On page 66 of Modern Cosmology by Dodelson, we have the formula for neutron to proton ratio as a function of time:$$\frac{dXn}{dt}=\lambda_{np}\left\{(1-X_n)e^{-\frac{Q}{T}}-X_n\right\}$$Everything in the derivation up to this point is pretty clear. Then there's an explanation that there are complicated time dependencies in the formula, which appears to be true because Mathematica is unable to numerically solve this formula, so we attempt to simplify the evolution parameter by:$$x=\frac{Q}{T}$$Now this is the part where I get lost. The author states that the left hand side becomes $\dot x \frac{dX_n}{dx}$ The equation then becomes:$$\frac{dx}{dt} \frac{dXn}{dx}=\lambda_{np}\left\{(1-X_n)e^{-\frac{Q}{T}}-X_n\right\}$$ So now all we need is an expression for $\frac{dx}{dt}$. Expanding the steps explicitly, we have: $$x\equiv \frac{Q}{T}$$ $$\frac{dx}{dt}=-\frac{Q\space\dot T}{T^2},Q=x\space T$$ $$\frac{dx}{dt}=-\frac{x\space\dot T}{T},T=\frac{T_0}{a}$$ $$\frac{dx}{dt}=\frac{x\space\dot a}{a},H=\frac{\dot a}{a}$$ $$\frac{dx}{dt}=x\space H$$Substituting, we get: $$x\space H \frac{dXn}{dx}=\lambda_{np}\left\{(1-X_n)e^{-\frac{Q}{T}}-X_n\right\}$$$$\frac{dXn}{dx}=\frac{\lambda_{np}}{x\space H}\left\{(1-X_n)e^{-\frac{Q}{T}}-X_n\right\}$$So how does the author get this formula for (3.27)?$$\frac{dXn}{dx}=\frac{x \lambda_{np}}{H(x=1)}\left\{(1-X_n)e^{-\frac{Q}{T}}-X_n\right\}$$How did the Hubble Function $H$ get moved into the denominator and why is it evaluated at $x=1$ instead of at $x$?

On page 66 of Modern Cosmology by Dodelson, we have the formula for the change in neutron to proton ratio as a function of time:$$\frac{dXn}{dt}=\lambda_{np}\left\{(1-X_n)e^{-\frac{Q}{T}}-X_n\right\}$$Everything in the derivation up to this point is pretty clear. Then there's an explanation that there are complicated time dependencies in the formula, which appears to be true because Mathematica is unable to numerically solve this formula, so we attempt to simplify the evolution parameter by:$$x=\frac{Q}{T}$$Now this is the part where I get lost. The author states that the left hand side becomes $\dot x \frac{dX_n}{dx}$ The equation then becomes:$$\frac{dx}{dt} \frac{dXn}{dx}=\lambda_{np}\left\{(1-X_n)e^{-x}-X_n\right\}$$ So now all we need is an expression for $\frac{dx}{dt}$. Expanding the steps explicitly, we have: $$x\equiv \frac{Q}{T}$$ $$\frac{dx}{dt}=-\frac{Q\space\dot T}{T^2},Q=x\space T$$ $$\frac{dx}{dt}=-\frac{x\space\dot T}{T},T=\frac{T_0}{a}$$ $$\frac{dx}{dt}=\frac{x\space\dot a}{a},H=\frac{\dot a}{a}$$ $$\frac{dx}{dt}=x\space H$$Substituting, we get: $$x\space H \frac{dXn}{dx}=\lambda_{np}\left\{(1-X_n)e^{-\frac{Q}{T}}-X_n\right\}$$$$\frac{dXn}{dx}=\frac{\lambda_{np}}{x\space H}\left\{(1-X_n)e^{-\frac{Q}{T}}-X_n\right\}$$So how does the author get this formula for (3.27)?$$\frac{dXn}{dx}=\frac{x \lambda_{np}}{H(x=1)}\left\{(1-X_n)e^{-\frac{Q}{T}}-X_n\right\}$$How did the Hubble Function $H$ get moved into the denominator and why is it evaluated at $x=1$ instead of at $x$?

