Formation of partial differential equation

Question

If $$z=f(x+ay)+g(x-ay)$$ ,$f$ and $g$ are arbitrary functions and a is constant, form a partial differential equations . I differentiate wrt $x$ and $y$ ..but how to eliminate functions...any one please help.. thanks in advance.

Answer 1

Note that you have: $$\frac {\partial z}{\partial {x}}=\frac {\partial f}{\partial {(x+ay)}}\frac {\partial (x+ay)}{\partial {x}}+\frac {\partial g}{\partial {(x-ay)}}\frac {\partial (x-ay)}{\partial {x}}$$ $$\frac {\partial z}{\partial {x}}=\frac {\partial f}{\partial {(x+ay)}}+\frac {\partial g}{\partial {(x-ay)}}$$ And also: $$\frac {\partial z}{\partial {y}}=\frac {\partial f}{\partial {(x+ay)}}\frac {\partial (x+ay)}{\partial {y}}+\frac {\partial g}{\partial {(x-ay)}}\frac {\partial (x-ay)}{\partial {y}}$$ $$\frac {\partial z}{\partial {y}}=a\frac {\partial f}{\partial {(x+ay)}}-a\frac {\partial g}{\partial {(x-ay)}}$$

Use chain rule and differentaite twice: $$\frac {\partial^2 z}{\partial {x^2}}=\frac {\partial^2 f(x+ay)}{\partial {(x+ay)^2}}+\frac {\partial^2 g(x-ay) }{\partial {(x+ay)^2}}$$ $$\frac {\partial^2 z}{\partial {y^2}}=a^2\frac {\partial^2 f(x+ay) }{\partial {(x+ay)^2}}+a^2\frac {\partial^2 g(x-ay)}{\partial {(x+ay)^2}}$$ Therefore: $$\boxed {\frac {\partial^2 z}{\partial {y^2}}=a^2\frac {\partial^2 z}{\partial {x^2}}}$$