05.2 Power Series

04/19/23Power Series Method Chapter 5 1

04/19/23Power Series Method Chapter 5 2

0 0

0

2

0 1 0 2 0 00

Differntial equation ,

Initial condition

has solution in form of a prowers seris in Powers of -

that is valid in some interval ab

n

nn

dyf x y

dxy x y

x x

y c c x x c x x c x x

0

0 1 2

out the point .

Here we are required to find the coefficients c ,c ,c ....

x x

04/19/23Power Series Method Chapter 5 3

Note.1. Power series about x = 0 is

0

33

2210

n

nn xcxcxcxccy

Note.2. Derivatives of Power series

2

224322

2

1

134

2321

13.4.2.3.1.2

432

n

nn

n

nn

xcnnxcxccdx

yd

xncxcxcxccdx

dy

04/19/23Power Series Method Chapter 5 4

Note.3. Important Power series (Maclaurin series)

n 0

2 1

0

2

0

1

0

0!

1 02 1!

1 02

1

-1

n

nn

n

x

n o

n

n

nn

n

xx

x

xx

n

xx

n!

x x

x

n

1 11

x

2 3 4

3 5 7

2 4

1 2 3

2 3 4

1 2! 3! 4!

sin 3! 5! 7 !

cos 1 2! 4!

1 1

ln 1 2 3 4

x x x xe x

x x xx x

x xx

x x x x

x x xx x

04/19/23Power Series Method Chapter 5 5

0)( 012 yxayxayxa

0 2 0 2 1

0 2 0 2 1

is ordinary point if a 0, are analytic

is singular point if a 0, is not analytic

x x a x and a x

x x a x or a x

1.Ordinary and Singular points For the linear-second order differential equation

Example: Find singular and ordinary points, if any, in the differential equation

2

2

x -1 2 6 0.

1 0, 1 and 1 are singular points

All other finite values of are ordinary points.

y y y

x x x

x

Note: Solution of the differential equation can be obtained in term of power series about the ordinary point.

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Example:Solve the differential equation

02 xyy .

Solution: A series solution of the differential equation about x = 0

as differential equation is analytic about x = 0.

by Power series method

2 30 1 2 3

0

2 11 2 3

1

2 3

nn

n

nn

n

y c c x c x c x c x

y c c x c x nc x

[1/5]

04/19/23Power Series Method Chapter 5 7

1 1

1 0

1 10 1

substituting these in the differential equation

2 0

1 1

1

( 1) 2 0

1

n nn n

n n

m mm m

m m

nc x c

n m n

x

m c x c

m

n m n

x

m

Powers of x are same

[2/5]

04/19/23Power Series Method Chapter 5 8

1 1 11 1

1 1 11

1

To make summation same, take out m 0 from first term.

c 1 2 0,

now summation is also same

c [ 1 2 ] 0

0 and

m

m mm m

m m

mm m

m

m c x c x

m c c x

c

1 1

m 1 1

1 2 0, 1, 2,3,

2 or c

1 1, 2,3,.... is called Recurrence Relation

m m

m

c c m

cm

m

[3/5]

04/19/23Power Series Method Chapter 5 9

2 0 0

3 1

2

4 2 0 0

5 3

3

6 4 0

2 1 c

22

2 c 032 1 1

3 c 1 .4 2 2!2

4 c 052 1 1

5 c 1 .6 3 2!

m c c

m c

m c c c

m c

m c c

0

1

3!c

m 1 1

2 c 1, 2,3,....

1 mc mm

[4/5]

04/19/23Power Series Method Chapter 5 10

2 3 4 50 1 2 3 4 5

2 4 60 0 0 0

2 3 42 2 2

20

4 62

0

1 1

Substituting these values in e

2! 3!

12! 3! 4!

quatio

n 1.

