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Power Series Method Chapter 5
1
Differntial equation
Initial condition
dy
 f  x, y 
dx
y  x0   y0
has solution in form of a prowers seris in Powers of
y  c0  c1  x  x0   c2  x  x0  
2

  cn  x  x0 
 x - x0 
n
n 0
that is valid in some interval about the point x  x0 .
Here we are required to find the coefficients
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Power Series Method Chapter 5
c0 ,c1 ,c 2 ....
2
Note.1.
Power series about x = 0 is

y  c0  c1 x  c 2 x  c3 x     c n x
2
3
n
n 0
Note.2.
Derivatives of Power series

dy
2
3
n 1
 c1  2c 2 x  3c3 x  4c 4 x     nc n x
dx
n 1

d2y
2
n2



2
.
1
.
c

3
.
2
.
c
x

4
.
3
c
x



n
n

1
c
x

2
3
4
n
2
dx
n2
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Power Series Method Chapter 5
3
Note.3.
Important Power series (Maclaurin series)
2
3
4
x
x
x
e  1 x    
2! 3! 4!
x3 x5 x 7
sin x  x   

3! 5! 7 !
x2 x4
cos x  1    
2! 4!
x
1  x 
1
 1 x  x  x 
2
3
x 2 x3 x 4
ln 1  x   x    
2 3 4
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
xn

n 0 x !


n 0
2 n 1
x
1n
2n  1!
 

x2 x
   1
2n!
n o

  xn
x 0
x 0
x 0
x 1
n 0
x n 1
   -1
n 1
n 0

Power Series Method Chapter 5
n
 1  x1
4
1.Ordinary and Singular points
For the linear-second order differential equation
a2 x  y   a1 ( x) y   a0 x  y  0
x0 is ordinary point if
a 2  x0   0, a2  x  and a1  x 
x0 is singular point if
a 2  x0   0, a2  x  or a1  x 
Example:
are analytic
is not analytic
Find singular and ordinary points, if any, in the differential equation


x 2 -1 y  2 y  6 y  0.
x 2  1  0,  x  1 and x  1 are singular points
All other finite values of x are ordinary points.
Note: Solution of the differential equation can be obtained in term of
power series about the ordinary point.
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Power Series Method Chapter 5
5
Example:Solve the differential equation
[1/5]
y   2 xy  0
by Power series method
.
Solution:
Power Series Method Chapter 5
A series solution of the differential equation about x = 0
as differential equation is analytic about x = 0.
y  c0  c1 x  c2 x 2  c3 x 3 
2

y  c1  2c2 x  3c3 x 
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


n
c
x
 n
n 0

 nc
n 1
n
x
n 1
6
[2/5]
substituting these in the differential equation


n 1
n 0
n 1
n 1
nc
x

2
c
x
0
 n
 n
n 1  m
n  m 1
n 1  m
n  m 1


m 0
m 1
m
m
(
m

1)
c
x

2
c
x

 m 1  0
m 1
Powers of x are same
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Power Series Method Chapter 5
7
[3/5]
To make summation same, take out m  0 from first term.


m 1
m 1
c1    m  1 cm 1 x m  2 cm 1 x m  0,
now summation is also same

c1   [ m  1 cm 1  2cm 1 ] x m   0
m 1

c1  0
and
 m  1 cm1  2cm1  0,
m  1, 2,3,
2
or
c m 1  
cm 1
m 1
m  1, 2,3,.... is called Recurrence Relation
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Power Series Method Chapter 5
8
2
c m 1  
cm 1
m 1
m 1
m2
m3
m4
m5
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m  1, 2,3,....
[4/5]
2
c 2   c0  c0
2
2
c3   c1  0
3
2
1
2 1
c 4   c2   1 .c0  c0
4
2
2!
2
c5   c3  0
5
2
1
3 1 1
c 6   c4   1 . c0   c0
6
3 2!
3!
Power Series Method Chapter 5
9
Substituting these values in equation 1.
[5/5]
y  c0  c1 x  c2 x 2  c3 x 3  c4 x 4  c5 x 5 
1 4
1 6
y  c0  c0 x  c0 x  c0 x 
2!
3!
4
6


x
x
2
y  c0 1  x    
2! 3!


2
     
2
3
4
2
2
2

x
x
x
2
 c0 1  x 




2!
3!
4!


