A) \[({{e}^{x}}+1)(y+1)=C{{e}^{y}}\]
B) \[({{e}^{x}}+1)|(y+1)|=C{{e}^{-y}}\]
C) \[({{e}^{x}}+1)(y+1)=\pm C{{e}^{y}}\]
D) None of the above
Correct Answer: A
Solution :
The given differential equation is \[({{e}^{x}}+1)\cdot y\,\,dy=(y+1){{e}^{x}}\,\,dx\] \[\Rightarrow \] \[\frac{y\,\,dy}{(y+1)}=\frac{{{e}^{x}}}{({{e}^{x}}+1)}dx\] \[\Rightarrow \] \[\left( \frac{y+1-1}{y+1} \right)dy=\frac{{{e}^{x}}}{{{e}^{x}}+1}dx\] \[\Rightarrow \] \[dy-\frac{1}{y+1}dy=\frac{{{e}^{x}}}{{{e}^{x}}+1}dx\] On integrating, we get \[\Rightarrow \] \[y-\log |y+1|=\log \,\,({{e}^{x}}+1)+\log k\] \[\Rightarrow \] \[y=\log |(y+1)({{e}^{x}}+1)|k\] \[\Rightarrow \] \[(y+1)({{e}^{x}}+1)={{e}^{y}}C\]
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