On page 66 of Modern Cosmology by Dodelson, we have the formula for neutron to proton ratio as a function of time:$$\frac{dXn}{dt}=\lambda_{np}\left\{(1-X_n)e^{-\frac{Q}{T}}-X_n\right\}$$Everything in the derivation up to this point is pretty clear. Then there's an explanation that there are complicated time dependencies in the formula, which appears to be true because Mathematica is unable to numerically solve this formula, so we attempt to simplify the evolution parameter by:$$x=\frac{Q}{T}$$Now this is the part where I get lost. The author states that the left hand side becomes $\dot x \frac{dX_n}{dx}$ The equation then becomes:$$\frac{dx}{dt} \frac{dXn}{dx}=\lambda_{np}\left\{(1-X_n)e^{-\frac{Q}{T}}-X_n\right\}$$ So now all we need is an expression for $\frac{dx}{dt}$. Expanding the steps explicitly, we have: $$x\equiv \frac{Q}{T}$$ $$\frac{dx}{dt}=-\frac{Q\space\dot T}{T^2},Q=x\space T$$ $$\frac{dx}{dt}=-\frac{x\space\dot T}{T},T=\frac{T_0}{a}$$ $$\frac{dx}{dt}=\frac{x\space\dot a}{a},H=\frac{\dot a}{a}$$ $$\frac{dx}{dt}=x\space H$$Substituting, we get: $$x\space H \frac{dXn}{dx}=\lambda_{np}\left\{(1-X_n)e^{-\frac{Q}{T}}-X_n\right\}$$$$\frac{dXn}{dx}=\frac{\lambda_{np}}{x\space H}\left\{(1-X_n)e^{-\frac{Q}{T}}-X_n\right\}$$So how does the author get this formula for (3.27)?$$\frac{dXn}{dx}=\frac{x \lambda_{np}}{H(x=1)}\left\{(1-X_n)e^{-\frac{Q}{T}}-X_n\right\}$$How did the $x$ get into the numerator and why is the Hubble function evaluated at $x=1$ instead of at $x$?

On page 66 of Modern Cosmology by Dodelson, we have the formula for neutron to proton ratio as a function of time:$$\frac{dXn}{dt}=\lambda_{np}\left\{(1-X_n)e^{-\frac{Q}{T}}-X_n\right\}$$Everything in the derivation up to this point is pretty clear. Then there's an explanation that there are complicated time dependencies in the formula, which appears to be true because Mathematica is unable to numerically solve this formula, so we attempt to simplify the evolution parameter by:$$x=\frac{Q}{T}$$Now this is the part where I get lost. The author states that the left hand side becomes $\dot x \frac{dX_n}{dx}$ The equation then becomes:$$\frac{dx}{dt} \frac{dXn}{dx}=\lambda_{np}\left\{(1-X_n)e^{-\frac{Q}{T}}-X_n\right\}$$ So now all we need is an expression for $\frac{dx}{dt}$. Expanding the steps explicitly, we have: $$x\equiv \frac{Q}{T}$$ $$\frac{dx}{dt}=-\frac{Q\space\dot T}{T^2},Q=x\space T$$ $$\frac{dx}{dt}=-\frac{x\space\dot T}{T},T=\frac{T_0}{a}$$ $$\frac{dx}{dt}=\frac{x\space\dot a}{a},H=\frac{\dot a}{a}$$ $$\frac{dx}{dt}=x\space H$$Substituting, we get: $$x\space H \frac{dXn}{dx}=\lambda_{np}\left\{(1-X_n)e^{-\frac{Q}{T}}-X_n\right\}$$$$\frac{dXn}{dx}=\frac{\lambda_{np}}{x\space H}\left\{(1-X_n)e^{-\frac{Q}{T}}-X_n\right\}$$So how does the author get this formula for (3.27)?$$\frac{dXn}{dx}=\frac{x \lambda_{np}}{H(x=1)}\left\{(1-X_n)e^{-\frac{Q}{T}}-X_n\right\}$$How did the Hubble Function $H$ get moved into the denominator and why is it evaluated at $x=1$ instead of at $x$?