
Frobenius method
Hi. I need some help with this problem. I have to solve
$\displaystyle x(1x)y''+2(12x)y'2y=0$ (1)
Using Frobenius method around zero. So proposing $\displaystyle y=\Sigma_{n=0}^\infty a_n x^{n+\alpha}$, differentiating and replacing in (1):
$\displaystyle x(1x)y''+2(12x)y'2y=\Sigma_{n=0}^\infty a_n(n+\alpha)(n+\alpha1)x^{n+\alpha1} \Sigma_{n=0}^\infty a_n(n+\alpha)(n+\alpha1)x^{m+\alpha} + 2 \Sigma_{n=0}^\infty a_n(n+\alpha)x^{n+\alpha1} 4 \Sigma_{n=0}^\infty a_n(n+\alpha)x^{n+\alpha}2 \Sigma_{n=0}^\infty a_n n^{n+\alpha}=0$
Regrouping and working a little bit this becomes:
$\displaystyle a_0\alpha(\alpha+1)+\Sigma_{n=1}^\infty a_n(n+\alpha)(n+\alpha+1)x^{n+\alpha1} \Sigma_{n=0}^\infty a_n [(n+\alpha)^2+2]x^{n+\alpha}$
I think this is fine, because the indicial equation I get gives the same roots from the other method: $\displaystyle \alpha(\alpha1)+p_0\alpha+q_0=0$
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.

Just use repeated iterations to get to the next expression:
$\displaystyle a_n = a_0 \frac{((n1)^2+2)((n2)^2+2)...2}{((n1)+1)((n1)+2)((n2)+1)((n2)+2)....1\cdot 2} $
You can rearrange the expression to get something more compact.

If you have a relation like
$\displaystyle a_{n+1}=p_{n}a_{n}$ for $\displaystyle n\geq m$,
which is a firstorder difference equation, then
$\displaystyle a_{n}=a_{m}\prod_{j=m}^{n1}p_{j}$ for $\displaystyle n\geq m$,
with the convention that the empty product is the unity.
In your case, you have
$\displaystyle a_{n}=a_{0}\prod_{j=0}^{n1}\frac{{j^{2}+2}}{(j+1)(j+2)}$
.....$\displaystyle =a_{0}\frac{1}{n!(n+1)!}\prod_{j=0}^{n1}(j^{2}+2)$
for $\displaystyle n\geq0$.