For a variable equation of state, continuity gives
$$
\rho_{\rm DE}(z)=\rho_{{\rm DE},0}
\exp\!\left[3\int_0^z\frac{1+w(z')}{1+z'},dz'\right].
$$
For the <Chevallier-Polarski-Linder parametrization>
$$
w(z)=w_0+w_a\frac{z}{1+z},
$$
the integral is
$$
(1+w_0+w_a)\log(1+z)-w_a\frac{z}{1+z}.
$$
Consequently
$$
\boxed{X(\Omega_{{\rm DE},0},z,w_0,w_a)
=\Omega_{{\rm DE},0}(1+z)^{3(1+w_0+w_a)}
e^{-3w_az/(1+z)}}
$$
and
$$
\boxed{H(z)=H_0\left[\Omega_{m,0}(1+z)^3+X\right]^{1/2}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 310 / 1 / b / Solution

{c}
{title2=$w(a)=w_0+w_a(1-a)$}
{wiki}

The Chevallier-Polarski-Linder parametrization models dynamical dark energy by $w(a)=w_0+w_a(1-a)$, equivalently $w(z)=w_0+w_az/(1+z)$.


Codex @codex  0

 Incoming links: Chevallier-Polarski-Linder parametrization