I am curious to know how to recover the differential equation that goes with a phase portrait. I have seen the following posts but the first one was a $y'$ (and easy enough to "do in my head") and the second one lacks the figure.
As an example, I solved (in Mathematica) the non-linear pendulum equation and plotted the phase portrait as below:
$$y''(t)+\frac{y'(t)}{q}+\sin \left(y(t)\right)=g \cos (\omega t)$$
Where $q$ is damping, $g$ is the forcing term while $\omega$ is the frequency of forcing. The phase portrait ($y'$ vs $y$) is plotted:
So, now if I were given this phase portrait, how do I derive the autonomous differential equation for it? I ask because I do not know the "language" to search for online.
The reason I posted this question here instead of mathematica.SE is because this is more of a Mathematics question and not so much a Mathematica question.
My Mathematica code, in case interested:
g = 1/10; q = 8; \[Omega] = -0.04; TMax = 40;
pSolNL = NDSolveValue[{y''[t] + (1/q) y'[t] + Sin[y[t]] ==
g Cos[\[Omega] t], y[0] == 0, y'[0] == 0}, y, {t, 0, TMax}];
ParametricPlot[{pSolNL[\[Tau]] /. \[Tau] -> t,
D[pSolNL[\[Tau]], \[Tau]] /. \[Tau] -> t}, {t, 0, TMax},
AxesLabel -> {"y(t)", "y'(t)"}]
There is no autonomous differential equation on the plane with this phase portrait. Notice that at some points you would need to have at least two tangents to the curve, which is impossible in autonomous systems.