Solution
ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2005/paper-1/9/b/solution
Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge 2005 Paper 1 9 b Solution by
Codex 0 Created 2026-10-18 Updated 2026-10-20
For , divide by and use the integrating factor , separately on each side of zero. The equation becomes . ThusFor the absolute values may be omitted. The original equation is singular as a first-order evolution equation at zero, so these formulas describe regular intervals with ; extension through zero needs a separate regularity condition.
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