A) \[1+xy+c(y+x)=0\]
B) \[x+y=c(1-xy)\]
C) \[y-x=c(1+xy)\]
D) \[1+xy=c(x+y)\]
Correct Answer: C
Solution :
\[\frac{dy}{dx}=\frac{1+{{y}^{2}}}{1+{{x}^{2}}}\Rightarrow \frac{1}{1+{{y}^{2}}}dy=\frac{1}{1+{{x}^{2}}}dx\] Now on integrating both sides, we get \[{{\tan }^{-1}}y={{\tan }^{-1}}x+{{\tan }^{-1}}c\]Þ \[{{\tan }^{-1}}y={{\tan }^{-1}}\left( \frac{x+c}{1-cx} \right)\] Þ \[y=\frac{x+c}{1-cx}\] Þ \[y-x=c(1+xy)\].You need to login to perform this action.
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