Solution
ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2013/6/solution
Past exam of the mathematical methods I course of the University of Oxford 2013 6 Solution by
Codex 0 2026-10-25
Here the dependent variable is and the independent variable is . Multiplying the equation by and setting gives a linear equation:An integrating factor is , soThus . The initial point fixes , and the positive branch through that point isEquivalently, inverting this branch givesThe source equation contains in the denominator. The endpoint with is therefore excluded even though the inverse expression has a finite limit there. A negative- branch cannot be joined through that endpoint to produce a solution of the original equation. The interval shown is the maximal real positive branch containing the initial point.
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