Solving a Second Order Homogeneous Differential Equation

[Kb1jij]

I just finished a final in my differential equations class. One of the problems had me solve a second order homogeneous differential equation using series. I boiled it down to this recursion relation:

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

I found that the even coefficients work out nicely to the following sum:

$$y=a_{0}+\Sigma^{\infty}_{n=2}\frac{(n+1)a_{0}}{4*6^{n-2}}x^{n}$$

I couldn't get a nice result for the odd coefficients and still can't find one. It's kind of bothering me now. Is it even possible? I can boil it down to this series:

$$\frac{1}{3},\frac{1}{20},\frac{1}{210},\frac{1}{3024},\frac{1}{55440},...$$

 

[Kb1jij]

Wow, never mind that recursion relation is wrong. Silly me.

If anyone feels like it they are still welcome to find an expression for that sequence

 

[Crosson]

Putting this recursion relation in closed form requires the use of the double factorial function (or equivalently, a half integer gamma function). Here is my solution for the odd coefficients:

$$a_{n,odd} = a_1 \frac{2^{-\frac{n}{2}-\frac{3}{2}} \left(-1+(-1)^n\right) (n+1) }{n (n-2)!}$$

You can see the definition of the double factorial here:

http://mathworld.wolfram.com/DoubleFactorial.html

It is just like a regular factorial, instead of steps of one it decrements in steps of 2.

The first time I solved a series problem with a double factorial it took me 4 hours of really difficult thinking. After a bit of practice you can get used to how the 2^n factors combine with the odd integer double factorial, but it's a mind bender at first.