Differential Equations Question 82
Question: The solution of $ (1+xy)ydx+(1-xy)xdy=0 $ is
Options:
A) $ \frac{x}{y}+\frac{1}{xy}=k $
B) $ \log ( \frac{x}{y} )=\frac{1}{xy}+k $
C) $ \frac{x}{y}+\frac{1}{xy}=k $
D) $ \log ( \frac{x}{y} )=xy+k $
Show Answer
Answer:
Correct Answer: B
Solution:
$ ydx+xdy+xy^{2}dx-x^{2}ydy=0 $
$ \frac{ydx+xdy}{x^{2}y^{2}}+\frac{dx}{x}-\frac{dy}{y}=0 $ .
On integrating, we get $ -\frac{1}{xy}+\log x-\log y=k $
Therefore $ \log \frac{x}{y}=\frac{1}{xy}+k $ .
BETA
