$If y = e^{s i n^{- 1} x}, t h e n (1 - x^{2}) y_{2} - x y_{1} =$

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Solution

The correct option is C
y

$y = e^{s i n^{- 1} x}, \Rightarrow y_{1} = \frac{y}{\sqrt{1 - x^{2}}}$

$\Rightarrow y_{1}^{2} (1 - x^{2}) = y^{2}$
$\Rightarrow 2 y_{1} y_{2} (1 - x^{2}) + y_{1}^{2} (- 2 x) = 2 y y_{1}$
$\Rightarrow (1 - x^{2}) y_{2} - x y_{1} = y$

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Similar questions

$If y = a c o s (l o g x) + b s i n (l o g x) t h e n x^{2} y_{2} + x y_{1} =$

$If y = s i n^{- 1} x, t h e n (1 - x^{2}) y_{2} - x y_{1}$

y = sin (a {sin}^{- 1} x)

(1 - x^{2}) y_{2} - x y_{1} + a^{2} y = 0

y = {sin}^{- 1} x

(1 - x^{2}) y_{2} - x y_{1} = 0

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(1 - x^{2}) y_{2} - x y_{1} - a^{2} y = 0

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