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If Y = (Sin−1 X)2, Prove that (1 − X2) D 2 Y D X 2 − X D Y D X + P 2 Y = 0 ? - Mathematics

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प्रश्न

If y = (sin−1 x)2, prove that (1 − x2)

\[\frac{d^2 y}{d x^2} - x\frac{dy}{dx} + p^2 y = 0\] ?

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उत्तर

Here,

\[y = \left( \sin^{- 1} x \right)^2 \]

\[\text { Now,} \]

\[ y_1 = 2 \sin^{- 1} x \frac{1}{\sqrt{1 - x^2}}\]

\[ \Rightarrow y_2 = \frac{2}{1 - x^2} + \frac{2x \sin^{- 1} x}{\left( 1 - x^2 \right)^{3/2}}\]

\[ \Rightarrow y_2 = \frac{2}{1 - x^2} + \frac{2x \sin^{- 1} x}{\left( 1 - x^2 \right)\sqrt{1 - x^2}}\]

\[ \Rightarrow y_2 = \frac{2}{1 - x^2} + \frac{x y_1}{\left( 1 - x^2 \right)}\]

\[ \Rightarrow y_2 \left( 1 - x^2 \right) = 2 + x y_1 \]

\[ \Rightarrow y_2 \left( 1 - x^2 \right) - x y_1 - 2 = 0\]

Hence proved.

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अध्याय 12: Higher Order Derivatives - Exercise 12.1 [पृष्ठ १७]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 12 Higher Order Derivatives
Exercise 12.1 | Q 20 | पृष्ठ १७

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