Solve system inhomogeneous differential equations with variable coefficients Given the system of differential equations
$$\frac{d\vec{y}}{dx} = \begin{pmatrix}
        0 & 1 \\
        -1 & 0 \\
        \end{pmatrix}\vec{y} \ + \begin{pmatrix}
        sin(wx) \\
        0\\
        \end{pmatrix} \ \ \ \ (w \neq \pm1) $$
There are two questions which I can't answer. 1. How can I find the general solution? 2. How can I find the periodic solutions (in general). I've tried to solve the following system$$\frac{d\vec{y}}{dx} = \begin{pmatrix}
        0 & 1 \\
        -1 & 0 \\
        \end{pmatrix}\vec{y} $$ to find the complementary solution, which was 
$$\vec{y} = c_1\begin{pmatrix}
        cos(x) \\
        -sin(x)\\
        \end{pmatrix} \ +c_2\begin{pmatrix}
        sin(x) \\
        cos(x)\\
        \end{pmatrix} $$
So how do I proceed?
 A: Try:
$$
\vec{y} = \left( \begin{matrix} y_1 \\ y_2 \end{matrix} \right) \quad\Rightarrow \quad \dfrac{d\vec{y}}{dx} = \left( \begin{matrix} y'_1 \\ y'_2 \end{matrix} \right)$$
Thus:
$$
\left( \begin{matrix} y'_1 \\ y'_2 \end{matrix} \right) = \left( \begin{matrix} 0 & +1 \\ -1 & 0 \end{matrix} \right) \left( \begin{matrix} y_1 \\ y_2 \end{matrix} \right) + \left( \begin{matrix} \sin(wx) \\ 0 \end{matrix} \right)
\quad\Rightarrow\quad
\left\{ 
\begin{align}
 y'_1 &= y_2 + \sin(wx) \\
 y'_2 &= - y_1
\end{align}
\right.
$$
Differentiating the second equation and replacing, you can get a second order differential equation:
$$
y_1 = - y'_2 \quad\Rightarrow\quad y'_1 = - y''_2 \quad\therefore\quad y''_2 + y_2 = - \sin(wx)
$$
The general solution for $y_2$ is:
$$
\forall A,B\in\mathbb{C}: \quad y_2 = A \cos(x) + B \sin(x) + \dfrac{1}{w^2-1}\sin(wx)
$$
(Note that $w\neq\pm 1$). For $y_1$:
$$
y_1 = - y'_2 \quad\Rightarrow\quad y_1 = A \sin(x) - B \cos(x) - \dfrac{w}{w^2-1}\cos(wx)
$$
Finally, for all $A,B\in\mathbb{C}$ and $w\in\mathbb{C}$ such that $w\neq\pm 1$:
$$
\vec{y} = \left( \begin{matrix} A \sin(x) - B \cos(x) - \dfrac{w}{w^2-1}\cos(wx) \\ A \cos(x) + B \sin(x) + \dfrac{1}{w^2-1}\sin(wx) \end{matrix} \right)
$$

Added: Periodicity Analysis. By definition of periodic function:
$$
y(x)\mbox{ is periodic in T}\quad\Leftrightarrow\quad\forall k\in\mathbb{Z},\,\exists T\in\mathbb{C}: \quad \vec{y}(x) = \vec{y}(x + Tk)
$$
The function $\vec{y}$ consists of functions having different periods, $T_1$ and $T_2$. They corresponds to $\cos(x),\sin(x)$ and $\cos(xw),\sin(xw)$ functions, respectively:
$$
\begin{align}
T_1 &= 2\pi & T_2 &= \frac{2\pi}{w} \\
\end{align}
$$ 
So this means that:
$$
\forall k\in\mathbb{Z},\,\exists k1,k2\in\mathbb{Z}: \quad Tk = T_1 k_1 = T_2 k_2 \quad\Rightarrow\quad \dfrac{T_1}{T_2} = w = \dfrac{k_2}{k_1} $$
Since $k1,k2\in\mathbb{Z}$, necessarily $\dfrac{k_2}{k_1}\in\mathbb{Q}$. Then:
$$
\vec{y}(x)\mbox{ is periodic}\quad\Leftrightarrow\quad w \in\mathbb{Q}
$$
