Explanation

(d)
Given differential equation is
$\frac{dy}{dx} = \frac{1}{x + y^{2}}$
$\Rightarrow$ $\frac{dx}{dy} - x = y^{2}$
Here, $P = - 1,\ \ Q = y^{2}$
$IF = e^{\int_{}^{}{- 1\ dy}} = e^{- y}$
$\therefore$ Solution is
$xe^{- y} = \int_{}^{}{e^{- y}\ \ y^{2}\ \ \ dy}$
$= - e^{- y}\ y^{2} + \int_{}^{}{2e^{- y}}\ y\ dy$
$= - e^{- y}\ y^{2} + 2\left\lbrack - e^{- y}\ y + \int_{}^{}e^{- y}\ dy \right\rbrack + c$
$\ = - e^{- y}\ y^{2} + 2\left\lbrack - e^{- y}\ y - e^{- y} \right\rbrack + c$
$\Rightarrow$ $xe^{- y} = e^{- y}\left( - y^{2} - 2y - 2 \right) + c$
$\Rightarrow$ $x = - y^{2} - 2y - 2 + ce^{y}$