$y''+\frac{y'}{x^2}-\frac{a^2}{x^2}y = 0$ by Frobenius, finding $s$ in $\sum a_n x^{n+s}$ I need to solve:
$$y''+\frac{y'}{x^2}-\frac{a^2}{x^2}y = 0, \ \ \ a\ge 0$$
by Frobenius method. So I did:
$$y = \sum_{n=0}^{\infty}a_nx^{n+s}$$
$$y' = \sum_{n=0}^{\infty}a_n(n+s)x^{n+s-1}$$
$$y'' = \sum_{n=0}^{\infty}a_n(n+s)(n+s-1)x^{n+s-2}$$
so 
$$y''+\frac{y'}{x^2}-\frac{a^2}{x^2}y  = \\  \sum_{n=0}^{\infty}a_n(n+s)(n+s-1)x^{n+s-2} + \sum_{n=0}^{\infty}a_n(n+s)x^{n+s-3} +  \sum_{n=0}^{\infty}(-a^2)a_nx^{n+s-2}$$
I couldn't find a way to make $\sum_{n=0}^{\infty}a_n(n+s)x^{n+s-3}$ as a series with coefficient $x^{n+s-2}$ so I had to change the other terms to have $x^{n+s-3}$. I eneded up with:
$$\sum_{n=1}^{\infty}a_{n-1}(n-1+s)(n+s-2)x^{n+s-3} + \\ a_0\cdot s \cdot x^{s-3} + \sum_{n=1}^{\infty}a_n(n+s)x^{n+s-3} + \\ -a^2\cdot a_0\cdot x^{s-2} + \sum_{n=1}^{\infty} a_{n-1}x^{n+s-3}$$
For $x^{s-2}$ we have: $x^{s-2}(a_0s-a^2a_0) = 0 \implies s = a^2$. But the term $x^{s-3}$ says that $a_0$ must be $0$ or $s$ must be $0$. I guess there's some problem here. What's happening?
Also, I ended up with the following relation:
$$x^{n+s-3}(a_{n-1}(n-1+s)(n+s-2) + a_n(n+s) + a_{n-1}) = 0$$
Do you think this relation is right, even though I had a problem determining the possible $s$ variables that can be used?

The coefficient of $y'$ is $1/x^2$, so $x=0$ is not a regular singular point.  The Frobenius method won't work here.
