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Question

The solution of the differential equation xdydx=y(logylogx+1) is

A
y=xecx
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B
y+xecx=0
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C
y+ex=0
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D
None of these
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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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