Question

If $\frac{d y}{d x}=\frac{2^{x} y+2^{y} \cdot 2^{x}}{2^{x}+2^{x+y} \log _{e} 2}, y(0)=0$, then for $y=1$, the value of $x$ lies in the interval:

• A. $(1,2)$
• B. $\left(\frac{1}{2}, 1\right]$
• C. $(2,3)$
• D. $\left(0, \frac{1}{2}\right]$

Answer 1

Answer: (A)

$\frac{d y}{d x}=\frac{2^{x}\left(y+2^{y}\right)}{2^{x}\left(1+2^{y} \ell n 2\right)}$
$\Rightarrow \int \frac{\left(1+2^{y}\right) \ell n 2}{\left(y+2^{y}\right)} d y=\int d x$
$\Rightarrow \ell n\left|y+2^{y}\right|=x+c$
$x=0 ; y=0 \Rightarrow c=0$
$\Rightarrow x=\ell n\left|y+2^{y}\right|$
$\Rightarrow $ at $y=1, x=\ln 3$
$\because 3 \in\left(e, e^{2}\right)$
$ \Rightarrow x \in(1,2)$