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Mathematics LibreTexts

2.3: 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 1: Pollution

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

\[\begin{align} \dfrac{\text{grams}}{\text{day}}  &= \dfrac{\text{gallons}}{\text{day}} \dfrac{\text{grams}}{\text{gallon}} \\ &= \dfrac{12,000}{1} \dfrac{2}{1} \\  &=  24,000 \text{ grams per day} \end{align}\]

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

\[\begin{align} \dfrac{grams}{day} = \dfrac{gallons}{day} \dfrac{grams}{gallon} &= \dfrac{10,000}{1} \dfrac{x}{500,000 + 2,000 \, t} \\ &= \dfrac{10x}{500 + 2t} \text{grams per day}. \end{align} \]

Putting this all together, we get

\[ \dfrac{dx}{dt} = 24,000 - \dfrac{10x}{500 +2t}.\]

This is a first order linear differential equation with 

\[ p(t) = \dfrac{10}{500 + 2t} \;\;\; \text{and} \;\;\;  g(t) = 24,000. \]

We have

\[\begin{align} \large \mu &= e^{\int \frac{10}{500 + 2t}dt} \\ &= e^{5\, \ln (500 + 2t)} \\ & = (500 + 2t)^5. \end{align} \]

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

\[  (500 + 2t)^5x  =  2,000(500 + 2t)^6 + C  \]

\[ \implies x = 2000(500+2t) + \dfrac{C}{(500+2t)^5}. \]

Now use the initial condition to get

\[ x = 2000(500)+\dfrac{C}{(500)^5} \]

\[\implies  C  =  -3.125 \times  10^{19}. \]

Now plug in 10 for \(t\) and calculate \(x\)

\[ \begin{align} x &= 2000(500+2(10)) + \dfrac{-3.125\,\text{x}\,10^{19}}{(500+2(10))^5} \\  &=\,218,072\,\text{grams}. \end{align}\]

A graph is given below

Example 2: Getting Lucky

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 

\[x = \text{the balance after $t$ years.}\]

The rate out is 

\[300,000\]

and the rate in is

\[0.04x.\]

We have the differential equation

\[ \dfrac{dx}{dt}  =  0.04x  -  300,000.\]

This is both linear first order and separable. We separate and integrate to obtain

\[\begin{align} \int \dfrac{dx}{0.04x - 300,000} &= \int dt   \\   \implies 25\, \ln\, (0.04\, x - 300,000)  &=  t + C_1 \\  \implies  0.04x  -  300,000  &=  C_2 e^{\frac{t}{25}} \\ \implies x   &=  Ce^{\frac{t}{25}} + 7,500,000.   \end{align}\]

Now use the initial condition that when \(t =  0\), \(x  =  5,000,000\)

\[  5,000,000  =  C + 7,500,000. \]

So that 

\[   x   =  -2,500,000\, e^{t/25} + 7,500,000. \]

Plugging in 20 for \(t\) gives 

\[   x  =  1,936,148. \]

You will have about $2 million left.

Contributors

  • Integrated by Justin Marshall.