Solve $x^{2}y''-3xy'+4y=x^{2}lnx$. 
Solve $x^{2}y''-3xy'+4y=x^{2}lnx$.

I wanted to try and follow my professor's notes but I was confused when she wrote $\eta=lnx$ and $\frac{dy}{dx}=\frac{1}{x}\frac{dy}{d\eta}$. Something about changing the equation to have constant coefficients.
Is there a step-by-step way of solving such a differential equation?

Let $D \equiv \frac{d}{dx}, D^2 \equiv \frac{d^2}{dx^2}$ and $\theta \equiv\frac{d}{d\eta}, \theta^2 \equiv\frac{d^2}{d\eta^2}$.
Once you substitute $\ln x = \eta \implies xD = \theta, x^2D^2 = \theta (\theta -1)$
In general 

$x^nD^n = \theta(\theta-1)\cdots(\theta -n +1)$

This works only in case you have differential equations with coefficients of the form $kx^nD^n$, where $k$ is a constant.
So the equation reduces to,
$\theta(\theta-1)y - 3\theta y + 4y = \eta e^{2\eta} \implies (\theta^2-4\theta+4)y =\eta e^{2\eta} \implies (\theta -2)^2y = 2 \eta e^{2\eta}$
You'll get a complementary solution:$C_y = (A\eta+b)e^\eta = (A\ln x + b)x$
and a particular solution, $$P_y = \frac{1}{(\theta-2)^2}\eta e^{2\eta} = e^{2\eta}\frac{1}{(\theta+2-2)^2}\eta = e^{2\eta}\frac{1}{\theta^2}\eta = \frac{1}{6}e^{2\eta}\eta^3= \frac{1}{6}x^2(\ln x)^3$$
So, $$ y =  (A\ln x + b)x + \frac{1}{6}x^2(\ln x)^3$$
