 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-1/9/b/solution

f

has no explicit

x

dependence, the chain rule gives

\frac{df}{d x} = f_{y} y^{'} + f_{y^{'}} y^{''} .

Consequently

\frac{d}{d x} (f - y^{'} f_{y^{'}}) = y^{'} (f_{y} - \frac{d}{d x} f_{y^{'}}) = 0

h = f - y^{'} \frac{\partial f}{\partial y ^{'}} = constant .

The opposite sign is an equally valid convention. The conserved value can be zero for particular integrands or paths; absence of explicit

x

alone does not guarantee a nonzero numerical value. The useful statement is that this expression is constant along an extremal.

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