Can't work out method of reduction question

Been struggling to work this method of reduction question out. Would appreciate explanation.
Given first solution of homogeneous y(x)=x. Derive second solution of homogeneous equation as Cx^-1
from x^2y" + xy' - y = 4xlnx

actually this is a Cauchy-Euler equation and the solution to the homogeneous are x^k

for some k values...but y_h = w(x) y1 (x) = w x

dy/dx = [ dw/dx ] x + w

d²y / dx² = [ d²w / dx² ] x + 2 [ dw/dx ] ...put into the DE

x² [ x w ' ' + 2 w ' ] + x [ w + x w ' ] - [ xw] = x³ w ' ' + 3 x² w ' = 0--->

x [ w ' ] ' + 3 w ' = 0 { 1st order in w ' }..IF of x³...{ x³ w ' } ' = 0 --->

w ' = a / x³ ---> w = b / x² ---> y_h = b / x as the 2nd homogeneous solution

Note ; if you set p = ln x this will convert the DE into a 2nd order constant coefficient DE in p

{ y = c1 x + c2 / x + [x/4] [ ln x ]² - x / 2 [ ln x ] }

Method of reduction and variation of parameters are actually the same method, for some reason they are given different names and may differ only at the outset. Anyway, the problem is that you asked this question about 20 minutes ago as a variation of parameters question, so I spent a good 15 minutes typing it out for you as a variation of parameters solution, wish you phrased it this way earlier. The details below are more expansive than your problem even asks for, can you follow it? If not, please be specific with questions.

let y ~ x^r, then your homogeneous DE becomes

(r(r-1) + r - 1)x^r = 0

since x^r =/= 0, we have

r(r-1) + r - 1 = 0

r^2 - 1 = 0 --> (r - 1)(r + 1) = 0, so r = 1 and -1, thus

y = Ax + B / x ............[This is your second solution of the form Cx^{-1}]

Now, vary the parameters A and B

y = A(x) x + B(x) / x
y' = A'x + A + B' / x - B / x^2 = A'x + B' / x + A - B / x^2