Suppose that a population grows according to a logistic model with carrying capacity 6000 and k=0.0015 per year.
Write a logistic differential equation for these data.

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- TuringTest

this terminology is the only thing confusing me here...

- TuringTest

I guess I don't know what a "logistic" DE is

- experimentX

|dw:1343418531683:dw|

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## More answers

- TuringTest

ok... that makes sense

- experimentX

something like population growth or bacteria growth ... etc etc that depends on initial population

- anonymous

\[y=6000e^{0.0015t}\]

- anonymous

?

- experimentX

|dw:1343508816527:dw|

- experimentX

these kids of equation, differentiate once .... remove these terms|dw:1343509003699:dw|

- anonymous

are those logistic functions? the 6000 is not your initial population, it's your carrying capacity....
the differential equation is: \(\large \frac{dy}{dt}=ky(6000-y) \)

- anonymous

What's the thought process...how could I come up with this on my own

- lgbasallote

...so what's this difference from the usual differential equation for exponential growth?

- anonymous

so we start with \[y=y_0e^kt\] and we take the derivative of this to find a rate of change?

- TuringTest

no, start with the idea that the current rate of change is proportional to the current population\[\frac{dy}{dt}\propto y=ky\]separate the variables\[\frac{dy}y=kdt\]integrate both sides:\[\ln y=kt+C\]\[y(t)=Ce^{kt}\]by plugging in t=0 we find that C the original population\[y(0)=Ce^0=C\implies C=y_0\]so we get\[y=y_0e^{kt}\]

- TuringTest

I just don't know all the terminology above lol, but I'm familiar with the problems

- anonymous

how do we go from
\[\ln y=kt+C\] to
\[y(t)=Ce^{kt}\] I presume it was doing e^.... on both sides but why is the C multiplied to e?

- TuringTest

\[e^{kt+C}=e^{kt}\cdot e^C\]and \(e^C\) is just another constant, so I just called it C again. I maybe should have numbered them for the sake of distinction, but that gets tedious in higher math\[e^{kt+C_1}=C_2e^{kt}\]

- anonymous

Oh I see.
Now that I have this equation
\[y(t)=Ce^{kt}\] why am I taking the derivative again?

- anonymous

or do I just plug in numbers at this point?

- TuringTest

just plug it in usually, but carrying capacity... not sure of that is at time t=0
sounds like a maximum to me
again the terminology is messing me up

- anonymous

@dpalnc statement looks legit...(don't know why)
\[\large \frac{dy}{dt}=ky(6000-y)\]

- anonymous

check this out:
\[\frac{dP}{dt}=kP(1-\frac{P}{K})\]

- anonymous

Now I'm really confused

- anonymous

I am soo sorry! My book calls the logistic equation little k
and carrying capacity big K....

- TuringTest

I'm just going to say what I'm sure of, which is the initial derivation

- anonymous

I think I'm kinda done with this problem LOL. I suck at word problems. But your steps in the initial derivative makes sense though.

- anonymous

It seems like my book just gave me this equation
\[\frac{dP}{dt}=kP(1-\frac{P}{K})\] and just wants me to plug and chug numbers. ....fine with me!

- TuringTest

I guess that's what I'd do as well :p

- anonymous

Sounds like a plan :P
Thanks Turing

- TuringTest

that equation makes sense too because when P=K, dP/dt=0
i.e. when the population = max capacity it stops changing
reasonable to me :)

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