## 17Calculus Differential Equations - Population Dynamics

This page builds on the discussion of exponential growth and decay and specifically applies it to the dynamics of population change, i.e. population growth and decline. In fact, we will touch on basic exponential growth and decay and then extend the discussion to more realistic models. Here are the four topics covered on this page. The initial condition for each equation is $$p(0)=p_0$$.

exponential growth

$$p' = rp$$

$$p(t) = p_0e^{rt}$$

exponential growth with critical threshold

$$\displaystyle{ \frac{dp}{dt} = -r \left( 1-\frac{p}{T} \right)p }$$

$$\displaystyle{ p(t) = \frac{p_0T}{p_0+(T-p_0)e^{rt}} }$$

logistic growth

$$\displaystyle{ \frac{dp}{dt} = r \left( 1-\frac{p}{L} \right)p }$$

$$\displaystyle{ p(t) = \frac{p_0L}{p_0+(L-p_0)e^{-rt}} }$$

logistic growth with critical threshold

$$\displaystyle{ \frac{dp}{dt} = -r \left( 1-\frac{p}{T} \right)\left( 1-\frac{p}{L} \right)p }$$

$$\displaystyle{ \frac{(L/p-1)^T}{(T/p-1)^L} = \frac{(L/p_0-1)^T)}{(T/p_0-1)^L}e^{-(L-T)rt)} }$$

Exponential Growth

plot 1 - exponential growth

$$p(t) = p_0e^{rt}$$

As discussed on the exponential growth and decay page, we start with the differential equation $$p' = rp$$, which solves to $$p(t) = p_0e^{rt}$$. In this equation, $$p(0) = p_0$$ is the initial population and $$r$$ is the growth rate. The graph of $$p(t)$$ with $$p_0 = 1$$ for various values of $$r > 0$$ is shown in plot 1.

There are two problems with this model when we are trying predict population growth.
1. Notice that as $$t$$ increases, $$p(t)$$ increases without bound. Although this model for population growth may be accurate over short time periods, it is not realistic as space, food supply and other resources would limit the growth in any reasonable situation.
2. Some populations cannot sustain itself if there are too few members. For example, our model shows that even when starting out with only one member, the population increases. Again, this is not realistic.

In order to handle the limitations of resources, we will use another model called logistic growth but first, let's look at how to handle the second issue, starting out with enough members to sustain a population.

Exponential Growth with Critical Threshold

autonomous equation

A special type of differential equation of the form $$y' = f(y)$$ where the independent variable does not explicitly appear in the equation.

plot 2 - exponential growth with threshold

$$\displaystyle{ p(t) = \frac{p_0T}{p_0+(T-p_0)e^{rt}} }$$

In order to handle the situation when a population needs more than one member to sustain itself or it will die out, we use another autonomous differential equation

 $$\displaystyle{ \frac{dp}{dt} = -r \left( 1-\frac{p}{T} \right)p }$$

where $$r$$ is the growth rate and $$T$$ is called the critical threshold (or just threshold).

The solution to this differential equation is $$\displaystyle{ p(t) = \frac{p_0T}{p_0+(T-p_0)e^{rt}} }$$. (See the practice problems for the derivation of this equation.) Let's discuss what is going on with the equation. We have plotted $$p(t)$$ for various values of $$p(0)=p_0$$ in plot 2.

When $$p_0=T$$, i.e. the initial population is exactly the threshold value $$T$$, the coefficient of the exponential term is zero, so $$p(t)=T$$ and the population stays exactly at the threshold value for all time $$t>0$$. However, if the initial population is either above or below the threshold value, the exponential term $$e^{rt}$$ drives the plots away from the threshold. This seems logical since, if the initial population is higher than the threshold value, the population will increase and if the initial population is below the threshold value the population will decrease, eventually dying out. In the second case, there are not enough individuals to sustain the population.

