/(f)
=
1
—
t.
Using
the formula of Problem 12,148, we thus have
i
JV
v{M)
1
s(
-t
i
s
2s
-e
" + -r e
e
(i-f) dt
1
2
-s
e
J
(i
-e
2
+ 1) + s -
s (l
-e
1
2s
)
Fig. 12.13
f°e-"f{x)dx
12.156
Prove that if
/( v
f (o) = -fix),
<f{f(x)} =
then
1
I
+e-°"
Since
/ (
fix)
is
v
+ 2w) = /[(x + to) + to] = -f(x + <o) = - [ -f(x)] = f(x)
periodic with period 2«). Then, using the formula of Problem 12.148 with x replacing t and
7 replaced
by 2io, we have
2<
J
o
y;/(vi:
Substituting
y = x — to
"
e
-
"fix) dx
-e
"
J™
*
sx
/(.x) dx
2<i>s
1
+
«
Jj"
—e
"/ (x) dx
2 era
into the second integral, we find that
£" e~"f{x)dx = J; e-f+^fiy + to)dy = e~~ f° e"\_ -fly)] dy = -e"- ^ e *>f(y)dy
If
—e'
we change the dummy variable of integration back to x, this last integral becomes
Then
2
12.157
:
m
(1
-e~n ge-"fix)dx
1
-<?
2tus
il-e-°")j°e sxflx)dx
+p-"
(1 -e"' )(l
os
Solve Problem 12.151 using the formula derived in the previous problem.
J%-"/(x)rfx
,s
)
1
+<?"""
sx
e' f(x)dx.
LAPLACE TRANSFORMS
I The square wave f(x) satisfies the equation f(x + 1) — —f(x), and on the interval (0, 1)
With to = 1, the formula of the previous problem becomes
f{x) = 1.
V»/(*)dx
J"
&{/(*)}
1
+e"
s
fie-V)dx
+e
(1 /,
1
1
I The function
(with
cd = n)
f(x) — sinx
s
tanh -.
f(x) = sin x.
f(x + n) = —f(x),
satisfies the equation
defined by
+e
1
Use the formula of Problem 12.156 to obtain the Laplace transform of
is
X i_
This is the same result as is obtained in Problem 12.151. It may also be simplified to
12.158
it
so the formula of Problem 12.156
becomes
e
nn*)}
sx
sin x dx
Jo*
1
+e~ ns
s
2
+
(
— ssin x — cosx)
s
1
1
+e~"
D
2
(e~
+
ns
+ 1)
1
s
1
+e
s
2
+
1
305
j
.
CHAPTER 13
Inverse Laplace Transforms and Their
Use in Solving Differential Equations
INVERSE LAPLACE TRANSFORMS BY INSPECTION
13.1
Develop a table of inverse Laplace transforms.
I
¥~ {F{s)\ —f(x)
x
Since
if
and onlv if
£f{f(x)} = F(s),
every formula generated in Chapter 12 for a
We have, for
= 1. We have from Problem 12.2
Laplace transform automatically provides us with a formula for an inverse Laplace transform.
example, from Problem
if {e
that
ax
}
—
s
that
J/^ [sin
2
—
s(s
j-
2
+ 4a 2
^or
5£
so it follows that
¥~
for any constant a, so it follows that
—a
ax) =
i^{l} = 1/s,
12.1 that
_1
{l/s}
x
(s
—a
an y constant a, so it follows that
if
\
'
= eax
>
We have from Problem 12.23
.
-1
[s{s
)
2
+ 4a 2 )\
Continuing in this manner, we generate Table 13.1. where all the inverse Laplace transforms are given as
functions of x. To obtain an inverse Laplace transform as a function of t instead, we simply replace x with t.
13.2
Find if " {2/s 3 } as a function of v.
1
I
13.3
It
Find
It
13.4
Find
follows from Table 13.1. entry 3, with
¥~
x
l
<-=—[s
It
13.5
Find
2
->
13.6
Find
follows from Tabic
— —
5£ "' <-=
1
3.1.
entrj
l
[(s
9,
with
T>
+ l) 2 J
1
1.
with
13.7
Find
It
13.8
Find
follows from Table 13.1. entry 12. with
if
x
<-^
¥~ \
x
306
}
+ 5]
^~ X-^
a = 5
that
Sf
a = 5
that
'J'
a=
that
i£
}
= x2
.
-> = sin 2x.
x
~1
'
<
-
2
>
= cos 5x.
'
— = cosh
\—2
>
5x.
1
,
f
2s
(s
+ iy2
\—^2
— I\ = xsmx.
as a function of x.
follows from Table 13.1. entry 9, with
[5
It
>
V + 3{
that
3
as a function of x.
I
It
a = 2
{2/s
as a function of x.
>
follows from Table 13.1, entry
&~ <—z2
if
as a function of x.
+ 25
V-25]
It
that
as a function of v.
>
{—^2
\s +4\
follows from Table 13.1, entry 8. with
'J'
_1
n = 3
S
-1
a = ^3
that
if
-5
that
<£~ x \
j
2
>
= cos V3x.
as a function of t
follows from Table 13.1. entry 7, with
a =
-X = e' 5
'.
INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS
Table 13-1
f(x) =
F(s)
-
1.
¥- {F(s)}
,
(*
> 0)
1
(8
> 0)
X
s
i
2.
x«-» (n = l,2, ...)
3.
s"
4.
^V^S-3/i
(s>0)
V*
5.
V7s-^2
(s>0)
l/V*
6.
(l)(3)(5)---(2«-l)V^ __ M _ I/9
(
2n
7.
C- "^
s
—a
S2
+ a*
("
'
a
8.
9.
a
S2
_ a2
S2
_ a2
(s 2
+ a2) 2
s2
— a2
(s
— a) n
13.
14.
(s
16.
17.
cos ax
c?
S
"
> Uh
W)
sinh ax
"N \n\\
l°l>
cosh ax
(~^
0\
S> °)
x sin ax
(
,
x cos ax
x n-l e ax ( n = 1,2, ...)
v
'
M
6
)
e bx sin ax
rr^-w
(«»- b
e bx cos ax
(-"
ft:
s ~~ 6
+ a2)2
}
(
_ 6) 2 +a 2
(s
2a "
"
l« 5
(ji
_ 6) 2 + a 2
(s 2
e ax
//\
> 0)
a
15.
= 1,2, ...)
(S
2(1 S
12.
a;n-l/2 (n
sin ax
S
11.
0)
^ °n\
(<•
o
2
s z -h a^
10.
_
~i>
>
r,-m
5 -°)
(
sin ax
— ax cos ax
307
—
308
CHAPTER 13
Table 13-1 (continued)
= i?" W«)}
/•(*)
F(s)
1
18.
+ as
1
a
- (e°* - 1)
1
19.
s(s
— a)
8(1
+ as)
a
1
20.
- e~ x a
'
1
1
21.
+ as) 2
(1
a2
1
22.
(8
a — b
— a)(s — b)
e — x/a
1
23.
(1
+ as)(l +
8
24.
(s
- a) 2
(1
+ as) 2
— g — r/b
a — b
6s)
(1
+ ax)e ai
8
25.
(s
28.
29.
30.
a*
ae az _ bgbj:
8
26.
27.
4;(a-x)e- l/a
— a)(s — b)
a- b
8
ae~ xlh — be~ z/a
(1 + as)(l + bs)
ab(a — 6)
1
_L ( e ax - i _ ax )
s 2 (s
— a)
a-
2a 2
s(s + 4a 2 )
sin 2 ax
2
s{s 2
2a 2
— 4a 2 )
a3
31.
s
4
+ a4
sinh 2 ai
1
/
ax
ax
ax
ax\
——[cosh——sin-— — sinh— cos
\/2
\/2
V2
V2/
\/2V
,
.
.
,
—
)
INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS
Table 13-1 {continued)
Ax) = ^-'{F(s)}
F(s)
a2s
32.
33.
34.
+
—
s4
39.
s4
s4
a(s*
s4
s3
42.
s4
+ 4a4
as 2
43.
(s 2
+ a2 2
(s 2
+ a2 2
(s 2
- a2 2
(s 2
- a2 2
2
)
as 2
47.
(s 2
- a2 2
(s 2
- a2 2
)
cos ax sinh ax
)
sin ax cosh ax
cos ax cosh ax
— (sin ax + ax cos ax)
cos ax
—
— ax sin ax
.
— (ax cosh ax — sinh ax)
— sinh ax
2
— (sinh ax + ax cosh ax)
)
a3
48.
sin ax sinh ax
)
as
46.
—ax
)
a3
45.
.
)
s3
44.
V2
\/2
—(cosh ax + cos ax)
- a4
+ 2a 2
+ 4a4
,
cosh
— (sinh ax + sin ax)
- a4
- 2a 2
+ 4a4
sin
— (cosh ax — cos ax)
2a 2 s
+ 4o4
a(s 2
a*
.
,
— (sinh ax — sin ax)
s3
38.
—+
—
-a*
s4
,
ax
as 2
37.
ax
,
.
cos — cosh
\/2
y/2
a2 s
s*
.
V2
a4
- a4
36.
ax
ok
ox
/
smh
[cos
V2V
a3
s4
41.
1
+ a4
s3
s4
— smh Vz
V2
as 2
s4
35.
40.
.
sin
+ a4
s4
)
cosh ax
+
— sinh ax
—
ax \
y/2/
309
>
CHAPTER 13
310
13.9
Find
if
-1
1
It
as a function of f.
+ 2s)
s(l
_
a = 2
follows from Table 13.1, entry 20, with
that
M
JSf
)
js(l
13.10
Find
if
-1
It
= 1 - e~" 2
.
as a function of /.
- 9) 4
(s
+ 2s)
a = 9
follows from Table 13.1, entry 14, with
n = 4
and
A = ^e
<e~ l \
that
9 '.
2s
13.11
Find
jSf
s
It
13.12
Find
4
a = sj2
follows from Table 13.1, entry 36, with
^~' ]^— t[ = . (cosh v 2f - cosV2r).
that
2s
a>-\
<e
(s
It
as a function of t.
+4
as a function of f.
- 4) 2
2
a = 2
follows from Table 13.1, entry 46, with
Y
y "
that
'
= -sinh It.
\-^— —=[
2
](.s--4)
2
J
LINEARITY
13.13
Find
It
13.14
Find
W
'f
-^
follows from Tabic 13.1. entry 7, with
<£
s
#
It
as a function of t.
2
a- 2
that
a - 3
that
S?
'
V
<
= 4y _1 {-
as a function of t.
+9
follows from Table 13.1, entry 8. with
.13
I
V + 9?
]s
3
2
+9
Is
3
2
= - sin 3f
+9
3
1
13.15
Find
<£
s
I
It
2
as a function of t.
-3
follows from Table 13.1. entrj
^-i
1
s
13.16
Find
_1
{1/.s
jS?
3
}
2
)
-3
10.
v 3
f
.
1
2
V3
that
)
V3
1
-3J
V3
ls
follows from Table 13.1, entry 3. with
n = 2
that
JSf'M—3
s
13.17
Find
It
S£
l
{\/yfs}
^ind
follows from Table 13.1, entry 5. that
f
<e~
x
{
*t
)
1
2
-(V3) i
v'3
2 1
2
s
1
-S"~M-r3
3
2
s
as a function of .x.
if" 1 <—
i-
*-0-U— l*- ^
i
iJ
.
ly/s)
13.18
2
as a function of t.
I
It
a = v 3
with
y/71
ly/s)
yjll
as a function of x.
We use Table 13.1, entries 2 and 3. to write
,
5s + 4 J
.
5s
4
^4> + 2j^-
1
5x + 2.x
2
.
i
r
INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS
3
—+—
s)
fs
13.19
Find
I
'<
J§?
^
as a function of x.
>
We use Table 13.1, entry 3, first with
3
+
s
.v
:
13.20
Find
.
s- 5 ^- >
s
Find
*
2!
J
js
+
3
4!
j
n = 4
^
to write
X
5
2
4!
21
j
J*
and entry 6 with
n = 3.
s
7T
s
->
'•{-
if
5
n = 4
then with
+
*
24
as a function of t.
f We use Table 13.1, entry 3 with
13.21
n = 2
- 30>/sl
("24
M
if ~
311
to write
^
as a function of t.
{25 -3j
a =§
f From Table 13.1, entry 7, with
13.22
Find
if
'
<-
->
13.23
It
Find
follows from Table 13.1, entry 7, with
a =
3s + 5J
(s-(-5/3)
.
if
Find
^
13.25
s
+ 9 j
\
+ 9 J
|s
_,x [2s + 18)
if
<—z
V+
a = 3,
2
9|
Is
(s
Find
,
2
-1
<^=2
\s
—+ ^\ + -r&
}
18
25j
5
s
f
\~22
|s
25]
if
a = 5,
.
(s
_
Find
I
if
,
'
,
f
16s
13.27
Find
l
2
a = J5,
we have
18
1
>
25j
—
.
sin5t.
5
8
<
8 - 6s
&- \—,
x
2cos5t +
5
as a function of x.
>
+ 5i
— 6s 1
->
^
2
(16s + 9J
f
we have
—+ 7 =
f
\~22
I From Table 13.1, entries 8 and 9, both with
13.26
= 2cos3x-
9
as a function of x.
>
—+ ^\ = ^
f2s
we have
25J
H- 18")
\-2
-<>
2
From Table 13.1, entries 8 and 9, both with
.
that
as a function of x.
>
2
—f
— = 2¥ ^L-i—
i-6^-'i^
+
+
\-52 =:
s
f
~ 18 1
f 2s
3
From Table 13.1, entries 8 and 9, both with
l
13.24
(3 s + 5/3 J
<^
(
f
we have
as a function of x.
(3s + 5 J
f
,
as a function of t.
-^ - ^7=^'
rhr
+
+
}
1
f 1
8
(16s
2
,
9]
6s
}
9/16j
2s + 3
£e~ {^,2
|4s
+ 20
\
as a function of t.
2
3
.
3/4
f
(s
,
2
.,.,2
]
+ (3/4) j
3
f
s
]
2
3f
3
3t
-^
^^—sin^-^cos
-x^"
8
(s + (3/4) j
3
4
8
4
,
1
,
2
2
J
312
1
1
CHAPTER 13
U
]4s
2
= - cos V5f H
—
+
2^
!4s 2 + 5(
+ 20f
=JSf
2
s^Cv^) )
V5
-
4^5
]
2
+ (n/5) J
s"
p sin V^t
4^5
1
13.28
Find
if" 1
2s
-3s
2
as a function of t.
-7
= 2>->
2>-U±^-\
-7
2s
2
{
2
2 s
—
Find
&
~ l
s
2
- (V7/2)
1>
+5
as a function of f.
2
+5
3 s
2
= -cos
3
if
-1
-(V7/2)
V2
cg-x
3s
Find
2
2s + 7
<J-^2
3s
13.30
s
/- 1
cosh
2
\]2
13.29
2^7/2
7/2 J
/- t
sinh
--if -
if- 1
-
3.v
^-
2
+ 5/3]
7
/5
3
[s
/-f
V
V3
15
+ (V5/3) i
3^5/3
= jy _
1 -l) +
5/3
1
if"
2
Is
2
+ (VV3) 2
/5
.
/-f + -=sin
3
+
2
+ 2
T\
as a function of t.
(s- 1)-
^- M Ji±li = ^(s
- 1)
i
p-D^ +
(s-1)
:
2
i
5
5
|3(
1
(s-1)
-fc^M*-' (s-1)
5
(s-1) 5
5
24
2
)
COMPLETING THE SQUARE AND TRANSLATIONS
13.31
Develop a method for completing the square for a quadratic polynomial.
f
Every real quadratic polynomial in s can be put into the form
a(s
+ k) 2 + h 2
To do so, we write
.
2
as
2
+ bs + c — a\s 2 + -s) + e — a
where
= b/2a
A:
-Y
2a J
+ {c
_
-^a) =
i
S
+
Ya)
+
C
{
-^a^
a{S
+ k)2 + h2
h = sjc — b /4a.
2
and
1
13.32
Find
tf'
1
[s-
- 2s + 9
as a function of .x.
# No function of this form appears in Table 13.1. But, by completing the square, we obtain
2
2
2
2
Hence,
s - 2s + 9 = (s - 2s + 1) + (9 - 1) = (s - l) + (y/%)
.
s
-2s + 9 ~
_
1
1
2
(S
2
-1) +(V8)
Then, using linearity and Table 13.1. entry 15, with
a = y/S
c/>-\
<£
s
2
2
V8 (s - l) + (v^)
and
2
b = 1,
-
2
;
Find
y/S
1
~
V«
<£
13.33
2
- 2s + 9j
7s
(s
- l) 2 + (V8)
>
we find that
= —= e x sin v8x
as a function of x.
+ 4s + 6
# No function of this form appears in Table 13.1. However, by completing the square of the denominator,
we obtain s 2 + 4s + 6 = (s 2 + 4s + 4) + (6 - 4) = (s + 2) 2 + 2. Then, from Table 13.1, entry 15, with a = yjl
=
INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS
b —
and
— 2, we have
s
13.34
if"
Find
313
1
—
'
<
[s
,
2
>
- 6s + 25J
2
+ 4s + 6j
[(s
J2
+ 2) 2 + 2j
Ji
as a function of x.
f Completing the square of the denominator, we obtain
2
2
2
Then, from Table 13.1, entry 16, with
s - 6s + 25 = (s - 6s + 9) + (25 - 9) = (s - 3) + 16.
a = 4
6 = 3,
and
——1 = ^-0-—^ + — =
if'j 2
we have
I
,
13.35
Find
if
_1 <-—
->
=
[2s
2
+ 4s + 7 J
[(s-3) 2
-6s + 25j
[s
,+1
2s
.
2
'
+ 4s +
[2(s
^
1
,
if
'
— - 8s
<
[4s
>
2
+ l)
l(s
5J
s+l
f
+
l)
2
s
(l
.
2
S
2
Find
Then, from Table 13.1, entry 16, with
1-g- L. s+ + \-*->
.
7J
=
13.36
cos 4x.
— 1, we get
b =
and
3x
as a function of x.
f Completing the square of the denominator yields
2
2
2
2
2s + 4s + 7 = 2(s + 2s) + 7 = 2(s + 2s + 1) + (5 - 2) - 2(s + l) + 5.
a — \j5/2
e
16j
[2 (s
^^ = -e
+ (VV2)
]
2
J
1
+
l
2
+ l) + 5/2
_x
/5
/-t
cos
2
V2
as a function of x.
J
I Completing the square of the denominator, we obtain
4s
2
- 8s = 4(s 2 - 2s) = 4(s 2 - 2s + 1) - 4 = 4(5 - l) 2 - 4. Then
a — I,
This last result follows from Table 13.1, entry 10, with
(see
Problem 12.106): Since
if{sinhx}=^
s
&~ \
x
Find
It
-,
we have
if {e'sinhx} =
— 1) —
- l) 2 -
&~ 1—=
l
as a function of x.
>
&\
follows from entry 5 of Table 13.1 that
5£ <e
ix
\
>
= —= S£ \—j=\ = -j= sins'
=
sjn yjx )
(.
^
" J
<(
!>
.
[y/2s + 3
—3
as a function of t.
.
lV2sT3j
+4
" \
+ 4s +
V2
l(5
s
13.39
Find
l
£e~ l-^2
s
as a function of x.
+ 3/2) 1/2 J
1 '2
so we conclude that
y/s
nx
'"feb}-**—
Find
so that
1
1
from the first translation property that
13.38
,
5
(s
i
5
[(s
13.37
—
coupled with the first translation property
V2
v^
V^
=—.
It
follows
—
-
)
1
CHAPTER 13
314
# No function of this form appears in Table 13.1. But completing the square in the denominator yields
s
2
+4s + 8 = (s 2 + 4s + 4) + (8-4) = (s + 2) 2 + (2) 2
*4
'**
=+ 4s + 8 (s + 2) 2 + (2)u2
5
Hence,
.
s
2
.
Table 13.1 does not contain this expression either. However, if we rewrite the numerator as
s + 4
and then decompose the fraction, we get
2
sr2
^ 98
+ A4s. +
s + 2
2
+ i, *>\22 r^2
->\i
2
n\22
(s + 2)
+ 2)
+ (2)
+ (2)
=
±
t*
(s
_,_
,
i
+ 4 = (s + 2) + 2
s
Then from entries 15
-
>
,
and 16 of Table 13.1,
+4
= JSf + 4s + 8
+2
)
+ &~
2
2
^(s + 2)
+ (2)
5
s
13.40
Find
J^
_1
s
2
2
s + 2
- 3s + 4
s
i
3
4)
+
(
and
2
'
s
s + 2
-3s + 4
2
- -
"4
4
We now rewrite the numerator as
s
y
(s
(
fyfiY
-TT- 1
I
I
s-2/ +
+ 2
2
2
S - 3/2) + (V7/2)
= if
r - 3s + 4
V
= |s--l + V7^'
2
+
^ -
-4
6s
&
>/'
V7/2
(s
3/2)
2
+ V7/2) 2
(
Y**-\(s-3/2) +
2
2
(V7/2)
/7
2
- 2) + 8
6(s
i
2
6 e?
Find
(
^,,3/2), sin Vi x
[(s-2) +16j
13.42
2
as a function of f.
- 4s + 20
y/
+2
2
S - 3/2) + (v^/2)
s
-4
6s
'
f
sin 2x
^^7
3\
2
y
s + 2
-^—
2
s -3s + 4
L
so that
,
- 3/2) 2 + (V7/2) 2
(s
/7
Find
2x
But completing the square in the denominator yields
= e<3/2>* cos V_ x
13.41
— e 2x cos 2x + e
+ 2) 2 + (2)-
as a function of x.
# No function of this form appears in Table 13.1.
- 3s +
2
f
,
2'
cos At + 2c
2'
s-2
= 6^ -
+ 2^-'
[(s-2) 2 + 16
(s
- 2) 2 + 16
= 2e 2 (3 cos 4f + sin 4f
sin 4r
'
4s + 12
i
s
2
as a function of t.
+ 8s + 16
4s + 12
/
s
2
As + 4) - 4
= &'
i
+ 8s 4
1
6
[
(
(s
+ 4)
= 4&
2
1
-
1
-4JSf _I
i
(s
J
+ 4)
2
= 4e- 4 '-4t<r 4 = 4e- 4 '(l - r)
'
13.43
Find
3s + 7
cp-\
¥
s
2
as a function of f.
- 2s - 3
3s + 7
j&r 1
-2s -3
3(s
= iT
- 1) + 10
2
-4
(s - l)
= 3JST
s-
1
1
(s- l) 2 -4
+ 5J2?-
1
(s
= 3e' cosh It + 5e' sinh It = e'{3 cosh 2f + 5 sinh It) = 4e 3 - e
- l) 2 - 4
'
'
13.44
Find
if
s+
-1
s-
s+
^o?l
s
2
1
+s+
as a function of t.
1
1
= y^
s
1
+
5
1
2
+s+ 1
(5
+ I) + I
1
f
V3^
-r/2
>/3f
/V3t
—— V3
cos -— + sin —
2
2
.
I
V3 V
+ i2^2
+4
2
-i) + !
>/3/2
—-- + -^= e f/it/2I2 sin—
= e " 2 cos —
>/3f
T
2^/3
1
_
-
>/3r
2
+
=5
=
'
INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS
13.45
£?~ l {-z—
Find
[s
I
2
-)
Completing the square of the denominator and rewriting yield
s
s
2
as a function of t.
