The solution of differential equation $\frac{dy}{dx} = \frac{1 + y^2}{1 + x^2}$ is
Answer & explanation
Correct answer: option 2
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
$\tan^{-1} y = \tan^{-1} x + C$
$\Rightarrow \tan^{-1} y - \tan^{-1} x = C$
$\Rightarrow \tan^{-1} \left( \frac{y - x}{1 + xy} \right) = C$
$\Rightarrow \frac{y - x}{1 + xy} = \tan C$
$\Rightarrow y - x = \tan C(1 + xy)$
$\Rightarrow y - x = K(1 + xy) \quad [∵\tan C = k]$