Explanation

(a)
Given differential equation is
$\left( 1 + y^{2} \right) + \left( x - e^{\tan^{- 1}y} \right)\frac{dy}{dx} = 0$
$\Rightarrow$ $\left( 1 + y^{2} \right)\frac{dx}{dy} = - x + e^{\tan^{- 1}y}$
$\Rightarrow$ $\frac{dx}{dy} + \frac{x}{1 + y^{2}} = \frac{e^{\tan^{- 1}y}}{1 + y^{2}}$
Which is a linear differential equation,
Here, $P = \frac{1}{1 + y^{2}},\ \ Q = \frac{e^{\tan^{- 1}y}}{1 + y^{2}}$
$IF = e^{\int_{}^{}{P\ dy}} = e^{\int_{}^{}{\frac{1}{1 + y^{2}}\ dy}} = e^{\tan^{- 1}y}$
$\therefore$ Solution is
$x.IF = \int_{}^{}{Q.IF}\ dy + c$
$xe^{\tan^{- 1}y} = \int_{}^{}\frac{e^{\tan^{- 1}y}}{1 + y^{2}}\ .e^{\tan^{- 1}y} + \frac{c}{2}$
$\Rightarrow$ $xe^{\tan^{- 1}y} = \frac{e^{2\tan^{- 1}y}}{2} + \frac{c}{2}$
$\therefore$ $2xe^{\tan^{- 1}y} = e^{2\tan^{- 1}y} + c$