Q:

Running into trouble with this differential equation

We're having trouble with this differential equation:
$xy'' + x^2y + y = 0$
We figured it is regular singular because there are no singular points. We assumed a frobenius solution:
$y = \displaystyle \sum_{n=0}^\infty c_n x^{n+r}$
and got:
$\displaystyle \sum_{n=0}^\infty c_n(n+r)(n+r-1)x^{n+r-1} + \sum_{n=0}^\infty c_n(n+r)x^{n+r+1} + \sum_{n=0}^\infty c_n x^{n+r} $
from here we subbed k values to match the indices and got:
$\displaystyle \sum_{k=-1}^\infty c_{k+1}(k+r)(k+r+1)x^{k+r} + \sum_{k+1}^\infty c_{k-1}(k+r-1)x^{k+r} + \sum_{k=0}^\infty c_k x^{k+r} = 0$
From here we're a little fuzzy on how to find the r values. 

A:

You have it right.  Now just equate powers of $x$.
$x^{r-1}$: $r(r-1) c_0 = 0 \implies r=1$ or $r=0$.  Assume for this solution that $r=1$.
$x^1$: $2 c_1+c_0 = 0$.  This depends on $y(0)$.
$x^n$: 
$$(n+1)(n+2) c_{n+1} + c_n + n c_{n-1} = 0$$
$c_0$ and $c_1$ from above.

