 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2002/paper-1/8/c/solution

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

y (x) = \int_{1}^{e} G (x, ξ) f (ξ) d ξ

. Split the integral at

ξ = x

: the portion

ξ > x

uses the first branch of the Green's function, and the portion

ξ < x

uses the second. This yields

y (x) = x ln x \int_{x}^{e} (ln ξ - 1) f (ξ) d ξ + x (ln x - 1) \int_{1}^{x} ln ξ f (ξ) d ξ .

For

f (ξ) = 1/ ξ

, put

t = ln x

. The two integrals are

\int_{x}^{e} \frac{l n ξ - 1}{ξ} d ξ = - \frac{1}{2} - \frac{t ^{2}}{2} + t, \int_{1}^{x} \frac{l n ξ}{ξ} d ξ = \frac{t ^{2}}{2} .

Substitution and cancellation give

y (x) = \frac{x}{2} [(ln x)^{2} - ln x] .

Indeed

y (1) = y (e) = 0

. For a direct differential check,

y^{'} = \frac{1}{2} (t^{2} + t - 1), y^{''} = \frac{2 t + 1}{2 x},

and therefore

y^{''} - \frac{y ^{'}}{x} + \frac{y}{x ^{2}} = \frac{( 2 t + 1 ) - ( t ^{2} + t - 1 ) + ( t ^{2} - t )}{2 x} = \frac{1}{x} .

The homogeneous uniqueness shown in part (a) makes this the unique solution.

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