I'm trying to solve $ f(k) = f(k−1 ) + 2f(k−2) + 2^k $ where $f(0)=4 ,f(1)=12$. I first get the following homogenous part
$$(r+1)(r-2)=0$$ where $$ r_{1}=-1 \text{ and } r_{2}=2 $$ then I get the homogenous part as $$ (-1)^k . C_{1} + 2^k . C_{2} $$
For the partial part, I try to do the following.
$$f_{partial}(k) = A.2^k.k \text { , times $k$ because $2^k$ is in homogenous solution} \\ f_{partial}(k-1) = A.2^{(k-1)}.(k-1) \\ f_{partial}(k-2) = A.2^{(k-2)}.(k-2) $$
To solve it, I write it as follows.
$$A \cdot 2^k \cdot k=\frac{A \cdot 2^k \cdot(k-1)}{2}+\frac{2 \cdot A \cdot 2^k \cdot(k-2)}{4}+2^k \\ \Rightarrow A = \frac{2}{3}$$
Then I find $C_{1}$ and $C_{2}$,
$$C_{1}+C_{2}= 4 \\ -C_{1}+2C_{2}+\frac{4}{3}= 12 \\ \Rightarrow C_{2} = \frac{44}{9} \\ \Rightarrow C_{1} = -\frac{8}{9}$$
When we sum all up,
$$2^k \cdot \frac{44}{9}-\frac{8}{9} \cdot(-1)^k+\frac{2}{3} \cdot 2^k \cdot k$$
However, it seems wrong because the solution book says $$a_k=\frac{38}{9} \cdot 2^k+\frac{2}{3} \cdot(k+1) 2^k-\frac{8}{9} \cdot(-1)^k$$
Where is my wrong? Can you help, please?