Undetermined Coefficients for solving non homogeneous equation I can't seem to figure out where I am going wrong in my steps. I checked the answer and it is different.
The question is:
$$y'' + 2y' - 3y = 3te^t$$
The roots are: -3,1.
Thus the general solution is:
$$y=C_1e^{-3t} + C_2e^t$$
The particular solution i am going with is:
$$y_p = Ate^t$$
$$y'_p = Ae^t + Ate^t$$
$$y''_p = 2Ae^t + Ate^t$$
Therefore:
$$2Ae^t + Ate^t + 2Ae^t + 2Ate^t - 3Ate^t = 3te^t$$
$$4Ae^t = 3te^t$$
Then solving for A:
$$4A=3$$
$$A=3/4$$
Thus, $y_p = \frac{3}{4}(e^t + te^t)$
Then the general solution would be:
$y = C_1e^{-3t} + C_2e^t  + \frac{3}{4}(e^t + te^t)$
Any guidance with my mistake would be greatly appreciated. As an aside, what does it mean when a question asks to use the stability
result to determine they will have a globally stable solution of the above question. Thank you.
 A: Your particular solution is not correct. Note that $4Ae^t = 3te^t$ does not imply that $A=3/4$. Moreover, what is $C$?
In the characteristic polynomial, the multiplicity of the root ${\bf 1}$ is $m=1$ . Therefore, since $f(t)=te^{{\bf1}\cdot t}$, it follows that the particular solution should have the form
$$y_p=t^m(At+B)e^{{\bf1}\cdot t}=(At^2+Bt)e^t$$
where $A$ and $B$ are real constants to be found. Can you take it from here?
A: The particular solution should be:
$$y_p=(At^2+Bt)e^t \implies y'_p=(At^2+(2A+B)t+B)e^t$$
Better approach
Substitute $y=ze^t$ The equation becomes
$$z''+4z'=3t$$
Use the variation of constant now $(y_p=at^2+bt)$..It looks easier..
Another approach
$$y'' + 2y' - 3y = 3te^t$$
$$(y''-y')+3(y'-y)=3te^t$$
$$(y'e^{-t})'+3(ye^{-t})'=3t$$
$$(y'e^{-t})+3(ye^{-t})=\frac 32t^2+K_1$$
$$(y'e^{3t})+3(ye^{3t})=(\frac 32t^2+K_1)e^{4t}$$
$$(ye^{3t})'=(\frac 32t^2+K_1)e^{4t}$$
$$ye^{3t}=\int (\frac 32t^2+K_1)e^{4t}dt$$
$$y(t)=K_2e^{-3t}+K_1e^t+\frac 32e^{-3t}\int t^2e^{4t}dt$$
Finally
$$\boxed{y(t)=K_2e^{-3t}+K_1e^t+\frac 38e^tt^2-\frac 3{16}e^tt}$$
where
$$
\begin{cases}
\displaystyle y_p=\frac 38e^tt(t-\frac 1{2}) \\
\displaystyle y_h=K_2e^{-3t}+K_1e^t
\end{cases}
$$
