 Solution

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/iii/solution

Here

F

has no explicit

x

dependence, so use the Beltrami identity. With

p = y^{'}

, one has

F_{y^{'}} = y p / 1 + p^{2}

and therefore

\frac{y}{1 + ( y ^{'} ) ^{2}} - λ y = C .

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