# Fröbenius method

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 $$a_{n}$$.

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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.

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HallsofIvy
Homework Helper
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 $$a_{n}$$.
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!

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!

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.