Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2009 / Paper 1 / 17 / a / ii

Put $u=x+y$. Then $u^{\prime }=1+\cos u=2{\cos }^2(u/2)$. Along the branch through $u(0)=0$, the chain rule yields $[\tan (u/2){]}^{\prime }=1$, hence $\tan (u/2)=x$. Choose the continuous branch through zero to obtainIndeed $y^{\prime }=(1-x^2)/(1+x^2)=\cos (2\arctan x)=\cos (x+y)$, and $y(0)=0$. The right-hand side has a bounded derivative with respect to $y$, so the usual uniqueness theorem for ordinary differential equations rules out another solution through the same point.