= Solution
Since $P=P(\rho)$ and $\rho=\bar\rho(1+\delta)$,
$$
\frac{c^2}{\rho}\nabla P
=\frac{c^2}{\rho}\frac{dP}{d\rho}\nabla\rho
\simeq c_s^2\nabla\delta.
$$
To first order, continuity and the divergence of Euler give
$$
\dot\delta+\theta=0,\qquad
\dot\theta+2\frac{\dot a}{a}\theta
=-\frac{4\pi G}{c^2}\bar\rho\,\delta-c_s^2\nabla^2\delta.
$$
Eliminating $\theta$ and using $\nabla\mapsto i\mathbf k/a$ yields
$$
\ddot\delta+2\frac{\dot a}{a}\dot\delta
-\left(\frac{4\pi G}{c^2}\bar\rho-\frac{c_s^2k^2}{a^2}\right)\delta=0.
$$
The second term is Hubble damping, gravity drives growth, and <pressure> produces oscillatory restoration. Their balance defines
$$
k_J^2=\frac{4\pi G\bar\rho\,a^2}{c^2c_s^2},
\qquad \lambda_J=\frac{2\pi a}{k_J}.
$$
Modes longer than the Jeans length are gravitationally unstable; shorter modes are <pressure> supported.
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