Modeling with First Order Differential Equations Whenever there is a process to be investigated, a mathematical model becomes a possibility.  Since most processes involve something changing, derivatives come into play resulting in a differential equation.  We will investigate examples of how differential equations can model such processes.   Example    A Polluted Pond A pond initially contains 500,000 gallons of unpolluted water has an outlet that releases 10,000 gallons of water per day.  A stream flows into the pond at 12,000 gallons per day containing water with a concentration of 2 grams per gallon of a pollutant.  Find a differential equation that models this process and determine what the concentration of pollutant will be after 10 days. Solution     We let x(t) be amount of pollutant in grams in the pond after t days. We use a fundament property of rates:          Total Rate       =       Rate In   -   Rate Out To find the rate in we use              grams                gallons      grams              12,000      2                              =                                      =                                          day                     day         gallon                  1           1           =  24,000 grams per day   To find the rate out we first notice that since there was initially 500,000 gallons of water in the lake and the water level is increasing at a rate of 2,000 gallons per day, the total number of gallons of water in the lake after t days is          gallons  =  500,000 + 2,000 t The units for the rate out is grams per day.  We write              grams                gallons      grams              10,000                 x                              =                                      =                                                                      day                     day         gallon                  1           500,000 + 2,000 t                    10x         =                    grams per day                500 + 2t Putting this all together, we get              dx                                   10x                          =  24000   -                                   dt                                500 + 2t This is a first order linear differential equation with                                  10              p(t)  =                             g(t)  =  24,000                             500 + 2t We have Multiplying by the integrating factor and using the reverse product rule gives             ((500 + 2t)5x)'  =  24,000(500 + 2t)5 Now integrate both sides to get             (500 + 2t)5x  =  2,000(500 + 2t)6 + C                                                           C             x   =  2000(500 + 2t)  +                                                                             (500 + 2t)5      Now use the initial condition to get                                                 C             x   =  2000(500)  +                                                             (500)5          C  =  -3.125 x 1019 Now plug in 10 for t and calculate x                                                           -3.125 x 1019             x   =  2000(500 + 2(10))  +                                                                                       (500 + 2(10)5              =  218,072 grams A graph is given below Example You just won the lottery.  You put your \$5,000,000 in winnings into a fund that has a rate of return of 4%.  Each year you use \$300,000.  How much money will you have twenty years from now? Solution This is also a          Total Rate  =  Rate In - Rate Out  problem.  Let          x  =  the balance after t years The rate out is          300,000 and the rate in is         .04x We have the differential equation         dx/dt  =  .04x  -  300,000 This is both linear first order and separable.  We separate and integrate to obtain 25ln(.04x - 300,000)  =  t + C1         .04x  -  300,000  =  C2 et/25         x   =  Cet/25 + 7,500,000 Now use the initial condition that when t  =  0, x  =  5,000,000         5,000,000  =  C + 7,500,000 So that          x   =  -2,500,000 et/25 + 7,500,000 Plugging in 20 for t gives          x  =  1,936,148 You will have about 2 million dollars left.   Back to the Differential Equations Home Page