Solve the recurrence relation $a_{k+2}-6a_{k+1}+9a_k=3(2^k)+7(3^k), k\geq 0, k_0=1,k_1=4.$ 
Solve the recurrence relation $a_{k+2}-6a_{k+1}+9a_k=3(2^k)+7(3^k), k\geq 0, k_0=1,k_1=4.$

I know that I need a general solution of the form $a_k=a^{(h)}_k+a^{(p)}_k$, where the first term is a general solution to the recurrence relation given on the LHS and the particular solution which depends on both the LHS and the RHS.
So far, I got the characteristic polynomial of the LHS, which is given by $t^2-6t+9=(t-3)^2=0.$ So I have double roots $t=3$. So the homogenous solution is given by $a^{(h)}_k=x3^k+ky3^k$. 
Now i'm having trouble determining the particular solution. How do I proceed? 

The $3(2^k)$ part is easy to deal with. Look for a solution of the shape $c(2^k)$.
The $7(3^k)$ part is made more complicated by the fact that $3$ is a root, indeed a double root, of the characteristic polynomial. Look for a solution of the shape $dk^2(3^k)$. There will be very nice cancellation.
For a particular solution, by linearity we can add the expressions obtained in the two paragraphs above.
