Answer to Question #320069 in Differential Equations for Sherry

Question

Solve the standard type of first order partial differential equation

p(1+q) = qz

Answer 1

p(1+q) =qz

This is of form F(z, p, q) =0

let z= f(x + ay) be the solution

put u =x + ay

p = ∂z/∂x = dz/du

q =∂z/∂y = adz/du

dz/du(1 + a dz/du) = az dz/du

1+ adz/du = az

integrating both sides

$\int$ adz/ (az-1) =$\int$ du

log(az-1) = u + c

log(az-1) = x + ay +c