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anonymous

  • one year ago

Differential equations question !

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  1. anonymous
    • one year ago
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    \[3\frac{d^3y(t)}{ dt }+\frac{ dy(t) }{ dt }+2y(t)=u(t)\]

  2. anonymous
    • one year ago
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    where u(t) has this form : |dw:1436021576568:dw|

  3. anonymous
    • one year ago
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    and I must calculate :\[\int\limits_{0}^{5}y(t) dt\]

  4. anonymous
    • one year ago
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    @oldrin.bataku never used that before but my question is: Does my diff eq has two solutions: y1(t) for t [0,2] y2(t) for t>2 ??

  5. anonymous
    • one year ago
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    \[\int\limits_{0}^{5}y(t)dt=\int\limits_{0}^{2}y1(t)dt+\int\limits_{2}^{5}y2(t)dt\]

  6. anonymous
    • one year ago
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    am I right ?

  7. anonymous
    • one year ago
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    there are distinct solutions, but you need ICs to 'glue' them together

  8. anonymous
    • one year ago
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    what's ICs ? Am I right with this distinct solutions ?

  9. anonymous
    • one year ago
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    Initial conditions do you mean right ?

  10. anonymous
    • one year ago
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    Can someone please guide me to all the steps needed(ICs,...) to help me understand this problem !

  11. beginnersmind
    • one year ago
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    "and I must calculate : ∫05y(t)dt" Maybe you don't need to solve the differential equation in general then

  12. beginnersmind
    • one year ago
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    If you express y(t) and integrate both sides from 0 to 5 you get a much simpler problem.

  13. anonymous
    • one year ago
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    can you be more specific please

  14. anonymous
    • one year ago
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    I mean how can I express something I don't know ? y(t) is the solution of that equation

  15. anonymous
    • one year ago
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    I forgot that the ICs are considered null.

  16. beginnersmind
    • one year ago
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    It's just an idea but I think it works |dw:1436022843809:dw|

  17. beginnersmind
    • one year ago
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    Assuming all derivates at 0 are zero, I get: |dw:1436023098647:dw|

  18. beginnersmind
    • one year ago
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    Where the integral with u(t) is known.

  19. beginnersmind
    • one year ago
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    Ok, not sure how to make progress from here. Maybe it doesn't work after all :(

  20. anonymous
    • one year ago
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    @ganeshie8 any idea ?

  21. ganeshie8
    • one year ago
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    is the order really 3 ?

  22. anonymous
    • one year ago
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    yup

  23. anonymous
    • one year ago
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    actually I don't need to solve this eq but rather I need to know if my presumption is right that the eq got two distinct solutins depending on u(t); am I right ?

  24. ganeshie8
    • one year ago
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    particular solution changes depending on u(t) but the homogeneous solution remains same

  25. anonymous
    • one year ago
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    well than I will reformulate the problem

  26. anonymous
    • one year ago
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    this eq needs to be solved using numerical analysis method RUNGE-Kutta in MatLab so I must transform this equation in a first degree equations

  27. ganeshie8
    • one year ago
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    try @dan815

  28. anonymous
    • one year ago
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    @dan815 are you familiar with matlab or numerical analysis ?

  29. dan815
    • one year ago
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    first order runge kutta = euler method right

  30. anonymous
    • one year ago
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    My question is if in my script of matlab do i need 2 functions: x1=Ax+5*b and x2=Ax+cos2t*b

  31. dan815
    • one year ago
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    and you have no initial values?

  32. dan815
    • one year ago
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    uh whats that for?

  33. anonymous
    • one year ago
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    Intial conditions are considered null

  34. dan815
    • one year ago
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    okay so first write a bunch of first order ODes

  35. anonymous
    • one year ago
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    That's for transforming the equation in a first order ODE

  36. dan815
    • one year ago
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    |dw:1436025643499:dw|

  37. dan815
    • one year ago
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    |dw:1436025758838:dw|

  38. dan815
    • one year ago
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    |dw:1436025814119:dw|

  39. dan815
    • one year ago
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    solve these in order

  40. dan815
    • one year ago
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    |dw:1436025883462:dw|

  41. anonymous
    • one year ago
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    Yeah right I think we should keep my notation as we are talking about vectors here: |dw:1436025828612:dw|

  42. dan815
    • one year ago
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    k thats looks fine

  43. anonymous
    • one year ago
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    the matrix A that I telled you behind looks like this 0 1 0 0 0 1 -2/3 -1/3 0

  44. dan815
    • one year ago
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    why are you putting it in a matrix

  45. anonymous
    • one year ago
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    x1=Ax+5*b and x2=Ax+cos2t*b |dw:1436026153233:dw|

  46. anonymous
    • one year ago
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    MATRIXlaboratory a.k.a MatLab

  47. dan815
    • one year ago
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    hmm let me think okay

  48. dan815
    • one year ago
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    if you want to solve it all in one matrix then, ud need it in the form like

  49. anonymous
    • one year ago
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    script: function x1=FCS(t,y) x1=A*x+b*5; end function x2=FCS(t,y) x2=A*x+b*cos(2t); end

  50. dan815
    • one year ago
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    |dw:1436026464161:dw|

  51. dan815
    • one year ago
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    |dw:1436026581332:dw|

  52. dan815
    • one year ago
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    im wondering if this is really possible in this case hmm

  53. anonymous
    • one year ago
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    dan I know that I'm just trying to understand my teacher method

  54. dan815
    • one year ago
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    what is his method?

  55. anonymous
    • one year ago
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    |dw:1436026645952:dw|

  56. anonymous
    • one year ago
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    than she makes a function : function xp=FCS(t,y) xp=A*x+b*u(t); end

  57. dan815
    • one year ago
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    so she got a new variable for x3

  58. dan815
    • one year ago
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    so that she doesnt have to worry about the extra vector

  59. anonymous
    • one year ago
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    then she does another script for the y function

  60. dan815
    • one year ago
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    |dw:1436026852588:dw|