<s>[INST] Solve the recurrence relation $a_n = 4a_{n-1} - 3a_{n-2} + 2^n $ 
Solve the recurrence relation
$$a_n = 4a_{n-1} - 3a_{n-2} + 2^n $$
With initial conditions:
$a_1 = 1$
$a_2 = 11$

I have done similar recurrence relation problems to this, but none that were a non-homogeneous recurrence relation such as this one.
So far I have:
$$r^n = 4^{n-1} - 3^{n-2} $$
Divide both sides by $$\frac{1}{r^{n-2}}$$
Giving me this as my Auxiliary Equation:
$$ r^n -  4r + 3 = 0  $$
I then solved for the $r$ values and got $r = -4$ and $r = 1$
I am stumped from here as to where the non-homogeneous piece comes into play, any help is appreciated.
 [/INST] Start by finding the general solution to the homogeneous recurrence relationship:
$$a_n = 4a_{n-1} - 3a_{n-2}$$
This has auxiliary equation $\lambda^2=4 \lambda-3$
$\lambda^2-4\lambda+3=0$
$\lambda_1=1, \lambda_2=3$ 
$$a_n = A(1)^n +B(3)^n$$
You want a particular solution to the non-homogeneous relationship.
Try $a_n=k(2)^n$
Then $a_{n-1}=\frac 12 k(2)^n$, $a_{n-2}=\frac 14 k(2)^n$
So $k(2)^n = 4\left (\frac 12 k(2)^n \right)-3 \left(\frac 14 k(2)^n \right)+(2)^n$
$k = 2k - \frac 34 k +1$
$k=-4$
Add this to the general solution to the homogeneous relationship to find the general solution to the non-homogeneous relationship.
$$a_n = A +B(3)^n-4(2)^n$$
Use the known values $a_1=1$ and $a_2=11$
$1=A+3B-8$
$11=A+9B-16$
gives $10=6B-8$
$6B=18$
$B=3$
$1=A+9-8$
$A=0$
$$a_n =3(3)^n-4 (2)^n$$
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