 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-1/13/a/solution

For

x \neq = 0

, divide by

x^{2}

to obtain

y^{'} + y / x = 1/ x^{2}

x

on either interval not crossing zero, since the product rule gives

(x y)^{'} = \frac{1}{x} .

Integration and division by

x

give the explicit general solution

y (x) = \frac{ln ∣ x ∣ + C}{x}, x \neq = 0.

The original equation would read

0 = 1

at zero, so no solution can include that point.

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