 Solution

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.