How to solve complex coupled differential equations I would like to solve the following differential equation,
$$\begin{align*}
\begin{pmatrix}
\dot{x}\\ \dot{y}
\end{pmatrix}=\begin{pmatrix}
0 & e^{i(\omega+\omega_{0})t}\\
e^{-i(\omega-\omega_{0})t} & 0
\end{pmatrix}\begin{pmatrix}
x\\ y 
\end{pmatrix}
\end{align*}$$
However I'm not entirely sure how to proceed as I've only encountered real ODEs before. Does anyone have any suggestion as to how I should proceed? thanks!
 A: Excuse-me for taking so long to answer, I wasn't really that busy after all, but I just procrastinated.
I imagine that you arrive at such equations doing physics, more precisally classical mechanics. Your equations look like the equations of motion of a system of 2 springs coupled together, damped by friction and forced by some force.
Such systems of coupled ODE's, can be reduced to an independent system of ODE's by substituting the equations into each other.
\begin{align}
\dot{x} &= e^{+i(w+w_0)t}y \\
\dot{y} &= e^{-i(w-w_0)t}x 
\end{align}
By taking the derivative of the first equation, and then substituting the second in it, we obtain
\begin{align}
\ddot{x} &= i(w+w_0)e^{+i(w+w_0)t}y + e^{+i(w+w_0)t}\dot{y} \\
         &= i(w+w_0)\dot{x} + e^{2iw_0t}x
\end{align}
Doing the same thing for the second equation, we obtain,
\begin{align}
\ddot{y} &= -i(w-w_0)e^{-i(w-w_0)t}x + e^{-i(w-w_0)t}\dot{x} \\
         &= -i(w-w_0)\dot{y} + e^{2iw_0t}y
\end{align}
which is a set of linear ODE's, which can be solved by ordinary methods.
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About the comment
What I had in mind in my comment was the following. Let suppose you have an ODE of the form
\begin{equation}
y'(x) = f(x)y(x)
\end{equation}
Then it is easy to see, that the general solution is,
\begin{equation}
y(x) = \exp(\int_{}^{x}f(x)dx) C
\end{equation}
Trying to generalize this statement to a system of ODE's lead us to think, that maybe the general solution for a system of equation of this form
\begin{equation}
\vec{y}(x)' = A(x) \vec{y}(x) 
\end{equation}
is
\begin{equation}
\vec{y}(x) = \exp(\int_{}^{x}A(x)dx)\vec{C} 
\end{equation}
(If you do not know what the exp of a matrix is, you can look here)
However, when thinking about it, I remembered that this solution may only be true if the matrix $A$ commutes with itself at different times. That is, for each $x_1, x_2$,
\begin{equation}
A(x_1)A(x_2) = A(x_2)A(x_1)
\end{equation}
which is not true for your matrix, and therefore do not apply :(.
