Question

If $\frac{dy}{dx}+\frac{2^{x-y}\left(2^y-1\right)}{2^x-1}=0,x,y>0,y(1)=1$, then $y(2)$ is equal to :

• A. $2+{\log }_{2}3$
• B. $2+{\log }_{2}2$
• C. $-2+{\log }_{2}3$
• D. $2-{\log }_{2}3$

Answer 1

Answer: (D)

$\frac{dy}{dx}+\frac{2^{x-y}\left(2^y-1\right)}{2^x-1}=0$
$x,y>0,y(1)=1,y(2)=?$
$\frac{dy}{dx}=-\frac{2^x\left(2^y-1\right)}{2^y\left(2^x-1\right)}$
$\int \frac{2^y}{2^y-1}dy=-\int \frac{2^x}{2^x-1}dx$
$\frac{1}{\ln 2}\int \frac{2^y\ln 2}{2^y-1}dy=-\frac{1}{\ln 2}\int \frac{2^x\ln 2}{2^x-1}dx$
$\frac{1}{\ln 2}\ln \left|2^y-1\right|=\frac{-1}{\ln 2}\ln \left|2^x-1\right|+C$
At $x=1,y=1$
Putting this values in above relation we get $C=0$
$\ln \left|2^y-1\right|+\ln \left|2^x-1\right|=0$
$\left(2^x-1\right)\left(2^y-1\right)=1$
$2^y-1=\frac{1}{2^x+1}$
$\text{ At }x=2$
$2^y=\frac{1}{3}+1=\frac{4}{3}$
$y={\log }_2\frac{4}{3}={\log }_{2}4-{\log }_{2}3=2-{\log }_{2}3$