Question

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

• A. $2 + log_2\; 3$
• B. $2 + log_3 \;2$
• C. $2 – log_3 \;2$
• D. $2 – log_2 \;3$

Answer 1

Answer: (D)

$2 – log_2 \;3$

Answer 2

$\frac{dy}{dx} + \frac{2x-y(2^y-1)}{2x-1} = 0, x,y > 0,y(1) = 1$

$\frac{dy}{dx} = - \frac{2x(2y-1)}{2x-1}$

$∫ \frac{2y}{2y-1}dy = - ∫\frac{2x}{2x-1}dx$

= $\frac{\log_e(2y-1)}{\log_{e^2}} = - \frac{\log_e(2^x-1)}{\log_{e^2}}+\frac{\log_{e^c}}{\log_{e^2}}$

= $I(2^y-1)(2x-)|=c$

$∴ y(1)=1$

$∴ c = 1$

= $|(2^y-1)(2^x-1)| = 1$

As $x = 2$

$|(2^y-1)^3| = 1$

$2^y-1 = \frac{1}{3}$

$⇒ 2y = \frac{4}{3}$

Taking log to base 2.

$∴ y = 2 – \log_2 3$

Hence, the correct option is (D): $2 – log_2 \;3$