Series solutions near a singular point, 2nd order linear

In summary, the conversation discusses solving 2nd order linear differential equations near a singular point. The indicial equation and its roots are found, leading to the need to find the first three nonzero terms in the series. The solution corresponding to the larger root is needed, and a recurrence relation is used to find the value of a_n. However, the presence of lnx complicates the process and the conversation ends with a question on how to handle it.
  • #1
Gridvvk
56
1
So, I'm trying to solve 2nd order linear differential equations (series solutions near a singular point).

(lnx)y" + 0.5y' + y = 0 around the regular singular point x = 1

I got the indicial equation,

r(r-0.5) = 0,

which leads to the roots...

r1 = 0.5, r2 = 0

The problem only asks us to find the first three nonzero terms in the series y1 = Ʃa_n * (x-1)^r+n from n = 0 to infinity. And we only need to find one solution, corresponding to the larger root.

So I took the first and second derivatives of the y1 they gave and plugged it into the differential equation. Now at this point I usually factor out all the x terms. And since the left side has to equal zero for all x, I can divide by that x term to get a recurrence relation (that involves a_n terms). From the relation I can figure out what a_n is. However, in this case, I can't factor out all the x terms because there's a lnx.

Anyone know how to get rid of the lnx?

Thanks!

EDIT: MOVED TO HOMEWORK AND CLASSWORK SECTION
 
Last edited:
Physics news on Phys.org
  • #2
Replace ln(x) with a power series about x= 1.
 
  • #3
HallsofIvy said:
Replace ln(x) with a power series about x= 1.

Thank you.
 

1. What is a series solution near a singular point for a 2nd order linear equation?

A series solution near a singular point for a 2nd order linear equation is a method used to solve differential equations when the coefficients of the equation are not constant. It involves expressing the solution as a power series with coefficients that can be determined by substituting the series into the equation.

2. How is a singular point defined in a 2nd order linear equation?

A singular point in a 2nd order linear equation is defined as a point where the coefficients of the equation become infinite or the equation becomes undefined. This usually occurs when the equation has a singularity or a point where the solution is not well-defined.

3. What is the difference between an ordinary point and a regular singular point?

An ordinary point in a 2nd order linear equation is a point where the coefficients of the equation are finite and the solution is well-defined. A regular singular point, on the other hand, is a point where the coefficients of the equation may be infinite, but the equation can still be solved using a series solution.

4. How do you determine the radius of convergence for a series solution near a singular point?

The radius of convergence for a series solution near a singular point can be determined by using the Cauchy-Hadamard theorem. This theorem states that the radius of convergence is equal to the reciprocal of the limit of the nth root of the coefficients of the series.

5. What are some applications of series solutions near singular points in science?

Series solutions near singular points have many applications in science, particularly in physics and engineering. They are used to solve differential equations that arise in the study of physical phenomena such as heat transfer, fluid mechanics, and quantum mechanics. They also have applications in signal processing and financial modeling.

Similar threads

  • Differential Equations
Replies
7
Views
1K
Replies
2
Views
2K
  • Differential Equations
Replies
2
Views
915
  • Differential Equations
Replies
4
Views
922
  • Differential Equations
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
2K
  • Differential Equations
Replies
7
Views
2K
  • Differential Equations
Replies
3
Views
2K
  • Differential Equations
Replies
9
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
169
Back
Top