The solution of the differential equation $x\, dx + y\, dy +\frac{x\, dy-y\, dx}{x^2+y^2}=0, $ is

$y = x\, tan \left(\frac{C-x^2-y^2}{2}\right)$

The correct answer is option (3) : $y = x\, tan \left(\frac{C-x^2-y^2}{2}\right)$

$x\, dx + y\, dy +\frac{x\, dy-y\, dx}{x^2+y^2}=0$

$⇒\frac{1}{2}d(x^2 + y^2)+d\left(tan^{-1}\frac{y}{x}\right) = 0 $

On integrating, we obtain

$\frac{1}{2}(x^2+y^2) + tan^{-1}\frac{y}{x}=\frac{C}{2}$

$⇒\frac{C-x^2-y^2}{2}= tan^{-1}\frac{y}{x}$

$⇒y=x\, tan \left(\frac{C-x^2-y^2}{2}\right)$