 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2006/paper-2/9/d/solution

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

Differentiate

z^{'} = yz

without dividing by

z

z^{''} = (y^{'} + y^{2}) z = [(a + b) y - ab] z = (a + b) z^{'} - ab z .

z^{''} - (a + b) z^{'} + ab z = 0.

(λ - a) (λ - b)

, explaining both the distinct-root exponentials and the repeated-root factor

A + B x

. Nonzero solutions reconstruct

y = z^{'} / z

between their zeros; the identically zero

z

also satisfies the linear equation.

 New to topics? Read the docs here!