Differential Equations (Cliffs Quick Review)

Cliffs Quick Revie w

Differential Equations by Steven A . Ledu c

Series Editor Jerry Bobrow, Ph .D .

Cliffs Quick Revie w

Differential Equations by Steven A . Ledu c

Series Editor Jerry Bobrow, Ph .D .

CliffsQuickReviewTM Differential Equations Published by: Wiley Publishing, Inc . 909 Third Avenue New'York, NY 1002 2 www.wiley.co m Copyright 0 1995 Wiley Publishing, Inc ., New York, New York Library of Congress Control Number : ISBN : 0-8220-5320-9 Printed in the United States of Americ a 1098765 4 1 B/ST/QR/QS/I N No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by an y means, electronic, mechanical, photocopying, recording, scanning, or otherwise, except as permitted under Section s 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, 222 Rosewoo d Drive, Danvers, MA 01923, 978-750-8400, fax 978-750-4470 . Requests to the Publisher for permission should b e addressed to the Legal Department, Wiley Publishing, Inc ., 10475 Crosspoint Blvd ., Indianapolis, IN 46256 , 317-572-3447, fax 317-572-4447, or e-mail permcoordinator@wiley . cor n LIMIT OF LIABILITY/DISCLAIMER OF WARRANTY : THE PUBLISHER AND AUTHOR HAVE USE D THEIR BEST EFFORTS IN PREPARING THIS BOOK . THE PUBLISHER AND AUTHOR MAKE NO REPRESENTATIONS OR WARRANTIES WITH RESPECT TO THE ACCURACY OR COMPLETENESS OF TH E CONTENTS OF THIS BOOK AND SPECIFICALLY DISCLAIM ANY IMPLIED WARRANTIES OF MERCHANTABILITY OR FITNESS FOR A PARTICULAR PURPOSE . THERE ARE NO WARRANTIES WHICH EXTEND BEYOND THE DESCRIPTIONS CONTAINED IN THIS PARAGRAPH . NO WARRANTY MAY B E CREATED OR EXTENDED BY SALES REPRESENTATIVES OR WRITTEN SALES MATERIALS . THE ACCU RACY AND COMPLETENESS OF THE INFORMATION PROVIDED HEREIN AND THE OPINION S STATED HEREIN ARE NOT GUARANTEED OR WARRANTED TO PRODUCE ANY PARTICULA R RESULTS, AND THE ADVICE AND STRATEGIES CONTAINED HEREIN MAY NOT BE SUITABLE FO R EVERY INDIVIDUAL . NEITHER THE PUBLISHER NOR AUTHOR SHALL BE LIABLE FOR ANY LOSS O F PROFIT OR ANY OTHER COMMERCIAL DAMAGES, INCLUDING BUT NOT LIMITED TO SPECIAL , INCIDENTAL, CONSEQUENTIAL, OR OTHER DAMAGES . FULFILLMENT OF EACH COUPON OFFE R IS THE RESPONSIBILITY OF THE OFFEROR . Trademarks : Wiley, the Wiley Publishing logo, Cliffs, CliffsNotes, the CliffsNotes logo, CliffsAP, CliffsComplete , CliffsTestPrep, CliffsQuickReview, CliffsNote-a-Day and all related logos and trade dress are registered trademarks o r trademarks of Wiley Publishing, Inc ., in the United States and other countries . All other trademarks are property o f their respective owners. Wiley Publishing, Inc ., is not associated with any product or vendor mentioned in this book . For general information on our other products and services or to obtain technical support, please contact our Custome r Care Department within the U .S . at 800-762-2974, outside the U .S . at 317-572-3993, or fax 317-572-4002 . Wiley also publishes its books in a variety of electronic formats . Some content that appears in print may not be avail able in electronic books .



CONTENTS

REVIEW AND INTRODUCTION Differentiation Partial Differentiation Integration Techniques of Indefinite Integration Integration by substitution Integration by parts Partial Integration Introduction to Differential Equations

FIRST-ORDER EQUATIONS Exact Equations Integrating Factors Separable Equations Homogeneous Equations Linear Equations Bernoulli's Equation

SECOND-ORDER EQUATIONS Linear Combinations and Linear Independence Linear Equations Homogeneous Equations Homogeneous Linear Equations with Constan t Coefficients The Method of Undetermined Coefficients Linear combinations of n functions Variation of Parameters The Cauchy-Euler Equidimensional Equation Reduction of Order Type 1 : Second-order equations with th e dependent variable missing Type 2 : Second-order nonlinear equations with th e independent variable missing

DIFFERENTIAL EQUATIONS

1 1 8 10 12 12 14 16 18

29 29 34 41 47 52 60

63 63 66 67 71 77 79 90 97 100 100 105

CONTENTS

Type 3 : Second-order homogeneous linear equation s where one (nonzero) solution is known

POWER SERIES Introduction to Power Series Taylor Series Power Series Solutions of Differential Equations First-order equations Second-order equations

THE LAPLACE TRANSFORM Linear Transformations The Laplace Transform Operator Using the Laplace Transform to Solv e Differential Equations

A FEW APPLICATIONS Applications of First-Order Equations Orthogonal trajectories Radioactive decay Newton's Law of Cooling Skydiving (part I) Applications of Second-Order Equations Skydiving (part II) Simple harmonic motion Damped oscillations Electric circuits and resonance

10 7

11 3 11 3 11 7 123 123 128

13 5 13 5 13 7 14 8

15 7 15 7 15 7 162 165 167 17 1 17 1 17 4 17 8 18 2

CLIFFS QUICK REVIEW

REVIEW AND INTRODUCTIO N

Before embarking on a study of differential equations, it i s essential to solidify your technical facility with differentiation an d integration . This section serves to review the definitions an d techniques of differentiation and integration and then to introduc e you to the study of differential equations .

Differentiatio n Given a function y = f (x), its derivative, denoted by y' or dy /dx , gives the instantaneous rate of change of f with respect to x . Geometrically, this gives the slope of the curve (that is, the slope o f the tangent line to the curve) y = f (x) . y = f (x )

tangent line to curv e atx=xo

■ Figure 1

■

The second derivative identifies the concavity of the curve y = f (x) . A portion of a differentiable curve y = f (x) from x = a to x = b is said to be concave up if the curve lies above its tangent line s between a and b, and concave down if it lies below its tangent lines .

DIFFERENTIAL EQUATIONS

REVIEW AND INTRODUCTIO N

A curve y = f (x) is concave up at those points x where the secon d derivative is positive, and concave down where the second derivative is negative . Points where the concavity changes are called inflection points and are located at those points x0 where f " (x0 ) = 0 but f"' (x 0)  0 . concave u p f " (x)> o curve abov e its tangent lines

■ Figure 2 ■ Table 1 below lists the most frequently used properties o f derivatives and, for later reference, the corresponding properties o f integrals .

CLIFFS QUICK REVIE W



REVIEW AND INTRODUCTIO N

Table 1 COMPUTATIONAL PROPERTIES O F DIFFERENTIATION AND INTEGRATIO N Integration

Differentiation Linearity :

Linearity :

d(ku) = k du

f k du = k f du = ku

d(u + v) = du +dv

(du + dz)) = u + v

Product rule :

Integration by parts :

d(uv) = u dv + v du

f u dv = uv — f v du

Quotient rule : u

d V

=

vdu — udv v2

Chain rule : (u v)' = (u.' v) v'

Integration by substitution : (u' . 0 • v'

u

o v

In addition to being familiar with the definitions and fundamen tal properties, you should, of course, be able to actually differentiat e a function . Although Table 2 below does not contain ever y differentiation formula, it will probably suffice to differentiat e almost every function you are likely to encounter in practice . Again , for later reference, integration formulas are listed alongside th e corresponding differentiation formulas . (Note : To avoid the repeti tion of writing "+ c" after every result in the right-hand column, th e arbitrary additive constant c has been omitted from each of th e integration formulas, as in Table 1 above . ) DIFFERENTIAL EQUATIONS



REVIEW AND INTRODUCTION

Table 2 DIFFERENTIATION AND INTEGRATION FORMULA S Differentials

Integral s

d(constant) = 0

ti n +

d(u") = nu"-' du

u"du

d (e") = e" d u

e" du = e"

n+1

1 d(ln u) = —d u u

f !du u

d(sin u) = cos u du

f cosudu=sin u

d (cos u) = –sin u du

d (cot u) = – csc 2 u du d (sec u) = sec u tan u d u d(csc u) = –csc u cot u du

f sec' u du = tan u f csc'- u du = —cot u f sec u du = In sec u + tan u csc u du = —ln l csc u + cot u j

du

du = arcsin u

d(arctan u) = + u du l -

, du = arctan u 1 + u-

4

—

in l u l

sin u du = —cos u

d(tan u) = sec u du

d(arcsin u) =

(nom—1 )

u2

CLIFFS QUICK REVIEW



REVIEW AND INTRODUCTIO N

Example 1 : Differentiate each of the following :

(b)

J' = xZe x

(c)

y = In x/x

(e)

y = \/x2

(f)

y = sin(x2 )

(g)

y = sin 2 x

(h)

tanx y = e

(i)

y = csc(sin

+

1

The solutions are as follows : (a)

y ' =6x— 5

(b)

Using the product rule, y' = x 2

(c)

By the quotient rule,

• e x + e x • 2x = xe v(x

+ 2)

1

x•x—lnx•1

y' =

1—ln x x2

All of the remaining parts use the chain rule (as embodied in the formulas in Table 2) . (d)

y' =4(x 3 +x— 1) 3 (3x2 + 1 )

DIFFERENTIAL EQUATIONS



REVIEW AN D INTRODUCTIO N

+ 1)1/2 ( e ) y = (x2

+ 1)-' /2 . 2x =

y' =

(f)

y' = 2x cos(x 2 )

(g)

y = (sinx) 2 = y' = 2 sinx cosx = sin 2x

(h)

y' = etan .r sect x

(i)

y' _ —csc(sin f) cot(sin

cos • 2\5c

x \/x '- + 1

■

Example 2 : What is the equation of the tangent line to the curv e y = e-' In x at the point (1, 0) ? The first step is to find the slope of the tangent line at x = 1 , which is the value of the derivative of y at this point : slope at point (1, 0) =

dy

dx

1

= e x . — + lnx•e k

x

= e x= l

Since the point-slope formula says that the straight line with slop e m which passes through the point (x0 , yo) has the equatio n Y — Yo =m (x—xo )

the equation of the desired tangent line is y = e(x — 1) .

Example 3 : Is the curve y = arcsin down at the point (4, 6) ?

■

concave up or is it concav e

Concavity is determined by the sign of the second derivative .

6

CLIFFS QUICK REVIEW



REVIEW AN D

INTRODUCTION

Since d(arcsin u) =

1 vl — u 2

du

the first derivative of y = arcsin V.- i s 1

1 v/-x-

\/1

1 (x — x-) - ' — 2 r—x2

Its second derivative is therefore 1 4 (x —

x 2 ) - 3/2 (1

2x — 1

2x)

4\/(x _ x- )

For x = 4'- , the denominator in the expression above for y" is positive (as it is for any x in the interval 0 < x < 1), but the numerator i s negative . Therefore, y"(4) < 0, and the curve is concave down at th e point (4, ) . ■

Example 4 : Consider the curve given implicitly by the equatio n 3x2y—y=x+

1

What is the slope of this curve at the point where it crosses the x axis ? To find the slope of a curve defined implicitly (as is the cas e here), the technique of implicit differentiation is used : Differentiat e both sides of the equation with respect to x ; then solve the resultin g equation for y' . 3xy-y3=x+ 1

(3x2y' + 6xy) — 3y y' = 1 y' (3x 2 —

3y2 )

= 1 — 6xy 1 -6xy

Y'

DIFFERENTIAL EQUATIONS

3(x2 _ Y )



REVIEW AN D INTRODUCTION

The curve crosses the x axis when y = 0, and the given equatio n clearly implies that x = -1 at y = 0 . From the expression directl y above, the slope of the curve at the point (—1, 0) i s 1

1 — 6xy

Y ' I(-1,0)

=

3(x2

y2 )

1u,

3

■

Partial Differentiatio n Given a function of two variables, f (x, y), the derivative with respec t to x only (treating y as a constant) is called the partial derivative off with respect to x and is denoted by either of /8x or f.. Similarly, th e derivative of f with respect to y only (treating x as a constant) i s called the partial derivative off with respect toy and is denoted b y either of /cry or f, . The second partial derivatives

off come

in four types : Notation

■

Differentiate f with respect tox twice . (That is, differentiate f with respect to x ; then differentiate the result with respec t to x again . )

a2f n►x- or j ,

■ Differentiate f with respect toy twice . (That is, differentiate f with respect toy ; then differentiate the result with respec t toy again . )

af ay

or f., .

Mixed partials : ■

8

First differentiate f with respect tox ; the n differentiate the result with respect toy .

of ay ax

or L.,.

CLIFFS QUICK REVIEW



REVIEW AND INTRODUCTIO N

■ First differentiate f with respect toy ; then differentiate the result with respect tox .

a2f axay

or

f .,_

For virtually all functions f (x, y) commonly encountered in practice , will be identical to f, .; that is, the order in which the derivative s are taken in the mixed partials is immaterial .

f,

Example 5 :

Iff (x, y) = 3x2y + 5x — 2y 2 +

1, find f., f, .,

an d

First, differentiating f with respect to x (while treating y as a constant) yields f,-=6xy+ 5 f

Next, differentiating constant) yields

with respect to y (while treating x as a f .=3x2—4y

The second partial derivative with respect to x ; therefore,

means the partial derivative o f

a

a

fit = (f,)-, =~x(f,-)=c~x(6xy+5)=6Y

The second partial derivative f, ., . means the partial derivative of f. with respect toy ; therefore , fn .= (f .),,=

The mixed partial toy ; therefore,

f

a

(f. ) =

Y

( 3x2—4y ) = - 4

means the partial derivative

off.

with respec t

(f,-), . = ~ (f = 1( 6xy +5) =6 x

DIFFERENTIAL EQUATIONS

9



REVIEW AN D INTRODUCTION

x

The mixed partial f, means the partial derivative off. with respec t to .x ; therefore, a

f«=(f3 = x (fr)= Note that

fa. = f

as expected .

a ax

3x'--4y)=6 x

■

Integratio n Indefinite integration means antidifferentiation ; that is, given a function f (x), determine the most general function F(x) whos e derivative is f (x) . The symbol for this operation is the integral sign , followed by the integrand (the function to be integrated) and a differential, such as dx, which specifies the variable of integration . On the other hand, the fundamental geometric interpretation o f definite integration is to compute an area . That is, given a functio n f (x) and an interval a < x < b in its domain, the definite integra l off from a to b gives the area bounded by the curve y = f (x), the x axis, and the vertical lines x = a and x = b . The symbol for thi s operation is the integral sign with limits of integration (a and b) , followed by the function and the differential which specifies th e variable of integration .

f,

r,

■ Figure 3 ■

10

CLIFFS QUICK REVIEW

REVIEW AND INTRODUCTIO N

From their definitions, you can see that the processes o f indefinite integration and definite integration are really very different . The indefinite integral of a function is the collection of function s which are its antiderivatives, whereas the definite integral of a function requires two limits of integration and gives a numerica l result equal to an area in the xy plane . However, the fact that bot h operations are called "integration " and are denoted by such simila r symbols suggests that there is a link between them . The Fundamental Theorem of Calculus says that differentiatio n (finding the slope of a curve) is the inverse operation of definit e integration (finding the area under a curve) . More explicitly, Part I of the Fundamental Theorem says that if a function is integrated (t o form a definite integral with a variable upper limit of integration) , and the result is then differentiated, the original function i s recovered ; that is, differentiation "undoes" integration . Part I I gives the connection between definite and indefinite integrals . I t says that a definite integral can be computed by first determining a n indefinite integral (so computing the area under a curve is done b y antidifferentiating) . The Fundamental Theorem of Calculus (Part I) : Iff is continuous, then d dx

f f (t) dt = f (x) .

The Fundamental Theorem of Calculus (Part II) : If f is continuous with antiderivative F, then Jf(x) dx = F(b) — F(a) .

DIFFERENTIAL EQUATIONS



REVIEW AND INTRODUCTIO N

Example 6 : Evaluate the integra l (x4 —3x2 +x— 1) dx Using the first integration formula in Table 2, every functio n whose derivative equals f (x) = x4 — 3x2 + x — 1 is given b y (x4 — 3x 2 +x— 1)dx= .k x5 —x3 +4x2 — x+ c where c is an arbitrary constant .

■

Techniques of Indefinite Integratio n

Integration by substitution. This section opens with integration by substitution, the most widely used integration technique, illustrate d by several examples . The idea is simple : Simplify an integral b y letting a single symbol (say the letter u) stand for some complicate d expression in the integrand . If the differential of u is left over in the integrand, the process will be a success .

Example 7 : Determine

f x,,/x2 + 1 dx Let u = x2 + 1 (this is the substitution) ; then du = 2x dx, and th e given integral is transformed int o

f ~x 2 + 1 xdx= f

. 'du =

which transforms back to 4 (x2 + 1) 3 / 2 + c .

u h/ 2 du =

• 72 u 3/2 + c

■

CLIFFS QUICK REVIEW



REVIEW AN D INTRODUCTION

Example 8 : Integrate sin 3 x cosx dx Let u = sin x; then du = cos x dx, and the given integral become s

f ll 3 du =

u 4 +c =

Example 9 : Evaluate

dx :

sin 4 x+ c

f tan x dx

First, rewrite tan x as sin x/cosx, then let u = cosx, du = –sin x

f tanxdx=

sm x cosx ~

f

duu

= —InIul +c = —ln~cosx~ + c ■

Example 10: Evaluate xe x- dx

Let u = x2 ; then du = 2x dx, and the integral is transformed int o fxe2dx=feh1du=eh1+c=e+c

■

Example 11 : Determine

f sect x tan x dx Let u = sec x; then du = sec x tan transformed int o

x dx,

and the integral i s

f sec t x tan x dx = f sec x • sec x tan x dx = f u du = u 2 + c = 2 sec t x + c

DIFFERENTIAL EQUATIONS

■



REVIEW AND INTRODUCTIO N

Integration by parts . The product rule for differentiation say s d (uv) = u dz► + v du . Integrating both sides of this equation give s uv = f u dv + f v du, or equivalentl y

f u dz► = uv — f v d u This is the formula for integration by parts . It is used to evaluate integrals whose integrand is the product of one function (u) and th e differential of another (dv) . Several examples follow .

Example 12: Integrate

f xe dx Compare this problem with Example 10 . A simple substitutio n made that integral trivial ; unfortunately, such a simple substitutio n would be useless here . This is a prime candidate for integration by parts, since the integrand is the product of a function (x) and th e differential (e x. dx) of another, and when the formula for integratio n by parts is used, the integral that is left is easier to evaluate (or, i n general, at least not more difficult to integrate) than the original . Let u = x and dz► = ex dx ; then v=ek

u=x du = dx

dv = et dx

and the formula for integration by parts yield s

5 u dv = uv — 5 v du 5 xe x dx = xe x — 5 e dx =xe x — e x, + c = e x (x

14

— 1)+c

■

CLIFFS QUICK REVIE W



REVIEW AND INTRODUCTIO N

Example 13 : Integrate

f x cos x dx Let u = x and dv = cos x dx ; the n u=x

v = sin x

du=dx

dzv=cosxdx

The formula for integration by parts give s

f

f u dv = uzv – f v d u x cosxdx =xsinx –

sin xdx

= x sin x + cos x + c

Example 14: Evaluate

f

Inxdx

Let u = inx and dzv = dx ; the n u=lnx

zv= x

1

du = – dx dv = dx x

and the formula for integration by parts yield s

f u dzv = uv – f v du

f lnXdx=xtnx— f

i x--dx

=xlnx –x + c

DIFFERENTIAL EQUATIONS

x

■



REVIEW AND INTRODUCTIO N

Partial Integratio n Suppose it is known that a given function f (x) is the derivative of some function F(x) ; how is F(x) found? The answer, of course, is to integrate f (x) . Now consider a related question : Suppose it is known that a given function f (x, y) is the partial derivative with respect to x of some function F(x, y) ; how is F(x, y) found? The answer is to integrate f(x, y) with respect to x, a process I refer to as partia l integration . Similarly, suppose it is known that a given function f(x, y) is the partial derivative with respect to y of some functio n F(x, y) ; how is F(x, y) found? Integrate f (x, y) with respect toy . Example 15 : Let M(x, y) = 2xy2 + x2 — y . It is known that M equal s f;_ for some function f(x, y) . Determine the most general suc h function f (x, y) . Since M(x, y) is the partial derivative with respect to x of som e function f(x, y), M must be partially integrated with respect to x t o recover f. This situation can be symbolized as follows : itv

f(x ,Y) f (x, y) `

f(

> M(x, y ) M(x, y)

)iL

Therefore, f(x,y)= fM(x,y)8x = (2xy 2 +x' — f(x,y)=x'v2+

16

y)ax

x3 —xy+tp(Y )

CLIFFS QUICK REVIEW



REVIEW AN D INTRODUCTION

Note carefully that the "constant" of integration here is an y (differentiable) function of y —denoted by i/r(y) since any suc h function would vanish upon partial differentiation with respect to x (just as any pure constant c would vanish upon ordinary differentia tion) . If the question had asked merely for a function f (x, y) fo r which f. = M, you could just take r(y) = 0 . ■ Example 16 : Let N(x, y) = sin x cos y — xy + 1 . It is known that N equals f,, for some function f (x, y) . Determine the most general suc h function f (x, y) . Since N(x, y) is the partial derivative with respect to y of som e function f (x, y), N must be partially integrated with respect to y t o recover f. This situation can be symbolized as follows :

> N(x, y )

f(x, y ) f (x, y)

f()i)y

N(x,y )

Therefore, f (x, y )

=

f N(x, y ) ay f (sinxcosy — xy + 1)ay

f (x, y) = sinx siny —

+ y + (x )

Note carefully that the "constant" of integration here is any (differentiable) function of x —denoted by 4(x) since any such function would vanish upon partial differentiation with respect toy . If the question had asked merely for a function f (x, y) for which f,. = N, you could just take (x) = 0 . ■

DIFFERENTIAL EQUATIONS

REVIEW AND INTRODUCTIO N

Introduction to Differential Equation s In high school, you studied algebraic equations lik e 3x-2(x-4)=5x, x 2 -8x+15=0,

and

16x-21= 4

The goal here was to solve the equation, which meant to find th e value (or values) of the variable that makes the equation true . Fo r example, x = 2 is the solution to the first equation because onl y when 2 is substituted for the variable x does the equation become a n identity (both sides of the equation are identical when and onl y whenx = 2) . In general, each type of algebraic equation had its own particular method of solution ; quadratic equations were solved by on e method, equations involving absolute values by another, and so on . In each case, an equation was presented (or arose from a wor d problem), and a certain method was employed to arrive at a solution, a method appropriate for the particular equation at hand . These same general ideas carry over to differential equations , which are equations involving derivatives . There are different type s of differential equations, and each type requires its own particula r solution method . The simplest differential equations are those o f the form y' = f (x) . For example, consider the differential equatio n dy dx = 2

x

It says that the derivative of some function y is equal to 2x . To solv e the equation means to determine the unknown (the function y ) which will turn the equation into an identity upon substitution . I n this case all that is needed to solve the equation is an integration :

CLIFFS QUICK REVIEW

18

REVIEW AND INTRODUCTION

Thus, the general solution of the differential equation y' = 2x is y = x' + c, where c is any arbitrary constant . Note that there are actually infinitely many particular solutions, such asy = x 2 + 1,y = x — 7, or y = x 2 + Tr, since any constant c may be chosen . Geometrically, the differential equation y' = 2x says that at each point (x, y) on some curve y = y(x), the slope is equal to 2x . Th e solution obtained for the differential equation shows that thi s property is satisfied by any member of the family of curvesy = x 2 + c (any only by such curves) ; see Figure 4 .

■ Figure 4

■

Since these curves were obtained by solving a differential equatio n which either explicitly or implicitly involves taking an integral the y are sometimes referred to as integral curves of the differentia l equation (particularly when these solutions are graphed) . If on e particular solution or integral curve is desired, the differentia l equation is appended with one or more supplementary conditions . These additional conditions uniquely specify the value of th e arbitrary constant or constants in the general solution . For example , consider the proble m dy

dx=2x

and y=2when x= 0

DIFFERENTIAL EQUATIONS

19



REVIEW AND INTRODUCTIO N

The initial condition "y = 2 when x = 0 " is usually abbreviate d "y(0) = 2," which is read "y at 0 equals 2 ." The combination of a differential equation and an initial condition (also known as a constraint) is called an initial value problem (abbreviated IVP) . For differential equations involving higher derivatives, two o r more constraints may be present . If all constraints are given at th e same value of the independent variable, then the term IVP stil l applies . If, however, the constraints are given at different values o f the independent variable, the term boundary value problem (BVP ) is used instead . For example , this is an IVP : y" + 2y' — 3y = 0, y(0) = 1, y'(0) = 5 T

same i l

but this is a BVP :

y" + 2y' — 3y = 0,

y(0) = 1, y'(l) = 5 T

different T

To solve an IVP or BVP, first find the general solution of th e differential equation and then determine the value(s) of th e arbitrary constant(s) from the constraints .

Example 17 : Solve the IVP y' = 2x y(0) = 2 As discussed above, the general solution of this differentia l equation is the family y = x 2 + c . Since the constraint says that y must equal 2 when x is 0 , y(0)=2

[x'+c],-_,►=2

so the solution of this IVP is y = x2 + 2 .

