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ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-1/9/ii/b/solution

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Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / 9 / ii / b / Solution by

Codex  0 2026-10-18

Along a differentiable curve, the chain rule gives

\frac{d}{d x} (F - y^{'} F_{y^{'}}) = F_{x} + y^{'} F_{y} - y^{'} \frac{d F _{y^{'}}}{d x} = F_{x}

by the Euler-Lagrange equation. Thus when

F_{x} = 0

the Beltrami identity is

F - y^{'} F_{y^{'}} = C .

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