How to solve this non-homogeneous recurrence? I've made a few threads recently asking how to solve non-homogeneous recurrences and I think I've gotten the hang of it, but now I want to try a complicated thing like this:
$T(n) = 4T(n-1) + 2T(n-2) - 32T(n-3) + 59T(n-4) - 44T(n-5) + 12T(n-6) + 2^n + 5n^4 - 3n^3 + 2n + n \log(n)$
Where $T(0)... T(6) = 0...6$ for simplicity.
I constructed this in a very specific way.
So the homogeneous part has characteristic polynomial $x^6 - 4x^5 - 2x^4 + 32x^3 - 59x^2 + 44x - 12 = 0$ which is also $(x-2)^2 (x-1)^3 (x+3) = 0$ 
I assume this means the homogeneous part has solution of form:
$H(n) = \alpha_1 2^n + \alpha_2 n 2^n + \alpha_3 1^n + \alpha_4 n 1^n + \alpha_5 n^2 1^n + \alpha_6 (-3)^n$
Is this right?
Next, do I have to split up the non-homogeneous part into several pieces:
$2^n$ as its own piece
$5n^4 - 3n^3 + 2n$ as another piece
$n \log(n)$ as yet another piece?
Am I on the right track so far? My next question is how to correctly set up the "trial" equations for each piece. 
 A: You are on the right track, however there is an unfortunate side-effect of having $1$ as a root of the characteristic polynomial.  Note that $5n^4 = 5n^4\cdot 1^n$.  The invisible $1^n$ throws several people off.
For each "piece," as you say, you will need to try to find a particular solution of the same order.
For $2^n$, you would have needed to find a particular solution of the form $c_1\cdot 2^n$, however, note that there is already a term of that order from the solutions to the homogenous equation.  As a result, we multiply by $n$ to make it linearly independent.  But wait! There is also one of the order $n2^n$, so we have to take it yet another step.  It will then be that we need to find a solution of the form $c_1\cdot n^22^n$
For $5n^4-3n^3+2n$, you would have needed to find an arbitrary polynomial of fourth degree: $d_4\cdot n^4+d_3\cdot n^3+d_2\cdot n^2+d_1\cdot n+d_0$, however, as with the case for $2^n$, we already have similarly ordered terms appearing in the homogeneous solution due to the invisible $1^n$'s appearing here.  As a result, what we should really find is a solution of the form $d_4\cdot n^7+d_3\cdot n^6+d_2\cdot n^5+d_1\cdot n^4+d_0\cdot n^3$
For $n\log n$, there is fortunately no similarly ordered terms appearing in the homogeneous solution, however it is of the form $\log n$ times a polynomial so we simply search for solution of the form $\log n$ times a similarly ordered polynomial.  In this case, $e_1\log n + e_2n\log n$.  (caught this mistake during chatting with OP)
