 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ia/paper-1/2a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

Invert the equation and regard

x

y

\frac{d x}{d y} - x = e^{2 y} .

e^{- y}

gives

\frac{d}{d y} (x e^{- y}) = e^{y},

x = e^{2 y} + C e^{y}

x = 1

y = 0

gives

C = 0

. Hence

x = e^{2 y}

and

y (x) = \frac{1}{2} lo g x, x > 0.

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