Answer
${Y(t)=Ae^t+(Bt+C)e^{-t}+te^{-t}(Dcos(t)+Esin(t))}$
Work Step by Step
Let $\;\;\;\;\;y=e^{rt}\\\\$
$y^{(4)}+2{y}'''+2{y}''=0 \;\;\;\;\Rightarrow \;\;\;\; r^4e^{rt}-r^3e^{rt}-r^2e^{rt}-e^{rt}=0\\\\$
$r^4-r^3-r^2-1=r^2(r^2+2r+1)=0 $
$ \rightarrow\;\;\;\;\; r_{1,2}=0\;\;\;\;\;\;\;or\;\;\;\;,r_{3,4}=-1\pm i\;\;\;\;\;\;\;\;\;\\\\$
$\boxed{y_{c}(t)= C_{1}+C_{2}t+C_{3}e^{-t}cos(t)+C_{4}e^{-t}sin(t)}$
$\;\;\;\;g=Ae^t+(Bt+C)e^{-t}+e^{-t}(Dcos(t)+Esin(t))$
$g_{1}=e^{-t}(Dcos(t)+Esin(t))\;\;\;\;\;$multiply this equation by $t$
$g_{1}=te^{-t}(Dcos(t)+Esin(t))$
$g_{2}=(Bt+C)e^{-t}$
$g_{3}=Ae^t\;\;\;\;\;$
$Y(t)=g_{1}+g_{2}+g_{3}$
$\boxed{Y(t)=Ae^t+(Bt+C)e^{-t}+te^{-t}(Dcos(t)+Esin(t))}$
