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 = 0.0347

So the equation is:     P =  1 x 10⁶e ^0.0347t

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

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