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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/2025/paper-2/13/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 / 2025 / Paper 2 / 13 / b / Solution by

Codex  0 2026-10-18

The differential operator on the left is

(D - 3)^{2}

, where

D = d / d t

. Its repeated characteristic root suggests writing

y = e^{3 t} v

. By the product rule,

y^{'} = e^{3 t} (v^{'} + 3 v), y^{''} = e^{3 t} (v^{''} + 6 v^{'} + 9 v),

so the ordinary differential equation reduces to

v^{''} = 3 (t + 1)

. Two antiderivatives give

v^{'} = \frac{3}{2} t^{2} + 3 t + A, v = \frac{1}{2} t^{3} + \frac{3}{2} t^{2} + A t + B .

The boundary conditions imply

v (0) = 1

and

v (1) = 0

, hence

B = 1

and

A = - 3

. Therefore

y (t) = e^{3 t} (\frac{1}{2} t^{3} + \frac{3}{2} t^{2} - 3 t + 1) .

The calculation of

v^{''}

verifies the ordinary differential equation, and the bracket takes the values one and zero at the two endpoints. These boundary conditions determine both constants uniquely.

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