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Same roots

  1. Oct 18, 2005 #1
    why is it that when you have the same roots to an O.D.E., you usually add an x or x^2 to get a basis?
     
  2. jcsd
  3. Oct 18, 2005 #2
    because when you multipy anything by a variable it will not be a scaler multiple of the original, and you need two unique so;utions (or more) for an ode
     
  4. Oct 18, 2005 #3

    cronxeh

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    Because if Wronskian=0 then you have linearly dependent solutions, and you need linearly independent solutions (i.e. W not equal 0).
     
  5. Oct 19, 2005 #4
    but why pick x or x^2 to multiply? why not something like xy or x^y or e^x or something?
     
  6. Oct 19, 2005 #5
    you could but why make it more complicated than you have to
     
  7. Oct 19, 2005 #6

    Tide

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    Let D be a differential operator such that a is a double root:

    [tex](D-a)^2 y = 0[/itex]

    Consider the slightly modified DE:

    [tex](D-a + \epsilon)(D-a-\epsilon)[/tex]

    Solve this second equation subject to the initial conditions [itex]y(0) = y_0[/itex] and [itex]\dot y(0) = \dot y_0[/itex] then pass to the limit of [itex]\epsilon[/itex] going to zero. You'll find the answer to your question in the result! :)
     
  8. Oct 20, 2005 #7
    :bugeye:
    could you explain that part a little clearer?
     
  9. Oct 20, 2005 #8

    Tide

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    This ODE

    [tex]\left( \frac {d^2}{dt^2} - 2 a \frac {d}{dt} + a^2 - \epsilon ^2 \right) y= 0[/tex]

    has repeated roots when [itex]\epsilon \rightarrow 0[/itex]. Solve the equation as it is subject to specific initial conditions. When you're all done, pass to the limit [itex]\epsilon \rightarrow 0[/itex] and you'll see why the basis functions are the way they are when you have repeated roots! (Sorry - I don't have the time to type in the algebra but it is straightforward.)
     
  10. Oct 21, 2005 #9
    thanks!!! :)
     
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