0 2 +c=2

c= 2

■

CLIFFS QUICK REVIE W

20



REVIEW AN D INTRODUCTION

Example 18: Consider the differential equation y" + 2y' — 3y = 0 . Verify that y = c 1e v + c2e -=x (where c l and c, are arbitrary constants) is a solution . Given that every solution of this differentia l equation can be written in the form y = c 1e v + c,e -3x, solve the IVP y" + 2y' -3y= 0 y(0)= 1 y'(0) = 5

To verify that y = c i e x + c2e -3x is a solution of the differentia l equation, substitute . Since y' = c 1e x" — 3c,e -=

and

y" = c 1 e x +

once c 1e - + c,e -3' is substituted for y, the left-hand side of th e differential equation become s y"+2y' -3y= (c i e x +9c,e -3x )+2(c i ex- -3c,e - ' ) — 3(c 1 e x + c,e - 3x- ) = (c i e x

+

2c 1 e x 3c i ex ) — 6c,e -3x — 3c,e -3x + (9c,e )

=0

3

Now, to satisfy the conditions y(0) = 1 and y ' (0) = 5, the constants c l and c, must be chosen so that y(0)

1 = [c 1 e x

+

c,e-3"]x._1) = 1

c 1 + c2 = 1

an d y'(0)=5

[c 1 e' -3c,e-3x]x.=1,=5

c 1 -3c,= 5

Solving these two equations yields c 1 = 2 and c, = -1 . Thus, th e particular solution specified by the given IVP isy = 2e `" — e- 3x . n

DIFFERENTIAL EQUATIONS



REVIEW AN D INTRODUCTION

The order of a differential equation is the order of the highest derivative that appears in the equation . For example, y' = 2x is a first-order equation, y" + 2y' – 3y = 0 is a second-order equation , and y" – 7y ' + 6y = 12 is a third-order equation . Note that th e general solution of the first-order equation from Example 1 7 contained one arbitrary constant, and the general solution of th e second-order equation in Example 18 contained two arbitrar y constants . This phenomenon is not coincidental . In most cases, th e number of arbitrary constants in the general solution of a differentia l equation is the same as the order of the equation .

Example 19 : Solve the second-order differential equation y" = x + cos x. Integrating both sides of the equation will yield a differentia l equation for y' : y'

=

f y" (x + cos x) dx

=2x2 +sinx+c1 Integrating once more will give y : y =

y'

(2x-+sinx+c,)dx =x3 –cosx+c 1x+c2 where c 1 and c 2 are arbitrary constants . Note that there are tw o arbitrary constants in the general solution, which you shoul d typically expect for a second-order equation . ■



REVIEW AND INTRODUCTIO N

Example 20: For the following IVP, find the solution valid fo r x>0 : y"'=105x—

2 x'

+ 6

y(1)= 7 y'(1) = 3 7 y"(1)=7 3 The general solution of a third-order differential equatio n typically contains three arbitrary constants, so an IVP involving a third-order differential equation will necessarily have three constraint equations (as is the case here) . As in Examples 17 and 19, th e given differential equation is of the form y( 't'

=

f(x )

where y ( " ) denotes the nth derivative of the function y . Thes e differential equations are the easiest to solve, since all they requir e are n successive integrations . Note how the first-order differentia l equation in Example 17 was solved with one integration, and th e second-order equation in Example 19 was solved with two integra tions . The third-order differential equation given here will be solve d with three successive integrations . Here's the first : y"=

(105x' /2 — 2x -3

+6)dx

= 70x312 + x -2 + 6x + c , The value of this first arbitrary constant (c1) can be found by applying the condition y " (1) = 73 : y"(1) = 73

[70x 312 + x -2 + 6x + c,],._, = 73 X70+1+6+c,=73~c,=— 4

Thus,y" = 70x312 + x -2 + 6x — 4. DIFFERENTIAL EQUATIONS



REVIEW AND INTRODUCTIO N

Now, perform the second integration, which will yield y' : y' = f(7ox3/2 +x 2 + 6x - 4) v =28x /

2–x

-'

+

3x222 –4x+c ,

The value of this arbitrary constant (c 2 ) can be found by applying the constraint y' (1) = 37 : y'(1) = 37 = [28x 52 –x-' + 3x2 – 4x +

=37

28– 1 +3–4+c,=3 7 =11 Therefore,y' = 28x' ' 2 –x-' + 3x2 – 4x + 11 . Integrating once mor e will give the solution y : y = f (28x" – x - ' + 3x 2 – 4x + 11) dx =8x 71 2–lnx+x3 –2x2 + 1lx+c 3 The value of this arbitrary constant ( c 3 ) can be found by applyin g the condition y(1) = 7 : y(l) = 7

[8 x712 – In x + x 3 – 2x 2 + 1 lx + c3 ] .,._ 1

=

7

8–0+1–2+11+c3 = 7 c3 = - 1 1 Thus, the solution is y = 8x 71 – In x + x3 – 2x 2 2

+

11x – 11 .

A few technical notes about this example : ■ The given differential equation makes sense only for x > 0 (note the f and 2/x3 terms) . To respect this restriction, th e problem states the domain of the equation and its solutio n [that is, the set of values of the variable(s) where th e equation and solution are valid] asx > 0 . Always be aware o f the domain of the solution .



REVIEW AND INTRODUCTIO N

■ Although the integral of x-' is usually written in Ix', the absolute value sign is not needed here, since the domain o f the solution is x > 0, and 'xi = x for any x > 0 . ■ Contrast the methods used to evaluate the arbitrary constants in Examples 18 and 20 . In Example 18, the constraints were applied all at once at the end . In Example 20, however , the constants were evaluated one at a time as the solutio n progressed . Both methods are valid, and each particula r problem (and your preference) will suggest which t o use . ■

Example 21 : Find the differential equation for the family of curve s x2 + y 2 = c 2 (in the xy plane), where c is an arbitrary constant . This problem is a reversal of sorts . Typically, you're given a differential equation and asked to find its family of solutions . Here , on the other hand, the general solution is given, and an expressio n for its defining differential equation is desired . Differentiating bot h sides of the equation (with respect to x) give s x2

+

y-' = c ' dy

2x+2y-= 0 dy x+y

— d x= 0 dy

x

dx

y

This differential equation can also be expressed in another form , one that will arise quite often . By "cross multiplying," the differential equation directly above becomes ydy= –xdx

DIFFERENTIAL EQUATIONS

11



REVIEW AN D INTRODUCTION

which is then normally written with both differentials (the dx and the dy) together on one side : xdx+ydy= 0

Either y' = —x/y or x dx + y dy = 0 would be an acceptable way o f writing the differential equation that defines the given family (o f circles) x2 + y2 = c2 . n

Example 22: Verify that the equation y = ln(x/y) is an implicit solution of the IVP ydx—x(y+1)dy= 0 y(e) = 1

First note that it is not always possible to express a solution i n the form "y = some function of x ." Sometimes when a differentia l equation is solved, the solution is most naturally expressed with y' s (the dependent variable) on both sides of the equation, as in y = ln(x/y) . Such a solution is called an implicit solution, as opposed t o an explicit solution, which has y all by itself on one side of the equation and a function of x only on the right (as in y = x 2 + 2, fo r example) . Implicit solutions are perfectly acceptable (in some cases , necessary) as long as the equation actually defines y as a function o f x (even if an explicit formula for this function is not or cannot be found) . However, explicit solutions are preferable when available . Perhaps the simplest way to verify this implicit solution is t o follow the procedure of Example 21 : Find the differential equatio n for the solution y = ln(x/y) . To simplify the work, first rewrit e ln(x/y) as In x — In y : y = lnx — ln y dy 1 1 dy dx x y dx

26

CLIFFS QUICK REVIEW



REVIEW AN D INTRODUCTION

dx

1 + Y/

x

dy

1/x

dx 1 + (1/y ) dy

1 /x

xy

d x 1 + (1 /y) xy dy

y

dx x(y + 1 ) ydx-x(y+ 1)dy= 0

Therefore, the differential equation given in the statement of the problem is indeed correct . The initial condition is also satisfied , since 1 = ln(e/ 1) implies y(e) = 1 satisfiesy = ln(x/y) . ■

Example 23: Discuss the solution to each of the differential equations

(y')-+x-=0

and (y')--+y'= 0

The first differential equation has no solution, since no real valued function y = y (x) can satisfy (y' )2 = —x 2 (because squares o f real-valued functions can ' t be negative) . The second differential equation states that the sum of tw o squares is equal to 0, so both y' and y must be identically 0 . Thi s equation does have a solution, but it is only the constant functio n y = 0 . Note that this differential equation illustrates an exception t o the general rule stating that the number of arbitrary constants in th e general solution of a differential equation is the same as the order o f the equation . Although (y' ) 2 + y2 is a first-order equation, it s general solutiony = 0 contains no arbitrary constants at all . ■

DIFFERENTIAL EQUATIONS



REVIEW AND INTRODUCTIO N

One final note : Since there are two major categories of derivatives, ordinary derivatives lik e dy

dy

►► and, ch . = y

y'

j

and partial derivatives such a s f —ax

.fx.

and

a2f

ay y ax

.f.,,.

there are two major categories of differential equations . Ordinary differential equations (ODEs), like the equations in all of th e examples above, involve ordinary derivatives, while partial differen tial equations (PDEs), such a s 8f

82f

~t=k cox-

and

8 2w ax-

82w

+ , = 0 ay

involve partial derivatives . Only ordinary differential equations ar e examined in this book .

28

CLIFFS QUICK REVIEW



FIRST-ORDER EQUATIONS

From the Introduction, you know that a first-order differentia l equation is one containing a first but no higher derivative of th e unknown function . For virtually every such equation encountered i n practice, the general solution will contain one arbitrary constant , that is, one parameter, so a first-order IVP will contain one initia l condition . There is no general method that solves every first-orde r equation, but there are methods to solve particular types .

Exact Equation s Given a function f (x, y) of two variables, its total differential df i s defined by the equation df=~f dx+ f ~ dy Y

Example 1 : If f (x, y) = x-y

+ 6x - y 3,

then

df=(2xy+6)dx+(x--3y 2 )dy

n

The equation f(x, y) = c gives the family of integral curves (that is , the solutions) of the differential equatio n df = 0

Therefore, if a differential equation has the form af

af

dx+ —dy=O (* ) ax dy

for some function f (x, y), then it is automatically of the form

DIFFERENTIAL EQUATIONS

df = 0 ,

29



FIRST-ORDER EQUATIONS

so the general solution is immediately given by f (x, y) = c . In thi s case, 8f of dx + ~ dy ax Y is called an exact differential, and the differential equation (*) i s called an exact equation . To determine whether a given differentia l equation M(x, y) dx + N(x, y) dy = 0 is exact, use th e Test for Exactness: A

differential equation M dx + N dy = 0 is exact i f

and only if aM

aN

ay

ax

Example 2: Is the following differential equation exact ? ( y2— 2x ) dx +(2xy+ 1 ) dy = 0 The function that multiplies the differential dx is denote d M(x,y), so M(x,y) = y2 — 2x ; the function that multiplies th e differential dy is denoted N(x, y), so N(x, y) = 2xy + 1 . Sinc e aM

y

=2y

and

8N

~►x

=2y

the Test for Exactness says that the given differential equation i s indeed exact (since M,. = NO . This means that there exists a function f(x, y) such tha t af _

-x

30

M(x,y)=-2x

af

and - = N(x,y)=2xy+ 1 Y

CLIFFS QUICK REVIE W



FIRST-ORDER EQUATIONS

and once this function f is found, the general solution of th e differential equation is simply f(x,y)= c

(where c is an arbitrary constant) .

■

Once a differential equation M dx + N dy = 0 is determined t o be exact, the only task remaining is to find the function f (x, y) suc h that f, = M and f, = N . The method is simple : Integrate M wit h respect to x, integrate N with respect toy, and then "merge" the tw o resulting expressions to construct the desired function f.

Example 3 : Solve the exact differential equation of Example 2 :

(y 2 - 2x) dx + ( ay + 1)dy= 0 First, integrate M(x, y) = 2x with respect to x (and ignore the arbitrary "constant" of integration) : y2 –

f M (x , y ) ax=

( y22

2x)axxy 2

x2

Next, integrate N(x, y) = 2xy + 1 with respect toy (and again ignore the arbitrary "constant " of integration) :

f N (x, y) ay =

(2xy + 1)ay = xy 2 + y

Now, to "merge" these two expressions, write down each ter m exactly once, even if a particular term appears in both results . Here the two expressions contain the terms xy 2 , and y, s o –x2,

f(x,y)

DIFFERENTIAL EQUATIONS

xy2 -

x2 + y



I

FIRST-ORDER EQUATION S

(Note that the common term xy 2 is not written twice .) The genera l solution of the differential equation is f(x, y) = c, which in this cas e becomes xy 2 –x 2 +y=c

■

Example 4: Test the following equation for exactness and solve it i f it is exact : x(1 – siny)

dy = (cosx – cosy – y) dx

First, bring the dx term over to the left-hand side to write th e equation in standard form : (y+cosy–cosx)cGr+(x–xsiny)dy= 0 Therefore, M(x, y) = y + cos y – cos x, and N(x, y) = x – x sin y . Now, since aM ay

aN

= 1 – siny and

ax

= 1 – sin y

the Test for Exactness says that the differential equation is indeed exact (since M,. = NX ) . To construct the function f (x, y) such tha t fX = M and f,. = N, first integrate M with respect to x :

f M(x, y) ax =

(y + cosy – cos x) ax = xy + x cos y – sin x

Then integrate N with respect toy :

f N(x, y) ay =

(x – x sin y) ay = xy + x cos y

Writing all terms that appear in both these resulting expression s without repeating any common terms gives the desired function : f (x, y) = xy + x

cosy – sin x

CLIFFS QUICK REVIEW



FIRST-ORDER EQUATIONS

The general solution of the given differential equation is therefore xy+xcosy — sinx=c

■

Example 5 : Is the following equation exact ? (3xy -y 2 )dx+x(x

- y)dy = 0

Since aM aY

a d (3xy—y2) = x — 2y Y

but 0N

a (x2— Ay

ax ax

)= — y

it is clear that M,. ~ Na , so the Test for Exactness says that this equation is not exact . That is, there is no function f (x, y) whose derivative with respect to x is M(x, y) = 3xy — and which at th e same time has N(x, y) = x(x — y) as its derivative with respec t toy . ■ y2

Example 6: Solve the IV P (3x2y — 1) dx + (x 3 + 6y — y 2 ) dy = 0 y(0) = 3

The differential equation is exact becaus e aM _ a (3x 2y— 1)=3x 2 ay ay

and

Integrating M with respect to x give s

aN 0x

a ax

(x3

+6y—y 2 )=3x 2

f M(x,y)ax= f (3x 2y — i)ax=x ;y— x

DIFFERENTIAL EQUATIONS

FIRST-ORDE R EQUATION S

and integrating N with respect toy yield s

f N (x , y ) iy= f(x 3 + ~y —y 2 )3y =x 3y + 3y : —

y3

Therefore, the function f (x, y) whose total differential is th e left-hand side of the given differential equation i s f(x, y) = x-y - x + 3y- -y 3

and the general solution i s x 3y — x + 3y 2 —

y3= c

The particular solution specified by the IVP must have y = 3 whe n x = 0; this condition determines the value of the constant c : i)v=3 = C

[x 3y—x+3y 2 —

0—0+27—9=c

18= c

Thus, the solution of the IVP i s x 3y—x+3y 2 — y3= 18

■

Integrating Factor s If a differential equation of the for m M(x,y)dx + N(x,y)dy = 0 (* ) is not exact as written, then there exists a function µ(x, y) such tha t the equivalent equation obtained by multiplying both sides of (* ) by µ, (,uM) dx + (µN) dy = 0 is exact . Such a functionµ is called an integrating factor of th e original equation and is guaranteed to exist if the given differentia l equation actually has a solution . Integrating factors turn nonexac t

34

CLIFFS QUICK REVIEW

FIRST-ORDE R EQUATIONS

equations into exact ones. The question is, how do you find an integrating factor? Two special cases will be considered . Case 1 :

Consider the differential equation Mdx + N dy = 0 . If thi s equation is not exact, then M, . will not equal N,. ; that is, M,.— N,-  0 . However, if M, . — N,. N

is a function of x only, let it be denoted by (x) . The n µ(x )

=

will be an integrating factor of the given differentia l equation . Case 2:

Consider the differential equation Mdx + N dy = 0 . If thi s equation is not exact, then M, will not equal N, . ; that is , M,. — N,-  0 . However, if —M

is a function of y only, let it be denoted by 0(y) . The n µ(Y) = of o y) (n.

will be an integrating factor of the given differentia l equation .

Example 7 : The equatio n

(3xy—y 2)dx+x(x—y)dv= 0 is not exact, sinc e

M,.= a (3xy — y 2 ) = 3x — 2y but N,.=a(x'—xy)_2x — y Y

DIFFERENTIAL EQUATIONS

FIRST-ORDER EQUATIONS

(recall Example 5) . However, note tha t M, .—N,.

(3x — 2y) — (2x — y)

x—y

1

N

x(x—y)

x(x — y)

x

is a function of x alone . Therefore, by Case 1 , ef(1/ .v)ILV

= e In .v = x

will be an integrating factor of the differential equation . Multiplyin g both sides of the given equation by ,u, = x yield s (3x 2y —xy 2 )dx+(x 3 —x 2y )dy= 0 µM— M

µN

which is exact because aM aN a = 3x-'— may = ax Y Solving this equivalent exact equation by the method described i n the previous section, M is integrated with respect to x ,

f M ax = f(3x2y — xy 2 )

ax

= x 3y — x 2y 2

and Nis integrated with respect toy :

5 Nay= 5 (x3 —x-y)ay=x 3y—3x'y (with each "constant" of integration ignored, as usual) . Thes e calculations clearly give xy— ;xy 2 = c as the general solution of the differential equation .

■

Example 8 : The equatio n (x + y) sin y dx + (x sin y + cosy) dy = 0

36

CLIFFS QUICK REVIE W



FIRST-ORDER EQUATIONS

is not exact, sinc e M,. = (x + y) cosy + sin y

but

N,. = sin y

However, note that M, . — Na

_

(x + y) cosy + sin y — sin y

—(x + y) sin y

—M

cos y sin y

is a function ofy alone (Case 2) . Denote this function by iff(y) ; since cos y dy

y) dy = —

sin = — ln(sin y ) y

the given differential equation will hav e eh(' = e —In(sin-v)

= e in(sinr) - i

=

(sin y) — '

as an integrating factor . Multiplying the differential equatio n through by ,u, = (sin y)-' yields

cos y1

(x+y)cLr+ x+ S ~ ny ldy= U .Y~ µM= M

µN= N

which is exact because M,. = 1 = N,.

To solve this exact equation, integrate M with respect to x an d integrate N with respect toy, ignoring the "constant" of integratio n in each case :

f Max= f(x +y)3x = x 2 +xy y f Nay= f (x+ cos )ay=y+lnsiny I sin y

DIFFERENTIAL EQUATIONS



FIRST-ORDE R EQUATIONS

These integrations imply tha t +xy+lnIsinyl= c is the general solution of the differential equation .

■

Example 9: Solve the IV P (3&y + x) dx + e-' dy = 0 y(0) = 1 The given differential equation is not exact, sinc e M, . =

a y

(3e-5) + x) = 3e- v

but

N,

a (e-'-) = e-' . = ~x

However, note that M,. — N,. N

3e-'- e-

= 2

which can be interpreted to be, say, a function of x only ; that is, thi s last equation can be written as (x) = 2 . Case 1 then says tha t 0-0 ctv = e ft -r = e -, t. of

cr

will be an integrating factor . Multiplying both sides of the differential equation by µ(x) = e ' yields (3e 3-'y + xe 2') dx + (e''-') dy = 0

which is exact because M,. = 3e 3 ' = N,.

38

CLIFFS QUICK REVIE W



FIRST-ORDER EQUATIONS

Now, since

Max=

(3e 3 5' +xe'`)ax=e3`y +axe-' – e-`

and Nay =

f e3`. ay = e; 'y

(with the "constant" of integration suppressed in each calculation) , the general solution of the differential equation i s e 3`y+xe-

-e- = c

The value of the constant c is now determined by applying the initia l condition y(0) = 1 :

[e3`y + xe ;

– le?` "v =u,,=~ = C

=

c

Thus, the particular solution i s e3xy+xe' –

=

which can be expressed explicitly a s 3e_3' .+e-(1–2x ) Y

4

Example 10: Given that the nonexact differential equatio n

(5xy2 –2y)dx+(3xy–x)dy= 0 has an integrating factor of the form µ(x, y) = for some positive integers a and h, find the general solution of the equation . is a n Since there exist positive integers a and h such that integrating factor, multiplying the differential equation through b y

DIFFERENTIAL EQUATIONS

39



FIRST-ORDER EQUATIONS

this expression must yield an exact equation . That is , (5.x"+ ly b+2 - 2x"yh+ 1) dx + (3x"+ .?yh+ 1 - x"+ lyh) dy µM=M

0

(* )

µN= N

is exact for some a and b . Exactness of this equation mean s M, . = N,. 5(b + 2)x"+'y''+' – 2(b + 1)x"y'' = 3(a + 2)x"+'y''+' – (a + 1)x"y h By equating like terms in this last equation, it must be the case tha t 5(b+2)=3(a+2)

and

2(b+ 1)=a+ 1

The simultaneous solution of these equations is a = 3 and b = 1 . Thus the integrating factor x"y'' is x 3y, and the exact equatio n .Mdx+ Ndy=Dread s (5x 4y' - 2x 3y') dx + (3x5y' - x 4y) dy = 0 Now, since (5x 4y 3 – 2x3y2)ax=x5y3 –x4y 2

5 Max= and

5

Nay

=

5

(3x 5y' – x4y) ay = x 5y 3 – , x4y ,

(ignoring the "constant" of integration in each case), the genera l solution of the differential equation (*) and hence the original differential equation is clearly x'y' – ;x 4y , = c

40

■

CLIFFS QUICK REVIEW



FIRST-ORDER EQUATIONS

Separable Equation s Simply put, a differential equation is said to be separable if th e variables can be separated . That is, a separable equation is one tha t can be written in the form F(y) dy =. G (x) dx

Once this is done, all that is needed to solve the equation is t o integrate both sides . The method for solving separable equation s can therefore be summarized as follows : Separate the variables and integrate .

Example 11 : Solve the equation 2y dy (x 2 + 1) dx . Since this equation is already expressed in "separated" form , just integrate : 2ydy=(x 2 +1)dx

f 2y dy = (x' + 1) dx y- - 'x 3 +x+c

■

Example 12 : Solve the equatio n x dx + secx siny dy = 0 This equation is separable, since the variables can be separated : sec x sin y dy = —x dx sin ydy = —xcosxdx The integral of the left-hand side of this last equation is simpl y fsinydy= —cos y

DIFFERENTIAL EQUATIONS

41



FIRST-ORDER EQUATION S

and the integral of the right-hand side is evaluated using integratio n by parts : —x cos x dx = — x cos x dx _ —[x sin x — sin xdx ] (xsinx+cosx+c ) The solution of the differential equation is therefor e cos y = x sin x + cosx + c

■

Example 13 : Solve the IVP dy x(ex- + 2 ) 6y 2 dx = y(0) = 1 The equation can be rewritten as follows : 6y 2 dy=x(e x- +2)dx Integrating both sides yield s

f 6y 2 dy = x(ex" + 2) dx 2Y 3 = ex' + x 2 + c Since the initial condition states that y = 1 at x = 0, the parameter c can be evaluated : [ 2y3 ]y=i = [4e

+x2+c]~._2=4+c

c= 4

2 The solution of the IVP is therefore 2)2 3 = 4e-t + x-- + 4 o r 4y 3

42

= e x' + 2x 2

+3

■

CLIFFS QUICK REVIEW



FIRST-ORDER EQUATIONS

Example 14: Find all solutions of the differential equatio n (x 2 -1)y 3 dx+x 2 dy=0 . Separating the variables and then integrating both sides give s x 2 dy = —(x' . — 1)y 3 dx dy 1—x 2 = x, dx Y

(t)

fy-3dy=f(x-2_ 1) dx —'2y

= — x -' — x — c

1 1 =— +x+ c x 2yAlthough the problem seems finished, there is another solution o f the given differential equation that is not described by the family ;y -2 = x-' + x + c . In the separation step marked (t), both side s were divided by y 3 . This operation prevented y = 0 from bein g derived as a solution (since division by zero is forbidden) . It just s o happens, however, that y = 0 is a solution of the given differentia l equation, as you can easily check (note : y = 0 dy = 0) . Thus, the complete solution of this equation must includ e 1 1 both the family 2y- , = + x + c and x the constant function y = 0 The lesson is clear : If both sides of a separable differential equation are divided b y some function f(y) (that is, a function of the dependen t variable) during the separation process, then a valid solutio n may be lost . As a final step, you must check whether the constan t function y = yo [where f(yo) = 0] is indeed a solution of the give n differential equation . If it is, and if the family of solutions foun d DIFFERENTIAL EQUATIONS

43



FIRST-ORDE R EQUATIONS

by integrating both sides of the separated equation does no t include this constant function, then this additional solution mus t ■ be separately stated to complete the problem .

