What is the method used to find the singular points of an ODE?

Question

Suppose you have an ODE, say $$ x^{2} (x+1) y'' + 2y' + xy = 0 $$ How would you find the singular points of this?

I've looked online for an explanation of the method used to do this, but have not found any that I understand very well. Can anyone outline the method used to do this?

Answer 1

Re-writing $$y''+\frac{2}{x^{2}(x+1)}y'+\frac{1}{x(x+1)}y$$ You see that the singular points are $x=0, -1$. See http://www.math.washington.edu/~toro/Courses/99-00/135/regular.pdf For a further discussion

Further Edit Write the most general ODE $$N(x)y''+P(x)y'+ Q(x) = 0$$ in standard form $$y''+p(x)y'+q(x)y=0$$

Let $x_{0}$ be a singular point, and, multiply by $(x-x_{0})^{2}/N(x)$ one can write $$(x-x_{0})^{2}y''+(x-x_{0})u(x)y'+v(x)y=0$$ where \begin{align} u(x) &= \frac{(x-x_{0})P(x)}{N(x)}\\ v(x) &= \frac{(x-x_{0})^{2}Q(x)}{N(x)} \end{align} $x_{0}$ is then regular if $u(x)$ and $v(x)$ have no singularity at $x_{0}$; otherwise it is an irregular point.

I have regurgitated a lot of the detail in the link I have posted, see the theory on page 1.

Answer 2

Cusp points when $ y''$ goes to infinity. Here at $ x=0,-1$there would be curve reversal/sudden direction change etc.