Answer to Question #218257 in Differential Equations for anuj

Question

Find the general integrals of the PDE

(Y+1)p+(x+1)q=z

Answer 1

$\frac{dx}{y+1}=\frac{dy}{x+1}=\frac{dz}{z}$

$\int (x+1)dx=\int (y+1)dy$

$x^2/2+x=y^2/2+y+c_1$

$\frac{dx-dy}{y-x}=\frac{dz}{z}$

$-ln(x-y)=ln(c_2z)$

$c_2=\frac{1}{z(x-y)}$

$F(x^2/2+x-y^2/2-y,\frac{1}{z(x-y)})=0$