Question: > 
> How do you determine singular points of the equation
> 
> 
> $$(x^2 - 4)^2y'' + (x - 2)y' + y = 0?$$
> 
> 
> 

It says $x = -2$ and $x = 2$ are singular points of the equation. Do you just put whatever is in the front of $y''$ and equal it to zero and find the $x$'s and those become the singular points of the equation?

> 
> Please show it in the simplest and easiest way. Thank you
> 
> 
>

Answer: 
You certainly know that in OE: $$y^{''}+P(x)y'+Q(x)y=0$$ the point $x=x\_0$ is said to be a *regular singular point* if bpth of: $$(x-x\_0)P(x),~~(x-x\_0)^2Q(x)$$ are analytic at $x\_0$. This means that these two functions have power series in $(x-x\_0)$ with $R>0$. A singular point that is not regular is called *irregular singular point* of that ODE. Here, you have $$(x^2 - 4)^2y'' + (x - 2)y' + y = 0$$ so $$P(x)=\frac{1}{(x+2)^2(x-2)},~~Q(x)=\frac{1}{(x+2)^2(x-2)^2}$$ Now if $x=-2$ wants to be regular singular point so the factor $(x+2)$ should be of power at most $1$ in $P$ and be of power $2$ at most in $Q$ which is not right. Its power in $P$ is $2\neq 1$, so this point is an irregular one.In fact, the function $P$ has not a [Taylor series](http://en.wikipedia.org/wiki/Taylor_series) at $x=-2$. Do the same way for another point $x=+2$ to see that it is a regular singular point. See the plot of $P(x)$ near $x=2$ in an neighborhood.

![enter image description here](https://i.stack.imgur.com/FbRup.png)
First, you should write the ode in the form

> 
> $$ y''+p(x)y'+q(x)y=0,$$
> 
> 
> 

then you look for the singularities. 

**Example:**

> 
> $$ x^2 y''+ x y'+ y=0 \implies y''+\frac{1}{x}y'+ \frac{1}{x^2}y=0 $$
> 
> 
> 

which shows that the point $x=0$ is a regular singular point.