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anonymous

  • 5 years ago

Does does d(xy)/xy mean in regards to differential equations?

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  1. anonymous
    • 5 years ago
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    That just means the differential with respect to (xy), like du/u.

  2. anonymous
    • 5 years ago
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    So how would I solve that on paper? Would I take the derivative of xy and then take the integral of the entire thing?

  3. anonymous
    • 5 years ago
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    The whole problem is d(xy)/xy+dy=0

  4. anonymous
    • 5 years ago
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    hang on a sec - need to do something

  5. anonymous
    • 5 years ago
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    You can make it something more palatable for you by expanding the differential. That might be best. So, you'd write,

  6. anonymous
    • 5 years ago
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    \[\frac{d(xy)}{xy}=\frac{ydx+xdy}{xy}=\frac{dx}{x}+\frac{dy}{y}\]

  7. anonymous
    • 5 years ago
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    The equation is \[\frac{dx}{x}+\frac{dy}{y}+dy=0 \rightarrow \frac{dx}{x}=-(\frac{1}{y}+1)dy\]

  8. anonymous
    • 5 years ago
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    You can integrate this now.

  9. anonymous
    • 5 years ago
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    \[\ln x = -(\ln y +y)+c \rightarrow x=e^{-\ln y - y-c}=\frac{1}{y}e^{-y}e^{-c}=\frac{Ce^{-y}}{y}\]

  10. anonymous
    • 5 years ago
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    +c in the exponentiation, but it doesn't matter since it just becomes the constant at the end.

  11. anonymous
    • 5 years ago
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    You didn't do any integrating or differentation on that last part right? Just moved things around algebraically?

  12. anonymous
    • 5 years ago
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    Note, this could have been obtained in a couple of steps from the first method I mentioned, namely,\[\frac{d(xy)}{xy}+dy=0 \rightarrow \ln xy +y = C \rightarrow xy.e^y=C \rightarrow x=\frac{C}{y}e^{-y}\]

  13. anonymous
    • 5 years ago
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    Which last part? Once the 'dx' and 'dy' differentials disappear, there's no more integration.

  14. anonymous
    • 5 years ago
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    Nevermind. I think I see. I basically get the derivative of xy, which is 1. Then integrate 1/xy which is ln|xy|

  15. anonymous
    • 5 years ago
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    Why do you say the derivative of xy is 1?

  16. anonymous
    • 5 years ago
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    Hmm .. doesn't one normally want the answer to a diff eq on the form y=... ? I see you wrote x=...

  17. anonymous
    • 5 years ago
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    If you saw \[\int\limits_{}{}\frac{du}{u}\]you'd know what to do. You'd recognize it as ln u.

  18. anonymous
    • 5 years ago
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    Well, this would be transcendental in y, which is why it wasn't solved explicitly for y.

  19. anonymous
    • 5 years ago
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    Re. the du/u integral, the d(xy)/(xy) thing is in the SAME FORM...so you do to it what you would do in the situation du/u.

  20. anonymous
    • 5 years ago
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    I'm just having trouble figuring out with d(xy)/xy means. Is that just something I memorize du/u=ln u?

  21. anonymous
    • 5 years ago
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    Okay, if I said, "The anti-derivative of the differential of *something*, divided by that *something* is..?" what would you say to me?

  22. anonymous
    • 5 years ago
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    the integral of f(x) / f(x) ?

  23. myininaya
    • 5 years ago
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    whatever we are integrating with repect to

  24. myininaya
    • 5 years ago
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    or im not sure lol

  25. anonymous
    • 5 years ago
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    lol, who's got a headache?

  26. myininaya
    • 5 years ago
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    i just got here and i do

  27. anonymous
    • 5 years ago
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    Me. I've been struggling with this for over 10 hours now. I hate online school.

  28. anonymous
    • 5 years ago
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    In mathematics, it's all about *forms*.

