A) \[1\]
B) \[7\]
C) \[-1\]
D) \[0\]
Correct Answer: D
Solution :
\[y=\cos \,(m\,{{\sin }^{-1}}x).\frac{m}{\sqrt{1-{{x}^{2}}}}\] \[\frac{dy}{dx}=-\sin \,\,(m\,{{\sin }^{-1}}x).\,\frac{m}{\sqrt{1-{{x}^{2}}}}\] \[(1-{{x}^{2}})\,\left( \frac{dy}{dx} \right)\,=\,{{m}^{2}}\,{{\sin }^{2}}\,(m\,{{\sin }^{-1}}x)\] \[(1-{{x}^{2}}){{\left( \frac{dy}{dx} \right)}^{2}}={{m}^{2}}[1-{{\cos }^{2}}\,(m\,{{\sin }^{-1}}x)]\] \[(1-{{x}^{2}})\,{{\left( \frac{dy}{dx} \right)}^{2}}={{m}^{2}}(1-{{y}^{2}})\] \[(1-{{x}^{2}})\,.2\left( \frac{dy}{dx} \right)\,\frac{{{d}^{2}}y}{d{{x}^{2}}}+{{\left( \frac{dy}{dx} \right)}^{2}}(-2x)\] \[=-{{m}^{2}}.2y\frac{dy}{dx}\] \[(1-{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}-x\frac{dy}{dx}=-{{m}^{2}}y\] \[\Rightarrow \] \[(1-{{x}^{2}})\,\frac{{{d}^{2}}y}{d{{x}^{2}}}-x\frac{dy}{dx}+{{m}^{2}}y=0\]
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