# If $\large\frac{dy}{dx}=\frac{x-y}{x+y}$ then

$\begin{array}{1 1}(1)2xy+y^{2}+x^{2}=c&(2)x^{2}+y^{2}-x+y=c\\(3)x^{2}+y^{2}-2xy=c&(4)x^{2}-y^{2}-2xy=c\end{array}$

$\large\frac{dy}{dx}=\frac{x-y}{x+y}$
$(x+y) dy=(x-y)dx$
$x \;dy +y \;dy = x \; dx -y\; dx$
$x \;dy +y \;dx + y \; dy = x\;dx$
$d(xy) +y\;dy =x\;dx$