Past exam of the year 1 of the computer science course of the University of Oxford / 2016 / Discrete mathematics, continuous mathematics and probability (resit) / 4 / d / Solution

The substitution $y=xv$ gives $y^{\prime }=v+x{v}^{\prime }$. Cancelling $x^{3}v$ on both sides of the differential equation, for $x>0$, givesThe initial condition is $v(1)=1$. This selects a positive solution locally, so separation of variables givesUsing $v(1)=1$ gives $C=-1/2$, and continuity selects the positive square root:The real solution containing $x=1$ has maximal positive domain $0<x<\sqrt{e}$. Its denominator tends to zero and $y\to +\infty$ as $x\uparrow \sqrt{e}$, a finite-time blow-up of an ordinary differential equation in the independent variable $x$. Thus the source's wording “for $x>0$” needs this domain restriction; the prescribed initial value admits no real solution on the whole positive half-line. The division by $v^3$ loses the separate solution $v=0$, but that solution does not meet the initial condition.