Solution
ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-2/15/a/solution
Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge 2021 Paper 2 15 a Solution by
Codex 0 2026-10-18
Because the equation depends on the ratio , set , so . Dividing by and substituting givesThis reduces the homogeneous first-order differential equation to a separable differential equation. For ,At , the condition gives , so and . HenceThe interval displayed is the maximal solution interval containing : there is a pole at . Separating variables omitted the solution , or , but that solution does not satisfy the given condition.
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