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Separable ODE

  1. Apr 25, 2005 #1
    I need help in solving the following ODE.
    [tex] y'(t) = \frac{k}{M}y(M-y) [/tex]

    Not quite sure what to do. I multiplied everything out so I was left without any parenthesis, but I don't know where to go from there. Any ideas/hints would be appreciated.
     
  2. jcsd
  3. Apr 25, 2005 #2
    i think you might see it better with dy/dt notation
    what you have is [tex] \frac{dy}{dt} = \frac{k}{M} y(M-y) [/tex]
    which will become
    [tex] \frac{dy}{y(M-y)} = \frac{k}{M} dt [/tex]
    integrate away!
    use partial fractions, seems to work just fine
     
    Last edited: Apr 25, 2005
  4. Apr 25, 2005 #3
    Thanks, I think I got it. :cool:
     
  5. Apr 25, 2005 #4
    i got this answer by teh way
    [tex] y(t) = \frac{MCe^{kt}}{1+Ce^{kt}} [/tex]
    [tex] C = e^{C_{1}} [/tex] from the integration
     
  6. Apr 26, 2005 #5

    dextercioby

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    It could be done without partial fractions,using a hyperbolic tangent substitution.

    Daniel.
     
  7. Apr 26, 2005 #6
    You really like those hyperbolics dont you
     
  8. Apr 26, 2005 #7
    Thanks everyone. Yes I did get the same answer as stunner.
     
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