This model illustrates predator prey interactions using real-life data of wolf and moose populations on the Isle Royale. It was "cloned" from a model that InsightMaker provides to its users, at https://insightmaker.com/insight/2068/Isle-Royale-Predator-Prey-Interactions Thanks Scott Fortmann-Roe.
Math Modeling Environment Ecology Populations
This model illustrates predator prey interactions using real-life data of wolf and moose populations on the Isle Royale. It was "cloned" from a model that InsightMaker provides to its users, at
https://insightmaker.com/insight/2068/Isle-Royale-Predator-Prey-Interactions
Thanks Scott Fortmann-Roe.

I've added in an adjustment to handle population.

I've created a Mathematica file that replicates the model, at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/Moose-n-Wolf-InsightMaker.nb

It allows one to experiment with adjusting the initial number of moose and wolves on the island.

I used steepest descent in Mathematica to optimize the parameters, with my objective data being the ratio of wolves to moose. You can try my (admittedly) kludgy code, at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/Moose-n-Wolf-InsightMaker-BestFit.nb

{WolfBirthRateFactorStart,
WolfDeathRateStart,
MooseBirthRateStart,
MooseDeathRateFactorStart,
moStart,
woStart} =
{0.000267409,
0.239821,
0.269755,
0.0113679,
591,
23.};

Isle Royale: Predator/Prey Model for Moose and Wolves, with Total Population
Andrew E Long
This simple model demonstrates logistic growth.The differential equation looks like y'(t)=by(t)(1-y(t)/K) where K is the carrying capacity of the quantity y. Alternatively, y'(t)=by(t) - b/K*y(t)^2 so the growth term suggests exponential growth, but there is a loss term is of the form b/K y(t
Math Modeling Mat375
This simple model demonstrates logistic growth.The differential equation looks like

y'(t)=by(t)(1-y(t)/K)

where K is the carrying capacity of the quantity y. Alternatively,

y'(t)=by(t) - b/K*y(t)^2

so the growth term suggests exponential growth, but there is a loss term is of the form b/K y(t) -- loss is proportional to population (crowding).

A comparable Mathematica file is available at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/LogisticGrowth-and-DecayModel.nb
Clone of Logistic Growth
Samuel Kaelin
The basic model of Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators ( Tyson, et al.) demonstrates logistic growth in prey, and in predator (with prey dependence for carrying capacity). But interestingly, one possibility is limit cycles, which mimic the
Math Modeling Mat375
The basic model of Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators (Tyson, et al.) demonstrates logistic growth in prey, and in predator (with prey dependence for carrying capacity). But interestingly, one possibility is limit cycles, which mimic the cycling of the populations in nature.

The differential equations for the population of hare (x) is

x'(t) = rx(1-x/K)
            - gamma x^2/(x^2+eta^2)
            - alpha y x/(x+mu)

where K is the logistic carrying capacity of the prey (hare), in the absence of predation; the second term is a "generalist predation" term;  and the third term is the "specialist predation" (in the limit as the prey gets big, this becomes simply proportional to y (the lynx population)).

The differential equations for the population of lynx (y) is

y'(t) = sy(1- qy/x) = sy - sqy^2/x

for the predator (lynx), which is essentially logistic growth. Its growth term suggests exponential growth, but there is a loss term of the form sqy^2/x -- loss is proportional to population (crowding), and inversely proportional to prey density. As the hare population goes to zero, so shall the lynx....

As one can see, the prey density won't change if y=x/q. If the prey density were not changing at the same time, the system would be at equilibrium.

In this InsightMaker model, I scaled the second equation by multiplying by q, then replace y by w=qy throughout both equations. This requires a slight change in the prey equation -- alpha replaced by the ratio of alpha/q.  (I used my favorite mathematical trick, of multiplying by the appropriate form of 1!)

So what we're really looking at here is the system

x'(t) = rx(1-x/K)
            - gamma x^2/(x^2+eta^2)
            - alpha/q w x/(x+mu)
w'(t) = sw(1- w/x)

where w(t)=qy(t).

Tyson, et al. took q to be about 212 for hare and lynx -- so that it requires about 212 hare to allow for one lynx to survive at "equilibrium".

