If (dy)/(dx) + 2 ^ x - y ( 2 ^ y - 1) 2 ^ x - 1 = 0, x, y > 0, y...

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

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

$⇒ ∫ \dfrac{{2}^{y}}{{2}^{y} - 1}dy = - ∫ \dfrac{{2}^{x}}{{2}^{x} - 1}dx$

$⇒ \dfrac{1}{\ln 2} ∫ \dfrac{{2}^{y}\ln 2}{{2}^{y} - 1}dy = - \dfrac{1}{\ln^{2}} ∫ \dfrac{{2}^{x}\ln 2}{{2}^{x} - 1}dx$

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

At $x = 1, y = 1$

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

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

$⇒ \left({2}^{x} - 1\right)\left({2}^{y} - 1\right) = 1$

$⇒ {2}^{y} - 1 = \dfrac{1}{{2}^{x} + 1}$

At $x = 2$

$⇒ {2}^{y} = \dfrac{1}{3} + 1 = \dfrac{4}{3}$

$y = \log_{2}\dfrac{4}{3} = \log_{2}4 - \log_{2}3 = 2 - \log_{2}3$