Differential equation of the form $y'=Ay+b(x)$ with $b(x)=(\sin{(\omega x)},0)$ I have a question regarding the following specific differential equation. 
$$y'=\left(\begin{matrix}
        0 & 1 \\
        -1 & 0 \\
        \end{matrix}\right)y+\left(\begin{matrix}
    \sin{(\omega x)} \\
    0 \\ 
\end{matrix}\right) \;\; \textrm{with}\;\; y(x_{0})=y_{0}$$
I need to show that for every value of $\omega \neq \pm1$ there is at least one periodic solution for $y$. I have found that the solution to the equation is:
$$y(x)=\left(\begin{matrix}
\cos{x} & \sin{x} \\
-\sin{x} & \cos{x} \\
\end{matrix}\right)y_{0} + \int^{x}_{x_{0}}\left(\begin{matrix}
\cos{(x-\xi)}\sin{(\omega \xi}) \\
-\sin{(x-\xi)}\sin{(\omega \xi)} \\
\end{matrix}\right)d\xi$$ 
I have a hard time evaluating this integral, but I believe it should be possible to show that for every value $\omega \neq \pm1$ there is at least one periodic solution without evaluating the integral. I fail to see how any irrational number for $\omega$ should give a periodic solution. 
Also how would one show that all solutions are periodic if and only if $\omega \in \mathbb{Q}$?

Let $\zeta =y_{1}+iy_{2}$ and $p=-i\partial _{x}$. Then
$$p\zeta (x)=-\zeta (x)-i\sin \omega x.$$
Next we note that
$$
(\exp [iap]f)(x)=f(x+a).
$$
Thus
$$
(\exp [iap]p\zeta )(x)=p\zeta (x+a)=-\zeta (x+a)-i\sin \omega (x+a)
$$
Assume $a=\frac{2\pi n}{\omega }$. Then
$$
\sin \omega (x+a)=\sin \omega (x+\frac{2\pi n}{\omega })=\sin \omega x,
$$
so $\zeta (x)$ and $\zeta (x+\frac{2\pi n}{\omega })$ satify the same
equation. In case there is a single solution then $\zeta (x)$ is periodic
with period $\frac{2\pi }{\omega }$. Assume that there are two solutions $%
\zeta _{1}(x)$ and $\zeta _{2}(x)$. Then their difference $\Delta \zeta (x)$
satisfies the homogeneous equation
$$
p\Delta \zeta (x)=-\Delta \zeta (x),\;\Delta \zeta (x_{0})=0,
$$
which only has the zero solution.