However, when alpha -- the hares/lynx/year -- gets sufficiently large (e.g. 1867 -- and that does seem like a lot of hares per lynx per year...:), limit cycles develop (rather than a stable equilibrium). This means that the populations oscillate about the equilibrium values, rather than stabilize at those values.

Author: Andy Long, Northern Kentucky University (2020)

Reference: Tyson, Rebecca, Sheena Haines,  Karen Hodges. Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators. Theoretical Ecology. 3, 97–111 (2010). https://doi.org/10.1007/s12080-009-0057-1

Resource: A comparable Mathematica model can be found at  http://ceadserv1.nku.edu/longa//classes/mat375/days/Mathematica/BasicModel.nb,
which allows one to experiment a little more easily than one can with this InsightMaker model.
Clone of Basic Model, Tyson Lynx and Hare
Proctor Mercer
This is a first example of a simple SIR (Susceptible, Infected, Recovered) model. There are three pools of individuals: those who are infected (without them, no disease!), the pool of those who are at risk (susceptible), and the recovered -- who may lose their immunity and become susceptible again
Mat375 SIR Math Modeling
This is a first example of a simple SIR (Susceptible, Infected, Recovered) model.

There are three pools of individuals: those who are infected (without them, no disease!), the pool of those who are at risk (susceptible), and the recovered -- who may lose their immunity and become susceptible again.

A comparable model in Mathematica is available at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/SIRModel.nb

Clone of A Simple SIR (Susceptible, Infected, Recovered) Example
Jake Moore
The basic model of Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators ( Tyson, et al.) demonstrates logistic growth in prey, and in predator (with prey dependence for carrying capacity). But interestingly, one possibility is limit cycles, which mimic the
Mat375 Math Modeling
The basic model of Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators (Tyson, et al.) demonstrates logistic growth in prey, and in predator (with prey dependence for carrying capacity). But interestingly, one possibility is limit cycles, which mimic the cycling of the populations in nature.

The differential equations for the population of hare (x) is

x'(t) = rx(1-x/K)
            - gamma x^2/(x^2+eta^2)
            - alpha y x/(x+mu)

where K is the logistic carrying capacity of the prey (hare), in the absence of predation; the second term is a "generalist predation" term;  and the third term is the "specialist predation" (in the limit as the prey gets big, this becomes simply proportional to y (the lynx population)).

The differential equations for the population of lynx (y) is

y'(t) = sy(1- qy/x) = sy - sqy^2/x

for the predator (lynx), which is essentially logistic growth. Its growth term suggests exponential growth, but there is a loss term of the form sqy^2/x -- loss is proportional to population (crowding), and inversely proportional to prey density. As the hare population goes to zero, so shall the lynx....

As one can see, the prey density won't change if y=x/q. If the prey density were not changing at the same time, the system would be at equilibrium.

In this InsightMaker model, I scaled the second equation by multiplying by q, then replace y by w=qy throughout both equations. This requires a slight change in the prey equation -- alpha replaced by the ratio of alpha/q.  (I used my favorite mathematical trick, of multiplying by the appropriate form of 1!)

So what we're really looking at here is the system

x'(t) = rx(1-x/K)
            - gamma x^2/(x^2+eta^2)
            - alpha/q w x/(x+mu)
w'(t) = sw(1- w/x)

where w(t)=qy(t).

Tyson, et al. took q to be about 212 for hare and lynx -- so that it requires about 212 hare to allow for one lynx to survive at "equilibrium".

However, when alpha -- the hares/lynx/year -- gets sufficiently large (e.g. 1867 -- and that does seem like a lot of hares per lynx per year...:), limit cycles develop (rather than a stable equilibrium). This means that the populations oscillate about the equilibrium values, rather than stabilize at those values.

Author: Andy Long, Northern Kentucky University (2020)

Reference: Tyson, Rebecca, Sheena Haines,  Karen Hodges. Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators. Theoretical Ecology. 3, 97–111 (2010). https://doi.org/10.1007/s12080-009-0057-1

Resource: A comparable Mathematica model can be found at  http://ceadserv1.nku.edu/longa//classes/mat375/days/Mathematica/BasicModel.nb,
which allows one to experiment a little more easily than one can with this InsightMaker model.
Clone of Basic Model, Tyson Lynx and Hare
Jake Moore
The basic model of Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators ( Tyson, et al.) demonstrates logistic growth in prey, and in predator (with prey dependence for carrying capacity). But interestingly, one possibility is limit cycles, which mimic the
Math Modeling Mat375
The basic model of Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators (Tyson, et al.) demonstrates logistic growth in prey, and in predator (with prey dependence for carrying capacity). But interestingly, one possibility is limit cycles, which mimic the cycling of the populations in nature.

