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Set $r=\sqrt{x^2+y^2}$. Wherever $r>0$, the chain rule and the given implicit differential equation giveThus $r=x+C$ and $C=r-x\ge 0$. Squaring without losing the square-root sign restriction giveswith 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 $y{y}^{\prime }=0$ while $-x+r=C>0$. The degenerate case $C=0$ gives the separate solution $\boxed{y=0\text{ for }x\ge 0}$ in the differential-form interpretation printed in the paper. It would be lost by dividing the equation by $y$.

Where $y\ne 0$ and $x^2\ne 1$, the separable differential equation givesThe requested point selects the implicit solution $2y^2+1=(x^2-1{)}^2$. The two real branches approaching that point areThere is an important initial value problem qualification: the differential equation as printed has no classical solution through the specified initial point. Its right side is undefined at $y=0$, and these branches have an infinite one-sided derivative as $x\downarrow \sqrt{2}$. They satisfy the differential equation on $x>\sqrt{2}$ and extend continuously to the requested point. Multiplying by the vanishing denominator produces an implicit differential equation with a larger domain; it does not repair the original initial value problem.