= Solution
Multiply the <Friedmann equation> by $a^2$ and differentiate:
$$
\dot a^2+kc^2=\frac{8\pi G}{3}\rho a^2,\qquad
2\dot a\ddot a=\frac{8\pi G}{3}(\dot\rho a^2+2\rho a\dot a).
$$
Insert the <cosmological continuity equation> to obtain
$$
2\dot a\ddot a=-\frac{8\pi G}{3}a\dot a(\rho+3P/c^2).
$$
On a nonstatic branch divide by $2a\dot a$; continuity extends the result across isolated turning points. The resulting <Friedmann acceleration equation> is
$$
\boxed{\frac{\ddot a}{a}=-\frac{4\pi G}{3}(\rho+3P/c^2).}
$$
An everywhere static branch needs the independent spatial <Einstein field equations>: the two differentiated constraints alone cannot determine the pressure when $\dot a\equiv0$. This is the qualification to deriving the acceleration equation solely by division.
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