Question

Need help guessing the particular solution form for y'''+y''-2y=te^(-t)+1

Answer 1

First the homogenous solution: You have the characteristic equation,
\[r^3+r^2-2=0\]
which has roots \(r=1, -1\pm i\).

The complex roots' solutions are of no consequence to your guess, since they contain factors of sine/cosine, and \(r=1\) contributes a \(e^t\), which also won't affect your guess solution.

I'd try \(y_p=Ate^{-t}+Be^{-t}+Ct+D\).
\[y'=(A-B)e^{-t}-Ate^{-t}+C\\
y''=(B-2A)e^{-t}+Ate^{-t}\\
y'''=(3A-B)e^{-t}-Ate^{-t}\]
\[\left[(3A-B)e^{-t}-Ate^{-t}\right]+\left[(B-2A)e^{-t}+Ate^{-t}\right]\\
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-2\left[Ate^{-t}+Be^{-t}+Ct+D\right]=te^{-t}+1\\
(A-2B)e^{-t}-2Ate^{-t}-2Ct-2D=te^{-t}+1\]
Matching up coefficients, you have
\[\begin{cases}A-2B=0\\
-2A=1\\
-2C=0\\
-2D=1\end{cases}\]
I've checked the answer, it all works out. Note that you can disregard the \(Ct\) part of the guess, I only used it to be safe.

Answer 2

Thanks a lot man!