= Solution
In the homogeneous background, $\mathbf v=0$ and the physical velocity is the <Hubble flow>
$$
\mathbf u=aH\mathbf x=\dot a\,\mathbf x.
$$
The <comoving coordinate> $\mathbf x$ is held fixed by the time derivative in the fluid equation, so
$$
\frac{\partial\mathbf u}{\partial t}=\ddot a\,\mathbf x.
$$
Moreover, spatial homogeneity gives $\nabla\bar P=0$. The <Euler equations for an inviscid fluid>[Euler equation] therefore reduces to
$$
\bar\rho a\ddot a\,\mathbf x=-\bar\rho\nabla\bar\Phi,
\qquad
\nabla\bar\Phi=-a\ddot a\,\mathbf x.
$$
Taking the <divergence> in three spatial dimensions gives
$$
\nabla^2\bar\Phi=-3a\ddot a.
$$
Consistency with the <Poisson equation>,
$$
\nabla^2\bar\Phi=\frac{4\pi G}{c^2}\bar\rho a^2,
$$
then requires
$$
\boxed{\frac{\ddot a}{a}
=-\frac{4\pi G}{3c^2}\bar\rho}.
$$
This is the pressureless, zero-cosmological-constant <Raychaudhuri equation>, obtained by the <Newtonian fluid derivation of the Raychaudhuri equation>.
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