Question

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

Answer 1

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

`dy/dx = k*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 `(x,y)` is equal to `k` times `y`. When we solve it, we end up with a function `y` of `x`:

`y = C * e^(kx)`

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:

`dy/dt = - 0.1 * y`

First step in solving is to separate the variables:

`-1/(0.1y) dy = dt`

Now, we will simply integrate:

`int -1/(0.1y) dy = int 1 dt`

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

`int -1/(0.1y) dy = t + C`

Note that we can pull `-1/0.1` out of the integrand on the left side:

`-1/0.1 int 1/y dy = t + C`

And now this is easily solved:

`-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 = -0.1t + C`

Exponentiate both sides:

`y = e^(-0.1t + C)`

This can be rewritten as:

`y = e^C * e^(-0.1t)`

Again, since `C` is an arbitrary constant, `e^C` is also an arbitrary constant. Therefore,

`y = C * e^(-0.1t)`

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 mm`, then we can solve for `C`:

`100 = C * e^(-0.1*0)`
`100 = C`

`y = 100 * e^(-0.1t)`

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 `(x, y)` the slope is equal to `-0.1*y`.