Okay, so far, so good. We have dealt with the second problem of the basic exponential growth model. Now let's address the first problem dealing with limited resources.

Logistic Growth

plot 3 - logistic growth

$$\displaystyle{ p(t) = \frac{p_0L}{p_0+(L-p_0)e^{-rt}} }$$

In the real world, populations are often limited by things like resources and space. To handle this situation, we model the population with the autonomous differential equation

 $$\displaystyle{ \frac{dp}{dt} = r \left( 1-\frac{p}{T} \right)p }$$

where $$r$$ is the growth rate and $$L$$ is called the saturation level.

If you look carefully at this equation and compare it to the equation for critical threshold above, you will see that they are almost identical except for a minus sign. Even though they are similar, the behaviour of this equation is quite different from the previous one. If you followed the derivation of the threshold equation (in a practice problem), you can derive the solution here as $$\displaystyle{ p(t) = \frac{p_0L}{p_0+(L-p_0)e^{-rt}} }$$. Notice that the exponential here is $$e^{-rt}$$.

As we did before, plot 3 shows the solution for various values of $$p(0)=p_0$$. Notice that the exponential in this equation drives the solution toward the saturation level (as compared to away from the threshold in the previous section).

Logistic Growth with Critical Threshold

plot 4 - logistic growth with threshold $$p(t)$$

In the last two sections we have addressed both problems with the basic exponential individually. Let's combine the two solutions into one equation. Our differential equation is

 $$\displaystyle{ \frac{dp}{dt} = -r \left( 1-\frac{p}{T} \right)\left( 1-\frac{p}{L} \right)p }$$

where $$r$$ is the growth rate, $$T$$ is the threshold and $$L$$ is the saturation level. In this discussion, we will assume that $$0 < T < L$$, i.e. the saturation level (limit on resources) is higher than the threshold.

The plot of $$p(t)$$ for various initial conditions is shown in plot 4. Notice how the plot tends to diverge away from the threshold but converge towards the saturation limit as we expected. The equation of the solution, in implicit form, is $$\displaystyle{ \frac{(L/p-1)^T}{(T/p-1)^L} = \frac{(L/p_0-1)^T)}{(T/p_0-1)^L}e^{-(L-T)rt)} }$$ (see the practice problems for a derivation of this equation)

Practice

Solve these differential equations using the initial conditions, if given.

Solve $$p' = rp$$, $$p(0)=p_0$$

Problem Statement

Solve $$p' = rp$$, $$p(0)=p_0$$

Final Answer

$$p(t) = p_0 e^{rt}$$

Problem Statement

Solve $$p' = rp$$, $$p(0)=p_0$$

Solution

$$\begin{array}{rcl} p' &=& rp \\ \displaystyle{ \frac{dp}{dt} } &=& rp \\ \displaystyle{ \frac{dp}{p} } &=& r~dt \\ \displaystyle{ \int{ \frac{dp}{p} } } &=& \displaystyle{ \int{ r~dt } } \\ \ln |p| &=& rt+C \\ e^{\ln |p|} &=& e^{rt+C} \\ p &=& ke^{rt} \\ \end{array}$$
general solution $$p=ke^{rt}$$
apply the initial condition $$p(0)=p_0$$
$$p_0=ke^0 \to k=p_0$$
particular solution $$p(t)=p_0e^{rt}$$

Final Answer

$$p(t) = p_0 e^{rt}$$

Log in to rate this practice problem and to see it's current rating.

