Sketch the phase flow for this system $\dot r = (1-r^2)\cosh(r)$ , $\dot \theta = 1$

Question

I am given the system $\\$ $\dot r = (1-r^2)\cosh(r)$, $\dot \theta = 1$ and am required to sketch the phase flow for it in both Cartesian co-ordinates $(x,y) = (r\cos \theta, r\sin \theta)$ and also polar co-ordinates $(r,\theta)$

So I know I need to solve differential equations which give me

$\theta = t + c$ where $c$ is a constant

but I am struggling to solve the other differential equation as separating the variable does not work, as the integration is difficult to solve.

$$\dfrac {dr}{(1-r^2)\cosh(r)} = 1 \cdot dt$$

Solving the differential equation is the only thing stopping me drawing the phase flow, but I am really struggling on how to solve it.

Answer 1

The above analysis in my comment shows that there is a stable limit cycle defined by $r=1$ i.e. all trajectories converge asymptotically towards this cycle. If we transform the system equation to cartesian coordinates we get $$\frac{dx}{dt}=-y+(1-x^2-y^2)\cosh(\sqrt{x^2+y^2})\frac{x}{\sqrt{x^2+y^2}}\\ \frac{dy}{dt}=x+(1-x^2-y^2)\cosh(\sqrt{x^2+y^2})\frac{y}{\sqrt{x^2+y^2}}$$

with streamplot showing in the following Figure which is as expected.

I suppose you can now complete the flow in the $(r,\theta)$ plane.