= Solution
<Differentiation> gives $u''+u=\frac32r_su^2$. With $u=b^{-1}(v_0+\varepsilon v_1+\cdots)$,
$$
v_0''+v_0=0,\qquad v_1''+v_1=\frac32v_0^2.
$$
The symmetry selects $v_0=\sin\phi$, the straight line at closest distance $b$. Since
$\frac32\sin^2\phi=\frac34(1-\cos2\phi)$,
$$
v_1=\frac34+\frac14\cos2\phi.
$$
The two asymptotic zeros shift from $0,\pi$ to $-\varepsilon,\pi+\varepsilon$, so the deflection is $2\varepsilon=2r_s/b=4GM/(bc^2)$, the <Schwarzschild light deflection>. It is tested by stellar deflection near the Sun and gravitational lensing.
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