A) \[(1-{{x}^{2}}){{y}_{2}}+x{{y}_{1}}-{{m}^{2}}y=0\]
B) \[(1-{{x}^{2}}){{y}_{2}}-x{{y}_{1}}+{{m}^{2}}y=0\]
C) \[(1+{{x}^{2}}){{y}_{2}}+x{{y}_{1}}-{{m}^{2}}y=0\]
D) \[(1-{{x}^{2}}){{y}_{2}}+x{{y}_{1}}+{{m}^{2}}y=0\]
E) \[(1-x){{y}_{2}}-x{{y}_{1}}+{{m}^{2}}y=0\]
Correct Answer: B
Solution :
\[y=\cos (m\,{{\sin }^{-1}}x)\] On differentiating w.r.t.\[x,\]we get \[{{y}_{1}}=-\sin (m{{\sin }^{-1}}x)\frac{m}{\sqrt{1-{{x}^{2}}}}\] \[\Rightarrow \]\[\sqrt{1-{{x}^{2}}}{{y}_{1}}=-m\sin (m{{\sin }^{-1}}x)\] \[\Rightarrow \]\[(1-{{x}^{2}}){{y}_{2}}-x{{y}_{1}}=-{{m}^{2}}y\] \[\Rightarrow \,\,\,\,(1-{{x}^{2}})\,{{y}_{2}}-x{{y}_{1}}+{{m}^{2}}y=0\]
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