CHAPTER 6

(Power Series Method)

Review of Power Series

2

210

0

)()()( axcaxccaxcn

n

nis a power series in power of (x-a),

where ,, 10 cc are real (complex ) constant and “a” is the centre of the

series.

2

210

0

xcxccxcn

n

nis a power series in x with centre at 0 (Maclaurin

series).

N

n

n

nN axcxS0

)()( is called nth partial sum.

If

LxSNN

)(lim series is convergent otherwise divergent.

Each Series is always convergent at its centre.

If

0

)(n

n

n axc converges for all points in Rax then

i. The region Rax is called region of convergence. R is called

radius of convergence.

ii. If R = 0 then series converges only at its centre. If R = then series

converges everywhere or x . In other situation R can be any finite

positive real number.

iii. Rax is called circle or interval of convergence.

x- Real or x-Complex

NOTE: If a series converges for Rax then it must be divergent

for Rax and for Rax , it may converge at every point or on

some points or may not converge at all.

HOW TO FIND RADIUS OF CONVERGENCE R?

Ratio Test:

Find Lc

c

axc

axc

n

n

nn

n

n

n

n

1

1

1

limlim)(

)(

i. The radius of convergence is L

R1

ii. If Rax || then the test is inclusive

iii. If Rax || so series diverges

iv. If Rax || so series converges

Example: Find radius and region of convergence for the series

(i)

1

2

n

nn

xn

(ii)

1

2)3(

)!(

)!2(

n

nxn

n

Sol: (i) 1

2,

2 1

1

n

cn

cn

n

n

n

2

1122

1

2

2)1(

2limlim

1

LRL

n

n

n

n

nn

n

n

and region of convergence is 2

1x

(ii) 212 ))!1((

)22(,

)!(

)!2(

n

nc

n

nc nn

4

114

)1(

)12)(22(

)!2())!1((

)!()!22(22

2

limlim

LR

n

nn

nn

nn

nn

and region of convergence is 4

13 x

Within region of convergence, series converges absolutely and uniformly.

It can be differentiated and integrated term by term and resulting series has

the same radius as well as region of convergence. In case R > 0 then the

corresponding power series always represents an analytic function.

A function f is analytic at a point a if it can be represented by a power

series in )( ax with a positive or infinite radius of convergence.

Term wise addition:

0

)(n

n

n axa

0

)(n

n

n axb =

0

))((n

n

nn axba

Term wise multiplication:

0

)(n

n

n axa

0

)(n

n

n axb = n

nnn

n

n axbabababa )}({ 02211

0

0

Series addition: in order o perform addition

ln

n

n axc )(

km

m

m axb )( given

series should satisfy nmkl and ,

If the series correspond to a function say 0

0( ) ( )n

n

n

f x c x x

Then the radius of convergence is the distance of the closest singular point

from the centre of the series.

Example: 111

1

0

2

Rxxxx n

n

POWER SERIES METHOD

(Basic method to solve LDE)

Let a differential equation be in standard form

) 0( ( )y yP x Q x y .

A point “x0” is called ordinary point if P(x) and Q(x) are analytic at “x0”,

otherwise the point “x0” will be called singular (Regular, Irregular).

Examples:

Standard form ) 0( ( )y yP x Q x y

1. '' ' s n 0ixy y ye x

2. '' ' ln 0xy y ye x

THEORM 6.1

Existence of power series solutions:

If 0xx is an ordinary point of the differential equation 0)()(" yxQyxPy ,

We can always find two linearly independent solutions in the form of power

series centered at x0 that is

0

0 )(n

n

n xxcy .

A series solution converges at least on some interval Rxx 0 , where R is

the distance from 0x to the closest singular point.

A solution of the form

0

0 )(n

n

n xxcy is said to be a solution about the

ordinary point x0

Example: Using Power series method solve 0 yy

Sol:

i. Assume

0n

n

n xcy is a solution then

1

1

n

n

nnxcy ,

2

2)1(n

n

n xnncy .

ii. 0)1(2 0

2

n n

n

n

n

n xcxnncyy

iii. Change k = n-2 in 1st term, k = n in 2nd term

0)1)(2(0 0

2

k k

k

k

k

k xcxkkc

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

0)2)(1(

2

2

kkk

cc

cckk

kk

kk

Recurrence formula

iv. Let thencandc 01 10 using Recurrence formula

0!4

1

0,!2

1

54

32

cc

cc

42

1!4

1

!21 x

xy

Next y2 :

Let thencandc 10 10 using Recurrence formula

11 c

6.5.4

10

3.2

1,0

54

32

cc

cc

53

26.5.4

1

3.2x

xxy

Hence, 2211 yCyCy that is basically

xCxCy sincos 21

Example: Using Power series method solve 022 yyxy

Sol:

i. Assume

0n

n

n xay is a solution then

1

1

n

n

nnxcy ,

2

2)1(n

n

n xnncy

022)1(222 1 0

12

n n n

n

n

n

n

n

n xcnxcxxnncyyxy

022)1(2 1 0

2

n n n

n

n

n

n

n

n xcnxcxnnc

Change k = n-2 in 1st term, k = n in 2nd and 3rd terms

022)1)(2(0 1 0

2

k k k

k

k

k

k

k

k xckxcxkkc

022)1)(2(1.21 1 1

022

k k k

k

k

k

k

k

k xcckxcxkkcc

0)22)1)(2(({221

202

k

k

kkk xckckkccc

0202 022 cccc (1)

,...3,2,12

2

0)22()1)(2(

2

2

kck

c

ckckk

kk

kk

(2)

Relation (2) is known as Recurrence Formula.

ii. Let thencandc 01 10 using Recurrence formula

07

2

6

1

6

1

05

2

2

1

2

1

02

3,1

5746

3524

1302

cccc

cccc

cccc

...6

1

2

11 642

1 xxxy

Now, let thencandc 10 10 using Recurrence formula, we get

2211

53

235

4

3

2yCyCyxxxy

Example: Find two series solution at x = 0 for 0 yyey x

i. Assume

0n

n

n xay is a solution then

1

1

n

n

nnxcy ,

2

2)1(n

n

n xnncy .

ii. 0)!2

1()1(2 1 0

22

n n n

n

n

n

n

n

n xcnxcx

xxnnc

0......432

...!4!3!2

1...201262

3

3

2

210

3

4

2

321

4323

5

2

432

xcxcxccxcxcxcc

xxxxxcxcxcc

tcoefficiencomparingxcxcxcc

xcccxccc

xcxcxcc

0...

