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

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

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

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

$$ \Rightarrow {\log _2}({2^y} - 1) = {2^x}{\log _2}e + C$$

$$\because$$ $$y(0) = 1 \Rightarrow 0 = {\log _2}e + C$$

$$C = - {\log _2}e$$

$$ \Rightarrow {\log _2}({2^y} - 1) = ({2^x} - 1){\log _2}e$$

put x = 1, $${\log _2}({2^y} - 1) = {\log _2}e$$

2^y = e + 1

y = log₂(e + 1) Ans.