# Separable D.E., Partial Fractions

1. Sep 28, 2009

### EtherealMonkey

1. The problem statement, all variables and given/known data

$$\frac{dP}{dt}=P-P^{2}$$

It seems that Partial Fractions should be used to solve this D.E., but I cannot find an example to go by.

I even tried to rewrite the equation as:

$$\frac{d}{dx}Y\left(x\right)=Y\left(x\right)-Y\left(x\right)^{2}$$

But, that isn't helping me either...

Anyone?

2. Sep 28, 2009

### EtherealMonkey

I know this may seem so obvious. And, in fact a great deal of my trouble in D.E. so far can be summarized by my "not seeing the forest for the trees".

So, any hint would probably get me going...

3. Sep 28, 2009

### EtherealMonkey

This is what I have done so far...

$$P\left(t\right)\frac{d}{dt}=P\left(t\right)-P\left(t\right)^{2}$$

$$P\left(t\right)\frac{d}{dt}=P\left(t\right)\left(1-P\left(t\right)\right)$$

$$\frac{1}{P\left(t\right)}dP=\left(1-P\left(t\right)\right)dt$$

$$\int \frac{1}{P\left(t\right)}dP= \int dt - \int P\left(t\right) dt$$

$$\ln \left(P\left(t\right)\right) = t - \int P\left(t\right) dt$$

4. Sep 28, 2009

### Staff: Mentor

I see you were able to get the 'y' in your user name!

One comment: you're still using the derivative operator incorrectly: d/dt requires something to the right of it, since it means to take the derivative with respect to t of something.

Now, let's look at your DE.
$$\frac{dP}{dt}~=~P~-P^2$$
After separation, you get this.
$$\frac{dP}{P - P^2}~=~dt$$

Now integrate both sides. You will want to use partial fractions for the integral on the left side, and should get two terms involving ln|P| and ln|1 - P|. Don't forget you'll need the constant of integration.

Is that enough to get you started?

5. Sep 28, 2009

### EtherealMonkey

Yes, thank you very much!

Sorry for the late response, I just got on campus.

Now, gotta go take a test, wish me much success!

Thanks again Mark!