Explanation

(d)
Given, $(x + y)^{2}\ \frac{dy}{dx} = a^{2}$
Put $x + y = v$
$\Rightarrow$ $\frac{dy}{dx} = \frac{dv}{dx} - 1$
$\therefore$ $v^{2}\left( \frac{dv}{dx} - 1 \right) = a^{2}$
$\Rightarrow$ $\frac{dv}{dx} = \frac{a^{2} + v^{2}}{v^{2}}$
$\Rightarrow$ $\left( 1 - \frac{a^{2}}{a^{2} + v^{2}} \right)dv = dx$
$\Rightarrow$ $v - a\tan^{- 1}\left( \frac{v}{a} \right) = x + c$ [integrating]
$\Rightarrow$ $(x + y) - a\tan^{- 1}\left( \frac{x + y}{a} \right) = x + c$
$\Rightarrow$ $y = a\tan^{- 1}\left( \frac{x + y}{a} \right) + c$