Given $\large\frac{dy}{dx}$$=2^{y-x}$
Clearly here the variables can be seperated to the form $f(x)dx=g(y)dy$
(ie) $\Large\frac{dy}{dx}=\Large\frac{2^y}{2^x}$
On seperating the variables we get,
$\Large\frac{dy}{2^y}=\frac{dx}{2^x}$
=>$2^{-y} dy =2^{-x} dx$
Integrating on both sides we get
$\int 2^{-y}dy=\int 2^{-x} dx$
$\int a^xdx=\Large\frac{a^x}{\log\;a}$$+c$
$\Large\frac{2^{-y}}{\log 2}=\frac{2^{-x}}{\log 2}$$+c$
=>$ 2^{-y}=2^{-x}+c$
or $ 2^{-y}-2^{-x}=c$