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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/2002/paper-2/7/b/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 / 2002 / Paper 2 / 7 / b / Solution by

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

Write

y (θ) = Y (x)

with

x = cos θ

. By the chain rule,

\frac{d y}{d θ} = - sin θ Y_{x}, \frac{d ^{2} y}{d θ ^{2}} = - cos θ Y_{x} + sin^{2} θ Y_{xx} .

The terms involving

Y_{x}

cancel in the original ordinary differential equation, leaving

sin θ y_{θθ} - cos θ y_{θ} + 2 y sin^{3} θ = sin^{3} θ (Y_{xx} + 2 Y) = 0.

On each interval where

sin θ \neq = 0

, the transformed linear ordinary differential equation is

Y_{xx} + 2 Y = 0

. Its characteristic equation has roots

\pm i 2

, yielding

y (θ) = C_{1} cos (2 cos θ) + C_{2} sin (2 cos θ) .

Direct substitution shows these expressions also extend smoothly through

θ = kπ

and satisfy the original equation there. The two constants describe the general solution on any interval where the coordinate substitution is invertible.

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