A) \[{{\sin }^{-1}}x+{{\sin }^{-1}}y=c\]
B) \[{{x}^{2}}+{{y}^{2}}={{(1+{{x}^{2}})}^{1/2}}+{{(1+{{y}^{2}})}^{1/2}}+c\]
C) \[{{(1+{{x}^{2}})}^{1/2}}+{{(1+{{y}^{2}})}^{1/2}}=c\]
D) \[{{\tan }^{-1}}x-{{\tan }^{-1}}y=c\]
E) \[{{\cos }^{-1}}x+{{\cos }^{-1}}y=c\]
Correct Answer: C
Solution :
\[x{{(1+{{y}^{2}})}^{1/2}}dx+y{{(1+{{x}^{2}})}^{1/2}}dy=0\] \[\Rightarrow \]\[\frac{x}{{{(1+{{x}^{2}})}^{1/2}}}dx+\frac{y}{{{(1+{{y}^{2}})}^{1/2}}}dy=0\] \[\int{\frac{x}{{{(1+{{x}^{2}})}^{1/2}}}}dx+\int{\frac{y}{{{(1+{{y}^{2}})}^{1/2}}}}dy=0\] \[\Rightarrow \]\[\sqrt{1+{{x}^{2}}}+\sqrt{1+{{y}^{2}}}=c\]
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