 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-1/6/solution

Separate variables near the nonzero initial condition:

y^{- 2} d y = d x

- 1/ y = x + C

, and

y (1) = 1

fixes

C = - 2

. Therefore

y (x) = \frac{1}{2 - x}, - \infty < x < 2.

This is the maximal connected interval containing the initial point. Direct differentiation gives

y^{'} = 1/ (2 - x)^{2} = y^{2}

; the pole at

x = 2

prevents continuation as a finite classical solution of a differential equation through that point.

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