How to find the general solution of $ \frac{d^2 y}{dt^2} + \frac{dy}{dt} - 6y = e^{2t} $? How to find the general solution of $$ \frac{d^2 y}{dt^2} + \frac{dy}{dt} - 6y = e^{2t} $$
My solution:
I solve for the auxiliary equation which is $ m^2 + m -6 = 0$ so that the complementary equation $y_c$ is $C_1 e^{2x} + C_2 e^{-3x} $ where $C_1$ and $C_2$ are arbitrary constants. And the particular equation is $$y_p = 0 $$ since if we let $y_p= Ae^{2t} $, take its first and second derivative, and substitute the result in our original equation, we we will get that $A=0$
Therefore, my answer is $y(t) = C_1 e^{2x} + C_2 e^{-3x} + 0 $. Is this correct?
 A: Not quite. You got the Homogenous solution correctly, but the particular solution doesn't work out. And if you substitute $y=0$ into the differential equation, you'll see it's not a solution.
When using the method of undetermined coefficients, you have to make sure that the form you're using for the particular solution is linearly independent of the homogenous solutions. That means, in practice, you find the form you'd usually use, and if it looks like the homogenous solution, you can just multiply it by a $t$. So, instead, you want to look at a particular solution that looks like $$ y_p = Ate^{2t}$$
Unlike if your forcing function had the $te^{2t}$, though, you don't have to add a constant times $e^{2t}$ term, which you'd get from differentiating your particular solution since that constant will be absorbed into the homogenous solution term.
Anyway, plugging that form into the ODE gives $$ 4A(t+1)e^{2t} + A(2t+1)e^{2t} - 6Ate^{2t} = e^{2t}$$ The $te^{2t}$ terms cancel out, and so your left with $$ 4Ae^{2t} + Ae^{2t} = 5Ae^{2t} = e^{2t}$$ so that $A = \frac{1}{5}$. Hence, the particular solution is $$ y_p = \frac{1}{5}te^{2t}$$ and the solution to the ODE is $$ \boxed{y(t) = \frac{1}{5}te^{2t} + C_1e^{2t} + C_2e^{-3t}} $$
