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Question

Form the differential equation corresponding to y=(sin1x)2+Acos1x+B where A, B are arbitrary constants.

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Solution

Let y=(sin1)2+Acos1x+B
so,
dydx=2sin1x11x2+A×11x2+0
dydx=2sin1x1x2A1x2
1x2dydx=2sin1xA
Again differenttating we get-
1x2d2ydx2+dydx121x22x=21x2
1x2d2ydx2x1x2dydx=21x2
(1x2)d2ydx2xdydx=2

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