Find the general solution of the differential equation $(1 + y^2) + (x - e^{\tan^{-1} y}) \frac{dy}{dx} = 0$.

$2x e^{\tan^{-1} y} = e^{2 \tan^{-1} y} + K$

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

Given, differential equation is

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

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

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

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

$\Rightarrow \frac{dx}{dy} + \frac{x}{1 + y^2} = \frac{e^{\tan^{-1} y}}{1 + y^2} \quad \text{[dividing throughout by } (1 + y^2)\text{]}$

Which is a linear differential equation. On comparing it with $\frac{dx}{dy} + Px = Q$, we get

$P = \frac{1}{1 + y^2}, Q = \frac{e^{\tan^{-1} y}}{1 + y^2}$

$\text{I.F.} = e^{\int P dy} = e^{\int \frac{1}{1 + y^2} dy} = e^{\tan^{-1} y}$

The general solution is $x \cdot \text{I.F.} = \int Q \cdot \text{I.F.} dy + C$

$x \cdot e^{\tan^{-1} y} = \int \frac{e^{\tan^{-1} y}}{1 + y^2} \cdot e^{\tan^{-1} y} dy + C$

$\Rightarrow x \cdot e^{\tan^{-1} y} = \int \frac{(e^{\tan^{-1} y})^2}{1 + y^2} dy + C$

Put $\tan^{-1} y = t \Rightarrow \frac{1}{1 + y^2} dy = dt$

$∴x \cdot e^{\tan^{-1} y} = \int e^{2t} dt + C$

$\Rightarrow x \cdot e^{\tan^{-1} y} = \frac{1}{2} e^{2t} + C$

$\Rightarrow 2x e^{\tan^{-1} y} = e^{2 \tan^{-1} y} + 2C$

$\Rightarrow 2x e^{\tan^{-1} y} = e^{2 \tan^{-1} y} + K \quad [∵K = 2C]$