Answer
$y'+y=\dfrac{2}{1+4e^{2x}}$ and $\dfrac{\pi}{2}$
Work Step by Step
Apply chain rule of differentiation.
Thus, $y'=-e^{-x} \tan^{-1} (2e^x) + e^{-x} (2e^x)[\dfrac{1}{1+(2e^x)^2}]$
or, $y'=-e^{-x} \tan^{-1} (2e^x) + \dfrac{2}{1+4e^{2x}}=-y+\dfrac{2}{1+4e^{2x}}$
or, $y'+y=\dfrac{2}{1+4e^{2x}}$
This is the required differential equation.
Apply the initial conditions.
Since we have $y=-e^{-x} \tan^{-1} (2e^x) $
Thus, $y(-\ln 2)=-e^{-(-\ln 2)} \tan^{-1} (2e^{(-\ln 2)}) =2 (\dfrac{\pi}{4})=\dfrac{\pi}{2}$
