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Question: If \(y = a x ^ { n + 1 } + b x ^ { - n }\) then \(x ^ { 2 } \frac { d ^ { 2 } y } { d x ^ { 2 } }\) ...

If y=axn+1+bx−ny = a x ^ { n + 1 } + b x ^ { - n } then x2d2ydx2x ^ { 2 } \frac { d ^ { 2 } y } { d x ^ { 2 } } equals to

A

n(n−1)yn ( n - 1 ) y

B

n(n+1)yn ( n + 1 ) y

C

ny

D

n2y

Answer

n(n+1)yn ( n + 1 ) y

Explanation

Solution

y=axn+1+bx−ny = a x ^ { n + 1 } + b x ^ { - n }

Differentiate with respect to x

dydx=a(n+1)xn−bnx−n−1\frac { d y } { d x } = a ( n + 1 ) x ^ { n } - b n x ^ { - n - 1 }

Again differentiate,

d2ydx2=an(n+1)xn−1+bn(n+1)x−n−2\frac { d ^ { 2 } y } { d x ^ { 2 } } = a n ( n + 1 ) x ^ { n - 1 } + b n ( n + 1 ) x ^ { - n - 2 }

⇒ x2d2ydx2=an(n+1)xn+1+bn(n+1)x−nx ^ { 2 } \frac { d ^ { 2 } y } { d x ^ { 2 } } = a n ( n + 1 ) x ^ { n + 1 } + b n ( n + 1 ) x ^ { - n }

⇒ x2d2ydx2=n(n+1)yx ^ { 2 } \frac { d ^ { 2 } y } { d x ^ { 2 } } = n ( n + 1 ) y.