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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/ia/paper-2/6a/a/ii/solution

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Past exam of the mathematics course of the University of Cambridge / 2016 / ia / Paper 2 / 6A / a / ii / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-06

The defining Wronskian identity gives the first-order linear differential equation

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

Work first on an interval where

y_{1} \neq = 0

. Dividing by

y_{1}

recognizes a derivative:

(\frac{y _{2}}{y _{1}})^{'} = \frac{W}{y _{1}^{2}}, y_{2} (x) = y_{1} (x) (C + \int_{x_{0}}^{x} \frac{W ( s )}{y _{1} ( s ) ^{2}} d s) .

This is reduction of order. Choose the nonzero Wronskian from the Abel identity to obtain linear independence. The constant

C

only adds a multiple of

y_{1}

. If

y_{1}

has zeros, this quotient formula is local; the resulting solution can be continued across ordinary zeros using the original second-order linear differential equation.

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