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

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

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Answer 1 · 2014-08-17T23:28:06

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

$\frac{d y}{d x} = k \cdot y$ $dy/dx = k*y$

$k$ $k$ 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 $(x, y)$ $(x,y)$ is equal to $k$ $k$ times $y$ $y$ . When we solve it, we end up with a function $y$ $y$ of $x$ $x$ :

$y = C \cdot e^{k x}$ $y = C * e^(kx)$

where $C$ $C$ is a constant due to integration. In this case, $C$ $C$ 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 $y$ $y$ 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, $y$ $y$ , which represents the size of the droplet at time $t$ $t$ .

This situation translates into the following differential equation:

$\frac{d y}{d t} = - 0.1 \cdot y$ $dy/dt = - 0.1 * y$

First step in solving is to separate the variables:

$- \frac{1}{0.1 y} d y = d t$ $-1/(0.1y) dy = dt$

Now, we will simply integrate:

$\int - \frac{1}{0.1 y} d y = \int 1 d t$ $int -1/(0.1y) dy = int 1 dt$

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

$\int - \frac{1}{0.1 y} d y = t + C$ $int -1/(0.1y) dy = t + C$

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

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

And now this is easily solved:

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

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

$ln y = - 0.1 t + C$ $ln y = -0.1t + C$

Exponentiate both sides:

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

This can be rewritten as:

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

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

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

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

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

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

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 $(x, y)$ $(x, y)$ the slope is equal to $- 0.1 \cdot y$ $-0.1*y$ .

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

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