1. CHAPTER 6
(Power Series Method)
Review of Power Series
 


2
210
0
)()()( axcaxccaxc
n
n
n is 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
xcxccxc
n
n
n is a power series in x with centre at 0 (Maclaurin
series).
 

N
n
n
nN axcxS
0
)()( is called nth partial sum.
 If 

LxSN
N
)(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.

2. HOW TO FIND RADIUS OF CONVERGENCE R?
Ratio Test:
Find L
c
c
axc
axc
n
n
n
n
n
n
n
n


 




1
1
1
limlim )(
)(
i. The radius of convergence is
L
R
1

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
n
n
x
n
(ii) 



1
2
)3(
)!(
)!2(
n
n
x
n
n
Sol: (i)
1
2
,
2 1
1




n
c
n
c
n
n
n
n
2
11
22
1
2
2)1(
2
limlim
1



 

 L
RL
n
n
n
n
n
n
n
n
and region of convergence is
2
1
x
(ii) 212
))!1((
)22(
,
)!(
)!2(


 
n
n
c
n
n
c nn
4
11
4
)1(
)12)(22(
)!2())!1((
)!()!22(
22
2
limlim 





 L
R
n
nn
nn
nn
nn
and region of convergence is
4
1
3 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:

3. 

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: 11
1
1
0
2





Rxxx
x 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 0ix
y y ye x  
2. '' ' ln 0x
y 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 .

4. 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
k
kk
c
c
cckk
k
k
kk
Recurrence formula
iv. Let thencandc 01 10  using Recurrence formula
0
!4
1
0,
!2
1
54
32




cc
cc
 4
2
1
!4
1
!2
1 x
x
y
Next y2 :

5. Let thencandc 10 10  using Recurrence formula
11 c
6.5.4
1
0
3.2
1
,0
54
32



cc
cc
 5
3
2
6.5.4
1
3.2
x
x
xy
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(22
2 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(1.2
1 1 1
022
k k k
k
k
k
k
k
k xcckxcxkkcc
 


 0)22)1)(2(({22
1
202
k
k
kkk xckckkccc
0202 022 cccc  (1)
,...3,2,1
2
2
0)22()1)(2(
2
2







kc
k
c
ckckk
kk
kk
(2)
Relation (2) is known as Recurrence Formula.

6. ii. Let thencandc 01 10  using Recurrence formula
0
7
2
6
1
6
1
0
5
2
2
1
2
1
0
2
3
,1
5746
3524
1302



cccc
cccc
cccc
...
6
1
2
1
1 642
1  xxxy
Now, let thencandc 10 10  using Recurrence formula, we get
2211
53
2
35
4
3
2
yCyCyxxxy 
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
2
2
  








n n n
n
n
n
n
n
n xcnxc
x
xxnnc
 
    







0......432
...
!4!3!2
1...201262
3
3
2
210
3
4
2
321
432
3
5
2
432
xcxcxccxcxcxcc
xxx
xxcxcxcc
 
  








tcoefficiencomparingxcxcxcc
xcccxccc
xcxcxcc
0...
...)32
2
1
()2(
...201262
3
3
2
210
3
321211
3
5
2
432
)(0123
2
1
)(026
)(02
4321
1213
012
iiicccc
iicccc
iccc




7. Let 

 ,0,
6
1
,
2
1
01 43210 ccccandc
 32
1
6
1
2
1
1 xxy
Now let 

 ,
24
1
,
6
1
,
2
1
10 43210 ccccandc
 432
2
24
1
6
1
2
1
xxxxy
Hence, 2211 yCyCy 

8. 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.instander 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,

9. 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
n xay 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
rn
nxcy .
Or
Then xp(x) = 3
1 and x2
q(x) = - 3
1 x. Hence p0 = 3
1 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
1
21
n
n
n
r
n
n
n
r
xbxcxaxcy
Case 2: 2121 rrandrr  is a positive integer then solution will be












 



 0
12
0
11
21
ln
n
n
n
r
n
n
n
r
xbxxCyyandxaxcy
𝑦 = 𝑐1 𝑦1 + 𝑐2 𝑦2

10. Case 3: rrr  21 then




















1
12
0
0
1
ln
0,
n
n
n
r
n
n
n
r
xbxxyy
axaxy
𝑦 = 𝑐1 𝑦1 + 𝑐2 𝑦2
NOTE: If 1y is known, method of reduction of order may be used to get




2
1
12
y
e
yy
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
2
1
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







11.  
  )(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 k
k k(k)(k ) (k) a x (k )(k) (k ) a x         11 3 1 1 1 0
k k
a a a
a a
   
1
1
0 1 2
r n
n
y x a x x ( x x x ) if x
x
          

0 2 3
1
1
1 1
1
Now
?2 y
Use the reduction formula:

12. 



2
1
12
y
e
yy
pdx
x(x )y ( x )y y     1 3 1 0
 




 2
)1)(ln(
1
21
)1(
13
)( xxpdx
xxxx
x
xP
x
x
dx
xx
dx
xx
x
x
dx
x
e
xy
e
yy
xxpdx
















1
ln1
1
1
)1(
)1(
1
1
)1(
11
1
2
2
2
)1)(ln(
2
1
12
2
2211 ycycy 