Example 15 : Solve the equation dy = (1 +

e-x)(Y2

–

Separating the variables give s dy

y 2 -1=(1+e-x)dx

(t )

(To achieve this separated form, note that both sides of the origina l equation were divided by y 2 – 1 . Thus, the constant functions y = 1 and y = -1 may be lost as possible solutions ; this will have to be checked later .) Integrating both sides of the separated equatio n yields

f

f

dy z— 1

+ y _ 1 ) dY =

f

(1 +

f+

e-A)

cLr

e -x )dy

– 1nfy+1~+1nIy– 1I =x – e -x + c In

y– 1 =2(x–e-x+c ) y+ 1

Now, both constant functions y = 1 and y = -1 are solutions of the original differential equation (as you can check by simply notin g that y = ± 1 dy/dx = 0), and neither is described by the famil y above . Thus the complete set of solutions of the given differentia l

44

CLIFFS QUICK REVIEW



FIRST-ORDER EQUATIONS

equation includes both the family In

y -1 y+ 1

=2(x—e ' + c )

and the constant functions y = ± 1

Example 16: Solve the differential equation xydx —

■

2 + 1)dy=0 .

Separate the variables , (x' + 1)dy =xydx .dy

xdx

y x- + 1

(t )

and integrate both sides : dy

x dx

y

x' + 1

Inly~ =

In(x'+1)+c '

2lnIy1 =ln(x 2 + 1)+In c In (y ' ) = In [c(x 2 + 1) ] y'=c(x 2 +1 )

Note that in the separation step (t), both sides were divided by y ; thus, the solution y = 0 may have been lost . Direct substitution o f the constant function y = 0 into the original differential equatio n shows that it is indeed a solution . However, the family y 2 = c(x 2 + 1 ) already includes the function y = 0 (take c = 0), so it need not b e separately mentioned . ■

DIFFERENTIAL EQUATIONS

45



FIRST-ORDE R EQUATIONS

Example 17 : Find the curve r = r(0) in polar coordinates that solve s the IVP dr d8 –r

tan 0

r(Tr) = 2

The given equation is separable, since it can be expressed in th e separated form dr — = –tan 8dO (t ) r

Now integrate both sides : dr

–sin Bd 9 COS

B

In Irl = In cos 91 + c ' 1nIrl =lnlcosB1 +lnlc ' InIrl = In c cos 0 1 r=ccos B

Since the solution curve is to pass through the point with pola r coordinates (r, 0) _ (2, Tr) , 2=ccosTr

c = -2

The solution of the IVP is therefor e r = -2 cos 0

This is a circle of diameter 2, tangent to they axis at the origin ; se e Figure 5 . Note : In the separation step (t), both sides were divide d by r (which is the dependent variable here) . However, even thoug h r = 0 formally satisfies the differential equation, it clearly does no t satisfy the initial condition r(-rr) = 2 . ■

CLIFFS QUICK REVIE W

46



FIRST-ORDER EQUATION S

(r,9)=(2,n )

■ Figure 5

■

Homogeneous Equation s A function equation

is said to be homogeneous of degree n if th e

f(x, y)

f(2x , zy) =

z`f(x , y )

holds for all x, y, and z (for which both sides are defined) .

Example 18: The function degree 2, sinc e f(

zx

, zy )

=

f(x, y) = x 2

+

y2

is homogeneous of

(zx ) 2 + (zy ) 2 = z2(x2 + y2) = z 2f(x, y )

•

Example 19 : The function f(x, y) = /x8 — 3x 2y h is homogeneous o f degree 4, since f(zv ,

zy ) = V ( ) 8 - 3 ( ) 2 (zy ) 6 = Vz 8 W - 3x2y6 ) =

DIFFERENTIAL EQUATIONS

— 3x 2y 6 = z 4f(x, y)

■

47



FIRST-ORDER EQUATIONS

Example 20 : The function f (x, y) = 2x + y is homogeneous of degree 1, sinc e f(zx , zy ) =2 (zx) + (zy ) =z 2x + y) = z' f (x, y)

■

Example 21 : The function f (x, y) = x 3 — y 2 is not homogeneous, since z ,y 2 f (zx, 2y ) _ (zx )3 — (z') 2 = z3x3 —

which does not equal z' f (x, y) for any n .

■

Example 22: The function f (x, y) = x 3 sin (y/x) is homogeneous o f degree 3, since zY f (zx, zy) = (zx) 3 sin ! = z-; x 3 sin x = z 3f (x, y)

■

A first-order differential equatio n M(x, y) dx + N(x, y) dy = 0 is said to be homogeneous if M(x, y) and N(x, y) are both homogeneous functions of the same degree .

Example 23 : The differential equatio n (x 2 — y 2 ) dx + xy dy = 0 is homogeneous because both M(x, y) = x 2 — y 2 and N(x, y) = xy are ■ homogeneous functions of the same degree (namely, 2) .

CLIFFS QUICK REVIE W 48



FIRST-ORDER EQUATION S

The method for solving homogeneous equations follows from thi s fact : The substitution y = xv (and therefore dy = xdv + vdx) transforms a homogeneous equation into a separable one .

Example 24 : Solve the equation

(x' — y 2 )

dx + xy dy = 0 .

This equation is homogeneous, as observed in Example 23 . Thus to solve it, make the substitutionsy = XI) and dy = x dv + v dx : [x 2 — (xv) 2]

dx + [x(xv)](x dv + v dx) = 0

(x- - x 'zv--)

dx + x 3z, dz~ + x-z►- dx = 0 x-dx+xvdv= 0 dx + xv dv = 0

This final equation is now separable (which was the intention) . Proceeding with the solution, dx

v dv = - x

dx x

,,dv _ -1-x 2 =

—In

+c'

Therefore, the solution of the separable equation involving x and v can be written C

x

DIFFERENTIAL EQUATIONS

49

FIRST-ORDER EQUATIONS

To give the solution of the original differential equation (whic h involved the variables x and y), simply note that v _ yx

y = xv

Replacing v by y/x in the solution above gives the final result :

l (12 – (n

2

c x

y2

= 2x 2 In

c

x

This is the general solution of the original differential equation . ■

Example 25 : Solve the IVP 2(x + 2y) dx + (y – x) dy = 0 y(1) = 0 Since the function s M(x, y) = 2(x + 2y)

and

N(x, y) = y – x

are both homogeneous of degree 1, the differential equation i s homogeneous . The substitutions y = xv and dy = x dv + v dx transform the equation into 2(x + 2xv) dx +

(xv –

x)(x dz) + v dx) = 0

which simplifies as follows : 2x dx + 4xzvdx +x2 vdzv –x 2 dv +xzv 2 dx – xvdx = 0 (2x + 3xv + xz) 2 ) dx + x(2 + 3v

+ z v 2)

(2 + 3v +

(x2z) – x2)

dv = 0

dx + x2(v – 1) dv = 0

z, 2 )

dx + x(v - 1) dzv = 0

CLIFFS QUICK REVIE W 50



FIRST-ORDER EQUATION S

The equation is now separable . Separating the variables and inte grating gives x(v — 1)dv= —(2+3v +v 2 )dx v — 1 ) + 3v + 2 dv

dx x

(I )

The integral of the left-hand side is evaluated after performing a partial fraction decomposition : v— 1 v 2 +3v+2 Therefore,

f

3 -2 1 + v+ 2 v+

v— 1 (v+ 1)(v+2)

f (

v— 1 v2+3v+2dv



)

-2 3 dz v v + 1 +v+2

= -2lnJz~+ 11 +31niv+2 1 =ln(v+ 1) - 22 (v+2)3 1 The right-hand side of (t) immediately integrates to cLr x

In+c ' =1nIcx - ' l

Therefore, the solution to the separable differential equation (t) i s (v + 1) -2 (v + 2) 3 = cx - ' Now, replacing v by y/x give s y

—+ 1 x

DIFFERENTIAL EQUATIONS

-2

y

(— X

3

+

2 =cx - '



FIRST-ORDE R EQUATION S

as the general solution of the given differential equation . Applying the initial conditiony(1) = 0 determines the value of the constant c : I

0

1-z I0

)3

Thus, the particular solution of the IVP i s 2

y x

1)_

IX

-+-

2 _ 8C- I

which can be simplified to (2x + y )3 = 8(x + y )2 as you can check . Technical note : In the separation step (t), both sides wer e divided by (v + 1)(v + 2), and v = -1 and v = -2 were lost a s solutions . These need not be considered, however, because eve n though the equivalent functions y = –x and y = -2x do indee d satisfy the given differential equation, they are inconsistent with th e ■ initial condition .

Linear Equation s A first-order differential equation is said to be linear if it can b e expressed in the form y' + P(x)y = Q(x ) where P and Q are functions of x . The method for solving suc h equations is similar to the one used to solve nonexact equations . There, the nonexact equation was multiplied by an integratin g factor, which then made it easy to solve (because the equation became exact) .

CLIFFS QUICK REVIEW



FIRST-ORDER EQUATIONS

To solve a first-order linear equation, first rewrite it (if necessary) in the standard form above ; then multiply both sides by th e integrating factor µ4x) = e 11

d

The resulting equation,

dy +µPy

=

µQ

(*)

is then easy to solve, not because it's exact, but because th e left-hand side collapses :

dy + µPy = dy +ef P c!x Py µ ms

k

dy

d

+y(e5Pdx

dy

)

d ,u,

dx + Y dx

Therefore, equation (*) become s d

Iv (AY)

= /I Q

making it susceptible to an integration, which gives the solution : µY =

(µQ) dx

Do not memorize this equation for the solution ; memorize the step s needed to get there .

DIFFERENTIAL EQUATIONS



FIRST-ORDER EQUATIONS

Example 26 : Solve the differential equatio n y'+2xy= x

The equation is already expressed in standard form, with P(x ) 2x and Q(x) = x . Multiplying both sides by µ(x ) = ef"v =

e

f2 (i =

transforms the given differential equation int o ex' y' + 2xe-" - y = xe-' d dx (ex Y) = xe'

Notice how the left-hand side collapses into (µy) ' ; as shown above , this will always happen . Integrating both sides gives the solution : ex - y

= fxe U dx

ex)? =+ c y= ;+ce -~ -

■

Example 27 : Solve the IVP 1 y' + —y = sin x x Y(n) = 1

Note that the differential equation is already in standard form . Since P(x) = 1/x, the integrating factor i s µ(x ) = efP`x = eft'

54

iv)( =

e in _ x

CLIFFS QUICK REVIE W



FIRST-ORDER EQUATIONS

Multiplying both sides of the standard-form differential equation b y µ = x gives xy' +y =x sin x

(xy)' = x sin x Again, note how the left-hand side automatically collapses int o (µy)' . Integrating both sides yields the general solution : xy = fxsinx v xy = –x cosx + sin x + c Applying the initial condition y(Tr) = 1 determines the constant c : Tr . 1 = –Tr cos Tr+sinTr+c = c= 0 Thus the desired particular solution i s xy = –x cosx + sin x or, since x cannot equal zero (note the coefficient P(x) = 1/x in th e given differential equation) , y=

sin x – cosx x

■

Example 28 : Solve the linear differential equation dy x- = -2y cbc First, rewrite the equation in standard form : dy

DIFFERENTIAL EQUATIONS

2 + x~Y =O

(* )



FIRST-ORDE R EQUATION S

Since the integrating factor here i s µ(x ) =

e fPd

=

e f(2

)th =

e_ 2 /

multiply both sides of the standard-form equation (*) by dy

e -2/x

+

2 .x 2

µ

= e-2/x,

e -2/x y = 0

collapse the left-hand side, d

~ (e -2/x y) = 0

and integrate : e -?iay— c

Thus the general solution of the differential equation can b e expressed explicitly as y = ce 2/x-

n

Example 29 : Find the general solution of each of the followin g equations : (a)

dy

dy

(b)

4 —xy= 0 4

• — x y =x 4

Both equations are linear equations in standard form, wit h -4/x . Since

P(x)

f PdY = f - -x ctx = 4

- 4 In = In (x -4 )

the integrating factor will b e = ,u(x) = efPdx = eln (x-4) x -4 CLIFFS QUICK REVIEW

56



FIRST-ORDER EQUATIONS

for both equations . Multiplying through by ,u, = x - 4 yield s d a _.

d

dx

(x- 4y) = 0

for equation (a )

(x -4y)

for equation (b )

=1

Integrating each of these resulting equations gives the general solutions : x -4y = c

y = cx 4

x -4y = x + c y = cx 4 + x 5

for equation (a ) for equation (b)

■

Example 30: Sketch the integral curve of (1 +x2 )y' =x ( 1 —y ) which passes through the origin . The first step is to rewrite the differential equation in standar d form : (1+x2)y'+xy=x .

x

x

Y'+l+x2y1+x2

(* )

Sinc e

f

Pdx=

x 2dx=ln(1 +x 2 ) +x

the integrating factor i s µ(x)

2 = e fPdx = e(1/2) In (1+x 2 ) = e ln (1+x ) 1/' =

DIFFERENTIAL EQUATIONS

(1

+x2)1/ 2



FIRST-ORDER EQUATIONS

Multiplying both sides of the standard-form equation (*) by (1 + x 22 ) ' /2 gives x

(1+x2)'/2y'+

µ =

x (1 + x2 )' / 2

(1 + x 2 )' /2y

As usual, the left-hand side collapses into (µy)' , d

[(1 + x2 )" 2y]

=



x 1+x2 'i 2

and an integration gives the general solution : +x2)1/2y

=

f x(1 +x 2 )

(1 + x 2 1 /2y

=

(1 +

(1

2)

2 dx

x2)1/2 +

c

y = 1 +c(1 +x 2 - ' /2 2)

2

To find the particular curve of this family that passes through th e origin, substitute (x, y) = (0, 0) and evaluate the constant c : 0= 1+c(1+0) -1/ ma c = — 1 2

Therefore, the desired integral curve i s y

1

1 ~l1+x 2

which is sketched in Figure 6 .

Example 31 : An object moves along the x axis in such a way that it s position at time t > 0 is governed by the linear differential equatio n dx + ~ - t_)x = t dt(t If the object was at position x = 2 at time t = 1, where will it be a t time t = 3?

CLIFFS QUICK REVIE W

58



FIRST-ORDE R EQUATION S

■ Figure 6

■

Rather than having x as the independent variable and y as th e dependent one, in this problem t is the independent variable and x i s the dependent one . Thus, the solution will not be of the form "y = some function of x" but will instead be "x = some function of t . " The equation is in the standard form for a first-order linea r equation, with P = t — t - ' and Q = t' . Since

f Pdt= f(t - t

1 )dt = . t 2 -

ln t

the integrating factor i s µ( t ) =

efPdt = e(t 2 /2)-In t = e t ,l2e -In t = et 2 /2eln t -1 = t -1et 2 / 2

Multiplying both sides of the differential equation by this integrating factor transforms it into dx

t-"2/2 dt +

t_2)et _,

(1

,/2

x = tet -

As usual, the left-hand side automatically collapses , d dt [t

, - 'e t -1 x1 = te' 2

DIFFERENTIAL EQUATIONS

59



FIRST-ORDE R EQUATIONS

and an integration yields the general solution : t - 'e t2/2x =

f te` 2/2 dt

=e t2/ 2-+ c x = t(1 + ce

- 1 '/2 )

Now, since the condition "x = 2 at t = 1 " is given, this is actually a n IVP, and the constant c can be evaluated : 2 = 1 • (1 + ce - ' 12 )

=

ce -1/2

=

1

c = e' / 2

Thus, the position x of the object as a function of time t is given b y the equation x = 41

+ e(' -' 2 )/2]

and therefore, the position at time t = 3 i s x I,_ 3

which is approximately 3 .055 .

= 3(1 + e -4 ) ■

Bernoulli's Equatio n The differential equatio n y ' + P (x )y = Q(x) y " is known as Bernoulli's equation . If n = 0, Bernoulli's equatio n reduces immediately to the standard form first-order linear equation : y' + P(x)y = Q(x ) If n = 1, the equation can also be written as a linear equation : y' + P(x) y = Q(x )y y' + [P (x ) — Q(x )] y = 0 CLIFFS QUICK REVIE W

60



FIRST-ORDE R EQUATION S

However, if n is not 0 or 1, then Bernoulli's equation is not linear . Nevertheless, it can be transformed into a linear equation (an d therefore solved by the method of the preceding section) by firs t multiplying through by n y -

y

,

-n

y'

+

P(x)y' -n

=

Q(x)

and then introducing the substitutions =

w w'

y'

-n

( 1

—

n )Y

-ny

The equation above then become s 1 1—n

w' + P(x)w = Q(x )

which is linear in w (since n  1) . Example 32 : Solve the equatio n Y ' + xY = xY 3

Note that this fits the form of the Bernoulli equation with n 3 . Therefore, the first step in solving it is to multiply through by y - n = Y -3 : y _3y' +x,-2 =x

(* )

Now for the substitutions ; the equation s w=y l-3 =y - 2

transform (*) int o

or, in standard form,

DIFFERENTIAL EQUATIONS

61



FIRST-ORDER EQUATIONS

Notice that the substitutions were successful in transforming th e Bernoulli equation into a linear equation (just as they wer e designed to be) . To solve the resulting linear equation, first determine the integrating factor : = e 2 µ(x) = e 1 = el- a& -x

Multiplying (* *) through the µ = e x` yields -

e -2w ' – 2xe -x`w = – 2xe -x ` (e' 2w)' = – 2xe -x ` And an integration gives e -x`w = e -x`w =

7

– 2xe -x` d.x e _x2

+ c

w = 1 + cex2 The final step is simply to undo the substitution w = solution to the original differential equation is therefore 1 y2

= 1 +ce

x

z

y- 2 .

Th e

■

CLIFFS QUICK REVIE W 62

SECOND-ORDER EQUATION S

Second-order differential equations involve the second derivativ e of the unknown function (and, quite possibly, the first derivative a s well) but no derivatives of higher order. For nearly every second order equation encountered in practice, the general solution wil l contain two arbitrary constants, so a second-order IVP must includ e two initial conditions .

Linear Combinations and Linear Independenc e Given two functions y 1 (x) andy Z (x), any expression of the form C LYI + c 2Y2 where c l and c 2 are constants, is called a linear combination of y l and y 2 . For example, if y 1 = ex and y 2 = x2 , the n 5ex + 7x 2 ,

ex – 2x 2 ,

and

–ex + x2

are all particular linear combinations of y l and y 2 . So the idea of a linear combination of two functions is this : Multiply the functions b y whatever constants you wish ; then add the products .

Example 1 : Is y = 2x a linear combination of the functions y, = x and y 2 = x 2? Any expression that can be written in the for m c lx+c 2x2

is a linear combination of x and x2 . Since y = 2x fits this form by taking c 1 = 2 and c2 = 0, y = 2x is indeed a linear combination of x and x2 . ■

DIFFERENTIAL EQUATIONS

63



SECOND-ORDER EQUATION S

Example 2 : Consider the three functions y, = sin x, y 2 = cosx, an d = sin(x + 1) . Show that y 3 is a linear combination of y, and y 2 .

y3

The addition formula for the sine function say s sin(x + 1) = cos 1 sin x + sin 1 cos x Note that this fits the form of a linear combination of sin x and cos x , c 1 sinx+c,cos x by taking c l = cos 1 and c, = sin 1 .

■

Example 3: Can the function y = x 3 be written as a linea r combination of the functions y l = x and y, = x2 ? If the answer were yes, then there would be constants c l and c 2 such that the equation x 3 =cox+c,x2

(* )

holds true for all values of x . Lettingx = 1 in this equation give s 1 = c~ +

c2

and lettingx = -1 gives -1=-c l +c , Adding these last two equations gives 0 = 2c 2, so c, = 0 . And since c 2 = 0, c, must equal 1 . Thus, the general linear combination (* ) reduces to x3 = x which clearly does not hold for all values of x . Therefore, it is not possible to write y = x3 as a linear combination of y l = x an d Y2 = x2 . ■

64

CLIFFS QUICK REVIEW



SECOND-ORDE R EQUATIONS

One more . definition : Two functions y, and y, are said to be linearly independent if neither function is a constant multiple of the other . For example, the functions y i = x 3 and y2 = 5x3 are not linearly independent (they're linearly dependent), since y2 is clearly a constant multiple ofy, . Checking that two functions are dependen t is easy ; checking that they're independent takes a little more work .

Example 4 : Are the functions y 1 (x) = sin x and y2 (x) = cos x linearl y independent ? If they weren't, then y 1 would be a constant multiple ofy, ; that is , the equation sinx=ccos x would hold for some constant c and for all x . But substituting x = 7r/2, for example, yields the absurd statement 1 = 0 . Therefore, the above equation cannot be true : y 1 = sin x is not a constant multipl e of y 2 cos x ; thus, these functions are indeed linearly indepen dent . ■

Example 5 : Are the functions y l = et- and y, = x linearly independent ? If they weren't, then y 1 would be a constant multiple of y2 ; that is, the equation = cx

would hold for some constant c and for all x . But this cannot happen, since substituting x = 0, for example, yields the absurd statement 1 = 0 . Therefore, y, = ex is not a constant multiple ofy, = x ; these two functions are linearly independent . ■

DIFFERENTIAL EQUATIONS

65



SECOND-ORDE R EQUATIONS

Example 6 : Are the functions y l = xee' and y 2 independent ?

=

ex linearl y

A hasty conclusion might be to say no because y l is a multiple o f But y, is not a constant multiple of y2 , so these functions truly ar e independent . (You may find it instructive to prove they're independent by the same kind of argument used in the previous tw o examples .) ■

y2 .

Linear Equation s Recall that the order of a differential equation is the order of th e highest derivative appearing in the equation . Thus, a second-orde r differential equation is one that involves the second derivative o f the unknown function but no higher derivatives . A second-order linear differential equation is one that can b e written in the form a(x)y" + b(x)y ' + c(x)y = d(x)

where a (x) is not identically zero . [For if a (x) were identically zero , then the equation really wouldn't contain a second-derivative term , so it wouldn't be a second-order equation .] If a(x)  0, then bot h sides of the equation can be divided through by a(x) and th e resulting equation written in the form y" + p(x)y' + q(x)y = r(x) It is a fact that as long as the functions p, q, and r are continuous o n some interval, then the equation will indeed have a solution (on that interval), which will in general contain two arbitrary constants (a s you should expect for the general solution of a second-orde r differential equation) . What will this solution look like? There is n o explicit formula that will give the solution in all cases, only variou s methods that work depending on the properties of the coefficient CLIFFS QUICK REVIE W

66



SECOND-ORDE R EQUATIONS

functions p, q, and r . But there is something definitive and very important that can be said about second-order linear equations .

Homogeneous Equation s There are two definitions of the term "homogeneous differentia l equation ." One definition (already encountered) calls a first-orde r equation of the form M(x, y) dx + N(x, y) dy = 0

homogeneous if M and N are both homogeneous functions of th e same degree . The second definition and the one which you'll se e much more often states that a differential equation (of any order ) is homogeneous if once all the terms involving the unknow n function are collected together on one side of the equation, th e other side is identically zero . For example , y " — 2y' + y = 0

is homogeneous

y" — 2y' + y = x

is no t

but

For the rest of this book, the term " homogeneous" will refer only t o this latter definition . The nonhomogeneous equatio n a(x)y" + b(x)y' + c(x)y = d(x)

(* )

can be turned into a homogeneous one simply by replacing th e right-hand side by 0 : a(x)y" + b(x)y' + c(x)y = 0

(** )

Equation (* *) is called the homogeneous equation corresponding t o the nonhomogeneous equation (*) . There is an important connec -

DIFFERENTIAL EQUATIONS

67



SECOND-ORDER EQUATIONS

tion between the solution of a nonhomogeneous linear equatio n and the solution of its corresponding homogeneous equation . Th e two principal results of this section are given below :

Theorem A . If y, (x) and y,(x) are linearly independent solutions o f the linear homogeneous equation (* * ), then every solution is a linea r combination of y, and Y, . That is, the general solution of the linea r homogeneous equation is Y = c 1YI + c ?Y2

Theorem B . If y(x) is any particular solution of the linear nonhomogeneous equation (*), and if y1, (x) is the general solution of th e corresponding homogeneous equation, then the general solution o f the linear nonhomogeneous equation i s Y = Y,, + y

That is , general solution of linear _ general solution of correspondin g homogeneous equatio n nonhomogeneous equation — + particular solution of give n nonhomogeneous equatio n [Note : The general solution of the corresponding homogeneou s equation, which has been denoted here by y,,, is sometimes calle d the complementary function of the nonhomogeneous equation (* ) . ] Theorem A can be generalized to homogeneous linear equations o f any order, while Theorem B as written holds true for linea r equations of any order . You can see Theorem B illustrated for a first-order equation by looking back at Example 29 on page 56 . Theorems A and B are perhaps the most important theoretical fact s about linear differential equations definitely worth memorizing .

CLIFFS QUICK REVIEW

68

SECOND-ORDE R EQUATIONS

Example 7: The differential equatio n y" – 2y' +y= 0 is satisfied by the function s y i = ek

and y, = xe'

Verify that any linear combination of y l and y, is also a solution o f this equation . What is its general solution ? Every linear combination of y i = ev and y, = xee' looks like this : y = c sex + c 2xev

for some constants c 1 and c, . To verify that this satisfies th e differential equation, just substitute . If y = c,et + c-axe', the n y ' =c~eA +c,e t (x+1 ) y"=c,ex

+c2

(x+2 )

Substituting these expressions into the left-hand side of the give n differential equation give s y" – 2y ' + y = [c i ev + c,e' (x + 2) ] – 2 [c i e' + c,e' (x + 1)] + [c 1 e' + c xe' ] = c l (e' – 2e' + e') + c,[e'(x + 2) – 2e'(x + 1) + xe' ]

=0

3

Thus, any linear combination of y, = e'- and y 2 = xe' does indee d satisfy the differential equation . Now, since y, = e' and y2 = xev are linearly independent (see Example 6), Theorem A says that th e general solution of the equation i s y = c 1 e' + c,xe

DIFFERENTIAL EQUATIONS

■

69



SECOND-ORDE R EQUATION S

Example 8 : Verify that y = 4x — 5 satisfies the equatio n y " +5y ' + 4y = 16x . Then, given that y, = e_a and y 2 = e-4x are solutions of th e corresponding homogeneous equation, write the general solution of the given nonhomogeneous equation .