  29. anonymous
    • 5 years ago
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    We always try to reduce more complicated problems into forms we already have solutions to.

  30. anonymous
    • 5 years ago
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    This is what we're doing here.

  31. anonymous
    • 5 years ago
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    We know that, whenever you have the form, \[\int\limits_{}{}\frac{d(something)}{(something)}\]the solution is \[\ln (something) + c\]

  32. anonymous
    • 5 years ago
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    Here the 'something' is (xy).

  33. myininaya
    • 5 years ago
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    oh yes i get the something thing now

  34. anonymous
    • 5 years ago
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    good!

  35. anonymous
    • 5 years ago
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    ok , so its just a form I need to memorize basically

  36. anonymous
    • 5 years ago
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    Yes.

  37. anonymous
    • 5 years ago
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    But even better, understand how we get that form.

  38. anonymous
    • 5 years ago
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    But, look, if it's a pain for you and you stress in an exam, just expand the differential like I did above.

  39. anonymous
    • 5 years ago
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    Unfortunetly thats just the tip of the iceberg in my lack of understanding. I'm in an 8 week online Calculus II class and struggling very badly.

  40. anonymous
    • 5 years ago
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    But thank you for the patience and the help. I think I understand a little bit better now.

  41. anonymous
    • 5 years ago
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    There's heaps of online resources. This site for one. Paul's Online Maths Notes are good, as well as www.khanacademy.org.

  42. anonymous
    • 5 years ago
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    You're welcome.

  43. anonymous
    • 5 years ago
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    I've watched all those videos and looked. They help somewhat, but when I work specific problems I guess recognizing what method to use is where I struggle.

  44. anonymous
    • 5 years ago
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    Yes, that's called 'the problem of fluency' in mathematics. You need to do two things: 1) understand the mathematics and 2) recognize what you're being shown so you can access what you know. Sounds like number 2 is hassling you. That's good, though, because it can be fixed.

  45. myininaya
    • 5 years ago
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    if I could become your fan again lokisan, i would

  46. anonymous
    • 5 years ago
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    Haha I think I'm also having difficulty in the understanding at some aspect as well. The bad part is I have little time to become fluent and understand before I move onto the next subject. I would also fan you 10 times.

  47. anonymous
    • 5 years ago
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    hey, thanks myininaya ;)

  48. anonymous
    • 5 years ago
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    Can I hire you over skype? lol

  49. anonymous
    • 5 years ago
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    lol, just log on here. I lurk around.

  50. anonymous
    • 5 years ago
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    The fluency part is 'easily' fixable because all it requires is doing problems.

  51. anonymous
    • 5 years ago
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    But

  52. anonymous
    • 5 years ago
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    there are two types of problem: (1) closed and (2) open.

  53. anonymous
    • 5 years ago
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    Well I'm doing them. For instance right now I'm trying to apply either seperation of variables or integrating combinations to this problem. I think I know what to do but I'm stuck

  54. anonymous
    • 5 years ago
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    good stuff lokisan....... you couldn't have said it any better!

  55. anonymous
    • 5 years ago
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    Closed problems are the ones you're probably used to, like, if the sides of a rectangle are 7 and 3 units, what's the perimeter? You can work that out easily since it's plug and play. But if someone says, "The perimeter of a rectangle is 20cm. What are it's dimensions?" people come unstuck.

  56. anonymous
    • 5 years ago
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    Cheers nadeem :)

  57. anonymous
    • 5 years ago
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    Hmm ok well in that same topic then... what if you didn't have d(something/(something) and it was just d(something)

  58. anonymous
    • 5 years ago
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    For instance d(xy)=-x/11 dx

  59. anonymous
    • 5 years ago
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    Well, Lokisan got one more fan for solving this differential eq :) I hadn't fanned him before (and is till now the only user here I am a fan of ;)

  60. anonymous
    • 5 years ago
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    thank you mstud.

