= Solution
Use <radiative homology with proton-proton burning and inverse-fourth-power opacity>, retaining the composition factors throughout. To distinguish the stellar <radius> from the <stellar gas constant>, write them as $R$ and $\mathcal R$. With <gas pressure> alone, introduce the scale factors
$$
\rho_* =\frac{M}{4\pi R^3},\qquad P_* =\frac{GM^2}{4\pi R^4},\qquad T_* =\frac{\mu GM}{\mathcal RR}.
$$
The <ideal gas> equation gives $\rho=\rho_*b$, $P=P_*p$ and $T=T_*p/b$. Substitution into <hydrostatic equilibrium> and <mass conservation> gives
$$
\frac{P_*}{R}\frac{dp}{dx}=-\frac{GM\rho_*}{R^2}\frac{qb}{x^2},\qquad \frac{M}{R}\frac{dq}{dx}=4\pi R^2\rho_*x^2b,
$$
so \b[$p'=-bq/x^2$ and $q'=x^2b$].
Set $\zeta=Z(1+X)$. The <radiative diffusion in a star> equation, with the stipulated <opacity>, becomes
$$
\frac{T_*}{R}\frac{d}{dx}\left(\frac pb\right)=-\frac{3\kappa_0\zeta\rho_*^2L}{16\pi acR^2T_*^7}\frac{b^9l}{x^2p^7}.
$$
Here $\rho^2T^{-7}=\rho_*^2T_*^{-7}b^9p^{-7}$. Thus
$$
\boxed{\frac{d}{dx}\left(\frac pb\right)=-D\frac{b^9l}{x^2p^7},\qquad D=\frac{3\kappa_0Z(1+X)\mathcal R^8LR}{256\pi^3ac\mu^8G^8M^6}.}
$$
The <luminosity> equation follows from the specific heating rate of the <Proton–proton chain>:
$$
\frac{L}{R}\frac{dl}{dx}=4\pi R^2x^2\epsilon_0X^2\rho_*^2T_*^4p^4b^{-2}.
$$
Consequently
$$
\boxed{\frac{dl}{dx}=Ex^2p^4b^{-2},\qquad E=\frac{\epsilon_0X^2M^6}{4\pi R^7L}\left(\frac{\mu G}{\mathcal R}\right)^4.}
$$
These give all four dimensionless <stellar structure equations> with their numerical coefficients.
The <stellar homology> assumption means the dimensionless profiles are shared across the models, not merely that an individual model can be written in dimensionless coordinates. The displayed equations then require $D$ and $E$ to be fixed. Holding the universal constants and $\kappa_0,\epsilon_0$ fixed gives
$$
L\propto\frac{\mu^8M^6}{\zeta R},\qquad L\propto\frac{X^2\mu^4M^6}{R^7}.
$$
Equating the two cancels $M^6$ and yields $R^6\propto X^2\zeta\mu^{-4}$. Therefore the <radius> and <mass-luminosity relation> are
$$
\boxed{R\propto X^{1/3}[Z(1+X)]^{1/6}\mu^{-2/3},\qquad L\propto\frac{M^6\mu^{26/3}}{X^{1/3}Z^{7/6}(1+X)^{7/6}}.}
$$
The mass-independent <radius> is a consequence of the particular heating and <opacity> exponents in this model; it is not a general <radius> law for main-sequence <stars>.
Initially $X=X_0$ is the same for all <stars>. The adopted <mean molecular weight> $\mu=4/(5X+3)$ also has the same initial value, neglecting its small direct dependence on $Z$. Thus $R\propto Z^{1/6}$. At fixed <luminosity>, the <Stefan–Boltzmann law> $L=4\pi R^2\sigma T_e^4$ gives
$$
\boxed{T_e\propto R^{-1/2}\propto Z^{-1/12}.}
$$
The <stars> compared here can have different <masses>; the <luminosity> law adjusts the <mass> to keep $L$ fixed.
For the evolution, take the stellar <mass> as constant and maintain homogeneous composition. The available energy in the remaining <hydrogen> is $ME_HX$, so <conservation of energy> in the nuclear-powered model gives
$$
L=-\frac{d}{dt}(ME_HX),\qquad\boxed{\frac{dX}{dt}=-\frac{L}{ME_H}.}
$$
This includes the entire mixed <hydrogen> reservoir, not just the <mass> in a central burning region.
To determine the <turnoff mass of a homogeneously mixed proton-proton-burning population>, write the <luminosity> law as $L=C M^6 Z^{-7/6}H(X)$, where
$$
H(X)=\frac{\mu(X)^{26/3}}{X^{1/3}(1+X)^{7/6}},\qquad \mu(X)=\frac4{5X+3}.
$$
Separating variables, the time to evolve from $X_0$ to $fX_0$ is
$$
t=\frac{E_H}{C}M^{-5}Z^{7/6}\int_{fX_0}^{X_0}X^{1/3}(1+X)^{7/6}\left(\frac{5X+3}{4}\right)^{26/3}dX.
$$
For a common $X_0$ and $0\leq f<1$, this integral is a positive finite constant independent of $M,Z,t$. It follows that \b[the turning-off <stars> satisfy]
$$
\boxed{M\propto Z^{7/30}t^{-1/5}.}
$$
The integration retains the evolving <luminosity> and <mean molecular weight>; replacing $L$ by its initial value is unnecessary.
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