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Higher Order Derivatives — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsHigher Order Derivatives3 Marks
Question
If $\text{y}=\text{ax}^{\text{n+1}}+\text{bx}^{-\text{n}}$ and $\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}=\lambda\text{y}$ then write the value of $\lambda$
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Answer

$\text{y}=\text{ax}^{\text{n}+1}+\text{b}\text{x}^{-\text{-n}}$
and $\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}=\lambda\text{y}$
Now,
$\frac{\text{dy}}{\text{dx}}=\text{a}(\text{n}+1)\text{x}^{\text{n}}-\text{bn x}^{-\text{n-1}}$
and $\frac{\text{d}^2\text{y}}{\text{dx}^2}=\text{an}(\text{n}+1)\text{x}^{\text{n}+1}-\text{bn}(-\text{n}-1)\text{x}^{-\text{n}-2}$
Now, $\text{x}^2\frac{\text{d}^2\text{y}}{\text{dx}^2}=\lambda\text{y}[\text{given}]$
$\Rightarrow\text{x}^2[\text{an}(\text{n}+1)\text{x}^{\text{n}-1}+\text{bn}(\text{n}+1)\text{x}^{-\text{n}-2}]=\lambda(\text{ax}^{\text{n+1}}+\text{b x}^{-\text{n}})$
$\Rightarrow\text{an}(\text{n}+1)\text{x}^{\text{n}+1}+\text{bn}(\text{n}+1)\text{x}^{-\text{n}}=\lambda\text{ax}^{\text{n+1}}+\text{b x}^{-\text{n}}$
$\Rightarrow\text{n}(\text{n}+1)\text{ax}^{\text{n}+1}+\text{bx}^{-n}=\lambda\text{ax}^{\text{n}+1}+\text{dx}^{\text{-n}}$
$\Rightarrow\lambda=\text{n}(\text{n}+1)$

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