-2s + 3]
s
- 1
s
(s-l) 2 + 2
-2s + 3
{s
_ x) + (s/2)
+ -=-
^{s-Xf + ^f
2
~l
s
s
2
-2s + 3j
l(s
¥ _ — 7s + 4
x
Find
[4s
f
2
+4
+ 4s + 9
7s/4 +
4s
s
2
-\
—-
7s/4 +
1
+ s + 9/4
7s/4 +
1
2
+ 1/2) + 2
(s
l
4
= if i- e~" 2 cos V2r +
4
I
S
if"
'
A
2
4
l s
+4s + 9j
+3
^ -A
N^
(4s + 4s +
s
13.47
Find
4s
—=
e~ tl2 sin v^r I
8^2
J
I4
—=
sin y/2t\
8V2
/
V4
I
—
—
8V2(5+l/2) 2 + (V2)
as a function of t.
>
9j
+ 3
s
2
+ 1/2) 2 + (V2) 2
)
2
I
(s
+ 1/2) +1-7/8
2
= e~ t/2 ( - cos yjlt +
1
_
\
2
—=- = JSf \- e'" cos V2~r 1 +
+ 1/2) 2 + (v^) 2
8 (s
J
(7/4)(s
1
+ 1/2 )2 + (x/2)
(s
= <£ \- e~ t/2 cos silt \ + - -
Therefore,
e' sin sjlt
as a function of t.
>
+ 4s + 9j
7s
2
_ l + -Ljy-vf
2
2
{(s - l) + (sjl)
V2
)
f
l
Then
.
— l) 2 + (V2) 2 J
= e' cos yjlt
13.46
—
—Jl
1
2
is + |
+ 4s + 9
s
2
(s
is + |
+s+l
(s
is + 4
+ i) + (! - i)
2
+ i)/4 + (3-i)/4
2
(s + i)
+ v^ 2
s
1
4 (s +
(s
+ |) 2 + 2
+i
2
i)
+
+ V2
2
13 -i
^
4
2
V2
(s
+ |) + (V2) 2
= - if {e"' /2 cos V2t} +• 1 -^= ^ {e~" 2 sin v^}
4
4
and therefore
13.48
Find
-1
+
= \e
/
|4s + 4s + 9j
J A
jSf
2
<£~ 1 <(—
\
2
4s +4s + 1
4s
~ "2
—=
cos sjlt +
is + |
+ 4s + 1
s
2
(s
+s+i
+ i)/4 + |
2
(s + i)
is + |
(s
+ i)
(
<j
if" 1 ^
3 e~
4s
+
4s
+{
4(s + i) 2
1
/+ 4s + 4 = ie"' + fr<T'
2
A
/2
1
S
2
s
= i^{e-" 2 }+l^{te-" 2 }
and therefore
e'"
2
sin yjlt.
8^2
as a function of f.
+ 3
s
2
>
A
2V2
/2
.
+ i)/4 + (3 - i)/4
2
(s + i)
5
1_
8(s + |)
_1_J_
2
4s + i
1
8(s + |)
2
2
3
CHAPTER 13
316
I
f2]
Sf~
13.50
'
Find
J -3
1
Sf
Since
\
Sf' 1
2
s
I
follows from the second translation property (Problem 12.135) that
'
as a function of x.
+4
I
Sf
Since
it
,
= (x - 4) 2 u(x - 4).
4s
e'
<-^> = x 2
<
—5
s
2
? )
+4
= cos 2x,
it
cos 2(x — 7r)
if" 1
s
2
follows from the second translation property that
x > re
x < re
+4
— u(x — re) cos 2(x — re) — u(x —
re)
(See Problem 12.144.)
cos 2x.
its/
13.51
Find
if
2
S
l
if
Since
+
as a function of t.
1
\^
s
13.52
Find
Sf~
l
}•
+
"**)_ f sin (t — re/3)
we have
f,
1
1
-i
(s - 2f
se
Sf~
s
2
1
~
J
t
> re/3
f<re/3'
JO
e
2'
1
6
3 „2(t-5)
lU-5) e
(s-2f
> 5
t
t<5
10
= i(f-5) 3 e
2,,
- 5)
u(f-5)
-4its/5
as a function of f.
+ 25
if
Since
t
3!
if
Find
+
—— = -fV\ we have
)
= e 2 'if-'<(-y =
5s
13.53
2
as a function of t.
3
v
if
Since
l
Sf
s
.)
<
= sin
s
2
= cos 5f,
+ 25
se
cp-\
tf
s
2
-*«/5"j
+ 25j
r
cos 5(f _ 4^5)
> 4?r / 5
r
f<4re/5
[0
f
[0
cos 5f
> 4^/5
< 4re/5
f
r
= cos 5f u(f - 4re/5)
13.54
Find
-J(s+l)g'
if"
s
2
+s+
as a function of t.
1
I Using the result of Problem 13.44, we conclude that
n) 2
,(l
_,J(s+lK
s?
2
s + s +
V3 cos
— — + — —
(t
re)
sin
(t
re)
t
>
re
f
<
re
V3~
1
<«
- »)/2
^3 cos
— (f
re)
+ sin
— —
(t
n)
u(t
-
re)
V3
,4 -3s
13.55
Find
Sf
(s
+ 4)
5
as a function of r.
2
)
We have
Sf~ x
+ 4) 5 2
'
(s
-4r
1
Thus,
3Vrc
,4 -3s
+ 4)
4e 4( r _3)3/2 e
is
^-1
(s
4f3,2
= c- 4 *jy-M4
? >=
.5/2
~ e*Sf~
=
5/2
f
1
\(s
j
4(t
- 3)
-4„-3)
+ 4)
r<3
3/2 g- 4(, - 4)
3Vre
>3
3V^
5 2
-3)
=
.
INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS
D
317
PARTIAL-FRACTION DECOMPOSITIONS
13.56
Develop the method of partial-fraction decomposition.
f
Every function of the form v(s)/w(s), where v(s) and w(s) are polynomials in s, can be reduced to the sum
of other fractions such that the denominator of each new fraction is either a first-degree or a quadratic
polynomial raised to some power. The method requires only that (1) the degree of v(s) be less than the degree
of w(s) (if this is not the case, first perform long division, and consider the remainder term) and (2) w(s) be
factored into the product of distinct linear and quadratic polynomials raised to various powers.
The method is carried out as follows: To each factor of w(s) of the form (s — a) m assign a sum of m
,
fractions, of the form
^1
^2
s-a
(s- a) 2
2
To each factor of w(s) of the form
+ bs + c) p
(s
B lS + C x
s + bs + c
(i
= 1, 2,
.
.
.
,
i
- a) m
(s
assign a sum of p fractions, of the form
,
B2s + C2
2
where A h Bj, and Ck
m
...
i
(s
m;j, k =
2
B ps + C p
2
+ bs + c)
1, 2,
.
.
.
,
(s
2
+ bs + c) p
are constants which still must be determined.
p)
Set the original fraction t;(s)/w(s) equal to the sum of the new fractions just constructed.
Clear the resulting
equation of fractions, and then equate coefficients of like powers of s, thereby obtaining a set of simultaneous
Ck
linear equations in the unknown constants A t , Bj, and
13.57
fractions to decompose
Use partial
V
V
r
.
.
+ l)(s 2 + 1)
(s
I To the linear factor
we assign the fraction
the fraction
We then set
s+l,
2
(Bs + C)/(s + 1).
Finally, solve these equations for A h Bj, and C k
.
A
1
2
(S+1)(5
to the quadratic factor
A/(s +1);
+
S+l
+1)
s
2
+ 1,
we assign
Bs + C
S
2
U)
+l
,
.
Clearing fractions, we obtain
1
2
or
1
= s (A + B) + s(B + C) + (A + C)
A + B — 0,
Equating coefficients of like powers of s, we conclude that
B = — \,
A — \,
solution of this set of equations is
——
1
the partial-fractions decomposition,
(s
+
(2)
2
+ s(0) +
s (0)
= A(s 2 + 1) + (Bs + C)(s + 1)
l)(s
2
—=
+ 1)
C = j.
and
B + C — 0,
A + C — 1.
and
The
Substituting these values into (/), we obtain
—
-s/2 + 1/2
1/2
s+l
H
r
s
2
+
1
The following is an alternative procedure for finding the constants A, B, and C in (7): Since (2) must hold
Substituting this value into (2), we immediately find
s = — 1.
A = \. Equation (2) must also hold for s = 0. Substituting this value along with A = \ into (2), we
B = —\.
Finally, substituting any other value of s into (2), we find that
obtain C = \.
for all s, it must in particular hold for
13.58
Use partial fractions to decompose
(s
f To the quadratic factors
(Cs + D)l(s
2
+ 4s + 8).
s
2
+
2
and
1
s
2
+ 4s + 8,
We then set
--=
7-7-5
+ l)(s 2 + 4s 4- 8)
2
2
=
1
(As + B)(s + 4s + 8) + (Cs + D)(s + 1), or
3
s (0)
+ s 2 (0) + s(0) +
2
we assign the fractions (As + B)/(s 2 + 1)
As + B
Cs + D
—— = —52
1
(s
obtain
+ l)(s 2 + 4s + 8)
s
+
+ -=2
1
s
....
+ 4s + 8
and clear fractions to
= s 3 (A + C) + s 2 (4A + B + D) + s(SA + 4/3 + C) + (8B + D)
1
Equating coefficients of like powers of s, we obtain
A + C=
The solution of this set of equations is
1
(s
2
+ l)(s 2 + 4s + 8)
-4s/65 + 7/65
s
2
+
SA + 4B + C =
4A + B + D =
1
A = -33,
B = 55,
4s/65 + 9/65
s
2
+ 4s +
C = 33,
D = 33.
and
SB + D =
I
Therefore,
318
CHAPTER 13
D
+ 3
(s-2)(s+\y
s
13.59
Use partial fractions to decompose
I To the linear factors
s + 3
™,
u
s
-2
s+l,
B
and
A
We then set
we assign the fractions
A/(s - 2)
B/(s +
and
1),
respectively.
-=— -+
and, upon clearing fractions, obtain s + 3 = A(s + 1) + B(s — 2).
—rrr
(s-2)(s+ 1)
s-2 s+l
- 1
To find A and B, we use the alternative procedure suggested in Problem 13.57. Substituting s
and
then s = 2 into the last equation, we immediately obtain A = 5/3 and
B = - 2/3. Thus,
s
(s
13.60
'
+3
- 2)(s + 1)
=
5/3
2/3
s-2 ~ s~+T
Use partial fractions to decompose
s
2
+ 6s + 5
i
This fraction has the partial-fraction expansion
s
and J 2 Multiplying by (s + 5)(s + 1),
5
shows that d,'I =— 4,
and setting s= — 1
2>
-3/2
5/2
constants d
—5
s-5
s
2
2
.
l
+ 6s + 5
s
+ 5
+
for some
H
+ 6s + 5
+ 5)(s +1) s + 5 s +
we obtain s
5 = d,(s + 1) + </,(s + 5).
j
shows that d 2 = -f.
Therefore,
2(s
l
Setting
-
s+l
3s + 7
13.61
Use partial fractions to decompose
-=
s
3s + 7
We let
s
3s + 7 = A(s +
1)
3s + 7
(s
2s
3s + 7
-2s -3
2
1
A
+ B(s - 3).
Setting
s+l
yields
1
Multiplying by
B = — 1;
setting
— 3)(s +1),
(s
s
= 3
we obtain
yields
,4
= 4.
Therefore.
1
+ 3)(s + 1) ~ s-3 ~ 7+T'
2s
13.62
B
+
s- 3
s = -
(s- 3)(s+ 1)
4
-3
Use partial fractions to decompose
2s
2
ABC
-4
2s
Then setting
=2
s
Therefore,
B = -f,
yields
2
-2
s
- 4 - A(s
2
-4
(s + l)(s
2)(s - 3)
2s
+
s+l
(s+ lH.s-2)(.s-3)
i„,_2)(s-3)
f
,,
We let
2
2)(s
+
s
-3
Clearing fractions, we obtain
- 3) + B(s + l)(s - 3) + C(s + l)(.s
setting
=3
s
-1/6
-4/3
7/2
s+l
s-2
s-3
yields
C = ];
and setting
2)
.s
;= -
1
yields
A = -,;.
8
13.63
Use partial fractions to decompose
3
s {s
2
-s-2)'
To the factor s 3 =(s — 0)\ which is a linear
2
3
To the linear factors
polynomial raised to the third power, we assign the sum AJs + A 2 /s + A 3 /s
(s — 2)
and (s+l), we assign the fractions B'(s — 2) and C/(s + 1). Then
I Note that
s
—s—2
2
factors into
(s
— 2)(s +1).
.
A,
s^s
— s — 2)
2
s
8
A2
A
2
3
s
s
B
s
+
—2
C
s +
or, after fractions are cleared.
1
= A^is - 2)(s + 1) + A 2 s(s - 2)(s + 1) + A 3 {s - 2)(s + 1) + Bs 3 (s + 1) + Cs^s - 2)
= — 1, 2, and consecutively, we obtain C = f, 6 = 5, and ,4 3 = —4. Then choosing
and s = — 2 and simplifying, we obtain the equations A + A 2 = —
and 2/1, — v4 2 = —8,
—
which have the solution A —
and A 2 — 2. Note that any other two values for s (not — 1, 2, or 0)
3
will also do; the resulting equations may be different, but the solution will be identical. Finally, we have
Letting
s =
s
1
1
x
x
2
3/^2
s\s
2
3
+
4
+
13
2)
8/3
+
2
- 15s 11
(s + l)(s - 2)
5s
13.64
Use partial fractions to decompose
s+ 1"
2
3
'
INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS
f
- 15s——
—
—- = A +
-
5s
-
We write
(s
2
+ l)(s
5s
Setting
s
= -
2
2)
s
- 15s -
and
1
3
s
+
1 1
=2
C—
- 2) 2
B—
11
- +
- 2) 3
(s
1
(s
D
-\
s
_
319
Clearing fractions, we obtain
-2
= A(s - 2) 3 + B(s + 1) + C(s + l)(s - 2) + D(s + l)(s - 2) 2
in turn yields
= -|
/I
B= -1.
and
This procedure does not determine C and D. However, since we know A and B, we have
5s
(s
2
-15s -11
+ l)(s - 2) 3
-1
C
+ ; - ^r3 + - r^r +
(s
(s
2)
2)
-1/3
+
C
7
1
D
C=4
and
21
= 1,
s
1
from which we have
~D
T=-6 + 7 + C
- 15s- 11 _ -1/3
-7
4
=
—
- +
+
+
r
3
3r
~ s+
(s + l)(s - 2)
(s - 2)
(s - 2)
5s
D = i.
and
=
s
^
¥=-3 + 8 + 4-2
Then
r
2
1
Now, to find C and D we can substitute two values for s, say
11
D
;
s
Thus
2
1/3
2
1
13.65
3s +
Use partial fractions to decompose
#
1
w 2 +
- l)(s
1)
(s
3s +
,4
Bs + C
——2
=
+ —2 -. Clearing fractions, we obtain
(s
l)(s
+ 1) s
s +
= A(s 2 + 1) + (Bs + C)(s - 1). Setting s =
3s +
yields
A = 2, so that
3s + 1
Bs + C
_ 2
+
2
(s - i)(s
+ i) ~ 7^1 s 2 + r
„_
1
We write
1
1
1
1
To determine B and C, we let
r
C=l
D
B=—
^
and
1
s
=
and
3s+
tu
u
Thus
we have
->
2.
Use partial fractions to decompose
(s
»
„,
We let
,
s
^
71.
2
(s
+ 2s
s
Then
Bo = i
i
2
+ l
2
2A + B + 2C + D = 1,
fl"o
m which
+ 2B + 2C + 2D = 2, and 5B + 2D = 3.
s + 2s + 3
1/3
2/3
——.
-^2
= -=
+ -j
(s
+ 2s + 2)(s 2 + 2s + 5) s 2 + 2s + 2 s 2 + 2s + 5'
Thus
5,4
/4
= 0,
2
and
2
D„ = |,
and
-
4s
—
4s + 4s ,
2
.
2
+ 2s + 3 = (As + B)(s 2 + 2s + 5) + (Cs + D)(s 2 + 2s + 2)
= (A + Qs 3 + (2A + B + 2C + D)s 2 + (5/1 + 2B + 2C + 2D)s + 5B + 2D
Use partial fraction to decompose
We let
s
As + B
+ 2s + 3
Cs + D
^
"9
~
^
^ + -72
^7712
T, = 2
+ 2)(s + 2s + 5) s + 2s + 2 s + 2s + 5?>
s
13.67
1
r
1
s-1
from which
2
4 + C = 0,
„ = 0,
«
C
2
-2s +
2
=
,
+ 2s + 3
+ 2s + 2)(s 2 + 2s + 5)
s
13.66
!
z
(s-l)(s 2 +l)
- = 2H
and
C
7 R 4-
7
— 1 = —2 + C
then
2;
=
3
(2s
2
+ 3
+4s- 3'
+ 3)(2s - 1)
=
+ 3 = A{2s - 1) + B(2s + 3). Setting s = |,
-3/8
s + 3
7/8
=
Therefore, —25
+
2s 4s + 4s - 3
2s + 3
2s -
1
B = £;
we find
s
Clearing fractions then yields
.
1-
2s + 3
setting
s
= -f, we find
A = -§.
1
13.68
Find
as a function of x.
>
z
J^~' <
[(s+l)(s 2 +l)j
f
Using
5 the result of Problem 13.57 and noting that
l(s
+ l)(s 2 + 1))
2
U+lJ
2
+
|—
—-~s
+
\
-*-
s
1«
2
\
2 Vs
1
+lj
(
= — -I -=2
2
+
2
ls
\
s
J
1
1
+-
/
+ lj
/
(
1
-^
2
\
1,
we find that
2
2
2\s + \)
2
1
1
CHAPTER 13
320
13.69
Find
+3
- 2)(s + 1)
s
-
if
(s
as a function of x.
I No function of this form appears in Table 13.1. However, by the result of Problem 13.59,
s + 3
5 2x
2
= -if if"
= rr-r«
(s - 2)(s + 1)1
3
s-2
3
)s +
3
3
.
1
1
13.70
Find
if
8
-1
V -s-2)
s
as a function of x.
f No function of this form appears in Table 13.1. However, by the result of Problem 13.63,
if
3&- t-\ + 2&-
- i
-s-2)
2
1
[
4 hi*"' —
1
l
3
s (s
s
+-if _1
,
21
3
s+l
= -3 + 2x - 2x 2 + - e 2x + -e~ x
13.71
Find
3s + 7
if
as a function of t.
- 2s - 3
2
s
a = 3
I Using the result of Problem 13.61 and Table 13.1, entry 7, first with
3s + 7
if" 1
(5
13.72
Find
= 4e M -e'.
S+l
2
as a function of t.
+ 6s + 5
,13
Using the result of Problem 13.60 and Table 13.1, entry 7, we have
s-5
y-X
s
2
5
,
+ 6s + 5(
= - <£~
Find
if"
x
2
2s
13.73
s-3
s-5
if
s
f
y-l
= 4if _1
- 3)(s + 1)
— 1,
a =
and then with
>
<
|s
+ 5|
if
-1
I
e
2
|s 4
~2
2
e
2
1
as a function of t.
{(s+ l)( 5 -2)(s-3)
I
Using the result of Problem 13.62 and Table 13.1, we determine
se~
x
2s
13.74
Find
2
= if"
1/6
-4/3
7/2
—2
s — 3
4
1
7
1
s+l
&
as a function of x.
s
I
Cfi~\
(s
I
-4
(s+l)(s-2)(s-3)
2
+ l)(s 2 + 4s + 8)
From Problem 13.58 we have
1
cp-\
tf
1
(s
2
-6*3 + 65
2
+ l)(s + 4s + 8)
s
2
+
4
1
^
<f
s
1
^s + ftT
2
s + 1
The first term on the right can be evaluated easily if we note that
5
g-;
2
+ -965
+ 4s + 8
4
65 s
s
2
+
1
+
65 s
1
2
+
1
To evaluate the second term, we must first complete the square in the denominator, writing
s
2
+ 4s + 8 = (s + 2) 2 + (2) 2
;
—
—s + g^
^s
then we note that
2
s
+ 4s + 8
4
s + 2
1
2
" 65 (s + 2) 2 + (2) 2 + T30(s + 2) 2 + (2) 2
'
Therefore,
^(s
2
+ l)(s 2 + 4s + 8)|
)s
1
J
130
=
2
65
4
+
4
|s
2
+
l
65
l(s
+ 2) 2 + (2) 2
2
}(s
+ 2) 2 + (2) 2
4
7
sinxH
cosx-l
65
1
-»J_ L_
65^
65
_,
e
65
cos 2x + —-- e
130
_,
.
„
sin 2x
we have
J
J
=
INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS
13.75
Find
f
It
2
_,i (5s
<-
if
- 15s- 11")
—
\{s +
—=-}
l)(s-2) 3 J
as a function of t.
follows from Problem 13.64 and Table 13.1, entries 7 and 14, that
(s
+ l)(s - 2) 3 j
\s+\
(s
- 2) 3
- 2) 3
(s
- -^-'--rV + 4fe 2 + -e 2
'
2
3
13.76
Find
——2
_1 <-
if
—>
5
+ l)j
l(s-l)(s
f
It follows from
Problem 13.65 and Table 13.1, entries 7 to 9, that
t
Find
if
_1
f
—+ ++-— +—+ + —\
s
{—.2
[(s
'
3
as a function of t.
= 2e' — 2 cos + sin
13.77
-2
s
2
2s
3
)
as a function of t.
=
2s
2)(s
2
t
2s
5) J
f Using the result of Problem 13.66 and Table 13.1, entry 15, we have
2
+ 2s + 2)(s 2 + 2s + 5) J
(s
|s
2
+ 2s + 2
= t^ _1 <:
3
s
2
+ 2s + 5
^ A + -&'
[(s
+ l) 2 +
1
= ^e" sin + |(i)e _t sin 2f = §
r
Find
if
_ l
\
—
2
+ 4) J
f
—+
=
s(s
x
13.79
Find
2
4)
= \ - 7 cos lx
A = \4 &' \-\ - \4 &~ 1-2^-rl
+ 4j
4
4
'
<?
(sin t
+ sin It)
+ 4) J
l/4)s
(
.
1
s
s
2
Thus,
4
l
1
*
[s(s
2
—— +
= 1/4
1
Using the method of partial fractions, we obtain
&- \
+ l) 2 + 4
as a function of x.
I
-.
|s(s
\{s
-
r
13.78
1
3
J
[s]
+
>
+ 4s - 3J
[s
2
-
S
5£~ x \
[4s
,
2
as a function of t.
I Using the result of Problem 13.67, we have
jy-J
[4s
*
2
+ 3
l
+ 4s - 3 J
=jy -i[-3/8
[2s + 3
16
13.80
Find
f s
7J**
»
^-r
4
(s
2
+ 2S - 4
>
+ 2s 35 + s 2? J
7/8
,
2s -
js + 3/2j
Jl
!
[2 s
1
+
-3/8
1
+ 3/2
2 s - 1/2 J
7/8
|
+
16^
Is -1/2
16
")
as a function of r.
I Using a partial-fraction decomposition, we obtain
s
s
4
2
+ 2s - 4
+ 2s 3 + s 2
D
+ 2s - 4 /I B
C
= — + -5-.+
r +
2
2
2
s (s + l)
s
s
s +
(s + l)
2
3
2
3
2
2
_ A(s + 2s + s) + fl(s + 2s + 1) + C(s + s + fls
2
2
s (s + l)
3
2
_ s (A + C) + s {2A + B + C + D) + s(A + 2B)+ 1(B)
2
2
s {s + l)
s
2
2
'
1
'
)
16
D
321
CHAPTER 13
322
Clearing fractions, we obtain
s
+ 2s - 4 = s 3 (A + C) + s 2 {2A + B + C + D) + s(A + IB) + 1(B).
