Answer
$y(t)=C_{1}+C_{2}t+C_{3}e^{2t}+C_{4}e^{-2t}+\frac{-1}{48}t^4+\frac{-1}{16}t^2+\frac{-1}{3}e^{t}$
Work Step by Step
Let $\;\;\;\;\;y=e^{rt}\\\\$
${y}^{(4)}-4{y}''=0 \;\;\;\;\Rightarrow \;\;\;\; r^4e^{rt}-4r^2e^{rt}=0\\\\$
$r^4-4r^2=r^2(r^2-4)=0 $
$ \rightarrow\;\;\;\;\; r_{1},r_{2}=0\;\;\;\;\;\;\;or\;\;\;r_{3}=2\;\;,\;\;r_{3}=-2\;\;\;\;\;\\\\$
$\boxed{y_{c}(t)= C_{1}+C_{2}t+C_{3}e^{2t}+C_{4}e^{-2t}}$
Let; $\;\;\;\;Y(t)=At^4+Bt^3+Ct^2+De^{t}$
${Y}'=4At^3+3Bt^2+2Ct+De^{t}$
${Y}''=12At^2+6Bt+2C+De^{t}$
${Y}'''=24At+6B+De^{t}$
${Y}^{(4)}=24A+De^{t}$
${Y}^{(4)}-4{Y}''=t^2+e^t$
$24A+De^{t}-48At^2-24Bt-8C-4De^{t}=t^2+e^t$
$-3De^{t}-48At^2-24Bt+(24A-8C)=t^2+e^t \;\;\;\;\;\;\Rightarrow \;\;\;A=\frac{-1}{48}\;\;\;,\;\;\;B=0\;\;\;,\;\;\;C=\frac{-1}{16}\;\;\;,\;\;\;D=\frac{-1}{3}$
$\boxed{Y(t)=cos(t)}$
The general solution :
$y(t)=\frac{-1}{48}t^4+\frac{-1}{16}t^2+\frac{-1}{3}e^{t}$
$y(t)=C_{1}+C_{2}t+C_{3}e^{2t}+C_{4}e^{-2t}+\frac{-1}{48}t^4+\frac{-1}{16}t^2+\frac{-1}{3}e^{t}$
