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Differential equation w/o an x variable

  1. Oct 8, 2007 #1
    1. The problem statement, all variables and given/known data
    [tex] y\prime = ay - by^{2} [/tex]


    2. Relevant equations



    3. The attempt at a solution

    Is this a linear DOE? If so when I use the integrating factor method, would the int. factor be
    [tex] e^{-a\int?}[/tex] ? would it by dy or dx? I think I'm missing the big picture with this process.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 8, 2007 #2
    Have you tried factoring out the y, dividing over, and partial fractions?
     
  4. Oct 8, 2007 #3
    Hmm... Well when I factor out y and divide I get

    [tex] \frac{y\prime}{y} = a - by [/tex]?

    can I integrate with that y on the other side still?
     
  5. Oct 8, 2007 #4
    No

    Try factoring more like this:

    [tex]
    ay-by^{2}=y(a-by)
    [/tex]

    Divide by the whole thing on the right and try partial fractions
     
  6. Oct 8, 2007 #5
    I don't see how that will work... I'm confused.
     
  7. Oct 8, 2007 #6
    Sooo...

    [tex]
    \int\frac{dy}{y(a-yb)}=\int dx
    [/tex]

    Do you understand how to do the left side? Use partial fractions
     
  8. Oct 8, 2007 #7
    OK so I did the partial fractions took the integral and then did some moving around of logs to get [tex] y = \frac{a^{2}}{e^{x} + ab}[/tex] Does that look close to anything you might have gotten. I'm not super confident in how I manipulated the logs.
     
  9. Oct 8, 2007 #8

    Maybe, you should get an e somewhere in your answer. If I have time later maybe I'll check it for you.
     
  10. Oct 8, 2007 #9

    Dick

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    Nope, doesn't look quite right. And you're missing a constant of integration. Can you show more work?
     
  11. Oct 8, 2007 #10
    If you even show your partial fractions that you created, it'd be very helpful
     
  12. Oct 8, 2007 #11
    So, when I did the seperation of variables I got
    [tex] \frac{1}{a} \int{\frac{dy}{y}} + \frac{b}{a} \int{\frac{dy}{a-by}} = \int{dx}[/tex]

    then when I took the integrals I get

    [tex] \ln{y^{-a}} - \ln{(a-by)^{-a}} = x + c [/tex]

    so then I messed around and combined the logs into a fraction, and since they were both to the -a I just switched to put the a-by on top and the a y on the bottom, then i took the whole thing to the power of e to get rid of the ln, and then brought the a out to the front.

    [tex] a(\frac{a-by}{y}) = e^{x+c} [/tex] then multiplying a into the fraction and solvinf for y I get

    [tex] \frac{a^{2}}{e^{x+c} + ab } = y [/tex]

    where'd I screw up? :)
     
  13. Oct 8, 2007 #12

    Dick

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    Put the a over on the dx side first. And I don't know how it turned into a '-a'. That 'a' should wind up inside the exponential.
     
  14. Oct 8, 2007 #13
    I didn't mean separation of variable, I mean partial fractions... sorry
     
  15. Oct 8, 2007 #14
    yeah multiplying that a makes it much simpler...

    So then I get

    [tex] \int{\frac{dy}{y}} + b\int{\frac{dy}{a-by}} = a \int{dx}[/tex]

    then I get

    [tex] \ln\{\frac{y}{a-by}} = ax + c [/tex]

    then raised to the e

    [tex] \frac{y}{a-by} = ce^{ax} [/tex]

    so then when i solve for y I get [tex] y = \frac{ace^{ax}}{1+bce^{ac}} [/tex]
     
  16. Oct 8, 2007 #15

    Dick

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    That looks much better.
     
  17. Oct 8, 2007 #16
    Thanks!
     
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