Answer
See below
Work Step by Step
We are given $y''-2y'+y=e^x\ln x$
In this case the substitution y(x)=erx yields the indicial equation
$r^2-2y+1=0$
It follows that two linearly independent solutions to the given differential equation are
$y_1(x)=e^{x}\\
y2(x)=xe^{x}$
so that the general solution is
$y(x)=u_1e^{x}+u_2xe^{x}$
We first compute the appropriate derivatives
$y′=u_1'(x)e^{-x}+u_2'(x)xe^{-x}\\
y''=u_1'(x)(e^{x})'+u_2'(x)(xe^{x})'$
Substituting these expressions into the given equation, we find that $u$ must satisfy
$u_2'(x)e^x=e^x\ln x\\
u_2'(x)=\ln x\\
u_2(x)=x\ln x-x$
and $u_1'(x)=-x \ln x\\
u_1(x)=-\frac{x^2}{2}\ln x+\frac{x^2}{4}$
Hence, the general solution is:
$y(x)=C_1e^x+C_2xe^x+\frac{x^2}{2}\ln x-\frac{3x^2}{4}$
