Higher-Order Differential Equations

Copyright © Jones and Bartlett；滄海書局 1

Higher-Order Differential Equations

CHAPTER 3

(3.1~3.6)

Copyright © Jones and Bartlett；滄海書局 Ch3_2

Chapter Contents

3.1 Theory of Linear Equations3.2 Reduction of Order3.3 Homogeneous Linear Equations with Constants C

oefficients3.4 Undetermined Coefficients3.5 Variation of Parameters3.6 Cauchy-Euler Equations


3.1 Preliminary Theory: Linear Equ.

Initial-value Problem An nth-order initial problem isSolve:

Subject to:

(1)

with n initial conditions.

)()()()()( 011

1

1 xgyxadxdy

xadx

ydxa

dxyd

xa n

n

nn

n

n

10)1(

1000 )(,,)(,)( n

n yxyyxyyxy


Let an(x), an-1(x), …, a0(x), and g(x) be continuous on I,

an(x) 0 for all x on I. If x = x0 is any point in this interval, then a solution y(x) of (1) exists on the interval and is unique.

Theorem 3.1.1 Existence of a Unique Solution


The IVP

possesses the trivial solution y = 0. Since this DE with constant coefficients, from Theorem 3.1.1, hence y = 0 is the only one solution on any interval containing x = 1.

Example 1 Unique Solution of an IVP

0)1(,0)1(,0)1(,0753 yyyyyyy


Example 2 Unique Solution of an VP

Please verify y = 3e2x + e–2x – 3x, is a solution of

This DE is linear and the coefficients and g(x) are all continuous, and a2(x) 0 on any I containing x = 0. This DE has an unique solution on I.

1)0(',4)0(,124" yyxyy


Boundary-Value Problem

Solve:

Subject to:

is called a boundary-value problem (BVP).See Fig 3.1.1.

)()()()( 012

2

2 xgyxadxdy

xadx

ydxa

10 )(,)( ybyyay


Fig 3.1.1 Colored curves are solution of a BVP


In example 4 of Sec 1.1, we see the solution of isx = c1 cos 4t + c2 sin 4t (2)

(a) Suppose x(0) = 0, then c1 = 0, x(t) = c2 sin 4tFurthermore, x(/2) = 0, we obtain 0 = 0, hence

(3)has infinite many solutions. See Fig 3.1.2.

(b) If(4)

we have c1 = 0, c2 = 0, x = 0 is the only solution.

Example 3 A BVP Can Have Money, Ine, or Not Solutions

016" xx

02

,0)0(,016

xxxx

08

,0)0(,016

xxxx


Example 3 (2)

(c) If

(5)we have c1 = 0, and 1 = 0 (contradiction).Hence (5) has no solutions.

12

,0)0(,016

xxxx


Fig 3.1.2 The BVP in (3) of Ex 3 has many solutions


Homogeneous Equations

The following DE

(6)

is said to be homogeneous;

(7)

with g(x) not identically zero, is nonhomogeneous.

0)()()()( 011

1

1

yxadxdy

xadx

ydxa

dxyd

xa n

n

nn

n

n

)()()()()( 011

1

1 xgyxadxdy

xadx

ydxa

dxyd

xa n

n

nn

n

n


Let dy/dx = Dy. This symbol D is called a differential operator. We define an nth-order differential operator as

(8)In addition, we have

(9)so the differential operator L is a linear operator.

Differential Equations We can simply write the DEs as

L(y) = 0 and L(y) = g(x)

Differential Operators

)()()()( 011

1 xaDxaDxaDxaL nn

nn

))(())(()}()({ xgLxfLxgxfL


Let y1, y2, …, yk be a solutions of the homogeneous

Nth-order differential equation (6) on an interval I.

Then the linear combinationy = c1y1(x) + c2y2(x) + …+ ckyk(x)

where the ci, i = 1, 2, …, k are arbitrary constants, is

also a solution on the interval.

Theorem 3.1.2 Superposition Principles – Homogeneous Equations


(a) y = cy1 is also a solution if y1 is a solution.

(b) A homogeneous linear DE always possesses the trivial solution y = 0.