The differential equations for the population of hare (x) is

x'(t) = rx(1-x/K)
            - gamma x^2/(x^2+eta^2)
            - alpha y x/(x+mu)

where K is the logistic carrying capacity of the prey (hare), in the absence of predation; the second term is a "generalist predation" term;  and the third term is the "specialist predation" (in the limit as the prey gets big, this becomes simply proportional to y (the lynx population)).

The differential equations for the population of lynx (y) is

y'(t) = sy(1- qy/x) = sy - sqy^2/x

for the predator (lynx), which is essentially logistic growth. Its growth term suggests exponential growth, but there is a loss term of the form sqy^2/x -- loss is proportional to population (crowding), and inversely proportional to prey density. As the hare population goes to zero, so shall the lynx....

As one can see, the prey density won't change if y=x/q. If the prey density were not changing at the same time, the system would be at equilibrium.

In this InsightMaker model, I scaled the second equation by multiplying by q, then replace y by w=qy throughout both equations. This requires a slight change in the prey equation -- alpha replaced by the ratio of alpha/q.  (I used my favorite mathematical trick, of multiplying by the appropriate form of 1!)

So what we're really looking at here is the system

x'(t) = rx(1-x/K)
            - gamma x^2/(x^2+eta^2)
            - alpha/q w x/(x+mu)
w'(t) = sw(1- w/x)

where w(t)=qy(t).

Tyson, et al. took q to be about 212 for hare and lynx -- so that it requires about 212 hare to allow for one lynx to survive at "equilibrium".

However, when alpha -- the hares/lynx/year -- gets sufficiently large (e.g. 1867 -- and that does seem like a lot of hares per lynx per year...:), limit cycles develop (rather than a stable equilibrium). This means that the populations oscillate about the equilibrium values, rather than stabilize at those values.

Author: Andy Long, Northern Kentucky University (2020)

Reference: Tyson, Rebecca, Sheena Haines,  Karen Hodges. Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators. Theoretical Ecology. 3, 97–111 (2010). https://doi.org/10.1007/s12080-009-0057-1

Resource: A comparable Mathematica model can be found at  http://ceadserv1.nku.edu/longa//classes/mat375/days/Mathematica/BasicModel.nb,
which allows one to experiment a little more easily than one can with this InsightMaker model.
Clone of Basic Model, Tyson Lynx and Hare
Jordan
This is a first example of a simple SIR (Susceptible, Infected, Recovered) model. There are three pools of individuals: those who are infected (without them, no disease!), the pool of those who are at risk (susceptible), and the recovered -- who may lose their immunity and become susceptible again
SIR Mat375 Math Modeling
This is a first example of a simple SIR (Susceptible, Infected, Recovered) model.

There are three pools of individuals: those who are infected (without them, no disease!), the pool of those who are at risk (susceptible), and the recovered -- who may lose their immunity and become susceptible again.

A comparable model in Mathematica is available at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/SIRModel.nb

Clone of A Simple SIR (Susceptible, Infected, Recovered) Example
Proctor Mercer
This simple model demonstrates logistic growth.The differential equation looks like y'(t)=by(t)(1-y(t)/K) where K is the carrying capacity of the quantity y. Alternatively, y'(t)=by(t) - b/K*y(t)^2 so the growth term suggests exponential growth, but there is a loss term is of the form b/K y(t
Mat375 Math Modeling
This simple model demonstrates logistic growth.The differential equation looks like

y'(t)=by(t)(1-y(t)/K)

where K is the carrying capacity of the quantity y. Alternatively,

y'(t)=by(t) - b/K*y(t)^2

so the growth term suggests exponential growth, but there is a loss term is of the form b/K y(t) -- loss is proportional to population (crowding).

