Identifying and Classifying Singular Points in Differential Equations

In summary, the conversation discusses identifying and determining the regular singular points of a differential equation, specifically x^3(x-1)y'' - 2(x-1)y' + 3xy =0. It is determined that there are no regular singular points, as none of the expressions (x-x_0)p(x) or (x-x_0)^2q(x) are finite as x approaches x_0. The conversation also notes a mistake in considering the limit as x goes to 1, as it should have been (x-1)\left(\frac{2}{x^3}\right).
  • #1
Ted123
446
0

Homework Statement



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

The Attempt at a Solution



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

Are the singular points [tex]x=0,\pm 1\;?[/tex]

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

Considering [tex]x \left( -\frac{2}{x^3} \right),\; (x-1) \left( -\frac{2}{x^3} \right) ,\; (x+1) \left( -\frac{2}{x^3} \right)[/tex] none of which are finite as [tex]x\to x_0[/tex] Does this mean there are no regular singular points?
 
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  • #2
Ted123 said:

Homework Statement



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

The Attempt at a Solution



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

Are the singular points [tex]x=0,\pm 1\;?[/tex]

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

Considering [tex]x \left( -\frac{2}{x^3} \right),\; (x-1) \left( -\frac{2}{x^3} \right) ,\; (x+1) \left( -\frac{2}{x^3} \right)[/tex] none of which are finite as [tex]x\to x_0[/tex] Does this mean there are no regular singular points?
What's wrong with [itex](x- 1)\left(\frac{2}{x^3}\right)[/itex] as x goes to 1?
 
  • #3
HallsofIvy said:
What's wrong with [itex](x- 1)\left(\frac{2}{x^3}\right)[/itex] as x goes to 1?

Doh - I've tended to 0 on every limit! :rolleyes:
 

1. What are singular points?

Singular points are points in a function or equation where the derivative does not exist or is undefined. They are also known as critical points or stationary points.

2. What is the significance of singular points?

Singular points play an important role in determining the behavior and properties of a function or equation. They can indicate the presence of extrema, inflection points, and other important characteristics of the function.

3. How do you identify singular points?

To identify singular points, you can set the derivative of the function or equation equal to zero and solve for the variable(s). The points where the derivative is undefined or does not exist are considered singular points.

4. Can a singular point be a critical point?

Yes, a singular point can also be a critical point. However, not all critical points are singular points. Critical points are points where the derivative is zero, while singular points are points where the derivative is undefined or does not exist.

5. How do singular points affect the graph of a function?

The presence of singular points can significantly alter the shape and behavior of a function's graph. For example, a function with a singular point may have a sharp turn or cusp at that point, causing the graph to be discontinuous or have a vertical tangent.

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