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

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

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

$$ = {{{{\log }_e}({2^y} - 1)} \over {{{\log }_e}2}} = - {{{{\log }_e}({2^x} - 1)} \over {{{\log }_e}2}} + {{{{\log }_e}c} \over {{{\log }_e}2}}$$

$$ = |({2^y} - 1)({2^x} - 1)| = c$$

$$\because$$ $$y(1) = 1$$

$$\therefore$$ $$c = 1$$

$$ = |({2^y} - 1)({2^x} - 1)| = 1$$

For $$x = 2$$

$$|({2^y} - 1)3| = 1$$

$${2^y} - 1 = {1 \over 3} \Rightarrow 2y = {4 \over 3}$$

Taking log to base 2.

$$\therefore$$ $$y = 2 - {\log _2}3$$