= Solution
Let $y=\beta/(1-\beta)=P_g/P_{\rm rad}$. The two pressure laws give
$$
P=\frac{a_{\rm rad}}3T^4(1+y),
\qquad
\rho=\frac{a_{\rm rad}\mu H}{3k_B}yT^3,
$$
and hence
$$
\kappa=\frac{\kappa_0a_{\rm rad}\mu H}{3k_B}
yT^{-1/2}.
$$
The radiative equation can be written
$$
\frac{dP_{\rm rad}}{dr}
=-\frac{\kappa\rho L_r}{4\pi cr^2}.
$$
Dividing it by <hydrostatic equilibrium> and using $L_r/M_r=L/M$ gives
$$
\frac{dP_{\rm rad}}{dP}
=\frac{\kappa L}{4\pi cGM}.
$$
Since $P=P_{\rm rad}(1+y)$ and $P_{\rm rad}\propto T^4$,
$$
\frac{dP}{dP_{\rm rad}}
=1+y+\frac T4\frac{dy}{dT}.
$$
With
$$
x=\frac{4\pi cGM}{\kappa_0L}
\frac{k_B}{\mu H}\frac3{a_{\rm rad}}T^{1/2},
\qquad
\frac{dx}{dT}=\frac{x}{2T},
$$
the reciprocal of the preceding moment equation becomes
$$
1+y+\frac{x}{8}\frac{dy}{dx}=\frac{x}{y}.
$$
Multiplication by $y$ proves
$$
\boxed{\frac18xy\frac{dy}{dx}=x-y(y+1)}.
$$
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