The solution of the differential equation xdydx=y(logy−logx+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
Noneofthese
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Solution
The correct option is Ay=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 dvvlogv=dxx On integrating, we get log (log v)=log x +log c ⇒log(yx)=cx⇒y=xecx