Given a solution [tex]y_1(x)=\ln x[/tex], we can attempt to find a solution of the form [tex]y_2(x)=v(x)y_1(x)[/tex]. We have derivatives
[tex]y_2=v\ln x[/tex]
[tex]{y_2}'=v'\ln x+\dfrac vx[/tex]
[tex]{y_2}''=v''\ln x+\dfrac{v'}x+\dfrac{v'x-v}{x^2}=v''\ln x+\dfrac{2v'}x-\dfrac v{x^2}[/tex]
Substituting into the ODE, we get
[tex]v''x\ln x+2v'-\dfrac vx+v'\ln x+\dfrac vx=0[/tex]
[tex]v''x\ln x+(2+\ln x)v'=0[/tex]
Setting [tex]w=v'[/tex], we end up with the linear ODE
[tex]w'x\ln x+(2+\ln x)w=0[/tex]
Multiplying both sides by [tex]\ln x[/tex], we have
[tex]w' x(\ln x)^2+(2\ln x+(\ln x)^2)w=0[/tex]
and noting that
[tex]\dfrac{\mathrm d}{\mathrm dx}\left[x(\ln x)^2\right]=(\ln x)^2+\dfrac{2x\ln x}x=(\ln x)^2+2\ln x[/tex]
we can write the ODE as
[tex]\dfrac{\mathrm d}{\mathrm dx}\left[wx(\ln x)^2\right]=0[/tex]
Integrating both sides with respect to [tex]x[/tex], we get
[tex]wx(\ln x)^2=C_1[/tex]
[tex]w=\dfrac{C_1}{x(\ln x)^2}[/tex]
Now solve for [tex]v[/tex]:
[tex]v'=\dfrac{C_1}{x(\ln x)^2}[/tex]
[tex]v=-\dfrac{C_1}{\ln x}+C_2[/tex]
So you have
[tex]y_2=v\ln x=-C_1+C_2\ln x[/tex]
and given that [tex]y_1=\ln x[/tex], the second term in [tex]y_2[/tex] is already taken into account in the solution set, which means that [tex]y_2=1[/tex], i.e. any constant solution is in the solution set.
