Find all solutions of the recurrence relation $a_n = 5a_{n-1} - 6a_{n-2} + 2^n + 3n$ Question:
Find all solutions of the recurrence relation $a_n = 5a_{n-1} - 6a_{n-2} +  2^n + 3n$ (Hint: Look for a particular solution of the form $qn2^n + p_1n + p_2$, where $q, p_1, p_2$ are constants).
Attempt:
The Hint indicates that the solution is of the form $qn2^n + p_1n + p_2$, thus
$$a_n = qn2^n + p_1n + p_2\\\iff a_n = qn2^n + p_1n + p_2= 5(qn2^{n-1} + p_1(n-1) + p_2) - 6(qn2^{n-2} + p_1(n-2) + p_2) + 2^n + 3n$$
I just need if this is a right approach, since I've already wasted hours solving this and I keep getting minute calculation error since my paper is not big enough (I write big apparently). I know for sure that $p_1 = \dfrac{3}{2}$. If my approach above is correct, I'll let this question rest.
Edit:
I think the hint is given due to the fact that $2^n + 3n$ looks nothing like linear homogenous recurrence nor does it even look like a "typical" linear nonhomogenous recurrence relation, since $2^n + 3n$. You might want to comment about that, but I think this book will cover this in future chapters.
 A: I love characteristic equation mathod, so
\begin{equation}\label{1}
a_n -5a_{n-1}+6a_{n-2}=2^n+3n
\end{equation}
If we decrease index by 1 and multiply equation by 2, we get 
\begin{equation}\label{2}
2a_{n-1}-10a_{n-2} = 2^n + 6(n-1) 
\end{equation}
Now if we substract the second equation from the first, we will get
\begin{equation}
a_n - 7a_{n-1} + 10a_{n-2} = 6-3n
\end{equation}
Now, we decrease the index agian and get
\begin{equation}
a_{n-1}-7a_{n-2} +10a_{n-3} = 6-3(n-1)
\end{equation}
After another substitution we get 
$$a_n-8a_{n-1}+17a_{n-2}-10a_{n-3} = 3$$
And again
$$a_{n-1}-8a_{n-2}+17a_{n-3}-10a_{n-4} = 3$$
and we finaly get
$$a_n-9a_{n-1}+25a_{n-2}-27a_{n-3} + 10a_{n-4} = 0$$
Characteristic equation of this recursion is
$$x^4-9x^3+25x^2-27x+10 = 0$$
od 
$$(x-1)^2\cdot(x-2)\cdot (x-5) = 0 $$
so, the solution of the recurrence is
$$a_n = C_1\cdot 1^n + C_2\cdot n\cdot 1^n + c_3\cdot 2^n + C_4\cdot 5^n.$$
or
$$a_n = C_1 + C_2\cdot n + c_3\cdot 2^n + C_4\cdot 5^n.$$
A: Rewrite the LHS as
$$ q(n-1)2^n+q2^n+p_1n+p_2$$
and the RHS as
$$ q(n-1)2^n+(\frac{3q}{2}+1)2^n+(3-p_1)n+(7p_1-p_2) $$ 
Then compare the coefficients of $2^n$, $n$ and the constant term to get
$$ q=\frac{3q}{2}+1,p_1=3-p_1,p_2=7p_1-p_2 $$
from which you can have
$$ q=-2,p_1=\frac{3}{2},p_2=\frac{21}{4}. $$
