If dy/dx = 2ˣy + 2^(y).2ˣ 2ˣ + 2^(x + y) log e 2, y(0) = 0, then for y = 1, the value of x lies in the interval:

Differential Equations: JEE Main 2021 Mathematics question

JEE Main 2021 · 31 Aug, Evening shift · Q3537

JEE Main 202131 Aug, Evening shiftMathematicsEasyDifferential Equations163× since 2002

{{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

JEE Main 202131 Aug, Evening shiftMathematicsEasyDifferential Equations163× since 2002

{{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 :