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Question

The solution of the differential equation xdydx=y(logylogx+1) is 
  1. y=xecx
  2. y+xecx=0
  3. y+ex=0
  4. None of these


Solution

The correct option is A y=xecx
Here dydx=yx(logyx+1)...(i)
It is homogeneous equation
So now put y =vx ~and dydx=v+xdvdx, then the equation (i) reduces to dvvlog v=dxx
On integrating, we get log (log v)=log x +log c
log(yx)=cxy=xecx

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