For <dynamical dark energy>, use $dz/dt=-(1+z)H$ in the continuity equation to obtain
$$
\frac{d\log\rho_{\rm DE}}{dz}
=\frac{3[1+w(z)]}{1+z}.
$$
Therefore the <variable dark-energy equation of state> gives
$$
\rho_{\rm DE}(z)=\rho_{{\rm DE},0}(1+z)^3
\exp\left[3\int_0^z\frac{w(z')}{1+z'}\,dz'\right].
$$
Combining this with $\rho_m=\rho_{m,0}(1+z)^3$ in the <Friedmann equation> yields
$$
\boxed{H(z)=H_0\left\{
\Omega_{m,0}(1+z)^3
+\Omega_{{\rm DE},0}(1+z)^3
\exp\left[\int_0^z\frac{3w(z')}{X(z')}\,dz'\right]
\right\}^{1/2}},
$$
where
$$
\boxed{X(z)=1+z}.
$$
This is the <Hubble parameter for matter and dynamical dark energy>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 310 / 1 / b / Solution

For pressureless matter and dark energy with equation of state $w(z)$,
$$
H(z)=H_0\left[\Omega_{m,0}(1+z)^3
+\Omega_{{\rm DE},0}(1+z)^3
\exp\left(3\int_0^z\frac{w(z')}{1+z'}\,dz'\right)
\right]^{1/2}.
$$


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 Incoming links: Hubble parameter for matter and dynamical dark energy