2
Equating
coefficients of like powers of s then yields
A + C=
2A + B + C + D = l
A = 10,
The solution to this set of equations is
2
+ 2s-4
4
s
+ 2s 3 + s
s
shows that
1
-4
2
if
s
-1
2
4
s
—1
C = — 10,
B = -4,
——
1
B= -4
A + 2B = 2
D = — 5,
and
so that
1
s+\
Taking the inverse Laplace transform of both sides then
(s+\) 2
+ 2s - 4
= 10 — 4f — 10«- _, -5te-'.
+ 2s 3 + s 2
12
13.81
Find
if"
1
as a function of f.
+ 20)(s 2 + 4)
(s
12 = A(s
have
2
— +C ^
A
12
Bs
Clearing
H
^w + 4)77 — s + ^
r6 fractions, we
(s + 20)(s
20
s + 4
in this equation yields
Setting s=-20
,4= T§T setting s = 0,
Using partial-fraction decomposition, we write
+ 4) + (Bs + C)(s + 20).
;
.
?
2
>
2
r
.
;
B = — Tor- Thus,
12
3/101
(-3/101)5 + 60/101
„
f1
p „
=
From the Fproperty
H
T^rr^2
J it follows that
F
J of linearity,
2
4
(s + 20)(s
+ 4) s + 20
s +
C = ^,
yields
s=l
and setting
—
yields
.
:
->
•
(s
2
+ 2)(s + 4)j
101
[s
+ 20
—
101
3
5+
if -l
Find
5
3
y-i J
30
-,
2
)s
+4
+
30
if
-1
,
101
]s
2
+4
-,
cos 2t h
sin 2f
101
101
101
13.82
As
3
if" 1
1
as a function of r.
+S
I The denominator may be factored into s 3 + s = s(s 2 +1). By the method of partial fractions, we then obtain
s +
-s + 1
=-+ 2
Therefore, using the linearity property, we have
3
s + 1
s + s
1
1
s
s
3
JSf"
+s
s
2
y-1
+ i^
+
2
s
=
+
1
— cos + sin
f
t.
1
1
13.83
Find
J?'
1
as a function of f.
(s- l)(s + 2)(s 2 + 1)
Using the partial-fraction expansion
d 2 = — T5,
^ 3 = —to,
dt
s-1
d 4 — ~yq.
and
1
if"
1
(s- l)(s + 2)(s 2 + 1)
1
(s- l)(s + 2)(s 2 + 1)1
d 3 s + dx
d2
+
s
+ 2
s
2
,
+
= 4^-1
1
s-\\
6
15
s
6
15
if
f(x) = e
Wlf
+ 2
3
1
.
sin t
t
10
10
and
g(x) = e
CONVOLUTIONS
Find the convolution
I
If
f(x) = e
3x
f{x) * g(x)
and
g(x)
= e2\
then
3x
= e 3v
f(v)
.
/(x) * </(x) = fif(v)g(x - v)dv =
= e2x {*e dv = e2x
Find
/(x) * gf(x)
if
/(x)
=x
and
fl<x)
= x2
.
2x
.
g(x
- v) = e 2{x ~ v\
J*
eV^dv = £ e ^" **
and
3
\
v=
13.85
d
10
s
x
Therefore.
— cos
13.84
we find
\
2
2
+
1
J
INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS
I
Here
= v
f(v)
#(x - v) = (x - v) 2 = x 2 - 2xv + v 2
and
Thus,
.
2
2
2
2
f{x) * g(x) = f* v(x - 2xv + v ) dv = x f * vdv -2x f * u di; + f
,
= x2
x
2
„
x
x
—
+—=—x
1
2
13.86
Prove
V
du
4
3
2x
323
4
3
.
4
12
f(x)*g(x) = g(x)*f(x).
I Making the substitution
x
= x — v,
we have
f(x) * g(x) = j*fiv)gix -v)dv =
f°
f(x - x)gix)i-dx) = -£° g(x)f{x - x)dx
= j* * db)f(x - t) dx = #(x) * /(x)
13.87
Determine
#(x) */(x)
for the functions defined in
Problem 13.84, and then use the result to verify that
convolutions are commutative.
I With f(x-v) = e i{x
' v)
g (x) * /( X ) =
and
J**
g{v)
ff („)/(
we have
,
x _ y)^ =
r e 2v e Mx ~
v)
dv = e
ix
P e"" dv = e 3x l -e~
v
J
which, from Problem 13.84, equals
13.88
= e 2v
/(x) * g(x).
Prove that
f(x) * [g{x) + h(x)] = f{x) * g(x) + f(x) * h{x).
I
f{x) * [g(x) + /i(x)] =
£
/(t;)[>(x
- y) + h{x -
i>)]
dv =
£ [/(%(x -
o)
+ /(t#(x - u)] du
f(v)g(x -v)dv + j* /(u)li(x -v)dv = fix) * g(x) + /(x) * /i(x)
=
J"*
13.89
Find
i"
-1
1
<—-=2
—>
as a function of x by convolution.
+ 4)
s(s
_1_ — =_1_1
We first note that -^
A
2
2
s s + 4
sis
+4)
Table 13.1, f(x) =
and gix) = 5sin2x.
Then, defining
.
It
1
*-
'
s(s
F(s)
= 1/s
and
Gis)
= l/(s 2 + 4),
we have, from
now follows from Problem 13.86 that
1 = ^-»{f(s)C(s)} = g(x),f(x) = f"9(ii)/(x - v)dv
J°
+ 4)J
= J**(isin2t;)(l)di; = £(l -cos2x)
Observe that in this problem it is easier to evaluate
——^f
than
gix) * fix)
See also Problem 13.78.
fix) * gix).
1
13.90
Find
I
If
S£
x
\
(s
-If
we define
F(s)
as a function of x by convolution
= Gis) = l/(s - 1),
^"Mr-TTjll =
13.91
Find
#
if
_1
Defining
{l/s
2
13.92
Find
if
_1
F(s)
<
[(s
/(x) = gix) = e
x
and
^~ {Fis)Gis)} =f(x) * gix) = £/(%(* -v)dv = £ e e x ~ dv = e x §*il)dv = xe*
v
1
v
as a function of x by convolution.
}
^ _1 fel =/(*>
then
= Gis) = 1/s,
* «(*)
we have from Table 13.1 that
= Jo /TO** -
- l)(s - 2)
>
^
=
(1)(1)
Jo*
^"x
as a function of x by convolution.
fix) — gix) = 1.
It
now follows that
*
CHAPTER 13
324
I
Defining
g(x)
=e
2x
= l/(s - 1)
F(s)
G(s) = l/(s
and
- 2), we have from Table 13.1, entry 7, that
f(x) = e
x
and
Then
.
* _1
{
- 1K5 - 2) 1
(5
= /W * ^ (X) =
~^ dv = Jo
\l m9{x
et' e2<A:
" ,,,
* = «*"
*"" df
Jo'
= e 2x(l - e~ x = e 2x - e x
)
13.93
Find
if" 1
as a function of x by convolution.
s(s+ 1)
I
Defining
(x)
= 2/s
F(s)
and
= if- {l/(s+ \) = e~
1
x
G(s)
= l/(s + 1),
/(x) = if
we have
_1
{2/s}
= 2i^ {l/s} = 2, and
1
Then
.
^ _1 I^TT)} = /(x) * 9(x) = 5o fiv)g{x ~ v dv = jo ^~ ~ ^ = 2e
lx
0)
~x
)
d«
Jo
= 2e" x (^ - 1) = 2 - 2e~ x
13.94
Find
S£
'
)
1
We can write
(s
2
+a
2
s
)
2
+a
2
s
2
Then since
+a
+ a2
— v)
cos
\
2
at' dv
J°
a
— - sin at
a
= cos at
+ a2
and
2
—
/•»
cos af
a
/
Jo
——
sin 2at;
au
2
- cos 2af
sin 2af\
f
(cos at)
4a
4a
a
sin at cos at\
t
- +
sin
1
z
af
t
sin af
(cos at)
2a
\2
a
1
1
at;
- +
\2
= -(sinaf)
(cos ai:)(sin af cos at — cos at sin at) dv
cos at f sin av cos at) di;
J°
(\ + cos2at;
t
a
—
a
Jo\
= - (sin at)
&~
2
a J°
)
= - sin at
rt
n
i
1
dv — -
2
Find
- i
s
sin a{t
^"Mr5
5T7f = P (cos aw)
2
2
J°
!(s + a
f
13.95
5£
we have
-,
s
2 2
sin at
s
if
as a function of f by convolution.
+ a2 2
2
(s
la
a
2a
1
x
2
We have
as a function of f by convolution.
+ 1)'
s (s
<£- x
\^\ = t
1
and
JST
te~'.
(s
^"
+ 1)'
WW} £ ^"^
=
'
Then
" V)dV = J"« (yf " PV "*
= [(t>f - 2 )(-e-") - (t - 2v)(e~ v + (-2)(-e-)]
= te~' + 2e'' + t-2
t;
)
1
13.96
Find
as a function of t using convolution.
<£
2
s (s
Since
if
_1
2
+ 3)
{l/s
2
}
=
f
and
¥~
l
<£
5
g(t)
=
—
a(f) */(f)
sin V3t,
2
+ 3j
1
i -4-
[s
73
2
— =—
and the required inverse Laplace transform is
= sin y/3t,
\
+ 3j
it
follows that
f(t)
=
/(f) * g(t).
In this case, it is easier to evaluate
(see Problem 13.86), so we have
(f
t,
^3
l
- v)[ --cos>/3i;
v =
f'o
(-l)(- -cosj3vjdv
.
INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS
325
SOLUTIONS USING LAPLACE TRANSFORMS
13.97
Solve
- 5y = 0;
y'
y(0)
= 2.
^{y'\ - 5if {y} = if {0}.
f Taking the Laplace transforms of both sides of this differential equation, we get
With
MO) = 2
[sY(s) - 2] - 5Y(s) = 0,
we have
&(y)=Y(s),
and
from which
Y(s)
=5
—
2
—5
Then, taking the inverse Laplace transform of Y(s), we obtain
y(x)
13.98
= Se~ {Y(s)} = <fl
Solve
- 5y = e 5x
y'
x
1^1 = 2if-» j^-4 = 2?
= 0.
y(0)
;
5 *.
¥{y'\ — 5 if {>'} = i^{e
f Taking the Laplace transforms of both sides of this differential equation, we find that
[sy(s)-0] - 5Y(s) =
so that
s
—5
,
Y(s) =
from which
(s
— 5)
Then, taking the inverse Laplace transform of Y(s), we obtain
5x
},
5-.
y(x)
= ¥~ {Y(s)\ = if _1
^> = xe
l
-J
5x
,
where we have used Table 13.1, entry 14.
13.99
Solve
/ - 5y = e 5x
= 2.
y(0)
;
Taking the Laplace transforms of both sides of this differential equation yields
[sY(s) — 2]
— 5 Y(s) =
-,
s
or
Y{s)
2
=
s
13.100
Solve
y'
1
+
^5 + (s~^5)
Taking inverse Laplace transforms, we then obtain
—j.
2
-5y = 0;
y(n) = 2.
I Taking the Laplace transforms of both sides of the differential equation, we obtain
Then, with
c
= y(0)
kept arbitrary, we have
[sY(s) — c ] — 5Y(s) =
or
Y(s)
—
^{y'} — 5if {y} = if {0}.
———
Laplace transforms, we find that
v(x)
— ¥
x
5x
> = c e
~ 5J
The result is c = 2e 5 ", so
{Y{s)}
— c Z£
'<
I
Now we use the initial condition to solve for c
Solve
I
y'
+ y = xe~ x
.
Taking inverse
-.
s
13.101
—5
5
.
s
y(x) — 2e
S(x
*\
.
Since no initial condition is given, we set
= c,
y(0)
where c denotes an arbitrary constant.
Taking the
Laplace transforms of both sides of this differential equation (using Table 13.1, entry 14), we obtain
1
[sY(s) — cl + Y{s) =
5-,
(s
+ 1)
from which
Y(s)
=-
c
s 4-
—
+
1
- + 1
(s
r.
Taking inverse Laplace transforms, we
iy
then have
The constant c can be determined only if an initial condition is prescribed.
13.102
Solve
y + y = sin x;
y(0)
=
of both sides of this differential equation, we obtain
transforn
f Taking the Laplace transforms
This yields
[sY{s) - 0] + Y{s) =
¥{y'} + ¥{y] = ^{sin x).
1
-^—-,
s + r
from which
2
Y{s) =
(S
1
+i)(s 2 + i)'
o both sides and using the result of Problem 13.68, we then obtain
Taking the inverse Laplace transforms of
^>- y "< rw >- y " i +
13.103
Solve
>-'
+ y = sin x;
y(0)
1
-e
i)!,'
= 1.
+
i)
}
2
*
1
2
1
.
cos x + - sin x.
2
326
CHAPTER 13
D
f Taking the Laplace transforms of both sides of the differential equation, we obtain
if {/} + if {y} = J^{sin x},
—
Y(s)
—4 |W +
(s
_2
t
Solving for 7(s), we find
.
1
1
1X
1)
<
l)(s
^2 +
_
—
1
i
1
[sY(s) - 1] + 7(s) =
which yields
+s~
+ 1
Taking the inverse Laplace transforms and using the result of Problem 13.68,
.
we then obtain
= J?- {Y(s)}
[
y(x)
/I
=
1
~ e
\
7L
2
\2
13.104
dN/dt = fcJV;
Solve
f
N(0) = 250,
fc
—\-
x
--Z£-
\
-1 + jsri'
(s+l)(s 2 + l)J
1
\
.
cos x + - sm .x + e
2
J
_x
(s+1
3
1
2
2
—-e x
cos x + - sin x
2
constant.
¥{N(t)} — n(s),
Taking the Laplace transforms of both sides of this differential equation and denoting
—
^S[dt i = Sf{kN} = k¥{N},
we get
from which
sn{s)
- 250 = kn(s),
so that
n(s)
=
s
)
N(t) =
Then, taking inverse Laplace transforms yields
^~ {n(s)} = <£~ \
x
x
>
—k
= 250 if"
.
1
<
I
= 250e*'.
(Compare with Problem 6.1.)
13.105
dP/dt = 0.05P;
Solve
P(0) = 2000.
I Taking the Laplace transforms of both sides of this differential equation and denoting if (P(f)} = p(s),
{dP)
2000
from which sp(s) - 2000 = 0.05p(s), so that p(s) =
} = y[0.05P} = 0.05i^{P},
—
dt
s
\
{-)Q00
~)
>
s- 0.05 J
f
1
= 2000^^ <
1
>
[s- 0.05 J
— 0.05
.
= 2000e 005
'.
(Compare with Problem 6.41.)
13.106
dQ/dt + 0.2Q = I;
Solve
0(0) = 0.
¥{Q\ — </(s),
I Taking the Laplace transforms of both sides of this differential equation and denoting
y
,UQ)
)
7( + °- 2<Sf ^} = -J
.,
.
'!
so that
!
I
[,v ( /(.s)
- 0] + 0.2^(s) =
_
1
This yields
.
g(s)
we have
1
=
Then, taking inverse Laplace transforms and using partial fractions to decompose the fraction on the right,
we obtain
Q = <£V
1
l
" = <e- \js(s +
1 = <f~
l
{q(s)}
XHV
5
'
|
[s
0.2)J
——
+ -^-1 = 5if " {-1 - S&- {
1 = 5 - 5^"° 2
5 + 0.2]
}sj
js + 0.2j
1
'
-
'
(Compare with Problem 6.80.)
13.107
Solve
dl/dt + 50/ = 5;
1(0)
= 0.
I Taking the Laplace transforms of both sides of this differential equation and denoting
obtain
&< — [ + 50¥{I} = 5i?{/},
'
y
so that
[si(s)
L
-0] + 50i(s) = 5(-),
or
/(s)
x
^{1} = i(s),
we
=
\sj
s(s + 50)
Then, taking inverse Laplace transforms and using partial fractions to decompose the fraction on the right, we
[dt]
'
'
have
/
jjP-i
= &- •{/(,-)}
{
lw; =
[s(s
L_l = if-' {— + ^-U = O.lif
{s
+ 50) J
s
+ 50j
" '
{-} - O.lif \s\
=
{—+^1
50j
o.i
- o.ie- 50
[s
(Compare with Problem 6.104.)
13.108
Solve
dT/dt + kT = 100/t;
7(0) - 50,
fc
constant.
f Taking the Laplace transforms of both sides of this differential equation and denoting
{it!
— + kSe{T} =
j>
100A-^{1},
so that
'1\
[-W-SOJ + hW-WO*!;!
«__ +
50
or
lOOit
£f{T) = t(s),
we
'
INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS
327
Then, taking inverse Laplace transforms, and using partial fractions, we obtain
s
= S£
X
\~P\* vl
+k
s(s
+ k))
= - 50jSf " 1
+k
[s
+100 27
"1
-
{-^-fcj
s
+ k
s
{-}= -50e "+ 100
(Compare with Problem 6.55.)
13.109
dQ/dt + 0.040 = 3.2e-° 04 ';
Solve
f
Q(0) = 0.
Taking the Laplace transforms of both sides of this differential equation and denoting
have
y\-*j=-l + 0.04y{Q} = 3.22>{e~ 004 '},
[sq{s)
¥(Q) = q(s),
we
^{1} — i(s),
we get
so that
- 0] + 0.04q(s) = 3.2
or
s
= 3.2
q{s)
v
^
+ 0.04
(s
+ 0.04) 2
Taking inverse Laplace transforms (see Table 13.1, entry 14), we then obtain
Q = 3.2^-
»
:
{
13.110
5-V
+ 0.04) 2
\(s
'
- 3.2te ~° 04
dl/dt + 20/ = 6 sin 2f;
Solve
(Compare with Problem 6.89.)
'.
= 6.
7(0)
I Taking the Laplace transforms of both sides of this differential equation and denoting
<£
|
— + 20i?
I
J
/ }
= 6^{sin It},
[si(s)
so that
2
- 6] + 20i(s) = 6 2,
or
A
+4
|
s
i(s)
+ ^ onw —
20
+
+^
+
12
6
=
t
s
(s
20)(s
2
4)
Then, taking inverse Laplace transforms and using the result of Problem 13.81, we obtain
f8f^
2
»'-
+ Jfi,-sin2l
Tit cos2(
(Compare with Problem 6.105.)
13.111
Solve
y" + 4y = 0;
y(0)
- 2,
= 2.
y'(0)
&{y"} + 4i?{y} = if{0},
I Taking Laplace transforms yields
2
Y(s)
[s
L
- 2s - 2]J + 4 Y(s) =
or
Y(s)
so that
= -*2
s
—+ = —+ + +
7
"2
4
s
7
2
~i2
4
s
7
4
Then, taking inverse Laplace transforms yields
= £~ {Y(s)} = 2^~
l
yix)
13.112
Solve
?' + 9y = 0;
y(0)
= 3;
l
\-^- -1 + tf-U-j- -l = 2cos2x + sin2x
y'(0)=-5.
f Taking the Laplace transforms of both sides of this differential equation, we obtain
Se{y") + 9&{y} = JSf{0},
so that
[s
2
- s(3) - (-5)] + 9Y(s) = 0,
Y(s)
or
=
Y(s)
^^. Taking
inverse Laplace transforms then gives
y =
13.113
Solve
5
^-J^Lil = ^-i{3 _i
*
[
s2
+ 9j
[
y" - 3y' + 4y = 0:
y(0)
s
2
+9
= 1,
2
Y(s)
5
3 s
y'(0)
I Taking Laplace transforms, we obtain
[s
^_l = ^-i{^^l_^-J_^_^l = 3cos3x- sin3x
2
3
+ 9]
!/ +
3
2
J
\s
3
2
+
2
3 J
= 5.
JSf{y"}
- s - 5] - 3[s Y(s) -
1
- 3^{y'j + 4j£?{y} = i^{0},
]
+ 4 Y(s) =
or
F(s)
from which
= -^
2
s - 3s + 4
3
—
328
CHAPTER 13
Then, taking inverse Laplace transforms and using the result of Problem 13.40, we obtain
Fj
y(x)
13.114
Fj
= e°' 2)x cos ^— x + yjl e i3l2)x sin ^- x.
-2y= 4x
y" - y'
Solve
= 1,
2
y(0)
;
= 4.
y'{0)
&{y") — &{y'\ — 2&{y] = 4if{x 2 },
I Taking Laplace transforms, we have
- s - 4] - [sY(s) - 1] - 2Y(s) =
2
[s Y{s)
or
Y(s)
so that
+3
+
— s — 2= s 3 (s 2 — s — 2)
s
=
s
2
Then, taking the inverse Laplace transform and using the results of Problems 13.69 and 13.70, we obtain
2_ — x
x
/^„2jc
e~
>'(*
13.115
y" + Ay' + By = sin x;
Solve
+ (-3 + 2x - 2x 2 + \e 2x + ie~ x = 2e 2x + 2e~ x - 2x 2 + 2x
)
)
= 1,
y(0)
S^{y") + 4if{y'} + 8if{y} = if {sin x},
f Taking Laplace transforms, we obtain
[s
2
Y(s)
= 0.
y'(0)
1
- s - 0] + 4[sY(s) - 1] + SY(s) =
s
2
or
+
Y(s)
+4
+ 4s + 8
s
=
s
1
so that
2
1
(s
2
2
+ l)(s +4s + 8)
Then, taking the inverse Laplace transform and using the results of Problems 13.39 and 13.74, we obtain
y(x) = (e~
2x
" 2x
+ (-^cosx + ^sinx + ^e~ 2 *cos2x + yyoe"
= e~ 2 *(H cos 2x + }§£ sin 2x) + ^ sin x - ^ cos x
13.116
Solve
cos2x + e
y + y-2y = sint;
y(0)
sin 2x)
= 0.
y(0)
2jc
sin2x)
= 0.
I Taking the Laplace transforms of both sides of this differential equation, we obtain
¥{y} + <£{y) - 2&{y) = J§f{sin t},
from which we write
[s
2
Y(s)
- s(0) - (0)] + [sY(s) - 0] - 2Y(s) =
s
Solving for Y(s) then yields
equation gives
13.117
Solve
= - 1.
1
Taking the inverse Laplace transform of both sides of this
(s- l)(s + 2)(s 2 + 1)
!
'
y(0)
t;
+
1
= %e' — ^e' 2 —
y{t)
y- y =
=
Y(s)
2
,
cos t — ,' sin t
(see
Problem 13.83).
= 1.
y(0)
I Taking the Laplace transforms of both sides of this differential equation, we obtain if {y} — Z£{y\ = if{r},
Solving for Y(s) gives (s 2 - l)Y(s) = -s+ 1+ 1/s 2
from which we write [s 2 Y(s) - s(-l) - 1] - Y(s) = 1/s 2
.
from which we eventually find that
Y(s)
= -=2
s
s
s
d 3 — j,
and
d4 =
— f,
2
13.118
Solve
3
+
= — + -^ +
d,
1
(s- l)(s+ 1)
2
1
x + 4x + 4x = 0;
s-\
d*
+
+
s
yields
d
x
= 0,
d2 =
2
2
1
and it follows that
- s3 +
2
s (s- l)(s + 1)[
s
1
(s- l)(s + 1)
s
The partial-fraction decomposition
- s3 +
2
x(0)
]
,
1
(
Is
= 10,
x(0)
)
3
1
2
I
1
,
1
(
2
)s
+
1
= 0.
f Taking the Laplace transforms of both sides of this differential equation, we obtain
if {x} + 4j^{x} + 4if {x} = if {0},
2
[s X(s)
from which
- s(10) - 0] + 4[sX(s) - 10] + 4X(s) -
10s + 40
or
X(s)
s
Then
10s + 40
x= <£'
(s
- 10e"
2'
+ 2) 2
+ 20fe"
(Compare with Problem 11.55.)
