= Solution
Integrate the reduced vertical <phase-space distribution function> over $v_z$. The Gaussian velocity integral cancels its normalization, giving
$$
\rho(z)=\rho_0 e^{-\Phi(z)/\sigma_z^2}.
$$
Here the constant vertical <velocity dispersion> is $\sigma_z$, and $\Phi(0)=0$ fixes the potential reference. Neglecting the radial derivatives in the <Poisson equation for Newtonian gravity> leaves
$$
\frac{d^2\Phi}{dz^2}=4\pi G\rho_0e^{-\Phi/\sigma_z^2}.
$$
With $\phi=\Phi/\sigma_z^2$, $\zeta=z/z_0$ and $z_0^2=\sigma_z^2/(8\pi G\rho_0)$, this becomes
$$
\frac{d^2\phi}{d\zeta^2}=\frac12e^{-\phi}.
$$
Thus the stated dimensionless equation follows with its factor of two.
For a direct solution, multiply by $2\phi'$ and integrate from the midplane. The initial conditions give
$$
(\phi')^2=1-e^{-\phi}.
$$
For $\zeta>0$ take the increasing branch and put $u=e^{\phi/2}$. Then $4(u')^2=u^2-1$, whose smoothly increasing solution consistent with the second-order equation is $u=\cosh(\zeta/2)$. Reflection about the midplane supplies the negative-$\zeta$ branch. Therefore
$$
\boxed{\phi(\zeta)=2\log\cosh(\zeta/2),\qquad
\rho(z)=\rho_0\operatorname{sech}^2\!\left(\frac{z}{2z_0}\right).}
$$
Substitution verifies $2\phi''=\operatorname{sech}^2(\zeta/2)=e^{-\phi}$ and both boundary conditions. The smooth second-order equation has a unique solution for those initial data, so no extra constant branch from the first-integral square root is admissible.
This is the <self-gravitating isothermal slab>. Its scale convention matters: if its profile is instead written $\rho=\rho_0\operatorname{sech}^2(z/H)$, then $H=2z_0=\sigma_z/\sqrt{2\pi G\rho_0}$. Its <surface density> is $\Sigma=4\rho_0z_0$. The force approaches $|\Phi'|=\sigma_z^2/z_0=2\pi G\Sigma$, consistent with the field of a plane sheet. These checks fix both the scale-height and Poisson factors.
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