 Solution

For constant $a$ and constant ${\rho }_m>0$, Hubble friction vanishes. Put ${\omega }_J=\sqrt{4\pi G{\rho }_m}$. The constant-coefficient ordinary differential equation is $\ddot{\delta }-{\omega }_J^2\delta =0$, so its characteristic polynomial has the real roots of a polynomial $\pm {\omega }_J$:These are the pressureless special case of the Static-universe Jeans modes. A positive growing-mode coefficient gives exponential gravitational instability, with time scale $(4\pi G{\rho }_m{)}^{-1/2}$. For initial data $\delta (t_{\ast })={\delta }_{\ast }$ and $\dot{\delta }(t_{\ast })=v_{\ast }$, the equivalent answer is ${\delta }_{\ast }\cosh [{\omega }_J(t-t_{\ast })]+(v_{\ast }/{\omega }_J)\sinh [{\omega }_J(t-t_{\ast })]$. This solves the stated perturbation model; a globally static dust-only FRW metric would require additional background support.