= Solution
For dust alone, write $\dot a^2=c^2(A/a-k)$, where
$$
\boxed{A=\frac{8\pi G\rho_{M,0}}{3c^2}=\frac{\Omega_{M,0}H_0^2}{c^2}.}
$$
Set $\theta=\sqrt{k}c\eta$ and $a=(A/k)\sin^2(\theta/2)$. Differentiation gives $\dot a=c\sqrt{k}\cot(\theta/2)$ once $dt=a\,d\eta$ is used, verifying the <Friedmann equation>. Integrating the time relation from the <Big Bang> yields the <closed matter-dominated Friedmann solution>
$$
a=\frac A{k}\sin^2\frac\theta2,\qquad
t=\frac{A}{2ck^{3/2}}(\theta-\sin\theta),\qquad0<\theta<2\pi.
$$
The <Hubble parameter> and the <cosmological density parameter> are
$$
\boxed{H(\theta)=\frac{ck^{3/2}}A\frac{\cos(\theta/2)}{\sin^3(\theta/2)},\qquad
\Omega_M(\theta)=\sec^2\frac\theta2.}
$$
The first decreases from positive infinity through zero at maximum expansion to negative infinity at the <Big Crunch>. The second approaches one near either singular endpoint and diverges at maximum expansion. This is a <density-parameter divergence at a cosmological turnaround>: $H=0$ makes the <critical density> vanish even though the physical dust density there is finite.
\Image[/past-exam-of-the-general-relativity-and-cosmology-course-of-the-university-of-oxford/2013-closed-dust-hubble-and-density.png]
{title=Closed dust universe: the Hubble parameter changes sign and the density parameter diverges at maximum expansion}
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