A comparable Mathematica file is available at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/LogisticGrowth-and-DecayModel.nb
Clone of Logistic Growth
Jon Ford
This simple model demonstrates logistic growth.The differential equation looks like y'(t)=by(t)(1-y(t)/K) where K is the carrying capacity of the quantity y. Alternatively, y'(t)=by(t) - b/K*y(t)^2 so the growth term suggests exponential growth, but there is a loss term is of the form b/K y(t
Math Modeling Mat375
This simple model demonstrates logistic growth.The differential equation looks like

y'(t)=by(t)(1-y(t)/K)

where K is the carrying capacity of the quantity y. Alternatively,

y'(t)=by(t) - b/K*y(t)^2

so the growth term suggests exponential growth, but there is a loss term is of the form b/K y(t) -- loss is proportional to population (crowding).

A comparable Mathematica file is available at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/LogisticGrowth-and-DecayModel.nb
Clone of Logistic Growth
Jake Moore
This is a first example of a simple SIR (Susceptible, Infected, Recovered) model. There are three pools of individuals: those who are infected (without them, no disease!), the pool of those who are at risk (susceptible), and the recovered -- who may lose their immunity and become susceptible again
SIR Math Modeling Mat375
This is a first example of a simple SIR (Susceptible, Infected, Recovered) model.

There are three pools of individuals: those who are infected (without them, no disease!), the pool of those who are at risk (susceptible), and the recovered -- who may lose their immunity and become susceptible again.

A comparable model in Mathematica is available at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/SIRModel.nb

Clone of A Simple SIR (Susceptible, Infected, Recovered) Example
Jon Ford
The basic model of Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators ( Tyson, et al.) demonstrates logistic growth in prey, and in predator (with prey dependence for carrying capacity). But interestingly, one possibility is limit cycles, which mimic the
Math Modeling Mat375
The basic model of Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators (Tyson, et al.) demonstrates logistic growth in prey, and in predator (with prey dependence for carrying capacity). But interestingly, one possibility is limit cycles, which mimic the cycling of the populations in nature.

The differential equations for the population of hare (x) is

x'(t) = rx(1-x/K)
            - gamma x^2/(x^2+eta^2)
            - alpha y x/(x+mu)

where K is the logistic carrying capacity of the prey (hare), in the absence of predation; the second term is a "generalist predation" term;  and the third term is the "specialist predation" (in the limit as the prey gets big, this becomes simply proportional to y (the lynx population)).

The differential equations for the population of lynx (y) is

y'(t) = sy(1- qy/x) = sy - sqy^2/x

for the predator (lynx), which is essentially logistic growth. Its growth term suggests exponential growth, but there is a loss term of the form sqy^2/x -- loss is proportional to population (crowding), and inversely proportional to prey density. As the hare population goes to zero, so shall the lynx....

As one can see, the prey density won't change if y=x/q. If the prey density were not changing at the same time, the system would be at equilibrium.

In this InsightMaker model, I scaled the second equation by multiplying by q, then replace y by w=qy throughout both equations. This requires a slight change in the prey equation -- alpha replaced by the ratio of alpha/q.  (I used my favorite mathematical trick, of multiplying by the appropriate form of 1!)

So what we're really looking at here is the system

x'(t) = rx(1-x/K)
            - gamma x^2/(x^2+eta^2)
            - alpha/q w x/(x+mu)
w'(t) = sw(1- w/x)

where w(t)=qy(t).

Tyson, et al. took q to be about 212 for hare and lynx -- so that it requires about 212 hare to allow for one lynx to survive at "equilibrium".

However, when alpha -- the hares/lynx/year -- gets sufficiently large (e.g. 1867 -- and that does seem like a lot of hares per lynx per year...:), limit cycles develop (rather than a stable equilibrium). This means that the populations oscillate about the equilibrium values, rather than stabilize at those values.

Author: Andy Long, Northern Kentucky University (2020)

Reference: Tyson, Rebecca, Sheena Haines,  Karen Hodges. Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators. Theoretical Ecology. 3, 97–111 (2010). https://doi.org/10.1007/s12080-009-0057-1

Resource: A comparable Mathematica model can be found at  http://ceadserv1.nku.edu/longa//classes/mat375/days/Mathematica/BasicModel.nb,
which allows one to experiment a little more easily than one can with this InsightMaker model.
Clone of Basic Model, Tyson Lynx and Hare
Jacob Adkins
This simple model demonstrates exponential growth or decay in a population. A comparable Mathematica file is at http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/ExponentialGrowth-and-DecayModel.nb
Mat375 Math Modeling
This simple model demonstrates exponential growth or decay in a population.

