Show that the PDE has vertical lines as a family of characteristic curves for $y=0$.

Question

Show that the PDE given by

$\dfrac{\partial ^2u}{\partial y^2}-y\dfrac{\partial ^2u}{\partial x^2}$ has vertical lines as a family of characteristics curves for $y=0$.

Taking $S=0,R=-y,T=1$ we get $S^2-4RT=4y>0$ if $y>0$ and $<0$ if $y<0$ i.e of elliptic type if $y<0$ and hyperbolic type if $y>0$ .

How to solve the problem using these information?

Answer 1

$$\frac{dy}{dx}=\frac{-S\pm \sqrt{S^2-4RT}}{2R}=\pm \frac{1}{-\sqrt{y}}$$ $$\implies \frac{2}{3}y^{3/2}=\pm x+c$$ if $y=0$ then $x=\pm c$ hence verticle lines as characteristics.