Y*-i^m--™\M
2'
= 10c"
2,
(l
+ 2r)
2
+ 4s + 4
+ 20if -1
(s
+ 2)'
— 1,
,
—
.
INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS
13.119
Solve
x + 4x + 4x = 0;
= 2,
x(0)
329
x(0)=-2.
I Taking the Laplace transforms of both sides of this differential equation, we obtain
if {x} + 4<£{x\ + ASe{x) = ^{0}, from which
2
- s(2) - - 2)] + 4[sA (s) - 2] + 4X(s) =
X(s) =
r
[s X(s)
or
(
s
+ 6] =
J25
x = ££'*{
Then
—,}
+ 2) 2
\(s
f2(s
S£
+ 4s + 4
+ 2) + 2]
+ 2)
(s
I
1
-?— } + 2JSr
1
.
{/
+ 2)
Is
2
+ 2) 2
{(s
(Compare with Problem 11.21.)
™ d/
^ 2/
13.120
Solve
JT2 + 20— + 2007 = 0;
-
dr
7(0)
= 0,
/'(0)
= 24.
<fr
# Taking the Laplace transforms of both sides of the equation,
which yields
-.
= -^
i(s)
—
[s
— s(0) — 24] + 20[si'(s) — 0] + 200i'(s) = 0,
2
i{s)
24
Completing the square of the denominator and taking inverse Laplace
rrpr.
s'
transforms then yield
;/
1 v
{(s
+ 20s + 100) + (200 - 100) J
2
5
+ 10) 2 + 10 2 J
\{s
5
(Compare with Problem 11.112.)
13.121
Solve
Y"
+Y =
7(0) = -2.
7(0) = 1,
t;
I Taking the Laplace transforms of both sides of the differential equation, we have £P{Y") + i?{7} = <¥{t},
from which s 2 y{s) — s + 2 + v(s) = 1/s 2
Thsn partial-fraction decomposition gives
.
1
= a,
2 ,2
2
>< S )
S (s
and
7 = if
i
[s
13.122
Solve
2
^s2
s
"I
5
+
s
1
7" - 3T + 27 = 4e 2
I We have
[s y(s)
2
3
s
1
Mt +
2
2
4
(s
- 3s +
Y^^-U ——7- + ^—
+
—2
Is
13.123
Solve
2
s
7" + 1Y + 57 = e _t sin t;
/ We have
[ S y(s)
1
-,
,
-.
2)
— 2y
^
= -7e' + 4e 2
7(0) = 0,
- s(0) - 1] + 2[sy(s) - 0] + 5>(s) =
T—
^= -7+
3s + 2
s
+ 2s + 5
(s
7 = &~ {y(s)} = i<T'(sin t + sin 2r)
l
1
2
4
s
4
* +
-2
(s
- 2) 2
'
7(0) = 1
=
+ /)2 + t
s
so that
2
+ ^+2
Then
-
2
+ 2s + 2)(s + 2s + 5)
(see Problem 13.77).
(s
2
2
+ 2s + 3
+ 2s + 2)(s 2 + 2s + 5)
s
1
1
s
-7
14 -3s
(g
and
Then partial-fraction decomposition yields
J^{7"} + 2if{7'} + 5if{7} = JS?{<?-'sinf},
2
1
from which
^ + "I
„w
2
2)(s s -
*
/
(s
+
4
+ 3s — 5] — 3[sy(s) + 3] + 2y(s) =
= 7-i2
S
1
2
7(0) = 5.
-3,
<e{Y'\ - 3^{Y'} + 2&{Y) '= 4<£{e '},
y( s )
3
s
1
1
7(0) =
';
-2
= f + cosr - 3sinf.
>
+
s
+ 2
+ TT—
7 = 2
2
+ ^
S + 1
S
S +
1)
.
330
13.124
CHAPTER 13
D
^
+8
+ 25Q = 150;
^-f
at
at
Solve
Q(0) = 0,
&\-M\ + %& \-f-\- + 25^{Q} = 150^{1},
Taking Laplace transforms yields
2
[s q(s)
= 0.
Q'(0)
- s(0) - 0] + S[sq(s) - 0] + 25q(s) = 150/s. Then
150
=
<?(*)
s(s
2
+ 8s + 25)
6s
s
s
2
6(s
s
(s
+ 4)
+ 4) 2 + 9
24
(s
+ 4) 2 + 9
= 6 - 6e _4 'cos 3r - 8e _4 'sin 3r.
<2
d2Q
+ X
dQ
= 50 sin 3r;
-yf- + 8 -£ + 25Q
Solve
+ 4) + 24
2
+9
(s + 4)
6
6(s
s
13.125
+ 48
+ 8s + 25
6
6
and
so that
Q(0) = 0,
I
Q'(0)
= 0.
—
150
Taking Laplace transforms yields
75
150
<?(*)
= _22
,
(s
Thus
+ 8s + 25)c/(s) = -^
(s
75
1
so that
-,
75
s
^ =^jT7-«T7
1
o2
2
n+^
,_
,i2
2
n
2
26 s + 9
52 s + 9
26 (s + 4) + 9
+ nw-2
9)(s + 8s + 25)
4
Q = §£ sin3r-|fcos3f + f|e~ 4 'sin3r + |f<?~ 'cos 3r
"
= ff (2 sin 3f — 3 cos 3f) + f|e 4 3 cos 3r + 2 sin 3f)
.
.
,
75
+ 52
(s
s + 4
+ 4) 2 + 9
'(
(Compare with Problem 11.130.)
J2y
13.126
Solve
2
> a2
X(0) = X
2
=-
at
co
Jy
— + 2a — + w X =
0;
..
V,S)
=
-2
2
*.._
)
..2
2
2as + co
.
4-
.
a,
,
and co are all constants and
Rework the previous problem for
In this case,
co
2
V + zX
+ x)X
+
3
-.2
2
2
2
(s +
(s + a)
+ co - i
+ a2 - a
„t2
,
.
y.)
K°
»
(5 + 1
co
2
< a2
+g
° g ~" sin % /ca 2 - g 2
Jco 2 — a 2
°
f
= a2
X = &~ '{x(s)} = JST i^- +
Rework Problem 13.126 for
so that
(s
= ,.
f
I
g *°
n
I = X Q e~
+ (F + a* )f<
+ ar J
"*"
(S
.
In this case.
*=*-{*} -*-N r
=X
cosh v a
2
+a
(
+
:2
a)
— co 2 f +
°
(s
(Compare with Problem 11.23 for
Solve
y"
- 3/ + 2y = e _I
y(l)
;
i = a'2m
= 0,
y'(l)
and
^°
- (a 2 - co 2
°
r sinh
y/a
13.129
,
X = jS? _i {x(s)} = Af g-" cos Jto 2 - g 2 +
and
f
where X Q V
,
— X s — V + 2a[sx(s) — X ] + co 2 x(s) = 0.
2
s x(s)
sX + (VQ + 2zX
s
13.128
V
.
I Taking Laplace transforms yields
13.127
X'(0) =
,
at
2
+w
co
F° +a *°
+
)
(s
+ if - (a 2 - co 2
Va 2 — u> 2
)
1.
2
— -Jk/m.)
= 0.
- 3jS?{/} + 2if{y} = ^{e~ x }, or
Here c and c, must remain arbitrary, since they
[s 7(s) - sc - cj - 3[s7(s) - c ] + 2Y(s) = 1 (s + 1).
I Taking Laplace transforms, we have
JS?(y"}
2
represent y(0) and y'(0). respectively, which are unknown. Thus.
y s = c o -52
(
)
s-3
1
1
+
+ c Ti - 3s
^TT^,
^V^
+2
-3s
+2
i
'
s
2
+ l)(s - 3s + 2)
2
(s
INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS
Using the method of partial fractions and noting that
s-1
x
s
- 2j
s
- 3s + 2 = (s -
2
s — 2
j
- \e x + \e 2x
(s
l)(.v
- 2),
|.s
1
+
331
we obtain
s
1
-
s-2
1
-*
= c (2e - e *) + c,(-e* + e + (Je
= (2c - c, - £)e* + (-c + c, + i)e 2 * + \e~ x = d e x + d e 2x + \e *
2x
2
)
)
x
where
d
= 2c - c, - |
^ = -c + c, +
and
5.
Applying the initial conditions to the last displayed equation, we find that
13.130
= —%e*~ + y
2
hence,
y(x)
Solve
y" - 2y' + y = f(x);
f
[s
2x ~ 3
= 0,
y(0)
2
7(s)
y'(0)
d x = \e~
3
;
= 0.
Solve
l
Jz?
y + y=f(x); y(0) =
{l/(s
or
— l) 2 } = xe*.
F(s)
=
y(s)
'
(s-1) 2
Taking the inverse transform of Y(s) and using
x
y(x) = xe * f(x) = \* te'f{x — t) dt.
= 0,
y'(0)
0,
/(x) =
if
X
T
I We note that
2
Y(s)
/(x) = 2u(x — 1).
^j.
x > 1
(2
Taking Laplace transforms, we obtain
- (0)s - 0] + Y(s) = Se{f(x)} = 2S£{u{x - 1)} =
—
,
so that
Y(s)
s
*~
'
it
follows that
y(x) =
fc?TT,} =
^~ \e'
1
s
Solve
^+
>
l
s(s^
I
13.132
and
.
- (0)s - 0] - 2[sY{s) - 0] + 7(s) = F(s)
convolution, we conclude that
[s
= -\e~ 2
In this equation /(x) is unspecified. Taking Laplace transforms and designating S^{f(x)} by F(s), we obtain
From Table 13.1, entry 14,
13.131
+ \e~
d
x
2
<i x
-yy + 16x = - 16 + 16u(f - 3);
™~
'
S
-
= ?" -^—.
+ 1)
s
Since
<?(s
™~
H>}
'
= 2 " 2 cos
*
= [2 - 2 cos (x - l)]u(x - 1).
1) I
x(0)
= x'(0) = 0.
~d?
I Taking the Laplace transforms of both sides of this differential equation, we obtain
2
[s X(s)
- s(0) - 0] + 16*(s) = - 16 - + 16 —
s
or
X(s) =
—+ —
—.z
s(s
s
- (e~
3s
- 1)
16)
Applying the method of partial fractions to the fraction on the right side and taking inverse Laplace
transforms, we get
= u(t - 3) -
13.133
Solve
y"
1
- u(t - 3) cos 4(f - 3) + cos 4f = - + cos 4t + u(t - 3)[1 - cos 4(? - 3)]
1
f We first note that
f(t)
<
°
- 3/ -4y = f(t) = i
<
;
^ (0) = °'
>' (0)
=a
J
= e' - u(t - 2)e' = e' - u(t - 2)e'~ 2 e 2
Then, taking the Laplace transforms of both
.
sides of the differential equation, we obtain
[s
ftr
2
Y(s)
- s(0) - 0] - 3[sY(s) - 0] - 4Y(s) =
v(s)
{
=
'
where we have used the fact that
(s
s
2
- l)(s + l)(s - 4)
e
2
e~ 2
- 3s - 4 = (s + l)(s - 4).
(s
—— -
e
2
j—- e
- 1)(5 + l)(s - 4)
2s
332
CHAPTER 13
D
-1/6
Now partial fractions yields
(s - l)(s + l)(s - 4)
1/10
H
s-1
1/15
+ 1
s
so it follows that
-,
H
s
-4
-^ + Ae~« + iV4 - e - 2)[-ie<" + ^-"- + ^4 ""
= ~¥ + fa'* + &" +
- 2)|>' - iV 4 *"' + Ts^ ^ 4
y =
2
'
2
«(r
2»
6
u(t
13.134
7" + 97 = cos 2t;
Solve
#
7(0) =
Since 7'(0) is not known, let
= c.
7'(0)
,
2
Then
s
- s(l) - c + 9y(s) =
s y(s)
s
=
v(s)
+c
+ 2
s + 9
(s
+ 9)(s 2 + 4)
s
2
4
c
1
5
3
5
To determine c, we note that
7 = f cos 3t + f sin 3r + § cos 2f.
Y" + a 2 Y = F(t);
Solve
f We have
y(s)
s
2
s
+
+9
5(s
2
2
s-2
=
s
+9
2
+ a2
s
2
f(s)
Thus,
.
s
^H
+9
5(s
2
s
+ 4)
5(s
2
+ 9)
2
— = — c/3 — 1/5
so that
1
+ a2
Find the general solution of
2
s y(s)
= 12/5. Then
2
from which
s y(s)
- s + 2 + a 2 ><s) = /(s),
and so
and
2
+a
2 sin at
= cos at
2
sin at
1-
F(t) *
r<
ft
i
1
1-
a
7(0) = c,
c
7'(0)=-2.
z
2 sin at
— cos at
Let
or
1
Then, using the convolution theorem, we find that
'
s
= 1,
7(0)
= —
- s7(0) - 7'(0) + a 2 y(s) = f(s),
v(s)
+
7(7r/2)
s
f
s
A
+4
+ 4)
A*)
13.136
+ -,2
,
2
Y = - cos 3f + - sin 3r + - cos 2r.
and
13.135
+
+ 9/
c
s
2
5Vs
]
']
- 1.
Y(n/2) =
1,
2)
F(v) sin a(t — v) <fo
a J°
7"-a 2 7 = F(f).
= c2
7'(0)
— sc — c 2 — a 2 y(s) = f(s),
l
Then, taking Laplace transforms, we find
.
so that
sc r
+c
2
—a
y{s)
s
2
2
As)
+
s
7 = c, cosh at + — sinh af + -
2
— a2
j
'
Thus,
F(u) sinh a(f
— y) dt;
a J°
a
= A cosh at + B sinh af + - r F(u) sinh a(t — v) dv
1
>
a J°
which is the required general solution.
13.137
Solve
y'"
+ y' = e x
:
y(0)
= y'(0) = y"(0) = 0.
I Taking Laplace transforms, we obtain
[s^is) - (0)s
2
- (0)s - 0] + [sY(s) - 0]
s-
or
7(5)
1
=
1
3
(s - l)(s
+ s)
Then, using the method of partial fractions, we obtain
y(x)
+
s
13.138
Solve
7'"
- 37" + 37' - 7 = tV;
I We have
3
1/2
1
= ^ _1
7
s —
+
1
7(0) = 1,
s
2
11
- 1/2
(l/2)s
1
+ 1
7'(0)
= 0,
2
2
1
.
sin x
2
7"(0) - -2.
^{ 7'"} - 3j^{ 7"} + 3J&P { 7'} - J^{ 7} = if {tV},
[s y(s)
+ - e x + - cos x
from which
- s 2 7(0) - s7'(0) - 7"(0)] - 3[s 2 y(s) - s7(0) - 7'(0)] + 3[sy(s) - 7(0)] - y{s) =
(s
- l)
2
INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS
Thus
(s
3
- 3s 2 + 3s - \)y(s) - s 2 + 3s - 1 =
—-j
— ~l)
333
2
,
a
(s
and
- 3s +
2
2
_ s 2 - 2s + 1 - s
=
+
+
"
3
3
l)
(s
l)
(s^T7
(s
(s^lF
2
2
(s
l)
2
(s
1)
=
+
vzz
7TT-:
rrr +
2
5-1r-:(S-1)
(S-1) 6
(S-1) 6
(S-1) 3
(S-1) 3
2
s
* S) "
1
"
1
1
1
1
:
Y = e* — te'
so that
1
.
60
2
13.139
Find the general solution of the differential equation in the previous problem.
f For the general solution, the initial conditions are arbitrary.
y"(0) = C,
If
Y(0) = A,
we let
Y'(0)
= B,
and
then the Laplace transform of the differential equation becomes
3
[s y{s)
- As 2 -Bs-C]- 3[s 2 v(s) - As - B] + 3[sy(s) - A] - ><s) =
or
(s-1)
As 2 + (B- 3A)s + 3A-3B + C
2
+
„,3
(s-l) (
(s-1)
=
>< s )
2
,
Since A, B, and C are arbitrary, so also is the polynomial in the numerator of the first term on the right.
can thus write
—
=
y(s)
(s
2
C
2
— 1) T
t^
t
-^- H
H
—
I
s
—
— 1)°K
(s
1
,
and invert to find the general solution
€*
Y = -^— e' + c 2 te' + c 3 e' + ——
where the c k are arbitrary constants. (The general solution is easier to find
,
60
2
— 1)
(s
We
than the particular solution, since we avoid the necessity of determining the constants in the partial-fraction
expansion.)
13.140
The differential equation governing the deflection Y(x) of a horizontal beam of length
d*Y
W
-r-r =
ax
—-,
<x<
/;
y"(0) = 0,
Y{0) = 0,
Y(l)
=
Y"(l) = 0.
/
is
known to be
W E, and
Find Y(x) if
,
/
denote positive
EI
constants.
I Taking the Laplace transforms of both sides of the differential equation, we have
s*y(s)
W
—
- s 3 y(0) - s 2 y'(0) - sy"(0) - r"(0) =
Letting the unknown conditions
.
y'(0)
=c
and
l
Els
y'"(0)
= c2
c
gives
y(s) ='-£
W^-,
c
+ -r
-\
and taking inverse Laplace transforms yields
Els
s
s
c
x3
WqX 4
W
——
= W 3 /24EI
and
-
2
nx) = cix+— + --=c lX+
From the last two given conditions, we find that
solution is
13.141
Solve
Y(x) =
£x —
24 EI
= ^,
^1
EI
ax*
3
(/
2/x
0<x</;
3
c,
l
c2x
+ x 4 = —±- x(l - x)(/ 2 + Ix - x 2
)
xa
3
+
c2
= - W l/2EI.
Thus, the required
).
24 EI
7(0) = 0,
7(0) = 0,
HO =
0,
r"(0 = 0.
# So as to apply Laplace transforms, we extend the definition of W(x) as follows:
w{x) =
(WQ
< x < //2
= wMx) _ u(x _
m
Now taking Laplace transforms yields
2
3
s y(s) - s Y(0) - s Y'(0) - sY"(0) - Y'"(0)
4
Letting the unknown conditions
r'(0)
= c,
and
r"(0) = c 2
,
W
1
- e~ sl12
s~
£7
we have
y(s)
c
c
s
s*
= -±3 + -| +
W
—
^
E/s
5
(1
- c" s,/2
).
334
CHAPTER 13
D
Inverting then yields
^\c x
YM
We now use the conditions
W x* W {x-lj2f (
3
2
=
Y"(l)
and
I
Q
=
Y'"(l)
to find
= W 2 /8EI
cx
and
1
WJ/2EI.
c2
Thus,
the required solution is
W
2
^/
16EI
\2E1
y (x) = ^oL x 2
z'
W0)= 1,
;
+ 4y =
I
wn
3,+ —^0 X 4
A
24EI\
24EI
X -2) U X
2
\
= -1.
Z(0)
# We denote &{y(x)} and y{z(x)} by Y(s) and Z(s), respectively.
Then, taking the Laplace transforms of
both differential equations, we obtain
s
1
0T(s)-l] + Z(s) = ^
sY(s) + Z(s)
2
+
1
=
s
s
4T(s) + sZ(s) = -
[iZ(i) + 1] + 47(s) =
1
The solution to this last set of simultaneous linear equations is
s
+ 5+1
2
Y(s)=
s
— 4)
s(s^
3
+ 4s 2 + 4
2
z
— 4)
s (5
Z(s)=
Using the method of partial fractions to solve each of these equations separately, we obtain
=^
y(x)
l
—— +r —-2— + —+— = —17,3.
+ -e + -e~
7/8
1/4
'/'-i
- 2>
=
{Y(s)}
l
y
-r = e*
v
1
I
We denote S?{w(x)},
'J
u-
l
7/4
f 1
,
jjr
4
7
3/4
- v
s-2
+ 2(
S
e
8
2,
8
3
£
H
4
4
= sin x
w' + y
;
= JSP
{Z(s)j
>
2\
s
2x
2x
I
-
s
s
.-,vl-
3/8
+ y =
\\\ x)},
w(0) = ().
;
i(0)
= 1.
z(0)
=
1.
1
and
by W(s),
S?{z(x)]
and Z(.s), respectively. Then, taking the Laplace
Y{s),
transforms of all three differential equations, we obtain
I
[sW(s) - 0] + Y(s) =
s
[s y( S)-i]-z(s)
sW{s) + VIM
+
s
1
!
=
[sZ(s)- 1] + W(s) + Y(s) =
2
S-
Z(s) =
sY{s)-
or
1
1
W(s) + Y(s) + sZ(s) =
s
The solution to this last system of simultaneous linear equations is
-1
W(s)
Y(s)
Z(s) =
(s-l)(s 2 + l)
s(s-l)
s
Using the method of partial fractions and solving each equation separately,
-
W{x) = 2'- 1 {w{s)} =se- x \[s
y(x)
z(x)
= Se~
x
{ Y{s)}
= if
= if- {Z(s)} = i?~
1
—
" !
\
\y
+z + y
z
+ y =0
3^0)
= 0.
v'(0)
- +S +
2
s
1
1
s
Dive
—
s
= 0,
2
+
1
COS X
1
z(0)
=1
2
+
1
2
+
1
s
ss+
1
1
—
)
INVERSE LAPLACE TRANSFORMS AND THEIR USE IN SOLVING DIFFERENTIAL EQUATIONS
335
f Taking the Laplace transforms of both differential equations, we obtain
[s
2
- (0)s - (0)] + Z(s) + Y(s) =
Y(s)
[sZ(s) - 1]
(s
+ \)Y(s) + Z(.v) =
2
+ [s Y(s) - 0] =
Y(s)
+ Z(s) = s
Solving this last system, we find that
Y(s)
—
=
and
j
=-+—
Z(.s)
s
y(x)
=
-W and W = + Wz
{z"
i
= cos x
— z — sin x
-I-
yv'
.
y
f
Then, taking inverse transforms yields
.
s
s
z(0)=-l,
;
z'(0)=-l,
y(0)=l,
=
v'(0)
Taking the Laplace transforms of both differential equations, we obtain
2
[s Z{s)
+ s + 1] + [sF(i-) - 1] = XTT
2
s +
2
-
s Z(s)
+ sY(s) =
s
1
2
+
1
or
[s
2
- 5 - 0] - Z(s) = -J—
T(5)
+
s^
+
s
+
s
Solving this last system, we find that
yields
z(x)
= — cosx — sinx
w" —
and
,
and
= -=2
Y(s)
s
1
1
s
Then, taking inverse transforms
•
«
.
4
= cosx.
v(x)
= 3e *
+ 2z
=0
-2w'+2y'+ z
2w'
2y + z' + 2z" =
t't
+
s
I
-=
-—
2
Z(s) =
1
- Z(s) + s 2 Y(s) = '
1
v
= 1,
vv(0)
;
w'(0)
= 1,
= 2,
y(0)
= 2,
z(0)
z'(0)=-2.
