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Loser66

  • one year ago

Question in comment Please, help

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  1. Loser66
    • one year ago
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    1) Find the equation for the periodic function and its period|dw:1436381101289:dw|

  2. Loser66
    • one year ago
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    2) find the Laplace transform of it

  3. Loser66
    • one year ago
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    3) find the inverse laplace transform \(L^{-1}(\dfrac{k}{as+b})\) where k, a, b in R/{0} and are constants

  4. dan815
    • one year ago
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    they are giving you the constants on your continous power series, u have to see what that gives you in t domain

  5. Loser66
    • one year ago
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    dan, how to know whether a differential equation linear or not?

  6. dan815
    • one year ago
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    no xy no y^2 and such

  7. Loser66
    • one year ago
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    why not? since the linear differential equation standard form is \(y^{(m)} +P_1(x) y^{(m-1)}+..... = Q(x)\) Like y" +y' +xy =0. ??

  8. dan815
    • one year ago
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    no xy^2

  9. dan815
    • one year ago
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    that equation is linear

  10. dan815
    • one year ago
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    i was confused with linear form there

  11. dan815
    • one year ago
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    by the way the question you asked earlier, what did u get for this 1) Find the equation for the periodic function and its period

  12. Loser66
    • one year ago
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    I didn't.

  13. dan815
    • one year ago
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    wheres your work so farq

  14. Loser66
    • one year ago
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    dan, remember the previous post ? http://math.stackexchange.com/questions/1354551/finding-differential-equation-from-the-solution

  15. Loser66
    • one year ago
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    Wow!!! they derive the problem into other way!! but it works also.

  16. Loser66
    • one year ago
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    dan, honestly, I don't know how to work on it.

  17. Loser66
    • one year ago
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    Since the slopes of the lines are equal, we have y'= 1 , that gives us y= x +a. But it doesn't make sense since y = x +a are family of vertical lines, not horizontal like this.

  18. Loser66
    • one year ago
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    what I don't understand is here|dw:1436404165210:dw|

  19. ybarrap
    • one year ago
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    What about starting with $$ x(t) = t - \lfloor t \rfloor = t - \operatorname{floor}(t) $$ Then scaling? http://www.wolframalpha.com/input/?i=t-floor%28t%29&dataset=

  20. ybarrap
    • one year ago
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    Then Laplace - http://www.wolframalpha.com/input/?i=lapace+transform+%28t-floor%28t%29%29

  21. Loser66
    • one year ago
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    how about initial value?|dw:1436405123893:dw|

  22. ybarrap
    • one year ago
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    What about multiplying everything by u(t), the step function: |dw:1436405297232:dw|

  23. Loser66
    • one year ago
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    Wow!! I am dizzy!! this is the beginning level of DE. The students cannot know that far!!

  24. ybarrap
    • one year ago
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    If you agree that this might work, then the Laplace would be http://www.wolframalpha.com/input/?i=lapace+transform+theta%28x%29%28t-floor%28t%29%29

  25. Loser66
    • one year ago
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    Thanks friend. I need time to digest it. :)

  26. ybarrap
    • one year ago
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    Ok - just some ideas, not sure if it will work

  27. ybarrap
    • one year ago
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    BTW, the period becomes clear if you use the function above, it's 1 second -- which you can scale by \(a\).

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