Question

solve

$(x^2-y^2)dx+2xydy=0$

Answer 1

$(x^2-y^2)dx+2xydy=0$

$\frac{dy}{dx}=\frac{y^2-x^2}{2xy}=\frac{1}{2}(\frac{y}{x}-\frac{x}{y})$ $y=vx$

$v+\frac{xdv}{dx}=\frac{1}{2}(v-\frac{1}{v})$ $\frac{dy}{dx}v\left(1\right)+\frac{xdv}{dk}$

$v+\frac{xdv}{dx}=\frac{v^2-1}{2v}$

$\frac{xdv}{dx}=\frac{v^2-1-2v^2}{2v}=\frac{-(1+v^2)}{2v}$

$\int \frac{2vdv}{(1+v^2)}=\int \frac{dx}{x}\Rightarrow ln(1+v^2)=-lnx+c$

$ln(1+\frac{y^2}{x^2})+lnx=c$

$ln(1+y^2)=c$