Variation of parameters method for differential equations.

Change the variable $x=e^t$ and then find the general solution for the following differential equation $2x^2y''-6xy'+8y=2x+2x^2lnx$

Here's my take so far. It seems a little suspicious that i can factor out 2 and x first. $x^2y''-3xy'+4y=x(ln(x)+1)$ and if we factor out $x^2$ now we end up with $y''-\frac{3}{x}y'+\frac{4}{x^2}y=\frac{xlnx+1}{x}$ (1) Therefore substituting $x=e^t$ (1) now becomes $y''-\frac{3}{e^t}y'+\frac{4}{e^(2t)}y=\frac{e^tt+1}{e^t}$ We need constant coefficients in order to use the variation of parameters method am i right? Please help.

After change the variable $x=e^t$ we get $$y''-4y'+4y=te^{2t}+e^t$$ Then $$y=\frac{t^3e^{2t}+6e^t}{6}+(C_1+C_2t)e^{2t}=$$ $$=\frac{x^2\ln^3x+6x}{6}+(C_1+C_2\ln x)x^2$$