4 series solutions

Power Series Solutions of Linear DEs

Chapter 4

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Learning Objective

AtAt thethe endend ofof thethe section,section, youyou shouldshould bebeableable toto solve DE with Power Series assolutions..


Power Series

A power series in is an infiniteseries of the form

ax

...)()()( 2210

0

axcaxccaxc n

nn

The above power series is centered at x = a.


Power Series

n

n

x )1(0

center x = -1

n

n

n x

0

12 center x = 0


Convergence

n

nn axc )(

0

A power series

is convergent at a specified value of x ifits sequence of partial sums converges.

NS

N

n

nn

NN

NaxcS

0

)(limlim exists.

If the above limit does not exists,the series is said to be divergent.

i.e.


Interval of Convergence

n

nn axc )(

0

Every power series

has an interval of convergence. The interval of convergence is the set of allreal number x for which the seriesconverges.


Radius of Convergence

n

nn axc )(

0

Every power series

has a radius of convergence R.

If R > 0, then the power series converges for and diverges

for

Rax

.Rax


Ratio Test

Given a power series ,)(0

n

nn axc

Laxc

cax

axc

axc

n

n

nnn

nn

n

11

1 lim)(

)(lim

the series converges if

the series diverges

the test is inconclusive if

1 Lax

1 Lax

1 Lax


A Power Series defines a function

A power series defines a function

n

nn axcxf )()(

0

whose domain is the interval of convergence of the series.


Examples


If the radius of convergence is R > 0, thenisf

• continuous • differentiable• integrable

over the interval (a-R, a+R).

Remark


Example

Given

0n

nn xcy

Find yy and


Example

...33

221

00

0

xcxcxcxcxcyn

nn

...320 2321 xcxccy

1

1

n

nn xncy

Since the first term is 0.

1

0

n

nn xncy


...32 2321

1

1

xcxccxncyn

nn

...620 32 xccy

2

2

)1(

n

nn xcnny

Example


Analytic at a Point

A function is analytic at a point aif it can be represented by a power seriesin x-a:

with a positive or infinite radiusof convergence.

n

nn axc )(

0

f


Adding two Power Series

Example

0

12

n

nn xcS

Write as a single summation.

2

21 )1(

n

nn xcnnS

21 SS


2 0

1221 )1(

n n

nn

nn xcxcnnSS

2 problems: exponents and starting indices

Example


2 0

12)1(n n

nn

nn xcxcnn

Let 2 nk 1 nk

Example


2

21 )1(

n

nn xcnnS

2 nk

021 )1)(2(S

k

kk xckk

Example

2 kn


0

12

n

nn xcS

1 nk

112

k

kk xcS

Example

1 kn


0 11221 )1)(2(

k k

kk

kk xcxckkSS

Now same exponentYet to solve: first term!

Example


1 112

0221 )1)(2()1)(2(

k k

kk

kk xcxckkxcSS

1122 ])1)(2[(2

k

kkk xcckkc

1 1122 )1)(2(2

k k

kk

kk xcxckkc

Example


12 2

2 3)1(2)1(n

nn

n n

nn

nn xncxcnnxcnn

Combine.

Exercise


12 2

2 3)1(2)1(n

nn

n n

nn

nn xncxcnnxcnn

Let nk 2 nk nk

1 2 3

Solution


2

)1(n

nnxcnn

nk

2

)1(k

kk xckk

1

Solution


2

2)1(2n

nnxcnn

22 knnk

02)1)(2(2

k

kk xckk

2

Solution


1

3n

nnxnc

nk

1

3k

kkxkc

3

Solution


1

3k

kkxkc

02)1)(2(2

k

kk xckk

2

)1(k

kk xckk

Solution


2

11 3)1(3

k

kk xkcxc

22

13

02 )1)(2(2)6(2)2(2

k

kk xckkxcxc

2

)1(k

kk xckk

Solution


222

2321

3)1)(2(2

)1(1243

k

kk

k

kk

k

kk

xkcxckk

xckkxccxc

Solution


kk

kkk xkcckkckk

xccc

]3)1)(2(2)1([

)123(6

22

312

Solution


Ordinary and Singular Point

A point is said to be an ordinarypoint of the DE if both and are analytic at A point that is not anordinary point is said to be a singularpoint of the DE.

