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anonymous

  • one year ago

Solution of y'+6y+3x=8e^t for y(0)=-1, x(0)=0 using Laplace transofrms

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  1. Jhannybean
    • one year ago
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    @Empty

  2. anonymous
    • one year ago
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    I am stuck when im at this stage: \[Y(s)(s+6)+3X(s)=\frac{ 8-s(s-1) }{ s-1 }\]

  3. anonymous
    • one year ago
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    i can't seem to use the initial condition of x(0)=0... i'm not sure if theres a unique solution to this question

  4. IrishBoy123
    • one year ago
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    you've got 2 different variables in there? y and x so y = x', yes?

  5. anonymous
    • one year ago
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    how does y=x'?

  6. anonymous
    • one year ago
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    they are just both functions of time

  7. IrishBoy123
    • one year ago
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    what i am saying is that this would make more sense if that were true but i do not have sight of the original question to confirm if they are independent functions, what are you solving for. don't you need another equation in x and y?

  8. anonymous
    • one year ago
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    Ive just never come across two independent variables in a Laplace transform before...

  9. ganeshie8
    • one year ago
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    there is no unique solution as we don't know any information about the function \(x(t)\)

  10. ganeshie8
    • one year ago
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    if you have another equation, then we can solve the system by elimination and get an unique solution \((x(t),y(t))\)

  11. anonymous
    • one year ago
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    actually, let me screen shot the probem, perhaps part a and b might be related. Give us a tick

  12. ganeshie8
    • one year ago
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    basically it is like solving an equation with many unknowns, there always exist infinitely many solutions when you have more variables but less equations

  13. anonymous
    • one year ago
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  14. ganeshie8
    • one year ago
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    are you doing control theory ?

  15. anonymous
    • one year ago
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    yep process control

  16. IrishBoy123
    • one year ago
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    you solve those as a pair laplace then normal algebra

  17. ganeshie8
    • one year ago
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    pretty sure the equations in "a" and "b" form the system

  18. anonymous
    • one year ago
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    ah, i thought they would be two seperate problems, but since the first didn't make sense, i was querying things.

  19. ganeshie8
    • one year ago
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    |dw:1440405030249:dw|

  20. ganeshie8
    • one year ago
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    i might be wrong though as idk abc of process control

  21. anonymous
    • one year ago
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    right, so when i've formed one equation, i get 4y'+12x=20e^t

  22. anonymous
    • one year ago
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    is that the way to go?

  23. Empty
    • one year ago
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    It seems awkward to put the ICs with the first equation like that

  24. ganeshie8
    • one year ago
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    but if those two equations form a single system, then you may use below method to solve it using laplace transform http://tutorial.math.lamar.edu/Classes/DE/SystemsLaplace.aspx

  25. Empty
    • one year ago
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    Still if this is a "system" then where's y'?

  26. anonymous
    • one year ago
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    yea true

  27. Empty
    • one year ago
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    Something doesn't seem right here haha

  28. anonymous
    • one year ago
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    i'll give that a miss then haha

  29. anonymous
    • one year ago
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    i'll follow the links process, and see if i get an answer.

  30. ganeshie8
    • one year ago
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    no wait, it cannot be a system because you cannot have x' = something and x' = something else

  31. anonymous
    • one year ago
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    perhaps the initial conditions are misspelt?

  32. anonymous
    • one year ago
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    in the mean time, i have another question that i think i've solved but just need someone to check it. \[\int\limits_{0}^{t}y(t)dt=\frac{ dy(t) }{ dt }\] with initial conditions y(0)=1

  33. anonymous
    • one year ago
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    i think its pretty straight forward, my answer was y(t)=cosh(t)

  34. anonymous
    • one year ago
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    my approach was simply taking the derivative of the integral

  35. anonymous
    • one year ago
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    \[y''-y=0\]

  36. anonymous
    • one year ago
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    if that approach is right, then im pretty confident. and well the derivative of cosh(x) is sinh(x) as well...hence that confirms

  37. ganeshie8
    • one year ago
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    simply plugin y=cosh(t) and check ?

  38. Empty
    • one year ago
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    Yeah I guess there are a couple different solutions you could have here though for the most general cases (second order equation needs 2 constants to be most general) \[y=A \sinh t + B \cosh t\] or \[y = A e^t + Be^{-t}\]

  39. anonymous
    • one year ago
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    but we are using laplace transforms here

  40. Empty
    • one year ago
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    Oh right I forgot about your initial conditions

  41. anonymous
    • one year ago
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    haha no problem

  42. ganeshie8
    • one year ago
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    y(0)=1 y'(0) = 0 y''-y=0

  43. anonymous
    • one year ago
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    yeah, i'm assuming you have to say that y'(0)=0...

  44. ganeshie8
    • one year ago
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    that follows trivially from the given equation : \[\int\limits_{0}^{t}y(t)dt=\frac{ dy(t) }{ dt } = y'(t)\] plugin \(t=0\)

  45. anonymous
    • one year ago
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    yeah fair call

  46. anonymous
    • one year ago
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    your simply integrating from 0 to 0

  47. ganeshie8
    • one year ago
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    right thats easy.. im still clueless about ur main question on the equation with two variables.. . gtg wil check in the night

  48. Empty
    • one year ago
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    I thought, what if I subtract the second from the first and you get: \[x+y=e^t\] Then you can see this is wrong because plugging in the intitial conditions will give: \[-1=1\] So weird hahaha

  49. anonymous
    • one year ago
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    yeah i'm baffled too...

  50. IrishBoy123
    • one year ago
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    if it helps \(x = \frac{5}{4}e^t - \frac{9}{4}e^{-3t}\) \(y = -\frac{1}{4}e^t + \frac{9}{4}e^{-3t}\)

  51. anonymous
    • one year ago
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    hmm, how did you get to that?

  52. anonymous
    • one year ago
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    perhaps i'll give it ago now, ive just finished writing the other minor problem down neatly

  53. IrishBoy123
    • one year ago
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    you'll kick yourself when you see this Example 4 at bottom of page

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  54. anonymous
    • one year ago
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    which example am i meant to be looking at?

  55. IrishBoy123
    • one year ago
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    example 4 shows the method here are my scribbles there is indeed a snag in this question. the iv for y on P3 of my scribbles is not met but the same iv is not used in the solution i would love to understand why this as the solution just feels right but has this issue is it a problem with the question or have i messed up some stupid algebra? i really hope this helps rather than hinders

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  56. anonymous
    • one year ago
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    yeah right, thats very interesting. its just the IV's that don't check hey

  57. anonymous
    • one year ago
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    @ganeshie8 i've tried this problem myself and it seems that there is a snap with the initial conditions, what do you think?

  58. anonymous
    • one year ago
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    snag*

  59. anonymous
    • one year ago
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    @Irishboy123 it seems that you don't even need to use the initial condition that y(0)=0 to solve the problem... so perhaps this is misleading information?

  60. IrishBoy123
    • one year ago
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    hi @chris00 i'm baffled. would love to know how you eventually come out on this

  61. ikram002p
    • one year ago
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    maybe since its defined by default in some books to apply Laplace on Certain conditions , u might check =)

  62. anonymous
    • one year ago
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    i'll let you guys know later on in the week when i have a tute on this. cheers guys

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