If y = sin(m sin^{-1} x), then (1-x^2) y_2 - | vedclass

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(B) Given $y = \sin(m \sin^{-1} x)$.First, differentiate with respect to $x$:$y_1 = \cos(m \sin^{-1} x) \cdot \frac{m}{\sqrt{1-x^2}}$.Multiply both si

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$-m^2 y$

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(B) Given $y = \sin(m \sin^{-1} x)$.First, differentiate with respect to $x$:$y_1 = \cos(m \sin^{-1} x) \cdot \frac{m}{\sqrt{1-x^2}}$.Multiply both sides by $\sqrt{1-x^2}$:$y_1 \sqrt{1-x^2} = m \cos(m \sin^{-1} x)$.Differentiate again with respect to $x$ using the product rule:$y_2 \sqrt{1-x^2} + y_1 \cdot \frac{1}{2\sqrt{1-x^2}} \cdot (-2x) = m \cdot (-\sin(m \sin^{-1} x)) \cdot \frac{m}{\sqrt{1-x^2}}$.$y_2 \sqrt{1-x^2} - \frac{x y_1}{\sqrt{1-x^2}} = -\frac{m^2 \sin(m \sin^{-1} x)}{\sqrt{1-x^2}}$.Multiply the entire equation by $\sqrt{1-x^2}$:$y_2(1-x^2) - x y_1 = -m^2 \sin(m \sin^{-1} x)$.Since $y = \sin(m \sin^{-1} x)$, we get:$(1-x^2) y_2 - x y_1 = -m^2 y$.

If $y = \sin(m \sin^{-1} x)$, then $(1-x^2) y_2 - x y_1$ is equal to (Here, $y_n$ denotes $\frac{d^n y}{dx^n}$)

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