Answer
$y(t)=C_{1}e^{t}+C_{2}te^{t}+C_{3}e^{-t}+\frac{1}{2}te^{-t}+3$
Work Step by Step
Let $\;\;\;\;\;y=e^{rt}\\\\$
${y}'''-{y}''-{y}'+y=0 \;\;\;\;\Rightarrow \;\;\;\; r^3e^{rt}-r^2e^{rt}-re^{rt}+e^{rt}=0\\\\$
$r^3-r^2-r+1=(r-1)(r^2-1)=0 $
$ \rightarrow\;\;\;\;\; r_{1},r_{2}=1\;\;\;\;\;or\;\;\;r_{3}=-1\;\;\;\;\;\\\\$
So the 3 roots are: $\;\;\;\;r_{1},r_{2}=1 \;\;\;,\;\;r_{3}=-1$
$\boxed{y_{c}(t)= C_{1}e^{t}+C_{2}te^{t}+C_{3}e^{-t}}$
Let; $\;\;\;\;Y(t)=Ate^{-t}+B$
${Y}'=Ae^{-t}-Ate^{-t}$
${Y}''=-2Ae^{-t}+Ate^{-t}$
${Y}'''=3Ae^{-t}-Ate^{-t}$
${Y}'''-{Y}''-{Y}'+Y=2e^{-t}+3$
$3Ae^{-t}-Ate^{-t}+2Ae^{-t}-Ate^{-t}-Ae^{-t}+Ate^{-t}+Ate^{-t}+B=2e^{-t}+3$
$4Ae^{-t}+B=2e^{-t}+3 \;\;\;\;\;\;\Rightarrow \;\;\;A=\frac{1}{2}\;\;\;,\;\;\;B=3$
$\boxed{Y(t)=\frac{1}{2}te^{-t}+3}$
The general solution :
$y(t)=y_{c}(t)+Y(t)$
$y(t)=C_{1}e^{t}+C_{2}te^{t}+C_{3}e^{-t}+\frac{1}{2}te^{-t}+3$
