Power Series Solutions
Recall, a power series in powers of is an infinite sum on the form
for example:
The set of all values of for which a power series converges is called the interval of convergence of the series.
,
0
0
n
n
n
x x
a
) ( x x
0x
)!. 1 2
) ( 1 ( ,
1 ) 1 3
, ( 3
) 2 (
0
1 2
1
1
0
1
n
n n
n
n
n
n
n
n x n
x x
Suppose that the power series
has a positive radius of convergence, that is there is a positive
number such that the power series converges for all
.
If for all , then
0
0 n
n
n
x x
a
0
1
0
n
n n
x x a
) ,
(
0
0
I x x x
I x
...
, 1 , 0
0
for all n
a
nTwo power series and can be
combined by addition or subtraction provided that:
(i) they start with the same power of
(ii) their summation indices start at the same value.
Definition:
A function is said to be analytic at a point if it can be represented by a power series on the form , with a positive radius of convergence.
For example, and are analytic functions everywhere, while is analytic except at .
i n
n
n x x
a 0
e
xf
0
0 n
n
n x x
a x0
j m
m
n x x
b 0
x sin
x
1 x 0
x
Remark. Every polynomial is analytic everywhere, and every rational function is analytic except at the zeros of its
denominator .
Consider the second order differential equation
Dividing both sides by , Eq.(1) can be written as
A point is called an ordinary point of equation (1) if both are analytic at .
) 1 ( .
0 )
( )
( )
( 2 1 0
2
2 a x y
dx x dy dx a
y x d
a
). (
) ) (
( ) ,
( ) ) (
( where
, 0 )
( )
(
2 0 2
1 2
2
x a
x x a
x q a
x x a
p
y x dx q
x dy dx p
y d
x0
) ( )
(x and q x p
x0
)
2(x a
A point which is not an ordinary point of the differential equation is called a singular point of the equation.
The point is an ordinary point of the DE
Because both functions are analytic at
since ,
and these series have the interval of convergence , while is a singular point of the DE
x x
q and e
x
p( ) x ( ) sin
0 0 x
. 0 )
(sin )
2 (
2 x y
dx e dy
dx y
d x
x0
0 !
n
n x
n
e x
0
1 2
)!
1 2
) ( 1 ( sin
n
n n
n x x
) ,
(
0 0 x
. 0 )
(ln 2
2
2 x y
dx x dy dx
y d
A singular point is called regular singular point of Eq.(1) if both and are analytic at . A
singular point which is not regular is said to be irregular singular point.
Example
Determine the ordinary points, the regular singular points and irregular singular points of the DE:
Let us put the equation on the form
x0
x ) ( )
(x x0 p x (x x0)2q(x) x0
0 )
1 (
' ) 1 2
( '' )
( x
4 x
2y x y x
2x y
where y
x q y
x p
y ' ' ( ) ' ( ) 0 ,
Thus all real numbers except 0,1,-1 are ordinary points and 0,-1,1 are singular points.
Now,
. 1, 1 )
1 )(
1 (
) 1 (
) 1 ) (
(
) , 1 )(
1 (
1 2
1 ) 2
(
2 2 2
4 2
2 2
4
factors common
canceling after
x x
x x
x x x
x x x x
q
x and x
x
x x
x x x
p
1 . ) ) (
( ) (
) , 1 )(
1 (
1 ) 2
( ) ( ) (
2 2 0
0
0 2 0
x x x x
q x
x
x and x
x x x x x
p x x
At we have
The first function is discontinuous at , therefore this is irregular singular point.
At we have
Both functions are analytic at , thus it is regular singular point.
At we have
Both functions are analytic at , thus it is regular singular point.
0 0 x
1 . )
( ) (
) , 1 )(
1 (
1 ) 2
( ) (
2 2
0
0
x
x x q x
x x and
x x x x
p x x
0 1 x
. 1 )
( ) (
) , 1 (
1 ) 2
( )
( 0 2 0 2
and x x q x x
x x x x
p x x
1 x
0 1 x
1 . ) 1 ) (
( ) (
) , 1 (
1 ) 2
( ) (
2 2
2 0
0
x
x x q x
x x and
x x x
p x x
1
x
0 x
Theorem
:If is an ordinary point of the DE
then there are two linearly independent power series solutions of this equation on the form
with an interval of convergence centered at and has a positive radius of convergence.
Suppose that are analytic at .
, ) ( )
( )
2 (
2
x f y x dx q
x dy dx p
y
d
x0
x0
,
,0
0 2
0
0
1
n
n n
n
n
n x x y b x x
a y
Find the general solution in power series form for the differential equation
about the ordinary point
Solution. Assume that the solution is given by
using the values of and in equation (1) we get
Example
.
