Differential Equations Question 262
Question: The solution of the differential equation $ x\frac{dy}{dx}=y(\log y-\log x+1) $ is
[IIT 1986; AIEEE 2005]
Options:
A) $ y=xe^{cx} $
B) $ y+xe^{cx}=0 $
C) $ y+e^{x}=0 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Here $ \frac{dy}{dx}=\frac{y}{x}( \log \frac{y}{x}+1 ) $
…..(i)
It is homogeneous equation
So now put
and $ \frac{dy}{dx}=v+x\frac{dv}{dx} $ , then the equation (i) reduces to $ \frac{dv}{v\log v}=\frac{dx}{x} $
On integrating, we get $ \log (\log v)=\log x+\log c $
Therefore $ \log ( \frac{y}{x} )=cx $
Therefore $ y=xe^{cx} $ .
BETA
