Answer
$y= [\;C_{1} +tC_{2}\;] cos(t)+ \;[C_{3} +tC_{4}\; ]sin(t)\;$
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=(r-i)^2(r+i)^2=0 $$ \rightarrow\;\;\;\; r_{1},r_{2}= i\;\;\;or\;\;\;r_{3},r_{4}=- i \;\;\;\;\;\;\\\\$
So the 8 roots are: $\;\;\;r_{1},r_{2}=\pm i \;\;\;,\;\;\;r_{3},r_{4}=\pm i $
The general solution for complex roots is:
$y= C_{1}e^{\alpha t}cos(\beta t)+C_{2}e^{\alpha t}sin(\beta t)\\$
$y= [\;C_{1}cos(t)+C_{2} sin(t)\;]+ t[\;C_{3}cos(t)+C_{4} sin(t)\;]$
$y= [\;C_{1} +tC_{2}\;] cos(t)+ \;[C_{3} +tC_{4}\; ]sin(t)\;$
