 Solution

Let $m_g=GM/c^2$, $u=1/r$, and use a prime for $d/d\varphi$. The null geodesic first integral obtained above becomesDifferentiating with respect to $\varphi$ and dividing by $2u^{\prime }$ away from a turning point givesThe same differential equation holds at the turning point by continuity. When $m_g=0$, the general solution is $u_0=C_1\cos \varphi +C_2\sin \varphi$. Choose the angular origin so that the incoming asymptote has $\varphi =0$ and the unperturbed closest approach has $\varphi =\pi /2$. The impact parameter then fixesThis is the straight reference path, not a curved orbit.

For perturbation theory with $m_g/b\ll 1$, write $u=u_0+u_1+O(m_g^2/b^3)$, where $u_1=O(m_g/b^2)$. Inserting this into the differential equation and discarding second-order terms givesA particular solution and the complementary solution areChoosing $A_1=B_1=0$ retains reflection symmetry about $\varphi =\pi /2$ and the specified impact parameter to first order. The two distant zeros of $u$ then occur at ${\varphi }_{\mathrm{in}}\simeq -2m_g/b$ and ${\varphi }_{\mathrm{out}}\simeq \pi +2m_g/b$. Their separation exceeds $\pi$ by $4m_g/b$, consistently giving the Schwarzschild light deflection $\delta \simeq 4GM/(b{c}^2)$.