Answer to Question #284554 in Differential Equations for redin rzgar

Question

Find a complete integral of x(1 + y)p = y(1 + x)q

Answer 1

$Let\ \ p=\frac{\partial z}{\partial x}\ \ \ and\ \ \ \ q=\frac{\partial z}{\partial y}\ \ \ \\ Then\ \ \ x\left(\mathrm{1}+y\right)\frac{\partial z}{\partial x}\ \ =\ \ y\left(\mathrm{1}+x\right)\frac{\partial z}{\partial y} \\ \\ x\left(\mathrm{1}+y\right)\frac{dz}{dx}\ \ =\ \ y\left(\mathrm{1}+x\right)\frac{dz}{dy} \\ \\ \frac{\left(\mathrm{1}+y\right)dy}{y}\ \ =\ \ \frac{\left(\mathrm{1}+x\right)}{x}dx \\ \\ \int{\left(\mathrm{1}+\frac{\mathrm{1}}{y}\right)}dy\ \ =\ \ \int{\left(\mathrm{1}+\frac{\mathrm{1}}{x}\right)}dx \\ \\ \left(y+\mathrm{ln}\left(y\right)\right)\ \ =\ \ \left(x+\mathrm{ln}\left(x\right)\right)\ \ +\ \ C$