# Separable ODE

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.

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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

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Thanks, I think I got it.

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

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

Daniel.

You really like those hyperbolics dont you

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