12! 3!

y c c x c x c x c x c x

y c c x c x c x

x x xc

x xy

x

c x

2

2

00

0

1

!

n n

x

x

xc

n

y c e

[5/5]

04/19/23Power Series Method Chapter 5 11

04/19/23Power Series Method Chapter 5 12

yy and

01 0

2

2

2

n

nn

n

nn xcxcnn

Substituting for in the differential equation, we obtained

m 2 mm 2 m

m 0 m 0

2 mm 2 m

m 0

Let 2

2

m 2 m 1 c x c x 0

m 2 m 1 c c x 0

m n

n m

22

2

m 2

Then, the Recurrence relation is

2 1 0 for m 0,1,2,3,

c , 0,1, 2,3, 2 1

m m

m

m m c c

c mm m

[2/4]

04/19/23Power Series Method Chapter 5 13

2

m 2

2 2

2 0 0

2 2

3 1 1

2 4 4

4 2 0 0

2 4

5 3

c , 0,1,2,3, 2 1

- m 0 c c c

2.1 2!

- m 1 c c c

3.2 3!

m 2 c c c c4.3 4.3.2.1 4!

m 3 c c 5.4 5.4.3.2.1

mc mm m

4

5 1c c5!

2 3 4 50 1 2 3 4 5

2 2 4 42 3 4 5

0 1 0 1 0 1

....

- - ...2! 3! 4! 5!

y c c x c x c x c x c x

c c x c x c x c x c x

[3/4]

04/19/23Power Series Method Chapter 5 14

2 4 62 4 5

0

2 4 63 5 7

1

10 0 2

12! 4! 6!

c3! 5! 7!

y cos sin cos sin

y c x x x

x x x x

cc x x c x c x

where 1

2

cc

.

[4/4]

04/19/23Power Series Method Chapter 5 15

Example:3. Solve the differential equation

2 0y x y Solution:

Differential equation is analytic at x = 0 ,we can consider a solution in the form of power series.

2 30 1 2 3

0

2 11 2 3

1

2 22 3

2

1

2 3

2 6 ( 1)

nn

n

nn

n

nn

n

y c c x c x c x c x

y c c x c x nc x

y c c x n n c x

[1/4]

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Substituting in the differential equation, we obtained 01

0

2

2

2

x

nn

n

nn xcxxcnn

2 2

2 0

1 0n nn n

n x

n n c x x c x

2 2

2 0

0 1 n nn n

n n

n n c x c x

Let n-2= m n+2 = m

m = n+2 m = n-2

Let n-2= m n+2 = m

n = m+2 n = m-2

2 20 2

2 3 2 22

2 2

3 3

2 2

2 1 0

2 6 2 1 0

2 0, c 0.

6 0, c 0.

2 1 0 for m 2,3,4,

is the Recurrence relation.

m mm m

m m

mm m

m

m m

m m c x c x

c c x m m c c x

c

c

m m c c

m 2 2

1 c

2 1 mcm m

[2/4]

04/19/23Power Series Method Chapter 5 17

0 1

4 0

5 1

6 2

7 3

8 4 0

considering c and c as arbitray constants

1m 2 c -

4.31

m 3 c -5.41

m 4 c - 06.51

m 5 c - 07.61 1

m 6 c -8.7 8.7.4.3

m 7

c

c

c

c

c c

9 5 1

1 1 c -

9.8 9.8.5.4 and so on

c c

[3/4]

04/19/23Power Series Method Chapter 5 18

substituting values of C’s in series

2 3 40 1 2 3 4

4 5 8 9

0 1 0 1 0 1

4 8 5 9

0 1

4 8

1

5 9

2

y c

12 20 672 1440

112 672 20 1440

112 672

20 1440

c x c x c x c x

x x x xy c c x c c c c

x x x xy c c x

x xy

x xy x

[4/4]

04/19/23Power Series Method Chapter 5 19

04/19/23Power Series Method Chapter 5 20

04/19/23Power Series Method Chapter 5 21

04/19/23Power Series Method Chapter 5 22

04/19/23Power Series Method Chapter 5 23

04/19/23Power Series Method Chapter 5 24

04/19/23Power Series Method Chapter 5 25

04/19/23Power Series Method Chapter 5 26

04/19/23Power Series Method Chapter 5 27

04/19/23Power Series Method Chapter 5 28

04/19/23

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04/19/23Power Series Method Chapter 5 30