Question

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

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

Answer 1

Answer: (B)

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