# Recurrence Relation With Non-constant Coefficient

I have a question that involves finding the closed form of the generating function for this sequence $$na_n = 3a_{n-1} -4a_{n-2}+ \frac{8.3^{n-2}}{(n-2)!}$$ with $$a_0=2, a_1=6$$ My lecturer told me to use the derivative of the generating function and solve the differential equation for it, but this comes out with an equation that can't be solved without using a lot of calculus, which I am not allowed to do. Any help would be appreciated.

• The last term strongly suggests trying $e^{3x}$. Fitting the first two terms then points to $2e^{3x}$ which works. I guess that is not an approved method! – almagest Apr 27 '16 at 8:30
• Yeah I have ended up with an $e^{3x}$ in my attempts to solve it, but it just doesn't end nicely. – Nico Monk Apr 27 '16 at 8:32

Let $A(x)=\sum_{n=0}^\infty a_n x^n$. To get to the differential equation, we first note that $A'(x)=\sum_{n=1}^\infty n a_n x^{n-1}$. With this in mind, we multiply both sides of the recurrence by $x^{n-1}$ and sum from $n=0$ to $\infty$ to obtain $$A'(x)=3A(x)-4x A(x)+8x e^{3x}$$ where we have used the fact that $\displaystyle\sum_{n=0}^\infty \frac{8x(3x)^{n-2}}{(n-2)!} =8x e^{3x}$ (since $1/n!=0$ for integer $n<0$).
One could solve this in a systematic way, but I'm not interested in such. Instead, I'll simply note that inspection of the ODE and the initial values of $A(x)$ (i.e. $A(0)=2$, $A'(0)=6$) suggests the guess $A(x)=2 e^{3x}$. This is readily confirmed to satisfy solve the initial value problem.