A) \[{{\sin }^{-1}}xy=c-x\]
B) \[xy=\sin (x+c)\]
C) \[\log (1-{{x}^{2}}{{y}^{2}})=x+c\]
D) \[y=x\sin x+c\]
Correct Answer: B
Solution :
\[xdy+ydx=\sqrt{1-{{x}^{2}}{{y}^{2}}}dx\] Þ \[\frac{xdy+ydx}{\sqrt{1-{{x}^{2}}{{y}^{2}}}}=dx\] \[\frac{dxy}{\sqrt{1-{{(xy)}^{2}}}}=dx\]. Integrating both side, we get \[{{\sin }^{-1}}xy=x+c\] Þ \[xy=\sin (x+c)\].
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