Explanation

(a)
Given equation can be rewritten as
$\frac{dx}{dy} + \frac{1}{(1 + y^{2})}x = \frac{e^{\tan^{- 1}y}}{(1 + y^{2})}$
$\therefore$ $IF = e^{\int_{}^{}{\frac{1}{1 + y^{2}}\ dy}} = e^{\tan^{1}y}$
$\therefore$ Required solution is
$xe^{\tan^{1}y} = \int_{}^{}\frac{e^{\tan^{- 1}y}e^{\tan^{- 1}y}}{1 + y^{2}}\ dy$
Put $e^{\tan^{- 1}y} = t \Rightarrow e^{\tan^{- 1}y}\ \frac{1}{1 + y^{2}}dy = dt$
$\therefore$ $xe^{\tan^{- 1}y} = \int_{}^{}t\ dt = \frac{t^{2}}{2} + c$
$\Rightarrow$ $2xe^{\tan^{- 1}y} = e^{2\tan^{- 1}y} + k$