Classical Zeeman & Lorentz effect
Stationary homogeneous magnetic field is directed along one axis ($OZ$)
An atom is a classical harmonic oscillator, then the Lagrangian for an electron will take the following form $$\mathcal{L}=T-U+V=\frac{m\dot{r}^2}{2}-\frac{kr^2}{2}+e(\vec{A} , \vec{\dot{r}})$$
Using the Euler-Lagrange equation $$\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial \dot{x_i}}-\frac{\partial\mathcal{L}}{\partial x_i}=0,$$
We obtain a system of ODLE $$ \ddot{x}(t)+\omega_0^2 x - \omega_H \dot{y}(t)=0$$
$$\ddot{y}(t)+\omega_0^2 y + \omega_H \dot{x}(t)=0 $$
$$\ddot{z}(t)+\omega_0^2z=0$$
I need to solve the first two, because third is independent of them. I've tried to differentiate the first and put in the second, but unsuccessfully
From
$$ \ddot{x}(t)+\omega_0^2 x - \omega_H \dot{y}(t)=0$$
$$\ddot{y}(t)+\omega_0^2 y + \omega_H \dot{x}(t)=0 $$
we obtain
$$ \ddot{x}(t)\dot x(t)+\omega_0^2 x \dot x - \omega_H \dot{y}(t)\dot x(t)=0$$
$$\ddot{y}(t)\dot y(t)+\omega_0^2 y\dot y + \omega_H \dot{x}(t)\dot y(t)=0 $$
and after addition
$$ \frac 12(\dot x^2+\dot y^2)'+\frac 12\omega_0^2(x^2+y^2)'=0 $$
obtaining a movement constant
$$ (\dot x^2+\dot y^2)+\omega_0^2(x^2+y^2)=C $$
Also applying the Laplace transform
$$ s^2\hat x(s) +\omega_0^2\hat x(s)-\omega_H s\hat y(s) = s x(0) + \dot x(0) - \omega_H y(0)\\ s^2\hat y(s) +\omega_0^2\hat y(s)+\omega_H s\hat x(s) = s y(0) + \dot y(0) + \omega_H x(0)\\ $$
or
$$ \cases{ \hat x(s) = \frac{\phi_1(s) \left(\omega_0 ^2+s^2\right)+\omega_H\phi_2(s) s}{\omega_0^4+s^2 \left(2 \omega_0 ^2+\omega_H^2\right)+s^4}\\ \hat y(s) = \frac{\phi_2(s) \left(\omega_0 ^2+s^2\right)-\omega_H\phi_1(s) s}{\omega_0^4+s^2 \left(2 \omega_0 ^2+\omega_H^2\right)+s^4} } $$
with $\phi_1(s) = s x(0) + \dot x(0) - \omega_H y(0),\ \ \phi_2(s) = s y(0) + \dot y(0) + \omega_H x(0)$ easily invertible.