The differential equation for which \[{{\sin }^{-1}}x+{{\sin }^{-1}}y=c\] is given by [Karnataka CET 2003]

\[\sqrt{1-{{x}^{2}}}\,\,dy\,\,-\sqrt{1-{{y}^{2}}}\,\,dx=0\]

\[\sqrt{1-{{x}^{2}}}\,\,dx\,\,+\sqrt{1-{{y}^{2}}}\,\,dy=0\]

\[\sqrt{1-{{x}^{2}}}\,\,dy\,\,+\sqrt{1-{{y}^{2}}}\,\,dx=0\]

\[\sqrt{1-{{x}^{2}}}\,\,dx\,-\sqrt{1-{{y}^{2}}}\,\,dy=0\]

The differential equation for which \[{{\sin }^{-1}}x+{{\sin }^{-1}}y=

1.	The differential equation for which \[{{\sin }^{-1}}x+{{\sin }^{-1}}y=c\] is given by [Karnataka CET 2003]
A.	\[\sqrt{1-{{x}^{2}}}\,\,dx\,\,+\sqrt{1-{{y}^{2}}}\,\,dy=0\]
B.	\[\sqrt{1-{{x}^{2}}}\,\,dy\,\,+\sqrt{1-{{y}^{2}}}\,\,dx=0\]
C.	\[\sqrt{1-{{x}^{2}}}\,\,dy\,\,-\sqrt{1-{{y}^{2}}}\,\,dx=0\]
D.	\[\sqrt{1-{{x}^{2}}}\,\,dx\,-\sqrt{1-{{y}^{2}}}\,\,dy=0\]
Answer» C. \[\sqrt{1-{{x}^{2}}}\,\,dy\,\,-\sqrt{1-{{y}^{2}}}\,\,dx=0\]

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The differential equation for which \[{{\sin }^{-1}}x+{{\sin }^{-1}}y=c\] is given by [Karnataka CET 2003]

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