= Solution
Since $\partial_x=\alpha\partial_\xi+\beta\partial_\eta$ and $\partial_y=\partial_\xi+\partial_\eta$,
$$
\boxed{\begin{aligned}
A(\alpha,\beta)&=a\alpha^2+b\alpha+c,\\
B(\alpha,\beta)&=2a\alpha\beta+b(\alpha+\beta)+2c,\\
C(\alpha,\beta)&=a\beta^2+b\beta+c.
\end{aligned}}
$$
If $\alpha,\beta$ are the distinct real roots of $as^2+bs+c$, then $A=C=0$ and $B=(4ac-b^2)/a\ne0$. Thus $v_{\xi\eta}=0$. For $s^2+3s+2=(s+1)(s+2)$, take $\xi=y-x$, $\eta=y-2x$, obtaining
$$
\boxed{u(x,y)=f(y-x)+g(y-2x)}
$$
for arbitrary twice differentiable functions $f,g$.
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