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Homework Help: Higher Order Homogeneous ODE (IVP)

  1. Apr 18, 2006 #1
    Higher Order Homogeneous ODE (IVP) [Solved]

    I am having problems with this IVP:

    y'''' + y' = 0

    y(0) = 5
    y'(0) = 2
    y''(0) = 4

    What I have done so far is:

    [tex] \lambda^3 + \lambda = 0 [/tex]

    [tex] \lambda(\lambda^2 + 1) = 0 [/tex]

    So one roots is [tex] \lambda = 0 [/tex]

    (though.. can there be a root that is zero??)

    And the other root is:

    [tex] \lambda^2 = -1 [/tex]

    [tex] \lambda = \pm\sqrt{-1} [/tex]

    [tex] \lambda = \pmi [/tex]

    Therefore [tex] \omega = 1 [/tex]

    [tex] y = ce^(0)x + Acos\omegax + Bsin\omegax [/tex]

    [tex] y = c + Acosx + Bsinx [/tex]

    [tex] y' = -Asinx + Bcosx [/tex]

    [tex] y' = -Acosx - Bsinx [/tex]

    So, putting in the initial values..

    c + A = 5 (so c = 9)
    B = 2
    A = -4

    and so the answer is:

    [tex] y = 9e^x - 4cosx + 2sinx [/tex]

    But it seems the answer is wrong..

    (though, considering my severe lack of math skills, it doesn't surprise me)

    I really need to be able to solve this problem.. can anyone help?
    Last edited: Apr 18, 2006
  2. jcsd
  3. Apr 18, 2006 #2


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    Science Advisor
    Homework Helper

    1.You put one prime more than was necessary in the first equation.

    2.Make the substitution

    [tex] y'(x)=f(x) [/tex]

    and then integrate the resulting ODE. Then you'll know what to do.

  4. Apr 18, 2006 #3


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    Science Advisor

    Yes, it is certainly possible for 0 to satisfy an equation!

    Well, yes, of course, that's wrong! You started by saying (correctly) that the solution must be of the form
    [tex] y = ce^{(0)x} + Acos \omega x + Bsin \omega x [/tex]
    [tex] y = c+ A cos x+ B sin x[/tex]

    But after solving (correctly) for c=9, A= -4, and B= 2, you give the solution as
    [tex] y = 9e^x - 4cosx + 2sinx [/tex]

    Do you see the difference?
    Last edited by a moderator: Apr 18, 2006
  5. Apr 18, 2006 #4
    oh!!! thank you!! i just understood! :D
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