 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-1/31e/v/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

The initial point has

x_{0} = 0

, so its orbit remains in

- 1 < x < 1

. Integrating the orbit equation from part (iv) and using

y (0) = y_{0}

gives

y (x) = y_{0} (1 - x^{2})^{k /2}, - 1 < x < 1.

This formula also covers

y_{0} = 0

and

k = 0

within the open strip. The endpoints are approached only as infinite time limits and have the behavior classified in part (iv).

 New to topics? Read the docs here!