Series Solution Near Ordinary Points for Second Order Differential Equations 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?
 A: If we set up $y = \displaystyle \sum^{\infty}_{n=1} a_n x^n$ like you did, plugging back to the original equation:
$$
(-2a_0+2a_2) + 6a_3 x +\sum^{\infty}_{n=2}\Big((n+2)(n+1)a_{n+2} - (n-1)(n+2)a_n\Big)x^n = 0
$$
simplify the expression for $a_{n+2}$ when $n\geq 2$:
$$
a_{n+2} = -\frac{n-1}{n+1} a_n \qquad (*)
$$
for the constant and $x$-term we have:
$$
a_0 = a_2, \text{ and } a_3 = 0
$$


*

*Now if the initial condition is $a_0 = 1, a_1 = 0$, then we have:
$$
a_2 = 1, \text{ and } a_{n+2} = (-1)^{n/2}\frac{n-1}{n+1}\cdot \frac{n-3}{n-1}\cdots\frac{1}{3} = \frac{(-1)^{n/2}}{n+1}
$$
and $n$ can be even numbers, let $n = 2k$ we have the solution is:
$$
y = 1+x^2 + \sum^{\infty}_{k=1}\frac{(-1)^k x^{2k+2}}{2k+1} = 1 + x\cdot \sum^{\infty}_{k=0}\frac{(-1)^k x^{2k+1}}{2k+1} = 1+ x\arctan x
$$

*Now if the initial condition is $a_0 = 0, a_1 = 1$, $a_2 = a_0 = 0$ implies all even powered $x$ coefficients are zero after $n=2$ because of the relation $(*)$, also by $(*)$ and $a_3 = 0$ we know that all odd powered $x$ coefficients are zero too after $n=3$, therefore the solution is just:
$$
y = x
$$
To sum up, the two linearly independent solutions are: 
$$y = x \;\text{ or }\; 1+ x\arctan(x)$$
