= Solution
{c}
Write the principal part as $e^{2y}u_{xx}-u_{yy}$. The <characteristic curves> satisfy
$$
e^{2y}(dy)^2-(dx)^2=0,
$$
or $dx/dy=\pm e^y$. Integration gives $x+e^y=\text{constant}$ and $x-e^y=\text{constant}$. Thus take $\xi=x+e^y$, $\eta=x-e^y$. Their <Jacobian determinant> is $-2e^y\ne0$, and their image is $\xi>\eta$.
For $u(x,y)=U(\xi,\eta)$, the <chain rule> gives
$$
u_{xx}=U_{\xi\xi}+2U_{\xi\eta}+U_{\eta\eta},\qquad
u_y=e^y(U_\xi-U_\eta),
$$
$$
u_{yy}=e^{2y}(U_{\xi\xi}-2U_{\xi\eta}+U_{\eta\eta})
+e^y(U_\xi-U_\eta).
$$
The first-derivative terms cancel exactly. The <canonical form of a second-order partial differential equation> is
$$
4e^{2y}U_{\xi\eta}=0,\qquad U=F(\xi)+G(\eta).
$$
On $y=0$, $F(x+1)+G(x-1)=x$. Differentiating this condition in $x$ and using $u_y=1$ gives
$$
F'(x+1)+G'(x-1)=1,\qquad F'(x+1)-G'(x-1)=1.
$$
Hence $F'=1$ and $G'=0$. Substitution into the value condition fixes their combined constant and yields
$$
\boxed{u(x,y)=x+e^y-1.}
$$
Direct substitution verifies the PDE and both initial data. The cancellation illustrates a <hyperbolic equation reduced by an exponential characteristic coordinate>.
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