= Solution
Write $A=8\pi G/(3c^2)$ and $H=\dot a/a$. Differentiating the <Friedmann equation> gives
$$
2H\dot H-2\frac{Kc^2}{a^2}H=A\dot\rho.
$$
The <cosmological perfect-fluid continuity equation> says $\dot\rho=-3H(\rho+P)$, so, first where $H\ne0$ and then everywhere by continuity,
$$
\dot H-\frac{Kc^2}{a^2}
=-\frac{4\pi G}{c^2}(\rho+P).
$$
Using $Kc^2/a^2=A\rho-H^2$ from the original equation gives
$$
\boxed{
\dot H+H^2
=-\frac{4\pi G}{3c^2}(\rho+3P)
}.
$$
Since $\dot H+H^2=\ddot a/a$ for the <Hubble parameter>, this is the <Friedmann acceleration equation>.
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