Answer
$y= [c_1+\ln (1+e^{-x} )] e^x +[c_2-e^{-x} +\ln (1+e^{-x} )] e^{2x}$
Work Step by Step
We are given that $y''-3y'+2y=\dfrac{1}{1+e^{-x}}$
The particular solution:
$y_p=u_1 e^x+u_2 e^{2x}$
$u'_1=\dfrac{-[e^{-x}]}{1+e^{-x}} $ and
$u_1=\int \dfrac{-[e^{-x}]}{1+e^{-x}}=\ln (1+e^{-x})$
and $u'_2=\dfrac{e^{x}}{e^{3x}+e^{2x}}$
and $u_2=\int \dfrac{e^{x}}{e^{3x}+e^{2x}}$
or, $=\ln (\dfrac{e^x+1}{e^x}) -e^{-x}$
or, $=\ln (1+e^{-x})-e^{x} $
We have $y_p=(e^x) \ln (1+e^{-x}) +(e^{2x}) [\ln (1+e^{-x})-e^{x}] $
Hence, $y= [c_1+\ln (1+e^{-x} )] e^x +[c_2-e^{-x} +\ln (1+e^{-x} )] e^{2x}$
