Given $(1+x^2)y''+2xy'-2y = 0$
The above equations obviously has analytic points everywhere except for $x=1$ and $-1$.
Find two linearly independent solutions $y_1$ and $y_2$ to the differential equation valid near $x_0=0$. To make life a little easier, choose the linearly independent equations:
$y_1$ $:$ $a_0$ = $y(x_0)$ = 1 and $a_1$ = $y'(x_0)$ = 0
$y_2$ $:$ $a_0$ = $y(x_0)$ = 0 and $a_1$ = $y'(x_0)$ = 1
After a mess of writing, I came up with the following:
$a_{n+2}$ = $[{-(n-1)(n)a_n - 2a_n(n-1)}]/[{(n+1)(n+2)}]$
I don't know if that monster is right, but that's where I need you help. Can somebody give this a sanity check, and then solve the rest?