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