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

  • 1
    $\begingroup$ 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

1 Answer 1


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.