Answer
$y=e^x [c_1+c_2 x-\dfrac{1}{2} \ln (1+x^2) +x \tan^{-1} x]$
Work Step by Step
The auxiliary solution .$r^2-2r+1=0 \implies r=1$
Thus, we have $y_c=c_1 e^x+c_2 xe^{x}$
The particular solution is:
$y_p=u_1 e^x+u_2 xe^{x}; y'_p=u'_1 e^x+u_1 e^{x}+u'_2 xe^x+u_2 e^{x}(1+x)$
This gives: $u'_1=--u'_2 x$
and $y''-3y'+2y=\dfrac{1}{1+e^{-x}}$
$u'_1+u'_2(1+x)=\dfrac{1}{1+x^2}$
$u'_2=\dfrac{1}{1+x^2}$
or, $u_2=\arctan x$
and $u'_1=-[\dfrac{x}{1+x^2}]$
or, $u_1=-\dfrac{1}{2} \ln (1+x^2)$
Hence, $y=y_c+y_p=e^x [c_1+c_2 x-\dfrac{1}{2} \ln (1+x^2) +x \tan^{-1} x]$
