= Solution
Let $\widehat{\mathbf n}$ point from the observer toward the emission point, so the photon propagation direction is $\widehat{\mathbf p}=-\widehat{\mathbf n}$ and
$$
\mathbf x(\eta)=\mathbf x_0+(\eta_0-\eta)\widehat{\mathbf n}.
$$
Along this path $D=d/d\eta=\partial_\eta-\widehat{\mathbf n}\cdot\nabla$. Substituting $\widehat{\mathbf p}=-\widehat{\mathbf n}$ into the <Photon Boltzmann equation with Thomson scattering> gives
$$
D(\Theta+\Psi)+\Gamma(\Theta+\Psi)
=\Phi'+\Psi'+\Gamma(\Theta_0+\Psi-\widehat{\mathbf n}\cdot\mathbf v_e).
$$
Because $\partial_\eta e^{-\tau}=\Gamma e^{-\tau}$, multiplication by the integrating factor gives
$$
D[e^{-\tau}(\Theta+\Psi)]=\widehat S,
$$
where
$$
\boxed{\widehat S=e^{-\tau}(\Phi'+\Psi')
+g(\Theta_0+\Psi-\widehat{\mathbf n}\cdot\mathbf v_e)}.
$$
Integrating along the ray and using $e^{-\tau}\to0$ at sufficiently early time yields the <Cosmic microwave background line-of-sight solution>
$$
\boxed{(\Theta+\Psi)(\eta_0,\mathbf x_0,\widehat{\mathbf n})
=\int_0^{\eta_0}d\eta',
\widehat S(\eta',\mathbf x_0+(\eta_0-\eta')\widehat{\mathbf n},
\widehat{\mathbf n})}.
$$
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