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Homework Help: What happens if you only get 1 solution to a 2nd Order Diff EQ?

  1. Feb 17, 2006 #1
    Hello everyone, i'm slightly confused on this problem, when i factored it and solved for r, i came out with only 1 answer, r = -13/72
    Here is my problem and work:
    http://img213.imageshack.us/img213/685/lastscan15uk.jpg [Broken]

    :biggrin:
     
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Feb 17, 2006 #2

    benorin

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    Homework Helper

    If you get repeated roots in your auxilary equation, e.g.

    [tex]y^{\prime\prime}-2y^{\prime}+1=0[/tex]

    has the auxilary equation

    [tex]r^{2}-2r+1=0[/tex] so [tex](r-1)^2=0[/tex]

    and hence r=1 (with multiplicity 2) so the general solution is of the form

    [tex]y=Ae^t+Bte^t[/tex]

    The roots of the equation

    [tex]5184r^2+1827r+169=0[/tex]

    are [tex]r=-\frac{13}{72}[/tex] (with multiplicity 2) so your general solution is of the form...
     
    Last edited: Feb 17, 2006
  4. Feb 17, 2006 #3

    Tide

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    One way to see why the method proposed by Benorin is valid would be to write your solution as

    [tex]y = A e^{r_1 t} + B e^{r_2 t}[/tex]


    and apply the initial conditions to determine the coefficients. Then let [itex]r_2[/itex] approach [itex]r_1 = r[/itex] (the root of your actual equation). Usually, l'Hopital's Rule can be applied to yield Benorin's solution.
     
  5. Feb 18, 2006 #4
    Thanks guys, i don't know if its right or not, but i think my method is right:
    http://img211.imageshack.us/img211/6219/lastscan0ku.jpg [Broken]
     
    Last edited by a moderator: May 2, 2017
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