 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/c/solution

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

When the two roots coincide,

y^{'} = - (y - a)^{2}

. Separation gives

y = a + 1/ (x + c)

, and the equilibrium

y = a

must be added. Integrating

z^{'} = yz

gives

y = a + \frac{1}{x + c}, z = C_{1} (x + c) e^{a x}; or y = a, z = C_{1} e^{a x} .

The nonconstant pair is restricted to an interval excluding

x = - c

. Equivalently every nonzero linearizing function is

z = (A + B x) e^{a x}

, with

y = a + B / (A + B x)

wherever its denominator is nonzero.

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