# Finding a solution in powers of $(x^2-1)y''+4xy'+2y=0$

I'm working on finding the general solution of $(x^2-1)y''+4xy'+2y=0$ in powers.

I assume the form: $$y(x)=\sum_{n=0}^\infty C_nx^n$$

My basic strategy is to first figure out each piece individually, shift the indexes if necessary, and then rewrite it into one series:

$$y(x)=\sum_{n=0}^\infty C_nx^n$$ $$y'(x)=\sum_{n=1}^\infty nC_nx^{n-1}$$ $$y''(x)=\sum_{n=2}^\infty n(n-1)C_nx^{n-2}$$

Working with these further: $$x^2y''(x)=\sum_{n=2}^\infty n(n-1)C_nx^{n} = \sum_{n=0}^\infty (n+2)(n+1)C_{n+1}x^{n+1}$$ $$-y''(x)=\sum_{n=0}^\infty -(n+2)(n+1)C_{n+2}x^n$$ $$4xy'(x)=\sum_{n=1}^\infty 4nC_nx^n$$ $$2y(x)=\sum_{n=0}^\infty 2C_nx^n$$

However, I'm not sure how to take these pieces and put them together in the form: $$\sum_{n=0}^\infty [.....]x^n$$

The reason I'm not sure how do do this is because if I shift all the indexes to zero, which is necessary to rewrite everything in one series, some pieces have an $x^n$ factor, while others have a $x^{n+1}$ factor, etc, so x remains as a term inside the series and I can't set it equal to zero to find the recurrence relation in only terms of n.

The ultimate goal is to find the general solution of the ODE.

$$x^2y''(x)= \sum_{n=1}^\infty (n+1)nC_{n}x^{n}$$
$$-y''(x)=\sum_{n=0}^\infty -(n+2)(n+1)C_{n+2}x^n$$ $$4xy'(x)=\sum_{n=1}^\infty 4nC_nx^n$$ $$2y(x)=\sum_{n=0}^\infty 2C_nx^n$$
$$\text{constante} + \sum_{n=1}^{+\infty} [...] x^n= 0$$