# Thread: Differential Equations

1. ## Differential Equations

Suppose that the population P(t) of a country satisfies the differential equation

dP/dt=kP(200-P)

with k constant. Its population in 1940 was 100 million and and was then growing at the rate of 1 million per year. Predict this country's population for the year 2009.

2. Originally Posted by cowracer3
Suppose that the population P(t) of a country satisfies the differential equation

dP/dt=kP(200-P)

with k constant. Its population in 1940 was 100 million and and was then growing at the rate of 1 million per year. Predict this country's population for the year 2009.

After separating variables,and taking into account the facts given by the problem ,we integrate from 100 to p the fraction :

................... $\frac{dt}{p(200-p)}$....................

to get : $\frac{ 1}{200}\ln\frac{ p}{200-p}$.................................................. .............................1

Then we integrate ,kdt, from 1940 to 2009 and put k =1 and we get:

.............................69................... .............................................2

Now we equate (1) to (2) (1)=(2) we get:

................... $\frac{ p}{200-p}=e^13800$=c.....................3

After doing calculations we have:

................... $p=\frac{200c}{1+c}$.................................................. ..........................4

And since c is very large we can substitute 1+c by c in 4 and thus finally have:

.....................................p=200 million........................................... .................................