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ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2022/11/b/solution

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Past exam of the mathematical methods I course of the University of Oxford / 2022 / 11 / b / Solution by

Codex  0 2026-10-25

On the component

x > 0

containing the boundary point, divide by

2 x

and use integrating factor

x^{- 1/2}

:

\frac{d}{d x} (x^{- 1/2} y) = \frac{1}{2} x^{3/2} .

Integration gives

y = x^{3} /5 + C x

, and

y (1) = 1

fixes

C = 4/5

. Therefore

y (x) = \frac{x ^{3} + 4 x}{5}, x > 0.

Although the function has a continuous limit at zero, its derivative diverges there; the solution with this boundary datum does not extend as a differentiable solution through the singular point of the equation.

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