= Solution
{c}
Choose characteristic parameter $t$ with initial value at $t=0$. The characteristic equations are
$$
\dot x=x,\qquad\dot y=2y,\qquad\dot z=xy,
\qquad(x,y,z)(0)=(s,s,s^2).
$$
The first two integrate immediately; substitution into the third gives
$$
\boxed{x=se^t,\qquad y=se^{2t},\qquad
z=\frac{s^2}{3}(2+e^{3t}),\quad t\in\mathbb R.}
$$
Eliminating $t$ gives the projected parabola
$$
\boxed{y=\frac{x^2}{s},\qquad x/s>0.}
$$
The sign restriction selects the half-parabola actually reached from the initial point.
The projection map $(s,t)\mapsto(se^t,se^{2t})$ has <Jacobian determinant> $se^{3t}\neq0$, so the initial curve is noncharacteristic for every allowed $s$. Its global inverse is
$$
s=\frac{x^2}{y},\qquad t=\log(y/x),
$$
defined precisely when $x,y$ have the same nonzero sign. Every such point is reached by exactly one characteristic. Hence the region of unique determination from these initial data projects onto
$$
\boxed{\{(x,y):xy>0\},}
$$
the first and third quadrants. The axes are not reached for finite parameter from $s\neq0$, and the other two quadrants have no <method of characteristics> issued from this initial curve.
As a direct check, substitution of the inverse parameters gives the solution
$$
z(x,y)=\frac{2x^4}{3y^2}+\frac{xy}{3}
\quad(xy>0).
$$
It has $z(s,s)=s^2$, and differentiating gives $xz_x+2yz_y=xy$. This is <quadrant uniqueness for unequal dilation characteristics>.
\Image[/past-exam-of-the-algebra-analysis-and-differential-equations-1-course-of-the-university-of-oxford/2007-characteristic-quadrants.png]
{title=Each initial point determines one half-parabola in the first or third quadrant}
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