<s>[INST] How to find solution to this ODE with powerseries I am working on solving this ODE 
$$ (1+x^2)y'' + xy'-y =0 $$ 
with  $1. y(0)=0,y'(0)=1 \\
     2. y(0)=1, y'(0)=1$
with $$ y= \sum_{n=0}^{ \infty} a_n x^n , y'= \sum_{n=1}^{ \infty}na_n x^{n-1}, y''=\sum_{n=2}^{ \infty} n(n-1)a_n x^{n-2}$$
plugged into the equation follows:
$(1+x^2) \sum_{n=2}^{ \infty} n(n-1)a_n x^{n-2}+x \sum_{n=1}^{ \infty}na_n x^{n-1}-\sum_{n=0}^{ \infty} a_n x^n =0 $
$ \leftrightarrow \sum_{n=2}^{ \infty} n(n-1)a_n x^{n}+\sum_{n=2}^{ \infty} n(n-1)a_n x^{n-2}+ \sum_{n=1}^{ \infty} na_n x^{n}-\sum_{n=0}^{ \infty} a_n x^n =0$
$ \leftrightarrow 2a_2-a_0 + \sum_{n=1}^{ \infty} (n(n-1)a_n + (n+2)(n+1)a_{n+2}+ na_n - a_n) x^n = 0 $
you get that 
$ a_{n+2}= \frac{a_n(1-n^2)}{(n+2)(n+1)} $
let now be $ a_0=1 $, $a_1=0 $ so that $a_2= \frac12, $( $2a_2 - a_0 = 0 \leftrightarrow a_2= \frac12 )$
for uneven $n$ such as $a_3, a_5,..$ it equals zero.
but for even $n$ i can not find any regularity:
$ n=2 , a_4 = \frac{a_2 (-3)}{12}= -\frac{1}{8}$
$ n=4 , a_6= \frac{a_4 (-15)}{30}= \frac{1}{16} $
$ n=6 , a_8= \frac{a_6 (-35)}{56}= - \frac{5}{128} $
i am totally stuck and do not know how to proceed :( 
I appreciate any help of you guys ! 
 [/INST] Note that by binomial theorem $n^2-1=(n-1)(n+1)$ so that the recursion cancels to
$$
a_{n+2}=-\frac{n-1}{n+2}a_n=\frac{(n-1)(n-3)}{(n+2)n}a_{n-2}=...
$$
For the even subsequence one can write
$$
a_{2(k+1)}=-\frac{2k(2k-1)}{4(k+1)k}a_{2k}=(-1)^{k}\frac{(2k)(2k-1)\cdots2\cdot1}{4^{k}(k+1)k^2(k-1)^2\cdots2^2\cdot1}a_2
\\=(-1)^k\frac{(2k)!}{2^{2k+1}(k+1)!k!}a_0
=\frac{(-1)^k}{2^{2k+1}(k+1)}\binom{2k}{k}a_0
$$
etc.
</s>