[Math] Find the singular points of the differential equation $x^3(x – 1)y” – 2(x – 1)y’ + 3xy = 0$.

ordinary differential equations

Consider the second order linear homogeneous equation
$$a_0(x)y'' + a_1(x)y'+ a_2(x)y = 0, x \in I \tag{1}$$
Suppose that $a_0$, $a_1$ and $a_2$ are analytic at $x_0 \in I$. If $a_0(x_0) = 0$, then $x_0$ is a singular point for $(1)$.
Definition: A point $x_0 \in I$ is a regular singular point for $(1)$ if $(1)$ can be written as
$$b_0(x)(x − x_0)^2y''+ b_1(x)(x − x_0)y'+b_2(x)y = 0, \tag{2}$$
where $b_0(x_0) \neq 0$ and $b_0$, $b_1$, $b_2$ are analytic at $x_0$.

The question is: Find the singular points of the differential equation $x^3(x – 1)y'' – 2(x – 1)y' + 3xy = 0$ and state whether they are regular singular points or irregular singular points.

I think, $x = 0$, irregular singular point, $x = 1$, regular
singular point. But, How can I prove this? Please proper guide me.

Best Answer

Consider the general homogeneous second order linear differential equation $$u''+P(x)u'+Q(x)u=0$$ where $z \in D \subseteq \mathbb{C}$.

The point $x_0 \in D$ is said to be an ordinary point of the above the 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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Here the given equation is $$x^3(x - 1)y'' - 2(x - 1)y' + 3xy = 0$$ $$\implies y''-\dfrac{2}{x^3}y'+\dfrac{3}{x^2(x - 1)}y=0$$ Here $~P(x)=-\dfrac{2}{x^3}~$and $~Q(x)=\dfrac{3}{x^2(x - 1)}~$.

Clearly $~x=0,~1~$ are singular points.

Again for the singular point $~x=0~$, $$(x-0)P(x)=-\dfrac{2}{x^2}\qquad \text{and}\qquad (x-0)^2Q(x)=\dfrac{3}{x-1}$$ Clearly $~(x-x_0)^2 Q(x)~$ is analytic at $~x_0=0~$ but $~(x-x_0)P(x)~$ is not.
So $~x=0~$ is irregular singular point.

For the singular point $~x=1~$, $$(x-1)P(x)=-\dfrac{2(x-1)}{x^3}\qquad \text{and}\qquad (x-1)^2Q(x)=\dfrac{3(x - 1)}{x^2}$$ Clearly both $~(x-x_0)P(x)~$ and $~(x-x_0)^2 Q(x)~$ are analytic at $~x_0=1~$.
So $~x=1~$ is regular singular point.

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