A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing


  • 5 years ago

Need a few pointers plz? Q. Find the first two positive eigenvalues of the following BVE : y" + (lambda)y = 0 BV(1) : 5y(0) - 2y'(0) = 0 BV(2) : 5y(1) + 2y'(1) =0

  • This Question is Closed
  1. anonymous
    • 5 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Solve for the characteristic eqn. So r^2+lambda=0 so r1,2= +/- sqrt(-lambda) Now since lambda can vary you have to do it for 3 cases because each case you would get a different solution to the differential equation. It's basically just solving 3 2nd order linear d.e. Case 1 would be if lambda is < 0 giving 2 real solutions. So since the solutions are real, you solve the 2nd order linear d.e accordingly. Differentiate to get y' since your BV involves y'. Now try to solve for c1 and c2 and lambda using your BV. Now if the only way to satisfie the BV is for c1=c2=0, then the trivial solution is obtained and there is no eigenvalue for that case. Repeat these steps for case 2 where lambda=0 to giving 1 root and case 3 where lambda>0 giving 2 non real roots. You want to find the specific instances where there is a way to satisfy the BV WITHOUT having c1=c2=0. For example unrelated to this problem, you could end up with c1lambda+ c2sinlambdapi and know for sure c1 equals 0 from one of the BV. Then you have to solve for when sinlambdapi=0. So lambdapi=n pi, pi's cancel leaving lambda=n for n=1,2,3... You get the idea.

  2. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...


  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.