Answer to Question #164071 in Differential Equations for mohit upadhyay

Question

dx/1+y = dy/1+x = dz/z

Answer 1

$\dfrac{dx-dy}{(1+y)-(1+x)} =- \dfrac{d(x-y)}{x-y}$

Also, since $d2=0, \dfrac{dx+dy}{1+y+1+x} = \dfrac{dx+dy+d2}{x+y+2} = \dfrac{d(x+y+2)}{x+y+2}$

Each of the fraction of the given DE is equal to

$-\dfrac{d(x-y)}{x-y}$ and $\dfrac{d(x+y+2)}{x+y+2}$

Solving

$-\dfrac{d(x-y)}{x-y} = \dfrac{dz}{z}$

$ln(x-y)^{-1} = lnz + c_1\\ c_1 = \frac{1}{z(x-y)}$

Also, solving

$\dfrac{d(x+y+2)}{x+y+2} = \dfrac{dz}{z}$

$ln(x+y+2) = lnz + c_2\\ c_2 = \frac{x+y+2}{z}$

So the solution is $F(\frac{1}{z(x-y)}, \frac{x+y+2}{z})=0.$