Q:

What is the series solution to $y''-3x^2y'-6xy=0$

In attempting this, I got that the answer was
$$y(x) = a_0 \sum_{n=0}^\infty \frac{3^n}{(3n)!!!}x^{3n} + a_1 \sum_{n=0}^\infty \frac{3^n}{(3n+1)!!!}x^{3n+1} + a_2 \sum_{n=0}^\infty \frac{3^n}{(3n+2)!!!}x^{3n+2}$$
but I can't find any way to verify that this is indeed correct.

A:

Let $\displaystyle y = \sum_{n=0}^{+\infty} a_n x^n$.  Then
\begin{align}
  y' &= \sum_{n=0}^{+\infty} na_n x^{n-1},\\[0.3cm]
  y''  &= \sum_{n=0}^{+\infty} n(n-1)a_n x^{n-2}.
\end{align}
Substituting these series into the original equation gives us
\begin{align}
  0 &= y'' - 3x^2y - 6xy\\[0.3cm]
    &= \sum_{n=0}^{+\infty}n(n-1)a_nx^{n-2} - 3x^2\sum_{n=0}^{+\infty}na_nx^{n-1} - 6x\sum_{n=0}^{+\infty} a_nx^n\\[0.3cm]
    &= \sum_{n=0}^{+\infty}n(n-1)a_nx^{n-2} - \sum_{n=0}^{+\infty}3na_nx^{n+1} - \sum_{n=0}^{+\infty} 6a_nx^{n+1}
\end{align}
Let's modify these series so we can combine them into one series to easily determine the coefficients.  Note that we will most likely need to have some "stray" terms sitting outside of the summation notation when we do this.  This is fine.
\begin{align}
  \sum_{n=0}^{+\infty}n(n-1)a_nx^{n-2} &= 0 + 0 + \sum_{n=2}^{+\infty}n(n-1)a_nx^{n-2}\\[0.3cm]
    &= \sum_{n=0}^{+\infty} (n+2)(n+1)a_{n+2}x^n
\end{align}
The second equality above is obtained by re-indexing the series.  Specifically, we replaced $n$ with $n+2$.
\begin{align}
  \sum_{n=0}^{+\infty}3na_nx^{n+1} &= 0 + \sum_{n=1}^{+\infty}3na_nx^{n+1}\\[0.3cm]
    &= \sum_{n=2}^{+\infty}3(n-1)a_{n-1}x^n
\end{align}
The second equality above comes from replacing $n$ with $n-1$.  The same with this equality below:
\begin{align}
  \sum_{n=0}^{+\infty} 6a_nx^{n+1} &= \sum_{n=1}^{+\infty} 6a_{n-1}x^n
\end{align}
So then we have:
\begin{align}
  0 &= y'' - 3x^2y - 6xy\\[0.3cm]
    &= \sum_{n=0}^{+\infty}n(n-1)a_nx^{n-2} - 3x^2\sum_{n=0}^{+\infty}na_nx^{n-1} - 6x\sum_{n=0}^{+\infty} a_nx^n\\[0.3cm]
    &= \sum_{n=0}^{+\infty}n(n-1)a_nx^{n-2} - \sum_{n=0}^{+\infty}3na_nx^{n+1} - \sum_{n=0}^{+\infty} 6a_nx^{n+1}\\[0.3cm]
    &= \sum_{n=0}^{+\infty} (n+2)(n+1)a_{n+2}x^n - \sum_{n=2}^{+\infty}3(n-1)a_{n-1}x^n - \sum_{n=1}^{+\infty} 6a_{n-1}x^n\\[0.3cm]
    &= (2)(1)a_2 + (3)(2)a_3x + \sum_{n=2}^{+\infty} (n+2)(n+1)a_{n+2}x^n - \sum_{n=2}^{+\infty}3(n-1)a_{n-1}x^n - \left[6a_0x + \sum_{n=2}^{+\infty} 6a_{n-1}x^n\right]\\[0.3cm]
    &= 2a_2 + 6(a_3-a_0)x + \sum_{n=2}^{+\infty} \left[(n+2)(n+1)a_{n+2} - 3(n-1)a_{n-1} - 6a_{n-1}\right]x^n
\end{align}
Since the only constant term is $2a_2$, we must have $a_2 = 0$.  Similarly, we must have $a_0 = a_3$.  Looking at the coefficients on the series, we see that we must have
$$(n+2)(n+1)a_{n+2} - 3(n-1)a_{n-1}-6a_{n-1} = 0$$
as well.  So then
\begin{align}
  0 &= (n+2)(n+1)a_{n+2} - 3(n-1)a_{n-1}-6a_{n-1}\\[0.3cm]
    &= (n+2)(n+1)a_{n+2} - [3(n-1)+6]a_{n-1}\\[0.3cm]
    &= (n+2)(n+1)a_{n+2} - 3(n+1)a_{n-1}\\[0.3cm]
  3a_{n-1} &= (n+2)a_{n+2}\\[0.3cm]
  3a_n &= (n+3)a_{n+3}
\end{align}
The last equality above comes from replacing $n$ with $n+1$.  So then we finally obtain
$$ a_{n+3} = \frac{3}{n+3}a_n.$$
Since we have a second-order ODE, we expect $a_0$ and $a_1$ to be the arbitrary constants by which all the other coefficients are determined.  And we already know that $a_2 = 0$.  So for the first few values of $n$, we have
\begin{align}
  a_3 &= \frac{3}{0+3}a_0 = \frac{3}{3}a_0\\[0.3cm]
  a_4 &= \frac{3}{1+3}a_1 = \frac{3}{4}a_1\\[0.3cm]
  a_5 &= \frac{3}{2+3}a_2 = 0\\[0.3cm]
  a_6 &= \frac{3}{3+3}a_3 = \frac{3}{6} \cdot \frac{3}{3} a_0\\[0.3cm]
  a_7 &= \frac{3}{4+3}a_4 = \frac{3}{7} \cdot \frac{3}{4} a_1\\[0.3cm]
  a_8 &= \frac{3}{5+3}a_5 = 0\\[0.3cm]
  a_9 &= \frac{3}{6+3}a_6 = \frac{3}{9} \cdot \frac{3}{6} \cdot \frac{3}{3}a_0\\[0.3cm]
  a_{10} &= \frac{3}{7+3}a_7 = \frac{3}{10} \cdot \frac{3}{7} \cdot \frac{3}{4} a_1\\[0.3cm]
  a_{11} &= \frac{3}{8+3}a_8 = 0\\[0.3cm]
\end{align}
So we have $a_0, a_1$ arbitrary and $a_2 = 0$, and for $n > 0$ we have
\begin{align}
  a_{3n} &= \frac{3^n}{(3n)!!!}a_0\\[0.3cm]
  a_{3n+1} &= \frac{3^n}{(3n+1)!!!}a_1\\[0.3cm]
  a_{3n+2} &= 0
\end{align}

