 Solution

Put $\Omega ={\Omega }_{m,0}$ and $b=1-\Omega$. Integrating the cosmological perfect-fluid continuity equation for the separately conserved cosmological fluids givesIn particular, ${\rho }_X{a}^2$ is constant. At the present epoch the flat Friedmann equation implies ${\Omega }_{X,0}=b$. Nonnegative fluid densities therefore require $0\le \Omega \le 1$.

The conformal time relation $dt=ad\tau$ gives the conformal Hubble parameter $\mathcal{H}=aH$. ConsequentlyOn the expanding branch, divide by $\mathcal{H}$ and use $a^{\prime }=a\mathcal{H}$:Eliminating the matter term yields the conformal Riccati equation for matter and a coasting fluid,The nonnegative square root fixes the convenient parameter convention; only ${\alpha }^2$ enters the differential equation.