Question

If $\frac{dy}{dx}=\frac{2^{x}y+2^y\cdot 2^x}{2^x+2^{x+y}{\log }_{e}2},y(0)=0$, 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)

$\frac{dy}{dx}=\frac{2^x\left(y+2^y\right)}{2^x\left(1+2^y\ell n2\right)}$
$\Rightarrow \int \frac{\left(1+2^y\right)\ell n2}{\left(y+2^y\right)}dy=\int dx$
$\Rightarrow \ell n\left|y+2^y\right|=x+c$
$x=0;y=0\Rightarrow c=0$
$\Rightarrow x=\ell n\left|y+2^y\right|$
$\Rightarrow$ at $y=1,x=\ln 3$
$\because 3\in \left(e,e^2\right)$
$\Rightarrow x\in (1,2)$