solve-PDE-py-3-qx-2-0-

Question Number 139159 by rs4089 last updated on 23/Apr/21

s o l v e P D E {p y}^{3} + {q x}^{2} = 0

Commented by Dwaipayan Shikari last updated on 23/Apr/21

\frac{\partial^{3} f (x, y)}{\partial y^{3}} + \frac{\partial^{2} f (x, y)}{\partial x^{2}} = 0

f (x, y) = Λ e^{μ (x + y)}

μ^{3} e^{μ (x + y)} + μ^{2} e^{μ (x + y)} = 0 \Rightarrow μ = 0 o r = - 1

f (x, y) = Λ e^{- (x + y)}