If the differential equation obtained by elim | vedclass

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(C) Given $y = (\sin^{-1} x)^2 + A \cos^{-1} x + B$ ... $(i)$ Differentiating with respect to $x$: $y' = 2(\sin^{-1} x) \cdot \frac{1}{\sqrt{1 - x^2}}

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$3$

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(C) Given $y = (\sin^{-1} x)^2 + A \cos^{-1} x + B$ ... $(i)$ Differentiating with respect to $x$: $y' = 2(\sin^{-1} x) \cdot \frac{1}{\sqrt{1 - x^2}} - \frac{A}{\sqrt{1 - x^2}}$ $y' \sqrt{1 - x^2} = 2 \sin^{-1} x - A$ ... $(ii)$ Differentiating again with respect to $x$: $y'' \sqrt{1 - x^2} + y' \cdot \frac{-2x}{2\sqrt{1 - x^2}} = \frac{2}{\sqrt{1 - x^2}}$ Multiply throughout by $\sqrt{1 - x^2}$: $y'' (1 - x^2) - x y' = 2$ Comparing this with the given equation $(a - x^2) y'' - x y' = b$,we get $a = 1$ and $b = 2$. Therefore,$\frac{b + a}{b - a} = \frac{2 + 1}{2 - 1} = \frac{3}{1} = 3$.

If the differential equation obtained by eliminating $A$ and $B$ from $y = (\sin^{-1} x)^2 + A \cos^{-1} x + B$ is $(a - x^2) y'' - x y' = b$,then $\frac{b + a}{b - a} =$

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