Question

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

• A. $n(n-1)y$
• B. $n(n+1)y$
• C. $ny$
• D. $n^{2}y$

Answer 1

The correct option is B $n(n+1)y$

$y=a{x}^{n+1}+b{x}^{-n}$
$x^2\frac{d^{2}y}{d{x}^2}=a{x}^2\frac{d^2\left(x^{n+1}\right)}{d{x}^2}+b{x}^2\frac{d^2\left(x^{-n}\right)}{d{x}^2}$
$x^2\frac{d^{2}y}{d{x}^2}=a{x}^{2}n(n+1)\left(x^{n-1}\right)+b{x}^2(-n)(-n-1)\left(x^{-n-2}\right)$
$x^2\frac{d^{2}y}{d{x}^2}=n(n+1)\left(a\right(x^{n+1})+b(x^{-n}\left)\right)$
$x^2\frac{d^{2}y}{d{x}^2}=n(n+1)y$
Hence the answer is option B