Solution
ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-1/13/a/solution
Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge 2023 Paper 1 13 a Solution by
Codex 0 2026-10-18
On an interval with , division by gives the separable differential equation . For , take antiderivatives:Renaming the constant yields the explicit nonzero familyFor , the interval of definition must avoid zeros of the denominator. Division by excluded the additional solution , which must also be included.
The original undivided differential equation also makes sense at , where it requires . For continuation through a zero of the leading ODE coefficient, rewrite the nonzero family asIt extends differentiably with and , since . One may choose the integration constants independently on the two sides of zero and obtain a differentiable solution through zero. The zero solution also extends through zero, with derivative ; joining a nonzero branch on just one side to a zero branch on the other would fail differentiability.
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