= Solution
For $\nu=1,\kappa=0$, the <first integral> is
$$
\boxed{E=\tfrac12y^2-\tfrac12x^2+\tfrac14x^4.}
$$
Its minimum $-1/4$ occurs at the two <centre equilibrium>[centres]. Levels between $-1/4$ and $0$ give a <periodic orbit> around each <centre equilibrium>; positive levels give larger <periodic orbits> surrounding both wells. The zero level is the figure-eight <separatrix> through the saddle, with two homoclinic loops.
On $E=0$, $y^2=x^2(1-x^2/2)$. The printed parametrization $x=\pm\sqrt2\operatorname{sech}z$, $y=\pm\sqrt2\operatorname{sech}z\tanh z$, taking matching signs, satisfies this identity because $1-\operatorname{sech}^2z=\tanh^2z$. It also has $dx/dz=-y$ and $dy/dz=-(x-x^3)$, so \b[its parameter is physical reversed time, $z=-t+\text{constant}$], rather than $z=t$. It is nonetheless exactly a parametrization of the requested <trajectory>. In physical forward time one can use $x(t)=\sqrt2\operatorname{sech}t$, $y(t)=-\sqrt2\operatorname{sech}t\tanh t$ on the right loop.
Approach to the origin requires $z\to\pm\infty$, hence infinite physical time. Equivalently $|y|\sim x$ near the right-loop endpoint, so the elapsed time contains the divergent integral $\int_0^\delta dx/x$. The point at $x=\sqrt2$ is reached in finite time from any interior point; it is the asymptotic saddle endpoints, not that turning point, that cause the infinite total traversal time.
\Image[/past-exam-of-the-topics-in-applied-mathematics-course-of-the-university-of-oxford/2009-duffing-separatrix.png]
{title=Double-well energy levels and the two homoclinic loops through the saddle}
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