## MathSofiya 3 years ago 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.

1. TuringTest

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

2. TuringTest

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

3. experimentX

|dw:1343418531683:dw|

4. TuringTest

ok... that makes sense

5. experimentX

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

6. MathSofiya

$y=6000e^{0.0015t}$

7. MathSofiya

?

8. experimentX

|dw:1343508816527:dw|

9. experimentX

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

10. dpaInc

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

11. MathSofiya

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

12. lgbasallote

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

13. MathSofiya

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

14. 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}$

15. TuringTest

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

16. MathSofiya

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?

17. 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$e^{kt+C_1}=C_2e^{kt}$

18. MathSofiya

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

19. MathSofiya

or do I just plug in numbers at this point?

20. 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

21. MathSofiya

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

22. MathSofiya

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

23. MathSofiya

Now I'm really confused

24. MathSofiya

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

25. TuringTest

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

26. MathSofiya

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.

27. MathSofiya

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!

28. TuringTest

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

29. MathSofiya

Sounds like a plan :P Thanks Turing

30. 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 :)