Problem

Find the

$\displaystyle (t^2-1)x''+(t+1)x=0$

$\displaystyle x(0)=1$

$\displaystyle x'(0)=0 $

Using expanded notation for series I can get an answer that matches up with maple:

$\displaystyle 1+(1/2)x^2+(1/6)x^3+(1/8)x^4+(1/12)x^5$

What I can't seem to do is find this with a closed form notation for series.

My attempt

$\displaystyle x=\sum_{n=0}^\infty a_n t^n$

$\displaystyle x'=\sum_{n=0}^\infty na_n t^{n-1}$

$\displaystyle x''=\sum_{n=0}^\infty n(n-1) a_n t^{n-2}$

First I substitute...

$\displaystyle \sum_{n=0}^\infty n(n-1) a_n t^{n} - \sum_{n=0}^\infty n(n-1) a_n t^{n-2}+\sum_{n=0}^\infty a_n t^{n+1}+\sum_{n=0}^\infty a_n t^n$

Then I try to get everything to be to power of n...

$\displaystyle \sum_{n=0}^\infty n(n-1) a_n t^{n} - \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} t^{n}+\sum_{n=1}^\infty a_{n-1} t^{n}+\sum_{n=0}^\infty a_n t^n$

Then I try to get everything to the same index by taking out n=0 elements where necessary...

$\displaystyle \sum_{n=1}^\infty n(n-1) a_n t^{n} - (2a_2) - \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} t^{n}+\sum_{n=1}^\infty a_{n-1} t^{n}+(a_0)+ \sum_{n=1}^\infty a_n t^n$

Using initial conditions:

$\displaystyle a_0=1 a_1=0$

The recurrence relationship I get is:

$\displaystyle a_{n+2}=\frac{n(n-1)a_n-a_{n-1}+a_n}{(n+2)(n+1)}$

But this is wrong because if I plug in n=2, I get

$\displaystyle a_{4}=\frac{3}{32}$

Where did I go wrong?