Answer
$y=\dfrac{e^{-x}+c}{e^x+1}$
Work Step by Step
The standard form of the given equation is:
$y'+(\dfrac{e^x}{1+e^x})y=-(\dfrac{e^{-x}}{1+e^x})$ ....(1)
The integrating factor is: $e^{\int (\dfrac{e^x}{1+e^x}) dx}=e^x+1$
In order to determine the general solution, multiply equation (1) with the integrating factor and integrate both sides.
$\int [(e^x+1)y]' =-\int e^{-x} dx$
or, $(e^x+1)y=e^{-x}+c$...(2)
Thus, equation (2) becomes:
$y=\dfrac{e^{-x}+c}{e^x+1}$
