If y=axn+1+bx−n then x2d2ydx2=

y=axn+1+bx−ndydx=a(n+1)xn−bnx−n−1xdydx=a.(n+1)xn+1−bx−n.nxdydx=(y−bx−n)(n+1)−bx−nnxdydx=(n+1)y−bx−n(n+1+n)xd2ydx2+dydx=(n+1)dydx−(2n+1)b(−n)x−n−1x2d2ydx2+xdydx=(n+1)dydxx+(2n+1)bnx−nx2d2ydx2+xdydx(1−n−1)=(2n+1)bnx−nx2d2ydx2−n((n+1)y)−bx−n(2n+1)=(2n+1)bnx−nx2d2ydx2−n(n+1)y+(2n+1)bn.x−n=(2n+1)bx−n.nx2d2ydx2=n(n+1)y

If y=axn+1+bx−n then x2d2ydx2=

y=axn+1+bx−ndydx=a(n+1)xn−bnx−n−1xdydx=a.(n+1)xn+1−bx−n.nxdydx=(y−bx−n)(n+1)−bx−nnxdydx=(n+1)y−bx−n(n+1+n)xd2ydx2+dydx=(n+1)dydx−(2n+1)b(−n)x−n−1x2d2ydx2+xdydx=(n+1)dydxx+(2n+1)bnx−nx2d2ydx2+xdydx(1−n−1)=(2n+1)bnx−nx2d2ydx2−n((n+1)y)−bx−n(2n+1)=(2n+1)bnx−nx2d2ydx2−n(n+1)y+(2n+1)bn.x−n=(2n+1)bx−n.nx2d2ydx2=n(n+1)y

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

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$n (n - 1) y$

$n (n + 1) y$

$n y$

$n^{2} y$

answer is B.

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Detailed Solution

$\begin{array}{l} y = a x^{n + 1} + b x^{- n} \\ \frac{d y}{d x} = a (n + 1) x^{n} - b n x^{- n - 1} \\ x \frac{d y}{d x} = a . (n + 1) x^{n + 1} - b x^{- n} . n \\ x \frac{d y}{d x} = (y - b x^{- n}) (n + 1) - b x^{- n} n \\ x \frac{d y}{d x} = (n + 1) y - b x^{- n} (n + 1 + n) \\ x \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} = (n + 1) \frac{d y}{d x} - (2 n + 1) b (- n) x^{- n - 1} \\ x^{2} \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} = (n + 1) \frac{d y}{d x} x + (2 n + 1) b n x^{- n} \\ x^{2} \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} (1 - n - 1) = (2 n + 1) b n x^{- n} \\ x^{2} \frac{d^{2} y}{d x^{2}} - n ((n + 1) y) - b x^{- n} (2 n + 1) = (2 n + 1) b n x^{- n} \\ x^{2} \frac{d^{2} y}{d x^{2}} - n (n + 1) y + (2 n + 1) b n . x^{- n} = (2 n + 1) b x^{- n} . n \\ x^{2} \frac{d^{2} y}{d x^{2}} = n (n + 1) y \end{array}$