= Solution
Use units $c=1$ and <cosmic time> $t$. Define the <Hubble parameter> $H=\dot a/a$, the <equation-of-state parameter> $\omega=P/\rho$, and the <cosmological density parameter>
$$
\Omega=\frac\rho{\rho_{\rm crit}},\qquad\rho_{\rm crit}=\frac{3H^2}{8\pi G}.
$$
Here $\rho_{\rm crit}$ is the <critical density>. The definition requires $H\ne0$; assume positive <energy density> and an expanding branch so that $\log a$ is a valid time coordinate. The <Friedmann equation> gives $\Omega-1=K/(a^2H^2)$. The <Friedmann acceleration equation> and <Hubble parameter identity> give
$$
\frac{\dot H}{H^2}=\frac{\ddot a}{aH^2}-1=-1-\frac12(1+3\omega)\Omega.
$$
Differentiating the <Friedmann equation> and combining with this identity yields the <cosmological perfect-fluid continuity equation> $\dot\rho=-3H(\rho+P)$. Consequently
$$
\frac{d\log\Omega}{d\log a}=\frac{\dot\rho}{H\rho}-2\frac{\dot H}{H^2}
=-3(1+\omega)+2+(1+3\omega)\Omega.
$$
Thus the <cosmological density parameter flow> is
$$
\boxed{\frac{d\Omega}{d\log a}=(1+3\omega)\Omega(\Omega-1).}
$$
This identity allows time-dependent $\omega$; constancy is needed only for the simple power-law integration in the next part.
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