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

1 Answer
Aug 17, 2014

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#.