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.