# Separation of Variables

1. Jan 19, 2012

### 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.

Last edited: Jan 19, 2012
2. Jan 19, 2012

### 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.

3. Jan 19, 2012

### Dustinsfl

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

Correct?

4. Jan 19, 2012

### Dick

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

5. Jan 19, 2012

### Dustinsfl

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.

6. Jan 19, 2012

### SammyS

Staff Emeritus
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.

7. Jan 20, 2012

### Dustinsfl

I know, I have fixed that part.

8. Jan 20, 2012

### Dustinsfl

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 $$N(0)=N_0$$

9. Jan 20, 2012

### 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.

10. Jan 20, 2012

### Dustinsfl

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. Jan 20, 2012

### Dick

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. Jan 20, 2012

### Dustinsfl

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. Jan 20, 2012

### Dick

Sure. Continue.

14. Jan 20, 2012

### Dustinsfl

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

Thanks.