$\tan^{-1} x + \tan^{-1} y = C$ is general solution of the differential equation
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $(1 + x^2)dy + (1 + y^2)dx = 0$ ##
Given that, $\tan^{-1} x + \tan^{-1} y = C$
On differentiating w.r.t. $x$, we get
$\frac{1}{1 + x^2} + \frac{1}{1 + y^2} \frac{dy}{dx} = 0$
$\Rightarrow \frac{1}{1 + y^2} \frac{dy}{dx} = -\frac{1}{1 + x^2}$
$\Rightarrow (1 + x^2) \, dy + (1 + y^2) \, dx = 0$