I have the following system:

$\frac{dx}{dt} = 3x + y - x(x^2+y^2)$

$\frac{dy}{dt} = -x +3y -y(x^2+y^2)$

Converting this to polar coordinates gives us:

$\frac{dr}{dt} = r(3-r^2)$

$\frac{d\theta}{dt} = -1$

This gives us a solution $\theta(t) = -t + \theta_0$. What would the solution for $r(t)$ be though?

  • $\begingroup$ You can avoid the nasty integration by recognizing the equation as Bernoulli and solving it as such. $\endgroup$ – Kaynex Dec 28 '16 at 17:51

The solution to $\dot{r} = r (3-r^2)$ is

$$ t = \int \frac{1}{\dot{r}} \,{\rm d}r + C = \int \frac{1}{r (3-r^2)} \,{\rm d}r + C $$

$$ t = \frac{1}{3} \ln\left(\frac{r}{r_0}\right) - \frac{1}{6} \ln\left(\frac{r^2-3}{r_0^2-3}\right) $$

with the initial condition $r(0)=r_0$

$$ r = \frac{\sqrt{3}\, r_0\, {\rm e}^{3 t} }{\sqrt{r_0 {\rm e}^{6 t}+3-r_0^2}} $$

  • $\begingroup$ PS. I used a CAS system for the integral. $\endgroup$ – ja72 Dec 28 '16 at 20:28

You have $$\frac{1}{r (\sqrt{3}-r)(\sqrt{3}+r)} dr = dt.$$ Use partial fractions on the left hand side and integrate.


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