If \[y=a{{x}^{n+1}}+b{{x}^{-n}},\] then \[{{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}\] equals to [RPET 2001]

If \[y=a{{x}^{n+1}}+b{{x}^{-n}},\] then \[{{x}^{2}}\frac{{{d}^{2}}y}{d

1.	If \[y=a{{x}^{n+1}}+b{{x}^{-n}},\] then \[{{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}\] equals to [RPET 2001]
A.	\[n(n-1)y\]
B.	\[n(n+1)y\]
C.	ny
D.	n2y
Answer» C. ny

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If \[y=a{{x}^{n+1}}+b{{x}^{-n}},\] then \[{{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}\] equals to [RPET 2001]

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