Question

Find the general integral of the partial differential equation using Lagrange's method
p(z+e^x)+q(z+e^y)=z^2-e^(x+y)

Answer 1

P=z+e^x. Q=z+e^y. R=z² - e^x+y

$$\frac {dx} {z+e^x}= \frac {dy} {z+e^y} = \frac {dz}{z^2-e^{x+y}}$$

Multiplyers: -z,e^x,1

$\frac {-zdx +e^xdy + dz} {-z^2 -ze^x + e^xz+ e^{x+y} + z^2-e^{x+y}} = \frac {-zdx +e^xdy + dz} {0}$

-zdx+e^xdy+dz=0

-zx+ye^x+z=c₁

Multiplyers: e^y,-z,1

$\frac {e^ydx -zdy + dz} {ze^y +e^{x+y}-z^2-ze^y+z^2-e^{x+y} } = \frac {e^ydx -zdy + dz} {0}$

e^ydx-zdy+dz=0

xe^y-zy+z=c₂

Answer: c₁=-zx+ye^x+z; c₂=xe^y-zy+z