A) \[{{m}^{2}}y\]
B) \[my\]
C) \[-{{m}^{2}}y\]
D) None of these
Correct Answer: C
Solution :
We have,\[y=\sin (m{{\sin }^{-1}}x)\] \[\frac{dy}{dx}=\cos (m{{\sin }^{-1}}x).\frac{m}{\sqrt{1-{{x}^{2}}}}\] \[\Rightarrow \]\[(1-{{x}^{2}}){{\left( \frac{dy}{dx} \right)}^{2}}={{m}^{2}}{{\cos }^{2}}(m{{\sin }^{-1}}x)\] \[={{m}^{2}}(1-{{y}^{2}})\] On differentiating w.r.t.\[x\], again, we get \[(1-{{x}^{2}})2\cdot \frac{dy}{dx}\cdot \frac{{{d}^{2}}y}{d{{x}^{2}}}+{{\left( \frac{dy}{dx} \right)}^{2}}\cdot (-2x)\] \[=-2{{m}^{2}}y\frac{dy}{dx}\] \[\Rightarrow \] \[(1-{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}-x\frac{dy}{dx}=-{{m}^{2}}y\] [\[\because \]Dividing by\[2dy/dx\]]
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