Question

D3-3D²D'+ D'³ z= 0

Answer 1

Given $(D^3-3D²D'+ D'³) z= 0$ where $D = \frac{\partial}{\partial x}, D' = \frac{\partial}{\partial y}$ .

Since, $D^3-3D²D'+ D'³$ is irreducible in linear factors,

assume solution is $z = \sum A e^{hx+ky}$ .

Now. $(D^3-3D²D'+ D'³)z =(h^3-3h^2k+k^3) \sum{Ae^{hx+ky}} = 0$ $\implies h^3 - 3h^2 k +k^3 = 0$ .

Hence, $z = \sum A e^{hx+ky}$ is solution of given differential equation where h and k holds the condition $h^3 - 3h^2 k +k^3 =0$ .