Series solution

  • Thread starter JohanL
  • Start date
  • #1
158
0
Im trying to solve the following equation
[tex]
y''(x) + \frac{y'(x)}{x} + \frac{y(x)}{x^2} = 0
[/tex]

Then with

[tex]y(x) = \sum_{i=0}^\infty a_i*x^{i+k}[/tex]

[tex]y'(x) = \sum_{i=0}^\infty a_i*(i+k)*x^{i+k-1}[/tex]

[tex]y''(x) = \sum_{i=0}^\infty a_i*(i+k)(i+k-1)*x^{i+k-2}[/tex]

I get

[tex]\sum_{i=0}^\infty a_i*(i+k)(i+k-1)*x^{i+k-2} + \sum_{i=0}^\infty a_i*(i+k)*x^{i+k-2} + \sum_{i=0}^\infty a_i*x^{i+k-2} = 0[/tex]

but how can i get a recurrence relation from this.
I need something like
[tex]a_{i+2} = f(i)*a_i[/tex]
But with only the same i+k-2 in all terms i dont know how to proceed.
 
Last edited:

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,847
964
Yes, go ahead and write out the formula you get:
[tex]a_i(i+k)(i+k-1)+ a_i(i+k)+ a_i= a_i((i+k)^2+ 1)= 0[/tex]
Either i+k= 0 or ai= 0 for all i.
The difficulty is that x= 0 is not a "regular" singular point for this equation. That's an "equipotential" (also called "Euler type") equation for which 0 is pretty much the "boundary" between regular singular point and irregular singular point. Multiplying on both sides of the equation by x2 you get x2y"+ xy'+ y= 0. Making the change of variable t= ln x or x= et reduces it to an equation with constant coefficients:
[tex]\frac{dy}{dx}= \frac{dx}{dt}\frac{dy}{dt}= e^t\frac{dy}{dt}= x\frac{dy}{dt}[/tex]
[tex]\frac{d^2y}{dx^2}= \frac{d }{dx}(x\frac{dy}{dt})= x^2\frac{d^2y}{dt^2}+ x\frac{dy}{dt}[/tex]
so [tex]x^2\frac{d^2y}{dx^2}+ x\frac{dy}{dx}+ y= \left(\frac{d^2y}{dt^2}+ \frac{dy}{dt}\right)- \frac{dy}{dt}+ y= \frac{d^2y}{dt^2}+ y= 0[/tex]
Solve that equation for y as a function of t and then replace t by ln x.
 
  • #3
158
0
Thx, I solved it!
 

Related Threads on Series solution

  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
3
Views
6K
J
  • Last Post
Replies
4
Views
891
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
1
Views
984
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
1
Views
976
Replies
1
Views
2K
  • Last Post
Replies
3
Views
926
Top