Closed form of recurrence relation $a_k=(n+k)a_{k-1}-ka_{k-2}$
Edit: Made a mistake in the resulting differential equation, so the problem is no longer so neat.
We consider the exponential generating function as Somos suggests to get
\begin{align}f(x)&=\sum_{n=0}^\infty\frac{a_k}{k!}x^k\\&=a_0+a_1x+\sum_{k=0}^\infty\frac{(n+k+2)a_{k+1}-(k+2)a_k}{(k+2)!}x^{k+2}\\&=a_0+[a_1-(n+1)a_0]x+\sum_{k=0}^\infty\frac{(n+k+1)a_k}{(k+1)!}x^{k+1}-\sum_{k=0}^\infty\frac{a_k}{(k+1)!}x^{k+2}\\&=a_0+[a_1-(n+1)a_0]x+xf(x)+\sum_{k=0}^\infty\frac{na_k}{(k+1)!}x^{k+1}-x\sum_{k=0}^\infty\frac{a_k}{(k+1)!}x^{k+1}\\y'&=a_0+[a_1-(n+1)a_0]x+xy'+(n-x)y\\0&=a_0+[a_1-(n+1)a_0]x+(n-x)y+(x-1)y'\end{align}
where we set $\displaystyle y=\sum_{k=0}^\infty\frac{a_k}{(k+1)!}x^{k+1}$ so that $y'=f(x)$.