Differential Equations Question 156
Question: The solution of the equation $ \frac{dy}{dx}=\frac{y}{x}( \log \frac{y}{x}+1 ) $ is
Options:
A) $ \log ( \frac{y}{x} )=cx $
B) $ \frac{y}{x}=\log y+c $
C) $ y=\log y+1 $
D) $ y=xy+c $
Show Answer
Answer:
Correct Answer: A
Solution:
Given $ \frac{dy}{dx}=\frac{y}{x}( \log \frac{y}{x}+1 ) $ . Put $ y=vx $
Therefore $ \frac{dy}{dx}=v+x.\frac{dv}{dx} $
$ \therefore v+x.\frac{dv}{dx}=v(\log v+1) $
$ v+x\frac{dv}{dx}=v\log v+v $
Therefore $ x\frac{dv}{dx}=v\log v $
Therefore $ \frac{dv}{v\log v}=\frac{dx}{x} $
Integrating both sides, $ \int _{{}}^{{}}{\frac{dv}{v\log v}}=\int _{{}}^{{}}{\frac{dx}{x}} $
$ \log \log v=\log x+\log c $
$ \Rightarrow $ $ \log v=xc $
Therefore $ \log (y/x)=xc $ .
BETA
