3-Higher Order de v-VI

V. Linear ODE of Higher Order.Variation of Parameters

Learning Objective

At the end of the module, you should be able to solve linear differential equations of higher order using Variation of parametersmethod.

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Derivation of the method

)()()()( 012 xgyxayxayxa

)(

)(

)(

)(

)(

)(

22

0

2

1

xa

xgy

xa

xay

xa

xay

)()()( xfyxQyxPy The standardform




)()()()( 2211 xyxuxyxuyp Let

where )(1 xy )(2 xyand

are the fundamental set of solutions on some interval I of the homogeneous DE.




)()()()( 2211 xyxuxyxuyp

22221111 uyyuuyyuyp

22221111 yuyuyuyuyp

22112211 uyuyyuyu

2211 uyuy



22221111 yuyuyuyu

)()()( xfyxQyxPy

22112211 uyuyPyuyuP 2211 yuyuQ

)(xf

)( 1111 QyyPyu )( 2222 QyyPyu

2211 yuyu 2211 uyuy 2211 uyuyP

2211 uyuy

2211 yuyu 2211 uyuy 2211 uyuyP



02211 uyuy

)(2211 xfuyuy

W

xfy

W

Wu

)(211

W

xfy

W

Wu

)(122 and

By the Cramer’s Rule:

where)(

0 ,

)(

0 ,

1

12

2

21

21

21

xfy

yW

yxf

yW

yy

yyW


Summary of the Variation of Parameters method

)()()()( 012 xgyxayxayxa

• Find the complementary solution

• Compute the Wronskian

2211 ycycyc

))(),(( 21 xyxyW


)()()( xfyxQyxPy

• Write the DE in standard form

• Compute

W

xfy

W

Wu

)(211

W

xfy

W

Wu

)(122 and



1u• Integrate and

• Finally

pc

p

yyy

yuyuy

2211

2u

The general solution:



Example

Solve

.)1(44 2xexyyy


Find 044 yyy

cy

0442 mm

221 mm

Example


xxc xececy 2

22

1

?py

Example


2211 yuyuyp

.)1(44 2xexyyy

xxc xececy 2

22

1

Standard form

xx xeyey 22

21 , From yc

f

Example


Examplexx xeyey 2

22

1 ,

21

21

yy

yyW

)2(2 222

22

xxx

xx

xeee

xeeW

Wronskian

))(2()2( 22222 xxxxx xeexeeeW

xeW 4


x

x

e

exx4

4)1(

xeW 4

xxxxu 21 )1(

23

23

1

xxu

Example

W

xfyu

)(21

xx xeyey 22

21 ,


x

x

e

ex4

4)1(

xeW 4

12 xu

xx

u 2

2

2

Example

W

xfyu

)(12

xx xeyey 22

21 ,


pc yyy

2211 yuyuyy c

xx

u 2

2

223

23

1

xxu

xxxx xexx

exx

xececy 22

223

22

21 )

2()

23(

Example

xxxx ex

ex

xececy 23

22

22

21

62


Solve

.3csc364 xyy

Example


Find 0364 yy

.cy0364 2 m

092 m

imim 3,3 21

3,0 ?cy

Example


xecxecy xxc 3sin3cos 0

20

1

xcxcyc 3sin3cos 21

?py

Example


2211 yuyuyp

xcxcyc 3sin3cos 21

xyxy 3sin,3cos 21

Example


xx

xxW

3cos33sin3

3sin3cos

xcxcyc 3sin3cos 21

)3sin3(3cos3 22 xxW

3]3sin3[cos3 22 xxW

Example


12

1

34

1

11

W

Wu

4

11 W3W

xu12

11

Example


12

3cot

33sin4

3cos

22

xx

x

W

Wu

x

xW

3sin4

3cos2 3W

xu 3sinln36

12

Example


pc yyy

2211 yuyuyy c

xxxxxcxcy 3sin)3sinln36

1(3cos)

12

1(3sin3cos 21

Example


Extending to Higher Order

Solve)()()()()( 0123 xgyxayxayxayxa

• Find the complementary solution

• Compute the Wronskian

332211 ycycycyc

))(),(),(( 321 xyxyxyW


)()()()( xfyxRyxQyxPy

• Write the DE in standard form

• Compute

W

Wu 1

1 W

Wu 2

2 and W

Wu 3

3



321

321

321

yyy

yyy

yyy

W

32

32

32

1

)(

0

0

yyxf

yy

yy

W

31

31

31

2

)(

0

0

yxfy

yy

yy

W

)(

0

0

21

21

21

3

xfyy

yy

yy

W



,1u• Integrate and

• Finally

pc yyy

2u


3u


332211 yuyuyuyp


Solve

xyy tan

Example


0 yy

cy 03 mm

0)1( 2 mm

imimm 321 ,,0

xcxccyc cossin 321

Example


xyy tan

py

xcxccyc cossin 321

xx

xx

xx

W

cossin0

sincos0

cossin1

Example

1cossin

sincos1

xx

xxW


xyy tan

xxx

xx

xx

W

cossintan

sincos0

cossin0

1

xxx

xxxW tan

sincos

cossintan1

Example


xyy tan

xx

x

x

W

costan0

sin00

cos01

2

x

x

xx

xW

cos

sin

costan

sin01

2

2

Example


xyy tan

xx

x

x

W

tansin0

0cos0

0sin1

3

xxx

xW sin

tansin

0cos13

Example


xx

W

Wu tan

1

tan11

xdxu tan1

Example

xu cosln1


x

xx

x

W

Wu

cos

sin

1cos

sin2

2

22

dxx

xdx

x

xu

cos

cos1

cos

sin 22

2

xxxu

dxxxu

sintansecln

)cos(sec

2

2

Example


xx

W

Wu sin

1

sin33

xdxu sin3

Example

xxu cos)cos(3


332211 yuyuyuyp

pc yyy

Complete it.

