 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-1/9/a/solution

h = f - y^{'} f_{y^{'}}

. For a density without explicit

x

dependence, the chain rule gives

\frac{d h}{d x} = f_{y} y^{'} + f_{y^{'}} y^{''} - y^{''} f_{y^{'}} - y^{'} \frac{d}{d x} f_{y^{'}} = y^{'} (f_{y} - \frac{d}{d x} f_{y^{'}}) .

h = f - y^{'} \frac{\partial f}{\partial y ^{'}} = constant, \frac{d h}{d x} = 0.

Where

y^{'} \neq = 0

this first integral is equivalent to the original equation. At a turning point it extends by continuity. On an entire constant segment, however, its derivative alone can vanish without imposing the full Euler-Lagrange equation, so such candidate paths must still be checked in the original equation.

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