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What should be the particular solution?

  1. May 13, 2005 #1
    Hi all!
    In this differentail equation, what particular should I try?
    Is it correct to try
    "[Asinx + Bcosx + C] exp(-2x)"?
     

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  3. May 13, 2005 #2

    Pyrrhus

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    Yes it's ok.
     
  4. May 13, 2005 #3

    dextercioby

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    Or you can do it by the method of variation of constants.I've always thought of it as being elegant...

    Daniel.
     
  5. May 14, 2005 #4
    what about this?
    p.s. please open the word file as i don't know how to type mathematical symbols here..
     

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  6. May 14, 2005 #5

    Pyrrhus

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    Yes you should try

    [tex] x^{2}e^{-2x}(Ax \cos 2x + Bx \sin 2x + C) [/tex]

    There are other methods for solving nonhomogenous constant coefficients ODEs such as Anulator Method by using the differential operator, and also Variation of parameters.

    Here's a preview of the Anulator Method.

    Imagine a ODE

    [tex] y'' + 3y' + 2y = 4x^2 [/tex]

    if we rewrite the above ODE with the differential operator D

    [tex] (D^2 + 3D +2)y = 4x^2 [/tex]

    Now if we use a differential operator [itex] D^3 [/itex] we could eliminate [itex] 4x^{2} [/itex]

    [tex] D^3(D^2 + 3D +2)y = 4 D^3 x^2 [/tex]

    [tex] D^3(D^2 + 3D +2)y = 0 [/tex]

    Thus making a fifth grade auxiliary equation. There are more steps it was just a preview.
     
    Last edited: May 14, 2005
  7. May 14, 2005 #6
    For the first differential equation that I firtsly raised, the answer of the particular solution should be something like "[1+xcosx] exp(-2x)" as given by the book. So I think the trial form should not be "[Asinx + Bcosx + C] exp(-2x)", right? So, what should be the correct form?
     
  8. May 14, 2005 #7

    dextercioby

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    I've already given you the hint to do it.It's up to you if u prefer other method,just as long as it leads to the result.The correct one.

    Daniel.
     
  9. May 14, 2005 #8
    um...you mean do it by variation of constants?
    Since this is an exercise in trying the particular solution, I would like to know the form of the trial particular solution.
     
  10. May 14, 2005 #9

    dextercioby

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    One of the solution of the homogenous one contains that e^{-2x},so that should give a hint on how to pick the trial function.

    Daniel.
     
  11. May 14, 2005 #10
    I previously thought that I should try "[Asinx + Bcosx + C] exp(-2x)"
    but the answer given by the book shows that this is wrong..
     
  12. May 14, 2005 #11

    dextercioby

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    Of course.You need an "x" to multiply the exponential,because the exponential is contained both in the solution of the hom.eq. and in the nonhomogeneity term.

    Daniel.
     
  13. May 14, 2005 #12
    So, it should be "[(Asinx + Bcosx)x + C] exp(-2x)" or "x[Asinx + Bcosx + C] exp(-2x)"?
     
  14. May 14, 2005 #13

    dextercioby

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    The first without the C,it's useless there.

    Daniel.
     
  15. May 14, 2005 #14

    Pyrrhus

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    If you're still refering to your ODE.

    [tex] y_{p_{1}} = Ce^{-2x} [/tex]

    [tex] y_{p_{2}} = xe^{-2x}(A \cos x + B \sin x) [/tex]

    [tex] y_{p} = Ce^{2x} + xe^{-2x}(A \cos x + B \sin x) [/tex]
     
  16. May 15, 2005 #15

    saltydog

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    Tony, where are you with this? Getting it?

    This is what I'd do (other than just plug it into Mathematica):

    I tell you what, since the RHS is a particular solution to the homogeneous ODE:

    [tex](D+2)(D^2+4D+5)y=0\quad\text{(1)}\quad[/tex]

    you can apply this operator to both sides of the original equation to aniliate the RHS, that is:

    If you have:

    [tex]y^{''}+4y^{'}+5y=e^{-2x}(1+Cos(x))\quad\text{(2)}\quad[/tex]

    Applying the differential operator in (1) to both sides yields:

    [tex](D+2)(D^2+4D+5)(D^2+4D+5)y=0\quad\text{(3)}\quad[/tex]

    Right?

    The solution of (3) will contain the solution of (2), just gets messy that's all.

    So the solution of (3) is:

    [tex]y(x)=c_1e^{-2x}+c_2e^{-2x}Cos(x)+c_3e^{-2x}Sin(x)+c_4xe^{-2x}Cos(x)+c_5xe^{-2x}Sin(x)[/tex]

    And thus a particular solution of (2) will be:

    [tex]y_p(x)=Ae^{-2x}+Bxe^{-2x}Cos(x)+Cxe^{-2x}Sin(x)[/tex]

    Now what? How about a plot of the solution (just make up some initial conditions)?
     
    Last edited: May 15, 2005
  17. May 15, 2005 #16

    saltydog

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    Just some follow up:

    Well, if you substitute [itex]y_p(x)[/itex] into (2) above, it gets a little messy, do this and that, equate coefficients, some things cancel, and we're left with:

    [tex]y_p(x)=e^{-2x}+\frac{1}{2}xe^{-2x}Sin(x)[/tex]

    So that the general solution is:

    [tex]y(x)=c_1e^{-2x}Cos(x)+c_2e^{-2x}Sin(x)+e^{-2x}+\frac{1}{2}xe^{-2x}Sin(x)[/tex]

    Letting:

    y(0)=1

    y'(0)=1

    we get:

    [tex]y(x)=3e^{-2x}Sin(x)+e^{-2x}+\frac{1}{2}xe^{-2x}Sin(x)[/tex]

    I've attached a plot of this solution. Tony, you can do all this right?
     

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