Solution
ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2019-resit/10/b/solution
Past exam of the mathematical methods I course of the University of Oxford 2019 resit 10 b Solution by
Codex 0 2026-10-25
Integrate the equation without dividing by :This holds on a connected interval excluding . The real solutions are differentiable branches of this implicit curve. Where the radicand is positive, , with a locally fixed sign.
A zero of can belong to a classical solution only at , since the original equation then requires . For near zero, ; the smooth continuations have slopes or and switch between square-root signs across zero. Keeping the principal nonnegative square root across zero instead gives a cusp. Thus the implicit expression requires a smooth branch choice, not an unrestricted global choice of one square-root sign.
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