= Solution
{c}
Let $m_g=GM/c^2$, $u=1/r$, and use a prime for $d/d\phi$. The <null geodesic> first integral obtained above becomes
$$
(u')^2+u^2-2m_g u^3=\frac1{b^2}.
$$
Differentiating with respect to $\phi$ and dividing by $2u'$ away from a turning point gives
$$
\boxed{u''+u=3m_g u^2=\frac{3GM}{c^2}u^2.}
$$
The same <differential equation> holds at the turning point by continuity. When $m_g=0$, the general solution is $u_0=C_1\cos\phi+C_2\sin\phi$. Choose the angular origin so that the incoming asymptote has $\phi=0$ and the unperturbed closest approach has $\phi=\pi/2$. The <impact parameter> then fixes
$$
\boxed{u_0=\frac{\sin\phi}{b},\qquad r\sin\phi=b.}
$$
This is the straight reference path, not a curved orbit.
For <perturbation theory> with $m_g/b\ll1$, 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 gives
$$
u_1''+u_1=3m_g u_0^2
=\boxed{\frac{3GM}{c^2}\frac{1-\cos2\phi}{2b^2}.}
$$
A particular solution and the complementary solution are
$$
u_1=\frac{m_g}{2b^2}(3+\cos2\phi)+A_1\cos\phi+B_1\sin\phi.
$$
Choosing $A_1=B_1=0$ retains reflection symmetry about $\phi=\pi/2$ and the specified <impact parameter> to first order. The two distant zeros of $u$ then occur at $\phi_{\rm in}\simeq-2m_g/b$ and $\phi_{\rm out}\simeq\pi+2m_g/b$. Their separation exceeds $\pi$ by $4m_g/b$, consistently giving the <Schwarzschild light deflection> $\delta\simeq4GM/(bc^2)$.
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