 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2013/discrete-mathematics-probability-and-continuous-mathematics/6/c/solution

Divide by

x^{2}

, which is allowed because

x > 0

y^{'} + \frac{1}{x} y = 4 e^{x^{2}} .

e^{\int d x / x} = x

. Multiplying through makes the left side a product rule derivative:

(x y)^{'} = 4 x e^{x^{2}} .

Part (b) now gives

x y = 2 e^{x^{2}} + C

. Therefore

y (x) = \frac{2 e ^{x^{2}} + C}{x}, x > 0.

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