Explanation

(c)
Given equation is
$\frac{dy}{dx} = 2^{y} \bullet 2^{- x}\ \Rightarrow 2^{- y}dy = 2^{- x}dx$
On integrating both sides, we get
$\frac{2^{- y}}{\log 2}( - 1) = \frac{2^{- x}}{\log 2}( - 1) + c_{1}$
$\Rightarrow \ - \frac{2^{- y}}{\log 2} = - \frac{2^{- x}}{\log 2} + c_{1}$
$\Rightarrow \ - 2^{- y} = - 2^{- x} + c_{1}\log 2$
$\therefore\frac{1}{2^{x}} - \frac{1}{2^{y}} = c_{1}\log{2 = c}$