Question

If $\frac{d y}{d x}+\frac{2^{x-y}\left(2^{y}-1\right)}{2^{x}-1}=0, x, y > 0, y(1)=1$, 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)

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