Modeling Population Growth Using Separation of Variables

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Homework Help Overview

The discussion revolves around modeling population growth using the separation of variables technique in differential equations. The original poster presents a differential equation related to population dynamics and attempts to manipulate it to find a solution.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the integration of a differential equation and the application of partial fraction decomposition. There are questions regarding the correctness of manipulating integrals and the steps taken to arrive at a solution.

Discussion Status

Participants are actively engaging with each other's reasoning, questioning assumptions about the integration steps and the manipulation of variables. Some guidance has been offered regarding the relationship between constants in different forms of the solution, but there is no explicit consensus on the final form of the solution yet.

Contextual Notes

There are indications of confusion regarding the application of partial fractions and the manipulation of constants in the context of the initial conditions provided. Participants are also working under the constraints of homework rules that may limit the extent of guidance provided.

Dustinsfl
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\displaystyle\frac{dN}{dt} = rN\left(1-\frac{N}{K}\right)\displaystyle\int\frac{KdN}{N\left(K-N\right)} = \int rdt

\displaystyle K\int\frac{dN}{N}-K\int\frac{dN}{K-N}=r\int dt

Now, I obtain:

K\ln\left(\frac{N}{K-N}\right) = rt+c

\left(\frac{N}{K-N}\right)^K=C_0r^{rt}

The final solution is N(t) =\frac{C_0Ke^{rt}}{K+C_0(e^{rt}-1)}

I don't see how I can manipulate my equation to that. Is there a mistake or am I not seeing something.
 
Last edited:
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On the left side, you seem to have changed \frac{K}{N (K-N)} into \frac{K}{N} - \frac{K}{K-N}. That's not right at all.
 
Dick said:
On the left side, you seem to have changed \frac{K}{N (K-N)} into \frac{K}{N} - \frac{K}{K-N}. That's not right at all.

It should be \frac{N}{K-N} = C_0e^{rt}

Correct?
 
Dustinsfl said:
It should be \frac{N}{K-N} = C_0e^{rt}

Correct?

You didn't do partial fractions correctly the first time. That looks like a step in the right direction, yes.
 
Dick said:
You didn't do partial fractions correctly the first time. That looks like a step in the right direction, yes.

There is still something wrong...

From that, I end up with:

N = KC_0e^{rt}-NC_0e^{rt} \Rightarrow N(t)=\frac{KC_0e^{rt}}{1+C_0e^{rt}}

I have a sign issue and I am missing a K.
 
Dustinsfl said:
\displaystyle\frac{dN}{dt} = rN\left(1-\frac{N}{K}\right)

\displaystyle\int\frac{KdN}{N\left(K-N\right)} = \int rdt

\displaystyle K\int\frac{dN}{N}-K\int\frac{dN}{K-N}=r\int dt

Now, I obtain:

K\ln\left(\frac{N}{K-N}\right) = rt+c

\left(\frac{N}{K-N}\right)^K=C_0r^{rt}

The final solution is N(t) =\frac{C_0Ke^{rt}}{K+N_0(e^{rt}-1)}

I don't see how I can manipulate my equation to that. Is there a mistake or am I not seeing something.
It's not true that
\displaystyle \int\frac{KdN}{N\left(K-N\right)} =K\int\frac{dN}{N}-K\int\frac{dN}{K-N}​
You cannot break up an integral that way.

\displaystyle\frac{1}{N}+\frac{1}{K-N}\ne\frac{1}{N(K-N)}\,,

However, \displaystyle \frac{1}{K}\left(\frac{1}{N}+\frac{1}{K-N}\right)=\frac{1}{N(K-N)}\,, which is a result that can be obtained using partial fraction decomposition.
 
SammyS said:
It's not true that
\displaystyle \int\frac{KdN}{N\left(K-N\right)} =K\int\frac{dN}{N}-K\int\frac{dN}{K-N}​
You cannot break up an integral that way.

\displaystyle\frac{1}{N}+\frac{1}{K-N}\ne\frac{1}{N(K-N)}\,,

However, \displaystyle \frac{1}{K}\left(\frac{1}{N}+\frac{1}{K-N}\right)=\frac{1}{N(K-N)}\,, which is a result that can be obtained using partial fraction decomposition.


I know, I have fixed that part.
 
Dustinsfl said:
It should be \frac{N}{K-N} = C_0e^{rt}

But I can't figure out how to manipulate that into

N(t)=\frac{N_0Ke^{rt}}{K+N_0(e^{rt}-1)}

I forgot to mention that N(0)=N_0
 
Use the form you derived before.
N(t)=\frac{KC_0e^{rt}}{1+C_0e^{rt}}.
That tells you
N_0=\frac{KC_0}{1+C_0}
That's the relation between the two constants in the two different forms.
 
  • #10
Dick said:
Use the form you derived before.
N(t)=\frac{KC_0e^{rt}}{1+C_0e^{rt}}.
That tells you
N_0=\frac{KC_0}{1+C_0}
That's the relation between the two constants in the two different forms.


I have all that, but I still can't get to the final form. I tried solving for C but that didn't go anywhere fruitful.
 
  • #11
Dustinsfl said:
I have all that, but I still can't get to the final form. I tried solving for C but that didn't go anywhere fruitful.

You seem to have exactly the right idea of what to do. Solve for C_0 in terms of N_0 and substitute that back into N(t). It works for me. If it doesn't go anywhere fruitful for you can you show why not?
 
  • #12
Dick said:
You seem to have exactly the right idea of what to do. Solve for C_0 in terms of N_0 and substitute that back into N(t). It works for me. If it doesn't go anywhere fruitful for you can you show why not?

Just to verify, this what you get when you substitute correct?

\displaystyle N(t)=\frac{K\left(\frac{N_0}{K-N_0}\right)e^{rt}}{1+\left(\frac{N_0}{K-N_0}\right)e^{rt}}
 
  • #13
Dustinsfl said:
Just to verify, this what you get when you substitute correct?

\displaystyle N(t)=\frac{K\left(\frac{N_0}{K-N_0}\right)e^{rt}}{1+\left(\frac{N_0}{K-N_0}\right)e^{rt}}

Sure. Continue.
 
  • #14
Dick said:
Sure. Continue.

I got it. I don't know what the problem was earlier.

Thanks.
 

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