= Solution
{c}
Introduce a characteristic parameter $s$ and label each point of the initial segment by $r\in[0,1]$. The <method of characteristics> gives
$$
\frac{dx}{ds}=x-2,\qquad\frac{dy}{ds}=y,\qquad\frac{dz}{ds}=y,
\qquad (x,y,z)|_{s=0}=(1,r,1).
$$
Integrating gives the parametric solution
$$
\boxed{x=2-e^s,\qquad y=re^s,\qquad z=1+r(e^s-1),\quad s\in\mathbb R,\quad0\leq r\leq1.}
$$
The projected <characteristic curves> satisfy $y=r(2-x)$, so each lies on a straight line through $(2,0)$. Its branch from the data always has $x<2$; the common vertex is approached only as $s\to-\infty$.
Eliminating $s,r$ using $e^s=2-x$ and $r=y/(2-x)$ yields
$$
\boxed{z(x,y)=1+y-\frac{y}{2-x}=1+\frac{y(1-x)}{2-x}.}
$$
Its derivatives give $(x-2)z_x+yz_y=y$, and substitution of $x=1$ gives the prescribed value one.
The <Jacobian determinant> of $(s,r)\mapsto(x,y)$ is $-e^{2s}\ne0$, and the inverse just displayed is unique. Hence the data determine the solution exactly on
$$
\boxed{\{(x,y):x<2,\ 0\leq y\leq2-x\}.}
$$
This is the <domain determined by a finite segment under a radial characteristic flow>. Its interior is $x<2$, $0<y<2-x$; the two boundary rays come from the endpoints of the initial segment. Every point there lies on exactly one characteristic from the data. Outside this sector, a characteristic either misses the initial segment or lies on the branch $x>2$, which cannot reach it. Thus the formula may be extended beyond this sector, but such an extension is not uniquely specified by the given data.
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