If y = (cos^{-1} x)^2,then (1-x^2) y_2 + p x | vedclass

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(C) Given $y = (\cos^{-1} x)^2$. Differentiating with respect to $x$,we get $y_1 = 2(\cos^{-1} x) \cdot \left(-\frac{1}{\sqrt{1-x^2}}\right)$. So,$\sq

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-$3$

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(C) Given $y = (\cos^{-1} x)^2$. Differentiating with respect to $x$,we get $y_1 = 2(\cos^{-1} x) \cdot \left(-\frac{1}{\sqrt{1-x^2}}ight)$. So,$\sqrt{1-x^2} y_1 = -2 \cos^{-1} x$. Squaring both sides,we get $(1-x^2) y_1^2 = 4 (\cos^{-1} x)^2 = 4y$. Differentiating again with respect to $x$,we get $(1-x^2) \cdot 2y_1 y_2 + y_1^2 (-2x) = 4y_1$. Dividing by $2y_1$ (assuming $y_1 
eq 0$),we get $(1-x^2) y_2 - x y_1 = 2$. Thus,$(1-x^2) y_2 - x y_1 - 2 = 0$. Comparing this with $(1-x^2) y_2 + p x y_1 + q = 0$,we get $p = -1$ and $q = -2$. Therefore,$p+q = -1 + (-2) = -3$.

If $y = (\cos^{-1} x)^2$,then $(1-x^2) y_2 + p x y_1 + q = 0$. Find the value of $p+q$.

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