ODE IVP: $y'-2y=e^t$, $y(0)=0$

Question

Question: $$y'-2y=e^t\quad ;\quad y(0)=0$$

When I tried to solve with integrating factor, I got $y=-e^t$, which can never satisfy the initial value condition.

Answer 1

$$y'-2y=e^t$$ $$(ye^{-2t})'=e^{-t}$$ Don't forget the constant of integration: $$ye^{-2t}=-e^{-t} +c$$ $$y(t)=-e^t+ce^{2t}$$ Apply initial condition: $$y(0)=0 \implies c=1$$ $$y(t)=e^{2t}-e^t$$