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Diff. eq.

  1. Apr 14, 2005 #1
    solve the following promblem
    i let y = e^kx
    y'= ke^kx

    so i got this



    now i have to find (K)

    how should i solve for (k) from this equation k^3-3K2+31k-37=0
    can i use synthetic division if yes how should i use it or which other method can i use

    is this right

    i solve the k by synthetic division
    5 1 1 -17 -65
    5 30 65
    1 6 13 0

    so the factor is (k-5) (K^2+6k+13)
    then i use this equation

    and got k = -3 +- 2i

    and my fianl answer is
    Last edited: Apr 14, 2005
  2. jcsd
  3. Apr 14, 2005 #2


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    Is this the equation

    [tex] \frac{d^{3}y}{dx^{3}}-3\frac{d^2y}{dx^{2}}+31\frac{dy}{dx}-37y=0 [/tex]

  4. Apr 14, 2005 #3


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    Since the equation is homogeneous and has constant coefficients, you can just solve its characteristic equation.

    You'll have an equation of the form:

    [tex]y(x) = c_1 e^{\lambda_1 x} + c_2 e^{\lambda_2 x} + c_3 e^{\lambda_3 x}[/tex]

    If the equation turns out to have two complex roots (I won't say if it does), you'll have a solution somewhat different:

    [tex]y(x) = c_1 e^{\lambda_1 x} + c_2 e^{\alpha x} \cos{\beta x} + c_ 3 e^{\alpha x} \sin{\beta x}[/tex]


    [tex]\lambda_2 = \alpha + \beta i[/tex]
    [tex]\lambda_3 = \alpha - \beta i[/tex]

    All of this should be in bold print (well, almost) in any ODE book.
    Last edited: Apr 14, 2005
  5. Apr 14, 2005 #4


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    I'd try starting with the rational root theorem, and simply check all the possibilities for a rational solution for k.

    (PS wasn't there a 9 in there before?)
  6. Apr 14, 2005 #5


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    There was. Otherwise, I would've mentioned the rational root theorem, as well. Don't you love it when you submit a (generous) post full of tips and someone changes the nature of the problem?
  7. Apr 14, 2005 #6


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    Here's the solution,courtesy of Maple.

    Last edited: Nov 22, 2006
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