<s>[INST] 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  $\,\,y=z^r$
$$e^{3}r(r^{2}-r-2)z^{r-3}+e^{z}rz^{r-1}-z^{r}=0$$
 [/INST] 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!
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