The solution of (y(1+x−1)+siny)dx+(x+logx+cosy)dy=0
A
(1+Y−1siny)+x−1logx=C
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B
(y+siny)+xylogx=C
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C
xy+ylogx+xsiny=C
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D
Noneofthese
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Solution
The correct option is Cxy+ylogx+xsiny=C The given equation can be written as (y(1+x−1)+siny)dx+(x+logx+cosy)dy=0 ⇒d(y(x+logx))+d(xsiny)=0⇒y(x+logx)+xsiny=C