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.
