Past exam of the mathematics course of the University of Cambridge / 2003 / ia / Paper 2 / 5D / b

After multiplication by ${\sec }^{2}x$, recognize both sides as derivatives:The nontrivial first integral is thereforeIt necessarily depends on the derivative, correcting the printed notation $h(x,y)$. A second integration giveson the specified interval. Since $y(0)=0$, $C_2=0$. A finite derivative at zero also implies $y{y}^{\prime }\to 0$ there, forcing $C_1=0$. A positive particular solution is consequentlyThe negative of this function is the other sign choice. Near zero, $-2\log (\cos x)=x^2+x^4/6+O(x^6)$, so the positive branch has $y=x+x^3/12+O(x^5)$ and $y^{\prime }(0)=1$; the negative branch has derivative $-1$.

For the positive branch, $y^{\prime }=\tan x/y>0$. Its second derivative has numerator $-2\log \cos x-{\sin }^{2}x$, multiplied by a positive factor. This numerator is zero at zero and has derivative $2{\sin }^{3}x/\cos x>0$, so the curve is convex for positive $x$. It starts tangent to $y=x$ and rises without bound as $x\uparrow \pi /2$, with $y\sim \sqrt{-2\log (\pi /2-x)}$. The third panel of the sketch above shows this branch.