Here's the question you clicked on:
Kaydian
3d2y/dt2-dy/dt +2y=e-3t Differential Equation
\[\frac{d^2 y}{dt^2}-\frac{1}{3} \frac{dy}{dt}+\frac{2 y}{3}=\frac{e^{-3t}}{3}\] Solve the homogenous part (=0). Then try a solution of the form: \[y_p(t)=Ae^{-3t}\] And differentiate and plug in. Be careful that your homogenous solution doesn't have a root of 3 or else you need something like: \[y_p(t)=Ate^{-3t}\]
the characteristic root of the homogeneous DE are complex numbers, different from the characteristic root of the RHS. writing the LHS as \(f(D)y\) you can determine right away the particular solution \[y_p=\frac{e^{-3t}}{f(-3)}\]
minor correction:\[\Huge y_p=\frac{\frac{e^{-3t}}{3}}{f(-3)}\]