 Solution

For overdamping, $\gamma =b/(2m)>\omega$. Put $\alpha =\sqrt{{\gamma }^2-{\omega }^2}>0$. Both characteristic roots $-\gamma \pm \alpha$ are negative real numbers. Replacing trigonometric functions by hyperbolic functions gives the exact trajectoryIt starts from $(c,c)$ with tangent $(-\omega c,\omega c)$, bends anticlockwise toward the origin, and approaches it without repeated revolutions. The right panel gives a representative choice $\gamma =1.5\omega$; the paper does not specify one particular value of $b$ in this regime.

One can make the qualitative shape precise. In radial and tangential directions relative to the initial radius, the angle turned from the starting direction obeysFor $c>0$, both coordinates stay positive, and the approach is along the fixed ray with $y/x\to (\gamma +\alpha +\omega )/(\gamma +\alpha -\omega )$. The initial velocity is tangential, so an immediately straight radial return would be incorrect. The absence of oscillatory characteristic roots prevents repeated revolutions. This is a non-spiralling overdamped central oscillator trajectory.