If (dy)/(dx) = 2 ^ x y + 2 ^ y · 2 ^ x 2 ^ x + 2 ^ x + y _ e 2 ,...

Given differential equation is

$\dfrac{dy}{dx} = \dfrac{{2}^{x} \cdot y + {2}^{y} \cdot {2}^{x}}{{2}^{x} + {2}^{x + y}\log_{e}2}$

$⇒ \dfrac{dy}{dx} = \dfrac{{2}^{x}\left(y + {2}^{y}\right)}{{2}^{x}\left(1 + {2}^{y}\log_{e}2\right)}$

$⇒ ∫ \dfrac{1 + {2}^{y}\log_{e}2}{y + {2}^{y}}dy = ∫ dx$

$⇒ ∫ \dfrac{d\left(y + {2}^{y}\right)}{y + {2}^{y}} = ∫ dx$

$⇒ \ln \left|y + {2}^{y}\right| = x + C\left[∵ ∫ \dfrac{{f}^{'}\left(x\right)}{f\left(x\right)}dx = \ln \left(f\left(x\right)\right) + C\right]$

Now $∵ y\left(0\right) = 0$

$⇒ C = 0$

$∴ \ln \left|y + {2}^{y}\right| = x$

Now for $y = 1$ we have

$x = \ln \left(1 + 2\right) = \ln 3 \in \left(1, 2\right)$