= Solution
Put $\Omega=\Omega_{m,0}$ and $b=1-\Omega$. Integrating the <cosmological perfect-fluid continuity equation> for the <separately conserved cosmological fluids> gives
$$
\rho_m=\rho_{m,0}a^{-3},\qquad \rho_X=\rho_{X,0}a^{-2}.
$$
In particular, \b[$\rho_Xa^2$ is constant]. At the present epoch the flat <Friedmann equation> implies $\Omega_{X,0}=b$. Nonnegative fluid densities therefore require $0\leq\Omega\leq1$.
The <conformal time> relation $dt=a\,d\tau$ gives the <conformal Hubble parameter> $\mathcal H=aH$. Consequently
$$
\mathcal H^2=H_0^2\left(\frac{\Omega}{a}+b\right),\qquad 2\mathcal H\mathcal H'=-H_0^2\frac{\Omega a'}{a^2}.
$$
On the expanding branch, divide by $\mathcal H$ and use $a'=a\mathcal H$:
$$
2\mathcal H'=-\frac{H_0^2\Omega}{a}.
$$
Eliminating the matter term yields the <conformal Riccati equation for matter and a coasting fluid>,
$$
\boxed{2\mathcal H'+\mathcal H^2-\alpha^2=0,\qquad \alpha=H_0\sqrt{1-\Omega_{m,0}}.}
$$
The nonnegative square root fixes the convenient parameter convention; only $\alpha^2$ enters the differential equation.
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