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.
Best Answer
There are various typos in your question and in what you entered into MATLAB. IF we're talking about $y''+4y=0$:
Yes, the general solution is $c_1e^{2it}+c_2e^{-2it}$. Do what you did and then also collect common terms: $$c_1e^{2it}+c_2e^{-2it}=(c_1+c_2)\cos(2t)+(ic_1-ic_2)\sin(2t)=d_1\cos(2t)+d_2\sin(2t).$$