Find the general solution of the differential equation $\frac{dy}{dx} = \frac{1 + y^2}{1 + x^2}$

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

The correct answer is Option (2) → $\tan^{-1} y = \tan^{-1} x + C$ ##

Since $1 + y^2 \neq 0$, therefore separating the variables, the given differential equation can be written as

$\frac{dy}{1 + y^2} = \frac{dx}{1 + x^2} \quad \dots (1)$

Integrating both sides of equation (1), we get

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

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

which is the general solution of equation (1).