A comparable Mathematica file is at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/ExponentialGrowth-and-DecayModel.nb
Clone of Exponential Growth
Travis Rhoades
The basic model of Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators ( Tyson, et al.) demonstrates logistic growth in prey, and in predator (with prey dependence for carrying capacity). But interestingly, one possibility is limit cycles, which mimic the
Math Modeling Mat375
The basic model of Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators (Tyson, et al.) demonstrates logistic growth in prey, and in predator (with prey dependence for carrying capacity). But interestingly, one possibility is limit cycles, which mimic the cycling of the populations in nature.

The differential equations for the population of hare (x) is

x'(t) = rx(1-x/K)
            - gamma x^2/(x^2+eta^2)
            - alpha y x/(x+mu)

where K is the logistic carrying capacity of the prey (hare), in the absence of predation; the second term is a "generalist predation" term;  and the third term is the "specialist predation" (in the limit as the prey gets big, this becomes simply proportional to y (the lynx population)).

The differential equations for the population of lynx (y) is

y'(t) = sy(1- qy/x) = sy - sqy^2/x

for the predator (lynx), which is essentially logistic growth. Its growth term suggests exponential growth, but there is a loss term of the form sqy^2/x -- loss is proportional to population (crowding), and inversely proportional to prey density. As the hare population goes to zero, so shall the lynx....

As one can see, the prey density won't change if y=x/q. If the prey density were not changing at the same time, the system would be at equilibrium.

In this InsightMaker model, I scaled the second equation by multiplying by q, then replace y by w=qy throughout both equations. This requires a slight change in the prey equation -- alpha replaced by the ratio of alpha/q.  (I used my favorite mathematical trick, of multiplying by the appropriate form of 1!)

So what we're really looking at here is the system

x'(t) = rx(1-x/K)
            - gamma x^2/(x^2+eta^2)
            - alpha/q w x/(x+mu)
w'(t) = sw(1- w/x)

where w(t)=qy(t).

Tyson, et al. took q to be about 212 for hare and lynx -- so that it requires about 212 hare to allow for one lynx to survive at "equilibrium".

However, when alpha -- the hares/lynx/year -- gets sufficiently large (e.g. 1867 -- and that does seem like a lot of hares per lynx per year...:), limit cycles develop (rather than a stable equilibrium). This means that the populations oscillate about the equilibrium values, rather than stabilize at those values.

Author: Andy Long, Northern Kentucky University (2020)

Reference: Tyson, Rebecca, Sheena Haines,  Karen Hodges. Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators. Theoretical Ecology. 3, 97–111 (2010). https://doi.org/10.1007/s12080-009-0057-1

Resource: A comparable Mathematica model can be found at  http://ceadserv1.nku.edu/longa//classes/mat375/days/Mathematica/BasicModel.nb,
which allows one to experiment a little more easily than one can with this InsightMaker model.
Clone of Basic Model, Tyson Lynx and Hare
Proctor Mercer
This simple model demonstrates logistic growth.The differential equation looks like y'(t)=by(t)(1-y(t)/K) where K is the carrying capacity of the quantity y. Alternatively, y'(t)=by(t) - b/K*y(t)^2 so the growth term suggests exponential growth, but there is a loss term is of the form b/K y(t
Math Modeling Mat375
This simple model demonstrates logistic growth.The differential equation looks like

y'(t)=by(t)(1-y(t)/K)

where K is the carrying capacity of the quantity y. Alternatively,

y'(t)=by(t) - b/K*y(t)^2

so the growth term suggests exponential growth, but there is a loss term is of the form b/K y(t) -- loss is proportional to population (crowding).

A comparable Mathematica file is available at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/LogisticGrowth-and-DecayModel.nb
Clone of Clone of Logistic Growth
Jacob Adkins
The basic model of Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators ( Tyson, et al.) demonstrates logistic growth in prey, and in predator (with prey dependence for carrying capacity). But interestingly, one possibility is limit cycles, which mimic the
Mat375 Math Modeling
The basic model of Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators (Tyson, et al.) demonstrates logistic growth in prey, and in predator (with prey dependence for carrying capacity). But interestingly, one possibility is limit cycles, which mimic the cycling of the populations in nature.

