 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-1/7/solution

For a nonzero solution, the separable differential equation gives

d y / y = 3 x^{2} d x

. Integrating yields

ln ∣ y ∣ = x^{3} + C

, hence

y = K e^{x^{3}}

K = 3/ e

, so

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

The zero solution omitted by division by

y

cannot satisfy the given initial condition. Differentiating the displayed solution verifies

y^{'} = 3 x^{2} y

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