First, to verify that y = 4x — 5 is a particular solution of th e nonhomogeneous equation, just substitute . If y = 4x — 5, then y' = 4 and y" = 0, so the left-hand side of the equation become s y"+5y' + 4y = 0 + 5(4) + 4(4x — 5 ) 3

= 16x

Now, since the functions y, = e_ x and y 2 = e-4x are linearly independent (because neither is a constant multiple of the other) , Theorem A says that the general solution of the correspondin g homogeneous equation i s yh

= c,e - .r

+

c2e

-4'

Theorem B then say s y=c,e -x

+c 2e -4v

+4x

— 5

is the general solution of the given nonhomogeneous equation .

■

Example 9: Verify that both y, = sin x and y 2 = cos x satisfy th e homogeneous differential equation y" + y = 0 . What then is th e general solution of the nonhomogeneous equation y" + y = x? If)), = sin x, then y ;' = —sin x, soy;' + y, does indeed equal zero . Similarly, if y2 = cos x, then y' = —cos x, so y + y2 is also zero, as desired . Since y, = sin x and y2 = cos x are linearly independent (se e

CLIFFS QUICK REVIE W 70



SECOND-ORDER EQUATIONS

Example 4), Theorem A says that the general solution of th e homogeneous equation y" + y = 0 i s yh = C 1

Sln x

+ C2

cos X

Now, to solve the given nonhomogeneous equation, all that i s needed is any particular solution . By inspection, you can see tha t y = x satisfies y" + y = x . Therefore, according to Theorem B, th e general solution of this nonhomogeneous equation i s y= c l sin x+ c 2 cos x+ x

■

Homogeneous Linear Equations with Constant Coefficient s The general second-order homogeneous linear differential equation has the form a(x)y"

+ b(x)y' + c(x)y

=0

[a(x) $ 0 ]

If a(x), b(x), and c(x) are actually constants, a(x) = a  0, b(x) = b , c(x) = c, then the equation becomes simpl y ay" + by' + cy = 0 This is the general second-order homogeneous linear equation wit h constant coefficients . Theorem A above says that the general solution of this equatio n is the general linear combination of any two linearly independen t solutions . So how are these two linearly independent solution s found? The following example will illustrate the fundamental idea .

DIFFERENTIAL EQUATIONS



SECOND-ORDE R EQUATIONS

Example 10: Solve the differential equationy " — y ' — 2y = 0 . The trick is to substitute y = e mx (m a constant) into th e equation ; you will see shortly why this approach works . If y = e"-' , then y' = me"'x and y " = m 2e"'-, so the differential equation become s m 2emx — me' — 2e"'x = 0 The term emx can be factored out and immediately canceled (sinc e e mx never equals zero) : emx(m2—m — 2) = 0 m 2 —m—2= 0 This quadratic polynomial equation can be solved by factoring : m 2 —m—2= 0 (m + 1)(m — 2) = 0 m=-1, 2 Now, recall that the solution began by writing y = e mu . Since th e values of m have now been found (m = -1, m = 2), bot h y = e -x

and y = e

are solutions . Since these functions are linearly independen t (neither is a constant multiple of the other), Theorem A says tha t the general linear combinatio n y

= c ie -x

+ c2e 2x

is the general solution of the differential equation .

■

Note carefully that the solution of the homogeneous differentia l equation ay" + by' + cy = 0

CLIFFS QUICK REVIEW



SECOND-ORDE R EQUATIONS

depends entirely on the roots of the auxiliary polynomial equatio n that results from substituting y = e' and then canceling out the e'm term . Once the roots of this auxiliary polynomial equation ar e found, you can immediately write down the general solution of th e given differential equation . Also note that a second-order linea r homogeneous differential equation with constant coefficients wil l always give rise to a second-degree auxiliary polynomial equation , that is, to a quadratic polynomial equation . The roots of any quadratic equatio n (a  0 )

am t +bm+c=0

are given by the famous quadratic formul a m=

–b ± \/b 2 –4a c 2a

The quantity under the square root sign, b 2 – 4ac, is called th e discriminant of the equation, and its sign determines the nature o f the roots . There are exactly three cases to consider . Case 1 : The discriminant is positive . In this case, the roots are real and distinct . If the two roots ar e denoted m 1 and m 2 , then the general solution of the differentia l equation is y= C

1 em

1 x+

c ge

m

2x

Case 2: The discriminant is zero . In this case, the roots are real and identical ; that is, th e polynomial equation has a double (repeated) root . If this doubl e root is denoted simply by m, then the general solution of th e differential equation is y=c l e'

DIFFERENTIAL EQUATIONS

+c 2xe"



SECOND-ORDE R EQUATIONS

Case 3: The discriminant is negative .

In this case, the roots are distinct conjugate complex numbers , r ± si . The general solution of the differential equation is the n y =e'=` (c~ cossx+c2sinsx )

So here's the process : Given a second-order homogeneou s linear differential equation with constant coefficients (a  0) , ay" + by' +cy= 0

immediately write down the corresponding auxiliary quadrati c polynomial equation am t +bm+c= 0

(found by simply replacingy" by m2, y' by m, and y by 1) . Determin e the roots of this quadratic equation, and then, depending o n whether the roots fall into Case 1, Case 2, or Case 3, write th e general solution of the differential equation according to the for m given for that Case .

Example 11 : Solve the differential equationy" + 3y' — 10y = 0 . The auxiliary polynomial equation i s m 2 +3m—10= 0 whose roots are real and distinct : m 2 + 3m — 10 = 0 (m + 5)(m — 2) = 0 m=—5, 2

This problem falls into Case 1, so the general solution of th e differential equation is y = c 1e -

74

5~

. +

c 2e2z-

■

CLIFFS QUICK REVIE W



SECOND-ORDER EQUATION S

Example 12 : Give the general solution of the differential equatio n y" —2y ' +y=0 . The auxiliary polynomial equation i s m 2 — 2m + 1 = 0 which has a double root : 2m+1= 0 (m 1) 2 = 0 m=

1

This problem falls into Case 2, so the general solution of th e differential equation is y = c+ c-ixe as you verified earlier in Example 7 .

■

Example 13 : Solve the differential equation y" + 6y ' + 25y = 0 . The auxiliary quadratic equation i s 2 +6m+25= 0

m22

which has distinct conjugate complex roots : m

-6± J6 2 —4 . 1 . 25 -6±~—64 2 .1 = 2

-6±8 i 2 = -3 ± 4i

This problem falls into Case 3, so the general solution of th e differential equation i s y =e -3x (c 1 cos 4x+c,sin4x )

DIFFERENTIAL EQUATIONS



SECOND-ORDER EQUATION S

Example 14 : Solve the IV P

First, rewrite the differential equation in standard form : 2y"+y' –y= 0

Next, form the auxiliary polynomial equatio n 2m22+m – 1 = 0

and determine its roots : 2m 2 + m – 1 = 0 (2m – 1)(m + 1) = 0 m= ;,– 1

Since the roots are real and distinct, this problem falls into Case 1 , and the general solution of the differential equation is therefor e y

=

c1e 2 +

c2e-a

(* )

All that remains is to use the two given initial conditions to determine the values of the constants c 1 and c2 : [c1ev/2 + c,e"] x= o = 1 =

[c1e2

– c

These two equations for c 1 and to yield 2c1

2e -A ID

c2

=o = 2

c1 + 2 – c2

1 =

2

can be solved by first adding them

=3c 1

= 2

then substituting c 1 = 2 back into either equation to find c, = -1 .

76

CLIFFS QUICK REVIEW



SECOND-ORDE R EQUATIONS

From (*), the solution of the given IVP is therefor e y=2e`/?_e

■

The Method of Undetermined Coefficient s In order to give the complete solution of a nonhomogeneous linea r differential equation, Theorem B says that a particular solutio n must be added to the general solution of the correspondin g homogeneous equation . Now that the method for obtaining th e general solution of a homogeneous equation (with constant coefficients) has been discussed, it is time to turn to the problem o f determining a particular solution of the original, nonhomogeneou s equation . If the nonhomogeneous term d(x) in the general second-orde r nonhomogeneous differential equatio n a(x)y" + b(x)y' + c(x)y = d(x)

(* )

is of a certain special type, then the method of undetermine d coefficients can be used to obtain a particular solution . The special functions that can be handled by this method are those that have a finite family of derivatives, that is, functions with the property tha t all their derivatives can be written in terms of just a finite number o f other functions. For example, consider the function d = sin x . Its derivatives ar e d' = cos x,

d" = —sin x,

d"' = —cos x,

d o "') = sin x ,

and the cycle repeats . Notice that all derivatives of d can be writte n in terms of a finite number of functions. [In this case, they are sin x and cosx, and the set {sinx, cosx} is called the family (of derivatives ) of d = sin x .] This is the criterion that describes those nonhomogeneous terms d(x) that make equation (*) susceptible to the metho d of undetermined coefficients : d must have a finite family. DIFFERENTIAL EQUATIONS



SECOND-ORDER EQUATION S

Here's an example of a function that does not have a finit e family of derivatives : d = tan x. Its first four derivatives ar e d'

= sec'' x,

d"' = 2 d O)

d" = 2

sec' x tan x ,

sec 4 x + 4 sec'x tan '- x ,

= 16 sec 4 x tan x + 8 sec' x tan' x

Notice that the nth derivative (n 1) contains a term involvin g tan''-' x, so as higher and higher derivatives are taken, each one wil l contain a higher and higher power of tan x, so there is no way that al l derivatives can be written in terms of a finite number of functions . The method of undetermined coefficients could not be applied i f the nonhomogeneous term in (*) were d = tan x . So just what are the functions d(x) whose derivative families are finite? See Table 3 below . Table 3 NONZERO FUNCTIONS WITH A FINITE FAMILY OF DERIVATIVE S Function

Family

k (k : a constant ) x" (n : a nonnegative integer ) ek `

Ixn ,xn-1 ,

,x, 1 1

Ieki }

sin kx

{sin kx, cos kx }

cos kx

{sinkx,coskx }

a finite product of any of the preceding types

all products of th e individual family members

CLIFFS QUICK REVIEW

78



SECOND-ORDE R EQUATIONS

Example 15 : If d(x) = 5x 2, then its family is {x2, x, 1 } . Note that any numerical coefficients (such as the 5 in this case) are ignored whe n determining a function's family . ■

Example 16 : Since the function d(x) = x sin 2x is the product of x and sin 2x, the family of d(x) would consist of all products of th e family members of the functions x and sin 2x . That is , family of x sin 2x = {x, 1} • {sin 2x, cos 2x } = {x sin 2x, x cos 2x, sin 2x, cos 2x}

■

Linear combinations of n functions . A linear combination of tw o functions y, and y2 was defined to be any expression of the form c j y i + c,y

2

where c l and c2 are constants . In general, a linear combination of n functions y,, y,, . . . , y,, is any expression of the for m d y +c I

1

2 y2

+

. .

.

+c„ y„

where c 1 , . . . , c,, are constants . Using this terminology, the nonhomogeneous terms d(x) which the method of undetermined coefficients is designed to handle are those for which every derivative ca n be written as a linear combination of the members of a given finit e family of functions . The central idea of the method of undetermined coefficients i s this : Form the most general linear combination of the functions i n the family of the nonhomogeneous term d(x), substitute thi s expression into the given nonhomogeneous differential equation , and solve for the coefficients of the linear combination .

DIFFERENTIAL EQUATIONS

79



SECOND-ORDER EQUATION S

Example 17 : Find a particular solution (and the complete solution ) of the differential equatio n y" + 3y' — 10y=5x2 As noted in Example 15, the family of d = 5x2 is {x 2 , x, 11 ; therefore, the. most general linear combination of the functions i n the family is y = Ax2 + Bx + C (where A, B, and C are th e undetermined coefficients) . Substituting this into the given differential equation give s (2A)+3(2Ax+B)— 10(Ax 2

+Bx+C) = 5x2

Now, combining like terms yield s (— 10A )x2

+

(6A — 1OB)x+(2A+3B— 10C)=5x2

In order for this last equation to be an identity, the coefficients o f like powers of x on both sides of the equation must be equated . Tha t is, A, B, and C must be chosen so tha t -10A = 5 6A — lOB= 0 2A +3B- IOC= 0 The first equation immediately gives A = - 4. Substituting this int o the second equation gives B = - 4 and finally, substituting both o f these values into the last equation yields C = Therefore, a particular solution of the given differential equation i s y =

-4x2 -

1(►

x

- ly

tO(►

According to Theorem B, then, combining thisy with the result o f Example 11 gives the complete solution of the nonhomogeneou s differential equation : y = c 1 e- 5-' + c2e 2x — 4- x2 — T}i x — ~<<~. ■

CLIFFS QUICK REVIEW

80



SECOND-ORDER EQUATION S

Example 18 : Find a particular solution (and the complete solution ) of the differential equatio n y" — 2y' + y = sin x Since the family of d = sin x is {sin x, cos x), the most genera l linear combination of the functions in the family is = A sin x + B cos x (where A and B are the undetermined coefficients) . Substituting this into the given differential equation give s (—A sin x — B cos x) — 2(A cos x — B sin x) + (A sin x + B cos x) = sin x Now, combining like terms and simplifying yield s (2B) sinx + (—2A) cosx = sin x In order for this last equation to be an identity, the coefficients A and B must be chosen so that 2B = 1

-2A = 0 These equations immediately imply A = 0 and B = A particula r solution of the given differential equation is therefor e y = ;cos x According to Theorem B, combining this y with the result o f Example 12 yields the complete solution of the given nonhomogeneous differential equation :y = c 1 ex + c2xe v + cosx . ■

DIFFERENTIAL EQUATIONS

81



SECOND-ORDER EQUATION S

Example 19 : Find a particular solution (and the complete solution ) of the differential equation y"+6y' +25y=8e -'x

Since the family of d = 8e-'x is just {e-'x }, the most general linear combination of the functions in the family is simply y = Ae - 'x (where A is the undetermined coefficient) . Substituting this into the given differential equation give s (49Ae -') + 6(– 7Ae - 'x ) + 25 (Ae - 'x) = 8e- 7x

Simplifying yields 32Ae-7x = 8e -7x

In order for this last equation to be an identity, the coefficient A must be chosen so that 32A = 8 which immediately gives A = A particular solution of the give n differential equation is therefor e Y = 4 e -7x

and then, according to Theorem B, combining y with the result o f Example 13 gives the complete solution of the nonhomogeneou s ■ differential equation : y = e_3x. (c 1 cos 4x + c 2 sin 4x) + e- 7-v.

Example 20 : Find the solution of the IV P y" – y' – 6y = – ex + 12x

y(0) = 1 y'(0) _ -2

CLIFFS QUICK REVIEW

82



SECOND-ORDER EQUATION S

The first step is to obtain the general solution of the correspond ing homogeneous equatio n y" — y' — 6y= 0 Since the auxiliary polynomial equation has distinct real roots , m 2 —m—6= 0

(m + 2)(m — 3) = 0 m = -2, 3 the general solution of the corresponding homogeneous equation i s Yh =

c 1 e-

+ c 2 e3x•

Now, since the nonhomogeneous term d(x) is a (finite) sum of functions from Table 3, the family of d(x) is the union of the familie s of the individual functions . That is, since the family of —e'" is {e'"} , and the family of 12x is (x, 11 , the family of —e`' + 12x = [the family of —e'' ) U [the family of 12x] = {e''} U {x, 1} = { e', x, 1 ) The most general linear combination of the functions in the famil y of d = —ex + 12x is therefore = Ae' + Bx + C (where A, B, and C are the undetermined coefficients) . Substituting this into the give n differential equation give s (Aex)—(Ae+B)—6(Ae+Bx+C)= —ex + 12x

Combining like terms and simplifying yield s (—bA)ex

+

(— 6B)x + (—B — 6C) =

+ 12x

DIFFERENTIAL EQUATIONS 83



SECOND-ORDE R EQUATIONS

In order for this last equation to be an identity, the coefficientsA, B , and C must be chosen so that - 6A=— 1 - 6B = 1 2 —B —

6C = 0

The first two equations immediately give A = 4 and B = - 2 , whereupon the third implies C = . A particular solution of th e given differential equation is therefor e Y = , e'—2x + According to Theorem B, then, combining this y with y,, gives th e complete solution of the nonhomogeneous differential equation : y = + c,e3r + -,,ea — 2x + . Now, to apply the initial conditions and evaluate the parameters c, and c, : y(0) = 1 = [c ) e -?~+ c,ev + y'(0) = -2

— 2x +

+ 3c,e 3x +

=1

c, + c, + = 1

— 2]X=O = - 2

-2c ) + 3c, — = - 2 Solving these last two equations yields c, = and c, = ~, . Therefore , the desired solution of the IVP i s y= (e' + 1)+yex +e')—2x

■

Now that the basic process of the method of undetermine d coefficients has been illustrated, it is time to mention that it isn' t always this straightforward . A problem arises if a member of a family of the nonhomogeneous term happens to be a solution of the corresponding homogeneous equation . In this case, that family must be modifie d before the general linear combination can be substituted into th e original nonhomogeneous differential equation to solve for th e undetermined coefficients . The specific modification procedure wil l be introduced through the following alteration of Example 20 . CLIFFS QUICK REVIE W

84



SECOND-ORDER EQUATION S

Example 21 : Find the complete solution of the differential equation Y"

Y' -6y= 10e

The general solution of the corresponding homogeneous equation was obtained in Example 20 : y1,

= c l e -'-~ + c,e k

Note carefully that the family {e3` } of the nonhomogeneous term d = 10e3; contains a solution of the corresponding homogeneou s equation (take c, = 0 and c, = 1 in the expression for yh ) . Th e "offending" family is modified as follows : Multiply each member of the family byx and try again . multiply eac h family membe r by x

{ e ~~ }

> {

Since the modified family no longer contains a solution of th e corresponding homogeneous equation, the method of undetermined coefficients can now proceed . (If xe 3- had been again a solution of the corresponding homogeneous equation, you woul d perform the modification procedure once more : Multiply each member of the family by x and try again .) Therefore, substituting = Axe = into the given nonhomogeneous differential equatio n yields (9Axe x + 6Ae x ) — (3Axe- + Ae 3~") — 6(Axe= ) = 10e = 5 Ae" = 10e = A= 2 This calculation implies that y = 2xe= is a particular solution of th e nonhomogeneous equation, so combining this with yh gives th e complete solution : y = c 1e _,x + c,e" + 2xe 3x

■

DIFFERENTIAL EQUATIONS

85



SECOND-ORDE R EQUATIONS

Example 22 : Find the complete solution of the differential equation y"–2y' =62–3e

2

First, obtain the general solution of the corresponding homogeneous equation y"–2y' = 0 Since the auxiliary polynomial equation has distinct real roots , m 2 – 2m = 0 m(m — 2) = 0 m=0, 2 the general solution of the corresponding homogeneous equation i s yh

= c i e~u+ c,eZ = c l + c,e '

The family for the 6x 2 term is {x2 , x, 11, and the family for the – 3ek term is simply {ex/2 } . This latter family does not contain a solution o f the corresponding homogeneous equation, but the family x, 1 } does (it contains the constant function 1, which matches yh whe n c l = 1 and c2 = 0) . This entire family (not just the "offending " member) must therefore be modified : "2

{x2,

multiply eac h family membe r by x

{x2,x, 1 }

> {x3 ,

x2,

x}

The family that will be used to construct the linear combination y i s now the union x l U frx/2} _ {x3 , x 2 , x, e- 2 } {x3, x2,

86

CLIFFS QUICK REVIE W

SECOND-ORDER EQUATIONS

This implies that y = Ax3 + Bx2 + Cx + Deg /2 (where A, B, C, and D are the undetermined coefficients) should be substituted into th e given nonhomogeneous differential equation . Doing so yield s (6Ax+2B+4De al ')–2(3Ax2

+2Bx+C+ 4Deg/2 ) =6x 22 –3e

/2

which after combining like terms read s ( – 6A)x2 +(6A – 4B)x+(2B–2C)+(–4De 2 )=6x2 –3e-12 In order for this last equation to be an identity, the coefficients A, B , C, and D must be chosen so tha t -6A = 6 6A – 4B = 0 2B – 2C = 0 4D

= -3

These equations determine the values of the coefficients : A = -1 , B = C = –z, and D = 4 . Therefore, a particular solution of th e given differential equation i s y= —x3 –4x--4x+4e4 According to Theorem B, then, combining this y with y', gives th e complete solution of the nonhomogeneous differential equation : y=c 1 + c 2e 2x — x 3 — 4x2 — 4x + 4e" . ■

DIFFERENTIAL EQUATIONS

87

SECOND-ORDE R EQUATIONS

Example 23 : Find the complete solution of the equatio n y" + 4y = x sin 2x + 8 First, obtain the general solution of the corresponding homogeneous equation y" + 4y = 0 Since the auxiliary polynomial equation has distinct conjugat e complex roots, m2 + 4 = 0 m = ±2i the general solution of the corresponding homogeneous equation i s yh

=c 1 cos2x+c2 sin 2x

Example 16 above showed that th e family of x sin 2x = {x, 1} • {sin 2x, cos 2x } = {x sin 2x, x cos 2x, sin 2x, cos 2x } Note that this family contains sin 2x and cos 2x, which are solution s of the corresponding homogeneous equation . Therefore, this entire family must be modified :

{x sin 2x, x cos 2x, sin 2x, cos 2x}

multiply eac h family membe r by x

{x 2 sin 2x, x 2 cos 2x, x sin 2x, x cos 2x } None of the members of this family are solutions of the correspond ing homogeneous equation, so the solution can now proceed a s usual . Since the family of the constant term is simply Ill, the famil y used to construct y is the union

CLIFFS QUICK REVIE W

88



SECOND-ORDE R EQUATIONS

{x 2

sin 2x, x 2 cos 2x, x sin 2x, x cos 2x} U {1 } = {x' sin 2x, x 2 cos 2x, x sin 2x, x cos 2x, 1 1

This implies tha t .-j; = Ax' sin 2x + Bx' cos 2x + Cx sin 2x + Dx cos 2x + E (where A, B, C, D, and E are the undetermined coefficients ) should be substituted into the given nonhomogeneous differentia l equation y" + 4y = x sin 2x + 8 . Doing so yield s [–8Bx+(2A -4D)]sin2x+ [8Ax + (2B + 4C)] cos 2x + 4E = x sin 2x + 8 In order for this last equation to be an identity, A, B, C, must be chosen so that

D,

and E

-8B = 1

2A–4D= 0 8A = 0 2B + 4C = 0

4E = 8 These equations determine the coefficients : A 0, B = –,, C = -116, D = 0, and E = 2 . Therefore, a particular solution of the give n differential equation i s y= --x'cos 2x+x sin2x+ 2 According to Theorem B, then, combining this y with y,, gives th e complete solution of the nonhomogeneous differential equation : ■ y = c 1 cos2x+c2sin2x–4-x'cos2x+ xsin2x+2 .

DIFFERENTIAL EQUATIONS

89



SECOND-ORDER EQUATIONS

Variation of Parameter s For the differential equatio n a(x)y " + b(x)y' + c(x)y = d(x)

(* )

the method of undetermined coefficients works only when th e coefficients a, b, and c are constants and the right-hand term d(x) i s of a special form . If these restrictions do not apply to a given nonhomogeneous linear differential equation, then a more powerful method of determining a particular solution is needed : th e method known as variation of parameters . The first step is to obtain the general solution of the correspond ing homogeneous equation, which will have the for m Yh =

C

IY1

+ c 2Y 2

where y, and Y 2 are known functions . The next step is to vary th e parameters ; that is, to replace the constants c, and c, by (as ye t unknown) functions 7' 1(x) and v 2(x) to obtain the form of a particular solution y of the given nonhomogeneous equation : Y

= z ► , (x )Y

1+

v 2(x )Y ,

The goal is to determine these functions v and v 2 . Then, since the functions y, and Y2 are already known, the expression above for y yields a particular solution of the nonhomogeneous equation . Combining y with y,, then gives the general solution of the non homogeneous differential equation, as guaranteed by Theorem B . Since there are two unknowns to be determined, v, and v 2, tw o equations or conditions are required to obtain a solution . One o f these conditions will naturally be satisfying the given differentia l equation . But another condition will be imposed first. Sincey will b e substituted into equation (*), its derivatives must be evaluated . Th e first derivative of y- i s Y'

90

=

7 '1Yi + 7 'iYi +

+ 7 ''L;Y 2 CLIFFS QUICK REVIEW



SECOND-ORDER EQUATION S

Now, to simplify the rest of the process and to produce the firs t condition on v l and v 2 set +

V zY2

=

0

This will always be the first condition in determining v 1 and v, ; th e second condition will be the satisfaction of the given differentia l equation (*) .

