I'm trying to solve the recurrence relation below by using generating function:

\begin{equation}
 a\_n=\begin{cases}
 0, & \text{if $n<0$}\\
 2, & \text{if $n=0$}\\
 1, & \text{if $n=1$}\\ 
 3a\_{n-1} + a\_{n-2}, & \text{otherwise}.
 \end{cases}
\end{equation}

The first thing I did was make the recurrence relation valid for all $n$ by using a kronecker delta:

$a\_0 = 3.(0) + 0 + 2.(\delta\_{n,0}) = 2$

$a\_1 = 3.(2) + 0 - 5.(\delta\_{n,1}) = 1$

The result I got was:

$$a\_n = 3a\_{n-1} + a\_{n-2} + 2\delta\_{n,0} - 5\delta\_{n,1}$$

Multiplying by $x^n$:

$$a\_n . x^n = 3a\_{n-1} . x^n + a\_{n-2} . x^n + 2\delta\_{n,0} . x^n - 5\delta\_{n,1} . x^n$$

Summing up both sides:

$$\sum\_{n\geq0} a\_n . x^n = \sum\_{n\geq0}3a\_{n-1} . x^n + \sum\_{n\geq0}a\_{n-2} . x^n + \sum\_{n\geq0}2\delta\_{n,0} . x^n - \sum\_{n\geq0}5\delta\_{n,1} . x^n$$

And making $F(x) = \sum\_{n\geq0} a\_n . x^n$, I got:

$$F(x) = 3xF(x) + x^2F(x) + 2 - 5x$$

which is:

$$F(x) = \frac{2 - 5x}{1-3x-x^2}$$

So far so good but from here on I can't find a way to calculate the $a\_n$

I've heard it has something to do with partials fractions but I'm a newbie in this subject and I have no idea how to follow through.

Does anyone can help me to finish the calculation?

Thanks in advance.