 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ia/paper-2/2a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

Differentiating the Wronskian gives

W^{'} = (y_{1} y_{2}^{'} - y_{2} y_{1}^{'})^{'} = y_{1} y_{2}^{''} - y_{2} y_{1}^{''} = y_{1} (- p y_{2}^{'} - q y_{2}) - y_{2} (- p y_{1}^{'} - q y_{1}) = - p W .

Thus

W^{'} + p W = 0

Also,

(\frac{y _{2}}{y _{1}})^{'} = \frac{y _{1} y _{2}^{'} - y _{2} y _{1}^{'}}{y _{1}^{2}} = \frac{W}{y _{1}^{2}} .

y_{2} (x_{0}) = 0

y_{2} (x) = y_{1} (x) \int_{x_{0}}^{x} \frac{W ( t )}{y _{1} ( t ) ^{2}} d t .

For the specified equation, division by

x^{2}

gives

p (x) = - 1/ x

, so

W = C x

. With

y_{1} = x^{3}

, an independent solution is

x^{3} \int x^{- 5} d x,

which is a nonzero multiple of

x^{- 1}

. Hence

y = A x^{3} + \frac{B}{x} .

The conditions at

x = 1

give

A + B = 0

and

3 A - B = 1

, so

y (x) = \frac{1}{4} (x^{3} - \frac{1}{x}) .

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