The solution of the differential equation x+y−1x+y−2dydx=x+y+1x+y+2, given that y = 1 when x = 1 ; is

x+y−1x+y−2dydx=x+y+1x+y+2 Put x+y=t⇒1+dydx=dtdxfrom (1)t−1t−2dtdx−1=t+1t+2⇒dtdx−1=t+1t+2t−2t−1⇒dtdx=(t+1)(t−2)(t+2)(t−1)+1dtdx=2t2−4t2+t−2By separating variables∫t2+t−2t2−2dt=∫2dx⇒∫1+tt2−2dt=2x+C⇒t+12log⁡t2−2=2x+C⇒(y−x)+12log⁡(x+y)2−2=C…2 Sub x=1,y=1⇒C=12log⁡2 From(2),required D.E is 2(y−x)+log⁡(x−y)2+22=0

The solution of the differential equation x+y−1x+y−2dydx=x+y+1x+y+2, given that y = 1 when x = 1 ; is

x+y−1x+y−2dydx=x+y+1x+y+2 Put x+y=t⇒1+dydx=dtdxfrom (1)t−1t−2dtdx−1=t+1t+2⇒dtdx−1=t+1t+2t−2t−1⇒dtdx=(t+1)(t−2)(t+2)(t−1)+1dtdx=2t2−4t2+t−2By separating variables∫t2+t−2t2−2dt=∫2dx⇒∫1+tt2−2dt=2x+C⇒t+12log⁡t2−2=2x+C⇒(y−x)+12log⁡(x+y)2−2=C…2 Sub x=1,y=1⇒C=12log⁡2 From(2),required D.E is 2(y−x)+log⁡(x−y)2+22=0

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The solution of the differential equation $\frac{x + y - 1}{x + y - 2} \frac{d y}{d x} = \frac{x + y + 1}{x + y + 2}$ , given that $y = 1$ when $x = 1$ ; is

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

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

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

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

answer is C.

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Detailed Solution

$\begin{aligned} (\frac{x + y - 1}{x + y - 2}) \frac{d y}{d x} = \frac{x + y + 1}{x + y + 2} \\ Put x + y = t \Rightarrow 1 + \frac{d y}{d x} = \frac{d t}{d x} \end{aligned}$
from (1)

$\begin{aligned} (\frac{t - 1}{t - 2}) (\frac{d t}{d x} - 1) = \frac{t + 1}{t + 2} \\ \Rightarrow \frac{d t}{d x} - 1 = (\frac{t + 1}{t + 2}) (\frac{t - 2}{t - 1}) \\ \Rightarrow \frac{d t}{d x} = \frac{(t + 1) (t - 2)}{(t + 2) (t - 1)} + 1 \\ \frac{d t}{d x} = \frac{2 t^{2} - 4}{t^{2} + t - 2} \end{aligned}$
By separating variables
$\begin{aligned} \int (\frac{t^{2} + t - 2}{t^{2} - 2}) d t = \int 2 d x \\ \Rightarrow \int (1 + \frac{t}{t^{2} - 2}) d t = 2 x + C \\ \Rightarrow t + \frac{1}{2} \log (t^{2} - 2) = 2 x + C \\ \Rightarrow (y - x) + \frac{1}{2} \log ((x + y)^{2} - 2) = C \dots (2) \end{aligned}$
$\begin{aligned} Sub x = 1, y = 1 \\ \Rightarrow C = \frac{1}{2} \log 2 \end{aligned}$
$From(2),required D.E is$
$2 (y - x) + \log |\frac{(x - y)^{2} + 2}{2}| = 0$