Corollaries to Theorem 3.1.2


The function y1 = x2, y2 = x2 ln x are both solutions of

Then y = x2 + x2 ln x is also a solution on (0, ).

Example 4 Superposition – Homogeneous DE

A set of f1(x), f2(x), …, fn(x) is linearly dependent on

an interval I, if there exists constants c1, c2, …, cn,

not all zero, such that c1f1(x) + c2f2(x) + … + cn fn(x) = 0If not linearly dependent, it is linearly independent.

Definition 3.1.1 Linear Dependence / Independence

0423 yyxyx


In other words, if the set is linearly independent, when c1f1(x) + c2f2(x) + … + cn fn(x) = 0then c1 = c2 = … = cn = 0

Referring to Fig 3.1.3, neither function is a constant multiple of the other, then these two functions are linearly independent.

If two functions are linearly dependent, then one is simply a constant multiple of the other.

Two functions are linearly independent when neither is a constant multiple of the other.


Fig 3.1.3 The set consisting of f1 and f2 is linear independent on (-, )


The functions f1 = cos2 x, f2 = sin2 x, f3 = sec2 x, f4 = tan2 x are linearly dependent on the interval (-/2, /2) since

c1 cos2 x +c2 sin2 x +c3 sec2 x +c4 tan2 x = 0

when c1 = c2 = 1, c3 = -1, c4 = 1.

Example 5 Linear Dependent Functions


Example 6 Linearly Dependent Functions

The functions are linearly dependent on the interval (0, ), since

f2 = 1 f1 + 5 f3 + 0 f4

,1)( ,5)( ,5)( 321 xxfxxxfxxf2

4 )( xxf


Suppose each of the functions f1(x), f2(x), …, fn(x)

possesses at least n – 1 derivatives. The determinant

is called the Wronskian of the functions.

Definition 3.1.2 Wronskian

)1()1()1(

21

21

1

21

), ,...(

nn

nn

n

n

n

fff

fff

fff

ffW


Let y1(x), y2(x), …, yn(x) be n solutions of the nth-order

homogeneous DE (6) on an interval I. This set of

solutions is linearly independent on I if and on if

W(y1, y2, …, yn) 0 for every x in the interval.

Theorem 3.1.3 Criterion for Linear Independence Solutions


Any set y1(x), y2(x),… , yn(x) of n linearly independent

solutions is said to be a fundamental set of solutions.

Definition 3.1.3 Fundamental Set of Solutions

There exists a fundamental set of solutions for (6) on an interval I.

Theorem 3.1.4 Existence of a Fundamental Set


Let y1(x), y2(x), …, yn(x) be a fundamental set of

solutions of homogeneous DE (6) on an interval I. Then

the general solution is y = c1y1(x) + c2y2(x) + … + cnyn(x)

where ci, i = 1, 2, …, n are arbitrary constants.

Theorem 3.1.5 General Solution – Homogeneous Equations


The functions y1 = e3x, y2 = e-3x are solutions of y” – 9y = 0 on (-, )Now

for every x. So y = c1e3x + c2e-3x is the general solution.

Example 7 General Solution of a Homogeneous DE

0633

),(33

3333

xx

xxxx

ee

eeeeW


Example 8 A Solution Obtained from a General Solution

The functions y = 4 sinh 3x - 5e3x is a solution of example 7 (Verify it). Observer

y = 4 sinh 3x – 5e-3x

xxx

xxx eee

eeey 333

333 52

4522


The functions y1 = ex, y2 = e2x , y3 = e3x are solutions of y’’’ – 6y” + 11y’ – 6y = 0 on (-, ). Since

for every real value of x. So y = c1ex + c2 e2x + c3e3x is the general solution on (-, ).

Example 9 General Solution of a Homogeneous DE

02

94

32),,( 6

32

32

32

32 x

xxx

xxx

xxx

xxx e

eee

eee

eee

eeeW


Nonhomogeneous Equations

Complementary Functiony = c1y1 + c2y2 +… + ckyk + yp = yc + yp

= complementary + particular

Any yp free of parameters satisfying (7) is called a particular solution. If y1(x), y2(x), …, yk(x) be a fundamental set of solutions of (6), then the general solution of (7) is

y= c1y1 + c2y2 +… + ckyk + yp (10) where ci, i = 1, 2, …, n are arbitrary constants.

