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anonymous

  • 5 years ago

how would i solve (x double dot) + x(x dot)+x=0

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  1. Owlfred
    • 5 years ago
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    Hoot! You just asked your first question! Hang tight while I find people to answer it for you. You can thank people who give you good answers by clicking the 'Good Answer' button on the right!

  2. anonymous
    • 5 years ago
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    help

  3. anonymous
    • 5 years ago
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    I think you can solve it by reduction of order

  4. anonymous
    • 5 years ago
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    how?

  5. anonymous
    • 5 years ago
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    you deleted that? it wasnt the same

  6. anonymous
    • 5 years ago
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    yea, i had it wrong

  7. anonymous
    • 5 years ago
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    that one can be solved differently

  8. anonymous
    • 5 years ago
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    i thought reduction of order was for coefficients which are random polynomials of in t

  9. anonymous
    • 5 years ago
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    \[x \prime \prime+xx \prime+x=0\]

  10. anonymous
    • 5 years ago
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    for this you say let u=dx/dt than d^2x/dt^2=u du/dt

  11. anonymous
    • 5 years ago
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    than the equation will be u du/dt + xu+x=0

  12. anonymous
    • 5 years ago
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    ahh, that is clever, i will try that

  13. anonymous
    • 5 years ago
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    thanks a ton

  14. anonymous
    • 5 years ago
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    and I guess you can use an integrating factor to solve this

  15. anonymous
    • 5 years ago
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    now I have to go to the shop. If you have a problem just post it

  16. anonymous
    • 5 years ago
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    sweet, thanks a ton you are amazing

  17. anonymous
    • 5 years ago
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    that integral doesn't make sense, it is with respect to the wrong variable

  18. anonymous
    • 5 years ago
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    you have to change u back to x

  19. anonymous
    • 5 years ago
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    I did not solve it, what did you get?

  20. anonymous
    • 5 years ago
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    ah, i'm saying even setting up the integration factor

  21. anonymous
    • 5 years ago
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    dot means that we differentiate with respect to t

  22. anonymous
    • 5 years ago
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    x is a function of t, but the integral should be with respect to t

  23. anonymous
    • 5 years ago
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    yea, i set up \[u \prime+ux+x=0\]

  24. anonymous
    • 5 years ago
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    u du/dt + xu+x=0 du/dt +u/x=-x

  25. anonymous
    • 5 years ago
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    u'=u du/dt by the chain rule

  26. anonymous
    • 5 years ago
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    e^integral1/x is the integrating factor that is just x

  27. anonymous
    • 5 years ago
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    so multiplying with x gives xdu/dt+u=-x^2

  28. anonymous
    • 5 years ago
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    (xu)dot=-x^2

  29. anonymous
    • 5 years ago
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    xu=-(x^3)/3 +C

  30. anonymous
    • 5 years ago
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    u=-(x^2)/3 +c/x

  31. anonymous
    • 5 years ago
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    u=dx/dt

  32. anonymous
    • 5 years ago
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    dx/dt=-(x^2)/3 +c/x

  33. anonymous
    • 5 years ago
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    so t= -(x^3)/9 +Clnx

  34. anonymous
    • 5 years ago
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    but when you set up that equation du/dt +u/x=-x we should be integrating with respect to t

  35. anonymous
    • 5 years ago
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    you are integrating with respect to x

  36. anonymous
    • 5 years ago
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    for the integrating factor no

  37. anonymous
    • 5 years ago
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    it is only for making it a full derivative

  38. anonymous
    • 5 years ago
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    the integrating factor makes sense

  39. anonymous
    • 5 years ago
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    I think it is like this, but I would not bet on it :-)

  40. anonymous
    • 5 years ago
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    the actual integratioin though does not and why is the equation u du/dt rather than du/dt

  41. anonymous
    • 5 years ago
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    u=dx/dt u dot= x double dot

  42. anonymous
    • 5 years ago
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    this part is not 100% clear for me :) but I am sure that is right and comes from the chain rule

  43. anonymous
    • 5 years ago
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    I just posted a question about it, maybe some1 is wise

  44. anonymous
    • 5 years ago
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    (I'm just a 1st year maths student)

  45. anonymous
    • 5 years ago
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    well, thank you for your help, i will check back later. i really appreciate it andras

  46. anonymous
    • 5 years ago
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    your welcome

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