 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/a/solution

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

Separate variables away from the equilibrium solutions. Partial fractions give

\frac{1}{a - b} ln \frac{y - a}{y - b} = - x + C, \frac{y - a}{y - b} = K e^{(b - a) x} .

y = \frac{a e ^{a x} - b K e ^{b x}}{e ^{a x} - K e ^{b x}}, K \in R, together with y = b,

on intervals where the denominator is nonzero. The value

K = 0

includes

y = a

. Setting

K = e^{(b - a) c}

recovers the printed expression after factoring out

e^{a c}

. For real

c

that substitution covers positive

K

only; negative

K

gives additional real branches between the equilibria, and the constant solutions must not be discarded when separating variables.

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