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Homework Help: ODE Problem with boundary conditions

  1. Apr 4, 2012 #1
    1. The problem statement, all variables and given/known data

    y'' - λy = 0

    where y(0)=y(1)=0, y=y(t)

    2. Relevant equations

    3. The attempt at a solution

    Hi everyone,

    This is part of a PDE question, I just need to solve this particular ODE. I know how to do it in the case for y'' + λy = 0, where you get the solution:

    λ = (nπ)^2,
    y = C.sin(nπt), C a constant

    However, I don't know how to go about doing it for this case, other than getting the trivial zero solution of course. Am I correct in thinking I should rewrite it as:

    y'' + (-λ)y = 0

    and then solve as for the positive case?

    I would be really grateful if you could please tell me if this is correct.
  2. jcsd
  3. Apr 4, 2012 #2


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    Science Advisor

    In other words, you are assuming [itex]\lambdas< 0[/itex], right? Okay, how about setting [itex]\lambda= -\alpha^2[/itex] where [itex]\alpha[/itex] is any non-zero real number?

    What is the characteristic equation for [itex]y''+ \alpha^2y= 0[/itex]. What characteristic roots do you get and what is the general solution?
  4. Apr 4, 2012 #3

    Thanks for your reply! So then I would have:

    Assume λ<0

    Let λ = -a^2, for some real, non-zero number a


    y'' + (a^2)y = 0

    y(t) = c1.sin(at) + c2.cos(at)

    y(0)=0, therefore c2 = 0

    y(1)=0, therefore c1 = 0 (trivial solution)

    or sin(a) = 0

    a = n∏

    so then λ = -(n∏)^2

    So then y(t) = c1.sin(n∏t)

    Then one of the conditions I was given in the question was that

    ∫(y^2)dt = 2, with limits of integration 0 to 1

    Integrating this out:

    (c1)^2.∫[sin(n∏t)]^2.dt =2

    so then (1/2)(c1)^2.∫(1-cos(n∏t))dt =2

    which gives us:

    (c1)^2 = 4

    c1 = ±2

    So then y(t) = c1.sin(n∏t), c1=±2

    I don't like that I've ended up with two possible values for c1, but this solution makes sense to me. Please could you tell me if I am correct?
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