Question

If $y = 2x^{n+1} + \frac {3}{x^n}$ ,then $x^2 \frac{d^2y}{dx^2}$ is

• A. 6n(n+1)y
• B. n(n+1)y
• C. $x \frac{dy}{dx}+y$
• D. $y$

Answer 1

Answer: (B)

$y=(2 x)^{(n+1)}+(3 x)^{(-n)}$
$\Rightarrow d y / d x=2(n+1) x^{n}-(3 n x)^{(-n-1)}$
$\Rightarrow \left(d^{2} y\right) /\left(d x^{2}\right)=2 n(n+1) x^{(n-1)}+3 n(n+1) x^{(-n-2)}$
$\Rightarrow x^{2}\left(d^{2} y\right) /\left(d x^{2}\right)=n(n+1)\left[(2 x)^{(n+1)}+3 / x^{n}\right]$
$\Rightarrow x^{2}\left(d^{2} y\right) /\left(d x^{2}\right)=n(n+1) y$