= Solution
For constant <barotropic equation of state> $P=w\rho$, the <cosmological perfect-fluid continuity equation> gives $\bar\rho\propto a^{-3(1+w)}$. In conformal time the flat <Friedmann equation> becomes $\mathcal H^2=(8\pi G/3)a^2\bar\rho$, where $\mathcal H=a'/a$. Integration, for $w\ne-1/3$, gives
$$
\boxed{a\propto|\eta-\eta_B|^{2/(1+3w)},\qquad a^2\bar\rho\propto(\eta-\eta_B)^{-2}.}
$$
Choose the conformal-time origin so that $\eta_B=0$ and work on the appropriate expansion branch. Write $\beta=2/(1+3w)$, so $\mathcal H=\beta/\eta$ and $2\mathcal H'+(1+3w)\mathcal H^2=0$.
Combining the supplied energy and momentum constraints gives the <comoving-gauge density contrast> Poisson relation
$$
\nabla^2\phi=4\pi Ga^2\bar\rho[\delta-3\mathcal H(1+w)v]
=4\pi Ga^2\bar\rho\Delta.
$$
For this barotropic fluid $\delta P=w\bar\rho\delta$. Substitute the energy constraint into the pressure equation to obtain
$$
\phi''+3(1+w)\mathcal H\phi'
+[2\mathcal H'+(1+3w)\mathcal H^2]\phi-w\nabla^2\phi=0,
$$
so the bracket vanishes and the coefficient of $\phi'/\eta$ is $A=6(1+w)/(1+3w)$.
Because $a^2\bar\rho=C\eta^{-2}$, the Poisson relation gives $\Delta=(\eta^2/(4\pi GC))\nabla^2\phi$. Apply the same potential equation to $F=\eta^2\nabla^2\phi$: differentiating $\eta^{-2}F$ gives
$$
F''+\frac{A-4}{\eta}F'+\frac{6-2A}{\eta^2}F-w\nabla^2F=0.
$$
Since $F$ is a constant multiple of $\Delta$, the <constant-equation-of-state comoving density equation> is
$$
\boxed{\Delta''+\frac{2(1-3w)}{(1+3w)\eta}\Delta'
-\frac{6(1-w)}{(1+3w)\eta^2}\Delta-w\nabla^2\Delta=0.}
$$
The primes here denote conformal-time derivatives, corresponding to the overdots in the question. A pure vacuum fluid $w=-1$ has zero enthalpy and no independent fluid velocity or nonzero-wave-number density perturbation; its formal equation should be interpreted with that degeneracy in mind.
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