= Solution
Assume a steady, locally plane-symmetric <galactic disk>, with negligible radial tilt term in the <vertical Jeans equation>. Write $\rho(z)$ for the stellar <mass density>, and $K_z=\partial_z\Phi$ for the upward gradient of the <Newtonian gravitational potential>; the actual vertical <gravitational acceleration> is $-K_z$. Constant vertical <velocity dispersion> gives
$$
\sigma^2\frac{d\rho}{dz}=-\rho K_z,\qquad
\rho(z)=\rho(0)\exp\left[-\frac{\Phi(z)-\Phi(0)}{\sigma^2}\right].
$$
Integrating the plane-parallel <Poisson equation for Newtonian gravity> from the symmetry plane gives $K_z(z)=4\pi G\int_0^z\rho(z')\,dz'$. Outside most of the mass, this tends to $2\pi G\Sigma$ for $z>0$, where the <surface density> includes both sides of the <galactic disk>. Treating this field as constant and integrating the <vertical Jeans equation> yields
$$
\rho(z)\propto e^{-|z|/h},\qquad
\boxed{h=\frac{\sigma^2}{2\pi G\Sigma}}.
$$
\b[This exponential is the thin-sheet approximation or the outer tail, not the exact profile at every height.] To see the distinction without an additional assumption, set $\rho=\rho_0e^{-u}$ with $u=(\Phi-\Phi(0))/\sigma^2$. The <Poisson equation for Newtonian gravity> becomes $u''=(4\pi G\rho_0/\sigma^2)e^{-u}$, with $u(0)=u'(0)=0$, and is solved by
$$
u=2\log\cosh(z/z_0),\qquad
\rho=\rho_0\operatorname{sech}^2(z/z_0),\qquad
z_0^2=\frac{\sigma^2}{2\pi G\rho_0}.
$$
Since $\Sigma=2\rho_0z_0$, the exact <self-gravitating isothermal slab> has $z_0=\sigma^2/(\pi G\Sigma)$. At large $|z|/z_0$, $\operatorname{sech}^2(z/z_0)\sim4e^{-2|z|/z_0}$, recovering precisely the boxed <scale height>; near the plane its slope vanishes, whereas an exponential has a cusp.
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