Find General Solution:
$y''+2y'+y=2e^{-t}$
$r^2+2r+r=0$
$(r+1)(r+1)=0$
$r=-1,-1$
$y=c_{1}e^{-t}+c_2te^{-t}$
$y_{p}(t)=Ae^{-t} \Rightarrow Ate^{-t} \Rightarrow At^2e^{-t}$
$ =$
$y_{p}'(t)=A2te^{-t}+At^2e^{-t}$
$y_{p}''(t)=2Ae^{-t}+2Ate^{-t}+2At-e^{-t}+At^2e^{-t}$
$y_{p}''(t)=2Ae^{-t}-e^t2At-e^{-t}2At+At^2e^{-t}$
$y_{p}''(t)=2Ae^{-t}-4Ae^{-t}+At^2e^{-t}$
$Y=2Ae^{-t}-4Ate^{-t}+At^2e^{-t}+2(A2te^{-t}+At^2(-1)(e^{-t}))+At^2e^{-t}$
$Y=2Ae^{-t}-4Ate^{-t}+At^2e^{-t}+4Ate^{-t}-2At^2e^{-t}+At^2e^{-t}=2e^{-t}$
$=2Ae^{-t}=2e^{-t}$
$A=1$
$Y=C_1e^{-t}+C_2te^{-t}+t^2e^{-t}$