0 0 x
) 1 ( ,
0 2
' xy y
, '
1
1
0
n
n n n
n
n
x y n c x
c y
y y'
0
1 1
1 2 0.
n
n n n
n
n x c x
c n
To make the powers of similar in both series, put
in the firs series and in the second one , to get
Since this true for all values of we get
k
n1 n1 k
. 0 ]
2 )
1 [(
, 0 2
) 1 (
, 0 2
) 1 (
1
1 1
1
1
1 1
1 1
1
1 0
1
k
k k
k
k
k k
k
k k
k
k k
k
k k
x c
c k
c
x c
x c
k c
x c
x c
k
x
) 2 ( ,...
3 , 2 , 1 ,
, 0
1 1 2 1
1
c for all k
c
and c
k k k
x
From the recurrence relation (2) we obtain
Now, from the assumption we have
...
, 5
, 0 4
, 3
, 0 2
, 1
6 0 1 3 4
1 6
5 3 2 5
2 0 1 2 2
1 4
3 1 2 3
0 2
c c
c k
c c
k
c c
c k
c c
k
c c
k
6
...
6 5
5 4
4 3
3 2
2 1
0 0
x c x
c x
c x
c x
c x
c c
x c y
n
n n
Or
Example
Find the general solution of the differential equation
about the ordinary point
.
...]
1 [
...
2 0
0 !
3
! 2
! 0 1
6 6 0
4 1 2 0
2 1 0 0
2
6 4
2
x k
k x
x x
x
e c
c
x c x
c x
c c
y
k
) 1 ( ,
0 '
' ' ) 1
( x y y
.
0
0
x
Solution
. First let us write equation (1) asAssume
using the values of and in equation (2) we get
To make the powers of similar in all series, put
in the firs series, in the second one , and in the last one, to get
, )
1 (
' ' , '
2
2 1
1
0
n
n n n
n n n
n
n x y nc x y n n c x
c y
) 2 ( ,
0 '
' ' '
' y y xy
'
, y
y
y''
1
1 2
2 2
1 ( 1) 0.
) 1 (
n
n n n
n n n
n
n x n n c x nc x
c n
n
x
k
n 1 n2 k
k n1
From the recurrence relation (3) we obtain
, 0 ]
) 2 )(
1 (
) 1 )(
1 [(
2
, 0 )
1 (
) 2 )(
1 (
2 )
1 (
, 0 )
1 (
) 2 )(
1 (
) 1 (
2 1
1 2
1
0
1 1
1
2 2
1
1
0
1 0
2 1
1
k k
k
k
k
k k
k
k k
k
k k
k
k k
k
k k
k
k k
x c
k k
c k
k c
c
x c k
c x
c k
k c
x c k
k
x c k
x c
k k
x c k
k
) 3 ( ,...
3 , 2 , 1 ,
, 0
2
2 1 1 2
2 1 1 2
2 1
c for all k
c
and c
c c
c
k k k k
, 1 c
3 32c
2 31c
1k
Now, from our assumption we have
...
, 3
, 2
5 1 1 5
4 1 1 4 3
3 4
c c
k
c c
c k
.
...]
[
...
...
1 1 0
5 5 4 1
4 3 1
3 2 1
2 1 1
0
5 5 1
4 1 4 1
3 1 3 1
2 1 2 1
1 1
0
5 5 4
4 3
3 2
2 1
0 0
n
k x n
n n
c
kc
x x
x x
x c c
x c x
c x
c x
c x
c c
x c x
c x
c x
c x
c c
x
c
y
Example
Find the general solution in power series form for the differential equation
about the ordinary point
Solution. Assume that the solution is given by
using the values of and in equation (1) we get
Put in the firs series and in the second one , to get
) 1 ( ,
3 2
'
' xy x
y
.
0 0 x
, )
1 (
' ' , '
2
2 1
1
0
n
n n n
n n n
n
n x y nc x y n n c x
c y
y y ''
0
1 2
2 2 3 .
) 1 (
n
n n n
n
n x c x x
c n
n k
n2 n1 k
Or
Which implies that
From the recurrence relation (2) we obtain
1
1 0
2 2 3
) 2 (
) 1 (
k
k k
n
k
k x c x x
c k
k
x x
c c
k k
x c c
c k k
k
k ] 2 3
) 2 (
) 1 [(
) 6
( 1
2
2 0
3
2
) 2 ( .
2 ,
0 )
2 (
) 1 (
, 3
6
, 2
1 2
6 0 1 3
0 3
2
c for k
c k
k
and c
c c
c c
k k
Which implies
and so on. Now, from the assumption we have
2
1, 2 , 3 , 4 ,...
) 1 (
1
2
c k
c
k k k k, 4
, 3
, 2
180 0 1 30 3
1 6
10 1 20 2
1 5
12 1 1 4
c c
c k
c c
k
c c
k
...
2 ...)
( ...)
1 (
...
2
...