Theorem 3.1.6 General Solution – Nonhomogeneous Equations


The function yp = -(11/12) – ½ x is a particular solution of

(11)From previous discussions, the general solution of (11) is

Example 10 General Solution of a Nonhomogeneous DE

xyyyy 36116

xecececyyy xxxpc 2

112113

32

21


Given(12)

where i = 1, 2, …, k. If ypi denotes a particular solution corresponding to the

DE (12) with gi(x), then(13)

is a particular solution of (14)

Theorem 3.1.7 Superposition Principles – Nonhomogeneous Equations

Any Superposition Principles

)()()()()( 01)1(

1)( xgyxayxayxayxa i

nn

nn

)()()(21

xyxyxyykpppp

)()()(

)()()()(

21

01)1(

1)(

xgxgxg

yxayxayxayxa

k

nn

nn


We find is a particular solution of

is a particular solution of

is a particular solution of

From Theorem 3.1.7, is a solution of

Example 11 Superposition – Nonmogeneous DE

824164'3" 2 xxyyy

xeyyy 224'3"

xx exeyyy 24'3"

321 ppp yyyy

)()(

2

)(

2

321

228241643xg

xx

xg

x

xg

exeexxyyy

241

xy p

xp ey 2

2


If ypi is a particular solution of (12), then

is also a particular solution of (12) when the right-hand member is

Note

,21 21 kpkppp ycycycy

)()()( 2211 xgcxgcxgc kk


3.2 Reduction of Order

IntroductionWe know the general solution of

(1)is y = c1y1 + c2y1. Suppose y1(x) denotes a known solution of (1). We assume the other solution y2 has the form y2 = uy1.Our goal is to find a u(x) and this method is called reduction of order.

0)()()( 012 yxayxayxa


Given y1 = ex is a solution of y” – y = 0, find a second solution y2 by the method of reduction of order.

Solution:If y = uex, then

And

Since ex 0, we let w = u’, then

Example 1 Finding s second Solution

eeueueyueuey xxxxx 2,

0)2(" uueyy x

uecw x 21

22

121

cecu x


Thus (2)

Choosing c1 = 0, c2 = -2, we have y2 = e-x. Because W(ex, e-x) 0 for every x, they are independent on (-, ).

xxx ecec

exuy 21

2)(

Example 1 (2)


General Case

Rewrite (1) as the standard form (3)Let y1(x) denotes a known solution of (3) and y1(x) 0 for every x in the interval.

If we define y = uy1, then we have

0)()( yxQyxPy

uyuyyuyuyyuy 11111 2,

0)2(][ 111

zero

111

uPyyuyQyyPyu

QyyPy


This implies that

or(4)

where we let w = u’. Solving (4), we have

or

0)2( 111 uPyyuy

02)2( 111 wPyywy

021

1

Pdxdxyy

wdw

cPdxwy ||ln 21

Pdxecwy 121


then

Let c1 = 1, c2 = 0, we find

(5)

221

1 cdxy

ecu

Pdx

dxxy

exyy

dxxP

)()( 2

1

)(

12


The function y1= x2 is a solution of

Find the general solution on (0, ).

Solution:The standard form is

From (5)

The general solution is

Example 2 A second Solution by Formula (5)

04'3"2 yxyyx

043

2 yx

yx

y

xxdxx

exy

xdx

ln24

/32

2

xxcxcy ln22

21


3.3 Homogeneous Linear Equation with Constant Coefficients

Introduction (1)

where ai, i = 0, 1, …, n are constants, an 0.

Auxiliary EquationFor n = 2,

(2)Try y = emx, then

(3)

is called an auxiliary equation.

