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If Y = Cosec−1 X, X >1, Then Show that X ( X 2 − 1 ) D 2 Y D X 2 + ( 2 X 2 − 1 ) D Y D X = 0 ? - Mathematics

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प्रश्न

If y = cosec−1 xx >1, then show that \[x\left( x^2 - 1 \right)\frac{d^2 y}{d x^2} + \left( 2 x^2 - 1 \right)\frac{dy}{dx} = 0\] ?

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उत्तर

Here,

\[y = {cosec}^{- 1} x\]

\[\text { Differentiating w . r . t . x, we get }\]

\[\frac{d y}{d x} = \frac{- 1}{x\sqrt{x^2 - 1}}\]

\[\text { Differentiating again w . r . t . x, we get }\]

\[\frac{d^2 y}{d x^2} = \frac{\sqrt{x^2 - 1} + \frac{x^2}{\sqrt{x^2 - 1}}}{x^2 \left( x^2 - 1 \right)}\]

\[ \Rightarrow \frac{d^2 y}{d x^2} = \frac{x^2 - 1 + x^2}{x^2 \left( x^2 - 1 \right)\sqrt{x^2 - 1}}\]

\[ \Rightarrow \frac{d^2 y}{d x^2} = \frac{2 x^2 - 1}{x^2 \left( x^2 - 1 \right)\sqrt{x^2 - 1}}\]

\[ \Rightarrow \frac{d^2 y}{d x^2} = \frac{2}{\left( x^2 - 1 \right)\sqrt{x^2 - 1}} - \frac{1}{x^2 \left( x^2 - 1 \right)\sqrt{x^2 - 1}}\]

\[ \Rightarrow \left( x^2 - 1 \right)\frac{d^2 y}{d x^2} = \frac{2}{\sqrt{x^2 - 1}} - \frac{1}{x^2 \sqrt{x^2 - 1}}\]

\[ \Rightarrow \left( x^2 - 1 \right)\frac{d^2 y}{d x^2} = - 2x\frac{dy}{dx} + \frac{1}{x}\frac{dy}{dx}\]

\[ \Rightarrow x\left( x^2 - 1 \right)\frac{d^2 y}{d x^2} = - \left( 2 x^2 - 1 \right)\frac{dy}{dx}\]

\[ \Rightarrow x\left( x^2 - 1 \right)\frac{d^2 y}{d x^2} + \left( 2 x^2 - 1 \right)\frac{dy}{dx} = 0\]

Hence proved.

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अध्याय 12: Higher Order Derivatives - Exercise 12.1 [पृष्ठ १८]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 12 Higher Order Derivatives
Exercise 12.1 | Q 42 | पृष्ठ १८

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