The differential equations for the population of hare (x) is

x'(t) = rx(1-x/K)
            - gamma x^2/(x^2+eta^2)
            - alpha y x/(x+mu)

where K is the logistic carrying capacity of the prey (hare), in the absence of predation; the second term is a "generalist predation" term;  and the third term is the "specialist predation" (in the limit as the prey gets big, this becomes simply proportional to y (the lynx population)).

The differential equations for the population of lynx (y) is

y'(t) = sy(1- qy/x) = sy - sqy^2/x

for the predator (lynx), which is essentially logistic growth. Its growth term suggests exponential growth, but there is a loss term of the form sqy^2/x -- loss is proportional to population (crowding), and inversely proportional to prey density. As the hare population goes to zero, so shall the lynx....

As one can see, the prey density won't change if y=x/q. If the prey density were not changing at the same time, the system would be at equilibrium.

In this InsightMaker model, I scaled the second equation by multiplying by q, then replace y by w=qy throughout both equations. This requires a slight change in the prey equation -- alpha replaced by the ratio of alpha/q.  (I used my favorite mathematical trick, of multiplying by the appropriate form of 1!)

So what we're really looking at here is the system

x'(t) = rx(1-x/K)
            - gamma x^2/(x^2+eta^2)
            - alpha/q w x/(x+mu)
w'(t) = sw(1- w/x)

where w(t)=qy(t).

Tyson, et al. took q to be about 212 for hare and lynx -- so that it requires about 212 hare to allow for one lynx to survive at "equilibrium".

However, when alpha -- the hares/lynx/year -- gets sufficiently large (e.g. 1867 -- and that does seem like a lot of hares per lynx per year...:), limit cycles develop (rather than a stable equilibrium). This means that the populations oscillate about the equilibrium values, rather than stabilize at those values.

Author: Andy Long, Northern Kentucky University (2020)

Reference: Tyson, Rebecca, Sheena Haines,  Karen Hodges. Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators. Theoretical Ecology. 3, 97–111 (2010). https://doi.org/10.1007/s12080-009-0057-1

Resource: A comparable Mathematica model can be found at  http://ceadserv1.nku.edu/longa//classes/mat375/days/Mathematica/BasicModel.nb,
which allows one to experiment a little more easily than one can with this InsightMaker model.
Clone of Clone of Basic Model, Tyson Lynx and Hare
Proctor Mercer
The basic model of Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators ( Tyson, et al.) demonstrates logistic growth in prey, and in predator (with prey dependence for carrying capacity). But interestingly, one possibility is limit cycles, which mimic the
Mat375 Math Modeling
The basic model of Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators (Tyson, et al.) demonstrates logistic growth in prey, and in predator (with prey dependence for carrying capacity). But interestingly, one possibility is limit cycles, which mimic the cycling of the populations in nature.

The differential equations for the population of hare (x) is

x'(t) = rx(1-x/K)
            - gamma x^2/(x^2+eta^2)
            - alpha y x/(x+mu)

where K is the logistic carrying capacity of the prey (hare), in the absence of predation; the second term is a "generalist predation" term;  and the third term is the "specialist predation" (in the limit as the prey gets big, this becomes simply proportional to y (the lynx population)).

The differential equations for the population of lynx (y) is

y'(t) = sy(1- qy/x) = sy - sqy^2/x

for the predator (lynx), which is essentially logistic growth. Its growth term suggests exponential growth, but there is a loss term of the form sqy^2/x -- loss is proportional to population (crowding), and inversely proportional to prey density. As the hare population goes to zero, so shall the lynx....

As one can see, the prey density won't change if y=x/q. If the prey density were not changing at the same time, the system would be at equilibrium.

In this InsightMaker model, I scaled the second equation by multiplying by q, then replace y by w=qy throughout both equations. This requires a slight change in the prey equation -- alpha replaced by the ratio of alpha/q.  (I used my favorite mathematical trick, of multiplying by the appropriate form of 1!)

So what we're really looking at here is the system

x'(t) = rx(1-x/K)
            - gamma x^2/(x^2+eta^2)
            - alpha/q w x/(x+mu)
w'(t) = sw(1- w/x)

where w(t)=qy(t).

