Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2013/6/solution

Here the dependent variable is and the independent variable is . Multiplying the equation by and setting gives a linear equation:
An integrating factor is , so
Thus . The initial point fixes , and the positive branch through that point is
Equivalently, inverting this branch gives
The 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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