Differential Equation, answers don't match I am new to differential  equations, to solve $y'+2y=3e^t$ I used the method of variation of constant and I get $y=e^t+Ce^{-2t}$ but when I use another method I get a different answer (I don't know the name of this method)
$(y'+2y=3e^t)*I(t) \Rightarrow Iy'+2yI=3Ie^t$
$(Iy)'=I'y+Iy'=Iy'+2yI \Rightarrow I'y=2yI \Rightarrow I'=2I$
$I=Ce^2t \Rightarrow \int(Iy)'=\int 3Ie^t \Rightarrow ye^{2t}=e^{3t} \Rightarrow y=e^t$
$y=e^t+Ce^{2t} ≠ y=e^t+Ce^{-2t}$
Please Correct me
 A: You essentially had it. It's correct that $I$ (the integrating factor, a slightly different but essentially equivalent method) is equal to $Ce^{2t}$. The arbitrary constant would just get cancelled (since it can be moved out of the derivative and exists on both sides), so we can ignore it. We then have
$$(ye^{2t})'=3e^{3t}$$
$$ye^{2t}=e^{3t}+C$$
Don't forget the constant of integration! This is the arbitrary constant you care about that will show up in the final solution. Just isolate for $y$:
$$y=e^{t}+Ce^{-2t}$$
Forgetting constants of integration is a common pitfall when solving ODEs. If something bizarre like this shows up again, in my experience "did I forget the constant?" is one of the first things you should be asking yourself.
A: The method you are trying to use here is called the integrationg factor method:
http://en.wikipedia.org/wiki/Integrating_factor
In this case, the integrating factor can be
$$
I(t)=\exp\left( \int_0^t2ds\right)=e^{2t}.
$$
You got this right.
Then multiply the ode by $I(t)$:
$$e^{2t}y'+2e^{2t}y=3e^{3t}\quad\Leftrightarrow\quad (e^{2t}y)'=3e^{3t}.
$$
Now integrate and don't forget the integration constant:
$$
e^{2t}y(t)=e^{3t}+C \quad\mbox{hence}\quad y(t)=e^t+Ce^{-2t}.
$$
Alternative: the method of undetermined coefficients.
Given the rhs, we know we can look for a particular solution of the form
$$
y_p(t)=Ce^t.
$$
Plugging this into the ode, we find that for $C=1$, we do get a solution:
$$
y_p(t)=e^t.
$$
Then add the general solution of the homogeneous equation $y_h(t)=Ae^{-2t}$ to get
$$
y(t)=Ae^{-2t}+e^t
$$
the general solution of the ode.
