Answer
$y=C_1e^x+ke^x \ln x$
Work Step by Step
Given $xy''+(1-2x)y'+(x-1)y=0$
We first compute the appropriate derivatives:
$y'=e^x(u'+u)\\
y''=e^x(u''+u'+u)$
Substitute:
$x[e^x(u''+u'+u)]+(1-2x)e^x(u'+u)+(x-1)e^x=0$
which simplifies to
$xe^xu''+e^xu'=0$
or equivalently,
$w'x+w=0$
where $w=u'$
Separating the variables yields
$-x=\frac{w}{w'}$
By integrating, we obtain
$\ln w=-\ln x +C\\
wx=k$
Therefore,
$u'=\frac{k}{x}$
Integration by parts gives
$u=k\ln x +C$
The general solution is
$y=C_1e^x+ke^x \ln x$