...)322

1()2(

...201262

3

3

2

210

3

321211

3

5

2

432

)(01232

1

)(026

)(02

4321

1213

012

iiicccc

iicccc

iccc

Let

,0,6

1,

2

101 43210 ccccandc

32

16

1

2

11 xxy

Now let

,24

1,

6

1,

2

110 43210 ccccandc

432

224

1

6

1

2

1xxxxy

Hence, 2211 yCyCy

Section 6.2:

Solution about Singular Points

Given

) 0( ( )y yP x Q x y

The point x0 is called regular singular point if the functions p(x) = (x – x0) P(x)

and 2

0( ) ( )) (q x x x Q x are both analytic at x = x0.

Otherwise irregular singular point

Definition

A function, f(x), is called analytic at x= x0

if the Taylor series for f(x) about x= x0

has a positive radius of convergence (series exist), if x0 = 0 we have Maclaurin

series.

The point x = x0 is a regular singular point if (x – x0) has at most power 1 in

the denominator of P(x) and at most power 2 in the denominator of Q (x).

) 0( ( )y yP x Q x y

Example:

3 2

3 2

3 3 3

3

0

first put theEq.in stander form

0

0

0 poin

4 3

4 3

4

.

3

t

y y y

y y y

y y y

x x

x

x

x

x x x

x x

irregular

…………………………………..

.int3,0

0)3(

poregularx

yyxx

………………………………….

iii. irregularxregularx

yxyxyxx

5&0

0)25(4)5( 222

NOTE: The power series method is applicable at ordinary points but fails at

regular singular points. At regular singular points, Frobenius’ method,

which is basically an extension of power series method, helps to find the

solution.

It possible to obtain the indicial equation in advance of substituting

0n

rn

nxay into the differential equation, the general indicial equation is

0)1( 00 qrprr where 00 qandp are given by the definition of

p(x) and q(x).

Example

Find the indicial equation and indicial roots of the differential equation 3xy'' + y' - y = 0.

Sol:

Since x = 0 is a regular singular point of the differential equation

3xy’’ + y’ – y =0

We try a solution of the form

0n

rnnxcy .

Or

Then xp(x) = 31 and x2q(x) = -

31 x. Hence p0 =

31 and q0 = 0.

Hence the indicial equation is

r(r – 1) + 3

1 r + 0 = 0 r (3r-2) = 0

NOTE: If r1 and r2 are the roots of indicial equation then

Case 1: 2121 rrandrr is negative integer then solution will be

0

2

0

121

n

n

n

r

n

n

n

rxbxcxaxcy

Case 2: 2121 rrandrr is a positive integer then solution will be

0

12

0

1121 ln

n

n

n

r

n

n

n

rxbxxCyyandxaxcy

𝑦 = 𝑐1𝑦1 + 𝑐2𝑦2

Case 3: rrr 21 then

1

12

0

0

1

ln

0,

n

n

n

r

n

n

n

r

xbxxyy

axaxy

𝑦 = 𝑐1𝑦1 + 𝑐2𝑦2

NOTE: If 1y is known, method of reduction of order may be used to get

2

1

12y

eyy

pdx

Example :

0)13()1( yyxyxxSolve

Sol: Assume

00 n

rn

n

n

n

n

r xaxaxy is a solution.

n r

n

n r

n

y a (x r)x &

y a (n r)(n r )x

1

21

x(x )y ( x )y y 1 3 1 0

0)()(3

)1)(()1)((

1

1

rn

n

rn

n

rn

n

rn

n

rn

n

xaxarnxarn

xarnrnxarnrn

)(0)()1)((

1)(3)1)((

1 Axarnrnrn

xarnrnrn

rn

n

rn

n

)(0)()1)((

1)(3)1)((

1 Axarnrnrn

xarnrnrn

rn

n

rn

n

Indicial Equation: Take n = 0 and select coefficient (except na ) of the smallest

power of x we get 0)1( rrr the indicial equation. Roots of indicial

equation are r = 0 , 0 (double root)

Now we need Recurrence Formula:

In (A), take r = 0

)(0)()1)((

1)(3)1)((

1 Axarnrnrn

xarnrnrn

rn

n

rn

n

0)()1)((1)(3)1)(( 1 n

n

n

n xannnxannn

Taking n = k in 1st term and n -1 = k in 2nd term, we get

k kk k(k)(k ) (k) a x (k )(k) (k ) a x 11 3 1 1 1 0

k ka a a

a a

1

10 1 2

r n

ny x a x x ( x x x ) if x

x

0 2 3

1

11 1

1

Now ?2 y

Use the reduction formula:

2

1

12y

eyy

pdx

x(x )y ( x )y y 1 3 1 0

2)1)(ln(

1

21

)1(

13)( xxpdx

xxxx

xxP

x

xdx

xx

dxxx

x

xdx

x

e

xy

eyy

xxpdx

1

ln1

1

1

)1(

)1(

1

1

)1(

11

12

2

2

)1)(ln(

2

1

12

2

2211 ycycy