05.2 power series
DESCRIPTION
study guide for power seriesTRANSCRIPT
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.
04/19/23
Pow
er S
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s M
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hapt
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6
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]
04/19/23
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16
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
Pow
er S
erie
s M
etho
d C
hapt
er 5
29
04/19/23Power Series Method Chapter 5 30