Solve $$\displaystyle{ \frac{dp}{dt} = -r \left( 1-\frac{p}{T} \right)p }$$, $$p(0)=p_0$$

Problem Statement

Solve $$\displaystyle{ \frac{dp}{dt} = -r \left( 1-\frac{p}{T} \right)p }$$, $$p(0)=p_0$$

Final Answer

$$\displaystyle{ p(t) = \frac{Tp_0}{p_0+(T-p_0)e^{rt} } }$$

Problem Statement

Solve $$\displaystyle{ \frac{dp}{dt} = -r \left( 1-\frac{p}{T} \right)p }$$, $$p(0)=p_0$$

Solution

This differential equation is separable.
$$\displaystyle{ \frac{dp}{(1-p/T)p} = -r~dt }$$
On the left side of the equation, we need to use partial fractions.

 $$\displaystyle{ \frac{1}{(1-p/T)p} = \frac{A}{1-p/T} + \frac{B}{p} }$$ $$1 = Ap+B(1-p/T)$$ $$1 = Ap+B-Bp/T$$ $$1 = p(A-B/T) + B$$

Remember that the variable in the above partial fraction expansion is $$p$$. Now we equate coefficients to find the constants A and B.
$$B=1$$
$$A-B/T=0 \to A=B/T \to A=1/T$$
So now our integrals are

 $$\displaystyle{ \int{\frac{1/T}{1-p/T} + \frac{1}{p} dp} = \int{-r~dt} }$$ $$\displaystyle{ \int{\frac{1}{T-p} + \frac{1}{p} dp} } = -rt+C$$ $$\displaystyle{ \ln \left| \frac{p}{T-p} \right| } = -rt+C$$ $$-\ln |T-p| + \ln |p| = -rt+C$$ $$\displaystyle{ \frac{p}{T-p} = e^{-rt+C} }$$

Before we apply the initial condition to find C, let's solve the last equation for $$p$$.
$$\begin{array}{rcl} \displaystyle{ \frac{p}{T-p} } &=& e^{-rt+C} \\ \displaystyle{ \frac{1}{T/p-1} } &=& \\ \displaystyle{ \frac{T}{p} - 1 } &=& e^{rt-C} \\ \displaystyle{ \frac{T}{p} } &=& 1+e^{rt-C} \\ p &=& \displaystyle{ \frac{T}{1+ke^{rt}} } \end{array}$$
In the last line, we set $$k=e^{-C}$$ just for convenience. Now we will apply the initial condition $$p(0)=p_0$$.
$$\begin{array}{rcl} p_0 &=& \displaystyle{ \frac{T}{1+ke^0} } \\ 1+k &=& \displaystyle{ \frac{T}{p_0} } \\ k &=& \displaystyle{ \frac{T}{p_0}-1 } \\ k &=& \displaystyle{ \frac{T-p_0}{p_0} } \end{array}$$
Our particular solution is
$$\displaystyle{ p(t) = \frac{T}{1+e^{rt}(T-p_0)/p_0} }$$
$$\displaystyle{ p(t) = \frac{Tp_0}{p_0+(T-p_0)e^{rt} } }$$

Final Answer

$$\displaystyle{ p(t) = \frac{Tp_0}{p_0+(T-p_0)e^{rt} } }$$

Log in to rate this practice problem and to see it's current rating.

Solve $$\displaystyle{ \frac{dp}{dt} = -r \left( 1-\frac{p}{T} \right)\left( 1-\frac{p}{L} \right)p }$$, $$p(0)=p_0$$

Problem Statement

Solve $$\displaystyle{ \frac{dp}{dt} = -r \left( 1-\frac{p}{T} \right)\left( 1-\frac{p}{L} \right)p }$$, $$p(0)=p_0$$

Final Answer

$$\displaystyle{ \frac{(L/p-1)^T}{(T/p-1)^L} = \frac{(L/p_0-1)^T)}{(T/p_0-1)^L}e^{-(L-T)rt)} }$$

Problem Statement

Solve $$\displaystyle{ \frac{dp}{dt} = -r \left( 1-\frac{p}{T} \right)\left( 1-\frac{p}{L} \right)p }$$, $$p(0)=p_0$$

