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

Table of contents
- Solution i

Solution

Near the initial point, division by

cos x

gives

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

. An integrating factor is

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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