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anonymous

  • 5 years ago

How would I use the Bernoulli Equation to solve t^2dy/dt+2ty-y^3=0, t>0?

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  1. anonymous
    • 5 years ago
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    The basic idea is to move y^3 to the other side. Then you might recognize Left Hand Side is just the derivative of y * t^2 with respect to t. Then you can integrate both sides with respect to t. On the Left Hand Side, you get y * t^2. On the Right Hand Side, you get t*y^3 + C. Then the solution will be curves y(t) that satisfy y*t^2 – t*y^3 = C for some arbitrary C. duongni asked that about 2 hrs ago. I responded to it about half an hour ago. If you want the details, you can scroll down there.

  2. anonymous
    • 5 years ago
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    Thank you very, very much, verifry. You're answer to duongi's curiously identical question was immensely helpful.

  3. anonymous
    • 5 years ago
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    I'm having a really hard time with the check on this one. would you be willing to walk me through it?

  4. anonymous
    • 5 years ago
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    You mean checking the solution to see if it solves the original equation?

  5. anonymous
    • 5 years ago
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    yeah

  6. anonymous
    • 5 years ago
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    yeah

  7. anonymous
    • 5 years ago
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    I keep ending up with a -3ty^2dy/dt that doesn't fit

  8. anonymous
    • 5 years ago
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    would you be willing to walk me through that part as well?

  9. anonymous
    • 5 years ago
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    You know what, I think what I wrote isn't the solution anymore. I'm sorry. I thought it checked out, but you're right when you say it doesn't fit.

  10. anonymous
    • 5 years ago
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    The problem is that when we integrated y^3, we shouldn't just get ty^3 + C, since y isn't a constant, it's actually a function of t.

  11. anonymous
    • 5 years ago
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    How should it look then?

  12. anonymous
    • 5 years ago
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    Let me work it out and check it this time before typing anything.

  13. anonymous
    • 5 years ago
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    cool beans. Thank you again for sticking with me through all of this. I really appreciate the help!

  14. anonymous
    • 5 years ago
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    any lucl?

  15. anonymous
    • 5 years ago
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    *luck, even

  16. anonymous
    • 5 years ago
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    I'm trying to check it right now, but the check is not pretty. The basic method of solving it is better explained here: http://tutorial.math.lamar.edu/Classes/DE/Bernoulli.aspx The idea is to convert it down to an equation in which we can apply the method of integrating factors without having a function of y on the right hand side.

  17. anonymous
    • 5 years ago
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    The reason the method of integrating factors didn't work above in my first explanation was because the Right Hand Side had a function of y, so when we integrated it, we can't solve the integral explicitly.

  18. anonymous
    • 5 years ago
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    that makes sense

  19. anonymous
    • 5 years ago
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    The reason the method of integrating factors didn't work above in my first explanation was because the Right Hand Side had a function of y, so when we integrated it, we can't solve the integral explicitly.

  20. anonymous
    • 5 years ago
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    So if we divide both sides by y^3 first, we can make the Right Side not a function of y anymore. Then we try a substitution to convert the differential equation to one in which it's easier to see how we can use the method of integrating factors.

  21. anonymous
    • 5 years ago
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    would you be willing to walk me through that?

  22. anonymous
    • 5 years ago
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    Sure.

  23. anonymous
    • 5 years ago
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    Starting from t^2 * y' + 2ty = y^3, we want to divide by y^3 to get the RHS to not be a function of y.

  24. anonymous
    • 5 years ago
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    So we get t^2 * (y'/y^3) + 2t / y^2 = 1.

  25. anonymous
    • 5 years ago
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    Now we want to make a change of variables so that it's easier to do the method of integrating factors. Maybe we'd like our new variable to be v, and we want v so that v' is some constant multiple of y'/y^3.

  26. anonymous
    • 5 years ago
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    Normally, after solving lots of these problems and generalizing the method, you'd just let v = y^(1-n) because you've seen the pattern. But if you haven't seen the pattern, then we might as well just let v' be equal to y'/y^3

  27. anonymous
    • 5 years ago
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    Why can't we use v=y^(1-n)?

  28. anonymous
    • 5 years ago
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    We can. We probably should to make it simplify. I was just thinking that if this was the first time you've seen this, letting v = y^(1-n) might seem to pop out of nowhere.

  29. anonymous
    • 5 years ago
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    I sort of read through the website explanation, but I just don't pick stuff up very well that way. We could use v=y^(1-n) though

  30. anonymous
    • 5 years ago
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    Yeah let's let v = y^(1-n). Here our n = 3, so let's let v = y^(-2). So then dv = (-2) y^(-3) dy. So dv/dt = (-2) y^(-3) dy/dt.

  31. anonymous
    • 5 years ago
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    If you aren't sure of the answer you should visit aplha math help its on google, and shows you the exact answer!

