Solving the logistic ODE y'(t) = (k/M)y(M-y)

[tandoorichicken]

I need help in solving the following ODE.
$$y'(t) = \frac{k}{M}y(M-y)$$

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.

 

[stunner5000pt]

i think you might see it better with dy/dt notation
what you have is $$\frac{dy}{dt} = \frac{k}{M} y(M-y)$$
which will become
$$\frac{dy}{y(M-y)} = \frac{k}{M} dt$$
integrate away!
use partial fractions, seems to work just fine

 

[tandoorichicken]

Thanks, I think I got it.

 

[stunner5000pt]

i got this answer by the way
$$y(t) = \frac{MCe^{kt}}{1+Ce^{kt}}$$
$$C = e^{C_{1}}$$ from the integration

 

[dextercioby]

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

Daniel.

 

[whozum]

You really like those hyperbolics don't you

 

[tandoorichicken]

Thanks everyone. Yes I did get the same answer as stunner.