Is it possible to show that $x''(t)-x'(t)²+x(t)²-x(t)=0$ has at least a periodic orbit?
I've made it a system by setting $y=x'$ and get $x'=y; y'=y²-x²+x$. I'm asking the question because I find a center (so we expect a periodic orbit). I use the change of variable $u=x+y$, $v=x-y$ and get the new simpler system $u'=u+v-uv;v'=uv$. I'd solve $\frac{du}{dv}=\frac{u+v-uv}{uv}$ to get something but I can't do it... Can you help me solving this? I'd like to know if my instinct's right...
The center is at $(x,y)=(1,0)$ and one stays on a curve $y=g(x)$ if $y'=g'(x)x'$, that is, if $g(x)^2-x^2+x=g'(x)g(x)$, or equivalently, if $(g^2)'(x)=2g^2(x)+2x-2x^2$, or equivalently, after integration, if $$g(x)^2=x^2-c\mathrm e^{2x}. $$ The center $(1,0)$ corresponds to $c=\mathrm e^{-2}$. If $c=\mathrm e^{-2-2\gamma^2}$, the curve $y=\pm\sqrt{x^2-c\mathrm e^{2x}}$ meets the axis $y=0$ at two points $x_\pm$ such that $x_\pm=1\pm\gamma+o(\gamma)$ when $\gamma$ is small and is homotopic to a circle around $(1,0)$.
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In terms of the second order differential equation, this means that the quantity $V(x(t),x'(t))$ does not depend on $t$, where $$V(x,y)=\mathrm e^{2-2x}(x^2-y^2),$$ and that, if $V(x(0),x'(0))$ is in $(0,1)$ and $x(0)\gt0$, then $t\mapsto x(t)$ is periodic.