# Question about Frobenius Method

I am having some confusion and looking for some help/suggestions about the following.

Consider the ODE; with regular singular point $x_0=0$

$$2x(x-1)y''+3(x-1)y'-y=0$$

And I am supposed to find the power series solution.

What I have tried:

First I took note that $$\lim_{x \to 0} \frac{x3(x-1)}{2x(x-1)}=3/2$$

and that $$\lim_{x \to 0} \frac{-1x^2}{2x(x-1)}=0$$

Which allows me to write the euler equation

$$r(r+\frac{1}{2})=0$$

that is $r_1=-1/2$ and $r_2=0$ which are real and distinct roots suggesting a solution my be of the form $$y=c_1+c_2x^{-1/2}$$ where $c_1,c_2 \in \mathbb{R}$

Now, I said,

assume that a power series solution does exist ,

$$y= \sum_{n=0}^{\infty}a_nx^{r+n}$$

$$y'=\sum_{n=0}^{\infty}(r+n)a_nx^{r+n-1}$$

$$y''=\sum_{n=0}^{\infty}(r+n-1)(r+n)a_nx^{r+n-2}$$

Then by using the original equation and expanding , I get

$$\sum_{n=0}^{\infty}2(r+n-1)(r+n)a_nx^{r+n}-\sum_{n=0}^{\infty}2a_n(r+n-1)(r+n)x^{r+n-1}+\sum_{n=0}^{\infty}3a_n(r+n)x^{r+n}-\sum_{n=0}^{\infty}3(r+n)a_nx^{r+n-1}-\sum_{n=0}^{\infty}a_nx^{r+n}=0$$

But now I am confused. Firstly, is it looking okay so far? And secondly, how can I proceed? what is the best way to do so? I know that my index and coefficient terms must be the same, but I don't see how I can get it all to work out well?

Any help/suggestions/hints/answer/etc?

Thank you all

## 1 Answer

Hint: Shift the bounds of the summation to to match the powers of $x$. You will have one starting at $n=0$ and one starting at $n=1$. Take out the $0 ^{th}$ term from the one that has the bound $n=0$ and combine the two summations which now have the same $x$ power. You will have a recursive relationship between $a_n$'s.

• Thanks, is what I have done so far look correct as well? – Quality May 22 '15 at 16:51
• It looks correct to me, but keep in mind that sometimes when you differentiate a series, the constant term must disappear. – grdgfgr May 22 '15 at 16:53