Solution of x+y-1x+y-2dydx=x+y+1x+y+2, given that y = 1 when x = 1, is

(x+y−1x+y−2)dydx= x+y+1x+y+2Let x+y=v1+dydx=dvdx⇒ dydx=dvdx−1⇒ (v−1v−2)(dvdx−1)=v+1v+2⇒ dvdx−1=v+1v+2×v−2v−1dvdx= v2−v−2v2+v−2+1⇒ dvdx= v2−v−2+v2+v−2v2+v−2= 2v2−4v2+v−2⇒ v2+v−2v2−2 dv= 2dx⇒ ∫(1+vv2−2) dv= 2∫dx⇒ ∫1 dv +12 ∫2vv2−2 dv =2∫dx⇒ v+12 log (v2−2)= 2x+c⇒ y+x+12 log ((x+y)2−2)= 2x+c⇒ y−x+12log ((x+y)2−2)=cPut x=1, y=1⇒ 0+12log (4−2)=cc= 12log 2⇒ y−x+12log ((x+y)2−2)=12log 2⇒ 12(log ((x+y)2−2)−log 2)=x−y⇒ log ((x+y)2−22)= 2(x−y)

Solution of x+y-1x+y-2dydx=x+y+1x+y+2, given that y = 1 when x = 1, is

(x+y−1x+y−2)dydx= x+y+1x+y+2Let x+y=v1+dydx=dvdx⇒ dydx=dvdx−1⇒ (v−1v−2)(dvdx−1)=v+1v+2⇒ dvdx−1=v+1v+2×v−2v−1dvdx= v2−v−2v2+v−2+1⇒ dvdx= v2−v−2+v2+v−2v2+v−2= 2v2−4v2+v−2⇒ v2+v−2v2−2 dv= 2dx⇒ ∫(1+vv2−2) dv= 2∫dx⇒ ∫1 dv +12 ∫2vv2−2 dv =2∫dx⇒ v+12 log (v2−2)= 2x+c⇒ y+x+12 log ((x+y)2−2)= 2x+c⇒ y−x+12log ((x+y)2−2)=cPut x=1, y=1⇒ 0+12log (4−2)=cc= 12log 2⇒ y−x+12log ((x+y)2−2)=12log 2⇒ 12(log ((x+y)2−2)−log 2)=x−y⇒ log ((x+y)2−22)= 2(x−y)

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Solution of $(\frac{x + y - 1}{x + y - 2}) \frac{dy}{dx} = (\frac{x + y + 1}{x + y + 2}),$ given that y = 1 when x = 1, is

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$\log |\frac{{(x + y)}^{2} + 2}{2}| = 2 (x - y)$

$2 (y - x) + \log |\frac{{(x - y)}^{2} - 2}{2}| = 0$

$\log |\frac{{(x - y)}^{2} - 2}{2}| = 2 (x + y)$

$\log |\frac{{(x - y)}^{2} + 2}{2}| = 2 (x - y)$

answer is D.

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Detailed Solution

$\begin{array}{l} (\frac{x + y - 1}{x + y - 2}) \frac{d y}{d x} = \frac{x + y + 1}{x + y + 2} \\ L e t x + y = v \\ 1 + \frac{d y}{d x} = \frac{d v}{d x} \Rightarrow \frac{d y}{d x} = \frac{d v}{d x} - 1 \\ \Rightarrow (\frac{v - 1}{v - 2}) (\frac{d v}{d x} - 1) = \frac{v + 1}{v + 2} \\ \Rightarrow \frac{d v}{d x} - 1 = \frac{v + 1}{v + 2} \times \frac{v - 2}{v - 1} \\ \frac{d v}{d x} = \frac{v^{2} - v - 2}{v^{2} + v - 2} + 1 \\ \Rightarrow \frac{d v}{d x} = \frac{v^{2} - v - 2 + v^{2} + v - 2}{v^{2} + v - 2} \\ = \frac{2 v^{2} - 4}{v^{2} + v - 2} \end{array}$

$\begin{array}{l} \Rightarrow \frac{v^{2} + v - 2}{v^{2} - 2} d v = 2 d x \\ \Rightarrow \int (1 + \frac{v}{v^{2} - 2}) d v = 2 \int d x \\ \Rightarrow \int 1 d v + \frac{1}{2} \int \frac{2 v}{v^{2} - 2} d v = 2 \int d x \\ \Rightarrow v + \frac{1}{2} \log (v^{2} - 2) = 2 x + c \\ \Rightarrow y + x + \frac{1}{2} \log ({(x + y)}^{2} - 2) = 2 x + c \\ \Rightarrow y - x + \frac{1}{2} \log ({(x + y)}^{2} - 2) = c \end{array}$

$\begin{array}{l} P u t x = 1, y = 1 \\ \Rightarrow 0 + \frac{1}{2} \log (4 - 2) = c \\ c = \frac{1}{2} \log 2 \\ \Rightarrow y - x + \frac{1}{2} \log ({(x + y)}^{2} - 2) = \frac{1}{2} \log 2 \\ \Rightarrow \frac{1}{2} (\log ({(x + y)}^{2} - 2) - \log 2) = x - y \\ \Rightarrow \log (\frac{{(x + y)}^{2} - 2}{2}) = 2 (x - y) \end{array}$