We want a solution in the form

with derivatives


Substituting
and its derivatives into the ODE,

gives

Shift the index on the second sum to have it start at
:

and take the first term out of the other two sums. Then we can consolidate the sums into one that starts at
:
![\displaystyle(2a_2+2a_0)+\sum_{n\ge1}\bigg[(n+2)(n+1)a_{n+2}+(2-2n)a_n\bigg]x^n=0](https://tex.z-dn.net/?f=%5Cdisplaystyle%282a_2%2B2a_0%29%2B%5Csum_%7Bn%5Cge1%7D%5Cbigg%5B%28n%2B2%29%28n%2B1%29a_%7Bn%2B2%7D%2B%282-2n%29a_n%5Cbigg%5Dx%5En%3D0)
and so the coefficients in the series solution are given by the recurrence,

or more simply, for
,

Note the dependency between every other coefficient. Consider the two cases,
- If
, where
is an integer, then





and so on, with the general pattern

- If
, then


and we would see that
for all
.
So we have
![y(x)=\displaystyle\sum_{k\ge0}\bigg[a_{2k}x^{2k}+a_{2k+1}x^{2k+1}\bigg]](https://tex.z-dn.net/?f=y%28x%29%3D%5Cdisplaystyle%5Csum_%7Bk%5Cge0%7D%5Cbigg%5Ba_%7B2k%7Dx%5E%7B2k%7D%2Ba_%7B2k%2B1%7Dx%5E%7B2k%2B1%7D%5Cbigg%5D)
so that one solution is

and the other is

I've attached a plot of the exact and series solutions below with
,
, and
to demonstrate that the series solution converges to the exact one.