 Solution

Let $H=\dot{a}/a$ be the Hubble parameter, and normalize the scale factor by $a(t_0)=1$. The cosmological continuity equation gives ${\rho }_M={\rho }_{M0}{a}^{-3}$ for pressureless matter, whereas the cosmological constant has constant energy density ${\rho }_{\Lambda }$. The spatially flat Friedmann equation therefore becomesThe present Hubble constant is $H_0=H(t_0)$. In terms of the present critical density ${\rho }_{\mathrm{crit},0}=3H_0^2/(8\pi G)$, the matter cosmological density parameter is ${\Omega }_M={\rho }_{M0}/{\rho }_{\mathrm{crit},0}$. Spatial flatness gives ${\rho }_{\Lambda }/{\rho }_{\mathrm{crit},0}=1-{\Omega }_M$, so