0012)1(

1)(

yayayayaya nn

nn

0 cyybya

0)( 2 cbmamemx

02 cbmam


From (3) the two roots are

(1) b2 – 4ac > 0: two distinct real numbers.(2) b2 – 4ac = 0: two equal real numbers.(3) b2 – 4ac < 0: two conjugate complex numbers.

aacbbm 2/)4( 21

aacbbm 2/)4( 22


Case 1: Distinct real rootsThe general solution is

(4)

Case 2: Repeated real roots and from (5) of Sec 3.2,

(5)

The general solution is

(6)

xmey 11

xmxmxm

xmxm xedxedx

ee

ey 11

1

1

1

2

2

2

xmxm xececy 1121

xmececy xm 2121


Case 3: Conjugate complex rootsWe write , a general solution is

From Euler’s formula: and (7) and

imim 21 ,

xixi eCeCy )(2

)(1

sincos iei xixe xi sincos xixe xi sincos

xee xixi cos2 xiee xixi sin2


Since is a solution then setC1 = C1 = 1 and C1 = 1, C2 = -1 , we have two solutions:

So, ex cos x and ex sin x are a fundamental set of solutions, that is, the general solution is

(8)

xixi eCeCy )(2

)(1

xeeeey xxixix cos2)(1

xieeeey xxixix sin2)(2

)sincos(sincos 2121 xcxcexecxecy xxx


Solve the following DEs:

(a)

(b)

(c)

Example 1 Second-Order DEs

03'5"2 yyy

3,1/2,)3)(12(352 212 mmmmmm

xx ececy 32

2/1

025'10" yyy

5,)5(2510 2122 mmmmm

xx xececy 52

51

07'4" yyy

imimmm 32,32,074 212

)3sin3cos(,3,2 212 xcxcey x


Solve

Solution:

See Fig 3.3.1.

Example 2 An Initial-Value Problem

2)0(',1)0(,017'4"4 yyyyy

,01744 2 mm im 21/21

)2sin2cos( 212/ xcxcey x

,1)0( y ,11 c ,2)0(' and y 3/42 c


Fig 3.3.1 Graph of a solution of IVP in Ex 2


Two Equations worth Knowing

For the first equation:

(9)For the second equation:

(10)Let

Then(11)

,02 yky 0 ,02 kyky

kxckxcy sincos 21

kxkx ececy 21

kxckxcy sinhcosh 21

kxeey kxkx cosh)(1/21

kxeey kxkx sinh)(1/22


Higher-Order Equations

Given

(12)

we have

(13)

as an auxiliary equation.If the roots of (13) are real and distinct, then the

general solution of (12) is

0012)1(

1)(

yayayayaya nn

nn

0012

21

1 amamamama n

nn

n


Example 3 Third-Order DE

Solve

Solution:

043 yyy

2223 )2)(1()44)(1(43 mmmmmmm

232 mmxxx xecececy 2

32

21


Solve

Solution:

Example 4 Fourth-Order DE

02 2

2

4

4

ydx

yddx

yd

0)1(12 2224 mmm

immimm 4231 ,

ixixixix xeCxeCeCeCy 4321

xxcxxcxcxc sincossincos 4321


If m1 = + i is a complex root of multiplicity k, then m2 = − i is also a complex root of multiplicity k. The 2k linearly independent solutions:

Repeated complex roots

xexxexxxexe xkxxx cos,,cos,cos,cos 12

xexxexxxexe xkxxx sin,,sin,sin,sin 12


3.4 Undetermined Coefficients

IntroductionIf we want to solve

(1)

we have to find y = yc + yp. Thus we introduce the method of undetermined coefficients.

)(01)1(

1)( xgyayayaya n

nn

n


Solve (2)

Solution: We can get yc as described in Sec 3.3. Now, we want to find yp.

Since the right side of the DE is a polynomial, we set

After substitution, 2A + 8Ax + 4B – 2Ax2 – 2Bx – 2C = 2x2 – 3x + 6

Example 1 General Solution Using Undetermined Coefficients

6322'4 2 xxyyy

,2 CBxAxy p ,2, BAxyy pp Ay p 2


Example 1 (2)

Then6242,328,22 CBABAA

9,5/2,1 CBA

9252 xxy p


Find a particular solution of

Solution: Let yp = A cos 3x + B sin 3xAfter substitution,

Then

Example 2 Particular Solution Using Undetermined Coefficients

xyyy 3sin2

xxBAxBA 3sin23sin)83(3cos)38(

16/73,6/73 BA

xxy p 3sin7316

3cos736


Example 3 Forming yp by Superposition

Solve (3)

