 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ib/paper-3/7a/b/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

y (x) = \int_{0}^{1} G (x; ξ) e^{2 ξ} d ξ = e^{2 x} [(x - 1) \int_{0}^{x} ξ d ξ + x \int_{x}^{1} (ξ - 1) d ξ] .

The bracket simplifies to

x (x - 1) /2

, so

y (x) = \frac{1}{2} x (x - 1) e^{2 x} .

It vanishes at both endpoints, and applying

(D - 2)^{2}

returns

e^{2 x}

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