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.


You write the first line as:

$$x^2y''(x)= \sum_{n=1}^\infty (n+1)nC_{n}x^{n}$$

and not modify the others lines as

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

You can now writhe the sum as:

$$\text{constante} + \sum_{n=1}^{+\infty} [...] x^n= 0$$


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