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2nd order Linear DE

  1. Jan 17, 2013 #1

    When solving a 2nd order Linear DE with constant coefficients ([itex]ay''+by'+cy=0[/itex]) we are told to look for solutions of the form [itex]y=e^{rt}[/itex] and then the solution (if we have 2 distinct roots of the characteristic) is given by
    [itex]y(t)=c_1 e^{r_1 t}+c_2 e^{r_2 t}[/itex]

    This is clearly a solution, but how do we know there are no other solutions?
    That is, how do we know this is the general solution?
  2. jcsd
  3. Jan 17, 2013 #2


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    Hi Apteronotus! :smile:
    It's easy to prove for the first-order case …

    if y' - ry = 0, put y = zert, then (z' + rz)ert = rzert

    so ert = 0 (which is impossible),

    or z' + rz = rz, ie z' = 0, ie z is constant :wink:

    and now try (y' - ry)(y' - sy) = 0, using the same trick twice :smile:
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