A) \[\left( \frac{dy}{dx}-1 \right)\left( y+x\frac{dy}{dx} \right)=2\frac{dy}{dx}\]
B) \[\left( \frac{dy}{dx}+1 \right)\left( y-x\frac{dy}{dx} \right)=\frac{dy}{dx}\]
C) \[\left( \frac{dy}{dx}+1 \right)\left( y-x\frac{dy}{dx} \right)=2\frac{dy}{dx}\]
D) None of these
Correct Answer: C
Solution :
[c] \[\frac{x}{c-1}+\frac{y}{c+1}=1\] ? (i) \[\Rightarrow \frac{1}{c-1}+\frac{y'}{c+1}=0\] ? (ii) \[\Rightarrow \frac{y'}{1}=\frac{c+1}{1-c}\Rightarrow \frac{y'-1}{y'+1}=c\] Put value of c in (1) \[\Rightarrow \frac{x}{\frac{y'-1}{y'+1}-1}+\frac{y}{\frac{y'-1}{y'+1}+1}=1\] \[\Rightarrow \frac{x(y')+1}{-2}+\frac{y(y'+1)}{2y'}=1\] \[\Rightarrow \frac{(y'+1)}{2}\left( \frac{y}{y'}-x \right)=1\] \[\Rightarrow \left( 1+\frac{dy}{dx} \right)\left( y-x\frac{dy}{dx} \right)=2\frac{dy}{dx}\]
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