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

The Correct Option is D

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

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

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

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

= \(I(2^y-1)(2x-)|=c\)

\(∴ y(1)=1\)

\(∴ c = 1\)

= \(|(2^y-1)(2^x-1)| = 1\)

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): \(2 – log_2 \;3\)