= Solution
For a plane-parallel slab, all equilibrium quantities depend only on height $z$. The vertical <hydrostatic equilibrium> equation and the <Poisson equation for Newtonian gravity> are
$$
\frac{dp}{dz}=-\rho\frac{d\psi}{dz},\qquad
\frac{d^2\psi}{dz^2}=4\pi G\rho.
$$
With the <polytropic equation of state> $p=K\rho^2$ and constant $K$,
$$
2K\rho\frac{d\rho}{dz}=-\rho\frac{d\psi}{dz}.
$$
In the gas interior, where $\rho>0$, divide by $\rho$ and differentiate once:
$$
2K\frac{d^2\rho}{dz^2}=-\frac{d^2\psi}{dz^2}=-4\pi G\rho.
$$
Consequently
$$
\boxed{\frac{d^2\rho}{dz^2}+\frac{2\pi G}{K}\rho=0.}
$$
The coefficient $2\pi G/K$ has units inverse square metres. The <index-one plane-parallel polytrope> illustrates the resulting structure: symmetry about $z=0$ gives $\rho(z)=\rho_c\cos(kz)$, with $k=\sqrt{2\pi G/K}$. A slab bounded by vacuum ends at its first zero, $|z|=\pi/(2k)$; the negative continuation of the cosine is not physical <mass density>.
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