Chapter 9: First-Order Differential Equations - Practice Exercises - Page 557: 13

$y=\dfrac{e^{-x}+c}{e^x+1}$

Here, we have $y'+(\dfrac{e^x}{1+e^x})y=-(\dfrac{e^{-x}}{1+e^x})$ The integrating factor is: $I=e^{\int (\dfrac{e^x}{1+e^x}) dx}=e^x+1$ Now, $(e^x+1) (y'+(\dfrac{e^x}{1+e^x})y)=(e^x+1) [-(\dfrac{e^{-x}}{1+e^x})]$ This implies that $\int [(e^x+1)y]' =-\int e^{-x} dx$ $(e^x+1)y=(e^{-x})+c \implies y=\dfrac{e^{-x}+c}{e^x+1}$