 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2011/paper-2/7/solution

e^{- y}

and integrate:

- e^{- y} = \frac{1}{2} ln (1 + x^{2}) + C .

C = - 1

. Hence

y (x) = - ln (1 - \frac{1}{2} ln (1 + x^{2})) .

For a real solution the outer logarithm requires a positive argument. The maximal interval containing the initial point is therefore

∣ x ∣ < e^{2} - 1

. At either endpoint

y

diverges, so the real solution cannot continue through it with a finite value.

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