how exactly does the standard form of the exponential function y[x] = k E^(r x) relate to this form of the logistic differential equation y'[t] == r y[t] (1 - y[t]/b) and it's counterpart y[t] = (b*E^(r*t + b*k))/(-1 + E^(r*t + b*k)) r is obvious k is not so obvious but probably the same between y[t] and y[x] but b???!!! b is a problem.

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how exactly does the standard form of the exponential function y[x] = k E^(r x) relate to this form of the logistic differential equation y'[t] == r y[t] (1 - y[t]/b) and it's counterpart y[t] = (b*E^(r*t + b*k))/(-1 + E^(r*t + b*k)) r is obvious k is not so obvious but probably the same between y[t] and y[x] but b???!!! b is a problem.

Mathematics
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omg I feel like I'm in Mathematical Biology again
lol, its awful, Im in hell

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Please come help
3 days I been trying to work this out
where there are two parameters in the logistic differential equation... r is the growth rate k is carrying capacity. ewww I don't want to do this again!
you any good with mathematica usuki?
too long ago.
but I know this logistic equation
In all the examples they provide b.. but then in the problem.. they dont.. I do have a table of data though.. an exponential sample.
is that required to work out b in the final equation?
my best guess is that the logistical curve is going to be an S .. at the middle of that S, in the 2nd derivative will be a 0 marking the point at which the concave changes.. this point is going to be at almost exactly 1/2 of b.. if there is proven to be a relationship between the corresponding exponential curve and the logistical curve then I can find b.
like the slope of the derivatives, might have some relationship that is propotional to r.
at that inflection point.
What exactly is the problem, this seems like logistic differential equation, I mean like does k represent the carrying capacity, and r,k >0 constant?
My problem is that I was given the top equation.. and asked to turn it into the bottom equation. after they showed me the middle equation.. and they're asking me to work it out.
if we name the equations 1,2,3. I know how to go from equation 2 to 3 .. but not from equation 1 to 2 or 1 to 3.
k as far as I can tell, is the value where x=0 y[0], I do have a value for it.. I have a data table for the exponential points.. and a formula that approximates those data points.
the formula is in the standard exponential form [x] = k E^(r x)
oops y[x] = k E^(r x)
The counterpart to me seems to be integrated, but I'm not really used to this stuff yet (haven't done much ODE atm), I've done similar but for word problems haha..mhm. I would take an attempt at it, but I don't want to lead you to the wrong place so I'll let the big guns handle it, so to speak haha @ganeshie8 @SithsAndGiggles
thank you astro for taking a look. appreciated.
Sorry about the inconvenience @hughfuve i dont want to make false promises but i will try and find someone to help :D @freckles
I've done logistic equation... but like I mentioned I don't want to go back to Mathematical Biology >_<
well I appreciate all help.. no matter how it turns out..
y'[t] == r y[t] (1 - y[t]/b) there are two parameters r and b. r is the growth rate and b is the carrying capacity.
you can't assign values to parameters.. the derivative is a constant.
so if b is the carrying capacity here.. could it just k from the first equation? The problem I have is when I tried that I got a very small S curve.. the data I have represents the US population, and according to a value of b=k, then the US population should not go over 80million.. as that is clearly not the reality.. I am concerned.
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
my Mathematical Biology book is still in my backpack since final exam week. I'm not taking it out XD!
touching it probably makes your fingers rash
my book is going to give me shingles one day.. I just know it.
well it's not a discrete model (that's for difference equations) it's a continuous model.. what I 'm saying may be a different point of view for the logistic equation
yes this is one of the first problems of my chapter introducing differential equations. what an intro.. geeez.
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.
I am on chapter 6.. integration isn't until chapter 10. I have a hunch that perhaps this course is not for teaching calculus, but instead for teaching mathematica to those who already know calculus.
your curriculum is backwards -_-
Populations tend to get larger until there is no longer enough food or space to support so many individuals. This type of growth is called logistic population growth What Is Logistic Population Growth? A group of individuals of the same species living in the same area is called a population. The measurement of how the size of a population changes over time is called the population growth rate, and it depends upon the population size, birth rate and death rate. As long as there are enough resources available, there will be an increase in the number of individuals in a population over time, or a positive growth rate. However, most populations cannot continue to grow forever because they will eventually run out of water, food, sunlight, space or other resources. As these resources begin to run out, population growth will start to slow down. When the growth rate of a population decreases as the number of individuals increases, this is called logistic population growth.
