Find the general solution of the differential equation $\frac{dy}{dx} = \frac{1 + y^2}{1 + x^2}$
Answer & explanation
Correct answer: option 2
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).