Hi. I need some help with this problem. I have to solve

$\displaystyle x(1-x)y''+2(1-2x)y'-2y=0$ (1)

Using Frobenius method around zero. So proposing $\displaystyle y=\Sigma_{n=0}^\infty a_n x^{n+\alpha}$, differentiating and replacing in (1):

$\displaystyle x(1-x)y''+2(1-2x)y'-2y=\Sigma_{n=0}^\infty a_n(n+\alpha)(n+\alpha-1)x^{n+\alpha-1}- \Sigma_{n=0}^\infty a_n(n+\alpha)(n+\alpha-1)x^{m+\alpha} + 2 \Sigma_{n=0}^\infty a_n(n+\alpha)x^{n+\alpha-1} -4 \Sigma_{n=0}^\infty a_n(n+\alpha)x^{n+\alpha}-2 \Sigma_{n=0}^\infty a_n n^{n+\alpha}=0$

Regrouping and working a little bit this becomes:

$\displaystyle a_0\alpha(\alpha+1)+\Sigma_{n=1}^\infty a_n(n+\alpha)(n+\alpha+1)x^{n+\alpha-1}- \Sigma_{n=0}^\infty a_n [(n+\alpha)^2+2]x^{n+\alpha}$

I think this is fine, because the indicial equation I get gives the same roots from the other method: $\displaystyle \alpha(\alpha-1)+p_0\alpha+q_0=0$

Then $\displaystyle \alpha_1=0,\alpha_2=-1$

So I have:

$\displaystyle a_0\alpha(\alpha+1)+\Sigma_{n=0}^\infty a_{n+1}(n+\alpha+1)(n+\alpha+2)x^{n+\alpha}- \Sigma_{n=0}^\infty a_n [(n+\alpha)^2+2]x^{n+\alpha}$

And then $\displaystyle a_{n+1}(n+\alpha+1)(n+\alpha+2)-a_n[(n+\alpha)^2+2]=0$

So for $\displaystyle \alpha_1=0$

$\displaystyle a_{n+1}=\frac{a_n[n^2+2]}{(n+1)(n+2)}$

The thing is I've tried some iterations but I can't get a expression for $\displaystyle a_k$ in terms of $\displaystyle a_0$. How should I do this?

Bye there, thanks for your help.-