= Solution
Write $\Omega=\Omega_{M0}$, $\beta^2=H_0^2(1-\Omega)$ and $\mathcal H=a'/a=\dot a$. The Friedmann equation is
$$
\mathcal H^2=H_0^2\left(\frac\Omega a+1-\Omega\right).
$$
The <Raychaudhuri equation> has only the matter component on its active-gravity side, so
$$
\mathcal H'=a\ddot a=-\frac{4\pi G}{3}a^2\rho_M=-\frac{H_0^2\Omega}{2a}.
$$
Combining these expressions proves
$$
\boxed{2\mathcal H'+\mathcal H^2-\beta^2=0}.
$$
This is the <conformal Riccati equation for matter and a coasting fluid>. The prime denotes <conformal time>, whereas $H_0$ is the physical present <Hubble parameter>.
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