1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Thanks, is what I have done so far look correct as well? $\endgroup$ – Quality May 22 '15 at 16:51
  • $\begingroup$ It looks correct to me, but keep in mind that sometimes when you differentiate a series, the constant term must disappear. $\endgroup$ – grdgfgr May 22 '15 at 16:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.