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johnny0929 Group Title

Solve the initial value problem\[u''+6u'+12u=0\]with \(u(0)=1\) and \(u'(0)=0\)

  • one year ago
  • one year ago

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  1. modphysnoob Group Title
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    use characteristic equation a^2 + 6a+12 solve for a

    • one year ago
  2. abb0t Group Title
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    this is a second order homogeneous differential equation. do you know the general solution? \[y_p = e^{rt}c_1+e^{rt}c_2\]

    • one year ago
  3. abb0t Group Title
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    @modphysnoob is correct. except he used "a" so your solution would be: \[y_p = e^{at}c_1+e^{-at}c_2\]

    • one year ago
  4. abb0t Group Title
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    solve the quadratic

    • one year ago
  5. johnny0929 Group Title
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    I actually have the answer but I am just a little confused about how to do it... It's a spring mass problem and the answer is \[u=e^{-3t}cos\sqrt{3}t+\sqrt{3}e^{-3t}sin\sqrt{3}t\]

    • one year ago
  6. abb0t Group Title
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    then, plug in the initial conditions. take a few derivatives and violê

    • one year ago
  7. abb0t Group Title
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    solve for u(0)=1 take the derivative of your particular solution and plug in u(0)=0

    • one year ago
  8. johnny0929 Group Title
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    ok. when I factor \(a^2+6a+12=0\) I get \((-i a+\sqrt{3}-3 i) (i a+\sqrt{3}+3 i)\)

    • one year ago
  9. modphysnoob Group Title
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    that mean it is sinosoid

    • one year ago
  10. johnny0929 Group Title
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    which means it's \((\sqrt{3}\pm(3+a)i)=0\) how do i make it look like the general solution?

    • one year ago
  11. johnny0929 Group Title
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    can you help? because I'm so confused

    • one year ago
  12. modphysnoob Group Title
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    hold on, my computer is apparently running on steam engine

    • one year ago
  13. modphysnoob Group Title
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    a=-3 (+-) i* sqrt(3) which is \[Ae^{(-3+ i \sqrt{3})t}+Be^{(-3- i \sqrt{3})t}\]

    • one year ago
  14. modphysnoob Group Title
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    use euler identity and you will get A sin( ) + B cos ()

    • one year ago
  15. johnny0929 Group Title
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    now i can see it's \[Ae^{-3}sin(\sqrt{3}t)+Be^{-3}cos(\sqrt{3}t)\]then all i have to do is plug in the initial values, right?

    • one year ago
  16. modphysnoob Group Title
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    you are missing t e^(-3t)

    • one year ago
  17. johnny0929 Group Title
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    oh yeah. ok there's a t.

    • one year ago
  18. johnny0929 Group Title
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    \[u=e^{−3t}cos\sqrt{3}t+\sqrt{3}e^{−3t}sin\sqrt{3}t\]one more question. how do i determine if this equation is over damped, underdamped, or critically damped?

    • one year ago
  19. modphysnoob Group Title
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    so sin() and cos(x) just go up and down at t go on

    • one year ago
  20. modphysnoob Group Title
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    |dw:1362965561866:dw|

    • one year ago
  21. modphysnoob Group Title
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    while e(-3t) |dw:1362965602529:dw|

    • one year ago
  22. modphysnoob Group Title
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    just read it http://en.wikipedia.org/wiki/Damping#Critical_damping_.28.CE.B6_.3D_1.29 it will explain much better than I ever can

    • one year ago
  23. johnny0929 Group Title
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    oh ok i am reading from my book that all i need to do is compare the coefficient in front of \(u'\) to \(\sqrt{4ac}\) from the equation \(a^2+6a+12=0\)

    • one year ago
  24. modphysnoob Group Title
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    crticle damp goes to zero without oscillating under dampm oscillate

    • one year ago
  25. johnny0929 Group Title
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    thank you!

    • one year ago
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