Solve $ x^{3}{y}'''+x{y}'-y = x\ln(x) \\ $

Solve $$ x^{3}{y}'''+x{y}'-y = x\ln(x) \\ $$ using shift $x=e^{z}$ and differential operator $Dz=\frac{d}{dz}$

What does $Dz = d / dz$ mean?

I did this but I don't know how to continue. Please help. $$ (e^{z})^{3}{y}'''+e^{z}{y}'-y = e^{z}\ln(e^{z}) \\\\(e^{3z}){y}'''+e^{z}{y}'-y = e^{z}{z}$$ And I tried $\,\,y=z^r$ $$e^{3}r(r^{2}-r-2)z^{r-3}+e^{z}rz^{r-1}-z^{r}=0$$

Let $Y(z) = y(x) = y(e^z)$. Then Chain Rule and Product Rule give \begin{align*} \frac{dY}{dz} & = \frac{dy}{dx}\frac{dx}{dz} = \frac{dy}{dx}e^z = x\frac{dy}{dx} \\ \frac{d^2Y}{dz^2} & = \frac{d}{dz}\left(\frac{dy}{dx}e^z\right) = \frac{dy}{dx}e^z + \frac{d^2y}{dx^2}e^ze^z = \frac{dy}{dx}e^z + \frac{d^2y}{dx^2}e^{2z} = \frac{dY}{dz} + x^2\frac{d^2y}{dx^2} \\ \frac{d^3Y}{dz^3} & = \frac{d}{dz}\left(\frac{dY}{dz} + \frac{d^2y}{dx^2}e^{2z}\right) = \frac{d^2Y}{dz^2} + 2\frac{d^2y}{dx^2}e^{2z} + \frac{d^3y}{dx^3}e^{3z} \\ & = \frac{d^2Y}{dz^2} + 2x^2\frac{d^2y}{dx^2} + x^3\frac{d^3y}{dx^3} = \frac{d^2Y}{dz^2} + 2\left(\frac{d^2Y}{dz^2} - \frac{dY}{dz}\right) + x^3\frac{d^3y}{dx^3} \\ & = 3\frac{d^2Y}{dz^2} - 2\frac{dY}{dz} + x^3\frac{d^3y}{dx^3}. \end{align*} Thus, the original DE transforms to \begin{align*} x^3y''' + xy' - y & = x\ln x \\ Y''' - 3Y'' + 2Y' + Y' - Y & = e^zz \\ Y''' - 3Y'' + 3Y' - Y & = ze^z. \end{align*} **Thanks to bjorn93 for pointing out my mistake!