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Separable ODE Problem

  1. Jan 19, 2007 #1
    Hi, I'm having trouble with this ODE problem:

    The prompt asks to find the solution of the ODE using separation of variables.

    dy/dt = (ab - c(y^2)) / a,

    where a, b and c are constants.

    I proceed to divide both sides by (ab - c(y^2)) and multiply both sides by dt, but I'm having trouble integrating and expressing the solution in terms of y.

    Thanks in advance.
  2. jcsd
  3. Jan 19, 2007 #2
    So you are trying to solve the following integration??

    [tex] \int \frac{dy}{b - c/ay^2} [/tex]

    one of the very standard way to do it is to use trig substitution.... i.e. [tex] y = C' sin \theta [/tex] determine what C' shall you use in order to cancel out the denumerater...
  4. Jan 19, 2007 #3
    Sorry, I must be missing something.. the integration that I was attempting to do was:

    [tex] \int \frac{dy}{ab - c(y^2)} [/tex]

    Will trig substitution also work with this?
  5. Jan 19, 2007 #4
    Yes... But if you use [tex]y = C' tan\theta [/tex], it might be easier to do....
    just try it and see where you go....
  6. Jan 19, 2007 #5
    ArcTanh[(Sqrt[c]*x)/(Sqrt[a]*Sqrt)]/ (Sqrt[a]*Sqrt*Sqrt[c])

    is the solution to the integral, but it seems too complex for this problem. I guess a better question would be: is there another way to tackle the ODE so that I will arrive at an easier solution?

    i.e. I need a formula in terms of y...
  7. Jan 19, 2007 #6
    So, you are saying [tex] t = tan^{-1} (C'' y) + C'''[/tex] is too complicated???
    I don't think you could simplify this further...
  8. Jan 19, 2007 #7
    how do I represent the equation in terms of y?

    btw, thanks for the help so far
  9. Jan 19, 2007 #8
    [tex] \int dt = \int \frac{1}{b}\frac{dy}{1-c/ab y^2} [/tex]
    [tex] t + C_0 = \frac {1}{2b} \int \frac{1}{1-\sqrt{c/ab}y} + \frac{1}{1+\sqrt{c/ab}y} dy [/tex]
    use [tex] z = \sqrt{c/ab}y [/tex]
    [tex] t + C_0 = \frac{1}{2b} \sqrt{ab/c}( ln(1+z) - ln(1-z) + C_1)[/tex]
    [tex] t + C_2 = 1/2\sqrt{a/bc} ln (1+z)/(1-z) [/tex]
    [tex] e^{2\sqrt{bc/a}t} + C_3 = (1+z)/(1-z) [/tex]
    [tex] z = (e^{2\sqrt{bc/a}t} + C_3 -1 ) / (e^{2\sqrt{bc/a}t} + C_3 + 1) [/tex]
    [tex] y = \sqrt{ab/c}(e^{2\sqrt{bc/a}t} + C_3 -1 ) / (e^{2\sqrt{bc/a}t} + C_3 + 1) [/tex]
    Last edited: Jan 19, 2007
  10. Jan 19, 2007 #9
    That actually helps me quite a bit.
    Thank you!
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