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 10^{6}

t = 20 yrs P = 2 x 10^{6}

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 10^{6}

1 x 10^{6} = C e ^{k(0)}

C = 1 x 10^{6}

Now use t = 20 yrs P = 2 x 10^{6}

2 x 10^{6} = 1 x 10^{6} e ^{k(20)}

k = ln2/20 = 0.03466

So the equation is: P = 1 x 10^{6}e ^{0.03466t}

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

P = 1 x 10^{6}e ^{0.03466(100) }= 32million

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