As a part of an H.W assignment, I'm requested to solve the following O.D.E using perturbation theory and compare to the analytical solution
$$\dot{y}=y+\epsilon$$
where the solution of this equation is given by
$$y(t)=Ae^t-\epsilon$$
Using perturbation theory I assume that
$$ y(t)=\sum_{i=0}^{\infty}y_i(t)\epsilon^i$$
by plugging in I obtain only term up to order of $O(\epsilon)$
$$\dot{y}_0+\epsilon \dot{y}_1 =y_0+\epsilon(y_1+1)$$
therefore
$$y_0(t)=Ae^t$$
$$y_1(t)=Ae^t-1$$
By plugging into $y(t)$ we obtain
$$y(t)=Ae^t+\epsilon(Ae^t-1+1)$$
and therefore the solution I get is
$$y(t)=Ae^t(1+\epsilon)$$
Can anyone point out why the solution in both cases is different, I've done this many times and can't figure out if this is a mistake using the perturbation method or something which is part of the problem.
Best Answer
You get $y_0(t)=A_0e^t$ and $y_1(t)=A_1e^t-1$. If you combine the solutions, you get $$ y(t)=y_0(t)+ϵy_1(t)=(A_0+ϵA_1)e^t-ϵ $$ The combination of the constants is just a new constant, so that this solution is already the exact solution.