Differential Equations Question 169
Question: Solution of the equation $ (x+\log y)dy+ydx=0 $ is
Options:
A) $ xy+y\log y=c $
B) $ xy+y\log y-y=c $
C) $ xy+\log y-x=c $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ xdy+ydx+\log ydy=0 $
Therefore $ xdy+ydx=-\log ydy $
$ y\frac{dx}{dy}+x=-\log y $
Therefore $ \frac{dx}{dy}+\frac{x}{y}=-\frac{\log y}{y} $
I.F. = $ {e^{\int{\frac{1}{y}dy}}}=y $
Hence solution is $ x.y=-\int{y.\frac{\log ydy}{y}+c} $
Therefore $ xy=-(y\log y-y)+c $
Therefore $ xy+(y\log y-y)=c $ .
BETA
