A) \[n\,(n-1)y\]
B) \[n\,(n+1)y\]
C) ny
D) \[{{n}^{2}}y\]
Correct Answer: B
Solution :
\[y=a{{x}^{n+1}}+b{{x}^{-n}}\Rightarrow \frac{dy}{dx}=(n+1)a{{x}^{n}}-nb{{x}^{-n-1}}\] Þ \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=n(n+1)a{{x}^{n-1}}+n(n+1)b{{x}^{-n-2}}\] Þ \[{{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}=n(n+1)y\].You need to login to perform this action.
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