I Taking the Laplace transforms of all three differential equations yields
[s
2
W(s) - s - 1] - Y(s) + 2Z(s) =
-2[sW(s) - 1] + 2[sY(s) - 2] + Z(s) =
2[sW(s) - 1] - 2Y(s) + [sZ(s) - 2] + 2[s Z(s) - 2s + 2] =
2
or
s
2
W(s)-
2Z(s) =
Y(s)+
s
2
—
4.
2s + 4
s
+
1
Z(s) = 2
-2sW{s) + 2sY(s) +
2sW(s) - 2Y(s) + (2s 2 + s)Z{s) = 4s
1
W(s) =
The solution to this system is
s
w(x) = e
x
—
-,
Y(s)
(s
1
y{x) = Se-'\
[s
dx dt = 2X ~ 3Y
''
13.147
Solve
\
lyy/d* =
x{0) = 8
.
r-2x
=
—
-
-
—2s
—
1 )(s
+
s
+
1
-,
s
1
+ - —l = ex + e-*
1
2
Z(s) =
,
and
,
z(x)
+
1
= 2e~ x
J
Y(0) = 3.
,
Sf {X} = x
f Taking Laplace transforms, we have, with
sx-8 = 2x-3y
sy — 3 = y — 2x
and
(s
J^{T}=>',
3y = 8
- 2)x +
or
2x + (s — 1 )y = 3
Solving this last system, we obtain
5
x =
s
which yield
X = 5e"' + 3e 4
fv"' j-
13.148
Solve
Y' 4- 3Y = 15<?~'
*
'
V
\Z„
7" - AX + 3Y = 15sin2f
t
•
<>
>
3
+
1
T = 5e
and
'
+
and
-4
s
'
s
- 2e 4
*<°> = 35
'
2
5
y =
+1
s
-
-4
'.
*' (0) =
" 48
'
y(0) = 27
'
HO) =-55.
so that
336
D
CHAPTER 13
f Taking Laplace transforms yields, in the notation of the preceding problem,
s
s
2
15
x - 5(35) - (-48) + sy - 27 + 3x =
30
y - s(27) - (-55) - 4(sx - 35) + 3y =
2
s
or
(s
2
2
+4
15
+ 3)x + sy = 35s - 21 +
+
s
-4sx + (s 2 + 3)y = 27s - 195 + -j2
1
U)
30
+4
s
Solving system (/). we then obtain
2
+ 300s - 63
15(s + 3)
+
2
2
+ l)(s + 9)
(s + l)(s
+ l)(s 2 + 9)
35s 3 - 48s
(s
2
2
45
30s
s
2
+
27s
s
1
- 55s
3
and
(s
2
2
s
s
+
s
'
s +
2
+4
60s
+
+ l)(s + l)(s + 9)
(s
2
2
+
s
1
2
+ 4
Thus
and
Y= y
Solve
2
+ 3)
+ l)(s + 4)(s 2 + 9)
30(s
2
2
2
3
1
+
(s
60
+ 9
1
+ l)(s + 9)
2
+ l)(s 2 + 4)(s 2 + 9)
X = ¥- {x} = 30cos/ - 15sin3f + 3e~' + 2cos2f
l
5/,
13.149
s +
+ 3s 585
7
2
30s
2
+9
2
'
2
2s
3
+
30s
(s
'{>'}
+ 2 -i+10/ =
--r
J/
dt
= 30cos3f -60 sin - 3?
f
'
+ sin li
2
<
:
/,(0)=/ 2 (0) = 0.
—- + 20I, + 15/ =
2
ilt
f Taking the Laplace transforms of both equations, we find
-5», - [si,
- /,(0)J + 2[s/ 2 - / 2 (0)] + 10/ 2 =
-/i(0)) +20/,
[sit
From the first equation,
Then inverting gives
1
= 2i 2
i,
=
2
\
—e
+ 15/ 2 =
(s
55
or
(s
55
'
2
and
/,
+ 20)i, + 15/ 2 =
= 2/ 2 = 2 — 2e
55
—
s
s
so that the second equation yields
,
+ 5)i, -(2s+ 10)i 2 =
55r/2
i
2
=
55
1
s(2s + 55)
s
2s + 55
CHAPTER 14
Matrix Methods
FINDING e At
14.1
Develop a method for calculating e A when A is a square matrix having numbers as its elements.
'
I
If
A is a matrix having n rows and n columns, then
,Ar
-
= a n _ A"- t"^ + a n _ 2 A"" 2 f"- 2 + •• + a 2 A 2 2 + o^Af + a I
1
1
where a
,
a l5
.
a„_ t are functions of t which must be determined for each A.
,
.
.
of At, then
e
= a,,-^"" + a„_ 2 A n ~ 2 +
Now if A, is an eigenvalue
+ a 2 A 2 + a,/. + a
Furthermore, if A, is an eigenvalue of multiplicity k, for k > 1, then the following
To determine the a's, we define
Xi
(/)
f
1
— r(A,).
r(X)
1
•
•
.
equations are also valid:
1
Ai
£
=
dX
k
r (A)
~ l
A = A,
When such a set of equations is found for each eigenvalue of At, the result is a set of n linear equations, all
a„. These values may then be substituted
containing e Xi on the left side, which may be solved for a
into (7) to compute e
14.2
Find e At for
Here
ot.
l
,
.
,
.
.
.
A=
'!•
14
I
,
At
Af =
-9j
with characteristic equation
A
2
.:]
-9t_,
-14f
+ 9rA + 14r 2 = 0.
-2r and — It. Since A has order 2x2, it follows from Problem 14.1 that
where a and a satisfy the equations
Then r(/.) = otjA + a
,
e
At
The eigenvalues are
= a,Af + a I.
t
e- 2
<r
Solving this set of equations, we obtain
vl
'
7'
= r(-2f) = a,(-2f) + a
= r(-7t) = a (-7f) + a
1
=
and
x
a
=
7e
-2«
2e"
Then
5t
e~
c
2t
— e- 1
r
'
le
1
f
~
_-14t
St
-9t_
-14e~
1
11
2'
- 2e~ lx
+ 14e~ 7
'
e~ 2
2e-
'
2'
-
e
+ 7e~
7r
7'
1"
14.3
de A for A =
'
Here
_-64
-20
t
Ar =
,
•64f
with characteristic equation
A
+ 20fA 4- 64r = 0.
— 4r and — I6t. Since A has order 2x2, it follows from Problem 14.1 that
where a, and a satisfy the equations
r{X) = a A + a
t
The eigenvalues are
-20f
e
At
= a, Af + a
I.
Then
,
e
-4
'
,- 16r
= r(-4t) = a (-4t) + a
= r( — 6t) = aj( — 16f) + a
1
1
337
"
338
-
CHAPTER 14
D
a: =
Solving this set of equations, we obtain
e
-4t_ e -l6t
and
=
a
Then
12*
4e" 4
t
— 64r
12*
14.4
Find e M
I
'
2)
20f
In that problem, we found e
A '.
f
'
s)
-e" 16
for the matrix
we replace
16e-
_
_ 64e
4 "- 2
A '.
I
Here,
1
2
4
3
A=
« = 2,
e
AI
+ 64e
— a,Af + a
+ 14^-
+ a
a,f
=
I
and
/.
2
— 5t,
'
Solving these equations, we find that
+ 16e -16(r-2)
— s,
we obtain
t
g-2(»-s)
7 «-s)
values and simplification yield
2a,f
and
3a, f + a
for
'
I
Since
_ g -7(r-s)
r(A)
= a,/. + a
a,
e
Ax
e ' = «i(5t)+ *o
= — (? - <?"')
and
5'
satisfy the equations
= - (? 5 + 5e~').
a
2e
5'
- 2e~''
4e
5'
+ 2e
5'
'
n = 2.
it
follows that
The eigenvalues of Af are
e
A
/.,
= 1l
'
"'
x,f
= a,Ar + a I =
and
e
Solving these equations for a, and ot
2
— — 4t,
r(/.)
we find that
substituting these values and simplifying yield
f
e
A
'
At
.
If
%l
= — (e — e~*')
2t
and
a
= - (2e 2 + e~ At
=
4e
2'
+ 2e>- 4
'
e
Se '-8e~*'
2e
2
we replace f with the quantity
4^2(1-1) _|_ 2^-4(1-1)
Se ni-D_ 8e
Find e A{
f
'
s)
In that problem we found e
At
.
If
„A(f-s)
Find e A for
'
2,
-e-*'
2'
+ 4^- 4
-«,-i)
e
2e
f
— 1,
2(I-l)
~ 1)
2i,
we obtain
4(1
+ 4e- 41! :.,]
for the matrix of Problem 14.7.
e
14.10
'
3
,A(r- 1)
14.9
f
A ==
16
we replace t with the quantity
_
"6
4 e 2(r
g e 2<'
- s)
- *>
.
which are both of multiplicity one. Thus, we have
n for the matrix of the previous problem.
In that problem we found e
= a,/. 4- a
e" 4' = a,( - 4f) + a
= «!(2t) + a
2t
,
/.
and
-2a
6f
Find e A "
Then substitution of these
A
8«,f
14.8
The eigenvalues of Af are
.
6
+ 4e~'
5
4e '-4e'
'2e
= -
"
5
= a,(-f) + a
(
A
'
_2e _2(,_s) + 7e~ 7(,-s)
6/
Find c
-4e" 4 + 16e
'
-4(,-2)
which are both of multiplicity one. Thus, a, and a
e
14.7
~|
_ e -16(l-2)
-4(f-2)
_ 4e
_ 2g-7(i-j)
2 "~ s|
4x,f
/
e
- 16(,-2)
we replace t with the quantity
If
- 14e~
for
+ 64<T 16
'
161
e~*'
— 2, we obtain
t
16( '- 2)
>-4e-
-4('-2,
7g-2(r-«)
'
'
A in Problem 14.2.
In that problem, we found e
Find e A
-64e" 4
with the quantity
t
„A(f-s)
14.6
ib
I6e~*' -4e~
'
12
If
12
Find e M
'
for the matrix of the previous problem.
,A(t-2)
14.5
+
_j_
2^ - 4(1
_ 8e " 4<
'
t
- s)
g2(«-*)
s>
_ s)
2(
2e
'
— s,
_ g-
we obtain
4(f -
+ 4e - 4( -•]
'
).
Then
MATRIX METHODS
on.
At =
Here
— lor
2x2,
has order
a t and a
it
with characteristic equation
,
follows from Problem 14.1 that
+ 16r = 0.
= a At + a
Then
I.
t
The eigenvalues are ±i4t. Since A
—
r{X)
ct
t
X+a
where
,
satisfy the equations
e''
4'
= r (/4t) = a (i4t) + a
4
= r( - /4t) = a - /4t) + a
'
,
(
e
1
and
sin 4t
1
0"
"1
t
1
ut
+ e~ i4
'
cos 4t.
ar
cos4f
£sin4t
-4fsin4t
cos 4t
=
+ cos 4f
-16f
At
Then
1
A=
Find e At for
''
=
a,
= — sin 4f
A,
e
e~
,
Solving this set of equations, we obtain
14.11
At
e
,
2
k
339
96
I
t
At =
Here
with characteristic equation
,
+ 96f 2 = 0.
2
k
The eigenvalues are ±i*j96t.
96f
Since A has order
where a, and a
2x2,
follows from Problem 14.1 that
it
l
^
b<
= r(iV96r) = a {i^96t) + a
e~
x
Solving this set of equations, we obtain
=
ax
e
iv96l
2
i
,
= cos yj%t.
<x
x
At + a I.
Then
r{k)
=z k+a
x
,
_ g/
l
^
= r(~i^96t) = a,(-iV96t) + a
b'
is 961
J
=
__
/
J%t
__ sin v96t
and
^96t
Then
f
sin J96t
cos J96t
0"
1
sin y/96t
(cos V96t)
96
96f
96f
Find e Al for
=
-i v "96f
an =
14.12
At
satisfy the equations
e
^i\ 96r
e
1
V% sin ^96r
cos J96t
I
A =
64
f
t
Ar =
Here
k 2 + 64r
with characteristic equation
,
-64t
Thus,
e
A
'
= a Af + a
t
and
I,
e
r(k)
= a^k + a
where a, and a
,
= r(iSt) = a,(i8f) + a
iSt
i8f
ax =
Solving this set of equations, we get
e*"*
e~''
„-i8r
e
—
—
tl
= — sin 8r
At
14.13
Find eA' for
The eigenvalues are ± /8f.
satisfy the equations
j
and
=
a
g
£8*
+ e -i8f
01
[1
+ cos 8t
= cos 8t.
Then
2
8r
64f
8i
= 0.
= r(-i8t) = a^-iSt) + a
= —sin St
;16f
e
8'
2
cos 8f
g-
sin 8t
3=
L°
x
-8 sin 8f
\
cos8f
1
A =
1
I
Here
Aj
=
it
n — 2;
and
A2 =
hence,
-it,
e
Ar
=
a
cc
x
Kit
At + a I =
and
/•(/.)
= aj/ + a
.
The eigenvalues of At are
which are both of multiplicity one. Thus,
e" = a^/'t) + a
Solving these equations for ol 1 and a
,
we find that
e"" = a,(-if) + a
o^
=
—
1
..
(e"
lit
Substituting these values above, we obtain
e
A
'
=
_'..
— e ") =
sin f
t
cos f
sin t
sint
cost
1
and
a
= - (e" + e ") = cos
t.
340
14.14
CHAPTER 14
D
Find e M,
#
~ n)
for the matrix of the previous problem.
In that problem, we found e
14.15
'~ s)
Find e A(
#
If
.
we replace t with the quantity
t
— n,
t
— s, we obtain
we obtain
sin (r- 7i)"
cos (f- 7i)
— sin(f —
Al
cos(f — n)
7t)
for the matrix of Problem 14.13.
At
In that problem we found e
If we replace t with the quantity
cos (t — s) sin (tt-s)l
.
,A(i - s)
— s)
sin(f
14.16
Find e A
A=
for
'
cos((f-s)J'
-255
-8j
I
t
Here
Af
-4f + /3r.
)} + 8r/. + 25r
with characteristic equation
,
-25*
-8f
2x2,
Since A has order
follows that
it
e
Kt
=
a.
x
\t + a I.
2
= 0.
Then
The eigenvalues are
r(A)
= a,/., + a
,
where ct
x
and a„ satisfy the equations
= r(-4t +
41 + i'3l
= a,(-4f + i3t) + a„
i"3r)
-*'-«3«
e
_ r(-4f _ ;3 = a ,(-4r - /3f) + a
f)
The solution to this set of equations is
a, =
e
-4i+.-3«_
e
-4.-.-3«
-4i^.-3i_
e
-.-3f^
e
"4
'sin3r
3t
j'6f
«o -
e
(-4f - Bt)e-*'
3f
+ ,i '
- (-4t + i3t)e
- i6t
-i3t
4-1
=e 4
=e
-i6r
'
- sin 3f 4- cos 3f
\3
Then
e
A(
_4
'sin3f
\
3t
14.17
Find e
At
A -
for
I
-25*
-8, J
-64
-12.8
where a, and
1
i sin 3f
2
The eigenvalues are
)} + 12.8U + 64f
with characteristic equation
— 12.8r
-64f
— 6.4f ± i4.8r.
+ cos 3*
-"sin 3f
* sin 3t
= e
t
Af =
Here
1
- sin 3( + cos 3r
(
Since A has order
2x2,
it
follows that
e
Al
+ 3t,Af + a
= 0.
Then
I.
r(/.)
= a, A + a
,
satisfy the equations
ac
e
-6M+i4.st = a]( _6.4f
g-6*'-«+-8i =
+ i4.it) + a
ai( _6.4f
- ,4.8f) + a
This solution to this system is
e
-6.4l + «4.8«
_ ^-6.4f-i4.8i
e
- 6 4'
=
S m4.%t
4.8f
;9.6?
a
-
(-6.4t-i4.8t)e- 6 *'
+ '+- 8t
'"' 4 8t
-(-6.4f + i4.8f e" 6 4
'
4
= e " 6AAf
-
'
-»9.6*
e~
64 sin 4.8*
r
'
f e
and
64*
4.8f
4
= e 6.4r
14.18
Find e A
'
64
I
1
_64
(
-sin 4.8* + cos4.8r
-12.8*
+ cos 4.8*
- #sin4.8r
4
1
*-
24 sin 4.8*
# sin 4.8* + cos 4.8*
-16
t
Here
At
,
-64*
eigenvalue
.
1
A=
for
sin 4.8*
/4
JO
- sin 4.8f
+ cos 4.8*
V3
with characteristic equation
'/}
+ 16*/ + 64f 2 = 0.
There is only one
-16t_
— 8r, of multiplicity two. Since A has order
2x2,
we have
e
At
= o^A* + a
I.
Then
-
MATRIX METHODS
= a,/. + a
r{X)
and
r'(X)
= a,,
e
The solution to this system is
e
where a, and a
-»'
a,
e
=e
8
Find ? A(
3)
'
8'
- r'(-8f) - a,
so that
',
+8f
1
t
+ (1 + 8f)*'
-16f
81
f
- 64f
!]
1
- 8f
for the matrix A of the previous problem.
In the previous problem, we found e Kt
f
= (1 + 8f)e
a
t
6At
14.19
and
= e -s>
*>
satisfy the equations
= r(-St) = a,(-8f) + a
8'
341
e
A(r -3)
If
we replace t with the quantity
-3
— 8(f — 3)
+ 8(r - 3)
1
_ e -8(, -3)
.
f
-64(r-3)
1
t
— 3,
we obtain
- 23
-64? +192
8f
8(f-3)
t
- 3
-8f + 25
1"
A=
d e Kt for
-4
-4
Af =
Here
I
Its
characteristic equation is
2x2,
eigenvalue of multiplicity two. Since A has order
=x a+a
and
,
r(X)
1
r\X)
—
<x
{
The solution to this system is
where a, and a
,
2t
e~
a,
=e
2i
and
e
Kt
which has
— a,Ar + a
I.
/.
= — It
as an
Then
e~
= (1 + 2t)e
a
= r'(-2t) = a,
2'
so that
,
(1
1
'
fe
+ 2t)e -2f
2
-4^"
(l-2f)e~ 2
'
'
1
Find f A " +1) for the matrix A given in the previous problem.
I
In the previous problem, we found e
Al
If
.
we replace t with the quantity
2,, + ,)
+ 1)>
+
[1 +2(f
rn
2( +1)
A(t + l)e-
,*«+»
'
Find <? A
'
for
#
A=
At =
Here
r
[-9n
6
o
n
|_-9r
(t+ l)e- 2(,+ 1)
-2(f + l)>-
[1
and
r'(x)
= a,,
Its characteristic equation is
where a, and a
The solution to this system is
e
f
e~
Kt
<x
= e3
x
3t
(1
=
X
_
— 6tX + 9t 2 = 0,
2
2x2,
we have
= r (3t) = a,(3t) + a
= e3
and
'
t
e
= (1 — 3t)e il
a
+ d - 3t)e 3
'
bt
4f-4
-2t
+
X = 3f
which has
as an
e
A
'
— a,At + a
I.
Then
r(X) == <XjA
+
oc
.
3'
= r'(3t) = a,
Then
"(1
1
- 3t)e il
te
3'
'
-9te
1
3'
(l+3f)f 3
'
A in the previous problem.
for the matrix
In the previous problem, we found e
Kt
f
satisfy the equations
-9f
~ Kt
+ 3
2f
2(1+ 1)
2 " +1)
6f
e
Find e
+ 1, we obtain
t
,
eigenvalue of multiplicity two. Since A has order
14.23
+ AtX + 4t 2 = 0,
satisfy the equations
+ (1 +2t)e~ 2
-At
At
14.22
2
we have
= r(-2t) = a,(-2f) + a
t
14.21
X
-4r_
At
Al
.
If we
replace t with
— we obtain
t,
+3f)e
9 te
-3«
(l-3t)e
0"
1
14.24
Find e
At
for
A
-2
-5
1
2
f
Here
At
-2f
-5f
f
2r
and ±it. Since A has order
,
3x3,
which has as its characteristic equation
we have
e
Kt
= a 2 A 2 r + a,Af + a
I.
A
3
+ Xt 2 — 0.
Then
/•(/)
Its eigenvalues
= a 2 A 2 + x,/. + a
are
.
CHAPTER 14
342
where x 2
,
ct
u and x
satisfy the equations
e° = r(0) - x 2 (0)
2
e" = r(it) = x 2 (/f)
+ x,(0) + x
2
+ otiO't) + a
-» = r( — if) = a (-/f) 2 + x,(-if) + a
e
2
+e
g»'«
x2 =
The solution to this set of equations is
-
2
-cost
1
-=
^
—
e" -e'"
=
a,
,
2
2
It
1
.
= - sin
lit
f
t,
ana
a
=
1.
t
Then
1
-5"
-2
— cos
t
0"
-1
f
Find c Ml
I
s)
for matrix
—5 + 5 cos
— 5 sin
t
t
cost + 1 sin t
in the result of the
previous problem yields
- 2 + 2 cos (f - s) + sin {t - s)
CQS ( f _ s) _ 2 s n (, _ s)
q
3
Find e
( f
)
1
A
for
-5 + 5 cos (t - s)
_5s n _ s
cos (t — s) + 2 sin (/ — s)
j
j
sin (f — s)
14.26
1
t
/
—s
t
1
gA« -s) _
'
1
1
A of the previous problem.
Replacing 1 with the quantity
K
+
2_
1
sin t
14.25
r
t
—1 + 1 cos + sin
cos — 2 sin
t
-5
-2
.
r + -(sin t)
-1
1
0"
1
1
1
,
3
1
3
I Here
n — 3,
so
9
A
t>
'
= a2A
2
2
/
+ x,A/ + x
I
6
- a,
[3
I
9
2
6
+ x,
/
3
9
l
)x 2
6x 2 r + x,/
r + 3x,f + K
x2f
9x 2 r + 3a, J + k
9x 2
Then
A,
r(A)
= a2A 2 + a A + a
t
= a 2 = A, = 3f,
,
r'(A)
= 2a 2 A + a
and
ls
f
= 2x 2
/•"(/.)
1
3
1
2
2
+ x,f
+ 3x,r + a
6a 2 f
2
1
°
L°
Since the eigenvalues of Af are
.
an eigenvalue of multiplicity three, it follows that
e
e
e
3i
= x 2 9f 2 + x,3f + x
3'
= x 2 6r + x,
3'
= 2a 2
The solution to this set of equations is x 2 = \c M
these values above and simplifying, we obtain
.
= (1 - 3t)e M
x,
2
"
t
1
t
1
,
and
x
= (1 - 3f + ft 2 )? 3
/l
t
1
14.27
Find e A for
'
A=
1
1
Here
n = 3,
0~
1
1
so
1
e
Kl
— x 2 A 2 2 + x,Af + x
f
I
=
x,f
x2f
2
+ x,f
x2f
2
+ x,f + x
'.
Substituting
MATRIX METHODS
= i 2 X 2 + a,/ + a
e° = r(0),
f ° = r'(0),
and
and
r{k)
The eigenvalues of Af are
.
e'
= r{t).
Since
=3
from which
a2
= (e' -
t
- l)/t 2
a,
,
r^A)
= A2 =
A,
= 2a 2 A + a„
= »i
f
= 1,
and
It
then follows that
these equations become
a,f + a,r + i.
=
a
/, = t.
and
343
Substituting these results above and simplifying, we
1.
obtain
0"
1
,Af
1
**_
14.28
4x4
Establish the equations required to find e Kt if A is a
constant matrix and Af has eigenvalues It and 3t,
both of multiplicity two.
