Answer
$y=e^{2t} +e^{t}$
Work Step by Step
We are given that the differential equation : at $ \dfrac{dy}{dt}-y=e^{2t}$
and the general solution can be written as:
$y=e^{-(-1)t} \int e^{2t} e^{(-1)t} \ dt $
Integrate to obtain:
$y=e^t \int e^{t} \ dt$
This implies that $y=e^{t}(e^{t}+C) \implies y=e^{2t}+Ce^t$
After applying the initial conditions, $y=2$ when $t=0$, we get $C=1$
Therefore, we have: $y=e^{2t} +e^{t}$