Solution: We can find

Let After substitution,

Then

xxexyyy 265432

xxc ececy 3

21

xxp EeCxeBAxy 22

xxx xexeECCxeBAAx 222 654)32(3323

4/3,2,23/9,4/3 ECBA

xxp exexy 22

34

2923

34

xxx exxececy 2321 3

42

923

34


Find yp of

Solution: First let yp = Ae2x

After substitution, 0 = 8e2x, (wrong guess)

Let yp = Axex

After substitution, -3Ae2x = 8e2x Then A = -8/3, yp = (−8/3)xe2x

Example 4 A Glitch in the Method

xeyyy 845


No function in the assumed yp is part of yc

Table 3.4.1 shows the trial particular solutions.

Case I


Example 5 Forms if Particular Solution – Case I

Find the form of yp of (a)

Solution: We have and tryThere is no duplication between yp and yc.

(b) y” + 4y = x cos x

Solution: We try There is also no duplication between yp and yc.

xx eexyyy 75258 3

xexxg )75()( 3 xp eECxBxAxy )( 23

xECxxBAxxp sin)(cos)(


Rule of Case I

If g(x) consists of a sum of m terms of the kind listed in the table, then (as in Ex 3) the assumption for a particular solution yp consists of the sum of the trial forms

The form of yp is a linear combination of all linearly independent functions that are generated by repeated differentiations of g(x).


Find the form of yp of

Solution: For 3x2:

For -5 sin 2x:

For 7xe6x:

No term in duplicates a term in yc

Example 6 Forming yp by Superposition – Case I

xxexxyyy 62 72sin53149

CBxAxy p 2

1

xFxEy p 2sin2cos2

xp eHGxy 6)(

3

321 pppp yyyy


Example 7 Particular Solution – Case II

Find a particular solution of

Solution:

The complementary function is yp = c1ex + c2xex.Assume yp = Aex will fail since it is apparent from yc that ex is a solution of the associated homogeneous equation

And we will not be able to find a particular solution of the form yp = Axex since the term xex is also duplicated in yc. We next try yp = Ax2ex, substituting into the given differential equation yields 2Axex = ex and so A = ½. The a particular solution is yp = ½x2ex.

xeyyy 2

.02 yyy


If g(x) consists of a sum of m terms of the kind given in Table 3.4.1, and suppose that the usual assumption for a particular solution is

where the are the trial particular solution forms corresponding to these terms.

If any consists terms that duplicates terms in yc, then that must be multiplied by xn, where n is the smallest positive integer that eliminates that duplication.

Multiplication Rule of Case II

mpppp yyyy 21

miyip ..., ,2 ,1 ,

ipy

ipy


Example 8 An Initial-Value Problem

Solve

Solution:

First trial: yp = Ax + B + C cos x + E sin x (5)However, duplication occurs. Then we try

yp = Ax + B + Cx cos x + Ex sin xAfter substitution and simplification,

A = 4, B = 0, C = -5, E = 0Then y = c1 cos x + c2 sin x + 4x – 5x cos xUsing y() = 0, y’() = 2, we have

y = 9 cos x + 7 sin x + 4x – 5x cos x

2)(',0)(,sin104" yyxxyy

xcxcyc sincos 21


Solve

Solution: yc = c1e3x + c2xe3x

After substitution and simplification,A = 2/3, B = 8/9, C = 2/3, E = -6

Then

Example 9 Using the multiplication Rule

xexyyy 32 12269'6"

21

32

ppy

x

y

p EeCBxAxy

xxx exxxxececy 32232

31 6

32

98

32


Solve

Solution: m3 + m2 = 0, m = 0, 0, -1yc = c1+ c2x + c3e-x yp = Aex cos x + Bex sin x

After substitution and simplification,A = -1/10, B = 1/5

Then

Example 10 Third-Order-DE CeasI

xeyy x cos"

xexeecxccyyy xxxpc sin

51

cos101

321


Find the form of yp of

Solution: yc = c1+ c2x + c3x2 + c4e-x

Normal trial:

Multiply A by x3 and (Bx2e-x + Cxe-x + Ee-x) by xThen

yp = Ax3 + Bx3e-x + Cx2e-x + Exe-x

Example 11 Fourth-Order-DE – Case II

xexyy 2)4( 1

21

2

pp y

xxx

yp EeCxeeBxAy


3.5 Variation of Parameters

Some Assumptions For the DE

(1)we put (1) in the form

(2)where P, Q, f are continuous on I.

