Frobenius method for x(1-x)y''+2(1-2x)y'-2y=0

[Telemachus]

Homework Statement

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

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

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

$$x(1-x)y''+2(1-2x)y'-2y=$$

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

Regrouping and working a little bit this becomes:

$$a_0\alpha(\alpha+1)+\Sigma_{n=1}^\infty a_n(n+\alpha)(n+\alpha+1)x^{n+\alpha-1}- \Sigma_{n=0}^\infty a_n [(n+\alpha)^2+2]x^{n+\alpha}$$
I think this is fine, because the indicial equation I get gives the same roots from the other method: $$\alpha(\alpha-1)+p_0\alpha+q_0=0$$

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

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

And then [tex]a_{n+1}(n+\alpha+1)(n+\alpha+2)-a_n[(n+\alpha)^2+2]=0]<br /> So for $$\alpha_1=0$$<br /> <br /> $$a_{n+1}=\frac{a_n[n^2+2]}{(n+1)(n+2)}$$<br /> <br /> The thing is I've tried some iterations but I can't get a expression for $$a_k$$ in terms of $$a_0$$. How should I do this?[/tex]

 

[Gib Z]

$$a_k = a_0 \left(\frac{a_k}{a_{k-1}} \cdot \frac{a_{k-1}}{a_{k-2}} \cdots \frac{a_2}{a_1} \cdot \frac{a_1}{a_0} \right) = \frac{(k-1)^2+2}{k(k+1)} \cdot \frac{ (k-2)^2+2}{(k-1)(k-2)} \cdots \frac{ 1^2+2}{2\cdot 3} \cdot \frac{0^2+2}{1\cdot 2} = a_0 \frac{ \prod_{m=0}^{k-1} m^2+2 }{k! (k+1)! }$$. There is no closed form expression for that product, so that's the best you can get.

 

[Telemachus]

Haha thank you verymuch :)

Maybe I've made some mistake