Find the singular points of the differential equation $x^3(x - 1)y'' - 2(x - 1)y' + 3xy = 0$.

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.

• Hi Harry, I've formatted your question with MathJax. I really encourage you to learn how to do this for yourself. I'd also like to remind you of your comment under your last question. – Theo Bendit Sep 23 '19 at 5:18
• @ Theo. Thanks again.I am trying latex but not done properly. I am on my way to working with MathJax. – user679406 Sep 23 '19 at 5:29
• That's good! Feel free to edit your question to see how I've formatted it. That might help you figure it out a bit faster. – Theo Bendit Sep 23 '19 at 5:31

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 $$~x=0~$$, $$(x-0)P(x)=-\dfrac{2}{x^2}\qquad \text{and}\qquad (x-0)^2P(x)=-\dfrac{2}{x}$$ Clearly both $$~(x-x_0)P(x)~$$ and $$~(x-x_0)^2 Q(x)~$$ are not analytic at $$~x_0=0~$$. So $$~x=0~$$ is irregular singular point.

For the singular point $$~x=1~$$, $$(x-1)P(x)=\dfrac{3}{x^2}\qquad \text{and}\qquad (x-0)^2P(x)=\dfrac{3(x - 1)}{x^2}$$ 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.

• @ nmsanta,I think this answer is correct and I am going with it. Am I right? – user679406 Sep 23 '19 at 5:32
• Have you faced any trouble in understanding my work ? I just doing things according to the definition. @Harry Richie – nmasanta Sep 23 '19 at 5:34
• No trouble. Clear like water..@ nmasanta – user679406 Sep 23 '19 at 5:37
• Only then my work will be well worth. – nmasanta Sep 23 '19 at 5:42
• Hello Mr. Down-voter, Would you like to explain the reason to give the down-vote in this answer ? – nmasanta Sep 24 '19 at 8:17