Question

Find the integral curves of the equations
dx/(y+zx) = dy/-(x+yz) = dz/(x^2 - y^2)

Answer 1

$\frac {dx}{y+zx} = \frac {dy}{-(x+yz)} = \frac {dz}{x^2 - y^2}$

$\frac {xdx+ydy}{x(y+zx)-y(x+zy)}= \frac {dz}{x^2 - y^2}$

$\frac {xdx+ydy}{z(x^2-y^2)}= \frac {dz}{x^2 - y^2}$

$xdx+ydy=zdz$

$x^2+y^2=z^2+C_1$

$\frac {xdx+ydy+zdz}{(z+1)(x^2-y^2)}= \frac {dz}{x^2 - y^2}$

$xdx+ydy+zdz=(z+1)dz$

$x^2+y^2+z^2=z^2+2z+C_2$

$x^2+y^2=2z+C_2$