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omg I feel like I'm in Mathematical Biology again

lol, its awful, Im in hell

Please come help

3 days I been trying to work this out

you any good with mathematica usuki?

too long ago.

but I know this logistic equation

is that required to work out b in the final equation?

like the slope of the derivatives, might have some relationship that is propotional to r.

at that inflection point.

the formula is in the standard exponential form
[x] = k E^(r x)

oops
y[x] = k E^(r x)

thank you astro for taking a look. appreciated.

well I appreciate all help.. no matter how it turns out..

you can't assign values to parameters.. the derivative is a constant.

hmmm mine had ecology situations

like for bird populations

lol, you studied at the height of environmental hysteria

this course one comes from a time of immigration hysteria

touching it probably makes your fingers rash

my book is going to give me shingles one day.. I just know it.

pfffffffft you should be doing integrating factor as the first topic

forgot to mention that the logistic equation is separable -_-!

Does that look right Hugh?

|dw:1435638707408:dw|

looks good to me.

your curriculum is backwards -_-

doing some reading

logistic equation is a 100% ODE though

Is this in james ste 7e?!

I got the larson hostetler, edwards video seriies too.. maybe theres something in there.

7e?

james stewart 7th edition

oh no.. its 2nd edition

OH THAT IS OLD !

2001, it was on our bookshelf.. my g/f used it way back when

:/

lol, no wonder Im struggling.. I need some new stuff..

I have pearsons college mathematics 13th edition if that helps

but they dont say much on this topic

actually this looks promising
http://reference.wolfram.com/language/tutorial/DSolveOverview.html

well well
y[n]/2 =f[n]+f[0]

@Michele_Laino is here!

crap.. I think it is the case

Hi michele.. welcome :)

now it looks like the integrating factor in the form of \[\frac{dy}{dx} =p(x)y=q(x)\]

Hi :) @hughfuve

NUGH NOT MY NIGHT!
\[\frac{dy}{dx} +p(x)y=q(x)\]

I can't do math in the heat D:!

so you do need integrating factor right?

right!
so we have:
\[\Large z\left( t \right) = c{e^{ - rt}} + \frac{1}{b}\]

now we have to know the initial conditions, since the formula above gives us infinite solutions

I assume
c =k
r = r
b = whatever

hence the idea that.. logistical growth is controlled growth ?

the coefficient b is called "carrying capacity"

I got that one too

ah gotcha, thanks.. the language helps

it is suffice to use this substitution:
\[\Large \frac{1}{c} = {e^{bk}}\]