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 equation is: $r^2-2r+1=0 ; r=1$
The complementary solution is: $y_c=c_1 e^x+c_2 xe^{x}$
The particular solution is:
$y_p=u_1 e^x+u_2 xe^{x}\implies y'_p=u'_1 e^x+u_1 e^{x}+u'_2 xe^x+u_2 e^{x}(1+x)$
This yields: $u'_1=-u'_2 x$
$u'_1+u'_2(1+x)=\dfrac{1}{1+x^2}$
$u'_2=\dfrac{1}{1+x^2}$
or, $u_2=\int \dfrac{1}{1+x^2}= \tan^{-1} x$
and $u'_1=\dfrac{-x}{1+x^2}$
or, $u_1=\int \dfrac{-x}{1+x^2}=-\dfrac{1}{2} \ln (1+x^2)$
Now, $y=y_c+y_p$
Hence, we have $y=e^x [c_1+c_2 x-\dfrac{1}{2} \ln (1+x^2) +x \tan^{-1} x]$
