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

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(A) Given $y=e^{m \sin^{-1} x}$.Differentiating with respect to $x$, we get $\frac{d y}{d x}=e^{m \sin^{-1} x} \cdot \frac{m}{\sqrt{1-x^{2}}}$.This im

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$m^{2}$

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(A) Given $y=e^{m \sin^{-1} x}$.Differentiating with respect to $x$, we get $\frac{d y}{d x}=e^{m \sin^{-1} x} \cdot \frac{m}{\sqrt{1-x^{2}}}$.This implies $\sqrt{1-x^{2}} \frac{d y}{d x}=m e^{m \sin^{-1} x} = m y$.Squaring both sides or differentiating again with respect to $x$ using the product rule:$\frac{d}{d x} \left( \sqrt{1-x^{2}} \frac{d y}{d x} \right) = \frac{d}{d x} (m y)$.$\sqrt{1-x^{2}} \frac{d^{2} y}{d x^{2}} + \frac{1}{2 \sqrt{1-x^{2}}} (-2 x) \frac{d y}{d x} = m \frac{d y}{d x}$.Multiplying throughout by $\sqrt{1-x^{2}}$:$(1-x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} = m \sqrt{1-x^{2}} \frac{d y}{d x}$.Substituting $\sqrt{1-x^{2}} \frac{d y}{d x} = m y$ into the equation:$(1-x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} = m(m y) = m^{2} y$.Rearranging gives $(1-x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - m^{2} y = 0$.Comparing this with the given equation $(1-x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - k y = 0$, we find $k = m^{2}$.

If $y=e^{m \sin ^{-1} x}$, then $(1-x^{2}) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}-k y=0$, where $k$ is equal to

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