Solving a weird recurrence relation using exponential generating functions and differential equations So the recurrence relation is:
$a(n) = (n-1)\cdot a(n-2) + (n-1)(n-2)\cdot a(n-3)$
I've tried several things, but the instructions are to use a differential equation and nothing I'm doing seems to work. Here's what I have so far:
$a(n) = (n-1)\cdot a(n-2) + (n-1)(n-2)\cdot a(n-3)$
$a(n)\frac{x^n}{n!} = (n-1)\cdot a(n-2)\frac{x^n}{n!} + (n-1)(n-2)\cdot a(n-3)\frac{x^n}{n!}$
$\sum \limits_{n\ge 0}a(n)\frac{x^n}{n!} = \sum\limits_{n\ge 0}(n-1)\cdot a(n-2)\frac{x^n}{n!} + \sum\limits_{n\ge 0}(n-1)(n-2)\cdot a(n-3)\frac{x^n}{n!}$
$A(x)= \sum\limits_{n\ge 0}(n-1)\cdot a(n-2)\frac{x^n}{n!} + \sum\limits_{n\ge 0}(n-1)(n-2)\cdot a(n-3)\frac{x^n}{n!}$
but now I'm stuck. Can anyone offer any advice?

Let $A(x)$ be the exponential generating function. For convenience I’ll write $a_n$ instead of $a(n)$.
$$\begin{align*}
A(x)&=\sum_{n\ge 0}a_n\frac{x^n}{n!}\\\\
&=\sum_{n\ge 0}(n-1)a_{n-2}\frac{x^n}{n!}+\sum_{n\ge 0}(n-1)(n-2)a_{n-3}\frac{x^n}{n!}\\\\
&=x^2\sum_{n\ge 0}\frac{(n-1)a_{n-2}}{n(n-1)}\cdot\frac{x^{n-2}}{(n-2)!}+x^3\sum_{n\ge 0}\frac{(n-1)(n-2)a_{n-3}}{n(n-1)(n-2)}\cdot\frac{x^{n-3}}{(n-3)!}\\\\
&=x^2\sum_{n\ge 0}\frac{a_n}{n+2}\cdot\frac{x^n}{n!}+x^3\sum_{n\ge 0}\frac{a_n}{n+3}\cdot\frac{x^n}{n!}\\\\
&=\sum_{n\ge 0}\frac{a_n}{n+2}\cdot\frac{x^{n+2}}{n!}+\sum_{n\ge 0}\frac{a_n}{n+3}\cdot\frac{x^{n+3}}{n!}
\end{align*}$$
Differentiating the righthand side with respect to $x$ yields
$$\sum_{n\ge 0}a_n\frac{x^{n+1}}{n!}+\sum_{n\ge 0}a_n\frac{x^{n+2}}{n!}=(x+x^2)A(x)\;,$$
so $A'(x)=(x+x^2)A(x)$. There’s your differential equation, and I’ll leave the rest to you.
