A) \[y+x+\log (x+y-2)=c\]
B) \[y+2x+\log (x+y-2)=c\]
C) \[2y+x+\log (x+y-2)=c\]
D) \[2y+2x+\log (x+y-2)=c\]
Correct Answer: C
Solution :
Given equation is \[\frac{dy}{dx}=-\left( \frac{x+y-1}{2x+2y-3} \right)\] Put \[x+y=t\]Þ\[\frac{dy}{dx}=\frac{dt}{dx}-1\] \[\therefore \frac{dy}{dx}=\frac{1-t}{2t-3}\]Þ\[\frac{dt}{dx}-1=\frac{1-t}{2t-3}\] Þ \[\frac{dt}{dx}=\frac{t-2}{2t-3}\] Þ \[\frac{2t-3}{t-2}dt=dx\]. Integrating both sides, we get \[\int_{{}}^{{}}{\frac{2t-4}{t-2}dt}-\int_{{}}^{{}}{\frac{3-4}{t-2}dt}=\int_{{}}^{{}}{1}dx\] Þ \[2t+\log (t-2)=x+c\] Þ \[2(x+y)+\log (x+y-2)=x+c\] Þ \[2y+x+\log (x+y-2)=c\].
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