Example PDE 1
Using the example as given by
Dmoreno (MSE)
and employing operator calculus we have:
$$
z_{xx}-z_{yy} =
\frac{\partial^2 z}{\partial x^2} - \frac{\partial^2 z}{\partial y^2} =
\left[ \left(\frac{\partial}{\partial x}\right)^2 -
\left(\frac{\partial}{\partial y}\right)^2 \right] z = 0
\quad \Longleftrightarrow \\
\left(\frac{\partial}{\partial x} + \frac{\partial}{\partial y}\right)
\left(\frac{\partial}{\partial x} - \frac{\partial}{\partial y}\right) z = 0
\\ \left(\frac{\partial}{\partial x} - \frac{\partial}{\partial y}\right)
\left(\frac{\partial}{\partial x} + \frac{\partial}{\partial y}\right) z = 0
$$
Let $z(x,y) = f(x-y)$ , then:
$$
\left(\frac{\partial}{\partial x} + \frac{\partial}{\partial y}\right) f(x-y)
= \frac{\partial f}{\partial (x-y)}\frac{\partial (x-y)}{\partial x}
+ \frac{\partial f}{\partial (x-y)}\frac{\partial (x-y)}{\partial y} = \\
= \frac{\partial f}{\partial (x-y)}(+1) + \frac{\partial f}{\partial (x-y)}(-1) = 0
$$
Let $z(x,y) = g(x+y)$ , then:
$$
\left(\frac{\partial}{\partial x} - \frac{\partial}{\partial y}\right) g(x+y)
= \frac{\partial g}{\partial (x+y)}\frac{\partial (x+y)}{\partial x}
- \frac{\partial g}{\partial (x+y)}\frac{\partial (x+y)}{\partial y} = \\
= \frac{\partial g}{\partial (x+y)}(+1) - \frac{\partial g}{\partial (x+y)}(+1) = 0
$$
It is concluded that the general solution is given by any linear combination
of $f$ and $g$ as :
$$
z(x,y) = \lambda\,f(x-y) + \mu\,g(x+y)
$$