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Homework Help: Higher Order Differential Equations, Solutions related

  1. Feb 13, 2010 #1
    1. The problem statement, all variables and given/known data
    Given the following differential equation
    t^3y''' - t^2y'' + 2ty' - 2y = 0; t > 0
    Find a solution that satisfies di fferential equation and the initial conditions
    y(1) = 3; y'(1) = 2; y''(1) = 1

    2. Relevant equations

    3. The attempt at a solution
    I tried plugging in 1 for y, y' and y'', but that gave me t^3y'''-t^2+2t-2=0, which doesn't give me anything... i looked in the book but still am positively stumped... Please help.
  2. jcsd
  3. Feb 13, 2010 #2


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    I don't understand why you would expect that to work! That would give you the value of y"' at x= 1 but no where else. The fact that they are equal to 1 at x= 1 tells you nothing about their values in general.

    That equation simply says that
    [tex]\frac{d^3y}{dt^3}= \frac{t^2- 2t+ 2}{t^3}= t^{-1}- 2t^{-2}+ 2t^{-3}[/tex]

    Integrate three times.
  4. Feb 13, 2010 #3
    I'm sorry, how did you get that equation?
  5. Feb 13, 2010 #4
    Are you familiar with constant coefficient ode's?

    If yes, then make the substitution x=ln(t), and find u(x)=y(e^x).
    (Remember that the derivatives of y wrt to t are different, and use the chain rule)
  6. Feb 13, 2010 #5


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    This type of DE is called an Euler or Euler-LaGrange equation. Look for a solution of the form y = tn. When you plug that into the DE you will get an equation that you can solve for n. It is similar to what happens for constant coefficient DE's. You can read about them, for example at:


    In particular, that location shows you what happens if you get a repeated root for n.
  7. Feb 13, 2010 #6


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    Sometimes we all type before we look closely. I don't think Halls really meant that.
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