For sure this kind of stuff has a name, but I can't remember.

So as to better understand what I mean, let's get concrete:

Consider the linear ODE: $$y''+4\,y = 0$$

the characteristic polynomial has complex solutions: $$\lambda^2+4 = 0 \quad \Rightarrow\quad\lambda\_{1/2} = \pm 2\,i$$

Now I just took it for granted that u can express such solutions as $$y = \mathrm{C\_1}\,\cos(2\,x)+\mathrm{C\_2}\,\sin(2\,x)$$

But if I were to plug the complex into the usual combination of exponentials:
$$\mathrm{C\_1}\,e^{+2\,i}+\mathrm{C\_2}\,e^{-2\,i} = \\\\ \mathrm{C\_1}\,\cos(2\,x)+\mathrm{C\_2}\,\cos(-2\,x)+\mathrm{C\_1}\,i\,\sin(2\,x)+\mathrm{C\_2}\,i\,\sin(-2\,x)$$

that doesn't look to me as close as the solution.

Furthermore to add even more confusion solving the ODE with MATLAB yields:
$$y = \mathrm{C\_1}+\mathrm{C\_2}\,e^{-4\,x}$$ **Apologies**: this is the solution of $y''+4\,y'= 0$

As I mentioned I didn't really keep myself busy with the complex part of differential equation. I hope u might explain to me how all these solutions come together.