If …, y(0) = 0, then for y = 1, the value of x lies in the interval :

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Differential Equations

JEE Main 2021 · 31 Aug, Evening shift · Q3537

MathematicsEasy38×since 2002Q3537

{{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 :

JEE Main 2021 · 31 Aug, Evening shift · Q3537

MathematicsEasy38×since 2002Q3537

{{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 :