Example 24: Give the general solution of the differential equation y" + y = tan x. Since the nonhomogeneous right-hand term, d = tan x, is not of the special form that the method of undetermined coefficients can handle, variation of parameters is required . The first step is t o obtain the general solution of the corresponding homogeneou s equation, y" + y = O . The auxiliary polynomial equation i s m2 + 1 = 0 whose roots are the distinct conjugate complex numbers m = ±i = 0 ± li . The general solution of the homogeneous equation is therefore y = c I sinx+c,cos x

Now, vary the parameters c l and c, to obtai n y = v sin x +

v2

cos x

Differentiation yield s = v, cos x + v I sin x – v, sin x + v ; cos x Next, remember the first condition to be imposed on v l and v ;Y 1

DIFFERENTIAL EQUATIONS

v2 :

+ v;y, = 0

91



SECOND-ORDER EQUATIONS

that is, I) ; sin x + v; cos x = 0

(1 )

This reduces the expression for y' t o y ' = v1 cos x — II2 sin x

so, then,

= — v1 sin x + vj cos x — v2

cos

x—

z)4

sin x

Substitution into the given nonhomogeneous equation y" + y = tan x yield s (— v 1 sin x + v ; cos x — v2 cos x — v ; sin x) + (v 1 sin x + v2 cosx ) = tan x v;

cos

x — v2 sin x = tan x (2 )

Therefore, the two conditions on vl and v 2 are v ; sin x + v ; cos x = 0 v cos x — v sin x = tan x

(Z )

To solve these two equations for v ; and v, first multiply the firs t equation by sin x; then multiply the second equation by cos x : v sin 2 x + v ; cos x sin x = 0 v ; co s 2 x - v2 sin x cos x = sin x Adding these equations yield s v ; (sin e x + cos2 x) = sin x

= sin x

Substituting v ; = sin x back into equation (1) [or equation (2)] the n gives z~2 =

92

—tan x sin x

CLIFFS QUICK REVIEW



SECOND-ORDER EQUATIONS

Now, integrate to find v> and integration in each case) :

V2

(and ignore the constant o f

= v;

=f

sin xdx

= –cos x and v2

=

cos x

dx

cos t x – 1 cos x

dx

= (cos x – sec x) dx = sin x — In Isecx + tan x Therefore, a particular solution of the given nonhomogeneou s differential equation i s y= V I

sin x + V 2 cos x

– cos x

sin x + [sin x – In ~ sec x + tan x ] cos x

= –cosx In ( sec x + tan x l Combining this with the general solution of the correspondin g homogeneous equation gives the general solution of the nonhomogeneous equation : y = c, sin x +

DIFFERENTIAL EQUATIONS

C2

cos x — cos x In I sec x + tan x~

■

93



SECOND-ORDE R EQUATIONS

In general, when the method of variation of parameters i s applied to the second-order nonhomogeneous linear differentia l equation a(x)y" + b(x)y' + c(x)y = d(x ) with y = v i (x)y, + v 2 (x)y2 (where y h = c, y, + c2y2 is the genera l solution of the corresponding homogeneous equation), the tw o conditions on v, and V2 will always b e v iYi + v 2' y 2 = 0 a (x )[v i y i + v 1Yfl = d(x)

(1) (2)

So after obtaining the general solution of the correspondin g homogeneous equation (yh = + c,y,) and varying the parameters by writingy = v,y, + v,y,, go directly to equations (1) and (2 ) above and solve for v ; and v .

Example 25 : Give the general solution of the differential equation y" — 2y' + y = e~ In x Because of the In x term, the right-hand side is not one of th e special forms that the method of undetermined coefficients ca n handle ; variation of parameters is required . The first step, obtainin g the general solution of the corresponding homogeneous equation , y" — 2y' + y = 0, was done in Example 12 : yh

=c,y, +c,y, =c l ev +c,xe '

Varying the parameters gives the particular solutio n Y

=z 'iYi +zv 2y; =v i e' +v,xe4

and the system of equations (1) and (2) become s v>e l

+ 7~2xe'

0

v eA+v ;ex (x+ 1) =eIn x

94

(1 ) (2 )

CLIFFS QUICK REVIEW



SECOND-ORDE R EQUATIONS

Cancel out the common factor of ex in both equations ; then subtrac t the resulting equations to obtai n v4 = In x Substituting this back into either equation (1) or (2) determine s v ; = –x In x Now, integrate (by parts, in both these cases) to obtain v l and from v ; and v 2' :

f xlnx=

1 1 (lnx)~x-) — I

f (!)(!

V2

)

=1x 2 (1 – 2 Inx ) v2

= f v l; = f Inx dx =

(ln x)(x)

1 (x)xdx =xlnx— 1

Therefore, a particular solution i s Y

+

=

7 '2Y 2

= v > eA + v,xe x

=4x 2 (1 – 21nx)e~ +x(lnx – 1)xe x = ~'-x 2ex (21n x – 3 ) Consequently, the general solution of the given nonhomogeneou s equation is y=

+ c 2xex +

DIFFERENTIAL EQUATIONS

x 2ea (21n x – 3)

■

95



SECOND-ORDER EQUATION S

Example 26 : Give the general solution of the following differential equation, given that y i = x and y 2 = x are solutions of its corresponding homogeneous equation :

3

2y " –3xy' + 3y = 12x 4 Since the functionsy 1 = x and = x 3 are linearly independent , x

y2

Theorem A says that the general solution of the correspondin g homogeneous equation i s yh

= Y I +c,y,=c 1 x+c,x

3

Varying the parameters c l and c2 gives the form of a particular solution of the given nonhomogeneous equation : .-Y- =

v 1YI + 1'2Y 2 = v 1 x + v,x 3

where the functions v i and V 2 are as yet undetermined . The two conditions on v> and V2 which follow from the method of variation o f parameters are v ;y 1 + v,y, = 0

(1 )

a (x) [v i y i + v ; y ;] = d(x) (2 ) which in this case (y, = x, y 2 = x 3 , a = x 2, d = 12x4 ) becom e v ;x + v ;x 3 = 0 x 2 [v j + v4 3x22 ] = 12x 4 (2) Solving this system for v ; and v ; yields v ; = – 6x

2

and v ; = 6

from which follow v,

v

96

(–6x 2) dx = - 2x

= 7, ; =

2=

f

v;=

f

3

6dx=6x

CLIFFS QUICK REVIEW



SECOND-ORDE R EQUATION S

Therefore, the particular solution obtained i s Y = z 'IY1 = V lx = (—

+ z '2Y 2

+ v,x 3

2e)x

+

(6 )x '

Y = 4x 4 and the general solution of the given nonhomogeneous equation i s y=c l x+c,x 3 +4x 4

■

The Cauchy-Euler Equidimensional Equatio n The second-order homogeneous Cauchy-Euler equidimensiona l equation has the form ax 2y "

+

bxy' + cy = 0

where a, b, and c are constants (and a  0) . The quickest way t o solve this linear equation is to substitute y = x"t and solve for m . I f y = x m , then y' = mxm-1 and y" = m (m — 1)x m- 2 so substitution into the differential equation yield s am(m —

1)m

+ bmxm + cx"' = 0

x m [am(m — 1) + bm + c] = 0 am(m — 1) + bm + c = 0 am' + (b — a)m + c = 0 (* )

DIFFERENTIAL EQUATIONS

97



SECOND-ORDE R EQUATIONS

Just as in the case of solving second-order linear homogeneou s equations with constant coefficients (by first setting y = e lm'. an d then solving the resulting auxiliary quadratic equation for m), thi s process of solving the equidimensional equation also yields a n auxiliary quadratic polynomial equation . The question here is, ho w isy = x m to be interpreted to give two linearly independent solution s (and thus the general solution) in each of the three cases for th e roots of the resulting quadratic equation ? Case 1 : The roots of (*) are real and distinct. If the two roots are denoted m 1 and m,, then the general solution of the second-order homogeneous equidimensional differential equation in this case i s Y

=

C 1 xm'

+

C Zx" ,

Case 2: The roots of (*) are real and identical. If the double (repeated) root is denoted simply by m, then th e general solution (for x > 0) of the homogeneous equidimensiona l differential equation in this case i s y = clx m +c,x"'ln x

Case 3 : The roots of (*) are distinct conjugate complex numbers . If the roots are denoted r ± si, then the general solution of th e homogeneous equidimensional differential equation in this case is y = x'"[c 1 cos(s In x) + c, sin(s In x) ]

Example 27 : Give the general solution of the equidimensiona l equation x 2y"—3xy' + 3y = 0

98

CLIFFS QUICK REVIEW



SECOND-ORDE R EQUATIONS

Substitution ofy = x m results i n x 2 m(m — 1).x"' -2 — 3x•mxm- '

+

3x"' = 0

x"'[m(m— 1)—3m+3]= 0 m(m — 1) — 3m + 3 = 0 m 2 — 4m + 3 = 0 (m — 1)(m — 3) = 0 m = 1, 3 Since the roots of the resulting quadratic equation are real an d distinct (Case 1), both y = x' = x and y = x 3 are solutions an d linearly independent (see Example 17), and the general solution o f this homogeneous equation i s y =cox+c,x 3 (Compare this result with the statement of Example 26 .)

■

Example 28 : For the following equidimensional equation, give th e general solution which is valid in the domain x > 0 : x 2y " —3xy ' + 4y = 0 Substitution ofy = x"' results i n x2 . m (m — 1)x"n — 3x • mx,m — 1 + 4x "' = 0 x"'[m(m— 1) — 3m + 4] = 0 m(m — 1) — 3m + 4 = 0 m 2 —4m+4= 0 (m—2) 2 = 0 m = 2, 2

DIFFERENTIAL EQUATIONS

99



SECOND-ORDE R EQUATIONS

Since the roots of the resulting quadratic equation are real an d identical (Case 2), both y = x 2 and y = x 22 In x are (linearl y independent) solutions, so the general solution (valid for x > 0) o f this homogeneous equation i s y=c l x '-+c,x 2 ln x

■

If the general solution of a nonhomogeneous equidimensional equation is desired, first use the method above to obtain the general solution of the corresponding homogeneous equation ; then apply variation of parameters (see Example 26) .

Reduction of Orde r Some second-order equations can be reduced to first-order equations, rendering them susceptible to the simple methods of solvin g equations of the first order . Three particular types of such secon d order equations will be discussed in this section . Type 1 : Second-order equations with the dependent variabl e missin g Type 2 : Second-order nonlinear equations with the independen t variable missing Type 3 : Second-order homogeneous linear equations where on e (nonzero) solution is know n

Type 1 : Second-order equations with the dependent variable missing. Examples of such equations includ e y" + y' = x

and xy" — 2y' = 12x 2

The defining characteristic is this : The dependent variable, y, doe s not explicitly appear in the equation . This type of second-orde r

100

CLIFFS QUICK REVIEW



SECOND-ORDE R EQUATIONS

equation is easily reduced to a first-order equation by the transformation = w

Y

This substitution obviously implies y" = w', and the origina l equation becomes a first-order equation for w . Solve for th e function w ; then integrate it to recovery .

Example 29: Solve the differential equation y" + y' = x . Since the dependent variable y .is missing, let y' = w and y" = w' . These substitutions transform the given second-order equation int o the first-order equation w'+w= x

which is in standard form . Applying the method for solving suc h equations, the integrating factor is first determined , µ=e fPd'- = e f ` x = e x and then used to multiply both sides of the equation, yieldin g + exw = xex d dx (e

xw) = xex e xw

f xex. dx = xe x — f ex dx

=

= xe x — (ea + c l )

Therefore, w=x — 1 —c 1 e

DIFFERENTIAL EQUATIONS

101



SECOND-ORDER EON PONS

Now, to give the solution y of the original second-order equation , integrate : y' = w may=

w=

(x - 1 —c l e -x )dx

This gives y=2x 2

—x+c l e -x +c 2

Referring to Theorem B, note that this solution implies that y = c l e -x + c 2 is the general solution of the corresponding homogeneou s equation and that y = 2x 2 – x is a particular solution of th e nonhomogeneous equation . (This particular differential equatio n could also have been solved by applying the method for solvin g second-order linear equations with constant coefficients .) ■

Example 30 : Solve the differential equatio n xy" – 2y' = 10x 4

Again, the dependent variable y is missing from this second order equation, so its order will be reduced by making th e substitutions y' = w and y" = w' : xw' — 2w = 10x 4

which can be written in standard form 2 w' -- w= lox = x The integrating factor here i s

µ

= efPdx = ef(-2/x)dx = e-21nx _ eln(x - `' ) = x - 2

which is used to multiply both sides of the equation, yieldin g

102

CLIFFS QUICK REVIEW



SECOND-ORDE R EQUATIONS

x -2w' – 2x -3w = 10x d dx (x

-2w) = 1Qx

x-2w=

f 10xdx

x - 2w =5x 2 +c ; w=5x 4 +c ;x Integrating w gives y : y' =w 'y= =

w +

-(5x4

= x5 +

c ;x 2 ) dx

cx 3

+c 2

Letting c l = 6, , the general solution can be writte n y=x5 +c lx 3 +c 2

n

Example 31 : Sketch the solution of the IV P y"+ (Y ' ) 2= 0 y(0) = 1 y' (0) = 1 Although this equation is nonlinear [because of the term (y') 2 ; neither y nor any of its derivatives are allowed to be raised to an y power (other than 1) in a linear equation], the substitutions y ' = w and y" = w' will still reduce this to a first-order equation, since th e variable y does not explicitly appear . The differential equation i s transformed into w' + w 2 = 0

DIFFERENTIAL EQUATIONS 103



SECOND-ORDER EQUATIONS

which is separable : dw

—w -

dw = dx wdw = d — x w1 — =x+c i w w =

1 x + c1

Since y' = w, integrating gives y

w

1 dx= ln(x+c I ) + c , x+c

Now apply the initial conditions to determine the constants c 1 an d c2: y(0) = 1 y' ( 0) = 1

[ln(x + c i ) + c,]a.

= 1 = In c 1 + c, = 1

[(x + c l) -' ]X_(► = 1 = c i = 1

Because c, = 1, the first condition then implies c_, = 1 also . Thus th e solution of this IVP (at least forx > -1) i s y=ln(x+ 1) + 1 whose graph is shown in Figure 7 .

104

CLIFFS QUICK REVIEW

SECOND-ORDE R EQUATIONS

■ Figure 7

■

Type 2 : Second-order nonlinear equations with the independen t variable missing . Here's an example of such an equation : y 2y" - (y' ) 3 = 0

The defining characteristic is this : The independent variable, x, doe s not explicitly appear in the equation . The method for reducing the order of these second-orde r equations begins with the same substitution as for Type 1 equations , namely, replacingy' by w . But instead of simply writingy" as w', th e trick here is to express y" in terms of a first derivative with respect to y . This is accomplished using the chain rule : d dy

y"

dz. dx

dw =

dx

dw dy =

dydx

dw = dy Y

Therefore, dw Y "=w d Y

This substitution, along with y' = w, will reduce a Type 2 equatio n to a first-order equation for w . Once w is determined, integrate to find y .

DIFFERENTIAL EQUATIONS

105



SECOND-ORDER EQUATIONS

Example 32 : Solve the differential equatio n y 22y"_ ( y' ) 3= 0

The substitutions y' = w and y" = w(dw/dy) transform this second-order equation for y into the following first-order equatio n for w : y2

dw w d –w'= 0 y

~dw

wy- d

y

– w-= 0

Therefore, w = 0 or

dw y2–w2=

dy

0

The statement w = 0 means y' = 0, and thus y = c is a solution fo r any constant c . The second statement is a separable equation, an d its solution proceeds as follows :

y

dw dy dw w2

= w2

dy – y 2

dw 1dy w2 –

y

1

1

W

106

CLIFFS QUICK REVIEW



SECOND-ORDE R EQUATIONS

Now, since w = dy/dx, this last result become s dx 1 dy =

+Cl

y

(y - ' + c,) dy = dx (y - ' + c j ) dy = dx

which gives the general solution, expressed implicitly as follows : lnIyl +cry=x+c 2 Therefore, the complete solution of the given differential equatio n is y=c

or

In I y~+cry=x+c,

■

Type 3 : Second-order homogeneous linear equations where one (nonzero) solution is known . Sometimes it is possible to determin e a solution of a second-order differential equation by inspection , which usually amounts to successful trial and error with a fe w particularly simple functions . For example, you might discover tha t the simple function y = x is a solution of the equation x 2 y " — xy' + y = 0 or that y = ex satisfies the equatio n .xy—(x+l)y'+y= 0 Of course, trial and error is not the best way to solve an equation , but if you are lucky (or practiced) enough to actually discover a solution by inspection, you should be rewarded .

DIFFERENTIAL EQUATIONS

107



SECOND-ORDE R EQUATIONS

If one (nonzero) solution of a homogeneous second-orde r equation is known, there is a straightforward process for determining a second, linearly independent solution, which can then b e combined with the first one to give the general solution . Let y , denote the function you know is a solution . Then let y = y 1 v(x) , where v is a function (as yet unknown) . Substitute y = y i z' into th e differential equation and derive a second-order equation for v . Thi s will turn out to be a Type 1 equation for v (because the dependen t variable, v, will not explicitly appear) . Use the technique describe d earlier to solve for the function v ; then substitute into the expressiony = y i v to give the desired second solution .

Example 33 : Give the general solution of the differential equatio n x2y" —xy' +y= 0 As mentioned above, it is easy to discover the simple solutio n y = x. Denoting this known solution by y l , substitute y = y l v = xv into the given differential equation and solve for v . If y = xv, the n the derivatives are y' =xv' + v y " =xv " +2v ' Substitution into the differential equation yield s x 2 (xv" + 2v') — x(xv' + v) + xv = 0 x 3 v" + x-'v ' = 0 71

108

'

- v = 0 x

+ -

CLIFFS QUICK REVIEW



SECOND-ORDE R EQUATIONS

Note that this resulting equation is a Type I equation for v (becaus e the dependent variable, v, does not explicitly appear) . So, letting v' = w and v " = w', this second-order equation for v becomes th e following first-order equation for w : W' +

— x

W

= 0

(*)

The integrating factor for this standard first-order linear equation i s

µ = e 1'

=

ef(1 /x)dx

=

'elnx = x

and multiplying both sides of (*) by ,u, = x give s xw' +w= 0 d dx (xw)

0

xw= c w=cx - ' Ignore the constant c and integrate to recover v : v'=w~v=

fw=

x - 'dx=lnIx '

Multiply this by y, to obtain the desired second solution , y, = y,v = x In I x The general solution of the original equation is any linear combination of y 1 = x and y 2 = x In lx l : y = c i x + c2x In 'x i This agrees with the general solution that would be found if thi s problem were attacked using the method for solving an equidimensional equation . ■

DIFFERENTIAL EQUATIONS 109



SECOND-ORDER EQUATIONS

Example 34 : Determine the general solution of the followin g differential equation, given that it is satisfied by the function y = ex : xy" — (x + 1)y' + y = 0 Denoting the known solution by y 1, substitute y = y l z) = exv int o the differential equation . With y = exv, the derivatives are y' = ex v ' + exv y ►► = e z

x ,"

+

2ex v' + exv

Substitution into the given differential equation yield s x(exv " + 2ex v ' + e xv) — (x + 1) (e xv ' + exv) + exv = 0

which simplifies to the following Type 1 second-order equation fo r v: xv " + (x — 1)v' = 0 Letting v ' = w, then rewriting the equation in standard form, yield s w' +

x— 1 w = 0 (* ) x

The integrating factor in this case is e fPdx

e

fIl-(l/x)]dx

e x-inx- =eXe -inx =

ex

x Multiplying both sides of (*) by p . = ex /x yield s d ex dx x w

w = cxe -x

110

CLIFFS QUICK REVIE W

SECOND-ORDER EQUATIONS

Ignore the constant c and integrate to recover v : v =

= fxedx = —xe -x —

— e' dx

= —xe -x — e ' e -x(x + 1 ) Multiply this by y 1 to obtain the desired second solution , Y2=y1v

= ex

• [_ e_x (x

+ 1)] = —(x + 1 )

The general solution of the original equation is any linear combination ofy, and y 2 :

DIFFERENTIAL EQUATIONS



POWER SERIES

It often happens that a differential equation cannot be solved i n terms of elementary functions (that is, in closed form in terms o f polynomials, rational functions, ex , sin x, cos x, In x, etc .) . A power series solution is all that is available . Such an expression is nevertheless an entirely valid solution, and in fact, many specifi c power series that arise from solving particular differential equation s have been extensively studied and hold prominent places in mathematics and physics .

Introduction to Power Serie s A power series in x about the point xo is an expression of the for m co + c 1(x

—

x0) +

c2(x — x0)2 +

where the coefficients c,, are constants . This is concisely written using summation notation as follows :

n= 0

cn (x — xo ) n

Attention will be restricted to x 0 = 0 ; such series are simply called power series in x : co + c 1 x + c 2 x2 +

=

n= 0

c n.x' n

A series is useful only if it converges (that is, if it approaches a finit e limiting sum), so the natural question is, for what values of x will a given power series converge? Every power series in x falls into on e of three categories :

DIFFERENTIAL EQUATIONS



POWER SERIE S

Category 1 : The power series converges only forx = 0 . Category 2 : The power series converges for G R and diverges (that is, fails to converge) for Ix' > R (where R i s some positive number) . Category 3 : The power series converges for all x. Since power series that converge only for x = 0 are essentiall y useless, only those power series that fall into Category 2 or Category 3 will be discussed here . The ratio test says that the power serie s

E Cnxf

n= 0

will converge if lim n-~x

n+ l Cn+l x C~ n

<1

(* )

and diverge if this limit is greater than 1 . But (*) is equivalent t o < lim n—x

Cn cn+ 1

so the positive number R mentioned in the definition of a Category 2 power series is this limit : R= lim

Cn

n->x

If this limit is co, then the power series converges for (xi < oo which means for all xand the power series belongs to Category 3 . R i s called the radius of convergence of the power series, and the set o f all x for which a real power series converges is always an interval , called its interval of convergence .

114

CLIFFS QUICK REVIEW



POWE R SERIES

Example 1 : Find the radius and interval of convergence for each o f these power series : 2"

x

(a ) ,1=

x

0n!

,1 =1

x

n3

3

(c)

,1

—

X"

,1=1 n

[Recall that n ! ("n factorial") denotes the product of the positiv e integers from 1 to n . For example, 4! = 1 • 2 • 3 • 4 = 24 . By definition, 0! is set equal to 1 . ] (a) In this power series, c,, = 2"/n !, so the ratio test says R = li m

it —~x

cil C r1 +

2'' (n +

1

I ) !

2,1+1

= li m n! • 11

n+ 1

= lim

2

=

x

Therefore, this series converges for all x . (b) The radius of convergence of the power series in (b) i s R = li m PI —~ x

C il

C"+ 1

li m it — x

n3

3 n+ 1

3" • (n +1) 3 = lim

n3

(n + l)

3,1+ 1

3"

=1 .3= 3

Since R = 3, the power series converges for Ix ~ < 3 and diverges for (x > 3 . For a power series with a finite interval of convergence, the question of convergence at the endpoints of the interval must b e examined separately. It may happen that the power series converge s at neither endpoint, at only one, or at both . The power serie s

±a

i1 =

1

il

Xf l

converges at neither the endpoint x = 3 nor x = -3 because the individual terms of both resulting serie s

DIFFERENTIAL EQUATIONS



POWER SERIES

1n

x

3

E

and

n=1

(-1)"n

3

it= 1

clearly do not approach 0 as n -> oc . (For any series to converge, it i s necessary that the individual terms go to 0 .) Therefore, the interva l of convergence of the power series in (b) is the open interval -3 < x < 3. (c) The radius of convergence of this power series is R = lim

„-max

c,, c„ + 1

1

n+1

n

1

= lim - • ►t-~x

= lim

,r->x

n+ 1

n

= 1

Since R = 1, the series

converges for Ix' < 1 and diverges for Ix' > 1 . Since this power series has a finite interval of convergence, the question of convergence at the endpoints of the interval must be examined separately . At the endpoint x = -1, the power series become s

I (-o

n= 1

nn

which converges, since it is an alternating series whose terms go t o 0. However, at the endpointx = 1, the power series become s

n=1 n

which is known to diverge (it is the harmonic series) . Therefore, th e interval of convergence of the power serie s x

1 _x”

n=1 n

is the half-open interval -1 < x < 1 .

116

■ CLIFFS QUICK REVIE W



POWER SERIE S

Taylor Serie s A Category 2 or Category 3 power series in x defines a function f by setting f (x) = co + c 1 x + c2 x 2 + c3x3 + c 4 x4

+ .. .=

c,,xn n=( )

(0)

for any x in the series' interval of convergence . The power series expansion for f (x) can be differentiated term by term, and the resulting series is a valid representation of f'(x) i n the same interval : f '(x) = c l + 2c 2 x + 3c 1x2

+

4c4 x3

+•••=

E nc„x" -'

,,= 1

(1 )

Differentiating again gives f " (x) = 2c2

+

6c3x+ 12c 4 x2

+ ••• =

n= 2

n(n —

1)c„x" -2

(2 )

and so on . Substitutin g x = 0 in equation (0) yields co = f (0) , x = 0 in equation (1) yields c l = f '(0) , f"(0 ) x = 0 in equation (2) yields c 2 = 2 , and in general, substitutingx = 0 in the power series expansion fo r the nth derivative off yields c„ _

DIFFERENTIAL EQUATIONS

f (0 0 )

n!



POWE R SERIES

These are called the Taylor coefficients off, and the resulting powe r series 'f(")(0) Yi

n=

o n .!

is called the Taylor series of the function

f.

