Modeling Population Growth Using Separation of Variables

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.
 
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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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