Tyson, et al. took q to be about 212 for hare and lynx -- so that it requires about 212 hare to allow for one lynx to survive at "equilibrium".

However, when alpha -- the hares/lynx/year -- gets sufficiently large (e.g. 1867 -- and that does seem like a lot of hares per lynx per year...:), limit cycles develop (rather than a stable equilibrium). This means that the populations oscillate about the equilibrium values, rather than stabilize at those values.

Author: Andy Long, Northern Kentucky University (2020)

Reference: Tyson, Rebecca, Sheena Haines,  Karen Hodges. Modelling the Canada lynx and snowshoe hare population cycle: The role of specialist predators. Theoretical Ecology. 3, 97–111 (2010). https://doi.org/10.1007/s12080-009-0057-1

Resource: A comparable Mathematica model can be found at  http://ceadserv1.nku.edu/longa//classes/mat375/days/Mathematica/BasicModel.nb,
which allows one to experiment a little more easily than one can with this InsightMaker model.
Clone of Basic Model, Tyson Lynx and Hare
Adheke Silas
This simple model demonstrates logistic growth.The differential equation looks like y'(t)=by(t)(1-y(t)/K) where K is the carrying capacity of the quantity y. Alternatively, y'(t)=by(t) - b/K*y(t)^2 so the growth term suggests exponential growth, but there is a loss term is of the form b/K y(t
Math Modeling Mat375
This simple model demonstrates logistic growth.The differential equation looks like

y'(t)=by(t)(1-y(t)/K)

where K is the carrying capacity of the quantity y. Alternatively,

y'(t)=by(t) - b/K*y(t)^2

so the growth term suggests exponential growth, but there is a loss term is of the form b/K y(t) -- loss is proportional to population (crowding).

A comparable Mathematica file is available at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/LogisticGrowth-and-DecayModel.nb
Clone of Logistic Growth
Kate Quinn
This is a first example of a simple SIR (Susceptible, Infected, Recovered) model. There are three pools of individuals: those who are infected (without them, no disease!), the pool of those who are at risk (susceptible), and the recovered -- who may lose their immunity and become susceptible again
SIR Math Modeling Mat375
This is a first example of a simple SIR (Susceptible, Infected, Recovered) model.

There are three pools of individuals: those who are infected (without them, no disease!), the pool of those who are at risk (susceptible), and the recovered -- who may lose their immunity and become susceptible again.

A comparable model in Mathematica is available at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/SIRModel.nb

Clone of A Simple SIR (Susceptible, Infected, Recovered) Example
luke vanlaningham
This simple model demonstrates logistic growth.The differential equation looks like y'(t)=by(t)(1-y(t)/K) where K is the carrying capacity of the quantity y. Alternatively, y'(t)=by(t) - b/K*y(t)^2 so the growth term suggests exponential growth, but there is a loss term is of the form b/K y(t
Mat375 Math Modeling
This simple model demonstrates logistic growth.The differential equation looks like

y'(t)=by(t)(1-y(t)/K)

where K is the carrying capacity of the quantity y. Alternatively,

y'(t)=by(t) - b/K*y(t)^2

so the growth term suggests exponential growth, but there is a loss term is of the form b/K y(t) -- loss is proportional to population (crowding).

A comparable Mathematica file is available at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/LogisticGrowth-and-DecayModel.nb
Clone of Logistic Growth
Proctor Mercer
This model illustrates predator prey interactions using real-life data of wolf and moose populations on the Isle Royale. It was "cloned" from a model that InsightMaker provides to its users, at https://insightmaker.com/insight/2068/Isle-Royale-Predator-Prey-Interactions Thanks Scott Fortmann-Roe.
Populations Mat375 Math Modeling Environment Ecology
This model illustrates predator prey interactions using real-life data of wolf and moose populations on the Isle Royale. It was "cloned" from a model that InsightMaker provides to its users, at
https://insightmaker.com/insight/2068/Isle-Royale-Predator-Prey-Interactions
Thanks Scott Fortmann-Roe.

I've created a Mathematica file that replicates the model, at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/Moose-n-Wolf-InsightMaker.nb

It allows one to experiment with adjusting the initial number of moose and wolves on the island.

