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Forth Order DiffEq

  1. Apr 26, 2008 #1
    [SOLVED] Forth Order DiffEq

    I've recently come across the following differential equations. y''''+y=0 and y''''-y=0. Can differential equations such as these be solved with any technique other than guessing for the particular solutions? They seem very simular to trig's equation but are still not quite the same.
     
  2. jcsd
  3. Apr 26, 2008 #2
    the only solution that i can see is y=0.
    from those equations y = -y and y''''=-y'''' which can only happen if y=0
     
  4. Apr 26, 2008 #3
    According to mathematica, the most general answer is:
    [tex]y= e^{-\frac{x}{\sqrt{2}}} \left(\left(e^{\sqrt{2} x} C[1]+C[2]\right) \text{Cos}\left[\frac{x}{\sqrt{2}}\right]+\left(C[3]+e^{\sqrt{2} x} C[4]\right) \text{Sin}\left[\frac{x}{\sqrt{2}}\right]\right)[/tex].
    The way I would solve this is to write down the characteristic equation to it.
    [tex] r^4+1=0[/tex]
    [tex]r=(-1)^{1/4}[/tex]
    I think you know how to do it from here.
     
  5. Apr 26, 2008 #4
    I don't see how thats a solution to both of the differential equations (although it is a solution to one of them)...
     
  6. Apr 26, 2008 #5

    lurflurf

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    I hope that was meant to be two seperate equations, not a system.
     
  7. Apr 26, 2008 #6
    It's only the solution to the first one (y''''+y=0).
     
  8. Apr 27, 2008 #7

    lurflurf

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    factor

    y''''+y=0
    (D^4+1)y=0
    (D^2+2cos(pi/4)D+1)(D^2-2cos(pi/4)D+1)y=0

    y''''-y=0
    (D^4-1)y=0
    (D+1)(D-1)(D^2+1)y=0

    Then either do partial fraction decomposition or note that since the factors are independent the solution of the product is the sum of the solutions of the factors
    ie
    D(D-1)y=0
    Du=0->u=A
    (D-1)v=0->v=Bexp(x)
    y=u+v=A+Bexp(x)
     
  9. May 1, 2008 #8
    Apply Laplace transformation.
     
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