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dfresenius

  • one year ago

xy' + y = −2xy^2 Bernoulli Differential equation.

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  1. dfresenius
    • one year ago
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    Yes, this is a bit strange because they want me to substitute u=y^(1-n)

  2. dfresenius
    • one year ago
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    so, im thinking to start off, i will get x and y on one side.

  3. dfresenius
    • one year ago
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    |dw:1362709077091:dw|

  4. dfresenius
    • one year ago
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    i suppose i should state what bernoulli is

  5. zepdrix
    • one year ago
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    \[\large xy' + y = −2xy^2\]Hmm it's been a lil while.. I'm trying to remember how to do these. I think you want to get the equation to this form. ~Divide both sides by x, Divide both sides by y^2. \[\large y^{-2}y' + \frac{1}{x}y^{-1} = −2\] Then let \(\large u=y^{-1}\). Hmm \(\large u'=-y^{-2}\)

  6. zepdrix
    • one year ago
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    Woops, \(\large u'=-y^{-2}y'\)

  7. dfresenius
    • one year ago
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    |dw:1362709205198:dw|

  8. abb0t
    • one year ago
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    Haven't seen bernoulli in some time.

  9. dfresenius
    • one year ago
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    ya im a bit shaky on this but im caught up to you zep, so the u'=y'y^-2 , can you explain that?

  10. abb0t
    • one year ago
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    u' = y^-2y' is what your going to substitute. essentially, you're going to get a linear differential euqation.

  11. abb0t
    • one year ago
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    You should get something like: u' + p(x)u=g(x)

  12. abb0t
    • one year ago
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    after you substitute everything.

  13. zepdrix
    • one year ago
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    \[\large y'+p(x)y=q(x)y^n\] Dividing both sides by y^n, \(\large \color{royalblue}{y^{-n}y'}+p(x)\color{orangered}{y^{1-n}}=q(x)\) Making the sub \(\large \color{orangered}{u=y^{1-n}}\) \(\large u'=(1-n)y^{-n}y' \qquad \rightarrow \qquad \color{royalblue}{\dfrac{1}{1-n}u'=y^{-n}y'}\) Giving us, yah something like what abbot said :)

  14. zepdrix
    • one year ago
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    I tried to use colors to make that clearer, but it's still rather confusing lol

  15. dfresenius
    • one year ago
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    ya, im just puzzled

  16. dfresenius
    • one year ago
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    so, with my equation,

  17. zepdrix
    • one year ago
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    \[\large xy' + y = −2xy^2\]Dividing each side by x gives us,\[\large y' + \frac{1}{x}y = −2y^2\]Dividing each side by y^2 (Because that is our y^n is this case) gives,\[\large y^{-2}y' + \frac{1}{x}\color{royalblue}{y^{1-2}} = −2\]Then we want to make that blue term our \(\large u\).

  18. dfresenius
    • one year ago
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    |dw:1362709954712:dw|

  19. zepdrix
    • one year ago
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    Our \(\large n\) in this case is 2, because our y^n was the term on the RIGHT side of the equation. \(\large u=y^{1-n} \qquad \rightarrow \qquad u=y^{2-1}\qquad \rightarrow \qquad u=y^{-1}\)

  20. dfresenius
    • one year ago
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    ok that makes sense

  21. dfresenius
    • one year ago
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    |dw:1362710284245:dw|

  22. zepdrix
    • one year ago
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    I think maybe you missed a negative when taking the derivative of your u. \(\huge u=y^{-1} \qquad \rightarrow \qquad u'=\color{red}{-}y^{-2}y'\)

  23. dfresenius
    • one year ago
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    oh the derivative gotcha

  24. zepdrix
    • one year ago
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    I assume we're taking the derivative with respect to \(\large x\). That's probably why we get a prime term on each side :d

  25. dfresenius
    • one year ago
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    |dw:1362710497399:dw|

  26. zepdrix
    • one year ago
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    Yes good! :)

  27. zepdrix
    • one year ago
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    Remember that before you can get an integrating factor and all that fun stuff, your leading term needs to be positive.

  28. dfresenius
    • one year ago
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    ok

  29. dfresenius
    • one year ago
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    should i just switch sides?

  30. zepdrix
    • one year ago
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    Or maybe a better way to say that would be, it needs to be in standard form* no coefficients or anything on the leading term.

