Recurrence Relations With Exponents

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Solve the following:

$$a_n = 2a_{n-1} + 2^{n-1} , a_0 = 3$$

Workings:

$a_n = 2a_{n-1} + \frac{1}{2} 2^n, a_o = 3$

$a_n^{(h)} = 2a_{n-1}$

The characteristic equation is:

$ch(x) = (x-2)$

$a_n^{(h)} = c(2)^n$

We have a particular solution of the form cn(2)^n.

$cn2^n = c(2)^{n-1} + 2^n$

$2cn = c + 2$

Now I'm not too sure if what I did so far is correct and I'm not sure on what to do next.

Any help will be appreciated.

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To make your life easier look at $$ a_{k+1} = 2 a_k +2 ^k $$ Divide through $2^k$ to get $b_{k+1} = \frac{a_{k+1}}{2^k}$. You have $$ b_{k+1} = b_k + 1 \\ b_k = b_{k-1} + 1\\ \Delta b_{k+1} = \Delta b_k = \ldots = \Delta b_1\\ $$ Now sum on both sides (telescoping sum on LHS) and plug in back expression for $a_n$.