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

Given,

$$\frac{dy}{dx}+\frac{2^{x-y}(2^y-1)}{2^x-1}=0$$

Rearranging,

$$\frac{dy}{dx}=-\frac{2^x(2^y-1)}{2^y(2^x-1)}$$

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

Integrating both sides,

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

Using

$$\int\frac{2^t}{2^t-1}\,dt=\frac1{\ln2}\ln(2^t-1)$$

we get

$$\ln(2^y-1)+\ln(2^x-1)=C$$

$$ (2^y-1)(2^x-1)=k $$

Using the condition

$$y(1)=1$$

we get

$$ (2^1-1)(2^1-1)=k $$

$$k=1$$

Hence,

$$ (2^y-1)(2^x-1)=1 $$

Now, at

$$x=2$$

$$ (2^y-1)(4-1)=1 $$

$$3(2^y-1)=1$$

$$2^y=\frac43$$

$$y=\log_2\frac43$$

$$y=2-\log_23$$

Hence,

$$\boxed{2-\log_23}$$