Sometimes we don’t quite know what kind of function we are dealing with exactly, but we know some basic things about its derivative. For example, consider how many people live in a town or city. We’ll call this [latex]P(t)[/latex]. One general principle that often holds true for population is that the rate it is growing is proportional to the size of the population. That is, the bigger the city, the faster the growth. This is just saying in terms of raw numbers, a city like Hong Kong has the ability to add people much faster than Dillon, Montana, since Hong Kong is a much bigger city. Now, this doesn’t have to hold: some big cities actually shrink, while some small towns explode in population overnight. But on average, this is true. It follows the equation
[latex]\frac{dP}{dt} = 0.03 \cdot P[/latex]
The left hand side represents how fast [latex]P[/latex] is growing, and the right hand side represents some fraction of [latex]P[/latex]. The value [latex]0.03[/latex] is called the growth rate. This is an example of a differential equation. A differential equation is an equation relating a unknown function and its derivatives. Another way to write the same differential equation is to use Newton’s notation.
[latex]P'(t) = 0.03 \cdot P(t)[/latex]
Again, this is just saying how fast [latex]P(t)[/latex] is growing is equal to some constant times the size of [latex]P(t)[/latex].
We’ll learn how to “solve” an differential equation later on. But for now, note that [latex]{\color{red} P(t) = e^{0.03 t}}[/latex] solves this differential equation. Why? We’ll, note that [latex]{\color{blue} P'(t) = 0.03 e^{0.03 t}}[/latex] by the chain rule. Hence, we can verify that [latex]{\color{red} P(t) = e^{0.03t}}[/latex] solves this differential equation by using substitution.
\begin{align*}
{\color{blue} P'(t)} & = 0.03 \cdot {\color{red} P(t)} \\
({\color{blue} 0.03 e^{0.03t}}) & = 0.03 \cdot ({\color{red} e^{0.03 t}}) \\
0.03 e^{0.03t} & = 0.03 e^{0.03t}
\end{align*}
Since we get the same thing on both sides via substitution, we know that the differential equation is verified! What does this mean? Well, this says that population follows the exponential function [latex]e^{0.03t}[/latex]. This means populations eventually start to grow extremely fast.
While exponential functions grow bigger and bigger forever, in practice, population growth will eventually slow or even stop due to geographic or other constraints. No exponential function lasts forever in real life!
Let’s see a couple other examples of creating differential equations.
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Let [latex]E(t)[/latex] be the earnings of a large company, measured in millions of dollars. This company’s growth in earnings is 0.07 times their current earnings. What is a differential equation that models this situation?
We see that growth in earnings is the same thing as derivative. Hence [latex]E'(t)[/latex] is the growth in earnings, and this is equal to [latex]0.07[/latex] times their current earnings, so [latex]\boxed{E'(t) = 0.07 \cdot E(t)}[/latex].
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The growth of a function is [latex]2.5[/latex] less than [latex]0.3[/latex] times the value of the function. What is the differential equation now?
We see that this can be translated as [latex]\boxed{f'(t) = 0.3 f(t) - 2.5}[/latex].
Here are some examples of verifying a given solution is correct.
[latex]f'(x) = f(x) - x[/latex]
Here, we see [latex]{\color{red} f(x) = e^x + x + 1}[/latex] is given, and we can compute this by taking the derivative of each piece.
\begin{align*}
{\color{blue} f'(x)} & {\color{blue} = \frac{d}{dx} e^x + x + 1} \\
& {\color{blue} = e^x + 1} \
\end{align*}
We can then verify the differential equation using subsitution.
\begin{align*}
{\color{blue} f'(x)} & = {\color{red} f(x)} – x \\
{\color{blue} e^x + 1} & = {\color{red} (e^x + x + 1)} – x \\
e^x + 1 & = e^x + 1
\end{align*}
[latex]q'(x) = \frac{1}{q(x)}[/latex]
Here, we see [latex]{\color{red} q(x) = \sqrt{2x}}[/latex] is given, and we can compute using the chain rule
\begin{align*}
{\color{blue} q'(x)} & {\color{blue} = \frac{d}{dx} \sqrt{2x}} \\
& {\color{blue} = \frac{d}{dx} (2x)^{1/2}} \\
& {\color{blue} = \frac{1}{2} (2x)^{-1/2} \cdot 2} \\
& {\color{blue} = (2x)^{-1/2}} \\
& {\color{blue} = \frac{1}{\sqrt{2x}}}
\end{align*}
We can then verify the differential equation using substitution.
\begin{align*}
{\color{blue} q'(x)} & = \frac{1}{{\color{red} q(x)}} \\
\left( {\color{blue} \frac{1}{\sqrt{2x}}} \right) & = \frac{1}{({\color{red} \sqrt{2x}})} \\
\frac{1}{\sqrt{2x}} & = \frac{1}{\sqrt{2x}}
\end{align*}