I used steepest descent in Mathematica to optimize the parameters, with my objective data being the ratio of wolves to moose. You can try my (admittedly) kludgy code, at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/Moose-n-Wolf-InsightMaker-BestFit.nb

{WolfBirthRateFactorStart,
WolfDeathRateStart,
MooseBirthRateStart,
MooseDeathRateFactorStart,
moStart,
woStart} =
{0.000267409,
0.239821,
0.269755,
0.0113679,
591,
23.};

Clone of Isle Royale: Predator/Prey Model for Moose and Wolves
Hardy Pfeifer
This simple model demonstrates logistic growth.The differential equation looks like y'(t)=by(t)(1-y(t)/K) where K is the carrying capacity of the quantity y. Alternatively, y'(t)=by(t) - b/K*y(t)^2 so the growth term suggests exponential growth, but there is a loss term is of the form b/K y(t
Mat375 Math Modeling
This simple model demonstrates logistic growth.The differential equation looks like

y'(t)=by(t)(1-y(t)/K)

where K is the carrying capacity of the quantity y. Alternatively,

y'(t)=by(t) - b/K*y(t)^2

so the growth term suggests exponential growth, but there is a loss term is of the form b/K y(t) -- loss is proportional to population (crowding).

A comparable Mathematica file is available at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/LogisticGrowth-and-DecayModel.nb
Clone of Logistic Growth
Jon Ford
Detta är en enkel SIR-modell (Susceptible, Infected, Recovered). I den enkla SIR-modellen finns tre kategorier av individer:  1. Grupp av individer som befinner sig i riskzonen för att bli smittade/sjuka -  Susceptible  (mottagliga). 2. Grupp av individer som är infekterade -  Infected
Math Modeling SIR
Detta är en enkel SIR-modell (Susceptible, Infected, Recovered).

I den enkla SIR-modellen finns tre kategorier av individer: 

1. Grupp av individer som befinner sig i riskzonen för att bli smittade/sjuka - Susceptible (mottagliga).

2. Grupp av individer som är infekterade - Infected.   

3. Grupp av individer som är under tillfrisknande men som kan förlora sin immunitet och bli mottagliga igen - Recovered.

Thomas Hedlund Leijon
SIR-model (Susceptible, Infected, Recovered)
Thomas Hedlund Leijon
This model illustrates predator prey interactions using real-life data of wolf and moose populations on the Isle Royale. It was "cloned" from a model that InsightMaker provides to its users, at https://insightmaker.com/insight/2068/Isle-Royale-Predator-Prey-Interactions Thanks Scott Fortmann-Roe.
Populations Mat375 Environment Ecology Math Modeling
This model illustrates predator prey interactions using real-life data of wolf and moose populations on the Isle Royale. It was "cloned" from a model that InsightMaker provides to its users, at
https://insightmaker.com/insight/2068/Isle-Royale-Predator-Prey-Interactions
Thanks Scott Fortmann-Roe.

I've created a Mathematica file that replicates the model, at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/Moose-n-Wolf-InsightMaker.nb

It allows one to experiment with adjusting the initial number of moose and wolves on the island.

I used steepest descent in Mathematica to optimize the parameters, with my objective data being the ratio of wolves to moose. You can try my (admittedly) kludgy code, at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/Moose-n-Wolf-InsightMaker-BestFit.nb

{WolfBirthRateFactorStart,
WolfDeathRateStart,
MooseBirthRateStart,
MooseDeathRateFactorStart,
moStart,
woStart} =
{0.000267409,
0.239821,
0.269755,
0.0113679,
591,
23.};

Clone of Isle Royale: Predator/Prey Model for Moose and Wolves
Hardy Pfeifer
This is a first example of a simple SIR (Susceptible, Infected, Recovered) model. There are three pools of individuals: those who are infected (without them, no disease!), the pool of those who are at risk (susceptible), and the recovered -- who may lose their immunity and become susceptible again
Mat375 Math Modeling SIR
This is a first example of a simple SIR (Susceptible, Infected, Recovered) model.

There are three pools of individuals: those who are infected (without them, no disease!), the pool of those who are at risk (susceptible), and the recovered -- who may lose their immunity and become susceptible again.

A comparable model in Mathematica is available at
http://www.nku.edu/~longa/classes/2018spring/mat375/mathematica/SIRModel.nb

Clone of A Simple SIR (Susceptible, Infected, Recovered) Example
Sean Field