EE2092-Chapter4

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_______________________________________________________________9/30/2014 Chapter 4 Second-Order Differential Equations 4-1

Chapter 4Second-Order Differential

Equations

A second-order differential equation is called linear

if it can be written

)()()(2

2

xryxqdxdyxp

dxyd

wherep(x), q(x) and r(x) can be any given functions

ofx.

If r(x) = 0 0)()(2

2 yxq

dx

dyxp

dx

yd

is called homogeneous.

If r(x) 0 )()()(2

2xryxq

dx

dyxpdx

yd

is called non-homogeneous.

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Contents

1Ldenotes Linear Differential Eq.

Cdenotes Constant-coefficient Linear Differential Eq.

Solving The Homogeneous Equations

Distinct, Real Roots L,C1

Equal Roots L,C

Complex Conjugate Roots L,C

Wronskian Test For Linear Independence L

Solving For A Particular Solution

Solution By Undetermined Coefficients L,C

Solution By Variation of Parameters L

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Second-Order HomogeneousEquations: Constant

CoefficientsIn this section, we are interested in solving

homogeneous equations of the form

0

2

2

bydx

dya

dx

yd

where coefficients aand bare constants. Recall that

in the first-order case,

has a solution of the form

0 kydx

dy

kxey

To solve Eq. (4.1), let us try, as a possible solution,

(4.1)

xey

Substituting Eq. (4.2) and its derivatives

xx edx

yde

dx

dy 22

2

(4.2)

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00

2

2

bae

beeae

x

xxx

Into Eq. (4.1)

Hence, Eq. (4.2) satisfies Eq. (4.1) if is a solution

of the quadratic equation

02 ba

which is known as the characteristic equation.

Let 1and 2 be the roots of the characteristic eq.

24

2

2,1 baa

and the two solutions to the 2ndorder homogeneous

Eq. (4.1) are

xxeyey 21

21 and

You may verify this by substituting each solution and

its derivative into Eq. (4.1)

(4.3)

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From Eq. (4.3), three possible cases may arise,

depending on the sign of the discriminant a24b.

Case 1: Two distinct, real roots if a24b 0

Case 2: Equal, real roots if a24b = 0

Case 3: Two complex conjugate roots if a24b 0

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Two Distinct, Real Roots

In this case,xx

eyey 21 21 and

xx BeAey 21

constitute a basis of solutions of Eq. (4.1) and the

general solution is

To solve 02

2

bydx

dya

dx

yd

0

2

ba

The characteristic equation is

For real, distinct roots,

2

42

2,1baa

042 ba

(4.1)

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Example 1:Solve the initial value problem

5)0(,4)0(,022

2

yyydxdy

dxyd

The characteristic equation is given by

1or20)1)(2(

022

The general solution:xx BeAey 2

Putting in the initial conditions:

52

|2)0(

4)0(

02

BA

BeAey

BAy

xxx

Solving, A= 3 and B= 1. Therefore the solution isxx eey 23

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Example 2:Solve the boundary value problem

0)1(,4)0(,0432

2

yyydx

dy

dx

yd


4or1

0)4)(1(

0432

The general solution:xx BeAey 4

41

1

41

4

41

44Solving

0)1(

4)0(

ee

eB

ee

eA

BeAey

BAy

Putting in the boundary conditions

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Equal Roots

In this case, one solution isxey 1

x

xx

eBxA

BxeAey

To obtain a second independent solution, let

xxexyy 12

You may proof thaty2=xexis also a solution by

substitutingy2and its derivatives into Eq. (4.1).

Sincey1= exandy2=xe

xare independent (we will

prove this later) they form the basis for the generalsolution, which is

To solve 02

2

bydx

dya

dx

yd

02 baThe characteristic equation is

For equal roots,

224

2

abaa

042 ba

(4.1)

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Example 3:Solve

01682

2

ydxdy

dx

yd


roots)double(Equal,40)4(

0168

2

2

Hence the basis is

xeBxAy 4

xx xee 44 and

and the general solution is

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Example 4:Solve the initial value problem

1)0(,3)0(,0442

2

yyydxdy

dxyd


2

0)2(

044

2

2

The general solution:xeBxAy 2)(

51)(2)0(

3)0(

BABy

Ay


By differentiatingxx eBxABey 22 )(2

Therefore the solution is

xexy 2)53(

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sin()cos(

sin()cos(

formulaEulergUsin

)(

)(

ninxe

ninxe

nxi

nxi

Complex, Conjugate Roots

To solve 02

2

bydx

dya

dx

yd

02 baThe characteristic equation is

For complex roots, 042 ba

1i

The general solution is

nxDCinxDCe

nxinxDnxinxCe

eDeeCe

DeCe

DeCey

mx

mx

inxmxinxmx

xinmxinm

xx

sin)(cos)(

)sin(cossincos

)()(

21

inmab

ia

baa

2

4

2

24

2

22,1

nxBnxAey mx sincos

(4.1)

