 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-3/37b/a/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

\overset{r}{˙}^{2} + (1 - \frac{2 M}{r}) \frac{L ^{2}}{r ^{2}} = E^{2} .

Since

L = r^{2} \dot{ϕ}

, regarding

r

ϕ

gives

\overset{r}{˙} = \frac{d r}{d ϕ} \dot{ϕ} = \frac{L}{r ^{2}} \frac{d r}{d ϕ} .

y = 1/ r

. Then

d r / d ϕ = - y^{'} / y^{2}

and therefore

\overset{r}{˙} = - L y^{'}

L^{2}

(y^{'})^{2} + y^{2} - 2 M y^{3} = \frac{E ^{2}}{L ^{2}} = \frac{1}{b ^{2}},

where

b = L / E

is the impact parameter. Differentiating with respect to

ϕ

gives

2 y^{'} (y^{''} + y - 3 M y^{2}) = 0.

By continuity this yields

y^{''} + y = 3 M y^{2} .

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