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$If x d y - y d x + x \cos (\ln x) d x = 0, y (1) = 1, then y (e)$

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Correct option is B

xdy−ydx=−xcos⁡(log⁡x)dxxdy−ydxx2=−cos⁡(log⁡x)dxxdyx=−cos⁡(log⁡x)dxx By integrating yx=−sin⁡(log⁡x)+cy=x(−sin⁡(log⁡x)+c)y(1)=1⇒c=1∴y(e)=e(−sin⁡(1)+1)

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If xdy−ydx+xcos⁡(ln⁡x)dx=0,y(1)=1, then y(e)

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$If x d y - y d x + x \cos (\ln x) d x = 0, y (1) = 1, then y (e)$