Modeling Population Growth Using Separation of Variables

[Dustinsfl]

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

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

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.

 

[Dick]

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.

 

[Dustinsfl]

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?

 

[Dick]

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.

 

[Dustinsfl]

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.

 

[SammyS]

$$\displaystyle\frac{dN}{dt} = rN\left(1-\frac{N}{K}\right)$$

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

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

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.

 

[Dustinsfl]

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]

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 $$N(0)=N_0$$

 

[Dick]

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.

 

[Dustinsfl]

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.

 

[Dick]

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?

 

[Dustinsfl]

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?

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

 

[Dick]

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.

 

[Dustinsfl]

Sure. Continue.

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

Thanks.