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

$y - x = k(1 + xy)$

The correct answer is Option (2) → $y - x = k(1 + xy)$ ##

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

$\Rightarrow \frac{dy}{1 + y^2} = \frac{dx}{1 + x^2} \quad \text{[apply variable separable method]}$

On integrating both sides, we get

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

$\Rightarrow \tan^{-1} y - \tan^{-1} x = C$

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

$\Rightarrow \frac{y - x}{1 + xy} = \tan C$

$\Rightarrow y - x = \tan C(1 + xy)$

$\Rightarrow y - x = K(1 + xy) \quad [∵\tan C = k]$