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

$x =\frac{1}{2}\left( e^{\tan^{-1} y} + e^{-\tan^{-1} y}\right)$

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

Given differential equation can be written as

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

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

Solution is given by

$xe^{\tan^{-1}y} = \int \frac{e^{\tan^{-1}y}}{1 + y^2} \times e^{\tan^{-1}y} dy + c$

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

or $xe^{\tan^{-1}y} = \frac{e^{2\tan^{-1}y}}{2} + c$

when $x = 1, y = 0$ or $c = \frac{1}{2}$

$∴$ Solution is given by $xe^{\tan^{-1}y} = \frac{1}{2} e^{2\tan^{-1}y} + \frac{1}{2}$

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