Differential equation $x^2\ln^2(x)y''-2x\ln x(\ln x+1)y'+(2\ln^2x+3\ln x+2)y=0$ $x^2\ln^2xy''-2x\ln x(\ln x+1)y'+(2\ln^2x+3\ln x+2)y=0$  (linear)
$A(x)y''+B(x)y'+C(x)y=0$ represents the equation.
What I've tried:


*

*Solving it using substitution $y=UV$, which as far as I know, is used only when the power of x isn't decreasing going from left to right, but I've tried. I won't go into detail. For V you get $V=x\ln x$. When I plug that in you don't get a differential equation with constant coefficients. 
$$x^2\ln^2(x)U''+2U=0$$ Which I can't solve also.


2.Using a $t=\displaystyle\int \sqrt{C(x)/A(x)} \, dx$ you get this. Used $z=\ln x$ to simplify, and $1/2$ as a allowed constant under the square. Result is not helpful.


*Final method, which I know of, is finding a particular solutions, $y_p$ and then using a $y=zy_p$ substitution. I've tried so many combinations. Form of a particular solutions is so that's it's derivative is somewhere in a $B(x)/A(x)$. I'm not 100% sure about this statement. 


Didn't want to write too much, I hope it's understandable.
 A: $$x^2\ln^2(x)y''(x)-2x\ln(x)(\ln(x)+1)y'(x)+y(x)(2\ln^2(x)+3\ln(x)+2)=0\Longleftrightarrow$$

Let $y(x)=xv(x)$, which gives $y'(x)=v(x)+xv'(x)$ and $y''(x)=xv''(x)+2v'(x)$:

$$x^3\ln^2(x)v''(x)-2x^2\ln(x)v'(x)+x(\ln(x)+2)v(x)=0\Longleftrightarrow$$

Let $v(x)=u(x)\ln(x)$, which gives $v'(x)=\frac{u(x)}{x}+\ln(x)u'(x)$ and 
$v''(x)=-\frac{u(x)}{x^2}+\ln(x)u''(x)+\frac{2u'(x)}{x}$:

$$x^3\ln^3(x)u''(x)=0\Longleftrightarrow$$
$$u''(x)=0\Longleftrightarrow$$
$$\int u''(x)\space\text{d}x=\int 0\space\text{d}x\Longleftrightarrow$$
$$u'(x)=\text{C}_1\Longleftrightarrow$$
$$\int u'(x)\space\text{d}x=\int\text{C}_1\space\text{d}x\Longleftrightarrow$$
$$u(x)=\text{C}_1x+\text{C}_2\Longleftrightarrow$$
$$\frac{v(x)}{\ln(x)}=\text{C}_1x+\text{C}_2\Longleftrightarrow$$
$$v(x)=\ln(x)\left(\text{C}_1x+\text{C}_2\right)\Longleftrightarrow$$
$$\frac{y(x)}{x}=\ln(x)\left(\text{C}_1x+\text{C}_2\right)\Longleftrightarrow$$
$$y(x)=x\ln(x)\left(\text{C}_1x+\text{C}_2\right)$$
Where $\text{C}_1\space\space\text{and}\space\text{C}_2$ are arbitrary constants.
