# Population Growth

1. Jan 22, 2006

### zoldman

I am quite sure the first approach is to use partial fractions but I am
unclear how to finish this equation

1/P*dP/dt=b+aP

2. Jan 23, 2006

### HallsofIvy

Staff Emeritus
How to finish it? You haven't started it!

$$\frac{1}{P}\frac{dP}{dt}= b+aP$$
is separable:
$$\frac{dP}{P(b+aP)}= dt$$
We can write
$$\frac{1}{P(b+ aP)}= \frac{A}{P}+ \frac{B}{b+ aP}$$
for some A and B. Multiply both sides by P(b+ aP):
$$1= A(b+ aP)+ BP$$

What do you get if P= 0? What do you get if P= -a/b? Put those values for A and B into the fractions and integrate.

3. Jan 23, 2006

### zoldman

Solve P=? or t=?

I appreciate the confirmation of the partial Fraction step. I arrive at the same situation:

A=1/b and B=-a/b. And I know both a and b from a linear regression.

So substituting back I get

((1/b)/P+((-a/b)/(b+aP)=dt then integrate both sides

(1/b)lnP +(-1/b)ln(b+aP)=t +C

Now my question is here how do I solve for P= f(t).

4. Jan 24, 2006

### HallsofIvy

Staff Emeritus
The rest is algebra:
$$\frac{1}{b}ln P- \frac{1}{b}ln(b+aP)= ln\left(\frac{P}{b+aP}\right)^\frac{1}{b}= t+ c$$
Take exponential of both sides:
$$\left(\frac{P}{b+aP}\right)^\frac{1}{b}= e^{t+ C}= C'e^t$$
(C= eC)
Take ath power of both sides:
$$\frac{P}{b+aP}=C"e^{at}$$
(C"= C'a)
multiply both sides by b+ aP and expand:
$$P= C"e^{at}(b+ aP)= C"be^{at}+ aC"e^{at}P$$
$$P- aC"e^{at}P= P(1- aC"e^{at})= C"be^{at}$$
$$P= \frac{C"be^{at}}{1- aC"e^{at}$$

5. Jan 24, 2006

### mathwonk

i thionk that logistic equation is more usefully written as dP/dt

= aP(1 - P/N).

of course it is separable and solving it, shows that the populaion approaches N as time goes on.

6. Jan 24, 2006

### saltydog

Starting with:

$$\frac{1}{b}ln(P)-\frac{1}{b}ln(b+aP)=t+c$$

multiplying by b and collecting logarithms:

$$ln\left[\frac{P}{b+aP}\right]=b(t+c)$$

Taking exponentials:

$$\frac{P}{b+aP}=e^{b(t+c)}$$

multiplying both sides by b+aP and collecting the P's:

$$P\left[1-ae^{b(t+c)}\right]=be^{b(t+c)}$$

Isolating the P and then multiplying the top and bottom of the rational expression by $e^{b(t+c)}$ to make it cleaner leaves:

$$P(t)=\frac{b}{e^{-b(t+c)}-a}$$

Now, how about completely characterizing the solutions in terms of a and b assuming some initial condition like P(0)=1. What would it look like whatever it was? (just a suggestion )

Last edited: Jan 25, 2006
7. Jan 25, 2007

### raiders06

I am working on the same problem; however I started out a bit different. Follwing the above; now all I need to do is substitue my a and b values; my initial condition is P sub 0 = 3.9
Am I going in the right direction?