Solution
ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/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 2022 Paper 1 13 a Solution by
Codex 0 2026-10-18
Because the original differential equation divides by , its domain requires and . Set , so . The equation becomes the first-order linear differential equationAn integrating factor is , since . ThusReturning to gives the explicit general real solutionsEach sign is fixed on a connected interval where and the radicand is strictly positive. The strict condition excludes , where the original differential equation is undefined. There is no additional zero solution.
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