$x^3 y^{′′′}−x^2 y^{′′}+2xy^′−2y=x^3$
Consider the substitution due to Euler:
$e^z = x$
You'll get
$y'=\dfrac{dy}{dx} =\dfrac{dy}{dz}\dfrac{dz}{dx}=e^{-z}\dfrac{dy}{dz}$
So $xy' = \dfrac{dy}{dz} $
Similarily you'll get
$x^2 y'' = \dfrac{d^2y}{dz^2}-\dfrac{dy}{dz}$ or $\mathcal{D}(\mathcal{D}-1)y$
And finally (as you might have guessed by now)
$x^3y''' =\mathcal{D}(\mathcal{D}-1)(\mathcal{D}-2)y$
Note that $\mathcal{D} = \dfrac{d}{dz}$ is our new operator.
Plugging this in gives
$\mathcal{D}(\mathcal{D}-1)(\mathcal{D}-2)y−\mathcal{D}(\mathcal{D}-1)y+2\mathcal{D}y−2y=e^{3z}$
Now factor $y$ and expand, then factor to get
$\eqalign{ & \left( {{\mathcal{D}^3} - 4{\mathcal{D}^2} + 5\mathcal{D} - 2} \right)y = {e^{3z}} \cr & {\left( {\mathcal{D} - 1} \right)^2}\left( {\mathcal{D} - 2} \right)y = {e^{3z}} \cr} $
So now solve the homogeneous equation:
${\left( {\mathcal{D} - 1} \right)^2}\left( {\mathcal{D} - 2} \right)y = 0$
This gives the complementary solution in terms of $z$ or $\log x$
${y_c} = {c_1}z{e^z} + {c_2}{e^z} + {c_3}{e^{2z}}$
${y_c} = {c_1}x\log x + {c_2}x + {c_3}{x^2}$
You can easily get the particular with the original equation by assuming a solution $y=Ax^3$ and finding $A$.
ADD: In general, the substitution $x = e^z$ will transform
$\sum\limits_{k = 0}^n {{a_k}{x^k}{D^k}} y = F$
into a linear equation of constant coefficients.