 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2003/ia/paper-2/1d/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

Separate variables and integrate on

∣ y ∣ < 1

\int \frac{d y}{1 - y ^{2}} = x + C, artanh y = \frac{1}{2} lo g \frac{1 + y}{1 - y} = x + C .

Hence the general solution in this region is

y (x) = tanh (x + C), C \in R .

Every such solution increases from values near

- 1

towards

1

, remains strictly between the two equilibria, and is defined for all real

x

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