Solve the differential equation (1 + y^2) tan^-1 x dx + 2y (1 + x^2)…

The correct answer is Option (4) → $\frac{1}{2} (\tan^{-1} x)^2 + \log (1 + y^2) = C$ ##

Given differential equation is

$(1 + y^2) \tan^{-1} x \, dx + 2y (1 + x^2) \, dy = 0$

$\Rightarrow \quad (1 + y^2) \tan^{-1} x \, dx = -2y (1 + x^2) \, dy$

$\Rightarrow \quad \frac{\tan^{-1} x}{1 + x^2} \, dx = -\frac{2y}{1 + y^2} \, dy \quad \text{[applying variable separable]}$

On integrating both sides, we get

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

Put $\tan^{-1} x = t$ in LHS, we get

$\frac{1}{1 + x^2} \, dx = dt$

and put $1 + y^2 = u$ in RHS, we get

$2y \, dy = du$

$\Rightarrow \quad \int t , dt = -\int \frac{1}{u} , du \Rightarrow \frac{t^2}{2} = -\log u + C $

$\Rightarrow \quad \frac{1}{2} (\tan^{-1} x)^2 = -\log (1 + y^2) + C $

$\Rightarrow \quad \frac{1}{2} (\tan^{-1} x)^2 + \log (1 + y^2) = C$