Solve the recurrence relation $a_n=4a_{n-1}-3a_{n-2}+2^n, a_1=1, a_2=11.$ First I solved $a_n=4a_{n-1} -3a_{n-2}$:
$$x^2-4x+3=0\Rightarrow (x-3)(x-1)=0\Rightarrow a_n=k_1(1)^n +k_2(3^n)=k_1+k_2(3^n)$$ 
The problem is, I have no idea how to handle that part which has made it non-homogeneous $(2^n)$.
 A: If the inhomogeneous term $2^n$ was absent from the recurrence relation, then the general solution (without incorporating the initial conditions) would be of the form:
$$
a_n=c_1\cdot 1^n+c_2\cdot 3^n,
$$
as $1$ and $3$ are the roots of the characteristic polynomial $p(\lambda)=\lambda^2-4\lambda+3$. 
Now that we do have the in-homogeneous term, we look for a particular solution, multiple of the in-homogeneous term, i.e., of the form $c_3 2^n$:
$$
c_3 2^{n}=4c_32^{n-1}-3c_3 2^{n-2}+2^n
$$
and hence $c_3=-4$.
Thus, before the incorporation of the initial conditions the solution is of the form
$$
a_n=c_1\cdot 1^n+c_2\cdot 3^n-4\cdot 2^n.
$$
Next, incorporation of the initial conditions yields a $2\times 2$ linear system.
A: Let $a_n=c2^n$(a guess of special solution), plug into the recurrence relation we have $c=4c/2-3c/4+1$ we get $c=-4$. Therefore $-4\cdot2^n$ is a special solution of this recurrence relation.
The solution of your question will be $c_1+c_23^n-4\cdot2^n$ now plug in the initial conditions to find $c_1$ and $c_2$.
