= Solution
For the <Einstein static universe> with $P=0$ and $\rho_0>0$, the acceleration equation gives $\Lambda=4\pi\rho_0$. The first <Friedmann equation> then gives $k/a_0^2=4\pi\rho_0$, so
$$
\boxed{k=1,\qquad a_0=(4\pi\rho_0)^{-1/2},\qquad\Lambda=4\pi\rho_0.}
$$
Keep $\Lambda,k$ fixed and use the specified perturbation closure $\delta P=\delta\rho/3$. Linearization of the constraint gives $\delta\rho=-3\rho_0\delta a/a_0$. In the acceleration equation the background terms cancel, leaving
$$
\delta\ddot a=-\frac{4\pi}3a_0(\delta\rho+3\delta P)
=8\pi\rho_0\delta a=\frac2{a_0^2}\delta a.
$$
Thus
$$
\boxed{\delta a=C_+e^{\sqrt{8\pi\rho_0}\,t}+C_-e^{-\sqrt{8\pi\rho_0}\,t}.}
$$
The growing mode establishes \b[linear instability of the static universe]; a specially chosen decaying perturbation does not remove instability. This uses the pressure-density relation explicitly supplied for the perturbation. If zero density is allowed, there is also the degenerate vacuum static case $k=0$, $\Lambda=0$, which is not the positive-density static universe considered here.
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