Given a function f, its Taylor coefficients can be computed by th e simple formula above, and the question arises, does the Taylo r series off actually converge to f (x)? If it does, that is, i f f(x)=E f rt=0

(YI)(0) n.

x

Y1

for all x in some neighborhood of (interval around) 0, then th e function f is said to be analytic (at 0) . [More generally, if you form the Taylor series off about a pointx = xo , x f (") (xo) n=o

n•

Y (x — xo) '

and if this series actually converges to f (x) for all x in some neighborhood of xo, then f is said to be analytic at xo .] Polynomials are analytic everywhere, and rational functions (quotients of polynomials) are analytic at all points where the denominator is not zero . Furthermore, the familiar transcendental (that is, nonalgebraic ) functions e x, sin x, and cos x are also analytic everywhere . The Taylo r series in Table 4 are encountered so frequently that they are wort h memorizing . For a general power series (like the ones in Example 1), it i s usually not possible to express it in closed form in terms of familia r functions .

118

CLIFFS QUICK REVIEW



POWE R SERIES

Table 4 TAYLOR SERIE S f(n)o

f(x)

1

x



x2

x3

Ix"

(for Ix'

"=()

n=()

(for all x )

i n.

x5

,.2n+ 1

+ — ••• = I(—1) " 3! 5 ! (2n + 1) ! ll=O 2

4

cos x = 1 — ~+ i 2.

< 1)

xn

x3

ex= 1+x+ i+ i+•• — 2. 3.

sinx=x —

n!

ll=O

1 — x = 1 +x +x 2 +x 3 + . . . =

) x"



4.

. .=E(—1)" x 2i t (2n) . n=( )

(for all x )

(for all x)

Example 2 : Use Table 4 to find the Taylor series expansion of eac h of the following functions : (a)

1 1 — x2

(d)

(b)

1 1 —

(e)

x cosx

x

2

(c) ln(1 + x) (f) sinx cos x

(g) arctan x

DIFFERENTIAL EQUATIONS

119



POWE R SERIES

(a) Replacing x by x 1/(1 — x) give s

2

in the Taylor series expansion o f

1 = 1 +x2+(x2)2+(x2)3+ ••• 1 —x

2

=1 +x +x 2

4

= 1 x 211

for

for

1x 2 I

< 1

+

+ x6

Ix'

< 1

Ix 2 I

<1.

11= U

since

Ixi

< 1 is equivalent to

2

(b) Differentiating 1/(1 — x) gives 1/(1 — x) , so differentiating the Taylor series expansion of 1/(1 — x) term by term will giv e the series expansion of 1/(1 — x)2 : 1

d

x)2

(

1

1 —

d

= — Ix" = x ) dv n=()

E—

x

d

X 1l)

1r=()

=

nxil 11 =

1

for

Ix'

< 1

(c) First, replacing x by —x in the Taylor series expansion of 1/(1 — x) gives the expansion of 1/(1 + x) : 1 1 1+x 1--(—x)

1 +(—x)+(—x)2+(—x)3+•••

= E (—1)'1 x"

for

Ix'

ford—xi < 1

< 1

11 =( )

Now, since integrating 1/(1 + x) yields ln(1 + x), integrating th e Taylor series for 1/(1 + x) term by term gives the expansion fo r ln(1 + x), valid for Ix' < 1 :

120

CLIFFS QUICK REVIE W



POWER SERIE S

n(1 +x)=

x

E ( — 1)„ x n

1 l +x dx=

„=O

=

x

f(_1)n x n

rt=O

=

E(— 1 )n r,=0

xn+ l

n +

Technical note : Integrating 11(1 + x) yields ln(1 + x) + c (where c is some arbitrary constant), so strictly speaking, the equation abov e should have been written xn + 1

ln(1 + x) + c

= E (—1)'' n + 1 ,t

=0

However, substitutingx = 0 into this equation shows that c = 0, s o the expansion given above for ln(1 + x) is indeed correct . (d) Replacing x by —x 2 in the Taylor series expansion of & yields the desired result :

_

x (—x2)”



n=0

(e) gives

'

n•

X

= n=0 E (— 1 ) n

x 2" n.

Multiplying each term of the Taylor series for cos x by x

± (_1)h1 ,

~ x ~xcos= ,, = o (2n) !

(f) One way to find the series expansion for sin x cos x is to multiply the expansions of sin x and cos x. A faster way, however , involves recalling the trigonometric identity sin 2x = 2 sin x cos x and then replacing x by 2x in the series expansion of sin x :

DIFFERENTIAL EQUATIONS



POWER SERIES

1

x

(2x?' ' (2n+1) .

2n=O

(_ )n 22n

n=O

(2n+1)!

x2n +

(g) Since arctan x is the integral of 1/(1 + x 2 ), integrate th e series expansion of 1 /(1 + x 2 ) term by term : arctan x

= f 1 + x z dx

for IxZ l < 1

= ,r =(, (–1)n

x 2n+ 1 2n + 1

for (x ( < 1

Recall the technical note accompanying part (c) above (which also involved the term-by-term integration of a power series) . Th e integral of 141 + is actually arctan x + c, and the equatio n above should read x2)

arctan x + c = I (–1)n n=0

x2n+ 1 2n + 1

However, substituting x = 0 into this equation shows that c = 0, s o the expansion given above for arctan x is indeed correct . ■

CLIFFS QUICK REVIEW



POWER SERIE S

Power Series Solutions of Differential Equation s First-order equations. The validity of term-by-term differentiatio n of a power series within its interval of convergence implies tha t first-order differential equations may be solved by assuming a solution of the form y=

c xn rt

n =( )

substituting this into the equation, and then determining th e coefficients cn .

Example 3 : Find a power series solution of the for m y=cnxn n=( )

for the differential equation y ' –xy= 0

Substituting c nx n

y= n =0

into the differential equation yields x

x

E nCnxn—1 — n= 0 Cnx n+1

=0

n=1

Now, write out the first few terms of each series , (c 1 +2c 2 x+3c 3 x 2 +4c 4x 3 +•••)–(cox +c l x 2 +c 2 x 3 +•••)= 0 and combine like terms : c l + (2c 2 – c 0)x + (3c 3 – c 1 )x 2

DIFFERENTIAL EQUATIONS

+

(4c 4 – c 2 )x 3

+ • • • =

0



POWER SERIES

Since the pattern is clear, this last equation may be written a s

E2 ( nC r — C,,_2)x 11— I = 0

+

c

n=

In order for this equation to hold true for all x, every coefficient on the left-hand side must be zero. This means c 1 = 0, and for all n > 2 , nC„ —

C,,_2

= 0

This last equation defines the recurrence relation that holds for th e coefficients of the power series solution : c „- 2 n

c„ =

for n

2

Since there is no constraint on co, co is an arbitrary constant, and it i s already known that c, = 0 . The recurrence relation above says c, = -co and c3 = c l , which equals 0 (because c 1 does) . In fact, it is eas y to see that every coefficient c,, with n odd will be zero . As for c 4, th e recurrence relation says c4

=

c2 _

4

co 2 .4

and so on . Since all c„ with n odd equal 0, the desired power serie s solution is therefor e y=co+c2x2+C4x4+cx 6 + . . . =co+ co x2+ co x4+

2.4

2

=co 1 +

co 2 . 4 .6

1 1 1 2 x2+ 2 4 x4+ 2 4

+• •

6 x6+

. ..

x

= co

1 2n E nx n=0 n n. !2

124

CLIFFS QUICK REVIEW





POWER SERIE S

Note that the general solution contains one parameter (co), as expected for a first-order differential equation . This power series i s unusual in that it is possible to express it in terms of an elementary function . Observe : co

1

1

1 -— X 2 n.

x 2n = co

n=O n!2

n=O

,

It is easy to check that y = coe`' 2/2 is indeed the solution of the give n differential equation, y ' = xy . Remember : Most power series canno t be expressed in terms of familiar, elementary functions, so the fina l answer would be left in the form of a power series . ■

Example 4: Find a power series expansion for the solution of th e IVP y' = x + y y(0) = 1

Substituting y=

cx „ n= O

into the differential equation yield s x

E nc„x" -1

n=1

x

= x + ± c„x " n= O

or, collecting all the terms on one side , x

E

n=1

DIFFERENTIAL EQUATIONS

x

nC,,x" -1 — x —

n= O

C,1X"

= 0



POWER SERIE S

Writing out the first few terms of the series yield s (c 1 + 2c,x+3c3 x 2 +4c4 x 3

+ —

.

•• )

x — (co+c 1 x+c,x-+c3 x 3

+ •••) =

0

or, upon combining like terms , (c 1 — c 11)

+

(2c, — I — c 1 )x + (3c3 — c,)x 2

+

(4c4 — c3)x 3

+ • . . =

0

Now that the pattern is clear, this last equation can be writte n (c 1 —co)+(2c,— 1 —c1)x+(nc,l—1)x,l-1 = 0 19= 3

In order for this equation to hold true for all x, every coefficient on th e left-hand side must be zero. This mean s c 1 -c„=0,

2c,— 1 —c 1 = 0,

and

nc,, —

1

= 0 forn3

(* )

The last equation defines the recurrence relation that determine s the coefficients of the power series solution : c,, = c ,,-1 n

forn

3

The first equation in (*) says c 1 = co, and the second equation say s = ; (1 + co) . Next, the recurrence relation says

c, = 1(1 + c 1)

c3

c4

c2 =3 c3 4

4(1 + co)

1 +

3

=2 3

co

1+c0 2. 3 . 4

and so on . Collecting all these results, the desired power serie s solution is therefore y = c 0 +c 1 x+c 2x 2 +c 3x 3 +c4x 4

126

+

• .

CLIFFS QUICK REVIEW



POWER SERIE S

=co+cox+

1+ co 2 l+ co 1+ co 4 + 2 x + 2 3 x + 2 3 4x

= co+cox+ (1 +co)

1

x = c(1 + C pX +

(1 +

1

2 x2+ 2 3x-

1 +2 .3 .4x4+ •• •

1

X„ Co) E n =2 n .

Now, the initial condition is applied to evaluate the parameter co : x

y(0) = 1

i

co + cox + (1 + co) E i x" n=2

n.

=co= 1

Therefore, the power series expansion for the solution of the give n IVP is x

y=1+x+

2

E ix " n.

n=2

If desired, it is possible to express this in terms of elementar y functions . Since x

2 x" = 2 x0 + 2 E 0. 1. n=0 n .

X1

x 2

+ n=2 E n . Xn

equation ** may be written x 2 : y=1+x+ fix" n=2 n .

2 2 2 =1+x+ E, ix" – x 0 – i 1. n=0 n . 0. = 1 + x + (2e x – 2 – 2x )

2ext— x — 1 which does indeed satisfy the given IVP, as you can readily verify . ■ DIFFERENTIAL EQUATIONS



POWER SERIE S

Second-order equations . The process of finding power series solutions of homogeneous second-order linear differential equations i s more subtle than for first-order equations . Any homogeneou s second-order linear differential equation may be written in the form y" + p (x )y ' + q (x )y = 0 If both coefficient functions p and q are analytic at x0 , then x0 i s called an ordinary point of the differential equation . On the othe r hand, if even one of these functions fails to be analytic at x0 , then x0 is called a singular point . Since the method for finding a solutio n that is a power series in x0 is considerably more complicated if x0 is a singular point, attention here will be restricted to power serie s solutions at ordinary points .

Example 5 : Find

a power series solution in x for the IV P v"—xy'+y= 0 y(0) = 2 y'(0) = 3

Substituting y =

n=( )

c,1 x

,1

into the differential equation yield s x

,1

=2

x

n (n –

–

x

nc11x,1 + /1=1

c,,x" = 0 n=( )

(* )

The solution may now proceed as in the examples above, writing ou t the first few terms of the series, collecting like terms, and the n determining the constraints on the coefficients from the emergin g pattern . Here's another method .

128

CLIFFS QUICK REVIEW



POWE R SERIES

The first step is to re-index the series so that each one involve s x" . In the present case, only the first series must be subjected to thi s procedure . Replacing n by n + 2 in this series yield s x

x

n ( n – 1 )c,1x ,1

,1

=

=2

,1+2=2

( n + 2)[(n + 2) –

11

=E

c,,+,x(,1+2)_ 2

(n + 2)(n + 1)c„ + 2 x"

,1=( 1

Therefore, equation (*) become s x n=()

(n + 2)(n +

1)c,,+2x,1 –

x

x

"=1

,1= 0

E nc,,x" + E c„x,1 =

0

The next step is to rewrite the left-hand side in terms of a single summation . The index n ranges from 0 to 0c in the first and thir d series, but only from 1 to cc in the second . Since the common range of all the series is therefore 1 to cc, the single summation which wil l help replace the left-hand side will range from 1 to cc . Consequently , it is necessary to first write (* *) a s [2c2

+E "=1

x

(n + 2) (n + 1)c, + ,

x

–±nc„x"+ [co + ,1=

1

c,,x" = 0 ,1= 1

and then combine the series into a single summation : (2c2 + co) +

[(n + 2)(n + 1)c„+ 2 – (n – 1)c„]x" = 0

In order for this equation to hold true for all x, every coefficient on th e left-hand side must be zero. This means 2c, + co = 0, and for n 1 , the following recurrence relation holds : c"+'

DIFFERENTIAL EQUATIONS

n– 1 (n + 1)(n + 2) c "

129





POWER SERIE S

Since there is no restriction on co or c,, these will be arbitrary, an d the equation 2c2 + co = 0 implies c 2 = -1-co. For the coefficient s from c 3 on, the recurrence relation is needed : C

1— 1 3 = c c' '+' —( 1 + 1)(1 + 2 )

C

4

C

5

c `'

C2+'

2—1 C2 (2 + 1)(2 + 2)

3+2

3—1 - 3+1 3+2 c3

=C

c4+

4—1 c4 (4 + 1)(4 + 2)

'

c -7 =c5+2= (5

0 1 - CO 3 . 4 2 2 4 .50

0

3 — co 5 .6 2 3 - 4

5—1 4 7 = 1)(5 + 2c5—6 •0 0

The pattern here isn't too difficult to discern : c,, = 0 for all odd n > 3, and for all even n > 4, n— 3 c „ _ — n, c o This recurrence relation can be restated as follows : for all n c

=

2,

2n—3 3—2 n o (2n)! co — (2n)! c

The desired power series solution is therefore y=c„+c ix+c,x 2 +c 3 x 3 +cox` +c 5 x 5 +cox') + co, CO, 3c, 0 ~~ =c„+c,x— 2 x-+0— 4i x +0— x +•• • 6!

130

CLIFFS QUICK REVIE W



POWER SERIE S

1 1 3 =c(► 1 _ 2 x2 _4xa_6x~~ 1 =c(► 1 __ x 2

2

+ E 3 . — 2n x

—

. . . + c1x

z,T

n=? (2n) .

As expected for a second-order differential equation, the genera l solution contains two parameters (co and c l ), which will be determined by the initial conditions . Since y(0) = 2, it is clear that co = 2 , and then, since y ' (0) = 3, the value of c 1 must be 3 . The solution o f the given IVP is therefore x2

x 2(3 — 2n) (2n)! x " ,T=2

+E

■

Example 6: Find a power series solution in x for the differentia l equation (x 2 + 1)y"+y' —x-2y= 0 Substituting

y=Ec, T x,T TT

=O

into the given equation yield s x

x

(x 2 + 1) E n(n — 1)c„x" ,T=?

+ E nc„x" -1 ,T=1

x

E c„x" +2 = 0

—

n=( )

or x

x

E n(n — 1)c„x" + E n(n — 1)c„x" + Enc, lx" - 1

n=2

n=2

x

E c nx" +2 = 0 "=( ) DIFFERENTIAL EQUATIONS

(X )





POWER SERIE S

Now, all series but the first must be re-indexed so that each involve s x" : x n=2

=

n(n — 1)c„x" -2

=

n+2= 2

n= O

(n + 2) (n + 1)c„ + 2x "

x

/t = 1

ncrtx"-' =

x

it+ l = 1

x

1x(n+ l )— l =

(n + 1 )c,, +

n= O

x

n=()

c,,xft

+2

=

1]c„+2x(n+2)- 2

(n + 2)[(n + 2) —

(n +

1)cn+ l x "

x

n—2=0

C n—2x

(n—2)+2 n= 2

Cn_2 x

n

Therefore, equation (*) become s x

n=2

x

n(n -1)c„x"

+E

n=O

(n + 2)(n + 1)c,, +2 x" +

n= O

(n + 1)c,, + l x"

x

c,, _,x /1 = 0

—

(** )

n= 2

The next step is to rewrite the left-hand side in terms of a single summation . The index n ranges from 0 to oc in the second and third series, but only from 2 to oc in the first and fourth . Since the common range of all the series is therefore 2 to 00, the single summation which will help replace the left-hand side will range from 2 to 0c . It i s therefore necessary to first write (**) a s x

n=2

x

n(n — 1)c„x" + 2c, + 6c 3x

+ n=E2 (n + 2)(n +

+ [c l + 2c 2x +

E (n + 1)c,, +i x"

,1=2

1)c,, +2 x " x —

n= 2

c„_ 2x11 = 0

CLIFFS QUICK REVIEW



POWE R SERIE S

and then combine the series into a single summation : (c 1 +2c,)+(2c 2 +6c 3 ) x + E [n(n – 1)c„ + (n + 2)(n + 1)c,, +2 + (n + 1)c„+1 –

2] x"

= 0

n =2

Again, in order for this equation to hold true for all x, every coefficient on the left-hand side must be zero . This means c 1 + 2c, = 0, 2c2 + 6c3 = 0, and for n 2, the following recurrence relatio n holds : c" _ 2 – n(n – 1)c„ – (n + (n + 1)(n + 2 )

c"+2

Since there is no restriction on co or c 1 , these will be arbitrary; th e equation c 1 + 2c 2 = 0 implies c 2 = – 2 c 1, and the equation 2c 2 + 6c 3 = 0 implies c 3 =- -C2 = - 4( — C 1 ) = c 1 . For the coefficient s from C4 on, the recurrence relation is needed : C2_2 – C4 = C 2+2 =

2(2 – 1)C2 – (2 + 1 )C 2+ 1 (2 + 1)(2 + 2)

co–2c 2 –3c 3

3.4 C3_2 – C5 c3+2

ci

co–2(–2c 1 )–3(, c 3.4

3(3 – 1)C 3 – (3 + (3 + 1)(3 + 2 )

6c3 — 4c4

4. 5 = — U) C 11 — 12O C1

DIFFERENTIAL EQUATIONS

cl

4C11 + 24 0 1

1)C 3+ 1

6(40

4(T12-co + 24 0 1 )

4 .5



POWE R SERIES

Co = C4+2 =

c4 _, – 4(4 – 1)c 4 – (4 + 1 )c 4+ 1 (4 + 1)(4+ 2 )

C,–12c 4 –5c 5 5 . 6 —z c l — 12(lcO + ~4 c 1 ) – 5 . 6 -

To CO

720C 1

The desired power series solution is therefore

y=c () +c 1x+c 2 x 2

+C 3 x 3 +C4x 4

= co + c ix + ( - 1- c 1) x2 +(

CO

= CO(1 + 12 x4

+c 5x 5 +cox ('+ . . .

+ (c1)x3 +

12OCl )x5 + (-1C1 f~()x5

3f~()xf~ -

(I2 c () + Ac l ) x4 2 OCI )x`'

+

. . .l

1x5 - 23x(,+C(x - 21x2+ 61 x 3 +. 271x 4 - 12O 1 72( 1

.)

Determining a specific pattern to these coefficients would be a tedious exercise (note how complicated the recurrence relation is) , so the final answer is simply left in this form . ■

134

CLIFFS QUICK REVIE W

THE LAPLACE TRANSFORM

If you are faced with an IVP that involves a linear differentia l equation with constant coefficients, you can proceed by the metho d of undetermined coefficients or variation of parameters and the n apply the initial conditions to evaluate the constants . However , what if the nonhomogeneous right-hand term is discontinuous ? There exists a method for solving such problems that can also b e used to solve less frightening IVP's (that is, ones that do not involve discontinuous terms) and even some equations whose coefficient s are not constants . One of the features of this alternative method fo r solving IVP's is that the values of the parameters are not found afte r the general solution has been obtained . Instead, the initial conditions are incorporated right into the initial stages of the solution, so when the final step is completed, the arbitrary constants hav e already been evaluated .

Linear Transformation s A function is usually introduced as a rule which acts on a number t o produce a unique numerical result . That is, a function accepts a number as input and produces a number as output . For instance , consider the function defined by the equation f (x) = x2 ; it specifies a particular operation to be performed on any given value of x . When this function acts on the number 3, for example, it gives the result 9 , a fact which can be symbolized as follows : 3

H9

However, functions are not restricted to acting only on numbers t o produce other numbers . A function can also act on a function t o produce another function . These "superfunctions" are often referred to as operators or transformations . Therefore, an operato r accepts a function as input and produces a function as output . DIFFERENTIAL EQUATIONS



THE LAPLAC E TRANSFORM

A familiar example is the differentiation operator, v

D:

f '(x)

f(x)

This operator acts on a (differentiable) function to produce anothe r function : namely, the derivative of the input function . For example , 1)

x3

1)

3x'

sin x

1)

H cos x,

e`

H e`

Another well-known example is the integration operator, which act s on an (integrable) function to produce another function : it s integral . Since an operator is a function, it must produce one an d only one output for each input ; therefore, it makes sense here to consider an integration operator I of the for m f .

f

f(t)

dt

[The letter t, a dummy variable, is chosen simply to distinguish i t from x, the upper limit of integration . It ' s usually considered ba d form to write, for example, Jc~ f (x) dx . ] To illustrate, if f (x) = x 2 , the n I [ f l = ~► f (t) dt = £ t 2 dt=x 3

that is, x2 -

Both of these operators, differentiation D and integration I, enjoy an important property known as linearity . Any operator or transfor-

CLIFFS QUICK REVIEW



THE LAPLACE TRANSFORM

mation T is said to be linear if both of the following conditions , T[c .f]=c .T[f ] T[f +g] = T[f] + T[g]

always hold for all constants c and all admissible functions f and g. Sinc e the derivative of a constant multiple of a function is equal to th e constant times the derivative of the function , and since the derivative of the sum of two functions is the sum of th e derivatives , differentiation is indeed linear . Furthermore, since both of th e previous statements remain true when "derivative" is replaced b y "integral, " integration is also a linear transformation .

The Laplace Transform Operato r A particular kind of integral transformation is known as the Laplac e transformation, denoted by L . The definition of this operator i s L [f(x)] =

f

e-l' f(x) dx

The result called the Laplace transform off will be a function o f p, so in general, f (x )

DIFFERENTIAL EQUATIONS

F( p)



THE LAPLAC E TRANSFOR M

Example 1 : Find the Laplace transform of the function f(x) = x. By definition, x

L [x]

= fe 1 xdx

Integrating by parts yield s L[x]

= fePx xdx x x — _ e -pxp c~

x —

0

1 x = 0 + — ~~ e-rx. dx

p 1 p

1 --e —na p

1 --

edx

p

(for p > 0)

X

0

1 p2 Therefore, the function F( p) = 1 /p2 is the Laplace transform of th e function f (x) = x . [Technical note : The convergence of the imprope r integral here depends on p being positive, since only then wil l (x/p)e-P and e-rx approach a finite limit (namely 0) as x Therefore, the Laplace transform of f(x) = x is defined only fo r p>0 .] ■ In general, it can be shown that for any nonnegative integer n , the Laplace transform of f(x) = x" is

F(p)

n!

=

n+ 1

P

Like the operators D and Iindeed, like all operators the Laplace transform operator L acts on a function to produce anothe r

138

CLIFFS QUICK REVIE W



THE LAPLACE TRANSFORM

function . Furthermore, sinc e L[cf] =

f

x e ra 'cf(x)dx=c

x

f e'

fix) dx=cL[f ]

and I,[f + g ]

=

f

e-~'x [f(x ) + g (x )] dx [e'f(x) + e ,'g (x )] dx

x

e Px f(x) dx +

x

e ,» g(x) dx

=L[f] +L[g]

the Laplace transform operator L is also linear . [Technical note : Just as not all functions have derivatives o r integrals, not all functions have Laplace transforms . For a function f to have a Laplace transform, it is sufficient that f (x) be continuou s (or at least piecewise continuous) for x 0 and of exponentia l order (which means that for some constants c and X, the inequalit y ~f(x)I < ce '

holds for all x) . Any bounded function (that is, any function f tha t always satisfies ~ f (x) < M for some M > 0) is automatically o f exponential order (just take c = M and A = 0 in the definin g inequality) . Therefore, sin kx and cos kx each have a Laplace transform, since they are continuous and bounded functions . Furthermore, any function of the form eke , as well as any polynomial , is continuous and, although unbounded, is of exponential order an d therefore has a Laplace transform . In short, most of the function s you are likely to encounter in practice will have Laplace trans forms . ]

DIFFERENTIAL EQUATIONS

139



THE LAPLAC E TRANSFOR M

Example 2 : Find the Laplace transform of the function f(x) = x3 — 4x + 2 . Recall from the first statement following Example 1 that th e Laplace transform of f(x) = x" is F(p) = n!/p" + ' . Therefore, sinc e the Laplace transform operator L is linear, L[x3 — 4x+ 2] =L[x 3 ] + L[—4x] +L[2 ] = L[x3 ] 4L [x' ] + 2L [x"] 3!

p

1!

3+1

— 4

p

6 4

F( p

0!

