Solving : $\frac{\mathrm{d}x}{y-z} = \frac{\mathrm{d}y}{z-x} = \frac{\mathrm{d}z}{x-y}$ over $\mathbf{V} = (y-z,z-x,x-y)$

Question

I can't seem how to proceed with finding two curves $u_1$ and $u_2$ by solving the integral problem $$\frac{\mathrm{d}x}{y-z} = \frac{\mathrm{d}y}{z-x} = \frac{\mathrm{d}z}{x-y}$$ over the vector field $\mathbf{V} = (y-z,z-x,x-y)$. This is a part of proving that an integral surface is contained within $\mathbf{V}$. I would really appreciate any help given. Also Wolfram Alpha doesn't yield a solution for the problem : $$(y-z)u_x + (z-x)u_y + (x-y)u_z = 0$$

Answer 1

$$\frac{\mathrm{d}x}{y-z} = \frac{\mathrm{d}y}{z-x} = \frac{\mathrm{d}z}{x-y}$$ $$\frac{\mathrm{d}x}{y-z} = \frac{\mathrm{d}y+dz}{z-y}$$ You deduce that $$x+y+z=C$$ You can use that to integrate another equation...

Edit

$$\frac{\mathrm{d}x}{y-z} = \frac{\mathrm{d}y}{z-x} $$ But you know that $$z=C-x-y$$ $$\frac{\mathrm{d}x}{-C+x+2y} = \frac{\mathrm{d}y}{C-2x-y} $$ $$({-C+x+2y})dy = dx({C-2x-y} )$$ note that $d(xy)=xdy+ydx$ so $$({-C+2y})dy+dxy = ({C-2x} )dx$$ And note that $2xdx=dx^2$

$$-Cd(x+y)+dxy+dy^2 +dx^2=0$$

Thats easy to ntegrate....