Guess the particular solution to an exponential function`? Solve this differential equation $y''+2y'+y = e^{-t}$.

I got the homogenous solution to be

$y_h= (Bt + C)e^{-t}$

But I don't know what my guess to the particular function should be?
$Ae^{-t}$ solves the homogenous function so it can't be it. Then my
next try was $Ate^{-t}$, but this is what I got:

$y = Ate^{-t}$
$y'=A(e^{-t} - t^2 e^{-t})$
$y'' = A(-te^{-t} - 2te^{-t} +t^3e^{-t})$

Then I when I put it in the equation I get.

$Ae^{-t}(t^3-2t^2-2t+2)=e^{-t}$

But I have two variables and therefore I can't solve it. I need help.

 A: $$
y''+2y'+y = e^{-t}.
$$
Characteristic equation relatively of the provided differential equation will be:
$$
e^{rt}\left(r^{2}+2r+1\right)=0,
\\
\left(r+1\right)^{2}=0,
\\
r=-1,-1.
$$
So our general solution will be a linear combination of our homogeneous solutions of our characteristic roots $r_1$ and $r_2$:
$$
y(t)=C_{1}e^{-t}+C_{2}te^{-t}.
$$
The particular solution will be:
$$
Ae^{-t}-Ae^{-t}+Ae^{-t}=e^{-t},
\\
Ae^{-t}=e^{-t},\implies{A=1}.
$$
The total solution will be:
$$
y(t)=C_{1}e^{-t}+C_{2}te^{-t}+1.
$$
A: You wrote correctly that the homogeneous solution is given by $y_{h}(t)=(Bt+C)e^{-t}$. Perhaps, it should be better write as $y_{h}(t)=Ce^{-t}+\color{blue}{Bte^{-t}}$. You said that "$Ae^{-t}$ sove the homogeneous so it can't be it" as assumption for the particular solution and it is correct argument. However, the next step it is wrong, because when you try with $\color{blue}{Ate^{-t}}$, it is already in the homogeneous solution set (I painted it blue) [c.f: see here for more details], so you can't try that one, so we try with $y_{p}(t)=At^{2}e^{-t}$ and noticed that we don't have that solution in the set of homogeneous solutions. If you try $y_p$, you will find $y_{p}(t)=\frac{1}{2}t^{2}e^{-t}$. Therefore, the general solution is given by
$$y(t)=Ce^{-t}+Bte^{-t}+\frac{1}{2}t^2e^{-t}.$$
Note: To find $A$ in $y_{p}(t)=At^2e^{-t}$ noticed that
$$y_{p}=At^2e^{-t}$$
$$y_{p}'=-Ae^{-t}t^2+2Ate^{-t}$$
$$y_{p}''=2Ae^{-t}+At^{2}e^{-t}-4Ate^{-t}$$
Substitution in $y_{p}''+2y_{p}'+y_{p}=e^{-t}$ and then simplify give
$$2Ae^{-t}=e^{-t}\implies A=\frac{1}{2}.$$
A: $$y''+2y'+y = e^{-t}$$
$$(ye^{t})''=1$$
$$z''=1$$
Integrate twice . Or try $z_p=A$ this won't work (it gives $0=1$). $z_p=At$ won't work either ( you also get $0=1$) so you have to try $z_p=At^2$ ( you end with $2A=1$ this can be solved)..
