
# 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 $$\PageIndex{1}$$: 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 $$\PageIndex{2}$$: 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.