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