Solution
ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-2/14/c/solution
Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge 2014 Paper 2 14 c Solution by
Codex 0 2026-10-18
Here , whereas . Thus the ratio in (b) is , and one may choose the nonzero integrating factor on either half-line. A negative constant multiple on a half-line is harmless.
The multiplied form isHence , a quadratic equation for . Solving it givesChoose a real branch with positive discriminant to obtain a differentiable graph; a zero discriminant produces a vertical tangent. Although the ratio derivation initially cancels , the displayed exact differential verifies the multiplying factor directly wherever . At the branches are and . Only has a finite smooth extension across , where the original equation requires .
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