1+I

+ 2

2

p

~~+ 1

■

Example 3 : Determine the Laplace transform off (x) = e l' . Apply the definition and perform the integration : x

L [e ll =

e. _," e " dx dx

(1

1 p—k

"

In order for this improper integral to converge, the coefficien t (p — k) in the exponential must be positive (recall the technica l note in Example 1) . Thus, for p > k, the calculation yields 1 L [e t ] _ p — k

140

■

CLIFFS QUICK REVIE W



THE LAPLAC E TRANSFORM

Example 4 : Find the Laplace transform of f (x) = sin kx. By definition, L [sin kx]

_

f

e -r'- sin kx dx

This integral is evaluated by performing integration by parts twice , as follows : 1 e -1"- sinkxdx = - - e -P"- sinkx p

1 - - e -~'a kcos kx dx p

1 - ak = - - e P sin kx + - e -P-' cos kx dx p p 1 _ - - e -/"- sin kx + p k 1 – - - e -'"- cos kx + p p

1

f p-e-I)x

(-k sin kx) dx

1 k -e -1"- sinkx - ;e -1"- coskx p P-

k-

e sin kx dx

so kp-

1 k e -,"- sin kx dx = -p- e -' sin kx -P-- e - P" cos kx

5 e -"" sin kx dx = 5 e -'"- sin kx dx = DIFFERENTIAL EQUATIONS

1 k - p- e -P"- sin kx - e -/" cos kx P-

- e -'"- (p sin kx + k cos kx ) p- + k -

141



THE LAPLACE TRANSFOR M

Therefore, L [sin kx]

=f

e —px

sin kx dx =

—CPI (psinkx+kcoskx) " p2

+

jo

k2

k = p

Z +k 2

for p > 0 . By a similar calculation, it can be shown tha t L [cos kx]

= p, p

k2

■

Example 5 : Determine the Laplace transform of the functio n f (x)

1 for0 <x < 2 0 for x > 2

pictured in Figure 8 :

1 0 2 ■ Figure 8

142

■

CLIFFS QUICK REVIEW



THE LA PLA C E TRANSFORM

This is an example of a step function . It is not continuous, but i t is piecewise continuous, and since it is bounded, it is certainly of exponential order . Therefore, it has a Laplace transform . o •

• •

L [ f (x)] =

e

Px

f (x) dx

+

fo2 e_Px f(x) dx 2 0

e -Px . 1dx+ e-Px

f

e_Px f(x ) dx

x

-Px 0dx 2 e

dx

1 – e -2P P

Table 5 (page 144) assembles the Laplace transforms of a few o f the most frequently encountered functions, as well as some of th e important properties of the Laplace transform operator L .

Example 6 : Use Table 5 to find the Laplace transform of f (x) = sin 2 x . Invoking the trigonometric identit y sin e x = 2(1

cos 2x )

linearity of L implies L[sin z x] =

[1 – cos 2x] =

DIFFERENTIAL EQUATIONS

11

p

2(p p z2 + +44

2 p(p2+4)

■

143



THE LAPLAC E TRANSFORM

Table 5 LAPLACE TRANSFORM S f (x )

F(p )

1

1

p n!

xn

pn 1

ekx

sin

p— k k

kx

cos kx

p2

+ k2



p

p2

+k

2

Linearity : L [c, f (x) + czg(x)] = c i L [ f (x)] + c 2L[g(x) ]

Shifting formula : L [e kxf (x)] = F(p — k),

where L [ f (x)] = F(p )

Laplace transform of derivatives :

144

L [Y'J =

PL IYJ

L [Y"l =

P ZL IYJ

— Y(o) — PY(o ) — Y '( o)

CLIFFS QUICK REVIEW



THE LAPLAC E TRANSFORM

Example 7 : Use Table 5 to find the Laplace transform of g(x) = 5.' .

x3e

The presence of the factor e 5-v suggests using the shifting formul a with k = 5 . Since 3! L [f (x)] = L[x ] =p + 3+1

1 =

6 a = F(p)

p4

the shifting formula says that the Laplace transform of f (x)e 5 i x 3e 5 i is equal to F(p — 5) . In other words, the Laplace transform o f x 3e 5 i is equal to the Laplace transform of x 3 with the argument p shifted top — 5 : -

=

-

L [x3e 5 i ] -

6 =

( p—5 )

4

■

Example 8 : Use Table 5 to find the Laplace transform of f (x) = e- 1. sinx — 3 . First, since L [sin x] = 1/(p 2 + 1), the shifting formula (wit h k = -2) says L[e -'-'sin

x]

=

(p+2) 2 + l

Now, because L [3] = 3 . L[l] = 3/p, linearity implie s L[e - '' sinx — 3] =

DIFFERENTIAL EQUATIONS

1 3 - (p+2)-+1 p

145



THE LAPLAC E TRANSFORM

Example 9: Use Table 5 to find a continuous function whose Laplace transform is F(p) = 12/p 5 . This example introduces the idea of the inverse Laplac e transform operator, L - I . The operator L -' will "un-do" the actio n of L . Symbolically, f (x) - F ( p )

-1

f (x)

F(p)

If you think of the operator L as changing f (x) into F(p), then th e operator L-' just changes F(p) back into f (x) . Like L, the invers e operator L -' is linear. More formally, the result of applying L -' to a function F(p) is to recover the continuous function f (x) whose Laplace transform is th e given F(p) . [This situation should remind you of the operators D and I (which are, basically, inverses of one another) . Each wil l un-do the action of the other in the sense that if, say, I changes f (x ) into F(x), then D will change F(x) back into f (x) . In other words , D = I-', so if you apply I and then D, you're back where yo u started . ] Using Table 5 (reading it from right to left) , 12 -1

p

146

s

24

-1

p

s

_

_1

24 p

4! =

= 4L -1 p

:4 4

CLIFFS QUICK REVIE W



THE LAPLACE TRANSFORM

Example 10: Find the continuous function whose Laplace trans form isF(p) = 1/(p 2 – 1) . By partial fraction decomposition , 1 p2 _

– p+ l + p– 1

1

Therefore, by linearity of L -' , L - 1 ~. [p 2

—1

= 1'

[p+ij

_4 e -X

+L - 1 [

Lp-1 =

Example 11 : Determine

+

2(ex – e')

■

12

L -'

First, note that p has been shifted to p + 2 = p – (–2) . Therefore, sinc e L - '

2 12

p + 9

=L - ' p2+ 4 .3

=4 .L-' 9

3

p +

32

=4sin3x

the shifting formula (with k = -2) implies L-'

12 -2x sin 3x 2 [(P + 2) + 9] = 4e

Example 12: Evaluate

p –

L -1 p2

■

1 6p+25

Although p 2 – 6p + 25 cannot be factored over the integers, it can be expressed as the sum of two squares : p2 —

6p + 25 =

DIFFERENTIAL EQUATIONS

(p 2 –

6p + 9) + 16=(p–3) 2 +42

147



THE LAPLACE TRANSFORM

Therefore, L -'

p — -6p +25

~

L_

p — (p

-3) 2 +42

(p—3)+ 2 + — 3) 3)2

.L



=L -,

p 3 L (p—3)--+4- +

=

L-1

p

3

(p—3)2+4- +

4

~L

( p—3 ) 2+4'

= e 3x cos4x + 4e 3x sin 4x =eNcos4x+ ;sin4x)

■

Using the Laplace Transform to Solve Differential Equation s In this section, you will learn how to use the Laplace transfor m operator to solve (first- and second-order) differential equation s with constant coefficients . In particular, the differential equation s must be IVP's with the initial condition(s) specified at x = 0 . The method is simple to describe . Given an IVP, apply th e Laplace transform operator to both sides of the differential equation . This will transform the differential equation into an algebraic equation whose unknown, F(p), is the Laplace transform of th e desired solution . Once you solve this algebraic equation for F(p) , take the inverse Laplace transform of both sides ; the result is th e solution to the original IVP .

148

CLIFFS QUICK REVIEW



THE LAPLAC E TRANSFOR M

Before this process is undertaken, it is necessary to see what th e Laplace transform operator does to y' and y " . Integration by parts yields e

L [Y'] =

y , (x) dx

I.

x

= e y(x) –

= –y(0) +p

–pe -Tx y(x) dx

e P'Y(x ) dx

( p > 0)

= –y(0) +pL[y ]

so L [Y ') = PL [Yl —y(o )

Replacingy byy' in this result gives the Laplace transform of y" : L [Y " I = pL LY ' J – Y ' ( 0 ) = p [ pL [Y] – Y(0 )] – Y ' ( 0 ) L [Y"l = p2L LYE – pY( 0 ) – Y ' ( 0 )

Example 13 : Use the Laplace transform operator to solve the IV P y 2y=e 3 , y(0) = - 5 Apply the operator L to both sides of the differential equation ; then use linearity, the initial condition, and Table 5 to solve fo r L [Y] :

DIFFERENTIAL EQUATIONS

149



THE LAPLACE TRANSFORM

L[y'] — 2L[y] = pL[y] — y(0) — 2L[y] = PL [Y] + 5 — 2L [Y]

p— 3

(P — 2)L[y] =

_ 3 - 5 p

-sn L [Y] — LP —

+ i6

2 )(P — 3 )

Therefore, y =L-'

—Sp + 1 6 (P — 2 )( p — 3 )

By partial fraction decomposition , -5p+16 ( p—2 )( p—3 )

-6

1

p - 2 +p - 3

so -6 1 y=L-2+ 3 = - 6e 2 +e ` _ +p _ p

is the solution of the IVP .

Usually when faced with an IVP, you first find the genera l solution of the differential equation and then use the initia l condition(s) to evaluate the constant(s) . By contrast, the Laplac e transform method uses the initial conditions at the beginning of th e solution so that the result obtained in the final step by taking th e inverse Laplace transform automatically has the constants evaluated .

150

CLIFFS QUICK REVIEW

THE LAPLACE TRANSFOR M

Example 14 : Use Laplace transforms to solv e y" -3y' -4y= -16x y(0) = - 4 y'(o) _ - 5 Apply the operator L to both sides of the differential equation ; then use linearity, the initial conditions, and Table 5 to solve fo r L [y]:

L[y"] – 3L[y'] – 4L [Y] = L[–1(ix ]

= L[–lfix ] 6 [p2L[y] + 4p + s1 - 3{pL[y] + 4] - 4L[y} = – 1 P– 16 ( p -3p -4)L [ y ] +4p -7= P16

––4p+ 7 L [ y ]=

p--3p– 4

But the partial fraction decomposition of this expression for L[y] i s 16

––4p+ 7 -3p–4

16

– –, -4p+7 p2 p

-16–4p 3 +7p 2

-3p–4 p' p2 ( p+ 1 )( p–4 )

-3

+

4

DIFFERENTIAL EQUATIONS

1

-2

;+ p+l + p– 4

p p2



THE LAPLAC E TRANSFOR M

Therefore, y=L-'

3

4 + +

p p2

1 2 + p+l p– 4

which yields x

y= -3+4x+e_ —2e 4x

■

Example 15: Use Laplace transforms to determine the solution of the IVP y"—2y' + 5y = 0 y(0) = - 1 y' (0) = 7 Apply the operator L to both sides of the differential equation ; then use linearity, the initial conditions, and Table 5 to solve fo r L [y] :

L[ y° - Zy' + Sy] = L[0]

[p2L[y] — PY( 0 ) — y '(0)] — 2 [PL [Y] — Y( 0 )] + SL [Y] = 0

[ p 2L[y] + p — 7] — 2[pL[y] + 1 ] + SL [Y] = 0 (p 2 -2p+5)L[y]+p–9=0

–p+ 9 L[Y]—p~—2p+ 5

CLIFFS QUICK REVIEW



THE LAPLAC E TRANSFORM

Now , –p+9

–p+ 9

p 2 –2p+5 (p 2 –2p+1)+ 4 –p+ 9

(p - 1) z + 4 —(p—1)+ 8 (P — 1 ) z + 2 2 —(p—1)

4 . 2

(P —1 ) 2 + 22

(P —1 ) 2+22

so

( P –1 ) 2

+

LP_1

22

2 +22 ]

= — e x cos 2x + 4ex sin 2x or more simply, y = ex (4 sin 2x – cos 2x)

Example 16: Use the fact that if f (x) = L - '[F(p)], then for any positive constant k, L' fe-knF (P))

0 for x < k =

f (x – k) for x

to solve and sketch the solution of the IV P y' –

y – cr (x )

y(0) = 0

DIFFERENTIAL EQUATIONS

k



THE LAPLAC E TRANSFORM

where v- is the step function 0 for0x< 2 1 forx 2 shown in Figure 9 :

1

2 ■ Figure 9 ■ As mentioned earlier, the Laplace transform method is particularly well-suited to solving IVP's that involve discontinuous functions such as the step function r shown above . As usual, begin by taking the Laplace transform of both sides o f the differential equation : L [y' — y] = L

Mx)] (* )

Since y(0) = 0, the left-hand side of (*) reduces t o L [Y' — Y~ = L [Y'l — L [yl = [P L IY l — Y( o )) — L [Yl = ( p — 1 )L [Y l

Using the definition of L, the right-hand side of (*) is no w evaluated : L [cr(x) ] =

154

x

1

e -PXQ'(x) dx = i e -px dx = - - e -px~ P

e -2n

x _ 2

P

CLIFFS QUICK REVIEW



THE LAPLAC E TRANSFORM

Therefore, the transformed equation (*) read s e (P - 1 )L [Y] =

P

so e -2P y=L -

But e -2P

p ( p - 1)

-1 e

e p ( p - 1)

p+P- 1

so -,

1

-2P

y L

e

- L -,

1

2

e- P

P

P

Now, since L -' [1 (p - 1)] = e-', the formula given in the statemen t of the problem says 1

L

e

-1

0 Hex-2 rr' p -1

forx< 2 forx

2

and since L -1 [1 p] = 1, applying the formula given in the statemen t of the problem again yield s 11

e L

2P

p

forx < 2 = 101 for x 2

Alternatively, simply notice that e- 2n L[o-(x)] =

1 L -1 e-2P -

P

DIFFERENTIAL EQUATIONS

p

0 forx < 2 _ cr(x) _

1 forx >_ 2



THE LAPLACE TRANSFORM

Substituting these results into (**) gives the solution of the IVP : 0 Y =

forx < 2 forx 2

0 forx < 2 1 forx >— 2

which becomes 0

forx < 2 e-`" -2 — 1 for x >_ 2 This function is sketched in Figure 10 :

1

2 ■ Figure 10 ■

156

CLIFFS QUICK REVIEW



A FEW APPLICATIONS

It can be argued that the greatest importance of studying differential equations lies in their applications to science . Since the process of predicting and describing the changes in a physical system is perhaps the raison d'etre of science, and because the rates at whic h changes occur are expressed by derivatives, it is not surprising tha t so many scientific laws are formulated as differential equations . It is not possible here to list the tremendous number of practica l applications of differential equations, but this book would b e incomplete if there weren't some acknowledgment of their rol e outside pure mathematics .

Applications of First-Order Equation s Orthogonal trajectories . The term orthogonal means perpendicular, and trajectory means path or curve. Orthogonal trajectories, there fore, are two families of curves that always intersect perpendicularly. A pair of intersecting curves will be perpendicular if th e product of their slopes is -1, that is, if the slope of one is the negative reciprocal of the slope of the other . Since the slope of a curve is given by the derivative, two familes of curves f1 (x,y, c) = 0 and f,(x, y, c) = 0 (where c is a parameter) will be orthogona l wherever they intersect if

DIFFERENTIAL EQUATIONS

df,

-1

di

df,/dx



A FEW APPLICATIONS

Example 1 : The electrostatic field created by a positive poin t charge is pictured as a collection of straight lines which radiate awa y from the charge (Figure 11) . Using the fact that the equipotentials (surfaces of constant electric potential) are orthogonal to th e electric field lines, determine the geometry of the equipotentials o f a point charge .

■ Figure l I

■

If the origin of an xy coordinate system is placed at the charge , then the electric field lines can be described by the family y = cx

The first step in determining the orthogonal trajectories is to obtai n an expression for the slope of the curves in this family that does not involve the parameter c . In the present case , y cx dx

c

x (

)

The differential equation describing the orthogonal trajectories i s therefore

158

CLIFFS QUICK REVIEW



A FE W APPLICATIONS

dv x d

since the right-hand side of (**) is the negative reciprocal of th e right-hand side of (*) . Because this equation is separable, th e solution can proceed as follows : dy

x

dx

y

ydy=—xdx

f y dy = f —x dx 0y-=—x-+ c x 2 +y 2 =c -

where c 2 = 2c ' . The equipotential lines (that is, the intersection of the equipotentia l surfaces with any plane containing the charge) are therefore th e family of circles x 2 + y 2 = c 2 centered at the origin . The equipotential and electric field lines for a point charge are shown in Figure 12 . electric field lines

equipotentia l lines ■ Figure 12

DIFFERENTIAL EQUATIONS

■

159



A FE W APPLICATIONS

Example 2 : Determine the orthogonal trajectories of the family of circles x 2 + (y -- c) 2 = c 2 tangent to thex axis at the origin . The first step is to determine an expression for the slope of th e curves in this family that does not involve the parameter c . By implicit differentiation, x2+ ( y—c ) 2=c 2 x2 + y 2 = 2cy dy dy 2x + 2y = 2c dx dx dy

x

dx c

Y

To eliminate c, note that x 2 + (y c )' c -

x ' + y' = 2cy

c

x2 + y-

2y

The expression above for dy/dx may now be written in the form dy

x

ay — y- (~ )

x

dx c — y —x- +

-

y2

—

Y

Therefore, the differential equation describing the orthogona l trajectories is dy y-' —x dx

ay

since the right-hand side of (**) is the negative reciprocal of th e right-hand side of (* ) .

160

CLIFFS QUICK REVIEW



A FE W APPL I PP/ICA PONS

If equation r'') is written in the for m (y 22–x2)dx—2xydy= 0

note that it is not exact (since M, . = 2y but N. = - 2y) . However , because M,. -- Nx N

2

2y– (–2y) -2xy

x

is a function of x alone, the differential equation ha s

µ = e f (-2/x)d =

e-21n_,-

=

eln(a- `)

= x -2

as an integrating factor . After multiplying through by µ = x- 2 , the differential equation describing the desired family of orthogona l trajectories become s (x -2 y 2

–

1) dx – 2x - 'y dy = 0

which is now exact (because M,. = 2x - 22y = Ni ) . Sinc e Max =

(x -2y 2 –

1)ax= –x - ' y 2 – x

and

f

N ay

=

f(–'y)ay =

–

x - ' y -2

the solution of the differential equation i s – x = -2c (The reason the constant was written as -2c rather than as c will b e apparent in the calculation below .) With a little algebra, th e equation for this family may be rewritten : –x - ' y 2 – x = -2 c x - ' y 2 +x=2c y2+x-2 =2cx (x—c ) 2+y2=c 2

DIFFERENTIAL EQUATIONS

161



A FE W APPLICATIONS

This shows that the orthogonal trajectories of the circles tangent to the x axis at the origin are the circles tangent to the y axis at th e origin! See Figure 13 .

■ Figure 13

■

Radioactive decay. Some nuclei are energetically unstable and ca n spontaneously transform into more stable forms by various processes known collectively as radioactive decay . The rate at which a particular radioactive sample will decay depends on the identity of the sample . Tables have been compiled which list the half-lives o f various radioisotopes . The half-life is the amount of time require d for one-half the nuclei in a sample of the isotope to decay; therefore, the shorter the half-life, the more rapid the decay rate . The rate at which a sample decays is proportional to the amoun t of the sample present . Therefore, if x(t) denotes the amount of a radioactive substance present at time t, the n dx dt

—kx

(k > o)

CLIFFS QUICK REVIEW

162



A FE W APPLICATIONS

(The rate dx/dt is negative, since x is decreasing .) The positive constant k is called the rate constant for the particular radioisotope . The solution of this separable first-order equation i s x = xoe -k'

(* )

where x0 denotes the amount of substance present at time t = 0 . Th e graph of this equation (Figure 14) is known as the exponential deca y curve :

■ Figure 14

■

The relationship between the half-life (denoted T1,2) and the rat e constant k can easily be found . Since, by definition, x = - x0 at t = T1,2, (*) becomes xo = xoe -kT' ; -

– kT 1 /2 =

In

2

=

k=

In 2 T1, 2

Because the half-life and rate constant are inversely proportional , the shorter the half-life, the greater the rate constant, and , consequently, the more rapid the decay .

DIFFERENTIAL EQUATIONS

163



A FE WAPPLICATIONS

Radiocarbon dating is a process used by anthropologists an d archaeologists to estimate the age of organic matter (such as wood or bone) . The vast majority of carbon on earth is nonradioactiv e carbon-12 ( 12 C) . However, cosmic rays cause the formation o f carbon-14 ( 14 C), a radioactive isotope of carbon which become s incorporated into living plants (and therefore into animals) through the intake of radioactive carbon dioxide ( 14CO2 ) . When the plant o r animal dies, it ceases its intake of carbon-14, and the amoun t present at the time of death begins to decrease (since the 14C decay s and is not replenished) . Since the half-life of 14C is known to be 5730 years, by measuring the concentration of 14C in a sample, its age ca n be determined .

Example 3 : A fragment of bone is discovered to contain 20% of th e usual 14 C concentration . Estimate the age of the bone . The relative amount of 14C in the bone has decreased to 20% o f its original value (that is, the value when the animal was alive) . Thus, the problem is to calculate the value of t at whichx(t) = 0 .20x , (wherex = the amount of 14 C present) . Sinc e In 2 T 1/2

k

In 2 573 0

the exponential decay equation (*) say s 0.20x0 = x0 e—[(In2)/57301 t In (0 .20) = —

In 2 573 0 t

t = 5730 t

164

—in (0 .20 ) In 2

13,300 years

■

CLIFFS QUICK REVIE W



A FE W APPLICATIONS

Newton's Law of Cooling . When a hot object is placed in a coo l room, the object dissipates heat to the surroundings, and it s temperature decreases . Newton's Law of Cooling states that the rat e at which the object's temperature decreases is proportional to the difference between the temperature of the object and the ambien t temperature . At the beginning of the cooling process, the differenc e between these temperatures is greatest, so this is when the rate o f temperature decrease is greatest . However, as the object cools, th e temperature difference gets smaller, and the cooling rate decreases ; thus, the object cools more and more slowly as time passes . T o formulate this process mathematically, let T(t) denote the temperature of the object at time t and let T, denote the (essentially constant) temperature of the surroundings . Newton's Law o f Cooling then says dT dt —k(T —

T.)

(k > 0 )

Since T,. < T (that is, since the room is cooler than the object), T decreases, so the rate of change of its temperature, dT/dt, i s necessarily negative . The solution of this separable differentia l equation proceeds as follows :

fT—T dT

ln(T —

—k dt S

T) =

— kt + c '

T — T,. = T

=

T.+ce -k '

(* )

DIFFERENTIAL EQUATIONS

165



A FE W APPLICATION S

Example 4: A cup of coffee (temperature = 190°F) is placed in a room whose temperature is 70°F . After five minutes, the temperature of the coffee has dropped to 160°F . How many more minutes must elapse before the temperature of the coffee is 130°F ? Assuming that the coffee obeys Newton's Law of Cooling, it s temperature T as a function of time is given by equation (*) wit h T,=70 : T(t) = 70 + ce -k t

Because T(0) be evaluated :

190, the value of the constant of integration (c) ca n T(0) = 190 = 190=70+c = c= 12 0

Furthermore, since information about the cooling rate is provided (T = 160 at time t = 5 minutes), the cooling constant k can be determined : 160 = T(5 ) 160 = 70 + 120e - Sk e -5k =

4

-5k=1n 4

k=5ln 4 Therefore, the temperature of the coffee t minutes after it is place d in the room is T(t) = 70 + 120e-1''5)1°

166

43)1 '

CLIFFS QUICK REVIEW



A FE W APPLICATION S

Now, setting T = 130 and solving for t yield s 130 = 70 + 120e—[(1 /5)In(4/3)] t e —[(1 /5)ln(4/3)]t = 1

(—i In 43)t = I n t= t

2

5 In 2 In 12 minute s

This is the total amount of time after the coffee is initially placed i n the room for its temperature to drop to 130°F . Therefore, afte r waiting five minutes for the coffee to cool from 190°F to 160°F, it i s necessary to then wait an additional seven minutes for it to coo l down to 130°F . ■

Skydiving (part I) . Once a sky diver jumps from an airplane, ther e are two forces that determine her motion : the pull of the earth' s gravity and the opposing force of air resistance . At high speeds, th e strength of the air resistance force (the drag force) can be expresse d as kz v 2 , where v is the speed with which the sky diver descends and k is a proportionality constant determined by such factors as th e diver's cross-sectional area and the viscosity of the air . Once th e parachute opens, the descent speed decreases greatly, and th e strength of the air resistance force is given by Kv . Newton's Second Law states that if a net force Fnet acts on a n object of mass m, the object will experience an acceleration a give n by the simple equation Fnet

DIFFERENTIAL EQUATIONS

= ma

167



A FE W APPLICATIONS

Since the acceleration is the time derivative of the velocity, this la w can be expressed in the form dzv Fnet = m dt

(* )

In the case of a sky diver initially falling without a parachute, th e drag force is Fdrag = kv 2, and the equation of motion (*) become s mg — kz v'- =

m

dzv dt

or more simply, dv bz, 2 dt = g — where b = k/m . [The letter g denotes the value of the gravitationa l acceleration, and mg is the force due to gravity acting on the mass m (that is, mg is its weight) . Near the surface of the earth, g i s approximately 9.8 meters per second 2 .] Once the sky diver's descent speed reaches v = \/g/b = \/mg/k, the equation above says dv/dt = 0 ; that is, v stays constant . This occurs when the speed is great enough for the force of air resistance to balance the weight of the sky diver ; the net force and (consequently) the acceleration drop to zero . Thi s constant descent velocity is known as the terminal velocity . For a sky diver falling in the spread-eagle position without a parachute, th e value of the proportionality constant k in the drag equation Fdrag = kzv 2 is approximately 4 kg/m. Therefore, if the sky diver has a total mass of 70 kg (which corresponds to a weight of about 150 pounds) , her terminal velocity is Z1 terminal (no parachute)

=

\/mg k =

/(7o)(9 .8) l

or approximately 120 miles per hour .

