University Calculus: Early Transcendentals (3rd Edition)

$y'+y=\dfrac{2}{1+4e^{2x}}$ and $\dfrac{\pi}{2}$
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, $y(-\ln 2)=-e^{-(-\ln 2)} \tan^{-1} (2e^{(-\ln 2)}) =2 (\dfrac{\pi}{4})=\dfrac{\pi}{2}$