To solve via variation of parameters, you first need two linearly independent solutions. You can find them by solving the homogeneous part.
[tex]y''-3y'+2y=0[/tex]
has characteristic equation
[tex]r^2-3r+2=(r-2)(r-1)=0[/tex]
which has roots [tex]r=1,2[/tex]. So the characteristic solution is
[tex]y_c=C_1e^x+C_2e^{2x}[/tex]
with two linearly independent solutions, [tex]y_1=e^x[/tex] and [tex]y_2=e^{2x}[/tex].
Now when using variation of parameters, you're looking for solutions [tex]y_1u_1[/tex] and [tex]y_2u_2[/tex] where
[tex]u_1=-\displaystyle\int\frac{y_2\frac{e^x}{1+e^x}}{W(y_1,y_2)}\,\mathrm dx[/tex]
[tex]u_2=\displaystyle\int\frac{y_1\frac{e^x}{1+e^x}}{W(y_1,y_2)}\,\mathrm dx[/tex]
with [tex]W(y_1,y_2)[/tex] denoting the Wronskian of the characteristic solutions.
You have
[tex]W(y_1,y_2)=\begin{vmatrix}e^x&e^{2x}\\e^x&2e^{2x}\end{vmatrix}=2e^{3x}-e^{3x}=e^{3x}[/tex]
[tex]u_1=-\displaystyle\int\frac{e^{-x}}{1+e^x}\,\mathrm dx=-x+\ln(1+e^x)[/tex]
[tex]u_2=\displaystyle\int\frac1{1+e^x}\,\mathrm dx=-e^{-x}+\ln(1+e^{-x})[/tex]
and so the particular solution is
[tex]y_p=y_1u_1+y_2u_2[/tex]
[tex]y_p=-xe^x+e^x\ln(1+e^x)-e^x+e^{2x}\ln(1+e^{-x})[/tex]
The general solution is then
[tex]y=y_c+y_p[/tex]
[tex]y=C_1e^x+C_2e^{2x}-xe^x+e^x\ln(1+e^x)+e^{2x}\ln(1+e^{-x})[/tex]
where the [tex]e^x[/tex] term from [tex]y_p[/tex] gets absorbed into the same term from [tex]y_c[/tex].
