## (Regular) Singular Points

1. The problem statement, all variables and given/known data

Locate the singular points of $$x^3(x-1)y'' - 2(x-1)y' + 3xy =0$$ and decide which, if any, are regular.

3. The attempt at a solution

In standard form the DE is $$y'' - \frac{2}{x^3} y' + \frac{3}{x^2(x-1)} y = 0.$$

Are the singular points $$x=0,\pm 1\;?$$

Regular singular points $$x_0$$ of $$y'' + p(x)y' + q(x)y =0$$ satisfy $$(x-x_0)p(x) ,\; (x-x_0)^2q(x)$$ both finite as $$x \to x_0.$$

Considering $$x \left( -\frac{2}{x^3} \right),\; (x-1) \left( -\frac{2}{x^3} \right) ,\; (x+1) \left( -\frac{2}{x^3} \right)$$ none of which are finite as $$x\to x_0$$ Does this mean there are no regular singular points?
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 Quote by Ted123 1. The problem statement, all variables and given/known data Locate the singular points of $$x^3(x-1)y'' - 2(x-1)y' + 3xy =0$$ and decide which, if any, are regular. 3. The attempt at a solution In standard form the DE is $$y'' - \frac{2}{x^3} y' + \frac{3}{x^2(x-1)} y = 0.$$ Are the singular points $$x=0,\pm 1\;?$$ Regular singular points $$x_0$$ of $$y'' + p(x)y' + q(x)y =0$$ satisfy $$(x-x_0)p(x) ,\; (x-x_0)^2q(x)$$ both finite as $$x \to x_0.$$ Considering $$x \left( -\frac{2}{x^3} \right),\; (x-1) \left( -\frac{2}{x^3} \right) ,\; (x+1) \left( -\frac{2}{x^3} \right)$$ none of which are finite as $$x\to x_0$$ Does this mean there are no regular singular points?
What's wrong with $(x- 1)\left(\frac{2}{x^3}\right)$ as x goes to 1?
 Quote by HallsofIvy What's wrong with $(x- 1)\left(\frac{2}{x^3}\right)$ as x goes to 1?