  31. zepdrix
    • one year ago
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    Just multiply everything by -1 :)

  32. dfresenius
    • one year ago
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    |dw:1362710598175:dw|

  33. zepdrix
    • one year ago
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    Woops, standard form looks like this,\[\large u'+P(x)u=Q(x)\]

  34. zepdrix
    • one year ago
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    You had it in your middle step.

  35. dfresenius
    • one year ago
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    |dw:1362710722310:dw|

  36. zepdrix
    • one year ago
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    heh XD

  37. dfresenius
    • one year ago
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    ok so, i integrate both sides?

  38. zepdrix
    • one year ago
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    No you need to find an integrating factor \(\large \mu=e^{\int P(x)\;dx}\) Look familiar I hope? :)

  39. dfresenius
    • one year ago
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    ya let me check real quick

  40. dfresenius
    • one year ago
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    |dw:1362710969094:dw|

  41. zepdrix
    • one year ago
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    Woops the negative sign is also part of your \(\large P(x)\). Recall that the standard form shows addition in the middle there.

  42. dfresenius
    • one year ago
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    dangit, i need to stop rushing

  43. zepdrix
    • one year ago
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    And your equation should be \(\large u'-\dfrac{1}{x}u=2\) Positive 2 on the right. I think we missed that earlier.

  44. zepdrix
    • one year ago
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    Yes you do need to stop rushing Chris Farley!! Slow down in life! Or you'll end up leaving us too soon......... :'C

  45. dfresenius
    • one year ago
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    haha nice

  46. dfresenius
    • one year ago
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    |dw:1362711281571:dw|

  47. zepdrix
    • one year ago
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    \[\huge e^{-\ln x} \qquad = \qquad e^{\ln (x^{-1})} \qquad = \qquad ?\]

  48. dfresenius
    • one year ago
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    1/x

  49. zepdrix
    • one year ago
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    Cool c:

  50. dfresenius
    • one year ago
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    my bad

  51. dfresenius
    • one year ago
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    ok so now that i have that, i multiply by x^-1

  52. zepdrix
    • one year ago
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    Yes, if you want to be thorough, then multiply through by your integrating factor. I usually skip that step. But it might be good not to skip it since you're learning it still.

  53. dfresenius
    • one year ago
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    ok,

  54. dfresenius
    • one year ago
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    |dw:1362711568703:dw|

  55. dfresenius
    • one year ago
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    now, i want to isolate u' right?

  56. zepdrix
    • one year ago
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    Looks good so far c: BTW, just a little note: You can scale the size of the drawing box if you don't want to take up that much space.|dw:1362711683197:dw|

  57. zepdrix
    • one year ago
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    From here, you want to recognize that the LEFT side is the result of the product rule.

  58. zepdrix
    • one year ago
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    \[\large \left[u(x)f(x)\right]' \qquad = \qquad u'(x)f(x)+u(x)f'(x)\] We do that have in this case don't we? Where our f(x)=x^{-1}

  59. dfresenius
    • one year ago
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    yes, so I can simplify the left

  60. zepdrix
    • one year ago
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    ya sounds good c:

  61. dfresenius
    • one year ago
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    |dw:1362711903145:dw|

  62. zepdrix
    • one year ago
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    Oh sorry I forget you were calling it \(\large I\). I hope I didn't make that more confusing c:

  63. zepdrix
    • one year ago
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    yah everything looks good so far.

  64. dfresenius
    • one year ago
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    |dw:1362712182784:dw|

  65. dfresenius
    • one year ago
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    Cx

  66. zepdrix
    • one year ago
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    Yes good. Then we have to do a little work changing our \(\large u\) back to \(\large y\)

  67. dfresenius
    • one year ago
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    ok, u=y^-1

  68. zepdrix
    • one year ago
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    Good.\[\large \frac{1}{y}=2x \ln x+Cx\]So how do we solve for y?

  69. dfresenius
    • one year ago
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    |dw:1362712446678:dw|

  70. zepdrix
    • one year ago
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    Yay good job! I threw it into Wolfram just a moment ago to make sure we didn't make any fatal errors. http://www.wolframalpha.com/input/?i=xy%27%2By%3D-2xy%5E2 Same result! Yay! :)

  71. dfresenius
    • one year ago
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    YESSS

  72. dfresenius
    • one year ago
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    thank you so much for your patience, i have been doing very well in my class because of your help

  73. zepdrix
    • one year ago
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    Oh that's good to hear! c: I'm sure it's not just me. You seem to understand these concepts very very quickly.

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