Question

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

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

Answer 1

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.