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Constructing a DE

  1. Sep 22, 2013 #1
    Hi

    I'm working through some example question and the memo leaves out this question. I was hoping someone can help me out.

    The question:
    Construct a differential equation of the form y'' + by' + cy = 0 which has y(t) = e^t cos(3t) as one of its solutions.

    What I did:
    First I found the y'(t) and y''(t)
    Plugged that into the DE.

    My answer:
    b = -2
    c = -6

    y'' - 2y' - 6y = 0
     
  2. jcsd
  3. Sep 22, 2013 #2

    Simon Bridge

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    Check your answer by plugging the solution into it from scratch.
    It will also help to show your working step-by-step.
     
  4. Sep 22, 2013 #3

    mfb

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    Staff: Mentor

    You can use WolframAlpha to check your result.
     
  5. Sep 22, 2013 #4

    pasmith

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    Hint: What is the relationship between the ODE
    [tex]
    y'' + by' + cy = 0
    [/tex]
    and the quadratic equation
    [tex]
    \lambda^2 + b\lambda + c = 0?
    [/tex]

    You may also find the identity
    [tex]
    \cos kt \equiv \frac{e^{ikt} + e^{-ikt}}{2}
    [/tex]
    helpful.
     
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