# Fröbenius method

1. Mar 27, 2007

### DevoBoy

I'm trying to solve the differential equation:

$$x^2y''+2xy'+(x^2-2)y=0$$

using the Fröbenius method.

So I want a solution on the form

$$y=\sum_{n=0}^\infty a_{n}x^{n+s}$$

After finding derivatives of y, inserting into my ODE, and after some rearranging:

$$\sum_{n=0}^\infty ((n+s)(n+s-1)+2(n+s)-2)a_{n}x^{n+s} + \sum_{n=2}^\infty a_{n-2}x^{n+s}$$

Looking at n=0, assuming $$a_{0}$$ different from zero, I get two possible values for s:

$$s_{1}=1$$
$$s_{2}=-2$$

Both giving $$a_{1}=0$$

Choosing s=1, I get

$$a_{n}=\frac{-a_{n-2}}{(n+s)(n+s-1)+2(n+s)-2}=\frac{-a_{n-2}}{n(n+3)}$$

I know $$a_{n}=0$$ for odd n, so I'm interrested in finding $$a_{n}$$ for even n, expressed by $$a_{0}$$. Best thing I can come up with is

$$a_{n}=\sum_{n=2,4,6,..}^\infty \frac{3(-1)^{n/2}a_{0}}{(n+1)!(n+3)}$$

which seems overly complicated given the simple recursive formula. Any ideas?

Second problem; I want to find another linear independent solution, but not quite sure where to start given my ugly expression for $$a_{n}$$.

Last edited: Mar 27, 2007
2. Mar 28, 2007

### ObsessiveMathsFreak

Your equation is almost certainly some kind of Bessel equation. You should seek solutions in terms of Bessel functions.

Edit:
In fact, it is a Bessel function. Specifically, a spherical Bessel function apparently. There are also general solutions on the page. Remember that in order to obtain the correct expansion, you must use the boundary values of the problem.

Last edited: Mar 28, 2007
3. Mar 28, 2007

### HallsofIvy

Staff Emeritus
I don't see how you could get a sum like that. Starting with
$$a_n= \frac{-a_{n-2}}{n(n+3)}$$
You get, for the first few terms, $a_2= -a_0/(2(5))$, $a_4= a_0/(2(5)(4)(7))$, $a_6= -a_0/(2(4)(5)(7)(6)(9)$

Okay, it looks like
$$a_{2n}= \frac{3(-1)^n a_0}{(2n)!(2(n+1)+ 1)}$$
but there is no sum!

4. Mar 28, 2007

### DevoBoy

You're absolutely right. My mistake.

After reading up on Bessel functions (thanks ObsessiveMathsFreak!!), I've rewritten my recursion formula as

$$a_{2n}=\frac{(-1)^{n}a_{0}\Gamma(1+3/2)}{2^{2n}n!\Gamma(n+5/2)}$$

which seems to expand beautifully.