Separable ODE

1. Apr 25, 2005

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.

2. Apr 25, 2005

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

Last edited: Apr 25, 2005
3. Apr 25, 2005

tandoorichicken

Thanks, I think I got it.

4. Apr 25, 2005

stunner5000pt

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

5. Apr 26, 2005

dextercioby

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

Daniel.

6. Apr 26, 2005

whozum

You really like those hyperbolics dont you

7. Apr 26, 2005

tandoorichicken

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