Past exam of the mathematical methods I course of the University of Oxford / 2015 / 3

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- Solution 3

Solution

An integrating factor for the first-order linear differential equation is

μ (x) = e^{s i n x}

. Multiplication gives

(y e^{s i n x})^{'} = 2 sin x cos x e^{s i n x} .

Substitute

u = sin x

in the antiderivative:

\int 2 u e^{u} d u = 2 (u - 1) e^{u}

. Therefore

y (x) = 2 sin x - 2 + C e^{- s i n x}, C \in R .

Differentiating verifies the equation everywhere on the real line; no division by

cos x

restricts the solution.

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