= Solution
{c}
For a second-order <linear partial differential equation> with principal part $Au_{xx}+2Bu_{xy}+Cu_{yy}$, the discriminant is $B^2-AC$. Here $A=1$, $B=0$, $C=-x^2$, so
$$
\boxed{B^2-AC=x^2>0\quad(x>0).}
$$
The equation is therefore a <hyperbolic partial differential equation> throughout its domain. The lower-order term $-u_x/x$ does not affect this classification.
The <characteristic curves> satisfy
$$
A(dy)^2-2B\,dx\,dy+C(dx)^2=0,
\qquad \frac{dy}{dx}=\pm x.
$$
Integrating gives the two families $y\mp x^2/2=\text{constant}$. Choose coordinates constant along these families:
$$
\xi=x^2-2y,\qquad\eta=x^2+2y.
$$
Their <Jacobian determinant> is $8x\ne0$, so they define an invertible <change of variables> on $x>0$, with image $\xi+\eta>0$. Write $u(x,y)=U(\xi,\eta)$. The <chain rule> gives
$$
\begin{aligned}
u_x&=2x(U_\xi+U_\eta),\\
u_{xx}&=2(U_\xi+U_\eta)+4x^2(U_{\xi\xi}+2U_{\xi\eta}+U_{\eta\eta}),\\
u_{yy}&=4(U_{\xi\xi}-2U_{\xi\eta}+U_{\eta\eta}).
\end{aligned}
$$
Both the pure second <partial derivatives> and the first <partial derivatives> cancel in the original equation, leaving
$$
16x^2U_{\xi\eta}=0,\qquad\boxed{U_{\xi\eta}=0.}
$$
This is the full <canonical form of a second-order partial differential equation>, including the effect of the lower-order term. Integrating shows $U_\xi$ is a function of $\xi$ alone, hence
$$
\boxed{u(x,y)=f(x^2-2y)+g(x^2+2y),}
$$
with arbitrary twice <differentiable functions> $f,g$. The representation is global on $x>0$: for each fixed $\xi$, the allowed $\eta>-\xi$ form an interval, so $U_\xi$ is independent of $\eta$ throughout that interval; after subtracting its antiderivative, the same reasoning in the other coordinate leaves only $g(\eta)$. These are <quadratic characteristic coordinates for a variable-speed wave equation>.
For the data on $x=1$, set $s=1-2y$, so the other argument is $2-s$. The two <initial conditions> become
$$
f(s)+g(2-s)=1,\qquad 2\bigl(f'(s)+g'(2-s)\bigr)=2.
$$
Differentiating the first equation with respect to $s$ also gives $f'(s)-g'(2-s)=0$. Solving these two linear equations yields $f'(s)=g'(2-s)=1/2$. Since $s$ ranges over all real numbers,
$$
f(s)=\frac{s}{2}+C,\qquad g(t)=\frac{t}{2}-C.
$$
Consequently
$$
\boxed{u(x,y)=x^2.}
$$
Indeed $u_{xx}=2$, $u_{yy}=0$, $u_x=2x$, so the equation and both <initial conditions> hold directly.
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