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  $\,\,y=z^r$
$$e^{3}r(r^{2}-r-2)z^{r-3}+e^{z}rz^{r-1}-z^{r}=0$$
 A: 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!
A: Treating the homogeneous equation as Euler-Cauchy equation, that is, trying $y=x^r$, gives the equation
$$
0=r(r-1)(r-2)+r-1=r^3-3r+3r-1=(r-1)^3,
$$
so that the basis solutions are $y=x$, $x=\ln(x)x$, $y=\ln(x)^2x$.
The right side is in resonance of degree 2, thus the particular solution has the form
$$
y_p(x)=\ln(x)^3(A+B\ln(x))x.
$$
A: Let $y=f[\ln(x)],$ hence $$y'=\frac{f'[\ln(x)]}{x},$$ $$y''=\frac{f''[\ln(x)]-f'[\ln(x)}{x^2},$$ and $$y'''=\frac{xf'''[\ln(x)]-xf''[\ln(x)]-2xf''[\ln(x)]+2xf'[\ln(x)]}{x^4}=\frac{f'''[\ln(x)]-3f''[\ln(x)]+2f'[\ln(x)]}{x^3}.$$ Therefore, $$\left(f'''[\ln(x)]-3f''[\ln(x)]+2f'[\ln(x)]\right)+\left(f'[\ln(x)]\right)-f[\ln(x)]=f'''[\ln(x)]-3f''[\ln(x)]+3f'[\ln(x)]-f[\ln(x)]=x\ln(x).$$ This is equivalent to $$f'''(x)-3f''(x)+3f'(x)-f(x)=x\exp(x).$$ This can be solved for $f,$ since it is linear with constant coefficients, and then solved for $y$ by knowing that $y=f[\ln(x)].$ This is equivalent to what they asked you to do, because $y=f[\ln(x)]$ implies $f=y[\exp(z)],$ and this transformation gets you to the equation I derived if you differentiate with respect to $z$ and not $x.$
