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 $y\left(2\right)$ is equal to

Given, $\frac{dy}{dx}+\frac{2^{x-y}\left(2^y-1\right)}{2^x-1}=0$  $x, y>0, y\left(1\right)=1$

Now rearranging and integrating both side of  $\frac{dy}{dx}=-\frac{2^x\left(2^y-1\right)}{2^y\left(2^x-1\right)}$, we get

$\Rightarrow \int \frac{2^y}{2^y-1}dy=-\int \frac{2^x}{2^x-1}dx$

$\Rightarrow \frac{1}{\ln 2}\int \frac{2^y\ln 2}{2^y-1}dy=-\frac{1}{{\ln }^2}\int \frac{2^x\ln 2}{2^x-1}dx$

$\Rightarrow \frac{1}{\ln 2}\ln \left|2^y-1\right|=\frac{-1}{\ln 2}\ln \left|2^x-1\right|+C$

At $x=1,y=1$

Putting this values in above equation we get $C=0$

So, $\ln \left|2^y-1\right|+\ln \left|2^x-1\right|=0$

$\Rightarrow \left(2^x-1\right)\left(2^y-1\right)=1$

$\Rightarrow 2^y-1=\frac{1}{2^x+1}$

At $x=2$

$\Rightarrow 2^y=\frac{1}{3}+1=\frac{4}{3}$

$y={\log }_2\frac{4}{3}={\log }_{2}4-{\log }_{2}3=2-{\log }_{2}3$