Solving the recurrence relation obtained from the power series method Assuming the solution to my differential equation is of the form $y=\sum_{n=0}^\infty a_nx^n$, I was able to get to the recurrence relation.
The recurrence relation is $$a_{n+2} = \dfrac {3n-3}{(n+2)(n+1)}a_n$$
From this I get
$$a_2 = \frac {-3}{2}a_0$$ $$a_3 = 0 \implies a_{\text{odd}} = 0 \tag{except maybe $a_1$} \\ a_4 = \frac 14a_2 = \frac {-3}8a_0 \\ a_6 = \frac 3{10}a_4 = \frac {-9}{80}a_0 \\ \vdots$$
Thus my solution should be of the form $$y= a_1(x) + a_0 (1-\frac 32x^2 -\frac 38x^4 -\frac 9{80}x^6 + \cdots)$$
How do I simplify this -- that is write the infinite series for the even terms?  I've been trying for a while, but I'm not sure what the standard technique is for this.


$$y(x) = \sum_{n=0}^{\infty} a_nx^n$$
  $$a_{n+2} = \dfrac {3n-3}{(n+2)(n+1)}a_n$$
  $$a_{2n+1} = 0$$

$$\begin{align}
%
a_{n} &= \frac {3(n-3)}{(n-1)(n-0)} \times \frac {3(n-5)}{(n-3)(n-2)} \times \frac {3(n-7)}{(n-5)(n-4)}  \\
      & \times \dots \\
      & \times \frac {3(3)}{(6)(5)} \times \frac{3(1)}{(4)(3)}\times \frac{3(-1)}{(2)(1)} \times a_{0}
\end{align}$$
And canceling out terms:
$$\begin{align}
%
a_{n} &= a_0~3^{n/2}~\frac {1}{(n-1)(n-0)} \times \frac {1}{(n-2)} \times \frac {1}{(n-4)}  \\
      & \times \dots \\
      & \times \frac {1}{6} \times \frac{1}{4}\times \frac{(-1)}{2}
\end{align}$$
$$a_{n} = a_0~3^{n/2}~\frac{-1}{n-1}~\frac{1}{(n-0)(n-2)(n-4) \dots (4)(2)}$$
And rewriting it in various ways:
$$a_{n} = a_0~3^{n/2}~\frac{-1}{2(n/2)-1}~\frac{2^{-n/2}}{(n/2-0)(n/2-1)(n/2-2) \dots (2)(1)}$$
$$f_n(x) = a_{2n}x^{2n} = a_0~(3/2)^{n}~\frac{-1}{2n - 1}~\frac{x^{2n}}{(n-0)(n-1)(n-2) \dots (2)(1)}$$
$$f_n(x) = \frac{-a_0}{2n - 1}~\frac{\left(3x^2/2\right)^n}{n!}$$
$$-a_0~g_n\left( \frac{3x^2}{2} \right) = f_n(x) \iff \color{darkred}{g_n(x) = \frac{1}{2n - 1}~\frac{x^n}{n!}}$$
This $g$ doesn't look like any common taylor series, so checking wolfram online:
$$\sum_{n=0}^{\infty} \frac{1}{2n - 1}~\frac{x^n}{n!} = \sqrt{\pi x}~~ {\rm erfi}(\sqrt{x}) - e^{x}$$
where erfi is the imaginary error function.  So while your recursion has the above closed form, your series does not have an elementary closed form.