#
4x4,
Since the order of the matrix is
r(A)
it
follows that
= a 3 A + a2 A + a,/ + a
3
2
e
A
'
and
= x 3 AV + a 2 A 2 2 + a, A? + a I.
Then, also,
f
= 3a 3 / + 2a 2 A + a,
2
r'(A)
We then use
2'
= r(2f) - a 3 (2f) 3 + a 2 (2f) 2 + a,(2f) + a
2
= r '(2f) = 3a 3 (2f) 2 + 2a 2 (2f) + a,
e
3
= r (3f) = a 3 (3f) 3 + a 2 (3f) 2 + a,(3t) + a
e
3
= r'(3f) = 3a 3 (3f) 2 + 2a 2 (3f) + a,
e
<?
= 8f 3 a 3 + 4f 2 a 2 + 2fa, + a
2
= 12f 2 a 3 + 4fa 2 + a,
e
3
= 27f 3 a 3 + 9f 2 a 2 + 3faj + a
e
3
= 27f 2 a 3 + 6fa 2 + a,
e
e
2'
'
'
or
'
'
'
'
to solve for the a's, substituting the results in the first equation above.
14.29
Establish the equations required to find e
Kt
if
A is a 4x4
constant matrix and Af has as its eigenvalues
—
t
of multiplicity three and At of multiplicity one.
I
4x4, the formulas for e Kl r(A), and r'(x) are identical to those in the previous problem.
r"(k) — 6a 3 A + 2a
Now we must solve
Since A is of order
In addition,
,
2
e
'
e~
l
.
- r(-t) = a 3 (-f) 3 + a 2 (-f) 2 + a,(-f) + a
3
f
= r'(-t) = 3a 3 (-f) + 2a 2(-t) + a,
2
2
a3 + f a2
fa,
+a
or
= r"(-t) = 6a 3 (-0 + 2a 2
3
2
e*' = r(4t) = a 3 (4f) + a (4i) + a x (4t) + a
e~< =
<?-'
e
2
4'
-6fa 3 + 2a 2
= 64f 3 a 3 + 16f 2 a 2 + 4fa + a
t
for the as.
14.30
Establish the equations needed to find e
Kt
if
1
2
3
4
5
6
1
2
3
4
5
3
4
5
2
3
4
2
A -
1
I
Here
n = 6,
so
e
Al
= asA
5
5
f
+ a4A
r(A)
4 4
f
+ a3A
3
3
f
2 2
+ a 2 A f + a,Af + a I
and
= a 5 5 + a 4 A 4 + a 3 A 3 + a 2 / 2 + a,/ + a
/i
= 5a 5 4 + 4a 4 A 3 + 3a 3 A 2 + 2a 2 A + a,
2
3
r"(k) = 20a 5 A + 12a 4 A + 6a 3 A + 2a 2
r'(k)
A t = A 2 = A 3 = f,
The eigenvalues of Af are
/.
A 4 = A 5 = 2f,
and
A 6 = 0.
It
now follows that
= r(2f) - a 5 (2f) 5 + a 4 (2f) 4 + a 3 (2f) 3 + a 2 (2f) 2 + a,(2f) + a
2
- r'(2f) - 53t 5 (2f) 4 + 4a 4 (2f) 3 + 3a 3 (2f) 2 + 2a 2 (2f) + a,
e
2
= r"(2f) = 20a 5 (2f) 3 + 12a 4 (2f) 2 + 6a 3(2f) + 2a 2
e
5
+ a 4 (f) 4 + a 3 (f) 3 + a 2 (f) 2 + a,(f) +
e' = r(t) = a 5 (f)
2
4
3
e = r'(t) = 5a 5 (f) + 4a 4 (f) + 3a 3 (f) + 2a 2 (f) + a,
2
4
3
5
e° = r(0) - a 5 (0) + a 4 (0) + a 3 (0) + a 2 (0) + a,(0) + a
e
2'
'
'
oc
1
r
e~' = 3f a 3 — 2fa 2 + a!
2
344
CHAPTER 14
D
or, more simply,
e
= 32t 5 a 5 + 16f 4 a 4 + 8t 3 a 3 + 4f 2 a 2 + 2ta, + a
2'
= 80f 4 x 5 + 32f 3 a 4 + 12r 2 a 3 + 4ta 2 + a,
2
= 160f 3 a 5 + 48r 2 a 4 + 12ta 3 + 2a 2
e
4
5
3
2
e' =
a5 +
a4 +
a3 +
a2 + x + a
4
3
2
e' = 5f a 5 + 4f a 4 + 3f a
2fa 2 + a
3 +
=a
e
2'
'
f
r
f
f
t
i
l
t
1
must be solved for the a's.
MATRIX DIFFERENTIAL EQUATIONS
14.31
x + 16x + 64x = 0;
Transform the initial-value problem
f
x(0)
x(0)
x,
=x
x 2 = x.
and
into matrix form.
x =
Solving the differential equation for its highest derivative, we obtain
order 2, so we define two new variables:
=
- 16x - 64x.
This equation has
Differentiating each of these equations once
yields
= x = x-,
v-, = x = — 16x — 64x = — 16x,
*i
X=
This system has the matrix form
=
*2
x,
or
64x,
-64
I"
^2
-16
Xi
=
+ lx,
x 2 = — 64x, — 16x 2
Xi
Ox,
The initial conditions may be written
"1/6"
c=x(o=hn!i=ra
[x (0)J
Lv(°)_
2
14.32
x + 8x + 25x = 0;
Transform the initial-value problem
#
2,
x(0) = 0;
x(0)
— x
x,
Xi = x.
and
X,
x,(0)
v(0)
2 (0)
_x(0)
«c
14.33
L
2_J
The initial conditions may be written
-8
25
Ox, -I- lx 2
x 2 = — 25x, — 8x 2
r
o
=
V,
=
4
Transform the initial-value problem
f
X]
X =
"0"
C = X(0) -
or
25v - -8x 2 - 25x,
has the matrix form
This equation has order
Differentiating each of these equation once yields
x = x,
X, — X —
into matrix form.
x= -8x-25x.
Solving the differential equation for its highest derivative, we obtain
so we define two new variables:
=4
x + 20x + 64x = 0;
x(0)
= J,
x(0)
x,
=x
and
into matrix form.
x = — 20x — 64x.
Solving the differential equation for its highest derivative, we obtain
equation has order 2, we define two new variables:
=
x 2 = x.
Since the differential
Differentiating each of these
equations once yields
X,
= X = X,
or
7
This system has the matrix form
X=
-64
*7
C = X(0)
x,(0)
x 2 (0)
14.34
=
x(0)
=
l
The initial conditions may be written
-20
1/6
*(0)
Transform the initial-value problem
I
=
0x + lx 2
—
=
64x, — 20x 2
x2
X,
x = -20x - 64x = -20x - 64x,
x + 16x = 0;
x(0)
= — \,
x(0)
Solving the differential equation for its highest derivative, we obtain
x, = x
has order 2, we define two new variables:
and
x 2 = x.
=
into matrix form.
x= — 16x.
Differentiating each of these equations once
yields
=x=x
x 2 = x = — 6.x
x,
x,
1
or
16x,
Since the differential equation
=
Oxj + lx 2
= -16x, + Ox,
MATRIX METHODS
_
C = X(0) = ~*i(0)~ =
_x 2 (0)_
14.35
~x(0)~
The initial conditions may be written as
ihi.
0_|[x 2 J
16
2_
"-1/2"
=
_x(0)
x + 96x = 0;
Transform the initial-value problem
f
i
=
X =
This system has the matrix form
345
= £,
x(0)
=
x(0)
into matrix form.
x= -96x.
Solving the differential equation for its highest derivative, we obtain
has order 2, we define two new variables:
x,
=x
x 2 = x.
and
Since the differential equation
Differentiating each of these equations once
yields
x = x-
= x = -96x = -96x,
has th< t m atrix f orrr
=
X
i
X
_ 2_
x 2 (0)
14.36
*1
-96
_
The initial conditions may be written as
X 2_
x(0)
x + 64x = 0;
Transform the initial-value problem
f
r
o
2
"1/6"
~*(0)~
C = X(0) =
=
Ox, + lx
x-, = — 96x, + 0x-
x,
or
x(0)
= £,
=2
x(0)
x = — 64x.
Solving the differential equation for its highest derivative, we obtain
has order 2, we define two new variables:
x,
=x
x 2 = x.
and
into matrix form.
Since the differential equation
Differentiating each of these equations once
yields
x,
= x
x 7 = x = — 64x
This system has the matrix form
C = X(0) =
t
x + x = 3;
— 1,
x(n)
x(7i:)
This system has the matrix form
x
t
=x
x 2 = x.
and
into matrix form.
Xlfa)
x(n)
x 2 (n)
x{n)
Transform the initial-value problem
x
-I-
{
x,
t
+
x(0)
x 2 = x.
The initial conditions may be written as
= 2, x(0)=-2
This system has the matrix form
"x,(0)
[x(0)1
x 2 (0)
U(0)J
into matrix form.
x = — 4x — 4x.
This equation has order 2,
Differentiating each of these equations once yields
=x
Ox, + lx 2
x = — 4x — 4x = — 4x 2 — 4x
C = X(0) =
Ox, + lx 2
t
Solving the differential equation for its highest derivative, we obtain
and
=
-lx + 0x 2 + 3
3
x + 4x + 4x = 0;
=x
This equation has order 2,
or
-1
x
x = — x + 3.
Differentiating each of these equations once yields
1
X
so we introduce two new variables:
14.39
= 2
Solving the differential equation for its highest derivative, we obtain
= x = xx 2 = x = —x + 3 = — Xj
I
0x 2
The initial conditions may be written as
-64
x,
14.38
-I-
ta-ra
so we introduce two new variables:
C = X(n) =
lx 2
1
X=
Transform the initial-value problem
f
64x,
-I-
x,(0)'
x 2 (0)
14.37
=
Ox,
x 2 = -^64x
x,
or
X=
or
-4x, - 4x 2
t
1
The initial conditions may be written as
4 -4
2"
-2
Transform the initial-value problem
x + 12.8x + 64x = 0;
x(0)
x(0)
=
into matrix form.
346
CHAPTER 14
D
#
x,
=x
X=
This system has the matrix form
C - X(0) =
"x,(0)"|
|"x(0)"|
x 2 (0)J
[x(0)
x, =
or
12.8x 2 — 64x,
t
lx 2
— 12.8x
2
1
[';]-[: •64
The initial conditions may be written as
-12.8
x + 9x + 14x = 0;
x(0)
= 0,
x(0)
=x
and
-9x- 14x = -9x, - 14x,
or
equation is of order 2, we define two new variables:
=x
=x
x,
= -1
x 2 = x.
Since the differential
Then differentiation yields
=
x,
Ox,
-I-
lx 2
x 2 = — 14x, — 9x-
X
This system has the matrix form
into the matrix form.
x = — 9x — 14x.
Solving the differential equation for its highest derivative, we obtain
x-,
Ox, +
x 2 = — 64x
|_
j
x,
This equation has
Differentiating each of these equations once yields
n/6
Transform the initial-value problem
f
x 2 = x.
and
x, = x = — 12.8x — 64x
14.40
x = — 12.8x — 64x.
Solving the differential equation for its highest derivative, we obtain
order 2, so we define two new variables:
The initial conditions may be written as
14
C = X(0) =
Y,((»
>c
14.41
x(0)
=
=
-1
x(0)
2 (0)
x -I- 9x + 14x — { sin f;
Transiorm the initial-value problem
f
= 0,
x(0)
x(0)
= —
into matrix form.
1
This problem is similar to the previous problem, except now the differential equation is nonhomogeneous.
— 9x — 14x + \ sin u
x =
Solving for the highest derivative, we obtain
We define x, and x 2 as in the previous
problem, and then differentiate to obtain
x,
— x — x— 9x 2 — 14x, +
x2 = x =
sin f
Ox, -I- lx 2
- 14x, - 9x 2
1
X =
This system has the matrix form
=
x2 =
x,
or
\
°
f r
-9
14
i
-I-
2
sin f
The initial conditions take the
J sin J
f
same form as in the previous problem.
14.42
x + 2x — 8x — e'\
Transform the initial-value problem
f
x(0)
= 1,
x(0)
= —4
x =
Solving the differential equation for its highest derivative, we obtain
order 2, so we introduce two new variables:
x,
=x
x 2 = x.
and
into matrix form.
— 2x — 8x 4- e'.
This equation has
Differentiating each of these equations
once yields
X
\
—X—A
= x = — 2x + 8x + e' — — 2x 2 + 8x, + e'
2
This system is equivalent to the matrix equation
X(r)
=
Xj(t)
At/) =
14.43
and
-~>
x 2 (t)
initial
conditions are given by
X(f
Transform the differential equation
f
where
f
x
,
and
x 2 = x.
\(t)
x 2 (t)
x — 6x — 9x + t.
,
9
6
and
F(t)
=
t
C=
-4
then the
Since this equation has order 2, so we
x, =
1
=
where
Differentiation then yields
=x
These equations are equivalent to the matrix equation
=
= A(f)X(f) + F(f),
into matrix form.
X- = x = 6x -9x + t = 6x 7 - 9x,
X(t)
1
= 0.
x — 6x + 9x = t
=x
x,
or
Furthermore, if we define
F(f)
= C,
)
X(f)
Solving for the highest-order derivative, we find that
introduce two new variables:
x t (t)
= Ox, + lx 2 -f
x 2 = 8x, — 2x 2 + e
x,
-j
I
+
r
X(r)
Ox, + lx 2 -I-
x 2 = -9x, -I- 6x 2 + t
= A(f)X(r) + F(f),
where
MATRIX METHODS
D
347
2
14.44
Put the initial-value problem
d x
d x
dx
—
^ + 2 — - 3 -- + 4x =
=-
dt
3
dt
2
2
+ 5;
t
x(2)
x(2)=ll,
10,
jc(2)=12
into
dt
matrix form.
f
d 3 x/dt i = -2'x + 3x - 4x + t
Solving this differential equation for its highest derivative, we obtain
equation has order 3, so we introduce three new variables:
= x,
x,
x 2 = x,
x 3 = x.
and
2
+ 5.
This
Differentiating each of
these equations once yields
x, = x
= Ox, + lx 2 + 0x 3
x2 =
Ox,
0x 2 + lx 3
2
x 3 = -4x, + 3x 2 - 2x 3 + (r + 5)
x,
x2 = x
or
-I-
x 3 = x = — 2x + 3x - 4x + 2 + 5
t
0"
1
This system has the matrix form
X=
~Xi'
4
-2
3
2
Put the initial-value problem
I
2
d3x
1
'dt
=
x(2)
x 3 (2)
14.45
"10"
"x(2)l
C = X(2) = x 2{2) =
+ 5
f
-*3_
f"x,(2)l
The initial conditions may be written as
+
*2
1
11
12
x(2)
2
d x
dt
%e
2
3r.
x(l)
= 2,
x,
= x,
x 2 = x,
x(
— 1) =
intu matrix form.
This equation has
x — 2x — \x + 4e~ 3t
x 3 = x. Differentiating each of these
Solving the differential equation for its highest derivative, we obtain
order 3, so we introduce three new variables:
= — 2,
x(l)
.
and
equations once yields
Xj
x2 = x
x3
Ox, + lx 2 + 0x 3
x2 =
or
=x = 2x — jx + 4e~
=
+ 0x 2 + lx 3
\x + 0x 2 + 2x 3 + 4e -3f
0x
3f
t
x
~*1~
This system has the matrix form
X
*3
+
x2
1
-1/2
_
p,(l)l
C = X(l)
The initial conditions may be written as
x 2 (l)
dA x
Transform the initial-value problem
e
d2x
'
dt
7
+ e't 2
dx
'
2
~x(l)~
=
-2
=
x(l)
x 3 (l)
14.46
4e~ 3
X 3_
x(l)
= 5e~';
x(l)
=2
x(l)
= 3,
= 4,
x(l)
- - + 5.
This
x(l)
It
into matrix form.
4
I
2
d x
d x
—
^= —
Solving the differential equation for its highest derivative, we obtain
dt
equation has order 4, so we introduce four new variables:
x,
= x,
=-
e'
x 2 = x,
2
t
e
2'
It
dt
x 3 = x,
dx
and
x4
Differentiating each of these equations once yields
=x
x,
X, = X
or
x4
e'x
+e
t
x + 5
These equations are equivalent to the matrix equation
"x,(ff
W) =
x 2(t)
where
f
= 1.
x 2 = Oxj +
0x 2 + lx 3 -I- 0x 4 +
x 3 = Ox, +
= A(f)X(r) + F(r),
X(r)
where
1
A(t)
1
=
F(0
x 3 (0
1
-t 2 e 2
_x 4 (t)_
Furthermore, if we define
lx 2 + 0x 3 + 0x 4 +
0x 2 + 0x 3 + lx 4 +
x 4 = Ox,
fV'x 2 + e'x 3 + 0x 4 + 5
*3
2, 2
x x = Oxj +
C = [2, 3, 4, 5] r
,
'
e<
then the initial conditions are given by
X(r
)
= C,
x.
= 5
CHAPTER 14
348
14.47
x = fx + x — y + r + 1;
Put the following system into matrix form:
= 2,
x(l)
#
x(l)
= 3,
x(l)
= 4,
y(l)
y — (sin t)x + x
—y+ 2
t
;
= 5, #1) = 6.
Since this system contains a third-order differential equation in x and a second-order differential equation in y,
x,(M = x;
we will need three new x-variables and two new y- variables. We therefore define
x 3 (r) = x, y,(M = y, and y 2 (t) = y. Then differentiation yields
x 2 (r) = x,
x2 = x 3
x 3 = x = tx + x — y + t+l = tx 3 + Xi— y2 + t+l
9x = yi
$2
— y = (sin t)x + x — y +
= (sin t)x 2 + x, — y,
lx 2 4- 0x 3 + Oy, + 0y 2 +
0x 2 + lx 3 + Oy,
0y 2
0x 2 + tx 3 + Oy, - ly 2 + (t + 1)
x, = Ox, +
x 2 = Ox, +
-I-
x 3 = lxj +
2
2
t
r
-I-
-I-
= Ox, +
0x 2 + 0x 3 + 0>*i + ly 2 +
=
lx, + (sin f)x 2 + 0x 3 - ly, + 0y 2
y2
y,
2
t
-I-
These equations are equivalent to the matrix equation
X(t)
= A(t)X(t) + F(t),
"0
"*,(')"
0"
1
x 2(t)
Mt) =
1
x 3(t)
A(t)
=
-1
f
1
v,(n
14.48
C = [2, 3, 4, 5, 6] r
-1
sinf
1
Put the following system into matrix form:
f
= t+
1
and
x =
— 1,
t
2
0_
f
then the initial conditions are given by
-2x - 5y + 3;
y = x + 2y;
x(0)
= 0,
= 0,
x(0)
X(f
y(0)
)
— C.
= 1.
Since the system contains a second-order differential equation in x and a first-order differential equation
in y, we define the three new variables:
x,(f)
x 2 (f) = x,
= x,
and
These equations are equivalent to the matrix equation
-I-
= Ox! + lx 2 + 0y, +
x 2 = Ox, — 2x 2 — 5y,
3
or
3
X(f)
X(f)
= A(r)X(r)
and
= x 2(t)
C=
-I-
-I-
2y,
+
where
F(t),
1
AIM -
-2
-5
1
2
y,(n
=
-I-
y, = Ox, + lx 2
x,(t)
f
Then differentiation yields
x,
x 2 — x — — 2x — 5y + 3 = — 2x 2 — 5y,
y, = y = x + 2y = x 2 + 2y,
we also define
y t (t) = y.
= v,
x,
If
F(f)
1
_y 2(t)_
Furthermore, if we define
where
F(M =
then the initial conditions are given by
X(f
x = x + y — z + f;
2
)
= C.
1
14.49
Put the following system into matrix form:
x(l)=l,
I
x(l)=15,
y(l)
= 0,
y(l)
=-7,
z(l)
y = fx + v — 2v + r
+ 1;
z
= x — y + y + z;
= 4.
Since this system contains second-order differential equations in x and y and a first-order differential equation in
z,
we define five new variables
z,
= z.
— two
for x, two for y, and one for z:
x,
= x,
x 2 = x,
y,
= y,
y 2 = y,
and
Differentiating each of these variables once and using the original set of differential equations, we obtain
x,
= x = x2
x2 = x = x + y fi
+ = x 2 + y 2 — z, +
f
=y = y 2
y 2 = y = tx + y - 2y + t
ii
t
2
+
1
= fx, + y 2 - 2y, +
= z = x - y + y + z = Xi - y + y 2 + Zi
x
2
t
+
1
MATRIX METHODS
or
x,
349
= Ox, + lx 2 + 0>', + 0_y 2 + Oz, +
x 2 = Ox,
4-
+ \y 2 - \z x + t
lx 2 + 0y t
= Ox, + 0x 2 4- Oj/, + \y 2 + Oz, +
= fx, + 0x 2 - 2y, + ly 2 + Oz, + (t 2 + 1)
2
v ,
,
= 1.x, + 0x 2 -
+ \y 2 + lz, +
v,
1
X = A(r)X(f)
These equations are equivalent to the matrix initial-value problem
-I-
X(l) = C,
F(t);
f
0"
x 2 (t)
\(t)
15
t
=
Mt) =
where
=
F(r)
1
c=
+
r-
y 2 (t)
-7
i
4_
14.50
x = x + y;
Put the following system into matrix form:
i
Since the system consists of two first-order differential equations, we define two new variables:
and
v,(r)
= y.
x,(f)
=x
Thus,
x
—x—
x
=
>'i
If
y — 9x + y.
matrix equation
-v,(f)
"
l
X(r) =
we now define
>'
+ j',
x,
= 9 ^ + V = 9x, + >'i
1
and
]\
x + y —
1
then this last set of equations is equivalent to the
A(f)
X(r) = A(f)X(f).
SOLUTIONS
14.51
Solve
f
x + 9x + 14x = 0;
= 0,
x(0)
1
A=
where
X
14.2, we
may write its solution as
Therefore,
Solve
I
1.
X = AX;
This homogeneous differential equation has the matrix form
x(f)
le- 2 '-2e- 7
x,
;
-1
e~ 2
2t
x
7
+e
A=
X
Problem 14.40),
Using the r
~5 _2e~ 2 -le- lf
-1
'
1
_
1.17.)
= 0.
x(0)
C
and
-64
(see
~-e- 2t + e~ ir
1
(Compare this with Problem
').
- \,
x(0)
*2 = x.
0"
'
'
X(0) = C
and
-e' 1
-2e~
+ 7e~ 7 '_
'
X = AX;
This homogeneous differential equation has the matrix form
where
- x
21
= x {t) — j( — e
x + 20x + 64x = 0;
C=
and
-9
-14
X = e A 'C =
14.52
= -
x(0)
x,
=x
X(0) — C
x, = x.
and
(see
Problem 14.33),
Using the result of
20
Problem 14.3, we write its solution as
Therefore,
14.53
Solve
#
x(t)
4
-4e~ 16
-64f- 4t + 64e" 16
16e"
X = e A 'C =
'
-*<- e -"»
-4e~
A'
+\6e~
and
A=
C = X(0)
u
7.(0)^J
L x 2l
=
"x(0)
x(t)
4'
+ le- 16 'J
X = AX
(see
Problem 14.33), where
Here the initial conditions are
Using the result of Problem 14.3, we can write the solution as
4
1
12
Therefore,
=
_0_
-1
=
L Jx(0)
X = e A 'C
x 2 = x.
and
20
_
~xi(0f
=x
x,
64
_
x(0) = 4.
