 Solution

With an approximately constant Hubble parameter $H$, the scale factor is $a(t)=e^{Ht}$ after a choice of normalization. Integrate $d\tau =dt/a(t)$ and choose the integration constant so that the future endpoint is $\tau =0$:Substituting $dt=ad\tau$ in the FLRW metric gives $d{s}^2=a^2(\tau )(-d{\tau }^2+d{\mathbf{x}}^2)$. This is Conformal time during de Sitter expansion.