Differential Equations Question 313
Question: The solution of the differential equation $ 3e^{x}\tan ydx+(1-e^{x}){{\sec }^{2}}ydy=0 $ is
[MP PET 1993; AISSE 1985]
Options:
A) $ \tan y=c{{(1-e^{x})}^{3}} $
B) $ {{(1-e^{x})}^{3}}\tan y=c $
C) $ \tan y=c(1-e^{x}) $
D) $ (1-e^{x})\tan y=c $
Show Answer
Answer:
Correct Answer: A
Solution:
It can be written in the form of $ \frac{{{\sec }^{2}}y}{\tan y}dy=-3\frac{e^{x}}{1-e^{x}}dx $
$ \int{\frac{{{\sec }^{2}}y}{\tan y}}dy=-3\int{\frac{e^{x}}{1-e^{x}}dx} $
Therefore $ \log (\tan y)=3\log (1-e^{x})+\log c $
Therefore $ \tan y=c{{(1-e^{x})}^{3}} $ .
BETA
