<s>[INST] Finding the recurrence relation This is actually a very simple problem, but I am going to type everything out in case I really overlooked something
I am trying to find a recurrence relation in the series solution I got. 
The ODE is $y'' + xy'- y = 0$. 
$\begin{align}
y'' + xy' - y & = \sum_{n=2}^{\infty} n(n - 1)a_nx^{n-2} + x\sum_{n=1}^{\infty} na_nx^{n-1} - \sum_{n=0}^{\infty} a_nx^n \\
& = \sum_{n=2}^{\infty} n(n - 1)a_nx^{n-2} + \sum_{n=1}^{\infty} na_nx^n - \sum_{n=0}^{\infty} a_nx^n \\
& = \sum_{n=0}^{\infty} (n+2)(n+1)a_nx^n + \sum_{n=0}^{\infty} na_nx^n - \sum_{n=0}^{\infty} a_nx^n\\
& = \sum_{n=0}^{\infty} x^n[(n+1)(n+2)a_{n+2} - a_n + na_n ] \\
& = 0 
\end{align}$ 
The power series is 0 iff $(n+1)(n+2)a_{n+2} - a_n(1- n)  = 0 \iff a_{n+2} = \dfrac{a_n (1 - n)}{(n+1)(n+2)}$
So testing out various values of n = 1, 2, 3... led me to
$a_2 = \dfrac{a_0}{2}$
$a_3 = 0$
$a_4 = \dfrac{-a_2}{12} = \dfrac{a_0}{24}$
$a_5 = 0$
$a_6 = -\dfrac{a_4}{10} = \dfrac{a_0}{240}$
Okay so clearly the odd terms all go to 0 and there will be no odd terms. I know that the even terms alternate back and forth in series (so I will get an alternating series). I have no idea how the denominator can jump from 2 to 24 and then to 240 and then to a stunning 240*56. I see no pattern in how the denominator jumps at all, it jumps too fast if you ask me
 [/INST] It is obvious that $a_{2n+1}=0$ for every $n \ge 1$. Set $b_n=a_{2n}$ for $n \ge 0$. Then
$$
b_k=a_{2(k-1)+2}=-\frac{2k-3}{(2k-1)(2k)}b_{k-1}=-\frac{1}{2}\cdot\frac{1}{k}\cdot\frac{2k-3}{2k-1}b_{k-1} \quad \forall k \ge 0.
$$
Taking the product on both sides for $1 \le k \le n$ we get
$$
b_nb_{n-1}\cdots b_1=\left(-\frac{1}{2}\right)^{n}\cdot\frac{1}{n!}\cdot\frac{-1.3.\cdots(2n-3)}{1.3.\cdots(2n-3)(2n-1)}b_{n-1}b_{n-2}\cdots b_1b_0.
$$
After simplication we get
$$
b_n=\left(-\frac{1}{2}\right)^{n}\cdot\frac{-1}{n!(2n-1)}b_0,
$$
i.e.
$$
a_{2n}=\frac{(-1)^{n+1}}{n!2^n(2n-1)}a_0.
$$
</s>