This homogeneous differential equation has the matrix form
X=
6
16t
(Compare with Problem 11.^ »6.)
'.
= - 1,
x(0)
e
'
— Yge -16
= x,(f) = \e'
x + 20x + 64x - 0;
'
= x,(f) = — e 4f
4
-4e~ l("
-64e~ 4 + 64e" 16
\6e~
e
'
'
'
-*>
"-1
- (,-'<"
-4e~*' + \6e~
l("
4
"_ -4l
e
_
4e' M
'
350
14.54
CHAPTER 14
Q
Solve
f
x + 20x + 64.x - 0;
= 0.
x(2)
x(2)
= 4.
This differential equation has the matrix form
X = AX,
with X and A as in the previous two problems.
"0"
C = X(2) =
Here, however, the initial conditions take the form
=
=
!
[x(2)J
L.x 2 (2)J
Using the result of
14
Problem 14.4, we write the solution as
_—
c=
,A(t-2)f-
12
14.55
= x,(f) = i e
Thus,
x(t)
Solve
x + 16x = 0;
f
'~ 2)
- 4e- 16 <'- 2
' 2)
M,
-64e~
+ 64e' l6i '- 2)
I6e~ 4{
- 4 "- 2
- k" 16 <'- 2
»
= -{,
x(0)
x(0)
X —
A =
Solve
f
x(f)
C
and
x + 96x = 0;
=I
x(0)
1
=
C=
and
Therefore.
Solve
x(t)
x 2 = x.
and
cos4fJ
(see Problem 14.34),
Using the result of
l^n4f"ir-|"|
x,
(1
=x
X(0) = C
r-|cos4t"|
2sin4rJ
0J
= 2,
x(0)
(see Problem 14.35), where
x 2 = x.
and
N 96)sin N 96r
1
Using the result of Problem 14.11.
6
(1
6)cos v 96f
x 96 6)sin N 96fJ
cos v 96f
(Compare uith Problem
l
x(G)
=x
cos4f
X - AX;
N %sin x 96f
= x,(?) — I cos N ^6/.
x + 4x + 4x = 0;
m
cos N 96f
X = e A, C =
x,
X = c A 'C
This differential equation has the matrix form
96
|:
X(0) = C
- 0.
x(0)
we can write its solution as
I
X = AX;
|_-4sin4f
(Compare with Problem 11.1.)
= x,(f) = -}cos4f.
=
<-[::} A
14.57
+ 16"
-±e- lb «- 2) 1
= 0.
Problem 14.10, we can write its solution as
Therefore,
ir°l
».
0^
16
14.56
-4e~ M
'- 2)
This homogeneous differential equation has the matrix form
where
± e - M '~ 2)
21
>
1
1.2.)
= -2.
This differential equation has the matrix form
X — AX;
X(0) = C
Problem 14.38). where
(see
X
I
and
_4
_4
C
x,
[J]
= x
and
x 2 = x.
Using the result of Problem 14.20. we write its
solution as
1
C
[
Therefore,
14.58
Solve
/
x(r)
= x,(f) = (2 + 2t)e' 2
'.
x(- 1) = 2,
x + 4x + 4x = 0;
(2 + 2t)e
']
{-2-4t)e
[l-2t)e- 2t]l-2J
-4te~ 2t
(Compare with Problem
1
1.21.)
x(- 1) - -2.
This differential equation has the same matrix form as the differential equation in the previous problem,
except now the initial time is
t
=—
1
rather than
X( — 1) — C.
t
= 0.
That is, now
2t
+ 23
,+i
4f
-44 — 2t — iJL— 2J
Using the result of
Problem 14.21. we write the solution as
X = e A[l
14.59
C — eM
""
'
- x,(r) = (2r + 4)e _2(t+1)
Then
x(f)
Solve
x + 64x = 0;
I
' ( " l )]
x(0)
'
2
'C = e ~ "
"*
'
.
X 2_
A=
It
:i
+4
4f
6
r
L~
x(0)
I
-64
i_.-.
.
This differential equation has the matrix form
xl
2
i
,
and
C=
X = AX;
1/4
Xj
2
—x
X(0) — C
and
x, = x.
(see
Problem 14.36). where
Using the result of Problem 14.12,
MATRIX MFTHODS
D
351
we write its solution as
cos8f
X = e K, C =
-8sin8f
Therefore,
14.60
Solve
I
jc
x(t)
-2 sin 8f + 2cos8f
= x,(r) = \ cos 8f + £ sin 8f. (Compare with Problem
1
1
.4.)
This differential equation is the same as that of the previous problem, and so it has the same matrix form
with X and A as defined in Problem 14.59.
Nonetheless, we may set
C = X(0) =
~x(0)~
—
x(0)
=k
x 2 (0)
x
Solve
x{t)
= x (t) =
x
/c,
x + 8x + 25x = 0;
i
Then, using the result of Problem 14.12, we write the solution as
cos 8?
| sin 8f
-8sin8f
cos St
cos St + k 3 sin St,
where
= 0,
x(0)
1
A=
C=
and
-i
k y cos8r + i/c 2 sin 8f
ft,
-8/c, sin8f + fe 2 cos8f
A:
3
= k 2 /S.
= 4.
x(0)
X(0) = C
X = AX;
This differential equation has the matrix form
25
The difference here is that there are no initial conditions.
where k and k 2 denote unknown numbers, and write
,
k,
X = e A 'C =
Therefore,
x(0) = k 2
and
V
~*i(0)~
x(0)
I
i cos 8f + { sin St
2
+ 64x = 0.
X = AX,
14.61
1/4
cos St
=x
x,
(see
x, = x.
and
Problem 14.32), where
Using the result of Problem 14.16,
we can write its solution as
X = e A 'C = e-*'
Therefore,
14.62
Solve
I
x(f)
-¥sin3f
= x,(r) = \e
and
x(0)
Therefore,
Solve
=
/c
where k
2,
x(0)
*i(0)~|
p,]
x(0)
x 2 (0)J
lk 2 \
X = e x 'C = e
x(t)
4/
{
1.19.)
and k 2 denote unknown numbers; then
Using the result of Problem 14.16, we may write the solution as
§ sin 3f + cos 3t
^sin 3t
= x (f) = k 3 e 4 sin 3r + k e *' cos 3f,
x
x(0)
= {,
x(0)
1
A =
64
-lis]"
[::}
14.17, we may write its solution as
X = e K, C = e -6.4t
Solve
3
_
x
t
=x
and
^sin4.8f
24
-
= £,
x(0)
1
and
16
C
)
]
X(0) = C
- f sin 4.8t + cos 4.8f
x(0)
+ ^/c 2 sin 3t + /c, cos 3f
(— ^fej - f/c 2 )sin3f + k 2 cos3t
= f^ + ^k 2
- f sin 4.8r
t
-64
3
sin4.8r + cos4.8r
:
.
(see
X-,
Problem 14.39), where
= x.
1/6
Using the result of Problem
=e
6.4(
isin4.8f + zcos4.8r
%Q sin4.8f
(Compare with Problem 11.24.)
- 0.
This differential equation has the matrix form
<-::}
/c
X = AX;
1/6
^ C = Lo_
.
= x (t) = e~ 6 4 '(f sin4.8f + £cos4.8r).
x + 16x + 64x = 0;
(f/c!
4-1
= 0.
This differential equation has the matrix form
x(t)
where
'
t
x + 12.8x + 64x = 0;
Therefore,
=e
-§ sin 3f + cos3r
-^sin3f
X =
#
1
t sin 3f + 4 cos 3f
with X and A as defined in Problem 14.61. Since no initial conditions are specified, we may set
C = X(0) -
14.64
(Compare with Problem
sin 3f.
f sin 3t
= e -At
This differential equation is the same as that of the previous problem, and so it has the matrix form
x
f
isin 3t
— f sin 3f + cos 3f
x + 8x + 25x = 0.
X — AX,
x(0) = k
14.63
f sin 3r + cos3r
X = AX;
1/6
Xj
=x
X(0) = C
and
(see
x 2 = x.
Problem 14.31), where
Using the result of Problem M.18,
352
CHAPTER 14
D
we write its solution as
Therefore,
14.65
0),
= x,(£) = |(1 + 8f)e~
x + 2x-8x = 0;
Solve
I
x(t)
l-8tj[0 J
-64f
(Compare with Problem 11.28.)
x(l)
= 3.
X = AX;
This differential equation has the matrix form
where
X(t)
x,(t)
=
A=
x 2 (t)
ro
,
C=
and
-2
L8
X(l) = C
;
x,
B,
lU-e4t)e
8 '.
= 2,
x(l)
+ 8f
1
X = e A, C = e
Problem 14.42 with e' replaced by
(see
=x
i
x 2 = x.
and
Using the result of
Problem 14.8, we may write its solution as
\r4e 2i
6l8e
+ 2e- M l)
'l)
-Se- 4( l)
'- 1)
2{ '
'~
The solution to the initial-value problem is
14.66
x + 2x
Solve
f
=e
ix
f
x(0)=l,
;
2e
Xi(t)
=
F(t)
x 2(t)_
X = e K 'C +
8
f/
e
M
s)
'
11 2(1-1)
u + 4e -4((- 1)
l
'
l)
22 2((-l)
6 e
=
X = AX + F(f);
C=
and
-2J
x, = x
=-
JV
so
F(s)
Ue + 2e
1
=
2'
2c
4e 2d-5) + 2e
2
M -8e
8c "
4ii
W : i
F(.v)(is
o
6
J'o
j [2e
'
{2,
and
14.67
x(f)
x 4- x = 3;
Solve
f
= x,(f) = ^"
,2(t
i)
2c
=
x(n) =
e
J'„
2
+ U'
'- 4^
4
4
c=
,A(t-s)
so
,A(J-s)
j:
and
sin (f
6
2e
<2 ' _,)
+ 4e
i„( -4f+5»)\s = l
I)
(2,-s
(-2<?<
?
{
4 '" 5,(
"1
(
+ 4 e ,-4,
+ 5s , s = ,
oJ
)
'
2'
4
31,,
'
41
L
,
21
_ u
-f|e-« + i«*
I
F(r)
=
\{n) = C
X = AX + F(f);
and
J**
C=
x,=x
(see
Problem 14.37), where
x, = x.
and
Its
solution is
F(s)
=
F(s)ds =
Using the results of Problems 14.14 and 14.15, we have
cos(f — n)
— sin(f — n)
—
cos(f — n)
cos(r — s)
sin(f
- sin (t — s)
sin(f
3 sin (t
1
-sin(f - it) + 2cos(f -n)]
3sin(f — s)
— s)
cos (r — s)
— s) ds
3 cos (f — s) ds
_J'„
J'„
cos(f — n) + 2sin(f — 70
7t)
~
X(f)
— sin (f —
7i)
3cos(r — s)
3
3cos(t-s)|*='„
3
j-3sin(t-s)|?=',
cos (t — n) + 2 sin (f — 7t)
Thus,
N
1
+ \e
2 '+ 2e
y
6
c
'
-V + c
1
"
_
x{n) = 2.
1,
M, ~ s) F(s)ds.
,A(f-n),
and
<7.s
i
2'
4"
]
1
L*2(0j'
X = e M '~ K) C +
" 4 4C
[ ()
+ i^« - |e
'
A =
Its solution is
\T g(2*-») _ g(~4<+5j)
si
(21
'J
This differential equation has the matrix form
X(r)
2il
_ e 4d
+ 4e -*t+5a)]ds_ ~6
+
-4e~ 4, _
if
41+5
(
"
4r
e
X4
x 2 = x.
and
-4e:.]
-4
+ 4e
i)
4ii
g
s)
-fe' + c
1
21
L
t)
6 _-fe' + 2e
Thus,
Problem 14.42), where
e
-8e
1
=
s,
(<
(see
e'
6 |_8e
,Mt -s)
.
X(0) — C
2t
and
]
Using the results of Problems 14.7 and 14.9, we have
F(s)ds.
K
e 'C
~ 46 e -4(r
n
4"
+ \e
l<,-4(f-l)
,
= 4.
x(0)
n
A =
,
2{t '
= x,(t) = ^e 21
x(f)
This differential equation has the matrix form
X(f)
-4(1- 1)
,2(1- 1)
+ 2 cos (t — n)
3
+
3
tt)
cos (f - n) =
7i).
f
ri)
7i)
— 3 cos (t — re)
3 sin (f —
Noting that
x(f) = x,(f) = 3 - 2 cos (t - n) + 2 sin (t — n) = —sin
we finally obtain x(f) = 3 + 2 cos — 2 sin
t,
— 3 cos (t —
3 sin (t —
t.
— 2 cos (t — n) + 2 sin (t — n)
2 cos (t - n) + 2 sin (f - n)
-cos f
and
'
MATRIX ME T HODS
14.68
x + 9x + 14x = \ sin t;
Solve
I
x(0) = 0,
X(r)
=
F(r) =
-9
X = e A 'C + j'
14
x 2 (t) = x(t).
solution is
We determined e A 'C in Problem 14.51.
e
X = AX + F(f);
1
A =
x 2 (t)
A(,_,)
- 1.
x(0) =
This differential equation has the matrix form
x,(t)
Its
Je
F(s) =
-2(t-s)_2 -~Ht-s)
e
— \4e~ 2{, ~ +\4e~ 1( '~
s)
so
J*V«-
s)
C=
and
Problem 14.41), where
(see
with
x,(f)
= x(f)
and
sin t
A( '~ s)
F(s) ds.
<?
Using the result of Problem 14.5, we have
— 2e'
s)
X(0) = C
-2«-s)_ e -Ht-s)
e
353
D
2it
~ s)
+ le~ 1(,
e
1
~ s)
To
i sin s
-2(r-s)
•2e~
sms _ e -7(r-s) sins
2( '~ s)
+ 7e" 7(r_5) sins
sin s
- e lx \'Q e ls sin s ds
ls
7
TO L-2e- 2 'f e 2s sin s ds + 7e" »f e sin s ds
2I
e
1
F(s)ds =
J'
e
2s
sin sds
e~ 2 \\e sin t - \e cos r + i) - e~
2«~ 2t{\e 2 sin ( - |e 2r cos t +
+ le' 11
TO |_—
'
M sin
1
7,
2t
2t
1
1)
±cost + ±e- 2
t
(^
7'
- ^e 1 cos + ^)
- ^e 7 cos + ^)
sin
sin t
^
11
'
t
'
t
t
-&-«-
'
13
- 2„-2l + 50"
"To Jfesinf + i§cosf
X(t) =
Then
2<T 2
'
_500 Mn '
14.69
9
99 „-7r'
90 „-2t
ws — 500
e " + 500^
^ 500 Lub ^ 500 e
500 e
'
x 2 (t)
,
-2
-5
1
2
(see
y(0)
= 1
Problem 14.48), where
F(s)
x,
cos t — 2 sin t
—5 + 5 cos t
— 5 sin
sin t
cos t + 2 sin t
t
t
x,
y t — y.
and
cos (t — s) — 2 sin (t - s)
sin (t
— 5 + 5 cos
— 5 sin
cos + 2 sin
— s)
t
t
t
- 2 + 2 cos (t - s) + sin (t — s)
=
=x,
Using the results of Problems 14.24 and 14.25, we have
-2 + 2 cos + sin
C =
Kt
"1
e
= 0,
x(0)
C=
and
F(r)
s)
"1
Mt ~ s)
= 0,
X(0) = C
X = AX + F(r);
X = e 'C + $' e M '- F(s)ds.
A
e
x(0)
1
A=
l
solution is
y = x + 2y;
3;
(Compare with Problem 11.38.)
'.
0"
"0
_y (t)_
and
i
'
x=-2x-5y +
x r {ty
Its
To
500 cos lt
t
This system has the matrix form
X(t)=
<J
= Xj(t) = ^sin — s§ocosf — ^e" 2 + ^e" 7
x(f)
Solve the system
I
+
7
-le
soosinf
and
^sinf-^cosf + \e 2t JU-7H
2t
+ ^e~ 7, J
josint + ^cosr -\e~
1
r
t
-5 + 5 cos (t - s)
- 5 sin (f — s)
cos (f — s) + 2 sin (t — s)
6 + 6 cos — s) + 3 sin — s)
(t
3 cos (f
(t
— s) — 6 sin(r — s)
— s)
3 sin
(t
- 6 + 6 cos (f — s) + 3 sin (t - s)] ds
— s) — 6 sin(f — s)] ds
Jo [3 cos (t
3 sin {t — s) ds
J'
[-6s - 6sin(r - s) + 3cos(r - s)]j=
[-3 sin (t - s) - 6cos(t - s)] z'
jo [
t
so
f
e *v-')Y{s)ds =
"
s
s
3 cos (r- 5)||=
Then
X(f) =
— 5 + 5cosrl
— 5sinr
+
cos + 2 sin
f
and, finally, we have
x(f)
|~
|~
t
— 6 + 3 sin + 6cosf
3 — 3 cos t
t
= x^t) = 2 cos + 6 sin - 2 - 6f
r
r
+ 3 + 6 sin t — 3 cos t
— 6 + 3 sin t + 6 cos f
3 — 3 cos
6r
f
— 6f + 3 + 6 sin — 3cosrl
f
-
and
=
— 2 — 6f + 2cosf + 6sinr
— 6 + 6cosf — 2sinf
3 — 2 cos + 2 sin
r
t
y(f)
= y^f) = -2 cos + 2 sin + 3.
t
f
CHAPTER 15
Infinite-Series Solutions
ANALYTIC FUNCTIONS
15.1
Define "analytic function."
I
°°
A function /(x) is analytic at x
Taylor series about x
if its
f
Y
,
in)
(x )(x
15.2
P(x) = b (x)/b 2 (x)
If either
.
y" + P(x)y' + Q(x)y = 0,
A point x
is
an ordinary point if both P(x) and Q(x) are
then x
is
a singular point.
P(x) or Q(x) is not analytic at x
,
Define "regular singular point" for the differential equation of the previous problem.
I The point x
if
x
is
a regular singular point for the differential equation in standard form
is
a singular point (see the previous problem) and the products
both analytic at x
.
— x )P(x)
and
b^x))*'
y'
(
1
A point x
is
an ordinary point if P(x) is analytic at x
f(x) = e
Find a Maclaurin-series expansion for
Since every derivative of e* is e
e° =
1.
y" + P(x)y' + Q(x)y —
(x
—x
2
)
Q(x)
are
+ b (x)y — 0?
Dividing by b (x), we transform the differential equation to the standard form
P(x) = b Q (x)/b {x).
I
(x
Singular points which are not regular are called irregular.
What is an ordinary point for the differential equation
I
15.5
Q(x) = b (x)/b 2 (x).
and
l
analytic at x
15.4
b 2 (x)y" + b^xty + b (x)y = 0?
we divide by b 2 (x), we transform the differential equation to the standard form
If
where
15.3
n\
.
What is an ordinary point for the differential equation
f
converges to f(x) in some
,
n = o
neighborhood of x
— X )"
x
,
it
+ P(x)y = 0,
where
.
x
.
x =
follows that /(0) and all the derivatives of / at
are equal to
The Maclaurin series is then
<
e
=/(0) +/(0)x H
T7- x
H
=
1
,.
3!
+ (l)x + - x 2 + -- x 3 +
2!
15.6
•
•
diverges; whereas if
L = 1,
£= a converges
n
n
+ - x" +
•
•
•
n
= £ -
(/)
b n\
3!
L — lim
if
1
= lx| lim
lim
(n
+ 1)! x"
n-oo n
every value of x, the Maclaurin series converges everywhere.
Find a Maclaurin-series expansion for
I We have
*n+
L >
L = + x,
< 1.
If
= 0.
Since this ratio is less than unity for
1
or
the series
no conclusion can be inferred.
Using the ratio test, we have
f(x) — e
+
1
The interval of convergence is (— 00, 00).
2x
.
f( x ) = e~
2x
f(0) =
f'(x)= -2e~
f"(x) = 2 e2
2x
2x
f'"(x)= -2 e~
3
354
•
:
Determine the interval of convergence for the Maclaurin series obtained in the previous problem.
By the ratio test, the series
15.7
——x
xJ 4
z
2!
1
/'(0)= -2
/"(0) = 2
2x
2
/"'(0)= -2 3
1
INFINITE-SERIES SOLUTIONS
o2
'
so
2*
=
^3
2!
15.8
94
2"
- 2x + ±- x 2 - ±7 * 3 + 4 x 4 -
1
+ (-l)n -x n +
4!
3!
355
n!
Determine the interval of convergence for the Maclaurin series obtained in the previous problem.
« + i
2
Using the ratio test, we have
lim
(n
n-* + 00
x n+i
n]
+ 1)!
= |x|
lim
2"x"
The series converges for every
0.
n+
1
value of x.
15.9
Solve Problem 15.7 using the result of Problem 15.5.
I
We have from Problem 15.5 that
,-2x
= L-r(- 2*)"= I
» =
15.10
1
1
.
00
1
2
=
f{x) = In x.
and
1
f{\) = In 1
/'(1)
Recalling that
0!
n~0
=
=—
1
f'{x)
X
=0
n\
.
/<">(x)
,
.
= n(n — 1)!,
[n
> 1)
X
«=
/ '(i)(x - l)" =
(n
«
„—
0!
—
(-1)""
1
n
»
(-ly-Hn-mx-ir
n\
n\
1
1
1
2
3
(x-l)" = (x-l)--(x-l) 2 +-(x-l) 3 ----
n
so that In x does possess a Taylor series about
Does In x possess a Taylor series about
(n>l)
we find that
n
«!
f \\) = (-\r\n-\)\
...,
/ w (i)(x-i) = /(i)(x-i)°
= Z
x = 1.
x = 0?
Neither In x nor any of its derivatives exists at
series at
(-IT'Hn - 1)'
= K—!—i
and
00
I No.
.
2,
= 1, /"(1)=-1,
and
1
8
Thus,
1
Therefore we have
6
x = 1.
Determine whether In x possesses a Taylor series about
x
4
2
»
fix) = -,
X
15.13
111
1
)"= I -x "=l+x +-x + -x + -x +
I^(*
= on\
2!
3!
4!
= o"!
2
=
n
I Here
— 2x, we obtain
which is the series obtained in Problem 15.7.
x",
00
15.12
Replacing x with
Replacing x with x 2 in (/) of Problem 15.5, we get
**
15.11
S
= Y — x".
n\
Find a Maclaurin series for e*
I
x
(-l) n 2"
«=o
n!
e
x =
therefore, In x cannot possess a Taylor
x = 0.
Use the ratio test to determine those values of x for which the Taylor series found in Problem 15.11 converges,
f For the series of Problem 15.1
1
we have
(-l)"(x- l) n+1
a»+i
lim
= lim
n + 1
(-l)"
-1
^-
= lim
1)"
n^co n
1
+
=lx-
11
1
n
or, equivalently when
< x < 2.
We conclude from the ratio test that the series converges when |x — <
and x = 2 must be checked separately, since the ratio test is inconclusive at these points.
The points x =
1|
For
x =
we obtain
£
n=l
- = — £ -;
n = 1
n
1
this is the harmonic series, which is known to diveige.
1
CHAPTER 15
356
x = 2
For
£
we obtain
= - £
n
n=i
< x < 2.
Thus, the series of Problem 15.11 converges for
15.14
x = 2
Find a Taylor-series expansion around
I We have
which converges by the alternating-series test.
,
n
„=\
/(x) = lnx.
for
= ln2
/*(2) = 1
/(x) = lnx
/'(x^x-
/(2)
1
r(x)=-x- 2
/'"(x)
r(2)=-i
= 2x- 3
iv
f"(x)=-6x-*
In x
so
_
,
/ (2)=-|
1
x - 2 2
4
2!
