I am having some trouble solving this particular equation: $y_{t+1}-2y_t=2^t$
Which brought me to realize that I don't even understand how to solve the general case of a difference equation $y_{t+1}-ay_t=b(t)$, where b is some function of t itself. So far my math course has only covered the cases where b is either a constant or equal to zero. I have tried to look online as to what the process for this is, but have only encountered videos/documents covering the two afformentioned cases. I would greatly appreciate some assistance or guidance as to what the process for solving such an equation is.
I am given no initial conidtions, however when I asked my professor he told me to use the stationary solution of the form $at2^t$, but I'm not quite sure what he means by this.
My apologies in advance if this is a stupid question, or if there are any formating errors present.
Considering the recurrence
$$ y(k+1)-2y(k) = 2^k $$
due to linearity, the solution $y(k)$ can be represented as $y(k) = y_h(k)+y_p(k)$ where
$$ \cases{ y_h(k+1) - 2y_h(k) = 0\\ y_p(k+1) - 2y_p(k) = 2^k } $$
The solution for $y_h(k)$ is $y_h(k) = 2^kc_0$. Now proposing $y_p(k) = 2^kc_0(k)$ after substitution we get
$$ 2^{k+1}c_0(k+1)-2\cdot 2^kc_0(k) = 2^k $$
and we have now the recurrence
$$ c_0(k+1)-c_0(k) = \frac 12 $$
with particular solution $c_0(k) = \frac k2$ and finally
$$ y(k) = 2^kc_0 + 2^k \frac k2 $$