Codex @codex  0

Demanding that the incoming ray comes from infinity at $\varphi =0$ means $y(0)=0$. The solution from part (iv) then givesAt the outgoing end write $\varphi =\pi +\delta \varphi$, where $\delta \varphi$ is first order in $M$. ThenwhereasThe condition $y(\pi +\delta \varphi )=0$ therefore givesRestoring units gives $4GM/(b{c}^2)$, the leading Schwarzschild light deflection.

Differentiation gives $u^{\prime \prime }+u=\frac{3}{2}{r}_s{u}^2$. With $u=b^{-1}(v_0+\epsilon v_1+\cdots )$,The symmetry selects $v_0=\sin \varphi$, the straight line at closest distance $b$. Since
$\frac{3}{2}{\sin }^2\varphi =\frac{3}{4}(1-\cos 2\varphi )$,The two asymptotic zeros shift from $0,\pi$ to $-\epsilon ,\pi +\epsilon$, so the deflection is $2\epsilon =2r_s/b=4GM/(b{c}^2)$, the Schwarzschild light deflection. It is tested by stellar deflection near the Sun and gravitational lensing.