Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2015/paper-2/16/b/solution

Put , so . On an interval with , substitution and cancellation give
Since , no constant- solution is lost when separating. Integrate
to obtain
After restoring , a local implicit general solution is
The inverse tangent is chosen continuously on a solution branch. Using the continuous polar angle of extends the expression to charts through when the original equation permits it. A differentiable graph cannot pass through , because there the original equation would require .

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