# What is the differential equation that models exponential growth and decay?

Aug 17, 2014

The simplest type of differential equation modeling exponential growth/decay looks something like:

$\frac{\mathrm{dy}}{\mathrm{dx}} = k \cdot y$

$k$ is a constant representing the rate of growth or decay. A negative value represents a rate of decay, while a positive value represents a rate of growth.

This differential equation is describing a function whose rate of change at any point $\left(x , y\right)$ is equal to $k$ times $y$. When we solve it, we end up with a function $y$ of $x$:

$y = C \cdot {e}^{k x}$

where $C$ is a constant due to integration. In this case, $C$ represents the initial value, since there's an infinite number of functions we could have with the same property, each possible function differing only by the initial $y$ value.

Just to demonstrate how this works, let's say that we have a droplet of water being absorbed into a piece of cloth. At any given moment, the droplet of water is shrinking by 10% of its current size. We want to find a function, $y$, which represents the size of the droplet at time $t$.

This situation translates into the following differential equation:

$\frac{\mathrm{dy}}{\mathrm{dt}} = - 0.1 \cdot y$

First step in solving is to separate the variables:

$- \frac{1}{0.1 y} \mathrm{dy} = \mathrm{dt}$

Now, we will simply integrate:

$\int - \frac{1}{0.1 y} \mathrm{dy} = \int 1 \mathrm{dt}$

The right side is fairly easy. Remember the constant of integration:

$\int - \frac{1}{0.1 y} \mathrm{dy} = t + C$

Note that we can pull $- \frac{1}{0.1}$ out of the integrand on the left side:

$- \frac{1}{0.1} \int \frac{1}{y} \mathrm{dy} = t + C$

And now this is easily solved:

$- \frac{1}{0.1} \ln y = t + C$

Now, we will multiply both sides by $- 0.1$. Note that since $C$ is an arbitrary constant, it is left unchanged after we distribute the $- 0.1$.

$\ln y = - 0.1 t + C$

Exponentiate both sides:

$y = {e}^{- 0.1 t + C}$

This can be rewritten as:

$y = {e}^{C} \cdot {e}^{- 0.1 t}$

Again, since $C$ is an arbitrary constant, ${e}^{C}$ is also an arbitrary constant. Therefore,

$y = C \cdot {e}^{- 0.1 t}$

And there is our equation for the size of the droplet at time $t$. If we had been given a condition, for instance, that at $t = 0$ the droplet is $100 m m$, then we can solve for $C$:

$100 = C \cdot {e}^{- 0.1 \cdot 0}$
$100 = C$

$y = 100 \cdot {e}^{- 0.1 t}$

There we go. If you graph this function on your calculator, you can verify that it does indeed have the property that at any point $\left(x , y\right)$ the slope is equal to $- 0.1 \cdot y$.