Einstein's light-deflection formula

Step 1 (the orbit equation for light). For a null geodesic confined to the equatorial plane, the same conserved quantities $E=(1-r_s/r)c^2\dot t$ and $L=r^2\dot\phi$ used for the photon sphere, substituted into $g_{\mu\nu}\dot x^\mu\dot x^\nu=0$ and converted from $\tau$-derivatives to a $\phi$-derivative of $u\equiv1/r$ (using $\dot r = -L\,du/d\phi$), give after one more differentiation the standard light-bending orbit equation $$\frac{d^2u}{d\phi^2}+u = \frac{3r_s}{2}u^2, \qquad u\equiv\frac1r$$

Step 2 (zeroth order: the straight line). Dropping the right-hand side (setting $r_s=0$, flat spacetime) gives $u_0''+u_0=0$, whose solution passing at closest distance $b$ at $\phi=\pi/2$ is the straight line $$u_0(\phi) = \frac{\sin\phi}{b}$$ — the undeflected path in polar form.

Step 3 (first-order perturbation). Write $u=u_0+r_s u_1$ and substitute into Step 1's equation, keeping only terms linear in $r_s$; the $u_0''+u_0=0$ part cancels and what remains is a driven linear oscillator for $u_1$: $$u_1''+u_1 = \frac{3}{2b^2}\sin^2\phi$$ A particular solution matching the symmetric bending expected on both sides of closest approach is $$u_1(\phi) = \frac{1}{2b^2}\left(1+\cos^2\phi\right)$$ so the full first-order trajectory is $$u(\phi) = \frac{\sin\phi}{b} + \frac{r_s}{2b^2}\left(1+\cos^2\phi\right)$$

Step 4 (extract the total bending angle). Flat space has $u\to0$ exactly at $\phi=0$ and $\phi=\pi$; with the $r_s$ correction, $u\to0$ at $\phi=-\delta_1$ and $\phi=\pi+\delta_2$ for small $\delta_1,\delta_2$. Expanding $u(-\delta_1)=0$ to first order in $\delta_1$ and $r_s$ gives $-\delta_1/b+r_s/b^2=0$, so $\delta_1=r_s/b$; the same computation at the other end gives $\delta_2=r_s/b$. The total angle by which the outgoing asymptote misses being exactly antiparallel to the incoming one is $$\Delta\phi = \delta_1+\delta_2 = \frac{2r_s}{b} = \frac{4GM}{c^2 b}$$ which, restoring $r_s=2GM/c^2$, is the claimed formula.