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Quick question on repeated roots when solving differential equations

  1. May 30, 2014 #1
    say we have gone through the steps and have...

    ##(\lambda - 2)^{2}(\lambda ^{2}-9) = 0##

    which we can write as...

    ##(\lambda - 2)(\lambda - 2)(\lambda ^{2}-9) = 0##

    we have value for lambda of 2, 2, 3, -3

    because we have a repeated root.

    now, say we have

    ##(\lambda^{2} - 1)(\lambda + 1) = 0##

    my question is would you have 2 values of -1 like the example above i.e. 1, -1, -1? or would you have values of lambda of just 1, -1? and if the former would you treat the -1 as a repeated root?

    it becomes important when writing the solution because for a repeated root there an x in the second term with the exponential.

    thanks in advance for any direction on this.
     
  2. jcsd
  3. May 30, 2014 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    This isn't really a question about differential equations, it is a question about basic algebra.

    Since [itex]\lambda^2- 1= (\lambda- 1)(\lambda+ 1)[/itex], which I am sure you already knew,
    [itex](\lambda^2- 1)(\lambda+ 1)= (\lambda- 1)(\lambda+ 1)(\lambda+ 1)= (\lambda- 1)(\lambda+ 1)^2[/itex]
    so, yes, -1 is a double root.
     
  4. May 30, 2014 #3
    ahh, never thought of it that way! thats great, many thanks!
     
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