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

Question

Scarlett Allison · Accepted Answer

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#.

Ezra Adams · Answer

undefined The differential equation that models exponential growth and decay is:

dy/dt = ky

Where:
- dy/dt represents the rate of change of the quantity y with respect to time t.
- k is a constant representing the growth rate (if positive) or decay rate (if negative) of the quantity y.