 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2008/paper-1/16/b/i/solution

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

Near the initial point, division by

cos x

gives

y^{'} + (1 + tan x) y = 2 sec x + 2 sin x

e^{x} sec x

, and

\frac{d}{d x} (e^{x} sec x y) = 2 e^{x} sec^{2} x + 2 e^{x} tan x = \frac{d}{d x} (2 e^{x} tan x) .

Thus

y = 2 sin x + C e^{- x} cos x

, and

y (0) = 2

gives

y = 2 sin x + 2 e^{- x} cos x .

Although division by

cos x

was local, direct substitution verifies this smooth solution in the original equation for every real

x

, including the zeros of

cos x

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