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

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.

The product rule givesThus this is an exact differential, and its zero level differential means $y{\sin }^{2}x=C$. The initial data set $C=1$, givingThe displayed interval is the maximal connected real interval containing $\pi /2$ on which this finite-valued solution is defined.