Answer to Question #131966 in Differential Equations for Nikhil

Question

Solve the equations

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

Answer 1

$\dfrac{dx}{x+y-z}=\dfrac{dy}{y+z-x}=\dfrac{dz}{z+x-y}\\ \dfrac{2dx+dy-dz}{4y-4z}=\dfrac{dx-dz}{2y-2z}\\ 2dx+dy-dz=2dx-2dz=>d(y+z)=0=>\\ y+z=C_1\\ z=C_1-y\\ using \space property\space of \space the\space fraction\space we \space get:\\ \dfrac{dx+dy}{2y}=\dfrac{dx+dz}{2x}\\ we \space know \space that \space z=C_1-y(it \space means \space that \space dz=-dy)\\ using \space this \space condition, \space we'll\space get\\ \dfrac{dx+dy}{2y}=\dfrac{dx-dy}{2x}\\ (y-x)dx=(x+y)dy\\ y'=\dfrac{y-x}{y+x}\\ Let \space u=\dfrac{y}{x}\\ xu'+u=\dfrac{u-1}{u+1}\\ xu'=-\dfrac{u^2+1}{u+1}\\ -\dfrac{u+1}{u^2+1}du=\dfrac{dx}{x}\\ -(\dfrac{1}{2}\ln(u^2+1)+\arctan(u))+C_2=\ln(x)\\ we \space remember, \space that \space u=\dfrac{y}{x}\\ So\space we'll \space have\space the\space next \space answer\\ C_2=\dfrac{1}{2}\ln((\dfrac{y}{x})^2+1)+\arctan(\dfrac{y}{x}))+\ln(x)\\$