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Homework Help: Differential equation problem

  1. May 28, 2005 #1
    i'm trying to find what the intermediate steps were used in solving this differential equation:

    [tex]y'' + y' -3y =0[/tex]

    [tex]m=\frac{-1\pm\sqrt{13}}{2}[/tex] now we have

    [tex]y=C_1e^{\frac{-.5+\sqrt{13}}{2}}t + C_2e^{\frac{-.5 -\sqrt{13}}{2}}t[/tex]

    note that the .5 are 1\2. i couldn't get the latex for this to work
     
  2. jcsd
  3. May 28, 2005 #2

    arildno

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    Do you know where "m" came from?
     
  4. May 28, 2005 #3

    that m came from the auxillary equation of the differential
    [tex]r^2 +r -3=0[/tex]
     
  5. May 28, 2005 #4

    arildno

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    So what's bugging you, then?
     
  6. May 28, 2005 #5

    BobG

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    As you noted, you created the auxiliary (or characteristic) equation. The auxiliary equation was solved via the quadratic formula.

    Note: You should have:
    [tex]y=C_1e^{(-.5 + \frac{\sqrt{13}}{2})t} + C_2e^{(-.5-\frac{\sqrt{13}}{2})t}[/tex]
     
  7. May 29, 2005 #6

    HallsofIvy

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    If you were wondering where the characteristic equation came from, assume a solution ofthe form y= ert, plug it into the differential equation and see what r must be in order to make the equation true.
     
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