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

Choose the tangent angle $\psi =\arctan y^{\prime }$ in $(-\pi /2,\pi /2)$, so $1/\sqrt{1+(y^{\prime }{)}^2}=\cos \psi$. At the maximum, the first integral has value $C=y_0(1-\lambda )$. HenceA nonconstant regular branch below its positive maximum has $y>0$ and $\psi \ne 0$, giving $\lambda >1$. For completeness, strictness follows from the differential equation rather than assuming it at every point: the Euler-Lagrange equation simplifies toAt the maximum, $y_0{y}^{\prime \prime }=1-\lambda$ implies $\lambda \ge 1$. If $\lambda =1$, the positive-height initial data $y=y_0$, $y^{\prime }=0$ give the constant solution uniquely, contradicting the existence of the lower-height branch. Thus $\lambda =1$ is only the degenerate horizontal solution. The first integral with $\lambda >1$ also excludes $y=0$ on this regular branch.