= Solution
For a nonzero second-order principal part, the discriminant is $Q^2-PR$: positive gives a <hyperbolic partial differential equation>, negative an <elliptic partial differential equation>, and zero the <parabolic type> when the principal part has rank one. If all three coefficients vanish, the equation is instead degenerate to lower order.
Here $P=x$, $2Q=x^2$, $R=0$, so $Q^2-PR=x^4/4>0$ for $x>0$. The characteristic equation is
$$
P\,dy^2-2Q\,dx\,dy+R\,dx^2=x\,dy(dy-x\,dx)=0.
$$
Thus choose
$$
\boxed{\xi=y,\qquad\eta=y-\frac{x^2}{2}.}
$$
Their Jacobian <determinant> with respect to $(x,y)$ is $x\neq0$, so they are valid coordinates on the specified half-plane. The level sets are horizontal lines and the right-hand halves of parabolas $y=x^2/2+\eta$.
\Image[/past-exam-of-the-algebra-analysis-and-differential-equations-2-course-of-the-university-of-oxford/2011-ac2-characteristics.png]
{title=The two characteristic families for the hyperbolic equation}
Writing $z(x,y)=Z(\xi,\eta)$, the <chain rule> gives
$$
z_x=-xZ_\eta,\quad z_{xx}=-Z_\eta+x^2Z_{\eta\eta},\quad
z_{xy}=-x(Z_{\xi\eta}+Z_{\eta\eta}).
$$
The original equation reduces to $-xZ_\eta-x^3Z_{\xi\eta}=-xZ_\eta+x^3$, hence the canonical form is $\boxed{Z_{\xi\eta}=-1}$. Integrating once in each characteristic variable gives
$$
Z=f(\xi)+g(\eta)-\xi\eta,
$$
or
$$
\boxed{z=f(y)+g(y-x^2/2)-y^2+yx^2/2.}
$$
On $x=1$, differentiation gives $z_x=y-g'(y-1/2)$. The specified <derivative> therefore forces $g'=0$ for every argument, so $g$ is a constant. The specified value then gives $f(y)+g=1+y^2-y/2$. Consequently
$$
\boxed{z(x,y)=1+\frac y2(x^2-1).}
$$
Its derivatives are $z_x=xy$, $z_{xx}=y$ and $z_{xy}=x$, which verify both the equation and the data directly.
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