If $m \sin^{-1} x = \log_e y$, then $(1 - x^2)y'' - x y' = $ - DASH

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Question

If $m \sin^{-1} x = \log_e y$, then $(1 - x^2)y'' - x y' = $

$m^2 y$

-$m^2 y$

$2y$

-$2y$

Solution

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