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.

Solution Using Method of Annihilators
Let $E$ denote the "shift" operator, and let $\langle x_n \rangle$ denote the sequence $x_1, x_2, x_3, \ldots$
Then our recurrence can be re-written as:
$$x_{n+2} - x_{n+1} - x_n = n - 1 \iff (E^2-E-1)\langle x_n\rangle = \langle n-1\rangle$$
The annihilator for $n-1$ is $(E-1)^2$, we have:
$$(E-1)^2(E^2-E-1)\langle x_n\rangle = (E-1)^2\langle n-1\rangle=\langle 0\rangle$$
So, we've now converted our problem from a non-homogenous problem to a homogeneous problem; the only difficulty is that we have also changed the recurrence from a $2$nd order recurrence to a $4$th order recurrence. Thus, we need to compute two more initial conditions:
\begin{align}
x_2 &= 1\\
x_3 &= 3
\end{align}
We can easily move from the annihilator operator form to the characteristic of this fourth-order recurrence:
$$(r-1)^2(r^2-r-1) \implies r=1, 1, \frac{1+\sqrt 5}{2}, \frac{1-\sqrt 5}{2}$$
Thus, our general solution has the form:
$$x_n = A + Bn + C\left(\frac{1+\sqrt 5}{2}\right)^n + D\left(\frac{1-\sqrt 5}{2}\right)^n$$
Using our four initial conditions, we find:
\begin{align}
x_1 &= 0 = A +  B + \left(\frac{1+\sqrt 5}{2}\right)C   + \left(\frac{1-\sqrt 5}{2}\right)D\\
x_2 &= 1 = A + 2B + \left(\frac{1+\sqrt 5}{2}\right)^2C + \left(\frac{1-\sqrt 5}{2}\right)^2D\\
x_3 &= 1 = A + 3B + \left(\frac{1+\sqrt 5}{2}\right)^3C + \left(\frac{1-\sqrt 5}{2}\right)^3D\\
x_4 &= 3 = A + 4B + \left(\frac{1+\sqrt 5}{2}\right)^4C + \left(\frac{1-\sqrt 5}{2}\right)^4D\\
\end{align}
This is a system of linear equations; using any method, we arrive at the solution $A = 0, B=-1, C=1, D = 1$.
Thus, the solution is:
$$x_n =\left(\frac{1+\sqrt 5}{2}\right)^n + \left(\frac{1-\sqrt 5}{2}\right)^n - n$$
For what it's worth, the first several values of the sequence are:
1 |  0
2 |  1
3 |  1
4 |  3
5 |  6
6 | 12
7 | 22
8 | 39
9 | 67

