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(D) Given, $\sqrt{y}=\cos ^{-1} x \Rightarrow y=(\cos ^{-1} x)^{2}$.On differentiating both sides with respect to $x$, we get:$\frac{dy}{dx} = 2(\cos

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$2$

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(D) Given, $\sqrt{y}=\cos ^{-1} x \Rightarrow y=(\cos ^{-1} x)^{2}$.On differentiating both sides with respect to $x$, we get:$\frac{dy}{dx} = 2(\cos ^{-1} x) 	imes \left(\frac{-1}{\sqrt{1-x^{2}}}ight)$.Multiplying both sides by $\sqrt{1-x^{2}}$, we get:$\sqrt{1-x^{2}} \frac{dy}{dx} = -2 \cos ^{-1} x$.Again, differentiating both sides with respect to $x$ using the product rule:$\sqrt{1-x^{2}} \frac{d^{2}y}{dx^{2}} + \frac{dy}{dx} 	imes \left(\frac{-2x}{2\sqrt{1-x^{2}}}ight) = -2 	imes \left(\frac{-1}{\sqrt{1-x^{2}}}ight)$.$\sqrt{1-x^{2}} \frac{d^{2}y}{dx^{2}} - \frac{x}{\sqrt{1-x^{2}}} \frac{dy}{dx} = \frac{2}{\sqrt{1-x^{2}}}$.Multiplying the entire equation by $\sqrt{1-x^{2}}$, we get:$(1-x^{2}) \frac{d^{2}y}{dx^{2}} - x \frac{dy}{dx} = 2$.Comparing this with the given equation $(1-x^{2}) \frac{d^{2}y}{dx^{2}} - x \frac{dy}{dx} = c$, we find $c = 2$.

If $\sqrt{y}=\cos ^{-1} x$, then it satisfies the differential equation $(1-x^{2}) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}=c$, where $c$ is equal to

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