= Solution
Demanding that the incoming ray comes from infinity at $\phi=0$ means $y(0)=0$. The solution from part (iv) then gives
$$
A+\frac{M}{2b^2}(3+1)=0,
\qquad
A=-\frac{2M}{b^2}.
$$
At the outgoing end write $\phi=\pi+\delta\phi$, where $\delta\phi$ is first order in $M$. Then
$$
y_0(\pi+\delta\phi)
=-\frac{\delta\phi}{b}+O(\delta\phi^3),
$$
whereas
$$
\Delta y(\pi)
=-A+\frac{M}{2b^2}(3+1)
=\frac{4M}{b^2}.
$$
The condition $y(\pi+\delta\phi)=0$ therefore gives
$$
\boxed{\delta\phi=\frac{4M}{b},
\qquad |\delta\phi|\simeq\frac{4M}{b}.}
$$
Restoring units gives $4GM/(bc^2)$, the leading <Schwarzschild light deflection>.
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