Question

If $\frac{dy}{dx}=\frac{2^{x}y+2^y\cdot 2^x}{2^x+2^{x+y}{\log }_e{}^2},$ $y\left(0\right)=0,$ then for $y=1$, the value of $x$ lies in the interval

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

Answer 1

The correct option is C $(1,2)$

$\frac{dy}{dx}=\frac{2^x(y+2^y)}{2^x(1+2^y{\log }_{e}2)}$
$\Rightarrow \int \frac{(1+2^y{\log }_{e}2)dy}{(y+2^y)}=\int dx$
For LHS put $y+2^y=t$
$\Rightarrow (1+2^y{\log }_{e}2)dy=dt$
$\Rightarrow {\log }_e(y+2^y)=x+c$
$\because y\left(0\right)=0$
$\therefore c=0$
$\therefore {\log }_e(y+2^y)=x$
If $y=1$
$\Rightarrow x={\log }_{e}3$
$x\in (1,2)$