Using Van der Pol oscillator as a placeholder.
$$ {\displaystyle {d^{2}x \over dt^{2}}-\mu (1-x^{2}){dx \over dt}+x=0,} $$
This post is a work in progress
June 30, 2019
Using Van der Pol oscillator as a placeholder.
$$ {\displaystyle {d^{2}x \over dt^{2}}-\mu (1-x^{2}){dx \over dt}+x=0,} $$
This post is a work in progress