If dydx+2x−y2y−12x−1=0,x>0,y>0,y1=1 then y(2) is equal to

dydx+2x−y(2y−1)2x−1=0 ⇒ dydx+2x2x−1×2y−12y=0 ⇒ dydx=−2x(2y−1)2y(2x−1) ⇒ ∫2y2y−1dy=−∫2x2x−1dx ⇒ 11n2∫2y1n2y−1dy=−11n2∫2x1n2x−1dx ⇒ 1n|2y−1|=−1n|2x−1|+cAt x=1, y=1, (given)1n 1=−1n 1+c 0=−0+c⇒c=0 ⇒ 1n|2y−1|=−1n|2x−1| ⇒ 1n((2x−1)(2y−1))=0⇒(2x−1)(2y−1)=1 At x=2, y=? (2y−1)(4−1)=1 2y=1+13=43 y=log24/3=log24−log23=2−log23

If dydx+2x−y2y−12x−1=0,x>0,y>0,y1=1 then y(2) is equal to

dydx+2x−y(2y−1)2x−1=0 ⇒ dydx+2x2x−1×2y−12y=0 ⇒ dydx=−2x(2y−1)2y(2x−1) ⇒ ∫2y2y−1dy=−∫2x2x−1dx ⇒ 11n2∫2y1n2y−1dy=−11n2∫2x1n2x−1dx ⇒ 1n|2y−1|=−1n|2x−1|+cAt x=1, y=1, (given)1n 1=−1n 1+c 0=−0+c⇒c=0 ⇒ 1n|2y−1|=−1n|2x−1| ⇒ 1n((2x−1)(2y−1))=0⇒(2x−1)(2y−1)=1 At x=2, y=? (2y−1)(4−1)=1 2y=1+13=43 y=log24/3=log24−log23=2−log23

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

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$2 - \log_{2} 3$

$2 + \log_{2} 3$

$2 - \log_{3} 2$

$2 + \log_{3} 2$

answer is D.

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Detailed Solution

$\frac{d y}{d x} + \frac{2^{x - y} (2^{y} - 1)}{2^{x} - 1} = 0 \Rightarrow \frac{d y}{d x} + \frac{2^{x}}{2^{x} - 1} \times \frac{2^{y} - 1}{2^{y}} = 0 \Rightarrow \frac{d y}{d x} = \frac{- 2^{x} (2^{y} - 1)}{2^{y} (2^{x} - 1)} \Rightarrow \int \frac{2^{y}}{2^{y} - 1} d y = - \int \frac{2^{x}}{2^{x} - 1} d x \Rightarrow \frac{1}{1 n 2} \int \frac{2^{y} 1 n}{2^{y} - 1} d y = \frac{- 1}{1 n 2} \int \frac{2^{x} 1 n}{2^{x} - 1} d x \Rightarrow 1 n | 2^{y} - 1 | = - 1 n | 2^{x} - 1 | + c$

At $x = 1, y = 1,$ (given)

$1 n 1 = - 1 n 1 + c 0 = - 0 + c \Rightarrow c = 0 \Rightarrow 1 n | 2^{y} - 1 | = - 1 n | 2^{x} - 1 | \Rightarrow 1 n ((2^{x} - 1) (2^{y} - 1)) = 0 \Rightarrow (2^{x} - 1) (2^{y} - 1) = 1 A t x = 2, y = ? (2^{y} - 1) (4 - 1) = 1 2^{y} = 1 + \frac{1}{3} = \frac{4}{3} y = \log_{2} 4 / 3 = \log_{2} 4 - \log_{2} 3 = 2 - \log_{2} 3$