The question I was working on said: determine the equation of the limit cycle to:
$x'' + (4x^2 + x'^2 -4x)x' + 4x = 0 $
What is it's period.
I have found the limit cycle:
$x^2 +4y^2 =4$
Where $y=x'$
I'd like to be directed to some examples that I can work through so that I can start. It's been many years since I've looked at problems like this and I fear my brain has turned to mush. :(
There appears to be an error in the equation, in the middle term it should not be "$-4x$" but "$-4$" to get an easily solvable tutorial problem. The phase portrait for both variants confirms this difference in difficulty.
Going by the corrected second case, the limit cycle coincides with the root of the middle factor. Using $V(t)=\frac12(x'(t)^2+4x(t)^2-4)$ one gets the differential equation $$ V'=-2V\cdot x'^2 $$ so that the curve $V=0$ is a stable equilibrium.
This means that the equation of the limit cycle is $4x^2+y^2=4$, writing it as a circle equation $(2x)^2+y^2=2^2$ one gets for the initial point $x(0)=\sinϕ$, $x'(0)=y(0)=2\cosϕ$.
Then differentiate through to get $$ 2x'(x''+4x)=0 $$ which either gives a constant solution $x=\pm1$, $x'=0$ or the harmonic oscillator with angular velocity $\omega=2$.
Combined with the initial conditions the solution is $$ x(t)=\sin(2t+ϕ) $$ giving $\pi$ as period.