Official solution
\[{{dy} \over {dx}} = {{{2^{x + y}} - {2^x}} \over {{2^y}}}\]
\[{2^y}{{dy} \over {dx}} = {2^x}({2^y} - 1)\]
\[\int {{{{2^y}} \over {{2^y} - 1}}dy = \int {{2^x}\,dx} } \]
\[{{\ln ({2^y} - 1)} \over {\ln 2}} = {{{2^x}} \over {\ln 2}} + C\]
\[ \Rightarrow {\log _2}({2^y} - 1) = {2^x}{\log _2}e + C\]
\[\because\] \[y(0) = 1 \Rightarrow 0 = {\log _2}e + C\]
\[C = - {\log _2}e\]
\[ \Rightarrow {\log _2}({2^y} - 1) = ({2^x} - 1){\log _2}e\]
put x = 1, \[{\log _2}({2^y} - 1) = {\log _2}e\]
2y = e + 1
y = log2(e + 1) Ans.