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Example 5:Solve the boundary value problem

1)4(,1)0(,052

2

2

yyydxdy

dxyd


i21

0522

The general solution: )2sin2cos( xBxAey x

44 12

sin2

cos4

110sin0cos)0(

eBBAey

ABAy


Therefore the solution is

)2sin2(cos

4

xexey

x

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Wronskian Test ForLinear Independence of

SolutionsConsider 0)()(

2

2

yxqdx

dyxp

dx

yd

Supposep(x) and q(x) are continuous functions ofx

on some intervalI, the Wronskian Testis used to

test the independence of the solutionsy1andy2. The

Wronskian is defined

dx

dyy

dx

dyy

dx

dy

dx

dy

yy

yyW 122

121

21

21,

Note:

1. Two solutionsy1andy2 are linearly

independent(and hence is a basis for the

solution of Eq. (1)), if for somexin intervalI,

the Wronskian W

0.2. Conversely,y1andy2 are linearly dependentif

and only if W = 0 for somexinI.

3. If W = 0, for somexinI, W 0 onI.

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Example 6:

022

2

ywdxyd

iw

w

022

The two solutions are:

wxywxy sincos 21


We want to check for independence of the two

solutions

w

wxwxw

wxwwxw

wxwx

dx

wxd

dx

wxdwxwx

wxwxW

)sin(cos

cossin

sincos

sincossincos

sin,cos

22

Hence the two solutions are linearly independent on

any interval when w0.

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Example 7: Solve the initial value problem

3)0(,1)0(1034 22

2

yyeydxdy

dxyd x

3or1

0342

First, solve the homogeneous equation.The

characteristic equation is given by

The general solution to the homogeneous equation is

xxh BeAey

3

Next, we find a particular solution to thenonhomogeneous equation. Noting that the RHS is of

the form e2x, we try

xp Cey

2

Find derivatives

xp

xp

Cedx

yd

Cedx

dy

22

2

2

4

2

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Substitutingypand its derivatives into the

nonhomogeneous eq.

xp

x

xx

xxxx

ey

Ce

eCe

eCeCeCe

2

2

22

2222

32Therefore

3

20

1015

10384

The general solution is given by

xxx

ph

eBeAe

yyy

23

3

2

Putting in the initial conditions:

33

43

|3

43)0(

1

3

2)0(

023

BA

eBeAey

BAy

xxxx

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Solving,

1,3

4

BA


xxx

ph

eee

yyy

23

3

2

3

4

Note: The arbitrary constantsAandBare solved

only after the general solutiony=yh+ypis found.

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Solution By Undetermined

CoefficientsThis approach is to find the particular solution if the

RHS r(x) is of simple form. Given

)(

2

2

xrby

dx

dya

dx

yd

where aand bare constants, the following rules

apply.

Rule 1: Choose the form of the particular solution

yp(x) according to the table given below.

If r(x) equals Chooseyp(x) equals

xke xCe

nkx nnnn CxCxCxC

1

121

xk

xk

sin

cosxDxC sincos

xke

xkex

x

sin

cos)sincos( xDxCe x

(4.4)

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Rule 2: If the choice ofyp(x) is one of the solutions of

the homogeneous part, multiplyyp(x) by x (see

Example 9 and 10).

Rule 3: If r(x) is a sum of expressions of the first

column, then letyp(x) be the sum of the corresponding

expressions in the second column.

Example 8: Solve

22

2

84 xydx

yd

i2

042

First, solve the homogeneous equation.The

characteristic equation is given by

The general solution to the homogeneous equation is

xBxAyh 2sin2cos

To find a particular solution, we refer to Rule 1 of

Table. Let

322

1 CxCxCyp

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Find derivatives

12

221

2

2

Cdx

yd

CxCdx

dy

p

p

Substitutingypand its derivatives into the

nonhomogeneous eq.

