Solve the partial equation $(y-u)u_x+(x-y)u_y=u-x$

Question

By Lagrange by adding all the equations we have $dx+dy+dz=0$ implies $x+y+z=c_1$

but i couldn't get any other useful equation .

Answer 1

$$(y-u)u_x+(x-y)u_y=u-x$$ $$\frac{dx}{y-u}=\frac{dy}{x-y}=\frac{du}{u-x}=\frac{dx+dy+du}{(y-u)+(x-y)+(u-x)}=\frac{dx+dy+du}{0}$$ A first characteristic from $dx+dy+du=0$ : $$x+y+u=c_1\quad\implies\quad u=-x-y+c_1$$ A second characteristic from $\frac{dx}{y-u}=\frac{dy}{x-y}=\frac{dx}{y-(-x-y+c_1)}$

Solving $\quad\frac{dy}{x-y}=\frac{dx}{2y+x-c_1}\quad$ is for you.

After solving replace $c_1$ by $(x+y+u)$.