= Solution
{c}
Set $H=\dot a/a$, the <Hubble parameter>. The first supplied component of the <Einstein field equations> gives the <Friedmann acceleration equation>
$$
\frac{\ddot a}{a}=-\frac{4\pi G}{3}(\varrho+3p).
$$
Substitution into the second supplied equation gives
$$
2\left(H^2+\frac{k}{a^2}\right)
=4\pi G(\varrho-p)+\frac{4\pi G}{3}(\varrho+3p)
=\frac{16\pi G}{3}\varrho.
$$
Therefore the first <Friedmann equation> is
$$
\boxed{\dot a^2+k=\frac{8\pi G}{3}\varrho a^2.}
$$
Differentiate this equation with respect to <cosmic time>:
$$
2\dot a\ddot a=\frac{8\pi G}{3}\left(a^2\dot\varrho+2a\varrho\dot a\right).
$$
Insert the <Friedmann acceleration equation> on the left and cancel $8\pi Ga/3$, which is nonzero for $a>0$. This gives
$$
-\dot a(\varrho+3p)=a\dot\varrho+2\varrho\dot a,
\qquad
\boxed{a\dot\varrho+3(\varrho+p)\dot a=0.}
$$
This is the <cosmological continuity equation>, expressing <stress-energy conservation> for the <perfect fluid in general relativity>. No division by $\dot a$ was made, so the derivation remains valid at a <cosmological bounce> or an instant of maximum expansion.
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