If $\frac{dy}{dx}+\frac{2^{x-y}\left(2^y-1\right)}{2^x-1}=0, x, y>0, y\left(1\right)=1$, then $\mathrm{y} \left(2\right)$ is equal to:

$\begin{array}{l}\frac{dy}{dx}+\frac{2^{x-y}\left(2^y-1\right)}{2^x-1}=0\\ \\ \mathrm{x},\mathrm{y}>0, \mathrm{y}\left(1\right)=1, \mathrm{y}\left(2\right)=?\\ \\ \frac{dy}{dx}=-\frac{2^x\left(2^y-1\right)}{2^y\left(2^x-1\right)}\\ \\ \int \frac{2^y}{2^y-1}dy=-\int \frac{2^x}{2^x-1}dx\\ \\ \int \frac{2^y \ln 2}{2^y-1}dy=-\int \frac{2^x \ln 2}{2^x-1}dx\\ \\ \ln \left|2^y-1\right|=-\ln \left|2^x-1\right|+C\\ \\ At \mathrm{x}=1, \mathrm{y}=1\end{array}$

Putting this values in above relation we get C = 0

$\begin{array}{l}\ln \left|2^y-1\right|+\ln \left|2^x-1\right|=0\\ \\ \left(2^x-1\right)\left(2^y-1\right)=1\\ \\ 2^y-1=\frac{1}{2^x+1}\\ \\ At x=2\\ \\ 2^y=\frac{1}{3}+1=\frac{4}{3}\\ \\ y={\log }_2 \frac{4}{3}={\log }_2 4-{\log }_2 3=2-{\log }_2 3\end{array}$