Answer
$y=(\frac{x-e^{-x}+C}{1+e^x})^2$
Work Step by Step
We are given:
$$\frac{dy}{dx}+\frac{2e^x}{1+e^x}y=2\sqrt ye^{-x}$$
Dividing both sides by $y^{\frac{1}{2}}$
$$y^{-\frac{1}{2}}\frac{dy}{dx}+y^{\frac{1}{2}}\frac{2e^x}{1+e^x}y=2e^{-x}$$
Assume that $u=y^{\frac{1}{2}} \rightarrow \frac{du}{dx}=\frac{1}{2}y^{-\frac{1}{2}}\frac{dy}{dx}$
The equation becomes:
$$2 \frac{du}{dx}+u\frac{2e^x}{1+e^x}=2e^{-x}$$
$$ \frac{du}{dx}+u\frac{e^x}{1+e^x}=e^{-x}$$
Integrating factor:
$$I=e^{\int \frac{e^x}{1+e^x}dx}=e^{\int \ln|1+e^x|dx}=1+e^x$$
The equation becomes:
$$\frac{d}{dx}(u(1+e^x))=e^{-x}(1+e^x)$$
Integrating both sides:
$$u(1+e^x)=-e^{-x}+x+C$$
$$y^{\frac{1}{2}}=\frac{x-e^{-x}+C}{1+e^x}$$
Hence general solution is $y=(\frac{x-e^{-x}+C}{1+e^x})^2$
