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Help Solving a System Of DEQ's

  1. Mar 23, 2009 #1
    Hi Guys,

    I'm tying to solve a system of equations. I know I need to operate on the top and the bottom both in order to isolate the X's and Y's, but I can't seem to figure what to operate on them with. Here are the equations, any help is appreciated. Thanks

    D2x - Dy=t
    (D+3)x + (D+3)y=2

    I should be able to finish solving it if I can just get them in the forms I need.

    Edit: I have a feeling this is going to seem really obvious and easy when I see it. But I am just getting in to Differential Equations, so I am new at this stuff.
     
    Last edited: Mar 23, 2009
  2. jcsd
  3. Mar 23, 2009 #2
    what about if you substitute Dy from (1) into (2), then you have a second order diffential equation in form of x and t. I believe x and y are function of t
     
  4. Mar 23, 2009 #3
    If I do that then I get D2x + Dx + 3x - t + 3y = 2

    Then I don't know how to deal with the -t + 3y

    Thanks for the idea though. I hadn't thought of it.



    Edit: After another look you could then write it as D2x + Dx + 3x = t - 3y + 2

    But I still don't know what to do with it from here. I can solve the associated homogeneous equation, but then I still don't know what to do with the y on the right side. I think I have to get rid of either y or x entirely before I can solve for x(t) or y(t).
     
  5. Mar 23, 2009 #4

    tiny-tim

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    Welcome to PF!

    Hi ajohncock! Welcome to PF! :smile:
    erm :redface: … can't you just solve this on its own? :wink:
     
  6. Mar 23, 2009 #5

    HallsofIvy

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    tiny-tim, are you asserting that he should be able to solve a single equation in two unknowns?

    D2x - Dy=t
    (D+3)x + (D+3)y=2

    If you "multiply" the first equation by D-3 and the second equation by D you get
    D2(D-3)x- D(D-3)y= (D-3)t= -3t
    D(D+3)x+ D(D-3)y= 0 and then adding eliminates y:

    D2(D-3)x+ D(D-3)x= -3t or D(D+1)(D-3)x= -3t.

    Solve that equation for x, then put that x into D2x- Dy= t and solve that equation for y.
     
  7. Mar 23, 2009 #6
    The problem with this is that, the part I BOLDED is actually (D+3) in the original equation. However that may be on the right track. I tried multiplying the top by (D+3) and the bottom by D.

    I then added the equations and come up with this:

    D2x(D+3) + D(D+3)x = 2D + 3t + tD

    Simplified that is D3x + 4D2x + 3Dx = 2D + 3t + Dt

    Which does get rid of the y, but now I am unsure what to do with the right side.

    This is frustrating.
     
    Last edited: Mar 23, 2009
  8. Mar 23, 2009 #7

    tiny-tim

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    solution … ? :smile:
     
  9. Mar 23, 2009 #8
    Haha, I wish I knew what you meant by that.
     
  10. Mar 23, 2009 #9

    tiny-tim

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    re-arrange it! :wink:
     
  11. Mar 23, 2009 #10
    I assume you mean to solve the equation for Dy and substitute that back into the other equation. But when I solve it for Dy I get:

    Dy = -Dx - 3x - 3y +2 Which is all fine and good except for the -3y. Which poses a problem when substituted back into the first equation.
     
  12. Mar 23, 2009 #11

    tiny-tim

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    how is that re-arranging it? :confused:
     
  13. Mar 23, 2009 #12
    Haha, maybe I'm not as math savvy as I thought. I guess I don't know what you mean by re-arrange it.
     
  14. Mar 23, 2009 #13
    Man i've been looking at this and manipulating it for too long. I've got nothing! It's the right side I don't know how to deal with.
     
  15. Mar 24, 2009 #14

    tiny-tim

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    just got up :zzz: …

    (D+3)(x + y) = 2 ? :wink:
     
  16. Mar 24, 2009 #15
    since y is a function of t and x then if Dy = (D^2)x -t, you can apply integration so that you now obtain what is called integro-differential equation in form of x and t
     
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