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Finding the Fundamental Solutions of a Third Order ODE

  1. Oct 18, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the fundamental solution to ty''' - y'' = 0

    2. Relevant equations


    3. The attempt at a solution
    I think I'm missing something really obvious, but I have the characteristic polynomial:
    [itex] t\lambda^3 - \lambda^2 = 0 [/itex]
    Solving the equation:
    [itex] \lambda^2 (t\lambda - 1) = 0 [/itex]
    I get zero repeated twice, so two of the fundamental solutions are 1 and t.
    The third solution is [itex] \frac {1}{t} [/itex] to solve the equation, but substituting it in, I get the fundamental solution as e.
    However, the right solution is [itex] t^3 [/itex].
    Did I miss something? Thank you in advance.
     
  2. jcsd
  3. Oct 18, 2014 #2

    ehild

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    It is not a constant-coefficient linear de, with solutions in form ##e^{\lambda t}##, with ##\lambda## a constant. Lambda can not be 1/t.

    You can replace y"=u, a new function and solve for u(t). Then integral twice.

    ehild
     
  4. Oct 18, 2014 #3

    SteamKing

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    Yes. The characteristic polynomial method only works for ODEs with constant coefficients. t is not a constant here.
     
  5. Oct 18, 2014 #4
    Thanks for all your help! So I ended up solving the first order linear, and getting u(t) = t, which means y'' = t. Taking the integral twice, I'm getting:

    y = [itex] \frac {t^3}{6} + C_1 t + C_2 [/itex]

    I kind of understand what's happening now, so [itex] \frac {1}{6} [/itex] is my [itex] C_3 [/itex]?
     
  6. Oct 18, 2014 #5

    LCKurtz

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    You didn't show your work so It's hard to tell, but I think you are missing a constant in your work to get ##u(t)=t##.

    There is another way to solve the problem that you might find interesting. Multiply your equation through by ##t^2## then look for a solution in the form ##y=t^n## for unknown ##n## and see what ##n##'s work.
     
  7. Oct 18, 2014 #6
    No. The original equation is a linear equation, so if f(t) is a solution, so is Cf(t). Your [itex] C_3 [/itex] is just [itex] C_3 [/itex].
     
  8. Oct 18, 2014 #7

    LCKurtz

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    If you mean what I think you mean, you are wrong. His ##t^3/6## should be ##C_3t^3##.
     
  9. Oct 18, 2014 #8

    ehild

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    The integration constant is missing here. U(t)=C t.

    It would be [itex]y = C\frac {t^3}{6} + C_1 t + C_2 [/itex]

    You can take C3=C/6, and then the general solution is y=C1t+C2+C3t3.
    Each term of the sum is a solution.
     
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