= Solution
A <characteristic curve> in $(x,y,z)$ follows the <vector field> defining the first-order <partial differential equation>. Along such a curve the <chain rule> turns the equation into
$$
\frac{dx}{ds}=xy,\qquad \frac{dy}{ds}=y^2,\qquad
\frac{dz}{ds}=-\frac13z^4.
$$
Its <characteristic projection> is the curve $(x(s),y(s))$ in the independent-variable plane.
Writing $x(0)=\xi$, $y(0)=\eta$, $z(0)=\zeta$, <separation of variables> gives
$$
\boxed{x(s)=\frac{\xi}{1-\eta s},\qquad
y(s)=\frac{\eta}{1-\eta s},\qquad
z(s)=\frac{\zeta}{(1+\zeta^3s)^{1/3}}.}
$$
These formulas hold on the maximal interval through $s=0$ where the denominators remain nonzero, using the real <cube root>. They include $\eta=0$ (a stationary <characteristic projection>) and $\zeta=0$ (the zero height). For $\zeta\ne0$, the last formula follows from $d(z^{-3})/ds=1$. The alternative zero solution must be retained when deriving this by division by $z^4$.
Back to article page
