I was studying the Zeeman effect when we arrived at the system of ODE: $$x''+\omega ^2 x + eBy'/m=0$$ $$y''+\omega ^2 y - eBx'/m=0$$ And the solution was given as: $$u(t)=x(t)+iy(t)=u_0 \cdot Exp(-it(\omega-eB/2m))$$ $$v(t)=x(t)-iy(t)=v_0 \cdot Exp(-it(\omega+eB/2m))$$ The solution does indeed check out, but I haven't understood how we get to the solution using that substitution. It was said that "the solution is straightforward with that substitution", but after some algebra I get to: $$u''+\omega ^2 u + ieBv'/m=0$$ $$v''+\omega ^2 v - ieBu'/m=0$$ which are pretty much the equations of the initial ODE, so I don't see how to solve the system "straightforwardly" using that substitution.
Also, trying to find back $x(t)$ and $y(t)$ we can use: $$x(t)= Re(u(t))= Re(u_0)\cdot cos(t(\omega-eB/2m))+Im(u_0)\cdot sin(t(\omega-eB/2m))$$ or $$x(t)= Re(v(t))= Re(v_0)\cdot cos(t(\omega+eB/2m))+Im(v_0)\cdot sin(t(\omega+eB/2m))$$ And those things are oscillating at different frequencies, so I'm starting to get really confused here.
Let $k=eB/m$. We have the coupled ODE \begin{align}x''+\omega^2x+ky'&=0\tag1\\y''+\omega^2y-kx'&=0\tag2.\end{align} Note $(1)+i\times(2)$ gives $(x+iy)''+\omega^2(x+iy)-ik(x+iy)'=0$ so we have the one-variable ODE $$u''-iku'+\omega^2u=0.$$ This can readily be solved for $u(x,y)$ and the solutions $x(t)$, $y(t)$ can be determined by equating real and imaginary parts.
The full solution is $$u(t)=e^{ikt/2}\left(Ce^{-it\sqrt{(k/2)^2+\omega^2}}+De^{it\sqrt{(k/2)^2+\omega^2}}\right)$$ so $$x(t)=C\cos\left(t(K-\sqrt{K^2+\omega^2})\right)+D\cos\left(t(K+\sqrt{K^2+\omega^2})\right)$$ where $K=k/2=eB/2m$ and similarly for $y(t)$.