Finding Eigen Values/functions

  • Thread starter ShowerHead
  • Start date
  • #1

Homework Statement


I have to find the eigenvalues&function of the eqn:
[tex]y''+\lambda y =0[/tex]
With the boundary conditions:
[tex]y(0)+y'(0) = 0[/tex] and [tex]y(\pi) =0[/tex]

Homework Equations





The Attempt at a Solution


I get the general equations, okay, but am having trouble due to the boundary conditions.
Assuming [tex]\lambda>0[/tex], then get the general solution:
[tex]y(x)=A\sin(x\sqrt{\lambda})+B\cos(x\sqrt{\lambda})[/tex]

The best i can do now is that:
[tex]y(\pi)=A\sin(\sqrt{\lambda}\pi)+B\cos(\sqrt{\lambda}\pi)=0[/tex]
Which can only be valid if B = 0 and [tex]\sqrt{\lambda} = n[/tex] where n is a positive integer.
Now I run into problems, the second bc gives:
[tex]y(0)=A\sin(n 0)=0[/tex]
[tex]y'(0)=nA\cos( n 0)=nA[/tex]
So:
[tex]y(0)+y'(0) = 0 + nA = 0[/tex]

Am I on the right track, or can anyone see where I am messing up?
Thanks.
 

Answers and Replies

  • #2
gabbagabbahey
Homework Helper
Gold Member
5,002
6
[tex]y(\pi)=A\sin(\sqrt{\lambda}\pi)+B\cos(\sqrt{\lambda}\pi)=0[/tex]
Which can only be valid if B = 0 and [tex]\sqrt{\lambda} = n[/tex] where n is a positive integer.
No, as a simple counterexample consider [itex]\lambda=\frac{1}{16}[/itex] and [itex]A=-B[/itex]. If [itex]\sqrt{\lambda}[/itex] had to be an integer, then [itex]B[/itex] would have to be zero; but there's no reason to assume [itex]\sqrt{\lambda}[/itex] must be an integer.
 

Related Threads on Finding Eigen Values/functions

  • Last Post
Replies
5
Views
1K
Replies
4
Views
9K
  • Last Post
Replies
2
Views
2K
Replies
6
Views
1K
  • Last Post
Replies
2
Views
1K
Replies
3
Views
3K
  • Last Post
Replies
10
Views
1K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
14
Views
3K
  • Last Post
Replies
2
Views
758
Top