Thread: 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.

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 :


................... ....................

to get : .................................................. .............................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:


................... =c.....................3


After doing calculations we have:



................... .................................................. ..........................4

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

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