Exponential form of general integral of a linear system of ODE Consider the following linear system of differential equations:
$$
  \begin{cases}
\dot{x}=-4y \\
\dot{y}=x
\end{cases}
$$
where $x(t)$ and $y(t)$ are unknown real functions.
One can simply verify that the general solution is 
$$
\begin{pmatrix}
x(t) \\
y(t)
\end{pmatrix}
 = c_1
\begin{pmatrix}
\cos(2t) \\
\frac{1}{2}\sin(2t)
\end{pmatrix}
+c_2
\begin{pmatrix}
-2\sin(2t) \\
\cos(2t)
\end{pmatrix}
$$
where $c_1$ and $c_2$ are real parameters.
Question: Which is the exponential form of this expression? 
It should by something like 
$$
\begin{pmatrix}
x(t) \\
y(t)
\end{pmatrix}
 = k_1 e^{i2t}
\begin{pmatrix}
P_{11}\\
P_{21}
\end{pmatrix}
+k_2 e^{-i2t}
\begin{pmatrix}
P_{12} \\
P_{22}
\end{pmatrix}
$$
where $k_1$ and $k_2$ should (?) be complex parameters. 
 A: Well, defining $k\equiv c_2-ic_1/2$, you get a real 
$$
\begin{pmatrix}
x(t) \\
y(t)
\end{pmatrix}
 = k ~e^{i2t}
\begin{pmatrix}
i\\
1/2
\end{pmatrix}
+k^* ~e^{-i2t}
\begin{pmatrix}
-i \\
1/2
\end{pmatrix}.
$$
You did not specify anything about the form of the Ps, but you have sufficient freedom to recast these into something of your choice.
Note the problem simplifies to a triviality if you defined $z\equiv x+i2y$,
so $\dot{z}= i2~z$... take the c.c. ... do you see the point?
A: I think the key here is to rewrite the linear system as the following matrix equation
$$\underbrace{\begin{pmatrix}
0& -4\\
1 & 0
\end{pmatrix}\begin{pmatrix}
x\\
y
\end{pmatrix}=\begin{pmatrix}
\dot{x}\\
\dot{y}
\end{pmatrix}}_{A\vec{x}=\dot{\vec{x}}}.$$
The characteristic equation is $\lambda^2+4=0,$ so you're correct in using $\cos(2t)$ and $\sin(2t).$ Now, to get at a complex exponential solution, you need to find eigenvectors of $A$ with eigenvalues $-2i,2i.$ These are $$\begin{pmatrix}
2i\\
1
\end{pmatrix}\quad \text{and}\quad \begin{pmatrix}
-2i\\
1
\end{pmatrix}$$
respectively.
Consequently, you can write 
$$\begin{pmatrix}
x(t)\\
y(t)
\end{pmatrix}=c
_1\begin{pmatrix}
2i\\
1
\end{pmatrix}e^{2it}+c_2\begin{pmatrix}
-2i\\
1
\end{pmatrix}e^{-2it}.$$
It is worth noticing that the eigenvectors, are attached to their associated eigenfunctions, and that the eigenvalue of the derivative operator is identical to that of the eigenvalue under the action of $A.$
