Frobenius method

**Ulysses** · May 8th 2011, 04:07 PM

Hi. 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 $a_{n+1}(n+\alpha+1)(n+\alpha+2)-a_n[(n+\alpha)^2+2]=0$
So for $\alpha_1=0$

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

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

Bye there, thanks for your help.-

**InvisibleMan** · May 9th 2011, 11:47 PM

Just use repeated iterations to get to the next expression:

$a_n = a_0 \frac{((n-1)^2+2)((n-2)^2+2)...2}{((n-1)+1)((n-1)+2)((n-2)+1)((n-2)+2)....1\cdot 2}$

You can rearrange the expression to get something more compact.

**bkarpuz** · May 10th 2011, 07:12 PM

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

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