 Solution

For a flat universe containing only the stated fluid, the cosmological continuity equation is $\dot{\rho }+3H(\rho +P)=0$. A constant equation-of-state parameter $w$ therefore gives $\rho ={\rho }_0(a/a_0{)}^{-3(1+w)}$. Combine this with the Friedmann equation $H^2=8\pi G\rho /3$ to findOn the expanding branch, specifying $H(t_0)=H_0>0$ yieldsThe solution is used only where its denominator is positive. For $w\ne -1$, integration also givesFor $w>-1$, choosing the zero of time at the initial singularity gives $H=2/[3(1+w)t]$ and $\rho =1/[6\pi G(1+w{)}^2{t}^2]$. For $w=-1$, $H=H_0$, $\rho$ is constant and $a=a_0{e}^{H_0(t-t_0)}$. For $w<-1$, the displayed initial-value solution instead grows towards a future finite-time singularity; it is not the ordinary positive-time Big Bang branch. With positive density, accelerated expansion requires $w<-1/3$.