Question

Find the integral curves of the equation:
dx/(x²+y²) = dy/2xy = dz/{z(x+y)}

Answer 1

$\frac{dx}{x^2+y^2}=\frac{dy}{2xy}=\frac{dz}{z(x+y)}$

1) $\frac{dx+dy}{x^2+y^2+2xy}=\frac{dz}{z(x+y)}$

$\frac{dx+dy}{(x+y)^2}=\frac{dz}{z(x+y)}, \quad \frac{dx+dy}{x+y}=\frac{dz}{z} ,\quad d(\ln |x+y|)=d(\ln |z|)$

$\ln |x+y|=\ln|z|+C$

$|x+y|=e^C|z|$

$x+y=\hat {C}z$

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

$\frac{dx-dy}{(x-y)^2}=\frac{dz}{ \hat{C} z^2}, \quad d(\frac{-1}{x-y})=d(\frac{-1}{\hat{C}z})$

$\frac{-1}{x-y}=\frac{-1}{\hat{C}z}+C’=\frac{-1}{x+y}+C’$

$\frac{1}{x+y}-\frac{1}{x-y}=\frac{-2y}{x^2-y^2}=C’$

Answer: $x+y=C_1z, \; \frac{-2y}{x^2-y^2}=C_2$