CUET Preparation Today
$\tan^{-1} x + \tan^{-1} y = C$ is general solution of the differential equation |
$\frac{dy}{dx} = \frac{1 + y^2}{1 + x^2}$ $\frac{dy}{dx} = \frac{1 + x^2}{1 + y^2}$ $(1 + x^2)dy + (1 + y^2)dx = 0$ $(1 + x^2)dx + (1 + y^2)dy = 0$ |
$(1 + x^2)dy + (1 + y^2)dx = 0$ |
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 $\frac{1}{1 + x^2} + \frac{1}{1 + y^2} \frac{dy}{dx} = 0$ $\Rightarrow \frac{1}{1 + y^2} \frac{dy}{dx} = -\frac{1}{1 + x^2}$ $\Rightarrow (1 + x^2) \, dy + (1 + y^2) \, dx = 0$ |
