Answer
See below
Work Step by Step
Given: $y''-4y=\frac{8}{e^{2x}+1}$
In this case, the auxiliary polynomial of the associated homogeneous equation is
$P(r) = r^2-4 =0\\
\rightarrow r=\pm 2$
so that three linearly independent solutions are:
$y_1(x)=e^{2x}\\
y_2(x)=xe^{-2x}$
A particular solution is:
$y_p(x)=u_1(x)e^{2x}+u_2(x)xe^{-2x}$
Obtain $u_1'(x)e^{2x}+u_2'(x)xe^{-2x}=0\\
u_1'(x)e^{2x}+u_2'(x)(xe^{-2x})'= \frac{8}{e^{2x}+1}$
Consequently $u_1'=\frac{2}{e^{2x}(e^{2x}+1)}\\
u_2'=\frac{-2e^{2x}}{e^{2x}(e^{2x}+1)}$
By integrating, we obtain
$u_1=\int\frac{2}{e^{2x}(e^{2x}+1)}dx=\ln(\frac{e^{2x}+1}{e^{2x}})-e^{-2x}+C_1\\
u_2=\frac{-2e^{2x}}{e^{2x}(e^{2x}+1)} dx=\ln(\frac{e^{2x}+1}{e^{2x}})+C_2 $
where $C_1,C_2$ are integration constants.
The general solution to the given differential equation is therefore:
$y(x)=e^{2x}(\ln\frac{e^{2x}+1}{e^{2x}}-e^{-2x}+C_1)+e^{-2x}(\ln \frac{e^{2x}+1}{e^{2x}}+C_2)$
