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प्रश्न
If y = (sin−1 x)2, prove that (1 − x2)
\[\frac{d^2 y}{d x^2} - x\frac{dy}{dx} + p^2 y = 0\] ?
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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