- #1

- 244

- 0

[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.

- Thread starter tandoorichicken
- Start date

- #1

- 244

- 0

[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

- 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

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

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Thanks, I think I got it.

- #4

- 1,444

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[tex] y(t) = \frac{MCe^{kt}}{1+Ce^{kt}} [/tex]

[tex] C = e^{C_{1}} [/tex] from the integration

- #5

- 13,023

- 576

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

Daniel.

Daniel.

- #6

- 2,209

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You really like those hyperbolics dont you

- #7

- 244

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

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