= Solution
Write $k=\sqrt\mu$. The <equilibrium points> are a <stable node> $(-k,0)$ and a <saddle equilibrium> $(k,0)$, whose unstable and stable <eigenvalues> are $2k$ and $-\lambda$. The <saddle equilibrium>'s local <stable manifold> is $x=k$. When the right-going unstable <separatrix> returns at $(\nu,h)$ with $\nu=k$, it lands exactly on that <stable manifold>, producing a <homoclinic orbit> of the saddle. Crossing this condition changes whether the returning trajectory is captured by the node or lies on the outgoing side of the <saddle equilibrium>.
Use an incoming section $y=h$ and outgoing section $x=h$. For entry $(x,h)$ with $k<x<h$, integrate the local flow:
$$
T_{\rm loc}(x)=\int_x^h\frac{d\xi}{\xi^2-k^2}
=\frac1{2k}\log\left[\frac{h-k}{h+k}\frac{x+k}{x-k}\right],
$$
so the outgoing height is
$$
Y(x)=he^{-\lambda T_{\rm loc}}
=h\left[\frac{h+k}{h-k}\frac{x-k}{x+k}\right]^{\beta},\qquad\beta=\frac\lambda{2k}.
$$
The global excursion is smooth and transverse. Its return coordinate has expansion $x_{\rm next}=\nu+cY+O(Y^2)$, with $c>0$ in these section orientations. Thus the <local passage map near a dissipative saddle-node> gives
$$
\boxed{P(x)=\nu+ch\left[\frac{h+k}{h-k}\frac{x-k}{x+k}\right]^{\lambda/(2k)}+O(Y^2).}
$$
For small $\mu>0$, $\lambda>2k$, hence $P(k^+)=\nu$ and $P'(k^+)=0$. If $\nu-k>0$ is small, the graph starts above the diagonal and, with slope less than one nearby, crosses it close to $x=\nu$. More explicitly, $P(x)-x=(\nu-k)-(x-k)+O((x-k)^\beta)$, giving a unique small <fixed point> with <Floquet multiplier> $0<P'<1$. It corresponds to an attracting <periodic orbit>. For $\nu<k$ no such local crossing exists on $x>k$; the unstable branch returns to the node's side instead. The <Poincaré return map> panel shows this geometric fixed-point test.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-85-saddle-node-loop.png]
{title=Local return map and six local phase portraits near a saddle-node separatrix-loop point; outer return arcs are schematic}
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