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

The homogeneous first-order differential equation suggests $v=y/x$, so $y^{\prime }=v+x{v}^{\prime }$. Wherever the original equation is defined,Thus the implicit solution is $x\sin (y/x)=C$. The printed initial values force $C=x_0=2/\pi$.

There is no classical solution through the printed initial point. There $y/x=\pi /2$, so the tangent on the right is undefined. The implicit relation nevertheless has two neighboring real branches on $x>x_0$ which approach that point continuously:Both tend to $y=1$ as $x\downarrow x_0$, but their derivatives become unbounded with opposite signs. They solve the equation on the open interval $x>x_0$, not at its endpoint. Other inverse-sine periods cannot approach the specified initial value. This is an endpoint obstruction in a homogeneous differential equation. Constant-ratio branches $y=k\pi x$ also solve the general equation where $x\ne 0$, but do not meet the initial data.

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.