$$\begin{align}&(y-x)dy-(y+x)dx=0\\&(y-x) \frac{dy}{dx}-(y+x)=0\\&(y-x)y'+(-y-x)=0\end{align}$$
es una ecuación diferencial exacta
M(x,y)=-x-y N(x,y)=y-x
$$\begin{align}&\Psi_x=M(x,y)\\&\Psi_y= N(x,y)\\&\\&\Psi_x+\Psi_yy'=0\\&\\&\Psi= \int Mdx\\&\Psi=\int (-x-y )dx\\&\Psi= -yx-\frac{x^2}{2}+h(y) \ \ \ (1)\\&\\&\Psi_y=-x+h'(y)=N\\&h'(y)=x+N=x+y-x\\&h'(y)=y\\&h(y)= \int y dy= \frac{y^2}{2} \ \ \ \ \ (2)\\&\\&\Psi(x,y)=-yx-\frac{x^2}{2}+\frac{y^2}{2}=C\\&\\&\frac{y^2}{2}-xy+(-\frac{x^2}{2}+C_1)=0\\&\\&y^2-2xy+(-x^2+C_2)=0\\&y=\frac{2x \pm \sqrt{4x^2-(-4x^2+C_3)}}{2}=\frac{2x \pm \sqrt{8x^2+C_3}}{2}\end{align}$$
Ahí están las dos soluciones