= Solution
Set $r=\sqrt{x^2+y^2}$. Wherever $r>0$, the <chain rule> and the given <implicit differential equation> give
$$
\frac{dr}{dx}=\frac{x+yy'}r=\frac{x-x+r}r=1.
$$
Thus $r=x+C$ and $C=r-x\ge0$. Squaring without losing the square-root sign restriction gives
$$
\boxed{y^2=2Cx+C^2,\quad C>0,\quad x>-C/2,}
$$
with either sign of $y$ on its <interval> of definition. These <parabolas> satisfy $x+C>0$. Their vertex $(-C/2,0)$ is a regular point as a parametrized curve but cannot belong to a differentiable graph $y(x)$ satisfying the equation: there $yy'=0$ while $-x+r=C>0$. The degenerate case $C=0$ gives the separate solution $\boxed{y=0\text{ for }x\ge0}$ in the differential-form interpretation printed in the paper. It would be lost by dividing the equation by $y$.
\Image[/past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2014-resit-differential-equation-branches.png]
{title=Singular initial-point branches and the parabolic integral curves, with their graph-domain restrictions}
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