Question
If xdydx=y(log y−log x+1), then the solution of the equation is
Answer: Option C
:
C
dydx=yx(logyx+1)
Put y=vx⇒dydx=v+xdvdx
∴v+xdvdx=vlogv+v⇒1vlogvdv=1xdx⇒∫1vlogvdv=∫1xdx⇒log(logv)=logx+logc
⇒logyx=cx
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:
C
dydx=yx(logyx+1)
Put y=vx⇒dydx=v+xdvdx
∴v+xdvdx=vlogv+v⇒1vlogvdv=1xdx⇒∫1vlogvdv=∫1xdx⇒log(logv)=logx+logc
⇒logyx=cx
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