# Population Growth

Applications of Differential Equations

Population Growth

Problem:    The population of a certain country is 2 million and has doubled in the last 20 years. Find the expected population in 80 years.

First summarize the given information:

When          t  = 0 yrs             P = 1 x 106

t   = 20 yrs         P = 2 x 106

t = 100 yrs        P = ?

And             dP/dt = kP

Now we separate the variables of our differential equation:

dP/P = k dt

Then integrate and solve for P:

ln P = kt + C

P = e kt + C

P = C e kt

Refer back to previous example if you are unclear on this step.

Now use the given information to solve for two constants, k and C.

First use    t = 0     P = 1 x 106

1 x 106 = C e k(0)

C = 1 x 106

Now use     t  = 20 yrs         P = 2 x 106

2 x 106 =  1 x 106 e k(20)

k = ln2/20 = 0.03466

So the equation is:     P =  1 x 106e 0.03466t

Now just plug in our value for time, t = 100.

P = 1 x 106e 0.03466(100) = 32million

Note: You need to carry out the exponent 0.03466 several places to get good accuracy.