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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/a/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2005 / Paper 1 / 9 / a / Solution by

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

Use separation of variables on intervals avoiding zeros of

cos x

\int \frac{d y}{1 + y ^{2}} = \int sec^{2} x d x, arctan y = tan x + C .

(1)

Taking the tangent gives the real general solution

y (x) = tan (tan x + C) .

(2)

Each solution is restricted to an interval where both the original equation and this expression are finite. There is no real constant solution because

1 + y^{2}

never vanishes for real

y

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