1,0,2

cients,coeffiComparing

842

321

232

211

CCC

xCxCxCC

12 2 xyp

122sin2cos 2

xxBxA

yyy ph

The particular solution is

and the general solution is

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Example 9:(Demonstrating Rule 2) Solve

xeydx

dy

dx

yd 23

2

2

2or1

The roots of the characteristic equation is given by

Hence xxh BeAey 2

To find a particular solution, Rule 1 suggests that we

letyp= Cex. But we see that exis a solution of the

homogeneous equation. Hence, by Rule 2, let

xp Cxey

Find derivatives

xxpxxp exeCdx

ydexeC

dx

dy2,

2

2

Substitution gives

xxxx eCxeexCexC 213)2(

xxx xeBeAey 2

Comparing coefficients of ex, a general solution is

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Example 10:(Demonstrating Rule 2) Solve

xeydx

dy

dx

yd 22

2

The characteristic equation has a double root = 1

and the solution to the homogeneous equation is

xh eBxAy

We needyp. Rule 1 suggests that we letyp= Ce-x. Butwe see that exis a solution of the homogeneous

equation. Hence, we letyp= Cxex. Again, this is a

solution of homogeneous equation. Finally, by Rule

2, we letyp= Cx2e-x.

Find derivatives

x

xxp

xxxp

exxC

exexxCdx

yd

exxCxeexCdx

dy

24

222

22

2

2

2

2

22

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Substitution gives

21

2

)2(2)24( 222

C

eCe

eeCxexxCexxC

xx

xxxx

xx exeBxAy 22

1


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Solution By Variation OfParameters

We have )()()(2

2

xryxqdx

dyxp

dx

yd

with arbitrary functionsp, qand rthat are continuous

on some intervalI. Lety1andy2be two independentsolutions (their Wronskian 0) of the homogeneous

equation

0)()(2

2

yxqdx

dyxp

dx

yd

Then, the particular solutionyp(x) can be expressed

in the form

)()()()()( 21 xyxvxyxuxyp

where u(x) and v(x) satisfy the constraints:

)(

0

21

21

xrdx

dy

dx

dv

dx

dy

dx

du

ydxdvy

dxdu

(4.4)

(4.5)

(4.6)

(4.7)

(4.8)

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rdx

dvdx

du

dx

dy

dx

dy

yy 021

21

Notice that Eq. (4.7) and (4.8) form 2 simultaneous

equations on du/dxand dv/dx. Both equations can be

expressed as a single matrix equation as

u(x) and v(x) can be solved as follows. Let

rdx

dyy

W

dx

dyr

yW

dx

dy

dx

dy

yy

W

1

1

22

2

1

21

21

00

Using Cramers rule, the solution to Eq. (4.9) is

W

WWW

dx

dvdxdu

2

1

(4.9)

(4.10)

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Integrating,

dxW

W

dxW

W

v

u

2

1

The particular solution is

dxW

Wxydx

W

Wxy

xyxvxyxuxyp

22

11

21

)()(

)()()()()(

(4.11)

(4.12)

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Proof: Given

Ify1andy2are two independent solutions of the

homogeneous equation, let the particular solution be

)()()(2

2

xryxqdx

dyxp

dx

yd

)()()()()( 21 xyxvxyxuxyp

Differentiating

dx

dvy

dx

dyv

dx

duy

dx

dyu

dx

dyp2

21

1

Let us impose a constraint of

021 dx

dvy

dx

duy Constraint Eq. (4.7)

Substituting into Eq. (4.14),

dx

dy

vdx

dy

udx

dyp 21

(4.4)

(4.13)

(4.14)

(4.15)

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Differentiating again,

22

22

21

21

2

2

dxydv

dxdy

dxdv

dxydu

dxdy

dxdu

dxyd p

Substituting Eq. (4.13), (4.15) and (4.16) into (4.4),

and collecting terms

rdx

dy

dx

dv

dx

dy

dx

du

qydxdyp

dx

ydvqydxdyp

dx

ydu

21

22

22

2

11

21

2

The terms in the square brackets are zero becausey1andy2are solutions of the homogeneous equation. So

Eq. (4.17) becomes

rdx

dy

dx

dv

dx

dy

dx

du 21

which is the second constraint, Eq. (4.8). (Proved)

(4.16)

(4.17)

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Example 11:Solve

xydxyd sec2

2

The characteristic equation is

xBxAyh sincos

i

1

01

2

2

and the solution to the homogeneous equation is

Hence, let xyxy sin,cos 21

xdx

dyx

dx

dycos,sin 21

x

dxdv

dxduxxxx

sec0

cossinsincosThen

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1seccossecsin

0cos0

tansinseccossec

sin00

1sincoscossinsincos

1

1

2

2

2

1

2221

21

xxxx

x

rdx

dy

y

W

xxxxx

x

dx

dyr

y

W

xxxxxx

dx

dy

dx

dy

yy

W

Let

1

tan

2

1

W

W

dx

dv

xW

W

dx

du

Integrating

xdxv

xxdx

dxx

xxdxu

1

cosln)(coscos

1

cos

sintan

Solving the simultaneous equations,

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EE2092-Chapter4

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