Help with recurrence relation problem

I just cannot get past the first part of this problem. Any help would be awesome!
Problem: Consider the following recurrence relation:
an = 6a(n-1) - 9a(n-2) + 2^n . a0 = 1, a1 = 6. Of course all the a's have n's as subscripts.

1. Find the general solution to the corresponding homogeneous relation.
2. Find a particular solution.
3. Solve the original nonhomogeneous relation.

Please help me :)

1) Assuming that a(n) = r^n, we obtain
r^n = 6r^(n-1) - 9r^(n-2)
==> r^2 - 6r + 9 = 0
==> r = 3, 3.

So, the general solution to the homogeneous relation is a(n) = (A + Bn) * 3^n.

2) Assume that a(n) = C * 2^n
==> C * 2^n = 6C * 2^(n-1) - 9C * 2^(n-2) + 2^n
==> 4C * 2^n = 12C * 2^n - 9C * 2^n + 4 * 2^n
==> C = 4.

So, the particular solution is 4 * 2^n = 2^(n+2).

3) By parts 1 and 2, the general solution is
a(n) = (A + Bn) * 3^n + 2^(n+2).

Solve for A and B by using the initial conditions.
n = 0 ==> 1 = A + 2^2 ==> A = -3
n = 1 ==> 6 = 3(A + B) + 8 ==> B = 7/3.

So, the desired solution is
a(n) = (-3 + 7n/3) * 3^n + 2^(n+2) = (7n - 9) * 3^(n-1) + 2^(n+2).

I hope this helps!