= Solution
Choose $X=x-2y$ and $T=x$, so $(\alpha,\beta,\gamma,\delta)=(1,-2,1,0)$. The <Jacobian determinant> of the coordinate change is two, so it is an <invertible linear map>. By the <chain rule>, $\partial_x=\partial_X+\partial_T$ and $\partial_y=-2\partial_X$. Consequently
$$
\partial_x^2+\partial_x\partial_y
=(\partial_X+\partial_T)(\partial_T-\partial_X)
=\partial_T^2-\partial_X^2.
$$
This is the <wave equation> with $c=1$. The <D'Alembert formula> is $F(X,T)=U(X+T)+V(X-T)$, with arbitrary twice <differentiable functions> $U,V$. Returning to the old coordinates and absorbing factors of two into the arbitrary functions gives
$$
\boxed{f(x,y)=G(x-y)+H(y).}
$$
Direct differentiation checks the cancellation. Equivalently, first integrating $\partial_x(f_x+f_y)=0$ produces the same two-function family.
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