[Math] Particular solution of $y” -3y’ + 2y = e^t$

Question

I'm trying to find a particular solution of $$y'' -3y' + 2y = e^t$$

My fundamental set is: $$y_1 = e^{2t}\\y_2 = e^t$$

So I chose $y_p = A t e^t$, which gives me:$$y_p' = Ae^t + Ate^t\\y_p'' = 2Ae^t + Ate^t$$

but when I substitute this into the equation $$y'' -3y' + 2y = te^t$$ I get $A = -t$, and I have trouble understanding what is going wrong.

Answer 1

You have changed the right-hand side of the equation to $te^t$ instead of $e^t$.