If y=ax^ n+1 +bx^ -n Then x^2 d^2y dx^2 = - Vidyadip

MCQ

If $\text{y}=\text{ax}^{\text{n+1}}+\text{bx}^{-\text{n}}$ Then $\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}$ =

A
$\text{n(n - 1) y}$
✓
$\text{n(n + 1) y}$
C
$ny$
D
$n^2y$

✓

Answer

Correct option: B.

$\text{n(n + 1) y}$

Here
$\text{y}=\text{ax}^{\text{n}+1}+\text{bx}^{\text{-n}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{a}(\text{n}+1)\text{x}^\text{n}-\text{bn}\text{x}^{-\text{n}-1}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{an}(\text{n}+1)\text{x}^{\text{n}-1}+\text{bn}(\text{n}+1)\text{x}^{-\text{n}-2}$
$\therefore\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{x}^2\{\text{an}(\text{n}+1)\text{x}^{\text{n}-1}+\text{bn}(\text{n}+1)\text{x}^{-\text{n}-2}\}$
$=\text{n}(\text{n}+1)(\text{ax}^{\text{n}+1}+\text{b x}^{-\text{n}})$
$=\text{n}(\text{n}+1)\text{y}$

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