A) \[ny\]
B) \[{{n}^{2}}y\]
C) \[n(n-1)y\]
D) \[n(n+1)y\]
Correct Answer: D
Solution :
\[y=a{{x}^{n+1}}+b{{x}^{-n}}\] \[\frac{dy}{dx}=(n+1)a{{x}^{n}}-bn{{x}^{-n-1}}\] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=n(n+1)a{{x}^{n-1}}+n(n+1)b{{x}^{-n-2}}\] \[\Rightarrow \] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{n(n+1)}{{{x}^{2}}}[a{{x}^{n+1}}+b{{x}^{-n}}]\] \[\Rightarrow \] \[{{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}=n(n+1)y\]
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