1
= In 2 + - x - 2„, - V
2
=i
f'"(2)
1
x - 2 3
3
x - 2 4
4
3!
8
4!
+-
+
•
•
= In 2 + -(x - 2) - -(x - 2) 2 + l(x - 2) 3 - l(x - 2) 4 +
15.15
Determine the interval of convergence of the power series obtained in the previous problem,
f
Using the ratio test, we have
- If
2"M
= 2 |X
n
2 "(»+ l)(x-2)
(X
lim
n-* + oo
Thus the series converges for
For x = 0, the series is
In 2+1-2 + \ ~ i +
15.16
'
"
">
|x
or
< x < 4.
- (harmonic series), which diverges. For
In 2
l
x = 4,
the series is
which converges. Thus the series converges on the interval
x —
I We have
/(x) = In (1
f(x) — In (1
for
1
/(0) =
+ x)
f'(0)
+x
1
•
1
(1
f (x) = -
2
f'"(0)
+ x) 3
1-2-3
iv
1
-
= 2!
iv
(1
ln(l + x) =
=
/"(0) =
(1+x) 2
r\x) =
< x < 4.
+ x).
1
fix) =
/ (0)= -3!
+x) 4
x-^ + 2!^-3!^ +
1,1.1
—
= X ——x +-x
15.17
n- + oc 1 + 1
— 2| < 2
Find a Taylor-series expansion around
Hence
X - 2
rTT = 2l
^
I™
2|
n
x
--
+
+ (-ir
•
•
_1
(n-l)!-r +
+ (-l)"- -x n +
1
n
Determine the interval of convergence for the power series obtained in the previous problem.
Using the ratio test, we have
lim
n-* + X
1
+
= |x|
1
X
lim
n -. + x
M+l
= |x|.
The series converges absolutely for
and diverges for |x| > 1. Individual tests are required at x = 1 and x = — 1.
and is conditionally convergent. For x = — 1,
x = 1, the series becomes 1 — i + 3 — i +
the series becomes
—(1 +2 + 3 + 4+') ar>d is divergent. Thus the given series converges on the interval
|x|
<
1
For
'
•
'
-1 <x< 1.
15.18
Find a Maclaurin-series expansion for
f(x) — arctan x.
INFINITE-SERIES SOLUTIONS
I We have
f(x) = arctan x
357
f(0) =
1
= - 2 + x4 - X6 +
+ X2
3
5
f"(x) = -2x + 4x -6x + -2
4
f'"(x) = -2 + 12x -30x +
f'"(0)= -2!
/ (x) = 24x - 120x +
Ax) = 24 - 360x 2 +
/ (0) =
/ (0) = 4!
/'(*)=
1
/'(0)
.X
=
1
1
/"(0) =
-
iv
3
iv
•
v
•
/ (x) = -720x + -/ (x) = -720 + •••
/ (0) =
vi
vi
vii
arctan x = x
and
2!
6!
x3 +
3!
15.19
,
X
+
X
7i"
X
.2n+ 1
lim
n-» + x
-1
= -6!
X
r
n-
2n-l
1
2n-
2n+
<x<
In- 1- — y-^
1
x
2n_1
x'
In- 1
—
lim
lim
— 1)"~
|(
=x
„^4 x In + 1
=
'|
+
1
— 1)"| = 1, we have
|(
The series is absolutely convergent on the interval
.
Find a Taylor-series expansion around
x —
# We have
f(x) = sin x
for
=
0,-1,0
— +—
+ lx + — x 2 +
f(x)
•
•
= sin x.
/(0) =
/'(0)= 1
/"(x)=-sinx
/"(0) =
f'"(x)= -cosx
f'"(0)= -1
x4 +
x3
—x +
1
X3
,
5
x
•
•
+
X7
X5
+•• +(-l)"~
,2n- 1
X'
1
(In - 1)!
K
5!
7!
'
+
Determine the interval of convergence of the power series obtained in the previous problem.
,2n+ 1
Using the ratio test, we have
(2n
lim
(2/i
+ 1)!
Y
x
f(t)
= cosh
t.
n-> + or
I
- 1)!
1
x2
2 "" 1
= 0.
lim
„^ + 00 2n(2n + 1)
The series converges for
every value of x.
Find a Maclaurin-series expansion for
= cosh
/'(f) = sinh
I We have
/(f)
t
f(0) =
f
/'(0)
= sinh
/'"(f)
1
-
/"(0)=1
/"(f)- cosh f
/'"(0)
f
1010 -
=
The values of the derivatives form the cycles 1,0; 1,0;...; hence
cosh '= 1+0, +
(2
+
,3
3!
2!
15.23
'
,
3!
15.22
or
form cycles of 0, 1,0, — 1; hence
x =
The values of the derivatives at
and for x = 1, it becomes 1 — 3 + 5 — 7 +
— 1 < x < 1 and diverges elsewhere.
•
= cosx
/'(x)
sin x
x2 < 1
1.
For x = — 1, the series becomes — 1 + \ — 5 + 7 —
Both series converge; thus the given series converges for
15.21
(0)
Determine the interval of convergence for the Maclaurin series obtained in the previous problem,
f Using the ratio test and noting that
15.20
vii
/
+ 4!'
4
+ 5!' +
4
=1+ f + r +
2
f
2!
f
4!
6
f"
f
«
+
-=
°°
.2n
v
.?„M
Determine the interval of convergence of the power series obtained in the previous problem.
f
Using the ratio test, we have
n *-*oc
on ( — 00, 00).
2n + 2
lim
(2/i
(2/i)!
+ 2)!
i
t
f
2n
= 2 lim
I
t
„^ x (2/i + 2)(2/i + 1)
0.
The series converges
CHAPTER 15
358
15.24
x =
Find a Taylor-series expansion around
I
f{x) = cos x.
for
Differentiating the expansion obtained in Problem 15.20 yields
.
cos x = 1 -
5x 4
3x 2
7x 6
(2m-1)x 2
"- 2
1
(2n
7!
5!
3!
/
x4
x2
-^ + -z^ -=- + • +(-l)_B-1 -^^ -^—rr.— +
x6
'-2! + 4!-6! +
1)!
This series converges within the interval of convergence of the Maclaurin series for sinx, which, as determined
in
15.25
Problem 15.21, is all x.
Find a Taylor-series expansion around
I
x = \t
Setting
=
r
for
/(f)
in the result of Problem 15.20 yields
sin] ,. (;i) _!i£ +
+
|! S£ + ... +( _, r .i£ll
(2n-
Find a Maclaurin-series expansion for
f
'
= sinh
/(f)
4r
^+^+^+
2r
-)2n-l/
2n - 1
2-,
(2«-l)!
2
nt- 1
1)!
t.
Problem 15.22 yields
Differentiating the expansion obtained in
sinh( =
,2n- 1
K
7!
15.26
= sin|f.
=_+_+_+
6f'
.
..
£__
,2n- 1
.
..
=
This series converges within the interval of convergence of the original series, which, as determined in Problem
15.23, is everywhere.
15.27
x —2
Find a Taylor-series expansion around
f(x) — e*
for
f(x) = e
I We have
= $e*
f»( x = ±e*i2
15.28
2
f{2)
=e
f'(2)
= \e
2
=&
f"(2)
)
c«
.
x' 2
f'(x)
so that
2
l(x-2r ...
= cl+ _ (x _ 2) + -L_J.
+
+
I
_L_£r +
(x-2)"-
1
1
...
Determine the interval of convergence of the power series obtained in the previous problem.
(x
Using the ratio test, we obtain
- 2)
n
2"/,!
- 1)!
2"
'(/j
(x
- If
lira
= - |x — 2|
2
'
lim
„
-
+ ao
- = 0.
The series converges
n
for every value of x.
15.29
Find a Taylor-series expansion around
=
t
for
1
/(f)
= 2r 2
3
/'(t) = (-2)(2r
4
f"(t) = (2)(3!«
I We have
= 2/t 2
=2
/'(l) = (-2)(2!)
/"(1) = (2)(3!)
/(l)
f(t)
)
l
= (-2)(4!f- 5
/'"(f)
Hence
2f-
2
= 2 + (-2)(2!)(f- 1) +
(2)(3!)
2!
= (-2)(4!)
/'"(!)
)
2
(f-D
+
"'
v
.
-2)(4!)
"
(t-l) 3 + --
3!
'
= 2 - (2)(2)(f - 1) + (2)(3)(f - l) 2 - (2)(4)(r - l) 3 +
15.30
•
•
+ (-1)"(2)(« + l)(f - If + •
•
Determine the interval of convergence of the power series obtained in the previous problem.
2(« + 2)(f
Since
- 1)"
lim
2(«
+
=
\t
— 1| lim
l)r"
n-»oc
n + 2
H + I
=
|f
— 1|,
the series converges for
\t
—
f
= 2,
1|
< 1
or
< t < 2.
For
t
= 0,
series becomes
the power series becomes
2(1 — 2 + 3 — 4 H
),
2(1
+ 2 4- 3 + 4 +
which also diverges.
•)»
which diverges. For
Thus, the interval of convergence is
the power
<
t
< 2.
INFINITE-SERIES SOLUTIONS
15.31
- l/(x 2 - I)
Q(x) =
Determine whether
P(x) = x/(x
and
2
- 1)
359
x = 0.
are analytic at
We need to determine whether the Maclaurin series for these functions converge in some interval around
f
x = 0.
Using the geometric-series expansion
y = x2
w tn
= Z y
'
we wr tc
'
'
n=
ow-F^r-rh?
P(x) =
and
X (x
—x
—
"
x2 - 1
" 1 - x2
2
=
)"
1
+ X 2 + X4 + X6 +
= —x £ (x 2 )" — —x — x 3 — x 5
n=
The geometric series converges when \y\ < 1, so the series expansions for P(x) and Q(x) converge when
2
|x
< 1 or when - < x < 1. Therefore, both functions are analytic at x = 0.
1
|
15.32
Q(x) = x/(x
Determine whether
f
2
+ 4)
x = 0.
analytic at
is
We need to determine whether the Maclaurin series for Q(x) converges in some interval around
1
Using the geometric-series expansion
— X >'"
-y
1
i
x
2
l-(-.v 2 /4)
+4
we have
n =
—4/4
= -x
I
Q(x)
— x 2 /4,
y =
w tn
4 n ^o
x3 H
x5
x
f(x) — x/(x +1)
Determine whether
is
7
+
-
256
64
16
The geometric series converges when |^| < 1, so this Maclaurin series converges when
— 2 < x < 2. Therefore, Q(x) is analytic at x = 0.
15.33
x = 0.
— x 2 /4| <
|
or when
1
x = 1.
analytic at
x = 1, f(x) is analytic
That is, it has a Taylor-series expansion around x — 1 that converges in some interval centered at
x = 1. In particular, it follows from the geometric-series expansion of the previous problem with
I
Since f(x) is the quotient of two polynomials with a denominator that is not zero at
there.
y = -j(x - 1)
that
(x
x+1
- 1) +
- 1)
(x
1
2 + (x-l)
(x-1)
I
2
^ =
+
n=
1
~2 +
x-1
~
(x-l) 2
^
2
2
Determine whether
I
3
(x-
l)
2
j(x — 1)1 <
= (t - l)/f
g(t)
+
is
(x
V
f
= 1.
2
z
t
A
2
„=,
n+1
x+
- = x
x+
1
1
— — XJ
1
= 1.
x = t — 1,
t
—
t
for g(t) converges in some interval centered
1
-1
so that
+
£ — x) = £= (— l)"x
t
We now seek a Maclaunn-
.
1
with convergence on the interval
,
|x|
< 1.
Substituting
n
—
for x then yields
< 2.
n
n
(
°°
1
-= £ (- l)"(f — l) n+
',
with convergence on the interval
|r— 1|<1
or
n =
f
<
x
—
x+1
Using the geometric-series expansion, we have
t
f-1
n =
1
n=
(
7 n+1
2-,
lX
t
series expansion for
- (-l)"(x-l) n
>n+ 1
— < x < 3.
or when
analytic at
To simplify the algebra, we set
i-Wx-iy
3
We need to show that the Taylor-series expansion around
at
-z
- 1)
n =
4
1,
21 _[_(l/2)(x-l)]
2
M
This series converges when
15.34
—
1
I
+
l-[-(l/2)(x-l)]
2
I
I
I
Thus g(t) is analytic at
t=l.
ORDINARY AND SINGULAR POINTS
15.35
Determine whether
#
x =
is
an ordinary point for the differential equation
This equation is in the standard form
of these functions
is its
y" + P(x)y' + Q{x)y = 0,
with
y" - xy' + 2y = 0.
P(x)
= -x
own Maclaurin series with infinite radius of convergence, so
x =
Q{x) - 2.
and
is
Each
an ordinary point.
—
360
CHAPTER 15
15.36
Determine whether
f
.x
=
is
an ordinary point for the differential equation
y" + P(x)y' + Q(x)y = 0,
This equation is in the standard form
y" — xy — 0.
P(x) =
with
15.37
x = 5
Determine whether
is
Q{x) = x.
and
x =
these functions is its own Maclaurin series with infinite radius of convergence, so
is
Each of
an ordinary point.
an ordinary point for the differential equation of the previous problem.
f Polynomials and constants are analytic everywhere.
P(x) =
Since
is
a constant and
Q(x) = x
is
x = 5;
x = 5
and
degree polynomial, both functions are analytic everywhere, and in particular at
thus
a first-
every other point are ordinary points.
15.38
x =
Determine whether
is
an ordinary point for the differential equation
I
Dividing the differential equation by
y" + P(x)y' + Q(x)y = 0,
Q(x)
15.39
is
(see
x =
Determine whether
P(x) =
with
x =
analytic at
is
x
2
+ 4,
and
Problem 15.32),
we obtain
Q(x) = x/(x
x =
y"
2
+ 4)y" + xy = 0.
2
x
y = 0. This is in the standard form
x + 4
Because P(x) is analytic everywhere and
+ —.l
+ 4).
an ordinary point.
is
an ordinary point for the differential equation
#
(x
(x
2
— \)y" + xy' — y = 0.
x
1
y'
y = 0. This has the
5
x — 1
x2 — 1
2
standard form y" + P(x)y' + Q(x)y = 0, with
P(x) - x/(x - 1)
and Q(x) = - l/(x 2 - 1). Since both these
functions are analytic at
x =
(see Problem 15.31), that point is an ordinary point.
Dividing the differential equation by
15.40
x =1
Determine whether
is
v
2
- 1.
we obtain
y" H
2
an ordinary point for the differential equation in the previous problem.
and Q(x) = - l/(x 2 - 1). Since P(l) and Q{\) are undefined, it follows that
neither function has a Taylor series about
x — 1. so neither function is analytic and x = 1
is not an
f As before,
P(x) = x/(x
ordinary point.
It is
and
15.41
- 1)
Since
a singular point.
— \) 2 Q{x) — —
(x
2
x =
Determine whether
also analytic at
is
1
is
1
(x — l)P(x) = x/(x +1)
is
x =
x — 1
1.
=
is
1
Taylor series around
15.42
Determine whether
x
(x
(see
Problem 15.33)
— l) 2 y" — 2xy' = 0.
P(x) =
with
— 2x/(x — l) 2
and
x =
and is not analytic there.
— l)P(x) = — 2x/(x — 1) also does not have a
Since P(l) in undefined, P(x) does not have a Taylor series at
x
x = 1
a regular singular point.
is
2x
—2=/ = 0,
(x - l)
y"
This differential equation has the standard form
Q(x) — 0.
analytic at
an ordinary point for the differential equation
f
Therefore,
follows that
it
1
not an ordinary point.
Furthermore,
x = 1,
not a regular singular point; it is an irregular singular point.
=
so
is
x =
is
1
(x
an ordinary point for the differential equation
e
2x y"
+ y — 0.
we obtain y" + e 2x y — 0, which is in standard form. Since e~ 2x has
x =
(see Problems 15.7 and 15.8),
a Maclaurin-series expansion that converges in some interval centered at
I
e~
15.43
Dividing the given equation by e
2x
is
analytic there and
x =
x =
is
Determine whether
f
2x
is
,
an ordinary point.
an ordinary point for
This differential equation is in standard form, with
y" — (In x)y
P(.x)
= In x
— 0.
and
P(x) does not possess a Maclaurin-series expansion and is not analytic at
an ordinary point. In addition, xP(x) is also undefined at
15.44
Determine whether
x = 2
is
x = 0,
so
Q(x) = 0.
x = 0.
x =
is
Since P(0) is undefined,
Accordingly,
x =
is
not
not a regular singular point.
an ordinary point for the differential equation of the previous problem.
x = 2 that converges in an interval centered at
P(x) does have a Taylor-series expansion around
is analytic everywhere, including
x = 2 (see Problems 15.14 and 15.15); thus, P(x) is analytic there. Q(x) =
at
x = 2, so x = 2 is an ordinary point.
I
15.45
Determine whether
„
2
—
x =
or
x = 1
is
an ordinary point for the differential equation
3
/ +V + -,
Try y = o.
x
x(x —
f
1)
Neither point is an ordinary point: At
P(x) is analytic, Q(x) is not.
x = 0,
both P(x) and Q(x) are undefined; at
All other points are ordinary points.
x = 1,
although
^
INFINITE-SERIES SOLUTIONS
x =
The point
X Q( X ) = :
(x
is
Determine whether
15.46
3
= -3x(l + 3x + 6x 2 +
•)
and
for
f
<
Ixl
x = 0.
are analytic at
1
'
'
an irregular singular point, however, because the product
x = 1
undefined at
xP(x) = 2
a regular singular point, since
= -3x(l -x)
ttt
J
— 1)
x= 1
The point
is
is
3x
->
361
(x
- \) 2 Q{x) =
3
x(x -
1)
and hence is not analytic there.
=
2
an ordinary point for
is
1
t
y + 2y + (t
I
f-1
2
This differential equation has the standard form
- t)y = 0.
2
y + -y y H
y = 0.
Since 2/t
2
and
(t
- \)/t
are
each a rational function (that is, a polynomial divided by a polynomial), and since neither denominator is
zero at
15.47
t
— 1,
both functions are analytic there. Thus,
=
expansions around
t
Determine whether
x =
=
an ordinary point. (The Taylor-series
for these functions are derived explicitly in Problems 15.29 and 15.34.)
1
I
xy = 0.
-I-
x
y" -I
^
(x
and
x/(x + l)
2
+ \y
x =
is
x =
Determine whether
is
an ordinary point for
(x
2
^
x z^-,
+ 2x
2
z
+ 1)
r y
and
Q(x) =
t
—x ++ — —
1
y" -\
-1
-y—^
x
2
1
— 0.
Since both
x = 0,
2/(x
+
2
1 )
both functions are
+ 2x)y" + (x + l)y' — y = 0.
are undefined at
+ 2x
1
,
y'
y
- 2z
x
x+1
(x
an ordinary point.
#_.,.„
This differential equation has the standard form
P(x) =
/H
are rational functions, and since neither denominator is zero at
analytic there. Thus,
is
+ 2/
2
This differential equation has the standard form
15.48
is
1
2
(x 4- \) y"
an ordinary point for
is
t
2x
x = 0,
x
,
2
L
+ 2x
y = 0.
Since both
'
x =
neither is analytic there and
i
not an ordinary point.
The products
(x
— 0)P(x) =
x+1
'-
and
x = 0.
denominators at
(x
— O) 2 0x) =
—x
are both rational functions with nonzero
x + 2
Thus, both products are analytic at x = 0,
x + 2
which implies that
x =
is
a regular
singular point.
x = — 1
Determine whether
15.49
is
an ordinary point for the differential equation of the previous problem.
I Both P(x) and Q(x) are rational functions with nonzero denominators at
15.50
analytic there, so
x = —1
is
an ordinary point.
Determine whether
x =
is
an ordinary point for
I Here
.
15.51
P(x) =
—x
value of x, in particular
Determine whether
x = — 1.
Thus both functions are
y" — xy' + 2y = 0.
and Q(x) = 2 are both polynomials and so are analytic everywhere. Therefore, every
x = 0, is an ordinary point.
x = 1
or
x = 2
is
an ordinary point for
(x
2
— 4)y" + y = 0.
x 2 — 4. Then P(x) =
and
x = 1, this point is an ordinary point. At
Since both P(x) and Q(x) are analytic at
Q(x) — l/(x — 4).
x = 2, however, the denominator of Q(x) is zero; hence Q(x) is not analytic there. Thus, x — 2 is not an
I We first put the differential equation into standard form by dividing by
2
ordinary point but a singular point.
Note that
(x
- 2)P(x) =
and
(x
- 2) 2 Q(x) = (x - 2)/(x + 2)
are analytic at
x = 2,
so that
x = 2
is a regular singular point.
15.52
Determine whether
f
Dividing by 2x
x =
15.53
yields
is
(x
2x 2 /' + 7x(x + I)/ — 3y = 0.
an ordinary point for
P(x) = 7(x + l)/2x
(both denominators are zero there),
Note that
is
2
x =
- 0)P(x) = |(x + 1)
and
x =
(x
Q{x) =
and
is
— 3/2x 2
.
As neither function is analytic at
not an ordinary point but, rather, a singular point.
- 0) 2 £>(x) = -|
are both analytic at
a regular singular point.
Determine whether
x =
is
an ordinary point for
x 2 y" + 2/ + xy = 0.
x - 0;
thus,
x =
:
362
=
CHAPTER 15
D
P(x) = 2/x
f Here
2
Q(x) = 1/x.
and
15.54
x = 0,
Neither of these functions is analytic at
an ordinary point but a singular point. Furthermore, since (x — O)P(x) = 2/x
is not a regular singular point either; it is an irregular singular point.
x =
x =
is not
x = 0,
so
not analytic at
is
Determine which points are not ordinary points for the differential equation
(x
2
- 4x + 3)y" + (x - 2)y' - y = 0.
f
— -- +
x
This differential equation has the standard form
P(x) = —
x^
x-2
— 4x + 3
x = 1
when
4x
1
y'
s
x 2 - 4x + 3
3
y = 0.
Both
are rational functions with denominators that are zero only
x — 4x + 3
These are then the only two points that are not ordinary points for the given
x = 3.
and
2
=
x2
-1
Q(x) = —z
and
y" H
differential equation.
RECURSION FORMULAS
15.55
Find a recursion formula for the coefficients of the general power-series solution near
x =
of
y" — xy' + 2y = 0.
I
It is
x =
shown in Problem 15.35 that
is
an ordinary point for this differential equation, so we may solve
We assume
by the power-series method.
+ a x + a 2 x 2 + a 3 x 3 + a 4 x 4 + •• + a n x" + a„ +l x n+l + a n+2 x" +2 +••
y = a
Differentiating termwise, we have
y'
= a, + 2a 2 x
(7)
x
-I-
+ 4a 4 x 3 +
2
3a 3 x
y" = 2a 2 + 6a 3 x + 12a 4 x
2
+
•
•
+ na„x n
•
+ n(n - l)a„x
~ x
n ~ 2
+ (n + l)a B+1 x" + (n + 2)a n + 2 x" +i
+
+ (n + l)(n)a n+ ,x"~ + (n + 2)(n
l)a n + 2X
'
-I-
(2)
n
+
•
•
•
•
(5)
Substituting (/), (2), and (3) into the differential equation, we find
~
+ n(n - \)a„x nl + (" + l)(n)an + x n + (n + 2)(« + \)a n + 2 x" + ]
-x[a, + 2a 2 x