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Separable ODE

  • #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.
 

Answers and Replies

  • #2
1,444
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:
  • #3
Thanks, I think I got it. :cool:
 
  • #4
1,444
2
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
 
  • #5
dextercioby
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It could be done without partial fractions,using a hyperbolic tangent substitution.

Daniel.
 
  • #6
2,209
1
You really like those hyperbolics dont you
 
  • #7
Thanks everyone. Yes I did get the same answer as stunner.
 

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