Solve Frobenius Method Homework: 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 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&#039;ve tried some iterations but I can&#039;t get a expression for a_k in terms of a_0. How should I do this?

 

[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