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Higher Order Derivatives — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsHigher Order Derivatives5 Marks
Question
If $\text{x}=\sin\Big(\frac{1}{\text{a}}\log\text{y}\Big),$ show that $(1-\text{x}^2)\text{y}_2-\text{xy}_1-\text{a}^2\text{y}=0$
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Answer

Here,
$\text{x}=\sin\Big(\frac{1}{\text{a}}\log\text{y}\Big),$
$\Rightarrow\frac{1}{\text{a}}\log\text{y}=\sin^{-1}\text{x}$
$\Rightarrow\text{y}=\text{e}^\text{a}\sin^{-1}\text{x}$
Differentiating w.r.t.x, we get
$\text{y}_1=\text{e}^{\text{a} \sin^{-1}\text{x}}\times\frac{\text{a}}{\sqrt{1-\text{x}^2}}$
$\Rightarrow\text{y}_1=\frac{\text{ay}}{\sqrt{1-\text{x}^2}}$
Differentiating w.r.t.x, we get
$\text{y}_2=\frac{\text{ay}_1\sqrt{1-\text{x}^2}+\frac{\text{x}\text{ay}}{\sqrt{1-\text{x}^2}}}{(1-\text{x}^2)}$
$\Rightarrow\text{y}_2=\frac{\text{ay}_1(1-\text{x}^2)+\text{xay}}{(1-\text{x}^2)\sqrt{1-\text{x}^2}}$
$\Rightarrow\text{y}_2=\frac{\text{ay}_1}{\sqrt{1-\text{x}^2}}+\frac{\text{xay}}{(1-\text{x}^2)\sqrt{1-\text{x}^2}}$
$\Rightarrow\text{y}_2=\frac{\text{a}^2\text{y}}{1-\text{x}^2}+{\frac{\text{xy}_1}{(1-\text{x}^2)}}$
$\Rightarrow(1-\text{x}^2)\text{y}_2-\text{xy}_1-\text{a}^2\text{y}=0$

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