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anonymous

  • 5 years ago

Need help with LaPlace transform y''-y=5*sin2t when given y(0)=0, y'(0)=6

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  1. anonymous
    • 5 years ago
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    I've gotten to \[2^2L(y)-6-y=5*\sin2t\]

  2. anonymous
    • 5 years ago
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    but I think I'm stuck... that should say s^2 not 2^2

  3. anonymous
    • 5 years ago
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    \[s^2L(y)-6-y=5*\sin(2t)\]

  4. anonymous
    • 5 years ago
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    do you have the value of y'(0)?

  5. anonymous
    • 5 years ago
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    yep y'(0)=6

  6. anonymous
    • 5 years ago
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    I messed that all up.... it should be\[s^2L(y)-6-L(y)=10/s^2+4\]

  7. anonymous
    • 5 years ago
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    yeah this is right

  8. anonymous
    • 5 years ago
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    Ok so what do I do from there?

  9. anonymous
    • 5 years ago
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    try to solve for L(y)

  10. anonymous
    • 5 years ago
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    I thikn I would go \[s^2(L(y)-1)-6=10/s^2+4\]

  11. anonymous
    • 5 years ago
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    err

  12. anonymous
    • 5 years ago
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    ok I prefer to write L(y) as Y(s)

  13. anonymous
    • 5 years ago
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    \[L(y)(s^2-1)-6=10/s^2+4\]

  14. anonymous
    • 5 years ago
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    it should be : \[Y(s)(s^2-1)={10 \over s^2+4} +6\]

  15. anonymous
    • 5 years ago
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    then you divide each side by s^2-1 and then look up the transform on a table?

  16. anonymous
    • 5 years ago
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    =>6s^2+34 = Y{s^2 - 1}

  17. anonymous
    • 5 years ago
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    Im confused on where you got that thinker

  18. anonymous
    • 5 years ago
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    take the LCM :)

  19. anonymous
    • 5 years ago
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    = \[Y(s)={6s^2+34 \over s^2+4} . {1 \over s^2 -1}\]

  20. anonymous
    • 5 years ago
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    sry :) missed the denominator :)

  21. anonymous
    • 5 years ago
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    are you follwoing Scotty?

  22. anonymous
    • 5 years ago
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    now i guess u may use partial fractions to solve further

  23. anonymous
    • 5 years ago
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    1 sec I think so im writing it out

  24. anonymous
    • 5 years ago
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    take your time.. I will be right back

  25. anonymous
    • 5 years ago
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    try a way to write it back in time domain

  26. anonymous
    • 5 years ago
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    Yep I got it, you were adding 6 into there.

  27. anonymous
    • 5 years ago
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    Can I look up the laplace transform inverse for each and then combine them?

  28. anonymous
    • 5 years ago
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    i guess its possible when there's addition there..here we have multiplication of terms...can u use partial fractions?

  29. anonymous
    • 5 years ago
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    Forgot how to do those, you're talking about breaking it up right?

  30. anonymous
    • 5 years ago
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    take , say, (A/s^ + 4)+(b/s^2 -1)

  31. anonymous
    • 5 years ago
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    sry it is (A/s^2 + 4)+(b/s^2 -1)

  32. anonymous
    • 5 years ago
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    mm ok I got it now. Thanks guys... I'm sure I'm gonna post another one here in a min lol

  33. anonymous
    • 5 years ago
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    I am back

  34. anonymous
    • 5 years ago
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    I think there is something wrong

  35. anonymous
    • 5 years ago
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    no I just ran what we had through my solver and it came up with the correct answer...

  36. anonymous
    • 5 years ago
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    I mean the partial fractions

  37. anonymous
    • 5 years ago
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    whats that?? u know it was 2 years back i studies Laplace transforms :):)

  38. anonymous
    • 5 years ago
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    :)

  39. anonymous
    • 5 years ago
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    Oh well the laplace transform before you get to the fractions are correct :)

  40. anonymous
    • 5 years ago
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    ^^ yeah I know :)

  41. anonymous
    • 5 years ago
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    omg!! did i forget solving these?? lol

  42. anonymous
    • 5 years ago
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    if that's all you want, then it's ok.. but if you want to fin the final solution for the differential equation by yourself, then you need to do the partial fractions

  43. anonymous
    • 5 years ago
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    right

  44. anonymous
    • 5 years ago
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    the point is that you want to write the expression we have for Y(s) in a form that is easy to be transformed back to time domain

  45. anonymous
    • 5 years ago
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    I didn't learn the notation in form of Y(s) which seems to be the way most people learn it. Very annoying.

  46. anonymous
    • 5 years ago
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    was my suggestion of no help ??

  47. anonymous
    • 5 years ago
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    No you helped thinker. Thank you :)

  48. anonymous
    • 5 years ago
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    it doesn't matter what notation you use as long as you understand it.

  49. anonymous
    • 5 years ago
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    thnx :) @scott

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