Differential Equations Question 3
Question: Solution of the equation $ (e^{x}+1)ydy=(y+1)e^{x}dx $ is
[AISSE 1988]
Options:
A) $ c(y+1)(e^{x}+1)+e^{y}=0 $
B) $ c(y+1)(e^{x}-1)+e^{y}=0 $
C) $ c(y+1)(e^{x}-1)-e^{y}=0 $
D) $ c(y+1)(e^{x}+1)=e^{y} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ (e^{x}+1)ydy=(y+1)e^{x}dx $
Therefore $ ( \frac{y}{y+1} )dy=( \frac{e^{x}}{e^{x}+1} )dx $
Therefore $ [ 1-\frac{1}{y+1} ]dy=( \frac{e^{x}}{e^{x}+1} )dx $
Therefore $ \int _{{}}^{{}}{{ 1-\frac{1}{y+1} }}dy=\int _{{}}^{{}}{\frac{e^{x}}{e^{x}+1}}dx $
Therefore $ y=\log (y+1)+\log (e^{x}+1)+\log c $ or $ e^{y}=c(y+1)(e^{x}+1) $ .
BETA
