If {{dy} \over {dx}} = {{{2^x}y + {2^y}{{.2}^x}} \over {{2^x} + {2^{x + y}}{{\log }_e}2}}, y(0) = 0, then for y = 1, the value of x lies in the interval :

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Answer: A. (1, 2) {{dy} \over {dx}} = {{{2^x}(y + {2^y})} \over {{2^x}(1 + {2^y}\ln 2)}} \Rightarrow \int {{{(1 + {2^y})\ln 2} \over {(y + {2^y})}}dy = \int {dx} } \Rightarrow \ln \left| {y + {2^y}} \right| = x + c x = 0; y = 0 \Rightarrow c = 0 \Rightarrow x = \ln \left| {y + {2^y}} \right| \Rightarrow at y = 1, x = ln3 \because 3 \in (e,{e^2}) \Rightarrow x \in (1,2)

If $${{dy} \over {dx}} = {{{2^x}y + {2^y}{{.2}^x}} \over {{2^x} + {2^{x + y}}{{\log }_e}2}}$$, y(0) = 0, then for y = 1, the value of x lies in the interval :

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