= Solution
Use <natural units> and <conformal time> $dt=a\,d\tau$. The <conformal Hubble parameter> is $\mathcal H=a'/a=aH$. For the <barotropic equation of state> $P=(\gamma-1)\rho$, the <Friedmann acceleration equation> gives
$$
\mathcal H'=a\ddot a=-\frac{4\pi G}{3}a^2(3\gamma-2)\rho.
$$
The <Friedmann equation> gives $8\pi Ga^2\rho/3=\mathcal H^2+k$. Hence the <constant-equation-of-state conformal Riccati equation> is
$$
\boxed{\mathcal H'+\alpha(\mathcal H^2+k)=0,\qquad\alpha=\frac{3\gamma-2}{2}.}
$$
For $\alpha\ne0$, separation and integration of this <Riccati equation> yield, after choosing the origin of <conformal time>,
$$
\boxed{\mathcal H=\begin{cases}
\cot(\alpha\tau),&k=1,\\
1/(\alpha\tau),&k=0,\\
\coth(\alpha\tau),&k=-1.
\end{cases}}
$$
The open positive-<energy density> branch uses the <hyperbolic cotangent>, rather than the <cotangent>. Integrating $a'/a=\mathcal H$ gives
$$
\boxed{a(\tau)=\begin{cases}
A_+\sin^{1/\alpha}(\alpha\tau),&k=1,\\
A_0(\alpha\tau)^{1/\alpha},&k=0,\\
A_-\sinh^{1/\alpha}(\alpha\tau),&k=-1,
\end{cases}}
$$
where the positive constants are fixed by the <energy density> normalization. These formulas apply on branches where the bases are positive. For $\alpha>0$, take the <Big Bang> at $\tau=0$. The <cosmic time> is then
$$
t_+(\tau)=\frac{A_+}{\alpha}\int_0^{\alpha\tau}(\sin u)^{1/\alpha}\,du,\qquad
t_-(\tau)=\frac{A_-}{\alpha}\int_0^{\alpha\tau}(\sinh u)^{1/\alpha}\,du,
\qquad
t_0(\tau)=\frac{A_0}{\alpha+1}(\alpha\tau)^{(\alpha+1)/\alpha}.
$$
The <integrals> also give the general answer after selecting a suitable reference time when a finite <Big Bang> endpoint does not exist. The exceptional <coasting fluid> $\gamma=2/3$ has $\alpha=0$, so $\mathcal H=h$ is constant: $a=Ae^{h\tau}$ and $t=(A/h)e^{h\tau}+t_*$. If $h=0$, instead $a=A$ and $t=A\tau+t_*$. A flat fluid with $\gamma=0$ is the constant-density <de Sitter spacetime> case, for which the flat time integral is logarithmic.
During <matter domination>, $\gamma=1$ and $\alpha=1/2$. Absorbing numerical factors into separate positive constants gives the <closed matter-dominated Friedmann solution>, the flat <Einstein-de Sitter universe> and the <open matter-dominated Friedmann solution>:
$$
\begin{array}{c|c|c}
k&a(\tau)&t(\tau)\\\hline
1&C_+(1-\cos\tau)&C_+(\tau-\sin\tau)\\
0&C_0\tau^2&C_0\tau^3/3\\
-1&C_-(\cosh\tau-1)&C_-(\sinh\tau-\tau).
\end{array}
$$
To compare the <age of a dust universe without a cosmological constant>, specify the normalization. At fixed present <Hubble parameter> $H_0$, with $x=a/a_0$ and present matter <cosmological density parameter> $\Omega_m>0$, the <Friedmann equation> gives
$$
H_0t_0=\int_0^1\frac{\sqrt{x}\,dx}{\sqrt{\Omega_m+(1-\Omega_m)x}}.
$$
The integrand decreases strictly with $\Omega_m$ for $0<x<1$. Therefore <age ordering of dust universes at fixed Hubble parameter> gives, on the expanding branches,
$$
\boxed{t_{\rm closed}<\frac{2}{3H_0}<t_{\rm open}<\frac1{H_0}.}
$$
The final bound is approached by the empty <Milne universe>. This compares models with the same $H_0$; fixing instead the present <scale factor> and <energy density> gives a different comparison.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2003/iii/paper-55-dust-age.png]
{title=Age of an expanding dust universe at fixed present Hubble parameter}
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