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Homework Help: Integral Question

  1. May 21, 2014 #1
    d4y/dx4 - 4 d3y/dx3 - 5 d2y/dx2 + 36 dy/dx - 36y = -8e^x

    Given that λ^4 - 4λ^3 - 5λ^2 + 36λ - 36 = 0 can be factored to obtain (λ-2)^2 (λ^2 - 9) = 0

    find the general solution to this equation

    I can answer the question no problem BUT where does the -8 on the RHS of the equation go??

    If i set y=e^λx i get λ^4 - 4λ^3 - 5λ^2 + 36λ - 36 = -8 ??

    so why hasnt he factorised λ^4 - 4λ^3 - 5λ^2 + 36λ - 28 = 0


    Totally confused. Thanks for any help.
  2. jcsd
  3. May 21, 2014 #2


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    Gold Member

    That is guiding you to the general solution of the corresponding homogenous equation only. To obtain the general solution to your original equation, you also need a solution solving the equation with the term on the RHS present.
  4. May 21, 2014 #3
    The differential equation you have there is an inhomogeneous differential equation.
    The characteristic polynomial that is provided in the hint provides the solution to the homogenous differential equation
    [tex]y^{(4)} - 4y^{(3)} -5y''+36y'-36y = 0[/tex]
    The final solution will be the sum of the homogenous solution and the particular solution for the imhomogenous part.
  5. May 21, 2014 #4
    the particular solution is given as e^x

    so the complete solution will be

    λ = 2, 2, 3, -3

    -> y = Ae^2x + Bxe^2x + Ce^3x + De^-3x + e^x

    are you saying that the particular solution given takes care of the RHS, so I dont need to worry about it? Or do I have to include something else in the complete solution above?

    thanks for the help.


    oh, i see, it doesnt want a solution to the original equation. thanks again for the help!!

    Just out of interest how would the solution to the original question differ from my answer?
    Last edited: May 21, 2014
  6. May 21, 2014 #5
    Yes, that is the role of the particular solution. You can think of it this way:
    We have a differential equation in the form
    [itex]\hat{D} y = f(x)[/itex]
    The general solution can be written in terms of the homogenous and particular solutions:
    [itex]y = y_{h} + y_{p}[/itex]
    which satisfy
    [itex]\hat{D} y_{h} = 0[/itex]
    [itex]\hat{D} y_{p} = f(x)[/itex]
    So, clearly,
    [itex]\hat{D} (y_{h}+y_{p}) = f(x)[/itex]
  7. May 21, 2014 #6
    excellent! got it! Many thanks!!
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