$$y''-2y'+y=\frac{e^x}{x}$$
The solution is $$y=xe^xln|x|-xe^x+c_{1}e^x+c_{2}e^x$$
It is said that $y_{1}=e^x$ and $y_{2}=xe^x$ are the solution to the homogeneous equation.
How can I infer it? why for example $y_{1}=ln|x|$ and $y_{2}=xe^x$ are the solutions?
EDIT by S.C.B
The original question asked for solutions to
$$y''-2y'+y=0$$
To those wondering why I wrote the answer below, and why the OP accepted the answer.
$$y''-2y'+y=0 \iff y''-y'-y'+y=0 \iff (y'-y)'-(y'-y)=0$$
Note that $$(y'-y)'-(y'-y)=0 \iff y'-y=ce^{x} \iff (y'-y)e^{-x}=c$$ For some constant $c$. Now note $$(y'-y)e^{-x}=(ye^{-x})'$$ So we have that $$(ye^{-x})'=c \iff ye^{-x}=cx+b$$ For some constant $c,b$. So your answer is incorrect. The answer should be $$y=cxe^{x}+be^{x}$$ This is corroborated through computation.
For the homogeneous solution you can solve the characteristic polynomial as the ODE is lineal. So you replace $y^{i}$ by $\lambda^{i}$ when the first is derivate and the second a power. The idea is to find roots so: $$\lambda^2-2\lambda+1=0 \iff (\lambda-1)^2=0$$ for each root $r_i$ we associate a solution $A_i e^{r_ix}$ where $A_i$ can be determinate by initial conditions (IC). When you have a multiple root, as in the example, the next solution is amplified by $x,x^2,x^3,...$ depending on the multiplicity; in this case is two so the homogeneous solution is $A_1e^x+A_2xe^x$. To solve the particular solution you need the no-homogeneous part of the ODE, but the general procedure is to suppose a solution with the functions involved in this part (can be exponential, polynomial or sinusoidal for this method) with constants, and to insert it in the ODE to find the constants. As you particular solution has a $ln$, I guess it's not the case so should've applied other method like Laplace transform. If you post the rest of the ODE I'll be more helpful.