finding ordinary and regular singular points I am a little confused by how I find ordinary and regular singular points of a differential equation. Given $$4xu'' + 2u' + \lambda u = 0$$
How would I go about finding whether $x=0$ is a singular or ordinary point. I understand the intuition behind it but am unsure how I can apply it particularly in showing a function is analytic at said point.
 A: 
Consider the general homogeneous second order linear differential equation $$u''+P(x)u'+Q(x)u=0$$
  where $x \in D \subseteq \mathbb{C}$. 
The point $x_0 \in D$ is said to be an ordinary point of the above given differential equation if $P(x)$ and $Q(x)$ are analytic at $x_0$. 
If either $P(x)$ or $Q(x)$ fails to be analytic at $x_0$, the point $x_0$ is called a singular point of the given differential equation.
A singular point $x_0$ of the given differential equation is said to be regular singular point if the function $(x-x_0)P(x)$ and $(x-x_0)^2 Q(x)$ are analytic at $x_0$ and irregular otherwise.

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$4xu'' + 2u' + \lambda u = 0\implies u'' + \frac{u'}{2x} + \frac{\lambda}{4x} u = 0$
Comparing the equation with $u''+P(x)u'+Q(x)u=0$ we have $P(x)=\frac{1}{2x} $ and $Q(x)=\frac{\lambda}{4x}$
Clearly $x=0$ is the only singular point of the given differential equation. 
Since $$(x-x_0)P(x)=x \frac{1}{2x}=\frac{1}{2}$$ and 
$$(x-x_0)^2 Q(x)=x^2 \frac{\lambda}{4x}=\frac{\lambda x}{4}$$
so both are analytic at $x=0$, and hence $x=0$ is regular singular point.
