If logey=3sin−1x, then (1−x2)y′′−xy′ at x=21 is equal to
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Solution
logey=3sin−1xy=e3sin−1xdxdy=e3sin−1x⋅1−x231−x2dxdy=3y Again differentiate 1−x2⋅y′′−21−x22xy′=3y′(1−x)2y′′−xy′=3y′(1−x2) So value of 3y′(1−x2) at x=213⋅1−x23esin−1x(1−x2)=9e36π=9e2π