)()()()( 012 xgyxayxayxa

)()()( xfyxQyxPy


Method of Variation of Parameters

We try

(3) After we obtain yp’, yp”, we put them into (2), then

(4)

)()()()( 2211 xyxuxyxuy p

ppp yxQyxPy )()(

][][ 22221111 QyyPyuQyyPyu

2211221122221111 ][ uyuyuyuyPyuuyyuuy

221122112211 ][][][ uyuyuyuyPuydxd

uydxd

)(][][ 221122112211 xfuyuyuyuyPuyuydxd


Making further assumptions: y1u1’ + y2u2’ = 0, then from (4),y1’u1’ + y2’u2’ = f(x)

Express the above in terms of determinants

and (5)

where

(6)

Wxfy

WW

u)(21

1 W

xfyWW

u)(12

2

)(

0,

)(

0,

1

12

2

21

21

21

xfy

yW

yxf

yW

yy

yyW


Solve

Solution: m2 – 4m + 4 = 0, m = 2, 2 y1 = e2x, y2 = xe2x,

Since f(x) = (x + 1)e2x, then

Example 1 General Solution Using Variation of Parameters

xexyyy 2)1(4'4"

022

),( 4

222

2222

x

xxx

xxxx e

exee

xeexeeW

x

xx

xx

xx

x

exexe

eWxex

xeex

xeW 4

22

2

24

22

2

1 )1()1(2

0,)1(

2)1(

0


From (5),

Then

u1 = (-1/3)x3 – ½ x2, u2 = ½ x2 + xAnd

1)1(

,)1(

4

4

22

4

4

1 xe

exuxx

exex

u x

x

x

x

xxxxp exexxexxexxx 222322223

21

61

21

21

31

xxxxpc exexxececyyy 22232

22

1 21

61

Example 1 (2)


Solve

Solution: y” + 9y = (1/4) csc 3xm2 + 9 = 0, m = 3i, -3i y1 = cos 3x, y2 = sin 3x, f = (1/4) csc 3x

Since

Example 2 General Solution Suing Variation of Parameters

xyy 3csc364

33cos33sin3

3sin3cos)3sin,3(cos

xx

xxxxW

xx

xx

xW

xx

xW

3sin3cos

41

3csc4/13sin3

03cos

,41

3cos33csc4/1

3sin0

2

1


Then

And

(7)

1211

1 WW

u

xx

WW

u3sin3cos

1212

2

,12/11 xu |3sin|ln36/12 xu

|3sin|ln)3(sin361

3cos121

xxxxy p

|3sin|ln)3(sin361

3cos121

3sin3cos 21 xxxxxcxc

yyy pc

Example 2 (2)


Solve

Solution: m2 – 1 = 0, m = 1, -1 y1 = ex, y2 = e-x, f = 1/x, and W(ex, e-x) = -2

Then

The low and up bounds of the integral are x0 and x, respectively.

Example 3 General Solution Using Variation pf Paramenters

xyy

1

x

x

tx

dtt

eu

xeu

021

,2

)/1(11

x

x

tx

dtte

uxe

u02

1,

2)/1(

22


Example 3 (2)

x

x

x

x

tx

tx

p dtte

edtt

eey

0 021

21

x

x

x

x

tx

txxx

pc dtte

edtt

eeececyyy

0 021

21

21


For the DEs of the form

(8)

then yp = u1y1 + u2y2 + … + unyn, where yi , i = 1, 2, …, n, are the elements of yc. Thus we have

(9)

and uk’ = Wk/W, k = 1, 2, …, n.

