If y=acos(lnx)+bsin(lnx), then x2d2ydx2+xdydx is equal to
A
0.0
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B
y
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C
−y
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D
none of these
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Solution
The correct option is C−y We have, y=acos(lnx)+bsin(lnx) Differentiating both sides w.r.t. x, we get ⇒dydx=−asin(lnx)x+bcos(lnx)x ⇒xdydx=−asin(lnx)+bcos(lnx) ⇒xd2ydx2+dydx=−acos(lnx)x+−bsin(lnx)x ⇒x2d2ydx2+xdydx=−y