5 10 2 1
4 12
1 1
6 180 3 1
6 1 0
6 180 0
5 1 10 4 1
12 1 3 1
6 0 2 1
1 0
6 6 5
5 4
4 3
3 2
2 1
0 0
x x
x x
c x
x c
x c x
x c x
c x
x c c
x c x
c x
c x
c x
c x
c c
x c y
n
n n
Example
Find the general solution in power series form for the differential equation
about the ordinary point
Solution. Assume that the solution is given by
Now, let us write equation (1) as
using the values of and in (2) we get
) 1 ( ,
0 '
' xy y
.
0 2 x
. ) 2 (
) 1 (
'' ,
) 2 (
' )
2 (
2
2 1
1
0
n
n n
n
n n
n
n
n x y nc x y n n c x
c y
) 2 ( ,
0 2
) 2 (
'
' x y y y
y y ''
Putting in the firs series, in the second one and in the last one we get
or
It follows that
k
n2 n1 k
0 0
1 2
2 ( 2) 2 ( 2) 0.
) 2 (
) 1 (
n n
n n
n n
n
n
n x c x c x
c n
n
k n
1 0
1 0
2 ( 2) ( 2) 2 ( 2) 0,
) 2 (
) 1 (
k k
k k
k k
n
k
k x c x c x
c k
k
1 1
0 1
1
2
2 ( 1)( 2) ( 2) ( 2) 2 2 ( 2) 0,
2
k k
k k
k k
n
k
k x c x c c x
c k
k c
0 )
2 ](
2 )
2 (
) 1 [(
) (
2
1
1 2
0
2
n
k k
k
k
c c x
c k
k c
c
which implies that
….
Using the values of these coefficients in the assumption we have
) 3 ( ,...
3 , 2 , 1 ,
,
) 2 )(
1 (
2 2
0 2
1
for k
c
and c
c
k k
c c k
k k
, 2
, 1
12 2 12
2 4
6 2 3
1 0 1
2 0 1
c c c
c c c
c k
c k
...
) 2 (
) 2 (
) 2 (
) 2
(
0 1 2 2 3 30
x c x
c x
c c
x c
y
n
n n
Remark
We can use the following change of variables to transform the ordinary point to the origin and proceed as before:
Thus the differential equation becomes
...].
) 2 (
) 2 [(
....]
) 2 (
) 2 (
1 [
...
) 2 )(
2 ( )
2 (
) 2 (
3 1 1
3 6 0
2 1 0
3 0
6 1 2 1
0 1
0
x x
c x
c x
c
x c
c x
c x
c c
y
. ,
2
2 2 2
2 2
2
dt y d dx
dt dt
y d dx
y d
dt dy dx
dt dt dy dx
x
dyt
. 0 )
2
2
(
2
t y
dt y d
0 2
x t0 0
Example
Find the solution of the initial value problem
using the power series method about the ordinary point Solution. Assume that the solution is given by
using the values of and in (1) we obtain:
Put in the firs series, in the second, and in the last one , to get
) 1 ( ,
2 )
0 ( ' ,
1 )
0 ( ,
0 '
'
' x
2y y y y y
.
0 0 x
. )
1 (
'' ,
'
2
2 1
1
0
n
n n n
n n n
n
n x y nc x y n n c x
c y
' , y
y y ''
. 0 )
1 (
0 1
1 2
2
n
n n n
n n n
n
n
x n c x c x
c n
n
k
n 2 n1 k
k n
Hence,
. ,....
3 , 2 ,
, 2 ,
0 )
1 (
) 2 )(
1 (
, 6
, 0
2
, 0 ]
) 1 (
) 2 )(
1 [(
) 6
( ) 2
(
, 0 )
1 (
) 2 )(
1 (
6 2
, 0 )
1 (
) 2 )(
1 (
) 2 )(
1 (
) 1 ( 2
1 2
6 1 1 3 1
3
2 0 1 2 0
2
1 1
2 1
3 0
2
2 1
0 2
1 2
2 3
2
0 2
1 0
2
1
k c
k for c
c k
c k
k
and c
c c
c
c c
c c
x c c
k c
k k
x c c
c c
x c x
c c
x c k
x c
k k
x c c
x c x
c k
x c
k k
k k
c k c k
k k
k
k k k
k
k
k
k k k
k k
k
k k
k
k k k
k k
k
k k
k k
], [
] 2 [
3
], [
] [
2
0 6 1
1 12
1 2
20 3 1 5
1 2 0
1 12
1 1
12 2 1 4
c c
c c
c K
c c
c c
c k
Now, from the assumption we have
...).
( 2 ...)
1 (
, 2 2
) 0 ( '
, 1 1
) 0 (
...).
( ...)
1 (
...
) (
...
4 12 3 1
6 4 1
24 2 1
2 1
2 0
4 12 3 1
6 1 1
4 24 2 1
2 1 0
4 1
2 0 1 12 3 1
6 1 2 1
2 0 1 1
0
6 6 5
5 4
4 3
3 2
2 1
0 0
x x
x x
x y
hence
c y
and
c y
Since
x x
x c x
x c
x c c
x c x
c x
c c
x c x
c x
c x
c x
c x
c c
x c y
n
n n
