[tex]\begin{cases}(D^2+2)x-2y=0\\-2x+(D^2+5)y=0\end{cases}[/tex]
Solve for [tex]y(t)[/tex] in the first ODE:
[tex]y=\dfrac{(D^2+2)x}2[/tex]
Substitute this into the second ODE:
[tex]-2x+(D^2+5)\dfrac{(D^2+2)x}2=0[/tex]
[tex]-4x+(D^4+7D^2+10)x=0[/tex]
[tex](D^4+7D^2+6)x=0[/tex]
This ODE is linear with characteristic equation
[tex]r^4+7r^2+6=(r^2+6)(r^2+1)=0[/tex]
with roots at [tex]r=\pm i\sqrt6[/tex] and [tex]r=\pm i[/tex], so that the characteristic solution [tex]x_c(t)[/tex] is
[tex]\boxed{x(t)=C_1\cos(\sqrt6\,t)+C_2\sin(\sqrt6\,t)+C_3\cos t+C_4\sin t}[/tex]
and from here we can solve for [tex]y(t)[/tex]:
[tex]y=\dfrac{(D^2+2)x}2[/tex]
[tex]\boxed{y(t)=-2C_1\cos(\sqrt6\,t)-2C_2\sin(\sqrt6\,t)+\dfrac{C_3}2\cos t+\dfrac{C_4}2\sin t}[/tex]
