= Solution
Pressureless matter obeys the <cosmological perfect-fluid continuity equation>, $\dot\rho_m+3H\rho_m=0$. Integrating gives $\rho_m=\rho_{m,0}(a_0/a)^3$. Define the present <critical density> and <cosmological density parameters> by
$$
\rho_{\mathrm{crit},0}=\frac{3H_0^2}{8\pi G},\qquad
\Omega_{m,0}=\frac{\rho_{m,0}}{\rho_{\mathrm{crit},0}},\qquad
\Omega_{\Lambda,0}=\frac{\Lambda}{3H_0^2},\qquad
\Omega_{k,0}=-\frac{k}{H_0^2a_0^2}.
$$
Here $c=1$ and the curvature contribution is an effective density parameter, not an additional matter fluid. Substitution into the <Friedmann equation> gives
$$
H^2=H_0^2\left[\Omega_{m,0}\left(\frac{a_0}{a}\right)^3+\Omega_{k,0}\left(\frac{a_0}{a}\right)^2+\Omega_{\Lambda,0}\right].
$$
Using the <Hubble parameter> $H=\dot a/a$ therefore yields
$$
\boxed{\dot a^2=H_0^2\left(\frac{\Omega_{m,0}a_0^3}{a}+\Omega_{k,0}a_0^2+\Omega_{\Lambda,0}a^2\right).}
$$
At $a=a_0$ the same equation implies $\Omega_{m,0}+\Omega_{k,0}+\Omega_{\Lambda,0}=1$. The power in the matter term is $a_0^3$; the converted TeX's $a_0^2$ is an error not present in the original PDF.
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