The solution of the differential equation  $\frac{dy}{dx}=\frac{1+y^2}{1+x^2}, $ is

$dy-x=C(1+xy)$

The correct answer i soption (2) : $dy-x=C(1+xy)$

We have,

$\frac{dy}{dx}=\frac{1+y^2}{1+x^2}$

$⇒\frac{1}{1+y^2}dy =\frac{1}{1+x^2}dx$

On integrating, we get

$tan^{-1} y = tan^{-1} x + tan^{-1} C$

$⇒tan^{-1} y - tan^{-1}x + tan^{-1}C$

$⇒tan^{-1} \left(\frac{y-x}{1+xy}\right) = tan^{-1}C$

$⇒\frac{y-x}{1+xy}=C$

$⇒y - x = C (1+ xy)$