0x

)(xP )(xQ.0x

0)()( yxQyxPy


A point that is not an ordinary pointis said to be a singular point of the DE.

Note: If at least one of the function and

fails to be analytic at then is a singular point.

0x)(xP

)(xQ0x

Ordinary and Singular Point


1) Every finite value of is an ordinary point of the DE

x

.0)(sin)( yxyey x

Examples

0x

.0)(ln)( yxyey x

2) is a singular point of the DE


Existence of Power Series Solutions

If is an ordinary point of the DE,we can always find two linearly independentsolutions in the form of a power seriescentered at ( ).

0xx

0xx

0

0 )(n

nn xxcy

Theorem

,0 Rxx

Each series solution converges at least on some interval defined by where R is the distance from to the closest singular point 0x


Example

Find a power series solution centered at 0 for the following DE

.0 xyy


.0 xyy

Ordinary points: All real numbers x.

Example

Since there are no finite singular points,The previous Theorem guarantees two power series solutions centered at 0, andconvergent for .x


.0 xyy

Let the solution be n

nn

n

nn xcxcy

00

)0(

1

1

n

nnnxcy

2

2

)1(

n

nn xnncy


0)1(

0

2 0

2

n n

nn

nn xcxxnnc

xyy

2 0

12 0)1(n n

nn

nn xcxnnc


0)1(2 0

12

n n

nn

nn xcxnnc

2 nk1 nk

0)1)(2(0 1

12

k k

kk

kk xcxkkc


0)1)(2(0 1

12

k k

kk

kk xcxkkc

0)1)(2(21 1

120

2

k k

kk

kk xcxkkcxc

0])1)(2([21

122

k

kkk xckkcc


0)2)(1(.2

02.1

12

2

kk cckk

c

for ,...3,2,1k

Using the Identity Property:

The (recursive) relation generate consecutivecoefficients of the solution.


0)2)(1( 12 kk cckk

)3)(2(0)3)(2( ,1 0

303

cccck

)4)(3(0)4)(3( ,2 1

414

cccck

0)5)(4(

0)5)(4( ,3 2525

cccck

)6)(5)(3)(2()6)(5(0)6)(5( ,4 03

636

ccccck


...55

44

33

2210 xcxcxcxcxccy

...6.5.3.2

04.33.2

0 60413010 x

cx

cx

cxccy

n

nn xcy

0

..

4.3

1..

6.5.3.2

1

3.2

11 4

163

0 xxcxxcy


...6.5.3.2

1

3.2

11)( 63

1 xxxy

...4.3

1)( 4

2 xxxy

)()()( 2110 xycxycxy

where


Example

.0)1( 2 yyxyx



.0)1( 2 yyxyx

.0)1(

1

)1( 22

y

xy

x

xy

The standard form:

Ordinary Points: All real numbers x.Singular point: None.

Example


Let the solution be n

nnxcy

0

1

1

n

nnnxcy

2

2

)1(

n

nn xnncy

.0)1( 2 yyxyx


.0

)1()1(

0

1

1

2

22

n

nn

n

nn

n

nn

xc

xncxxcnnx

.0)1( 2 yyxyx


0

)1()1(

01

2

2

2

n

nn

n

nn

n

nn

n

nn

xcxnc

xcnnxcnn

nk 2 nk

nk nk


0

)1)(2()1(

01

02

2

k

kk

k

kk

k

kk

k

kk

xcxkc

xckkxckk


0][1

)1)(2(62)1(

21

00

21

22

23

02

k

kk

k

kk

k

kk

k

kk

xcxcxcxkcxc

xckkxcxcxckk


0

)1)(2()1(

62

22

22

2

32110

k

kk

k

kk

k

kk

k

kk

xcxkc

xckkxckk

xccxcxcc


0

)1)(2()1(

62

22

22

2

320

k

kk

k

kk

k

kk

k

kk

xcxkc

xckkxckk

xccc

combine


0]