Solution

The differential equation is separable.
$$\displaystyle{ \frac{dp}{dt} = -r \left( 1-\frac{p}{T} \right)\left( 1-\frac{p}{L} \right)p }$$
$$\displaystyle{ \frac{dp}{(1-p/T)(1-p/L)p} = -r~dt }$$
Using partial fraction expansion, we have
$$\displaystyle{ \int{ \frac{L}{T(L-T)}\frac{1}{1-p/T} + \frac{T}{L(T-L)}\frac{1}{1-p/L} + \frac{1}{p}dp } = \int{ -r~dt } }$$
$$\displaystyle{ \frac{-L}{L-T}\ln |T-p| - \frac{T}{T-L} \ln|L-p| + \ln|p| = -rt+C }$$
Multiply both sides of the equation by $$L-T$$ to get
$$\displaystyle{ -\ln|T-p|^L + \ln|L-p|^T + \ln|p|^{L-T} = (L-T)(-rt+C) }$$
Combining the natural log terms and doing some algebra, we get
$$\displaystyle{ \ln \left|\frac{p^{L-T}(L-p)^T}{(T-p)^L} \right| = (L-T)(-rt+C) }$$
$$\displaystyle{ \frac{p^{L-T}(L-p)^T}{(T-p)^L} = exp( (L-T)(-rt+C) ) }$$
$$\displaystyle{ \frac{(L/p-1)^T}{(T/p-1)^L} = k e^{-(L-T)rt} }$$

Now we apply the initial condition, $$p(0)=p_0$$ to get
$$\displaystyle{ k=\frac{(L/p_0-1)^T}{(T/p_0-1)^L} }$$
$$\displaystyle{ \frac{(L/p-1)^T}{(T/p-1)^L} = \frac{(L/p_0-1)^T)}{(T/p_0-1)^L}e^{-(L-T)rt)} }$$

We can leave this equation in implicit form, i.e. not explicitly solve for $$p(t)$$.

Final Answer

$$\displaystyle{ \frac{(L/p-1)^T}{(T/p-1)^L} = \frac{(L/p_0-1)^T)}{(T/p_0-1)^L}e^{-(L-T)rt)} }$$

Log in to rate this practice problem and to see it's current rating.

You CAN Ace Differential Equations

### Topics You Need To Understand For This Page

 precalculus - logarithms integration basics of differential equations exponential growth and decay

### Related Topics and Links

external links you may find helpful

population dynamics youtube playlist

### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

To bookmark this page and practice problems, log in to your account or set up a free account.

### Search Practice Problems

Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.

free ideas to save on books

 The 17Calculus and 17Precalculus iOS and Android apps are no longer available for download. If you are still using a previously downloaded app, your app will be available until the end of 2020, after which the information may no longer be available. However, do not despair. All the information (and more) is now available on 17calculus.com for free.

When using the material on this site, check with your instructor to see what they require. Their requirements come first, so make sure your notation and work follow their specifications.

DISCLAIMER - 17Calculus owners and contributors are not responsible for how the material, videos, practice problems, exams, links or anything on this site are used or how they affect the grades or projects of any individual or organization. We have worked, to the best of our ability, to ensure accurate and correct information on each page and solutions to practice problems and exams. However, we do not guarantee 100% accuracy. It is each individual's responsibility to verify correctness and to determine what different instructors and organizations expect. How each person chooses to use the material on this site is up to that person as well as the responsibility for how it impacts grades, projects and understanding of calculus, math or any other subject. In short, use this site wisely by questioning and verifying everything. If you see something that is incorrect, contact us right away so that we can correct it.

Links and banners on this page are affiliate links. We carefully choose only the affiliates that we think will help you learn. Clicking on them and making purchases help you support 17Calculus at no extra charge to you. However, only you can decide what will actually help you learn. So think carefully about what you need and purchase only what you think will help you.

We use cookies on this site to enhance your learning experience.

Copyright © 2010-2020 17Calculus, All Rights Reserved         [Support]     [About]

17Calculus
We use cookies to ensure that we give you the best experience on our website. By using this site, you agree to our Website Privacy Policy.