2n

1
x
 
x 0
n!
 c0 
y  c0e
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



n
 x2
Power Series Method Chapter 5
10
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Power Series Method Chapter 5
11
Substituting for
y and y 
in the differential equation, we obtained


n2
n 0
[2/4]
n2
2
n


n
n

1
c
x


c
x

 n 0
n
m  n2
n  m2
Let


m 0
m 0
m
2
m
m

2
m

1
c
x


c
x

 m 2

 m 0

2
m


m

2
m

1
c


c
x
0




m 2
m
m 0
Then, the Recurrence relation is
2
m

2
m

1
c




 m 2 cm  0
cm 2
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for m  0,1,2,3,
 2

cm , m  0,1, 2,3,
 m  2  m  1
Power Series Method Chapter 5
12
c m 2 
m0
m 1
m2
m3
 2
cm , m  0,1, 2, 3,
 m  2  m  1
 2
c2 
c0
2.1
 2
c3 
c1
3.2
 2
c4 
c2
4.3
 2
c5 
c3
5.4
[3/4]
- 2

c0
2!
- 2

c1
3!

4
4.3.2.1

c0 
4
5.4.3.2.1
4
4!
c5 
c0
4
5!
c1
y  c0  c1 x  c2 x 2  c3 x 3  c4 x 4  c5 x 5  ....
 c0  c1 x 
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2
2!
c0 x 2 -
2
3!
c1 x 3 -
4
4!
c0 x 4 
Power Series Method Chapter 5
4
5!
c1 x 5  ...
13
[4/4]
.
 2 2 4 4 6 5
y  c0 1 
x 
x 
x 
4!
6!
 2!



 2 3 4 5 6 7
 c1  x 
x 
x 
x 
3!
5!
7!

y  c0 cos  x 
where
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c2 
c1
c1




sin   x   c0 cos  x  c2 sin  x

Power Series Method Chapter 5
14
Example:3.
Solve the differential equation
[1/4]
y  x y  0
2
Solution: Differential equation is analytic at
x=0,
we can consider a solution in the form of power series.
y  c0  c1 x  c2 x  c3 x 
2
y  c1  2c2 x  3c3 x 2 
y  2c2  6c3 x 
2
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3

  cn x n  1
n 0

  ncn x n 1
n 1

  n(n  1)cn x n  2
n2
Power Series Method Chapter 5
15


n
Letnn  1c x n  2  x 2 n-2=
cn x m
0


n
= n+2
n2
x m
0
n+2 = m
m = n-2
Substituting in the differential equation, we obtained


n2
x 0
[2/4]
Power Series Method Chapter 5
n2
2
n
n
n

1
c
x

x
c
x



 n 0
n


n2
n2
n
n

1
c
x

c
x
  n
 n 0
n2
Let
n 0
n+2 = m
n = m-2
n-2= m
n = m+2

  m  2  m  1 c
m 2
m 0

x   cm  2 x m  0
m
m2

2c2  6c3 x    m  2  m  1 cm  2  cm  2 x m  0
m2
2c2  0,
c 2  0.
6c3  0, c3  0.
 m  2  m  1 cm  2  cm  2
0
for m  2,3,4,
is the Recurrence relation.
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cm2  
1
 m  2  m  1
cm  2
16
considering c 0 and c1 as arbitray constants
m2
m3
m4
m5
m6
m 7
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and so on
[3/4]
1
c4  c0
4.3
1
c5  c1
5.4
1
c6  c2  0
6.5
1
c7  c3  0
7.6
1
1
c8  c4 
c0
8.7
8.7.4.3
1
1
c9  c5 
c1
9.8
9.8.5.4
Power Series Method Chapter 5
17
substituting values of C’s in series
[4/4]
y  c0  c1 x  c2 x 2  c3 x 3  c4 x 4 
x4
x5
x8
x9
y  c0  c1 x  c0
 c1
 c0
 c1

12
20
672
1440
8
5
9
 x4


x
x
x
y  c0 1  
    c1  x  

20 1440
 12 672


4



8
x
x
 y1  1  

12 672
5
9
x
x
y2  x  

20 1440
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Power Series Method Chapter 5
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Power Series Method Chapter 5
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Power Series Method Chapter 5
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