Find the solution of $\frac{dy}{dx} = 2^{y-x}$.

$2^{-x} - 2^{-y} = K$

The correct answer is Option (1) → $2^{-x} - 2^{-y} = K$ ##

Given that, $\frac{dy}{dx} = 2^{y-x}$

$\Rightarrow \frac{dy}{dx} = \frac{2^y}{2^x} \quad \left[ ∵a^{m-n} = \frac{a^m}{a^n} \right]$

$\Rightarrow \frac{dy}{2^y} = \frac{dx}{2^x} $ [Applying variable separable method]

On integrationg both sides, we get

$\int 2^{-y} dy = \int 2^{-x} dx$

$\Rightarrow \frac{-2^{-y}}{\log 2} = \frac{-2^{-x}}{\log 2} + C$

$\Rightarrow -2^{-y} + 2^{-x} = +C \log 2$

$\Rightarrow 2^{-x} - 2^{-y} = K $ [where, $K = +C \log 2$]