= Solution
The vertical <hydrostatic equilibrium> equation balances the gradient of the total <pressure> against gravity. In a <thin disk> with <Keplerian rotation>, expanding the central gravitational field about the midplane gives vertical acceleration $-\Omega^2z$, so $p_z=-\rho\Omega^2z$. Disk self-gravity is neglected in this approximation.
The flux-gradient equation is local thermal balance: the upward <radiative flux> increases with height because <differential rotation> dissipates <energy>. For a Newtonian effective <dynamic viscosity> $\mu$, the volumetric heating rate is $\mu(r\Omega')^2$, which becomes $9\mu\Omega^2/4$ for <Keplerian rotation>. The symbol $\mu$ here is <dynamic viscosity>, not <kinematic viscosity>; the latter is $\nu=\mu/\rho$.
The <radiative diffusion> equation applies to an <optically thick medium> in <local thermodynamic equilibrium>. Its <diffusion> coefficient is $c/(3\kappa\rho)$, and the equilibrium radiation <energy density> is $u_{\rm rad}=4\sigma T^4/c$. Hence
$$
F=-\frac{c}{3\kappa\rho}\frac{du_{\rm rad}}{dz}=-\frac{16\sigma T^3}{3\kappa\rho}\frac{dT}{dz}.
$$
The minus sign means <energy> flows down the <temperature> gradient. Here $\sigma$ is the <Stefan-Boltzmann constant>, $\kappa$ the <opacity> per unit <mass>, and $c$ the <speed of light>. Constant <Thomson scattering> <opacity> is the adopted transport approximation; <local thermodynamic equilibrium> is an additional assumption behind the radiation <equation of state>.
Finally the <equation of state> adds <ideal gas> <gas pressure> $p_g=\rho kT/(\mu_m m_H)$ and isotropic <radiation pressure> $p_r=u_{\rm rad}/3=4\sigma T^4/(3c)$. The <Boltzmann constant> is $k$, and $\mu_m m_H$ is the mean <mass> per gas particle. Both contributions supply the total <pressure> entering vertical <hydrostatic equilibrium>.
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