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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

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    $\begingroup$ You need to solve $$y''y=-w(w \ln y +C)$$$$y'dy'=-w\int \frac {(w \ln y +C)dy}{y}$$ $\endgroup$ Nov 30, 2019 at 17:18
  • $\begingroup$ Is it solvable? Because then I have to take the square root in order to find $y'$ $\endgroup$ Dec 2, 2019 at 20:26
  • $\begingroup$ I don't think so. Not with elementary functions. Note that the integral is easy to evaluate it's the step after that is hard $$\int \frac {\ln y } y dy=\frac 1 2 \ln^2 y+K$$ $\endgroup$ Dec 2, 2019 at 20:29
  • $\begingroup$ Notice that $\dot x\ddot x+\dot y\ddot y=0$ and $\dot x^2+\dot y^2=v^2$. The motion has constant speed. $\endgroup$
    – user1010241
    Jan 7 at 17:28

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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 $x(t)$. $$ \boxed{t - t_0 = \pm \int_{y(t_0)}^{y(t)} \frac{\mathrm{d}y}{\sqrt{\omega \log(y) \left(\omega\log y + 2c_1\right) + c_2}}.} $$

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