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Homework Help: 3rd Order Linear Homogenous DE from Solutions

  1. Apr 12, 2012 #1
    1. The problem statement, all variables and given/known data

    Find a third order, linear, homogeneous DE which has the following solutions:

    e[itex]\pi[/itex]t, te[itex]\pi[/itex]t and e-t

    2. Relevant equations

    Standard form of a third-order linear homogenous ODE:

    Ay''' + By'' + Cy' + Dy = 0

    3. The attempt at a solution

    I tried deriving the characteristic equation given the r values of the solutions.

    For e[itex]\pi[/itex]t:
    r = [itex]\pi[/itex], therefore one part of the equation is (r - [itex]\pi[/itex])

    For e-t:
    r = -1, therefore another part of the equation is (r+1)

    But I can't figure out what to do for the second one.
    Last edited: Apr 12, 2012
  2. jcsd
  3. Apr 12, 2012 #2


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    Remember that a repeated root ##r## of the characteristic equation gives rise to the solution pair ##\{e^{rt},te^{rt}\}##.
  4. Apr 12, 2012 #3
    Ah, so it just has a repeated root.

    So the characteristic equation would be:


    Factor that out and we get:
    (r2 - 2r[itex]\pi[/itex] + [itex]\pi[/itex]2)(r+1)

    r3 - 2r2[itex]\pi[/itex] + r[itex]\pi[/itex]2 + r2 - 2r[itex]\pi[/itex] + [itex]\pi[/itex]2

    r3 - 2r2[itex]\pi[/itex] + r2 + r[itex]\pi[/itex]2 - 2r[itex]\pi[/itex] + [itex]\pi[/itex]2

    r3 + r2(-2[itex]\pi[/itex] + 1) + r([itex]\pi[/itex]2 - 2[itex]\pi[/itex]) + [itex]\pi[/itex]2

    Therefore the DE would be:

    y''' + (1-2[itex]\pi[/itex])y'' + ([itex]\pi[/itex]2 - 2[itex]\pi[/itex])y' + [itex]\pi[/itex]2y = 0

    Thanks for the help!
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