Consider the following linear system of differential equations: { <mtable columnalign="

I think the key here is to rewrite the linear system as the following matrix equation
$\underset{A\overrightarrow{x}=\dot{\overrightarrow{x}}}{\underbrace{\left(\begin{array}{cc}0& -4\\ 1& 0\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}\dot{x}\\ \dot{y}\end{array}\right)}}.$
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
$\left(\begin{array}{c}2i\\ 1\end{array}\right)\text{and}\left(\begin{array}{c}-2i\\ 1\end{array}\right)$
respectively.
Consequently, you can write
$\left(\begin{array}{c}x(t)\\ y(t)\end{array}\right)=c_1\left(\begin{array}{c}2i\\ 1\end{array}\right)e^{2it}+c_2\left(\begin{array}{c}-2i\\ 1\end{array}\right)e^{-2it}.$
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 $A.$.

Well, defining $k\equiv c_2-i{c}_1/2$, you get a real
$\left(\begin{array}{c}x(t)\\ y(t)\end{array}\right)=k{e}^{i2t}\left(\begin{array}{c}i\\ 1/2\end{array}\right)+k^{\ast }{e}^{-i2t}\left(\begin{array}{c}-i\\ 1/2\end{array}\right).$
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 $\dot{z}=i2z$... take the c.c. ... do you see the point?