Is the process of determining whether a singular is regular or not same for the homogenous and non-homogenous ODEs? I have this second-order non-homogeneous ODE:
$$0.5x^5A''(x)+\left(2x^4+x^3\right)A'(x)+\left(x^3+x^2+x-1\right)A(x)=-1$$
I have determined that $x=0$ is a singular point and divided both sides by $0.5x^5$. What I am now trying to do is determine whether it is regular or not.
I am aware of the process with homogenous ODEs, about coefficients being divisible by $(x-x_0)$. But I am wondering, does the same property retain with non-homogeneous ODEs? I have sought the answer on the internet but all examples are of homogeneous DEs.

$0.5x^5A''(x)+(2x^4+x^3)A'(x)+(x^3+x^2+x-1)A(x)=-1$
$A''(x)+\left(\dfrac{4}{x}+\dfrac{2}{x^2}\right)A'(x)+\left(\dfrac{2}{x^2}+\dfrac{2}{x^3}+\dfrac{2}{x^4}-\dfrac{2}{x^5}\right)A(x)=-\dfrac{2}{x^5}$
$\lim\limits_{x\to 0}\left(x\left(\dfrac{4}{x}+\dfrac{2}{x^2}\right)\right)=\lim\limits_{x\to 0}\left(4+\dfrac{2}{x}\right)=\infty$
$\lim\limits_{x\to 0}\left(x^2\left(\dfrac{2}{x^2}+\dfrac{2}{x^3}+\dfrac{2}{x^4}-\dfrac{2}{x^5}\right)\right)=\lim\limits_{x\to 0}\left(2+\dfrac{2}{x}+\dfrac{2}{x^2}-\dfrac{2}{x^3}\right)=\infty$
$\therefore$ the finite singular point $x=0$ is irregular.
That's not the end. We should also check the singularities at infinity. Because these also act as one of the key points of distinguishing the ODE type.
Let $t=\dfrac{1}{x}$ ,
Then $\dfrac{dA}{dx}=\dfrac{dA}{dt}\dfrac{dt}{dx}=-\dfrac{1}{x^2}\dfrac{dA}{dt}=-t^2\dfrac{dA}{dt}$
$\dfrac{d^2A}{dx^2}=\dfrac{d}{dx}\left(-t^2\dfrac{dA}{dt}\right)=\dfrac{d}{dt}\left(-t^2\dfrac{dA}{dt}\right)\dfrac{dt}{dx}=\left(-t^2\dfrac{d^2A}{dt^2}-2t\dfrac{dA}{dt}\right)\left(-\dfrac{1}{x^2}\right)=\left(-t^2\dfrac{d^2A}{dt^2}-2t\dfrac{dA}{dt}\right)(-t^2)=t^4\dfrac{d^2A}{dt^2}+2t^3\dfrac{dA}{dt}$
$\therefore t^4\dfrac{d^2A}{dt^2}+2t^3\dfrac{dA}{dt}-(4t+2t^2)t^2\dfrac{dA}{dt}+(2t^2+2t^3+2t^4-2t^5)A=-2t^5$
$t^4\dfrac{d^2A}{dt^2}-(2t^2+2t)t^2\dfrac{dA}{dt}-(2t^5-2t^4-2t^3-2t^2)A=-2t^5$
$\dfrac{d^2A}{dt^2}-\left(2+\dfrac{2}{t}\right)\dfrac{dA}{dt}-\left(2t-2-\dfrac{2}{t}-\dfrac{2}{t^2}\right)A=-2t$
$\lim\limits_{t\to 0}\left(-t\left(2+\dfrac{2}{t}\right)\right)=\lim\limits_{t\to 0}(-2t-2)=-2$
$\lim\limits_{t\to 0}\left(-t^2\left(2t-2-\dfrac{2}{t}-\dfrac{2}{t^2}\right)\right)=\lim\limits_{t\to 0}(-2t^3+2t^2+2t+2)=2$
$\therefore$ the singularities at $x=\pm\infty$ are regular.
