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