Differential Equations Question 46
Question: The solution of the differential equation $ 3{e^{x\tan }}ydx+(1-e^{x}){{\sec }^{2}}ydy=0 $ is
Options:
A) $ e^{x}\tan y=C $
B) $ Ce^{x}={{(1-\tan y)}^{3}} $
C) $ C\tan y={{(1-e^{x})}^{2}} $
D) $ \tan y=C{{(1-e^{x})}^{3}} $
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Answer:
Correct Answer: D
Solution:
[d] $ 3e^{x}\tan ydx+(1-e^{x}){{\sec }^{2}}ydy=0 $
$ \Rightarrow \frac{3e^{x}}{1-e^{x}}dx+\frac{{{\sec }^{2}}y}{\tan y}dy=0 $ Integrating we get $ \int{\frac{3e^{x}}{1-e^{x}}dx+\int{\frac{{{\sec }^{2}}y}{\tan y}dy=D}} $ ,
$ \Rightarrow -3\ell n(1-e^{x})+\ell ntany=D $
$ \Rightarrow -\ell n{{(1-e^{x})}^{3}}+\ell ntany=D $
$ \Rightarrow \ell n\frac{\tan y}{{{(1-e^{x})}^{3}}}=\ell nC\Rightarrow \tan y=C{{(1-e^{x})}^{3}} $
BETA