Higher-Order Equations

)()()()( 01)1(

1)( xfyxPyxPyxPy n

nn

02211 nnuyuyuy

02211 nnuyuyuy

)()1(2

)1(21

)1(1 xfuyuyuy n

nn

nn


For the case n = 3,

(10)

,11 W

Wu ,2

2 WW

u WW

u 33

321

321

321

21

21

21

3

31

31

31

2

32

32

32

1

and

)(

0

0

,

)(

0

0

,

)(

0

0

yyy

yyy

yyy

W

xfyy

yy

yy

W

yxfy

yy

yy

W

yyxf

yy

yy

W


3.6 Cauchy-Eulaer Equation

Form of Cauchy-Euler Equation

Method of SolutionWe try y = xm, since

)(011

11

1 xgyadxdy

xadx

ydxa

dxyd

xa n

nn

nn

nn

n

k

kk

k dxyd

xa kmkk xkmmmmxa )1()2)(1(

mk xkmmmma )1()2)(1(


An Auxiliary Equation

For n = 2, y = xm, thenam(m – 1) + bm + c = 0, oram2 + (b – a)m + c = 0

(1)

Case 1: Distinct Real Roots

(2)

2121

mm xcxcy


Solve

Solution:We have a = 1, b = -2 , c = -4

m2 – 3m – 4 = 0, m = -1, 4,y = c1x-1 + c2x4

Example 1 Distinct Roots

0422

22 y

dxdy

xdx

ydx


Using (5) of Sec 3.2, we have Then

(3)

Case 2: Repeated Real Roots

xxcxcy mm ln1121

xxy m ln12


Example 2 Repeated Roots

Solve

Solution:We have a = 4, b = 8, c = 1

4m2 + 4m + 1 = 0, m = -½ , -½

084 2

22 y

dxdy

xdx

ydx

xxcxcy ln2/12

2/11


Higher-Order: multiplicity

Case 3: Conjugate Complex Rootsm1 = + i, m2 = – i, y = C1x( + i) + C2x( - i)

Sincexi = (eln x)i = ei ln x = cos( ln x) + i sin( ln x)x-i = cos ( ln x) – i sin ( ln x)

Then y = c1x cos( ln x) + c2x sin( ln x) = x [c1 cos( ln x) + c2

sin( ln x)] (4)

Case 3: Conjugate Complex Roots

12 )(ln,,)(ln,ln, 1111 kmmmm xxxxxxx


Solve

Solution:We have a = 4, b = 0 , c = 17

4m2 − 4m + 17 = 0, m = ½ 2i

Apply y(1) = -1, y’(1) = 0, then c1 = -1, c2 = 0,

See Fig 3.6.1.

Example 3 Am Initial-Value Problem

21

)1(',1)1(,0174 2 yyyyx

)]ln2sin()ln2cos([ 212/1 xcxcxy

)ln2cos(1/2 xxy


Fig 3.6.1 Graph of solution of IVP in Ex 3


Example 4 Third-Order Equation

Solve

Solution:Let y = xm,

Then we have xm(m + 2)(m2 + 4) = 0m = -2, m = 2i, m = -2iy = c1x-2 + c2 cos(2 ln x) + c3

sin(2 ln x)

0875 2

22

3

33 y

dxdy

xdx

ydx

dxyd

x

33

3

22

21

)2)(1(

,)1(,

m

mm

xmmmdx

yd

xmmdx

ydmx

dxdy


Solve

Solution:We have (m – 1)(m – 3) = 0, m = 1, 3 yc = c1x + c2x3 , use variation of parameters,

yp = u1y1 + u2y2, where y1 = x, y2 = x3 Rewrite the DE as

Then P = -3/x, Q = 3/x2, f = 2x2ex

Example 5 Variation of Prarmeters

xexyxyyx 42 23'3"

xexyx

yx

y 22 2

33


Thus

We find

xx

x

xex

ex

xWex

xex

xW

xx

xxW

322

5

22

3

1

3

2

3

221

0,2

32

0

,231

,2

2 23

5

1x

x

exxex

u xx

exex

u 3

5

2 22

,2221

xxx exeexu xeu 2

Example 5 (2)


Example 5 (3)

Then

xx

xxxxp

xeex

xexexeexyuyuy

22

)22(2

322211

xxpc xeexxcxcyyy 22 23

21

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Higher-Order Differential Equations