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anonymous

  • one year ago

the laplace tranform of -tcosx=-cosx//s^2 am i right???

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  1. amistre64
    • one year ago
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    doesnt look right to me, might be your choice of notation tho

  2. amistre64
    • one year ago
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    do we agree that: L(y') = sL(y) - yo

  3. anonymous
    • one year ago
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    my question is different

  4. amistre64
    • one year ago
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    is it about bubble gum? because it looks to be about a laplace transform to me ... can you clarify it any?

  5. amistre64
    • one year ago
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    here is how Im reading you question since you do not wish to clarify your responses using the product rule of derivatives L(tg' + g) = sL(tg) - 0go, by linearity of the laplace we get L(tg') + L(g) = sL(tg) let g = -cos(x) g' = sin(x) sL(-tcos(x)) = L(tsin(x)) + L(-cos(x)) s^2 L(-tcos(x)) = sL(tsin(x)) + sL(-cos(x)) using the same construct we can determine the sL(tsin(x)) ---------------- sL(tsin(x)) = L(tcos(x)) + L(sin(x)) s^2 L(-tcos(x)) = L(tcos(x)) + L(sin(x)) + sL(-cos(x)) s^2 L(-tcos(x)) + L(-tcos(x)) = L(sin(x)) + sL(-cos(x)) (s^2+1) L(-tcos(x)) = L(sin(x)) + sL(-cos(x)) L(-tcos(x)) = [L(sin(x)) + sL(-cos(x))]/(s^2+1)

  6. amistre64
    • one year ago
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    your questions is in x and t ... is that a typo? or on purpose?

  7. amistre64
    • one year ago
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    assuming cos(x) is not a typo, then its simply a constant with respect to t

  8. anonymous
    • one year ago
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    sory for bieng late rep but i have some personal problems so thats y.

  9. anonymous
    • one year ago
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    yes cosx is treat constant.

  10. anonymous
    • one year ago
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    then the laplace is -cosx/s^2 is right?

  11. math&ing001
    • one year ago
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    Wolfram says it's right.

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  12. anonymous
    • one year ago
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    wow its amazing.

  13. amistre64
    • one year ago
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    late is better then never :)

  14. math&ing001
    • one year ago
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    You can use the tool from here http://www.wolframalpha.com/widgets/view.jsp?id=3272010f63d3145699ca78bbe0db05a7

  15. anonymous
    • one year ago
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    thank u nice to meet u.

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