= Solution
The terms in the <Boltzmann equation> respectively describe dilution by expansion, pair annihilation, and pair production from the thermal bath. The annihilation rate per volume is $\langle\sigma v\rangle n\bar n$; $P(t)$ is the corresponding production rate. Antiparticles satisfy the same equation with $n$ replaced by $\bar n$. Subtracting gives
$$
\frac{d(n-\bar n)}{dt}=-3H(n-\bar n),\qquad \boxed{(n-\bar n)a^3=\hbox{constant}.}
$$
Initial particle-antiparticle symmetry therefore persists. <Detailed balance> in the thermal bath gives $P=\langle\sigma v\rangle n_{\rm eq}^2$, so
$$
\frac{d(na^3)}{dt}=\langle\sigma v\rangle(n_{\rm eq}^2-n^2)a^3.
$$
During <radiation domination>, with fixed effective relativistic degrees of freedom, $T\propto a^{-1}$ and $H=H_m/x^2$. Thus $\dot x=Hx$ and the dilution terms cancel in $Y=n/T^3$:
$$
\dot Y=-\langle\sigma v\rangle T^3(Y^2-Y_{\rm eq}^2),\qquad \boxed{\frac{dY}{dx}=-\frac{m^3\langle\sigma v\rangle}{H_mx^2}(Y^2-Y_{\rm eq}^2).}
$$
After <cosmological particle freeze-out>, neglect $Y_{\rm eq}$. For constant $\lambda$ the exact reduced solution is
$$
\frac1{Y(x)}=\frac1{Y_f}+\lambda\left(\frac1{x_f}-\frac1x\right),\qquad \boxed{Y_\infty=\left(\frac1{Y_f}+\frac\lambda{x_f}\right)^{-1}.}
$$
For the <post-freeze-out relic abundance>, the familiar answer $Y_\infty\simeq x_f/\lambda$ additionally neglects $1/Y_f$ compared with $\lambda/x_f$. \b[It is an approximation, not an exact consequence of the post-freeze-out equation without initial data.] A smaller annihilation cross-section means smaller $\lambda$, earlier decoupling and less depletion, so weakly interacting particles normally have a larger relic abundance.
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