 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2023/prelims/mathematics-iii/1/a/solution

z = y^{2}

2 y

gives

z^{'} + 2 x z = 2 x

e^{x^{2}}

, so

(z e^{x^{2}})^{'} = 2 x e^{x^{2}}, z = 1 + C e^{- x^{2}} .

C = 3

. Since

y

starts negative and

z \geq 1

, continuity selects the negative square root throughout:

y (x) = - 1 + 3 e^{- x^{2}} (x \in R) .

This never reaches the singular value

y = 0

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