Factorial question in a power series solution

[elegysix]

Hello, I've been working on solving the equation y''-2xy'+2py=0. where p is a positive integer.

I've assumed y=$\sum a_{n}x^{n}$ for n=0 to inf

I'm getting two formulas for $a_{n}$
One is for odd n, the other for even n, related to $a_{0}$ and $a_{1}$

However, the relation involves something that looks like a factorial but it skips every other number, such as p(p-2)(p-4)(p-6)...

My question is whether this is valid:

$n!=n(n-1)(n-2)(n-3)...$

$2^{n}(n)!=2n(2n-2)(2n-4)(2n-6)...$

letting $p=2n$ then

$2^{p-1}(\frac {p}{2})!=p(p-2)(p-4)(p-6)...$

 

[HallsofIvy]

Yes, that is perfectly valid. I would have analyzed it the other way:
$$2*4*6*...*(2n)= (2*1)(2*2)(2*3)...(2*n)= 2^n(n!)$$
You can also say that
$$1*3*5*...*(2n+1)= \frac{1*2*3*4*5*...*(2n)*(2n+1)}{2*4*...*(2n)}= \frac{(2n+1)!}{2^n n!}$$

 

[elegysix]

I am supposed to find the solution in summation form, would you mind looking over it? I'm not sure if the factorials give the correct terms based on n. I'll post my equations for the coefficients, and what I think is the formula for them.

n even:
$a_{2}=-pa_{0}$

$a_{4}=\frac{(-1)^{2}2^{2}p(p-2)a_{0}}{4!}$

$a_{6}=\frac{(-1)^{3}2^{3}p(p-2)(p-4)a_{0}}{6!}$

$a_{8}=\frac{(-1)^{4}2^{4}p(p-2)(p-4)(p-6)a_{0}}{8!}$

n odd:
$a_{3}=\frac{(-1)2(p-1)a_{1}}{3!}$

$a_{5}=\frac{(-1)^{2}2^{2}(p-1)(p-3)a_{1}}{5!}$

$a_{7}=\frac{(-1)^{3}2^{3}(p-1)(p-3)(p-5)a_{1}}{7!}$my guess at the 'summation form of the complete solution' is this:

$y(x)=-pa_{0} + \sum a_{n}x^{n}$ for n = 1 to inf

where $a_{n} = \frac {-1^{\frac{n}{2}}2^{\frac{n}{2}}(\frac{p}{2})!a_{0}}{n!(\frac{p-n}{2})!}$ for n even

and $a_{n} = \frac {-1^{\frac{n-1}{2}}2^{\frac{n-1}{2}}(\frac{p-1}{2})!a_{1}}{n!(\frac{p-n-1}{2})!}$ for n odd

I know that's a nightmare, but I'd appreciate it if you could criticize or tell me if this is wrong / if its acceptable in form. I don't know that the ratio of the factorials gives the terms I need above, I think it does.