How to Solve Advanced Recurrence Relations
Here are two references which might be useful when learning recurrence relations.
Some interesting techniques are presented in Mathematics for the Analysis of Algorithms by D. E. Knuth and D. H. Greene. One of them is the so-called repertoire method. A real life application is presented in this post .
Recurrences are also a prominent theme in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik.
Hint: Sometimes we are lucky and can find a solution also by elementary means. Assuming initial values $a_0, a_1$ are given, we derive from \begin{align*} a_n=na_{n-1}-(n-3)a_{n-2}\qquad n\geq 2 \end{align*} successively \begin{align*} \color{blue}{1}a_2&=2a_1+a_0\\ a_3&=\color{blue}{3}a_2\\ a_4&=4a_3-a_2=\color{blue}{11}a_2\\ a_5&=5a_4-2a_3=\color{blue}{49}a_2\\ a_6&=6a_5-3a_4=\color{blue}{261}a_2\\ &\cdots\\ \end{align*}
Looking for the values $1,3,11,49,261$ in OEIS we find the sequence A001339 with representation \begin{align*} a_{n+2}=\sum_{k=0}^n\binom{n}{k}(k+1)!\qquad\qquad n\geq 0 \end{align*}
Hint.
Considering $S(x) = \sum_k a_k x^k$ we have
$$ S(x) -x\frac{d}{dx}\left(x S(x)\right)+x^4\frac{d}{dx}\left(\frac{S(x)}{x}\right) = 0 $$
and solving for $S(x)$ we obtain
$$ S(x) = C_0 (1-x)e^{-\frac 1x} $$