= Solution
{c}
For $\beta=0$, the equation for the <characteristic curves> is
$$
x(dy)^2-2y\,dx\,dy=dy\,(x\,dy-2y\,dx)=0.
$$
The two families are therefore $y=\text{constant}$ and the <characteristic projections> from part (a), $y=Kx^2$, together with the vertical characteristic $x=0$. On either region $y>0$ or $y<0$, the <hyperbolic partial differential equation> admits the <characteristic coordinates>
$$
\boxed{\varphi=y,\qquad \psi=\frac{x}{\sqrt{|y|}}.}
$$
Indeed $\varphi$ is constant along the first family, while $(x\partial_x+2y\partial_y)\psi=0$. Their <Jacobian determinant> is
$$
\frac{\partial(\varphi,\psi)}{\partial(x,y)}=-\frac1{\sqrt{|y|}}\ne0.
$$
The coefficients of the two pure second <partial derivatives> in these coordinates vanish, and the coefficient of the mixed <partial derivative> is $2y/\sqrt{|y|}\ne0$. Hence they give the hyperbolic <characteristic normal form>. They also remain regular at $x=0$, provided $y\ne0$.
The alternative $\varphi=y$, $\psi=y/x^2$ is valid on patches with $xy\ne0$. On $y=0$, it is a <parabolic partial differential equation> away from the fully degenerate origin, so two independent hyperbolic <characteristic coordinates> are unavailable.
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