Answer
$y=\frac{\ln (e^{2x}+1)-x+C}{1+e^{2x}}$
Work Step by Step
We are given:
$$\frac{dy}{dx}+\frac{2e^{2x}}{1+e^{2x}}y=\frac{1}{e^{2x}-1}$$
Integrating factor:
$$I=e^{\int \frac{2e^{2x}}{1+e^{2x}}dx}$$
To solve this, let
$$u=1+e^{2x} \rightarrow du=2e^{2x}$$
then
$$I=e^{\ln u}=e^{\ln (1+e^{2x})}=1+e^{2x}$$
The equation becomes:
$$\frac{d}{dx}[y(1+e^{2x})]=\frac{e^{2x}-1}{e^{2x}+1}=\frac{e^{2x}}{e^{2x+1}}-\frac{1}{e^{2x+1}}$$
Integrating both sides:
$$y(1+e^{2x})=\frac{1}{2}\ln (e^{2x}+1)-[-\frac{1}{2}\ln(e^{2x}+1)+x]+C=\ln (e^{2x}+1)-x+C$$
Hence general solution is $y=\frac{\ln (e^{2x}+1)-x+C}{1+e^{2x}}$