I have the textbook calculus concepts and contexts by james stewart.. I just found a section in there on this formula.. Its almost word for word
doing some reading
James Stewart? I had a Single Variable Calculus Book from that author, but it was only Calculus I and II.
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
might have found a possible solution, just have to work it backwards. if y[x] = logarithmic S curve f[x] = corresponding exponential curve with same r then y''[n]==0 solve for n finds the mid point of the S curve. Then take that n of midpoint of S curve and f[n]+f[0] = value for b
well well y[n]/2 =f[n]+f[0]
might have found a possible solution, just have to work it backwards. if y[x] = logarithmic S curve f[x] = corresponding exponential curve with same r then y''[n]==0 solve for n finds the mid point of the S curve. Then take that n of midpoint of S curve and f[n]+f[0] = value for b well well y[n]/2 =f[n]+f[0] so (y[n]/2)-f[0] =f[n]
You are right about the logistic equation producing scenarios that can't happen. Like the one for your 80 million people in the US. Obviously to sustain 80 million, you have to force people to not have plenty of kids. It's like the logistic equation is saying ok your world must sustain exactly 80 million people or less and if it goes 1 over, you'll have a shortage of supplies. Same with the birds. Either 100 birds or less...otherwise if there are 1000 birds there will be less resources and eventually the birds will die.
so does that mean then that b is arbitrary? and you can set it to whatever you want? say if you determine that you are going to put some cap on on growth and force it?
you can try to force it .. but in reality that's not possible... humans are going to mate anyways... same with birds.
I was told that Mathematical Biology is harder ... it actually takes Calculus III, IV, ODE, and Linear Algebra... puts it in a blender and messes everything up.
plus I had to use Matlab for assignments and a project... I still think my project about Killer Zombies saved my grade.
maybe I got thrown off because the examples show exponential plots vs logarithmic plots that are proportional to each other in respect to the value for b. They created a log plot, then used that to make the exponential plot. But now I am thinking that maybe the two plots are always proportional in this way no matter what the value for b.. and b is arbitrary, and it depends on what controls you want to place on the model. So Im looking for a shadow
crap.. I think it is the case
if we try this substitution: \[y\left( t \right) = \frac{1}{{z\left( t \right)}}\] where z is the new variable, then our starting equation, namely equation #2 can be rewritten as follows: \[\frac{{dz}}{{dt}} + rz = \frac{r}{b}\]
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:!
the solutions to that ODE are: \[\Large z\left( t \right) = {e^{ - rt}}\left( {c + \int {\frac{r}{b}{e^{rt}}dt} } \right)\]
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 return to our old variable y(t): \[\Large y\left( t \right) = \frac{b}{{bc{e^{ - rt}} + 1}}\] where c is an arbitrary real constant
@hughfuve didn't you mention earlier that integrating factor was not until real later in your book?
"hughfuve Best Response Medals 1 I am on chapter 6.. integration isn't until chapter 10. I have a hunch that perhaps this course is not for teaching calculus, but instead for teaching mathematica to those who already know calculus."
yes, but that's okay.. I can still use the Iintegrate[] mathematica command in this problem at least..
I would complain to your instructor if I were you. How can you do this problem without integrating factor?
Its okay.. Im learning for now.. this will at least put me in a place where I can put this one to rest..
.___. last time I checked all of differential equations come first...then the Mathematical Biology comes second
now we can multiply both numerator and denominator by \[\Large {{e^{rt}}}\] so we get: \[\Large y\left( t \right) = \frac{{b{e^{rt}}}}{{bc + {e^{rt}}}}\]
now we have to know the initial conditions, since the formula above gives us infinite solutions
so the exponential function I started with.. r = 0.01300265909581666; k = E^(4.382968909053888); yExp[x] = k E^(r x) I can basically pull the variables here as the starting conditions and then b is whatever we want to set the limit of the function too?
I assume c =k r = r b = whatever
and I will have an exponential version and a logarithmic version that start out the same, but end up going in different directions.
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: \[\Large \frac{1}{c} = {e^{bk}}\]
I have integrated that ODE using the separation of variable method, and I got this solutions: \[\Large y\left( t \right) = \frac{{b{c_1}{e^{rt}}}}{{1 + {c_1}{e^{rt}}}}\] where c_1 is the integration constant. Now it is suffice apply this substitution: \[\Large {c_1} = {e^{bk}}\]
yes that's correct... though these days I prefer integrating factor... I know it's more steps but taking the antiderivative is so much easier than integrating on both sides with separation of variables and end up getting stuck on the left side.

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