Behold the following system of differential equations:
$\dfrac{dy}{dx}=\begin{pmatrix}0&1\\-1&0 \end{pmatrix}y + \begin{pmatrix}\sin(\omega x)\\0 \end{pmatrix} , (\omega \neq \pm 1)$
a) Give the general solution.
b) Show that for all values of $\omega$ there is at least one periodic solution.
I found the solution for the following:
$\dfrac{dy}{dx}= \begin{pmatrix}0&1\\-1&0 \end{pmatrix}y\\ y_p(x)=\begin{pmatrix}\cos(x) & \sin(x)\\ - \sin(x) & \cos(x) \end{pmatrix}$
But then I don't know what to do next
The $\begin{pmatrix} \sin(\omega x)\\0\end{pmatrix}$ tells you that the solution is a column vector. Let's express it as $\begin{pmatrix} y_1\\y_2\end{pmatrix}$ and try to find the results for $y_1$ and $y_2$. If you expand the expression with the matrix, you find 2 equations:
$\displaystyle \frac{dy_1}{dx} = y_2 + \sin(\omega x)$,
$\displaystyle \frac{dy_2}{dx} = -y_1$.
If you derive wrt $x$ on both equations, you get
$\displaystyle \frac{d^2y_1}{dx} = \frac{dy_2}{dx} + \omega \cos(\omega x) = -y_1 + \omega \cos(\omega x)$,
$\displaystyle \frac{d^2y_2}{dx} = -\frac{dy_1}{dx} = -y_2 - \sin(\omega x)$.
Rearranging these expressions,
$\displaystyle \frac{d^2y_1}{dx} + y_1 = \omega \cos(\omega x)$,
$\displaystyle \frac{d^2y_2}{dx} + y_2 = - \sin(\omega x)$.
For $y_1$, the solution of the homogeneou equation is
$y_1^h = A \sin(x) + B \cos(x)$
A particular solution of the inhomogeneous equation has the form
$y_1^p = \kappa_1 \cos(\omega x)$.
You can find that $\kappa_1$ must be
$\displaystyle \kappa_1 = \frac{\omega}{1-\omega^2}$.
Therefore
$\displaystyle y_1 = A \sin(x) + B \cos(x) + \frac{\omega}{1-\omega^2} \cos(\omega x)$
To obtain $y_2$ you only need to use one of the first equations in this answer to find that
$\displaystyle y_2 = \frac{dy_1}{dx} - \sin(\omega x) = A \cos(x) - B \sin(x) - \frac{1}{1-\omega^2} \sin(\omega x)$