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APPLICATION OF DERIVATIVES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAPPLICATION OF DERIVATIVES5 Marks
Question
$\text{If y = e}^{\text{m}\sin^{-1}\text{x}},$ then show that $\text{(1 - x}^{2}) \frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} - \text{x}\frac{\text{dy}}{\text{dx}} - \text{m}^{2}\text{y} = 0.$
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Answer

$\text{y = e}^{\text{m}\sin^{-1}\text{x}},$ differentiate w.r.t. "x" we get $\frac{\text{dy}}{\text{dx}} = \frac{\text{m e}^{\text{m}}\sin^{-1}\text{x}}{\sqrt{1 - \text{x}^{2}}}$
$\Rightarrow\sqrt{1- \text{x}^{2}} \frac{\text{dy}}{\text{dx}} =$ my Differentiate again w.r.t. "x".
$\Rightarrow \sqrt{1 - \text{x}^{2}} \frac{\text{d}^{2}{\text{y}}}{\text{dx}^{2}} - \frac{\text{x}}{\sqrt{1 - \text{x}^{2}}} \frac{\text{dy}}{\text{dx}} = \text{m} \frac{\text{dy}}{\text{dx}}$
$\Rightarrow(1 - \text{x}^{2})\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}} - \text{x} \frac{\text{dy}}{\text{dx}} = \text{m}\bigg(\sqrt{1 - \text{x}^{2}}\frac{\text{dy}}{\text{dx}}\bigg) = \text{m (my)}$
$\Rightarrow( 1 - \text{x}^{2})\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\text{x} \frac{\text{dy}}{\text{dx}} - \text{m}^{2}\text{y} = 0$

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