A) \[2{{\sin }^{-1}}y=x\sqrt{1-{{x}^{2}}}+{{\sin }^{-1}}x+C\]
B) \[{{\cos }^{-1}}y=x{{\cos }^{-1}}x+C\]
C) \[{{\sin }^{-1}}y=\frac{1}{2}{{\sin }^{-1}}x+C\]
D) \[2{{\sin }^{-1}}y=x\sqrt{1-{{y}^{2}}}+C\]
Correct Answer: A
Solution :
Given, \[{{\left( \frac{dy}{dx} \right)}^{2}}=1-{{x}^{2}}-{{y}^{2}}+{{x}^{2}}{{y}^{2}}\] \[\Rightarrow \] \[{{\left( \frac{dy}{dx} \right)}^{2}}=(1-{{x}^{2}})-{{y}^{2}}(1-{{x}^{2}})\] \[\Rightarrow \] \[\frac{dy}{dx}=\sqrt{(1-{{x}^{2}})}\sqrt{(1-{{y}^{2}})}\] \[\int{\frac{dy}{(1-{{y}^{2}})}}=\int{\sqrt{(1-{{x}^{2}})}dx}\] (on integrating) \[\Rightarrow \] \[{{\sin }^{-1}}y=\frac{x}{2}\sqrt{1-{{x}^{2}}}+\frac{1}{2}{{\sin }^{-1}}x+\frac{C}{2}\] \[\Rightarrow \] \[2{{\sin }^{-1}}y=x\sqrt{1-{{x}^{2}}}+{{\sin }^{-1}}x+C\]
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