  32. anonymous
    • 5 years ago
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    Wow, alpha math help is pretty cool. I didn't know about that either.

  33. anonymous
    • 5 years ago
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    what's the url?

  34. anonymous
    • 5 years ago
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    http://blog.wolframalpha.com/2009/12/01/step-by-step-math/ But I don't know if it can do differential equations yet.

  35. anonymous
    • 5 years ago
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    fair enough

  36. anonymous
    • 5 years ago
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    would it be alright if we kept working on my question?

  37. anonymous
    • 5 years ago
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    yeah. From what we had about v' = -2y^(-3) y', we see that (-1/2)v' = y^(-3) * y'

  38. anonymous
    • 5 years ago
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    Now we can rewrite our original equation in terms of v's.

  39. anonymous
    • 5 years ago
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    We get (-1/2) t^2 * v' + 2t * v = 1.

  40. anonymous
    • 5 years ago
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    if you are rewriting the equation in terms of v's wouldn't it be v=1 and then the orginal equation....you are intentfying the v as a variable given already.

  41. anonymous
    • 5 years ago
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    i meant the orginal answer!

  42. anonymous
    • 5 years ago
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    I don't think you totally understand the question, ena, but thank you for trying to help.

  43. anonymous
    • 5 years ago
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    Now we want to get rid of the coefficient on v' to be able to use the method of integrating factors easily.

  44. anonymous
    • 5 years ago
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    Omnideus, do you know the method of integrating factors?

  45. anonymous
    • 5 years ago
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    Kinda. I missed one of the lectures on it, so I am a little behind.

  46. anonymous
    • 5 years ago
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    It's basically what I tried to do when I first responded to your question. Here it's a little harder since the integrating factor isn't as obvious.

  47. anonymous
    • 5 years ago
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    So we have v' - (4/t) v = -2/t^2 after multiplying both sides by (-2/t^2) to get rid of the coefficient on v'.

  48. anonymous
    • 5 years ago
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    this is starting to look a little more approachable

  49. anonymous
    • 5 years ago
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    Good. Haha.

  50. anonymous
    • 5 years ago
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    In the method of integrating factors, we want to express the left hand side as the derivative of the product of some function phi with the function v.

  51. anonymous
    • 5 years ago
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    Because the derivative of phi * v is phi * v' + phi' * v

  52. anonymous
    • 5 years ago
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    Let me call the function phi, let me call it Y. Then we want to find some Y such that when we multiply Y to the Left Hand Side and get Yv' - (4/t) Y*v, we want that to equal Y v' + Y' v.

  53. anonymous
    • 5 years ago
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    So what we really want is some function Y such that -(4/t)Y = Y'

  54. anonymous
    • 5 years ago
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    Solving this, we get Y = t^(-4), and this is the integrating factor.

  55. anonymous
    • 5 years ago
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    Multiplying both sides of v' - (4/t) v = -2/t^2 by t^(-4), we see that the left hand side is the derivative of v*t^(-4), while the right hand side is -2/t^6.

  56. anonymous
    • 5 years ago
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    We're pretty much done with solving for v, and once we solve for v, we've solved for y since v = y^(-2).

  57. anonymous
    • 5 years ago
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    When we have the left hand side is the derivative of v*t^(-4) and the right hand side is -2/t^6, we can integrate both sides with respect to t.

  58. anonymous
    • 5 years ago
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    You think you got it from there?

  59. anonymous
    • 5 years ago
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    Yeah, I got an answer that looks correct. Did you do the check? It looks ugly.

  60. anonymous
    • 5 years ago
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    No, but we can try wolfram alpha.

  61. anonymous
    • 5 years ago
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    kay

  62. anonymous
    • 5 years ago
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    I'm not really that tech savvy. Should I go to the site that other guy posted and you gave me the URL to?

  63. anonymous
    • 5 years ago
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    http://www.wolframalpha.com/input/?i=t^2+*+y%27+%2B+2ty+%3D+y^3

  64. anonymous
    • 5 years ago
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    You might have written it a bit differently. http://www.wolframalpha.com/input/?i=v%27+-+%284%2Ft%29+v+%3D+-2%2Ft^2 That's the solution to our equation for v. Then v = y^(-2) gives our solution in terms of y.

  65. anonymous
    • 5 years ago
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    You're a fluttering godsend, man. I would have struggled with that all night long and not made any progress.

  66. anonymous
    • 5 years ago
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    I can't thank you enough

  67. anonymous
    • 5 years ago
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    You're welcome. Thanks for letting me know I made a mistake. I'm gonna tell duongni now.

  68. anonymous
    • 5 years ago
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    Good luck with your hw

  69. anonymous
    • 5 years ago
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    haha, I hope duongni isn't in bed by now. THANKS A MILLION!

  70. anonymous
    • 5 years ago
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    Other guy? hehee I am a girl! if you are refering to me for giving you the website.

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