Solving Differential Equation with Fröbenius Method

[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}.

 

[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.

 

[HallsofIvy]

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!

 

[DevoBoy]

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.