recurrence relation concrete way to solve it I have the following recurrence relation:
$$a_n = 2a_{n-1} + 2^n; a_0 = 0$$ I used the characteristic equation method and some method I found online by calculating the $n+1$ th term and subtracting accordingly the equation with $a_{n+1}$ minus the equation with $a_{n}$:
$$a_{n+1} = 2a_n + 2^{n+1} \\ a_{n+1} - 2a_n = 2a_n - 4a_{n-1} + 2^{n+1}- 2^{n+1} \\ a_{n+1} - 4a_n + 4a_{n-1} = 0$$ Using $x^n$ as a solution: 
$$x^{n-2}(x^2 - 4x+4) = 0$$ I obtained the multiple solution $x= 2$. So we have that: $$a_n = 2^n A_1 + n 2^n A_2$$
How do I finish?

Given: $a_n = 2a_{n-1} + 2^n; a_0 = 0$ and you can find $a_n = 2^n A_1 + n \cdot 2^n A_2$. Now we have to find $A_1$ and $A_2$.
We put $n=1$ in $a_n = 2a_{n-1} + 2^n$. So $a_1 = 2a_0 + 2^1=2$.
Now, let's put $n=0$ and $n=1$ in $a_n = 2^n A_1 + n\cdot 2^n A_2$. Hence
$2^0 A_1 + 0\cdot 2^0 A_2 = 0 \implies A_1 =0$
$ 2^1 A_1 + 1\cdot 2^1 A_2 = 2 \implies A_2 = 1$
Therefore we find that $a_n=n\cdot2^{n}$.
