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Is this a correct solution ? to ax'' = 0

  1. Jan 18, 2010 #1
    [itex]ax^{''} = 0[/itex]

    characteristic eqation
    [itex]ar^{2} = 0[/itex]

    [itex]r_{1,2} = 0[/itex]

    since we have double result

    [itex]x(t) = c_{1}e^{0t} + c_{2}t e^{0t}[/itex]

    so [tex] e^0 = 1 [/tex]

    [tex]x(t) = c_{1} + c_{2}t[/tex]

    is this a correct result :S ? i'm learning integration by myself and i'm not shure about this case it's not cover in my book if results can be zero or it doesn't matter :S ?
     
    Last edited: Jan 18, 2010
  2. jcsd
  3. Jan 18, 2010 #2

    HallsofIvy

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    Yes, that's correct. The simpler way to see that is to first divide by a to get just x"= 0. That is the same as saying d(x')/dt= 0 so x' is a constant: x'(t)= c2. Now integrate that: the integral of a constant, c1 is, of course, x(t)= c2t+ c1.

    (I'm surprised you would know about "characteristic equations" for linear differential equations with constant coefficients if you are just learning integration!)
     
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