= Solution
Set $\tau=\varepsilon t$ and use the stated weighted spatial and parameter scales. Dividing the two equations by $\varepsilon^3$ and $\varepsilon^4$ respectively gives
$$
\boxed{u'=v,\qquad v'=u^2-\alpha+\varepsilon(\beta+u)v,}
$$
where prime now means differentiation with respect to $\tau$. The <Hamiltonian function> satisfies
$$
\frac{dH}{d\tau}=(\alpha-u^2)u'+vv'
=\varepsilon(\beta+u)v^2.
$$
Thus $H=v^2/2+\alpha u-u^3/3$ is conserved for $\varepsilon=0$.
For $\alpha>0$, let $s=\sqrt\alpha$. The conservative system has a center at $(-s,0)$ with $H_c=-2\alpha^{3/2}/3$ and a saddle at $(s,0)$ with
$$
\boxed{H_h=\frac23\alpha^{3/2}.}
$$
Closed contours in the left potential well have $H_c<H<H_h$. At $H_h$ the left-hand separatrix is a <homoclinic orbit>, since
$$
v^2=\frac23(u-s)^2(u+2s),\qquad -2s\le u\le s.
$$
It leaves and returns to $(s,0)$, with its left turning point at $u=-2s$. The right separatrix branches are unbounded. The energy levels above the saddle value are open, as are the separate right-hand components of lower energy levels. For $\alpha\le0$ there is no nondegenerate potential well and no such saddle loop.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2005/iii/paper-65-hamiltonian.png]
{title=Conserved-energy contours, center and saddle loop of the leading Hamiltonian system}
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