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Tough Diff EQ problem

  1. Apr 28, 2008 #1
    [SOLVED] Tough Diff EQ problem

    1. The problem statement, all variables and given/known data
    y^(4)-8y^(2)-9y=7e^(-3x)

    y^(4)=fourth derivative, y^(2)=second deriv.
    2. Relevant equations
    Need to find complimentary solution and particular solution using variation of parameters



    3. The attempt at a solution
    r^4-8r^2-9=0
    (r^2-9)(r^2+1)=0, r=9,-1

    Y(c)=Ce^(9x)+Ce^(-x)

    I think the comp soln is right, but I am unsure on how to start on Y(p)
     
  2. jcsd
  3. Apr 28, 2008 #2

    Hootenanny

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    Are you sure about your homogenous solution? I suggest that you re-check your solutions of the indicial equation.
     
    Last edited: Apr 28, 2008
  4. Apr 28, 2008 #3
    e^(-3x) right?

    I tried taking the 4th deriv and 2nd deriv and plugging in which would equal

    81e^-3x - 8(9e^-3x) - 9e^(-3x)=7e^(-3x)
    but the left side = 0??
     
  5. Apr 28, 2008 #4

    Hootenanny

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    In reference to the PM, perhaps it would help if I draw your attention to the following lines
    Good :approve:
    Not so, good. Notice the powers of r.
     
  6. Apr 28, 2008 #5

    Hootenanny

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    I'm quoting from the PM you sent me. As I said via PM, please note that in order to keep our assistance as transparent as possible, it is the policy of Homework Helpers not to provide tuition via PM.
    Now, in reference to your question,
    Better :approve:. However,
    You need to be careful with your constant coefficients here. The coefficients of the two exponentials need not be equal. Furthermore, you are missing the coefficients of the trigonometric terms.
    Yes, since e-3x has already been found to be a solution, you cannot use e-3x as your particular solution. Axe-3x would be the usual alternative.
     
  7. Apr 28, 2008 #6
    dumb mistakes!
    r=+/- 3, and 1+i
    Y(c)=Ce^3x + Ce^-3x + Ce^(x)cosx + Ce^(x)sin(x)
    Y(p) =Axe^-3x
    y'(p) =Ae^-3x - 3Axe^-3x
    y''(p) =-3Ae^-3x - 3Ae^-3x + 9Axe^-3x
    y'''(p) =9Ae^-3x + 9Ae^-3x + 9Ae^-3x - 27Axe^-3x
    y''''(p)=-27Ae^-3x -27Ae^-3x - 27Ae^-3x - 27Ae^-3x + 81Axe^-3x=
    81AXe^-3x - 108Ae^-3x

    substitute:
    (81Axe^-3x - 108Ae^-3x)-8(9Axe^-3x - 6Ae^-3x) - 9(Axe^-3x)=7e^-3x

    -60Ae^-3x=7e^-3x
    -60A=7, A=-7/60

    Answer:
    y=C(1)e^3x + C(2)e^-3x + C(3)e^(x)cos(x) + C(4)e^(x)sinx - 7/60xe^-3x

    does this look correct,
     
    Last edited: Apr 28, 2008
  8. Apr 28, 2008 #7

    HallsofIvy

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    Back up! Your original equation was [itex]r^4- 8r^2- 9= (r^2- 9)(r^2+ 1)= 0[/itex]
    so [itex]r^2= 9[/itex] and [itex]r^2= -1[/itex]. The roots of the first of those are [itex]\pm 3[/itex] but the roots of the second are not [itex]1\pm i[/itex]!
     
  9. Apr 28, 2008 #8
    the sqrt of -1= 1i correct-----ohhhhh wow that changes things:
    e^(0)cos(x)+e^0sinx= C(3)cosx + C(4)sinx(x)
    y=c(1)e^3x + c(2)e^-3x + c(3)cosx + c(4)sinx -7/60xe^-3x ??
     
  10. Apr 28, 2008 #9

    Hootenanny

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    :redface::shy::blushing:

    That's really really really embarrassing.
     
    Last edited: Apr 28, 2008
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