)1)(2()1([

62

22

320

kkk

kkk

xckc

ckkckk

xccc

0])1)(2()1)(1[(

62

22

320

k

kkk xckkckk

xccc


0])1)(2()1)(1[(

62

22

320

k

kkk xckkckk

xccc

0)1)(2()1)(1(

006

2

102

2

33

0220

kk ckkckk

cc

cccc


kk

kk

ck

kc

ckk

kkc

c

cc

)2(

)1(

)1)(2(

)1)(1(

0

2

1

2

2

3

02

,...5,4,3,2k


kk ck

kc

ccc

)2(

)1(

0;2

1

2

302

,...5,4,3,2k

0244.2

1

4

1,2 ccck

05

2,3 35

cck

0466.4.2

3

6

3,4 ccck

07

4,5 57

cck


...55

44

33

2210 xcxcxcxcxccy

...6.4.2

3.1

4.2

1

2

1 60

40

2010 xcxcxcxccy

)()(

][...]6.4.2

3.1

4.2

1

2

11[

2110

1642

0

xycxycy

xcxxxcy

n

nn xcy

0


xxy

xxxxy

)(

...6.4.2

3.1

4.2

1

2

11)(

2

6421


Example

.0)1( yxy



Example

0)1( yxy

0)1()1(02

2

n

nn

n

nn xcxxcnn

0)1(0

1

02

2

n

nn

n

nn

n

nn xcxcxcnn


Example

0)1)(2(1

100

2

k

kk

k

kk

k

kk xcxcxckk

0)1(0

1

02

2

n

nn

n

nn

n

nn xcxcxcnn

2 nknk

1nk


Example

0)1)(2(1

100

2

k

kk

k

kk

k

kk xcxcxckk

0

)1)(2(2

11

1

00

12

02

k

kk

k

kk

k

kk

xc

xcxcxckkxc


Example

0

)1)(2(2

11

11202

k

kk

k

kk

k

kk

xc

xcxckkcc

0])1)(2[(2 11

202

k

kkk

k xccckkcc


Example

,...3,2,1,)2)(1(

0)2)(1(

2

102

12

12

0202

kkk

ccc

ccckk

cccc

kkk

kkk

Using Identity Property:


Example ,...3,2,1,)2)(1(

2

1

12

02

k

kk

ccc

cc

kkk

3.2,1 01

3

ccck

4.3,2 12

4

ccck

5.4,3 23

5

ccck

6.5,4 34

6

ccck


Example02

2

1cc

3.23.2,1 001

3

cccck

4.3.24.34.3,2 0212

4

ccccck

5.3.25.4.25.4.3.25.4,3 00023

5

cccccck

6.4.3.26.5.3.26.5.4.3.26.5,4 00034

6

cccccck

Case 1: 0,0 10 cc


Example0

2

102 cc

3.23.2,1 101

3

cccck

4.34.3,2 112

4

cccck

5.4.3.25.4,3 123

5

cccck

6.5.46.5.3.26.5.4.36.5,4 01134

6

cccccck

Case 2: 0,0 10 cc


Example

...55

44

33

2210 xcxcxcxcxccy

From case 1:

...5.3.24.3.23.22

504030200 x

cx

cx

cx

ccy

...]5.3.2

1

4.3.2

1

3.2

1

2

11[ 5432

0 xxxxcy

...5.3.2

1

4.3.2

1

3.2

1

2

11)( 5432

1 xxxxxy


Example

...55

44

33

2210 xcxcxcxcxccy

From case 2:

...6.5.44.33.2

5141311 x

cx

cx

cxcy

...]6.5.4

1

4.3

1

3.2

1[ 543

1 xxxxcy

...6.5.4

1

4.3

1

3.2

1)( 543

2 xxxxxy


End chapter 3


4 series solutions

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