Answer
$y(t)=C_{1}+C_{2}t+C_{3}t^2+C_{4}e^{-t}+\frac{1}{40}cos(2t)+\frac{1}{20}sin(2t)$
Work Step by Step
Let $\;\;\;\;\;y=e^{rt}\\\\$
${y}^{(4)}+{y}'''=0 \;\;\;\;\Rightarrow \;\;\;\; r^4e^{rt}+r^3e^{rt}=0\\\\$
$r^4+r^3=r^3(r+1)=0 $
$ \rightarrow\;\;\;\;\; r_{1},r_{2},r_{3}=0\;\;\;\;\;\;\;or\;\;\;\;,r_{4}=-1\;\;\;\;\\\\$
$\boxed{y_{c}(t)= C_{1}+C_{2}t+C_{3}t^2+C_{4}e^{-t}}$
Let; $\;\;\;\;Y(t)=Acos(2t)+Bsin(2t)$
${Y}'=-2Asin(2t)+2Bcos(2t)$
${Y}''=-4Acos(2t)-4Bsin(2t)$
${Y}'''=8Asin(2t)-8Bcos(2t)$
${Y}^{(4)}=16Acos(2t)+16Bsin(2t)$
${y}^{(4)}+{y}'''=sin(2t)$
$16Acos(2t)+16Bsin(2t)+8Asin(2t)-8Bcos(2t)=sin(2t)$
$(16A-8B)cos(2t)+(16B+8A)sin(2t)=sin(2t) \;\;\;\;\;\;$$\Rightarrow \;\;\;16A-8B=0\;\;\;,\;\;\;16B+8A=1\;\;\;,\;\;\;\Rightarrow \;\;\;\;\;A=\frac{1}{40}\;\;\;\;\;B=\frac{1}{20}$
$\boxed{Y(t)=\frac{1}{40}cos(2t)+\frac{1}{20}sin(2t)}$
The general solution :
$y(t)=y_{c}(t)+Y(t)$
$y(t)=C_{1}+C_{2}t+C_{3}t^2+C_{4}e^{-t}+\frac{1}{40}cos(2t)+\frac{1}{20}sin(2t)$
