(Solvable?) System of differential equations

I was trying to solve a physics problem which was about a charged particle moving in a variable magnetic field. I ended up with this system of two differential equations:

$$ \left\{ \begin{array}{c} \ddot x = \omega {\dot y \over y} \\ \ddot y = -\omega {\dot x \over y} \end{array} \right. $$

Where $x$ and $y$ are functions of time $t$ and $\omega$ is a constant.

I am posting this problem here because it's the first time I come up with a system of differential equations and I don't know how to approach such a thing.

I have tried by equating the $\omega \over y$ term in the equations and integrating different times, but at the end I come up with:

$$ t + C = \pm \int {dy \over \sqrt {A-\mathrm{(B\pm \log y)}^2 }} $$

However, how can I solve the system? Is it possible to explicit the solutions $x$ and $y$ in terms of elementary functions?

Thanks in advance, Dave

As already commented, you can integrate your first equation, where $\dot{x} = \omega(\log y + c_1)$. Substituting this result into your second differential equation, $$ \ddot{y}y = \frac{\mathrm{d}y}{\mathrm{d}t} \frac{\mathrm{d}\phantom{t}}{\mathrm{d}y}\left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)y = -\omega(\omega\log y + c_1) \Rightarrow \frac{\dot{y}^2}{2} = -\omega\int \frac{\omega\log y + c_1}{y} \mathrm{d}y = \frac{\omega}{2}\log(y) \left(\omega\log y + 2c_1\right) + c_2. $$

Therefore, $$ \dot{y} = \pm \sqrt{\omega \log(y) \left(\omega\log y + 2c_1\right) + c_2}. $$

You can try to solve it but, in general, it does not seem possible to express it in terms of elementary functions. The same reasoning goes for $x(t)$. $$ \boxed{t - t_0 = \pm \int_{y(t_0)}^{y(t)} \frac{\mathrm{d}y}{\sqrt{\omega \log(y) \left(\omega\log y + 2c_1\right) + c_2}}.} $$