Example


VI-The Cauchy-Euler Equation-

At the end of the section, students should be able to identify and solve a

Cauchy-Euler equation.

Learning Outcome


The Cauchy-Euler Equation

A linear DE of the form

)(... 011

11

1 xgyadx

dyxa

dx

ydxa

dx

ydxa

n

nn

nn

nn

n

where the coefficients areconstants.

01,...,, aaa nn


2nd–order Cauchy-Euler equation

)(2

22 xgcy

dx

dybx

dx

ydax


cba ,,


Homogeneous

02

22 cy

dx

dybx

dx

ydax


cba ,,


Method of Solution

02

22 cy

dx

dybx

dx

ydax

Let be the form ofsolution.

mxy


02

22 cy

dx

dybx

dx

ydax

mxy

1 mmxy2)1( mxmmy

Substituteinto theDE.

Method of Solution


The Auxiliary Equation

0)1( cbmmam

0)(2 cmabam


Case I- Distinct Real Roots

If are two distinct real rootsof the auxiliary equation, then thegeneral solution is

21,mm

.21

21mm xcxcy


Case II- Repeated Real Roots

If are two repeated real rootsof the auxiliary equation, then thegeneral solution is

21,mm

.ln11

21 xxcxcy mm


Case III- Conjugate Complex Roots

If are two conjugate complex roots of the auxiliary equation, then thegeneral solution is

imim 21 ,

)].lnsin()lncos([ 21 xcxcxy

.21ii xcxcy

or


Example

Solve

.0422

22 y

dx

dyx

dx

ydx


.0422

22 y

dx

dyx

dx

ydx

0)(2 cmabam

The Auxiliary Equation:

04)12(1 2 mm

4 ,2 ,1 cba

0432 mm

Example


0432 mm

0)1)(4( mm

1 ,4 21 mm

Example

.12

41

xcxcy


Solve

.0842

22 y

dx

dyx

dx

ydx

Example


.0842

22 y

dx

dyx

dx

ydx

The Auxiliary Equation:

01)48(4 2 mm

0144 2 mm

0)12( 2 m

Example

1 ,8 ,4 cba


2

1 ,

2

121 mm

Example

.ln11

21 xxcxcy mm

xxcxcy ln2

1

22

1

1


Solve

.2

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

Example


0174 2 yyx

The Auxiliary Equation

017)40(4 2 mm

01744 2 mm

)4(2

)17)(4(416)4( m

Example

17 ,0 ,4 cba


)4(2

)17)(4(416)4( m

8

16)4( im

imim 22

1 ,2

2

121

Example


imim 22

1 ,2

2

121

)].lnsin()lncos([ 21 xcxcxy

)]ln2sin()ln2cos([ 212

1

xcxcxy

Example



1

xcxcxy

.2

1)1( ,1)1( yy

1)1( y

1))]1ln(2sin())1ln(2cos([)1( 212

1

cc

11 c

Example



1

xcxcxy

2

1)1( y

)]ln2cos(2

)ln2sin(2

[

)]ln2sin()ln2cos([2

1

212

1

212

1

xx

cx

x

cx

xcxcxy

Example


2

1)1( y

2

1)]1ln2cos(

1

2)1ln2sin(

1

2[)1(

)]1ln2sin()1ln2cos([)1(2

1

212

1

212

1

cc

cc

Example

2

12

2

12 c

02 c


)]ln2[cos(2

1

xxy


1

xcxcxy

The solution is

Example


Solve

.0875 23 yyxyxyx

Example


.0875 23 yyxyxyx

Let the solution bemxy

1 mmxy

2)1( mxmmy

3)2)(1( mxmmmy

Example


0875 23 yyxyxyx

087

)1)((5)2)(1)((1

2233

mm

mm

xxmx

xmmxxmmmx

Example

0]87

)1)((5)2)(1)([(

m

mmmmmxm


0]842[ 23 mmmxm

0 0842 23 mmm

Example

0)4)(2( 2 mm

imimm 2 ,2 ,2 321


The general solution


1 xcxcxxcy

)ln2sin()ln2cos( 322

1 xcxcxcy

Example imimm 2 ,2 ,2 321


NON HOMOGENEOUS

)(2

22 xgcy

dx

dybx

dx

ydax


cba ,,


)(2

22 xgcy

dx

dybx

dx

ydax


pc yyy

NON HOMOGENEOUS


py

Find by solving the homogeneous DE

Find using Variation of Parametersmethod

cy

pc yyy


Example

.42 24

2

22 xxy

dx

dyx

dx

ydx

.12

41

xcxcyc

py


Example .12

41

xcxcyc

41 xy 1

2 xy

23

14

4

xx

xxW 222 54 xxx


Example

.42 24

2

22 xxy

dx

dyx

dx

ydx

Standard form

142 2

22

2

xyxdx

dy

xdx

yd

1)( 2 xxf


Example4

1 xy 12

xy

22

1

11

0

xx

xW 1 xx

1)( 2 xxf

)(5

1

5

)( 31

2

11'

1

xx

x

xx

W

Wu

)2

1(ln

5

121x

xu


Example4

1 xy 12

xy

14

023

4

2

xx

xW 46 xx

1)( 2 xxf

)(5

1

524

2

462'

2 xxx

xx

W

Wu

)35

(5

1 35

2

xxu


Example 41 xy

12

xy

)35

(5

1 35

2

xxu )

2

1(ln

5

121x

xu

2211 uyuyyp

352ln

5

1 2424 xxxxxyp

56

5ln

5

1 424 xxxxyp


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