 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2005/paper-1/9/c/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

Rearranging gives

y^{'} = 2 x (y - 1)^{2}

y = 1

. For every other solution, separation of variables gives

\int (y - 1)^{- 2} d y = \int 2 x d x

, so

- 1/ (y - 1) = x^{2} + C

. Hence

y = 1, or y = 1 - \frac{1}{x ^{2} + C} .

A nonconstant solution is defined on any interval where its denominator is nonzero. Dividing by

(y - 1)^{2}

without retaining the equilibrium would lose one solution.

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