General Solution for a non-Homogeneous Recurrence 
What is the general solution to the recurrence:
  $$x(n + 2) = x(n + 1) + x(n) + n - 1$$
  where $n\geq 1$ with $x(1) = 0, x(2) = 1$?

It's a question on a practice exam I'm reviewing and I'm not quite sure why my solution is incorrect.
So I went through the characteristic equation and found the constants, and the solution to the homogeneous part of this question is:
$$x(n) = \frac{(1+\sqrt{5})^n - (1-\sqrt{5})^n}{\sqrt{5}}$$
This is the Fibonacci expression, right? So, what do I do with the non-homogeneous part of the expression? My professor mentioned something about solving them separately but I'm not sure what to do with it now.

Generating Function Approach
The below recurrence is equivalent to the recurrence in the OP, but has been extended to allow $n=0$:

$$a_{n+2} = a_{n+1} + a_n + n - 1\tag{$\star$}$$
  for $n\geq 2$, and $a_0=2, a_1=0$.

Define $A(x) = \sum_{n=0}^\infty a_n x^n$.  Then, multiplying $(\star)$ by $x^n$ and summing over $n$, we have: (note: all sums are to be taken as $\sum_{n=0}^\infty$, but I'm leaving off the limits for ease of typing.)
$$\sum_n a_{n+2}x^n = \sum_na_{n+1}x^n+\sum_n a_n x^n + \sum_n(n-1)x^n\tag{$\star\star$}$$
We can rewrite $(\star\star)$ as:
$$\frac{A(x)-2}{x^2}-\frac{A(x)-2}{x} -A(x) = \sum_n (n-1)x^n = \frac{2x-1}{(1-x)^2}$$
Rearranging the above, we have:
$$A(x)\frac{1-x-x^2}{x^2}=\frac{2-2x}{x^2}+\frac{2x-1}{(1-x)^2}$$
Thus, 
\begin{align}A(x) &= \frac{2(1-x)}{1-x-x^2} + \frac{2x^3-x^2}{(1-x-x^2)(1-x)^2} \\
&= \frac{5 x^2-6 x+2}{(1-x-x^2)(1-x)^2} \\
&= \frac{x-2}{x^2 + x - 1} -\frac{1}{(x-1)^2} - \frac{1}{x-1}
\end{align}
The Taylor Series for each of the above can be found (albeit the first term takes some work), then the coefficient of $x^n$ is the $n$th term.
