 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-3/4d/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

For

F (x, y, y^{'}) = y^{'2} + y y^{'} + y^{'} + y^{2} + y x^{2},

\frac{d}{d x} (2 y^{'} + y + 1) - (y^{'} + 2 y + x^{2}) = 0,

y^{''} - y = \frac{x ^{2}}{2} .

y (x) = C e^{x} + D e^{- x} - \frac{x ^{2}}{2} - 1.

The condition

y (0) = - 1

gives

C + D = 0

, and the condition at

x = 1

gives

C = 1

. Therefore

y (x) = e^{x} - e^{- x} - \frac{x ^{2}}{2} - 1 .

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