Question

# The solution of the differential equation xdydx=y(logy−logx+1) is y=xecxy+xecx=0y+ex=0None of these

Solution

## The correct option is A y=xecxHere 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)=cx⇒y=xecx

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