Word Count : 374

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यदि (y = e^{m \sin^{-1} x}), तो इसे (x) के अनुसार अलग करें। पहली अवकल: [ y_1 = \frac{dy}{dx} = e^{m \sin^{-1} x} \cdot \frac{m}{\sqrt{1 - x^2}} = \frac{m y}{\sqrt{1 - x^2}} ] दूसरी अवकल: [ y_2 = \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{m y}{\sqrt{1 - x^2}}\right) = m \left(\frac{y_1}{\sqrt{1 - x^2}} + y \frac{d}{dx}\left(\frac{1}{\sqrt{1 - x^2}}\right)\right) ] [ \frac{d}{dx}\left(\frac{1}{\sqrt{1 - x^2}}\right) = \frac{x}{(1 - x^2)^{3/2}} ] इसलिए: [ y_2 = m \left(\frac{m y}{1 - x^2} + ____ _________ _________ __________ _______ ________ ___ ________ ____ ________.
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