  61. anonymous
    • 5 years ago
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    re. your question scotty, what would you do if you saw,\[\int\limits_{}{}dx\]?

  62. anonymous
    • 5 years ago
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    My current problem is 11xdy+11ydx+xdx=0

  63. anonymous
    • 5 years ago
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    I know, I'm trying to make it easy on you.

  64. anonymous
    • 5 years ago
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    I can move the xdx over, factor the 11 and then group xdy +ydx to be d(xy)

  65. anonymous
    • 5 years ago
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    I would integrade which would be x

  66. anonymous
    • 5 years ago
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    Just give me a sec. to sort something non-mathematical out.

  67. anonymous
    • 5 years ago
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    No problem

  68. anonymous
    • 5 years ago
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    Okay, back. The reason I asked you about \[\int\limits_{}{} dx\]is because I wanted you to tie this idea of 'forms' from the last situation to this one. You've made life hard for yourself by going the route you did.

  69. anonymous
    • 5 years ago
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    \[\int\limits_{}{}d(xy)=\int\limits_{}{}-\frac{x}{11}dx \rightarrow xy=-\frac{x^2}{22}+c \rightarrow y=-\frac{x}{22}+\frac{c}{x}\]

  70. anonymous
    • 5 years ago
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    There's that 'form' thing going on again in d(xy).

  71. anonymous
    • 5 years ago
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    I think the route I went was the completely wrong way because that isn't the answer lol. According to my book its 22xy+x^2=C

  72. anonymous
    • 5 years ago
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    Yeah, you can rearrange what I gave you.

  73. anonymous
    • 5 years ago
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    11xdy+11ydx+xdx=0 My first instinct is to want to move the xdx to the other side, but I already know this isn't seperable. So I need to use another method.

  74. anonymous
    • 5 years ago
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    Yes, you'll need a different method.

  75. anonymous
    • 5 years ago
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    It's a lot messier than just recognizing the form of what you've been given and just exploiting it.

  76. anonymous
    • 5 years ago
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    Ok so I think I can combine differentials to integrate them as a unit right? the 11xdy+11ydx can combine to form 11d(xy) right?

  77. anonymous
    • 5 years ago
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    Yes

  78. anonymous
    • 5 years ago
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    That's right.

  79. anonymous
    • 5 years ago
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    Ok so from this point, can I just integrate each term?

  80. anonymous
    • 5 years ago
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    Just like what I did above.

  81. anonymous
    • 5 years ago
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    Ok so you're probably going to kill me, but d(xy) basically means the differential of our "function"?

  82. anonymous
    • 5 years ago
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    or do I just leave it alone? d(xy) just becomes xy..

  83. anonymous
    • 5 years ago
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    lol, d(xy) means the differential of xy.

  84. anonymous
    • 5 years ago
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    \[\int\limits_{}{}d(xy)=(xy)+c\]

  85. anonymous
    • 5 years ago
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    I think I'm getting hung up on the definition of a differential

  86. anonymous
    • 5 years ago
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    \[\int\limits_{}{}dw=w+c\]\[\int\limits_{}{}d(\cos \theta)=\cos \theta + c\]\[\int\limits_{}{}d(sty)=sty+c\]\[\int\limits_{}{}d(gp^4.78-\frac{q}{a}+x^2e^{xy})=gp^4.78-\frac{q}{a}+x^2e^{xy}+c\]

  87. anonymous
    • 5 years ago
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    I think that last thing you posted makes sense. I think I have it losesly in my mind what it means now

  88. anonymous
    • 5 years ago
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    See what I keep doing, no matter how complicated it gets?

  89. anonymous
    • 5 years ago
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    If you sleep on it, it will sink in.

  90. anonymous
    • 5 years ago
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    So it basically negates it. I got it. Thanks dude I'm going to try a few more examples

  91. anonymous
    • 5 years ago
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    Okay...good luck!

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