Answer to Question #294425 in Calculus for Lucky

Question

How do you obtain a partial differential equation by eliminating arbitrary function from z=f (x+iy) + g (x-iy)?

Answer 1

Given $z = f(x+iy)+g(x-iy)$

$P=\delta z/\delta x = f`(x+iy)+g`(x-iy)$

$q=\delta z/\delta y = i f`(x+iy)-ig`(x-iy)$

$r=\delta^2 z/\delta x^2=f``(x+iy)+g``(x-iy)$

$t=\delta^2z/\delta y^2 = -f``(x+iy)-g``(x-iy)$

r + t = 0 is the required p.d.e.