<s>[INST] solve $a_n=5a_{n-1}-4a_{n-2}+3\cdot2^n$ with initial conditions $a_0=1, a_1=10$ so i am pretty sure that i have solve the homogeneous solution correctly. $a_n^h = B\cdot 4^n+C\cdot1^n$ however I am not so confident on the particular solution. Here was my attempt.
Since $f(n)=3\cdot2^n$ I knew that the particular solution would be in the form of $D*2^n$ so i plugged that in and got $D 2^n=5 D 2^{n-1}-4 D 2^{n-2} +3\cdot2^n $ I then did some work to find D=-6. The way I did that was divide everything by $2^{n-2}$ and got the equation $4D=10D-4D+12$. 
So my question am i doing this correctly or was I wrong to starting out by saying the equation was of the form $D\cdot2^n$. In class we always did it by finding $s$, $t$, and $m$ where $s =2$, $t$ would be the degree so t=0 and m=multiplicity which i always get confused about. Thanks for the help 
 [/INST] As a way of checking, you could always fall back on the generating function route:
$$\begin{align*} f(z) &= \sum_{n=0}^\infty a_n z^n \\
&= a_0 + a_1 z + \sum_{n=2}^\infty (5a_{n-1} - 4a_{n-2} + 3(2^n))z^n \\ 
&= 1 + 10z + 5z(-1 + f(z)) - 4z^2 f(z) + 3\sum_{n=2}^\infty (2z)^n \\
&= 1 + 5z + (5z - 4z^2) f(z) + \frac{12z^2}{1-2z}. \end{align*}$$
Thus, $$\begin{align*} f(z) &= \frac{(1+z)(1+2z)}{(1-2z)(1-z)(1-4z)} \\
&= \frac{2}{1-z} - \frac{6}{1-2z} + \frac{5}{1-4z} \\
&= \sum_{n=0}^\infty 2z^n - 6(2z)^n + 5(4z)^n \\
&= \sum_{n=0}^\infty (2-6(2^n)+5(4^n))z^n, \end{align*}$$ giving $$a_n = 2-6(2^n)+5(4^n).$$
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