Answer
$y(t)=C_{1}cos(t)+C_{2}sin(t)+C_{3}tcos(t)+C_{4}tsin(t)+3+\frac{1}{9}cos(2t)$
Work Step by Step
Let $\;\;\;\;\;y=e^{rt}\\\\$
${y}^{(4)}+2{y}''+y=0 \;\;\;\;\Rightarrow \;\;\;\; r^4e^{rt}+2r^2e^{rt}+e^{rt}=0\\\\$
$r^4+2r^2+1=(r^2+1)^2=0 $
$ \rightarrow\;\;\;\;\; r_{1},r_{2}=i\;\;\;\;\;\;\;or\;\;\;r_{3},r_{4}=-i\;\;\;\;\;\\\\$
$\boxed{y_{c}(t)= C_{1}cos(t)+C_{2}sin(t)+C_{3}tcos(t)+C_{4}tsin(t)}$
Let; $\;\;\;\;Y(t)=A+Bcos(2t)+Csin(2t)$
${Y}'=-2Btsin(2t)+2Ccos(2t)$
${Y}''=-4Bcos(2t)-4Csin(2t)$
${Y}'''=8Btsin(2t)-8Ccos(2t)$
${Y}^{(4)}=16Bcos(2t)+16Csin(2t)$
${y}^{(4)}+2{y}''+y=3+cos(2t)$
$16Bcos(2t)+16Csin(2t)-8Bcos(2t)-8Csin(2t)+A+Bcos(2t)+Csin(2t)=3+cos(2t)$
$9Bcos(2t)+9Csin(2t)+A=3+cos(2t) \;\;\;\;\;\;\Rightarrow \;\;\;A=3\;\;\;,\;\;\;C=0\;\;\;,\;\;\;B=\frac{1}{9}\;\;\;\;\;\;$
$\boxed{Y(t)=3+\frac{1}{9}cos(2t)}$
The general solution :
$y(t)=y_{c}(t)+Y(t)$
$y(t)=C_{1}cos(t)+C_{2}sin(t)+C_{3}tcos(t)+C_{4}tsin(t)+3+\frac{1}{9}cos(2t)$
