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Homework Help: Initial value problem

  1. Feb 1, 2006 #1


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    I have a initial value problem that I have no idea how to start! Ahhh :(

    The equation is:

    [tex]xy' + 6y = 3xy^3 [/tex]

    How do I start? I have no idea! ahhh :cry: I don't like DE's!
  2. jcsd
  3. Feb 2, 2006 #2
    Seperation of vaiables.
  4. Feb 2, 2006 #3
    I can give you a link that should help, but I can't explain it very well because I'm still learning this too. Start by dividing everything by x then check out this link http://en.wikipedia.org/wiki/Bernoulli_differential_equation
  5. Feb 2, 2006 #4
    I'm pretty sure that isn't going to work on this equation.

    OK I was wrong it'll work it just wasn't obvious to me at first sorry.
  6. Feb 2, 2006 #5


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    So all x and y variables will be seperated? I can't seem to get them to seperate...
  7. Feb 2, 2006 #6
    Try dividing everything by y and then isolating the y' term that might make it more obvious.
  8. Feb 2, 2006 #7


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    Can i even do this???

    \frac{{xy' + 6y}}{y} = \frac{{3xy^3 }}{y} \\
    \frac{{xy'}}{y} + 6 = 3xy^2 \\
    \frac{{xy'}}{y} = 3xy^2 \\
    xy' = 3xy^3 \\
    y' = 3y^3 \\
    \frac{{dy}}{{dx}} = 3y^3 \\
    \frac{{dy}}{{3y^3 dx}} = 1 \\
    \frac{1}{{3y^3 }}dy = dx \\

    If thats the case, I guess i know exactly what to do next!
  9. Feb 2, 2006 #8
    You dropped a 6 out somewhere in teh middle and I realized that I screwed up when I tried to separate the variables so I guess it isn't separable after all, try the method in the link I posted above.
  10. Feb 2, 2006 #9


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    Wow i still can't get this...
  11. Feb 2, 2006 #10
    [tex] y' + \frac{6y}{x} = 3 y^3 [/tex]

    divide by [tex] y^3 [/tex]

    [tex] y'y^{-3} + \frac{6y^-2}{x} = 3 [/tex]

    Make a change of variables:

    [tex] w = \frac{1}{y^2} [/tex]


    [tex] w' = \frac{1-3}{y^3} y' = \frac{-2}{y^3} y' [/tex]

    This leads to:

    [tex] - \frac{w'}{2} + \frac{6w}{x} = 3[/tex]

    we-write this in standard form:

    [tex] \frac{dw}{dx} - \frac{12w}{x} = -6 [/tex]

    solve using an integrating factor of:

    [tex] M(x) = e^{\int P(x)dx } = e^{\int \frac{-12}{x}dx} = e^{-12ln(x)} = x^{-12}[/tex]

    multiply both sides by M(x):

    [tex] \frac{w'}{x^{-12}} - \frac{12w}{x^{-13}} = -\frac{6}{x^{-12}} [/tex]

    but that is equal to:

    [tex] \frac { d( \frac{w} {x^{-12}})} {dx} = -\frac{6}{x^{-12}} [/tex]

    integrate both sides:

    [tex] \frac{w} {x^{-12}} = \frac {6}{11} x^{-11} + c [/tex]

    Solve in terms of w:

    [tex] w = \frac{6}{11} x + cx^{12} [/tex]

    Now plug back in for w:

    [tex] \frac{1}{y^2} = \frac{6}{11} x + cx^{12} [/tex]
    Last edited: Feb 2, 2006
  12. Feb 2, 2006 #11


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    Now i gotta figure out how to even use Bernoulli's equation.
  13. Feb 2, 2006 #12
    Ok, I hope the anwser is correct. I probably have some mistakes in there. Im REALLY TIRED right now, and I HAVE to go to sleep. Please check it over someone:

    The solution is:

    [tex] \frac{1}{y^2} = \frac{6}{11} x + cx^{12} [/tex]
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