168

52 m/ s

Z

CLIFFS QUICK REVIE W



A FE W APPLICATIONS

Once the parachute opens, the air resistance force become s F air resist = Kv, and the equation of motion (*) become s mg — Ky

=m

dv

dt

or more simply, dv By dt = g — where B = Kim . Once the parachutist's descent speed slows to v = g/B = mg/K, the equation above says dv/dt = 0 ; that is, v stays constant . This occurs when the speed is low enough for the weigh t of the sky diver to balance the force of air resistance ; the net force and (consequently) the acceleration reach zero . Again, this constant descent velocity is known as the terminal velocity. For a sk y diver falling with a parachute, the value of the proportionalit y constant K in the equation Fairresist = Ky is approximately 110 kg/s . Therefore, if the sky diver has a total mass of 70 kg, the termina l velocity (with the parachute open) is onl y vterminal (with parachute)

mg

(70)(9 .8 )

K

11 0

6 .2 m/ s

which is about 14 miles per hour . Since it is safer to hit the groun d while falling at a rate of 14 miles per hour rather than at 120 mile s per hour, sky divers use parachutes .

DIFFERENTIAL EQUATIONS

169

A FE W APPLICATIONS

Example 5 : After a free-falling sky diver of mass m reaches a constant velocity of v 1 , her parachute opens, and the resulting ai r resistance force has strength Kv . Derive an equation for the spee d of the sky diver t seconds after the parachute opens . As stated above, once the parachute opens, the equation of motion is dv g – By dt –

where B = Kim . The parameter that will arise from the solution o f this first-order differential equation will be determined by the initia l condition v(0) = v 1 (since the sky diver's velocity is z ► 1 at the momen t the parachute opens, and the "clock" is reset to t = 0 at thi s instant) . This separable equation is solved as follows : dv – g – Bv

f dt

1 –Bln(g–Bv)=t+c " ln(g – By) = –Bt + c ' g – By = ce -B '

Now, since v(0) = v, g – Bz), = c, the desired equation for th e sky diver's speed t seconds after the parachute opens is g–B7 = (g–Bv>)e-'' By = g – (g – Bv, )e-'3 ' B (1 - - v1 e - "' g

v

170

mg K

K vl i + (_ mg

)e-(K/m) t

CLIFFS QUICK REVIEW



A FE W APPLICATIONS

Note that as time passes (that is, as t increases), the term e -(KIm)t goes to zero, so (as expected) the parachutist's speed v slows t o mg/K, which is the terminal speed with the parachute open . ■

Applications of Second-Order Equation s Skydiving (part II) . The principal quantities used to describe th e motion of an object are position (s), velocity (v), and acceleratio n (a) . Since velocity is the time derivative of the position, an d acceleration is the time derivative of the velocity, acceleration is th e second time derivative of the position . Therefore, the position function s(t) for a moving object can be determined by writin g Newton's Second Law, Fne , = ma, in the for m F'net = m

d '-s

di"

and solving this second-order differential equation for s . [You may see the derivative with respect to time represented b y For example, s ("s dot") denotes the first derivative of s with respect to t, ands ("s double dot") denotes the second derivative o f s with respect to t . The dot notation is used only for derivatives with respect to time . ] a dot.

DIFFERENTIAL EQUATIONS



A FE W APPLICATIONS

Example 6: A sky diver (mass m) falls long enough without a parachute (so the drag force has strength kv 2 ) to reach her firs t terminal velocity (denoted v l ) . When her parachute opens, the ai r resistance force has strength Kv . At what minimum altitude must her parachute open so that she slows to within 1% of her new (muc h lower) terminal velocity (v 2 ) by the time she hits the ground ? Let y denote the vertical distance measured downward from the point at which her parachute opens (which will be designated tim e t = 0) . Then Newton's Second Law (Fnet = ma) becomes mg – Kv = ma, or, since v = y and a = y , mg – Ky=my This situation is therefore described by the IV P y

K

+m y= g y(0) = 0 ,y(0) = v l

The differential equation is second-order linear with constan t coefficients, and its corresponding homogeneous equation i s y+By= 0 where B = Kim . The auxiliary polynomial equation, r2 + Br = 0, has r = 0 and r = –B as roots . Since these are real and distinct, th e general solution of the corresponding homogeneous equation i s Yh = cl + c2e-Bt

= cl + c 2e -(K/m)t

The given nonhomogeneous equation has y = (mg/K)t as a particular solution, so its general solution i s y=c l +c 2e -(K/m)t

+ mg t

K

()

CLIFFS QUICK REVIEW



A FE W APPLICATIONS

Now, to apply the initial conditions and evaluate the parameters c , and c2 : c2e-(K/m)t+

y(0)=0

cl +

)) ( 0 ) = v 1

K [_c 2_ e_ 1m )t +

mg K t t- o

m

K t= o

+c2= 0

=0

c1

= z, l

–c2

K

m+

mg K = vi

These two equations imply cl

m

= K

Vi –

mg K

and

c2

m

= – K z> > –

mg K

Once these values are substituted into (*), the complete solution t o the IVP can be written a s

=

m

g k)( 1 mgL [ (v'

J

1

The derivative of this expression gives the velocity of the sky diver t seconds after the parachute opens : v

mg y

K

1

Note that equation (***) is equivalent to the result of Example 5 (a s it should be) . The question asks for the minimum altitude at which the sky diver's parachute must be open in order to land at a velocity o f (1 .01)x 2 . Therefore, set v equal to (1 .01)v 2 in equation (* * *) an d solve for t ; then substitute the result into (**) to find the desire d altitude . Omitting the messy details, once the expression in (* * *) i s set equal to (1 .01)v 2, the value of t is found to b e

DIFFERENTIAL EQUATIONS



A FE W APPLICATIONS

v 1K - 1 m mg t =KIn 1 .01 v 2K 1 mg

v1K m mg K I n 0 .0 1

and substituting this result into (* *) yield s v 1K mg m mg v~ y — K K In 0 .01 + g

m K

0 .0 1 v iK 1 mg I

1

To evaluate the numerical answer, the following values from th e discussion in Skydiving (part I) are used : mass of sky diver : m = 70 k g gravitational acceleration :

g = 9 .8 m/s 2

air resistance proportionality constant : K = 110 kg/ s terminal velocity without parachute : v 1 = terminal velocity with parachute :

v2

/mg k

52 m/ s

mg = K 6 .2 m/ s

These substitutions give a descent time t [the time interval betwee n the parachute opening to the point where a speed of (1 .01)z 2 i s attained] of approximately 4 .2 seconds, and a minimum altitude a t which the parachute must be opened of y 55 meters (a littl e higher than 180 feet) . ■ Simple harmonic motion . Consider a spring fastened to a wall, wit h a block attached to its free end at rest on an essentially frictionles s horizontal table . The block can be set into motion by pulling o r pushing it from its original position and then letting go, or b y striking it (that is, by giving the block a nonzero initial velocity) . Th e

174

CLIFFS QUICK REVIEW



A FE W APPLICATION S

force exerted by the spring keeps the block oscillating on th e tabletop . This is the prototypical example of simple harmoni c motion . The force exerted by a spring is given by Hooke's Law; this state s that if a spring is stretched or compressed a distance x from it s natural length, then it exerts a force given by the equatio n F = —kx

The positive constant k is known as the spring constant and is directl y related to the spring's stiffness : The stiffer the spring, the larger th e value of k . The minus sign implies that when the spring is stretche d (so that x is positive), the spring pulls back (because F is negative) , and conversely, when the spring is compressed (so that x i s negative), the spring pushes outward (because F is positive) . Therefore, the spring is said to exert a restoring force, since it always tries to restore the block to its equilibrium position (the positio n where the spring is neither stretched nor compressed) . The restoring force here is proportional to the displacement (F = —kx « x) , and it is for this reason that the resulting periodic (regularl y repeating) motion is called simple harmonic. Newton 's Second Law can be applied to this spring-bloc k system . Once the block is set into motion, the only horizontal forc e that acts on it is the restoring force of the spring . Therefore, th e equation F1et = ma becomes —kx = mz, o r k z+—x= 0 m This is a homogeneous second-order linear equation with constan t coefficients . The auxiliary polynomial equation is r 2 + = 0, whic h has distinct conjugate complex roots ± i \lk/m . Therefore, the general solution of this differential equation is x = c l cos

DIFFERENTIAL EQUATIONS

mt + c 2 sin

k

t

(* )



A FE W APPLICATIONS

This expression gives the displacement of the block from it s equilibrium position (which is designated x = 0) .

Example 7 : A block of mass 1 kg is attached to a spring with forc e constant k = N/m . It is pulled m from its equilibrium positio n and released from rest . Obtain an equation for its position at an y time t ; then determine how long it takes the block to complete on e cycle (one round trip) . All that is required is to adapt equation (*) to the presen t situation . First, since the block is released from rest, its intia l velocity is 0 : X(0) = 0

–c l

k sin m

k k t+c — cos NiT ±t =0 m - m m t=0

c, = 0

Since c 2 = 0, equation (*) reduces to x = c, cos \/k/m t . Now, sinc e x(0) = +m, the remaining parameter can be evaluated : x(o)

=~~I

c, cos Vm t

3

t=o

10

c

=

3 10

Finally, since k = 45 N/m and m = 1 kg, \/k/m = \/25/4 = Therefore, the equation for the position of the block as a function o f time is given by x = ,3 ,

cos 5 t

where x is measured in meters from the equilibrium position of th e block . This function is periodic, which means it repeats itself a t regular intervals . The cosine and sine functions each have a period of 27r, which means every time the argument increases by 27r, th e function returns to its previous value . (Recall that if, say, x = cos 0 , then 0 is called the argument of the cosine function .) The argumen t

176

CLIFFS QUICK REVIEW



A FE W APPLICATION S

here is ;t, and ;t will increase by 2'rr every time t increases by 4'7 . Therefore, this block will complete one cycle, that is, return to it s original position (x = 10 m), every 5'rr 2 .5 seconds . ■

The length of time required to complete one cycle (one round trip) is called the period of the motion (and denoted by T) . It can b e shown in general that for the spring-block oscillator , \

T=2'rr

/-Wi

k

Note that the period does not depend on where the block started , only on its mass and the stiffness of the spring . The maximu m distance (greatest displacement) from equilibrium is called th e amplitude of the motion . Therefore, it makes no difference whethe r the block oscillates with an amplitude of 2 cm or 10 cm ; the period will be the same in either case . This is one of the definin g characteristics of simple harmonic motion : the period is independent of the amplitude . Another important characteristic of an oscillator is the numbe r of cycles that can be completed per unit time ; this is called th e frequency of the motion [denoted traditionally by v (the Gree k letter nu) but less confusingly by the letter f] . Since the perio d specifies the length of time per cycle, the number of cycles per unit time (the frequency) is simply the reciprocal of the period : f = 1/ T. Therefore, for the spring-block simple harmonic oscillator ,

f

1

Vic

2Tr

Frequency is usually expressed in equals 1 cycle per second .

DIFFERENTIAL EQUATIONS

m hertz

(abbreviated Hz) ; 1 H z



A FE W APPLICATIONS

The quantity s/k/m (the coefficient of t in the argument of th e sine and cosine in the general solution of the differential equatio n describing simple harmonic motion) appears so often in problem s of this type that it is given its own name and symbol . It is called th e angular frequency of the motion and denoted by w (the Greek lette r omega) . Note that w = 2Trf.

Damped oscillations . The spring-block oscillator discussed above i s an idealized example of a frictionless system . In real life, however , frictional (or dissipative) forces must be taken into account, particu larly if you want to model the behavior of the system over a lon g period of time . Unless the block slides back and forth on a frictionless table in a room evacuated of air, there will be resistanc e to the block's motion due to the air (just as there is for a falling sk y diver) . This resistance would be rather small, however, so you ma y want to picture the spring-block apparatus submerged in a larg e container of clear oil . The viscosity of the oil will have a profoun d effect upon the block's oscillations . The air (or oil) provides a damping force, which is proportional to the velocity of the object . (Again, recall the sky diver falling with a parachute . At th e relatively low speeds attained with an open parachute, the force du e to air resistance was given as Ky, which is proportional to th e velocity . ) With a restoring force given by –kx and a damping force give n by –Ky (the minus sign means that the damping force opposes th e velocity), Newton's Second Law (Fnet = ma) becomes –kx – Ky = ma, or, since v = .X and a = I , –kx – K.z=ml

This second-order linear differential equation with constant coefficients can be expressed in the more standard for m mx+KX+kx= 0

178

CLIFFS QUICK REVIEW



A FEW APPLICATIONS

The auxiliary polynomial equation is mr2 are r=

+ Kr +

k = 0, whose root s

–K ± JK2 – 4mk 2m

The system will exhibit periodic motion only if these roots ar e distinct conjugate complex numbers, because only then will th e general solution of the differential equation involve the periodi c functions sine and cosine . In order for this to be the case, th e discriminant K2 – 4mk must be negative ; that is, the damping constant K must be small ; specifically, it must be less than 2Jmk . When this happens, the motion is said to be underdamped, becaus e the damping is not so great as to prevent the system from oscillating ; it just causes the amplitude of the oscillations to gradually die out . [If the damping constant K is too great, then the discriminant i s nonnegative, the roots of the auxiliary polynomial equation are rea l (and negative), and the general solution of the differential equatio n involves only decaying exponentials . This implies there would be n o sustained oscillations . ] In the underdamped case (K < 2 Jmk), the roots of the auxiliary polynomial equation can be written a s

/4mk_K2

K 2m

r= – ±

2m

=–

K 1k K– ± 2m i m 4m2

and consequently, the general solution of the defining differentia l equation is X=e-(ki?m)t

Cocos

DIFFERENTIAL EQUATIONS

1k K2 -,t+c, sin m 4m-

k K2 ., t -m 4m -

179



A FE W APPLICATION S

Example 8 : (Compare to Example 7 .) A block of mass 1 kg i s attached to a spring with force constant k = N/m . It is pulled m from its equilibrium position and released from rest . If this spring-block apparatus is submerged in a viscous fluid mediu m which exerts a damping force of -4v (where v is the instantaneou s velocity of the block), sketch the curve that describes the position o f the block as a function of time . The net force on the block is — 4x — 4v = — x — 4.x, so Newton's Second Law become s

x+4.z+4x= 0 because m = 1 . Since the roots of the auxiliary polynomial equation , r2 = O, ar e r_—

-4±N/422 —4 . 1•--~ -4±3i -2±4 i 2 . 1 — 2 _— -

the general solution of the differential equation i s x=e - ''(c,

cos4t+c2sin4t)

Because the block is released from rest, v(0) = di(0) = 0 : [e- 2 '(-4c,

sin 4t + 4c2 cos 4t) — 2e-''(c, cos t +

c2

sin 4 t )],-,) = 0

This implies 4c2 — 2c l = 0. And, since x(0) = m , [e- 2'(c,

cos 4t +

c2

sin 4t)], =0 =

c,

c2 = (2c,) = 1(-h) = and the equation =Therfo, that gives th e position of the block as a function of time i s x = Tbe -2'(3 cos 4t + 4 sin 4t )

where x is measured in meters from the equilibrium position of th e block .

180

CLIFFS QUICK REVIEW



A FE W

APPLICATION S

This expression for the position function can be rewritten usin g the trigonometric identity cos(a — /3) = cos a cos /3+ sin a sin [3, a s follows: x = -Me -2'(3

cos 4t + 4 sin 4t)

;e-''[(cos _ ;e - ''[(cos x = ;e

cos

40(0 + (sin 3t)(O] 4t)(cos

(4t —

4) +

(sin 4t)(sin ~) ]

cf) )

The phase angle, 0, is defined here by the equations cos (P = and sin 4 = 5, or, more briefly, as the first-quadrant angle whose tangen t is 4 (it's the larger acute angle in a 3-4-5 right triangle) . The presence of the decaying exponential factor e -2 ' in the equation for x(t) means that as time passes (that is, as t increases), the amplitud e of the oscillations gradually dies out . See Figure 15 . envelope : x = Ae_R 2m) t position of block x

•,

time, t

■ Figurc 15

DIFFERENTIAL EQUATIONS

■

181



A FE W APPLICATION S

The angular frequency of this periodic motion is the coefficien t of t in the cosine, w' = rad/s, which implies a period o f 1 27 2-rr T ' =—_ =err 4.2 second s f' w' _ Compare this to Example 7, which described the same spring, block , and initial conditions but with no damping . The position functio n there was x = cos ;t, it had constant amplitude, an angula r frequency of w = rad/s, and a period of just 5 Tr 2 .5 seconds . Therefore, not only does (under)damping cause the amplitude t o gradually die out, but it also increases the period of the motion . But this seems reasonable : Damping reduces the speed of the block, so it takes longer to complete a round trip (hence the increase in th e period) . This will always happen in the case of underdamping, sinc e w' = \/k/m – K 2 /4m 2 will always be lower than w = \/k/m . ■

Electric circuits and resonance. When an electric circuit containin g an ac voltage source, an inductor, a capacitor, and a resistor i n series is analyzed mathematically, the equation that results is a second-order linear differential equation with constant coefficients . The voltage v(t) produced by the ac source will be expressed by th e equation z' = V sin wt, where V is the maximum voltage generated . An inductor is a circuit element that opposes changes in current , causing a voltage drop of L (di /dt), where i is the instantaneou s current and L is a proportionality constant known as the inductance . A capacitor stores charge, and when each plate carries a magnitud e of charge q, the voltage drop across the capacitor is q/C, where C i s a constant called the capacitance . Finally, a resistor opposes the flo w of current, creating a voltage drop equal to iR, where the constant R is the resistance . Kirchhoff's Loop Rule states that the algebraic su m of the voltage differences as one goes around any closed loop in a circuit is equal to zero . Therefore, if the voltage source, inductor , capacitor, and resistor are all in series, the n

182

CLIFFS QUICK REVIEW



A FE W

APPLICATIONS

di

q

t C –iR= 0

Vsinwt–Ld –

which can be rewritten a s L

di

dt

+Ri+

1

C q=

Vsin cut

Now, if an expression for 40—the current in the circuit as a function of time is desired, then the equation to be solved must b e written in terms of i . To this end, differentiate the equation directl y above, and use the definition i = dq/dt : L

d2i

di

1

dt , + Rdt + Ci =

wVcos cat

(* )

This differential equation governs the behavior of an LRC serie s circuit with a source of sinusoidally varying voltage . The first step in solving this equation is to obtain the genera l solution of the corresponding homogeneous equatio n d2i di 1 Ldt'-+Rdt+Ci=0

(** )

But notice that this differential equation has exactly the sam e mathematical form as the equation for the damped oscillator , m

d2x

dx

dt2 +Kdt

+kx=0

(*** )

By comparing the two equations, it is easy to see that the current (i ) is analogous to the position (x), the inductance (L) is analogous t o the mass (m), the resistance (R) is analogous to the dampin g constant (K), and the reciprocal capacitance (1/C) is analogous t o the spring constant (k) . Since the general solution of (* * *) wa s found to b e

DIFFERENTIAL EQUATIONS

183



-(x/ -"' f' c, cos

K K , t + c-) sin mkx=e — ,t m 4m

V4

ifK<2 m k the general solution of (* *) must be, by analogy ,

r~,_ e -(R/ 21-)r

Cl

1 _ R-t + c, sin cos LC 4L'

V

1 R-' _ t LC 4L' if R < 2 ~Ll c

But the solution does not end here . The original differentia l equation (*) for the LRC circuit was nonhomogeneous, so a particular solution must still be obtained . The family of th e nonhomogeneous right-hand term, c'V cos a, is {sin wt, cos a l , so a particular solution will have the for m i =A sin

a + B cos at

where A and B are the undetermined coefficients . Given thi s expression for i, it is easy to calculat e wA cos

a — (vB si n

—w '-Asinak—

(0 2 Bcosa t

Substituting these last three expressions into the given nonhomogeneous differential equation (*) yield s

I

—Ico ZL — 1

1

C~A — ujRB

l

J

sinwt

I 11 l + (wR~4 — I ce zL — C B cos at = ceV cos

1 J

Therefore, in order for this to be an identity, A and the simultaneous equations

B

a

must satisfy

CLIFFS QUICK REVIEW



I — w-L—C, A—(wR)B= 0 (wR)A—

1

W2L--

B=w V

The solution of this system i s w'R

A=

-WV

V B=

and

1 a) L — C, + (wR )'

w2L —

1 1W

2L —

C,

1 C

+ ((vR) 2

These expressions can be simplified by invoking the followin g standard definitions : ■ wL is called the inductive reactance and denoted X1. l

■, is called the capacitive reactance and denoted X( , ■ XI. — X( , is simply called the reactance and denoted X ■ ~Xj, - X( )' +

R2 =

+

R '- is called the impedance and

denoted Z Therefore, 1 '-

w2L—C

~

1

+(wR)_= wwL —wC + ((()R ) =wz

COL

= w2 [(XI,

1 WC +R-' XC,)2 + R2 ]

= w2 [X2 + R' ] =

DIFFERENTIAL EQUATIONS

.2 -z 2

M



and the expressions for the coefficientsA and B given above can b e written as A=

w-R V 1 C, + (wR ) w-L —

1 — wV w'L — C B= 1 w-L — C, + ((,)R)2

w'RV

RV

w'Z

Z-'

1 — wV • w wL -- wC

,

w-Z-

- w2 V(X1, — X( , ) VX — w'Z- - Z-' These simplifications yield the following particular solution of th e given nonhomogeneous differential equation : RV

VX

V/R

X

i = Z, sinat— Z, cos(t = Z~Zsinut—Zcosu t Combining this with the general solution of the correspondin g homogeneous equation gives the complete solution of the nonhomo geneous equation : i= ih + i o r i=

e-(R1'r_)t

c, cos

1 R' , t + c-) sin LC — 4L

1 R2 LC — ,t 4L V ~R

X

1

+Z Zsinuat —Zcoscet ~

Despite its rather formidable appearance, it lends itself easily t o analysis . The first term [the one with the exponential-decay facto r e -(Rl'1 )t] goes to zero as t increases, while the second term remain s indefinitely . For these reasons, the first term is known as th e transient current, and the second is called the steady-state current :

M

CLIFFS QUICK REVIEW



A FE W APPLICATIONS

i=

e-(R~ '--1.)' CI

1

cos

1

R'

LC, — 4 , t + c-) sin

—

R'

LC 4L

L

t

transient curren t (goes to 0)

vR

+ Z Z sin

ak

x

cos wt

steady-state curren t

Example 9: Consider the underdamped LRC series circuit discussed above . Once the transient current becomes so small that i t may be neglected, under what conditions will the amplitude of th e oscillating steady-state current be maximized? In particular, assuming that the inductance L, capacitance C, resistance R, and voltag e amplitude V are fixed, how should the angular frequency w of the voltage source be adjusted to maximize the steady-state current i n the circuit ? The steady-state curent is given by the equatio n V /R

1 =

X

Z I Z—Slll a — Z cos wt

1

By analogy with the phase-angle calculation in Example 8, thi s equation is rewritten as follows : V /R X 1 i =Z~Zsincot —Zcoscet l V R Z

X -+R'-

sin

V

wt —

X ~_ X '+R-)

= Z (cos ¢sin a — sin 0 cos V = Z—sin (cot —

DIFFERENTIAL EQUATIONS

~)

ut )

cos cot



(where cos 0 = R/ X-" + R 2 and sin 0 = X/ X--+R2) . Therefore , the amplitude of the steady-state current is VIZ, and, since V i s fixed, the way to maximize VIZ is to minimize Z . Becaus e Z= J -2+R2, Z will be minimized if X = U . And because w i s necessarily positive, X

OaX, .—Xc=OawL=

1

<-*

&J-

=

1 LC

WC

p w =

1

JC

This value of to is called the resonant angular frequency . When th e underdamped circuit is "tuned" to this value, the steady-stat e current is maximized, and the circuit is said to be in resonance . Thi s is the principle behind tuning a radio, the process of obtaining th e strongest response to a particular transmission . In this case, th e frequency (and therefore angular frequency) of the transmission i s fixed (an FM station may be broadcasting at a frequency of, say, 95 . 5 MHz, which actually means that it's broadcasting in a narrow band around 95 .5 MHz), and the value of the capacitance C or inductance L can be varied by turning a dial or pushing a button . According to the calculation above, resonance is achieved whe n

Therefore, in terms of a (relatively) fixed w and a variabl e capacitance, resonance will occur whe n C=

1

1

w'L

(2-rrf )2 L

(where f is the frequency of the broadcast) . Or in terms of a variabl e inductance, the circuitry will resonate to a particular station when L is adjusted to the value 1 L w-C

1 ( 21Tf) "C

CUFFS QUICK REVIEW