Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 2 / 5C / a

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- Solution a

Solution

The Wronskian is

W = y_{1} y_{2}^{'} - y_{2} y_{1}^{'}

. Abel's identity gives

W^{'} = - p W .

If

W \neq = 0

, the two column vectors in the displayed representation form a basis of

R^{2}

at each

x

. Differentiating the first component and comparing it with the second gives

u^{'} y_{1} + v^{'} y_{2} = 0.

Substitution in the differential equation gives

u^{'} y_{1}^{'} + v^{'} y_{2}^{'} = f .

Solving this two-by-two system yields

u^{'} = - \frac{y _{2} f}{W}, v^{'} = \frac{y _{1} f}{W} .

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