$\begin{aligned} & \frac{d y}{d x}=2^{y-x} \\ & \therefore \frac{d y}{d x}=\frac{2^y}{2^x} \Rightarrow \int \frac{d y}{2^y}=\int \frac{d x}{2^x} \\ & \therefore \int 2^{-y} d y=\int 2^{-x} d x \Rightarrow \frac{2^{-y}}{-\ell n 2}=\frac{2^{-x}}{-\ln 2}+c_1 \Rightarrow \frac{2^{-y}}{\ln 2}=\frac{2^{-x}}{\ln 2}-c_1 \\ & \therefore \frac{1}{\ln 2}\left[2^{-x}-2^{-y}\right]=c_1 \\ & \therefore \frac{1}{2^x}-\frac{1}{2^y}=c, \text { where } c=\left(c_1\right)(\ln 2)\end{aligned}$