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If $y=2 x^{n+1}+\frac{3}{x^{n}}$, then $x^{2} \frac{d^{2 y}}{d x^{2}}$ is
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Verified Answer
The correct answer is:
$n(n+1) y$
We have, $y$
$\begin{aligned}
y &=2 x^{n+1}+\frac{3}{x^{n}} \\
&=2 x^{n+1}+3 x^{-n}...(i)
\end{aligned}$
On differentiating Eq. (i) both sides w.r.t. $x$,
we get
$\frac{d y}{d x}=2(n+1) x^{n}+3(-n) x^{-n-1}$
Again differentiate w.r.t. to $x$, then we get
$\frac{d^{2} y}{d x^{2}}=2(n+1)(n) x^{n-1}+3(-n)(-n-1) x^{-n-2}$
$=n(n+1)\left(2 x^{n-1}+3 x^{-n-2}\right)$
$\Rightarrow \quad x^{2} \frac{d^{2} y}{d x^{2}}=n(n+1)\left(2 x^{n+1}+\frac{3}{x^{n}}\right)$
$\Rightarrow \quad x^{2} \frac{d y}{d x}=n(n+1) y$ [using Eq. (i)]
$\begin{aligned}
y &=2 x^{n+1}+\frac{3}{x^{n}} \\
&=2 x^{n+1}+3 x^{-n}...(i)
\end{aligned}$
On differentiating Eq. (i) both sides w.r.t. $x$,
we get
$\frac{d y}{d x}=2(n+1) x^{n}+3(-n) x^{-n-1}$
Again differentiate w.r.t. to $x$, then we get
$\frac{d^{2} y}{d x^{2}}=2(n+1)(n) x^{n-1}+3(-n)(-n-1) x^{-n-2}$
$=n(n+1)\left(2 x^{n-1}+3 x^{-n-2}\right)$
$\Rightarrow \quad x^{2} \frac{d^{2} y}{d x^{2}}=n(n+1)\left(2 x^{n+1}+\frac{3}{x^{n}}\right)$
$\Rightarrow \quad x^{2} \frac{d y}{d x}=n(n+1) y$ [using Eq. (i)]
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