# Modeling the populations of foxes and rabbits given a baseline

• JessicaHelena
JessicaHelena
Homework Statement
The population of foxes and rabbits on Nantucket Island has been studied by biologists. They measure the populations relative to a baseline, in hundreds of animals. (So ##x(2)=5## means that there are 500 more foxes than the baseline value, and ##y(2)=−5## means that there are 500 fewer rabbits than the baseline value.)

The biologists have established the following relationship between ##x(t)## (foxes' population) and ##y(t)## (rabbits' population): ##x' = 0.5x + y## ##y' = -2.25x + 0.5y##

Suppose that at ##t=0## there are ##100## more foxes than the baseline: ##x(0) = 1##; the rabbit population is at the baseline value, ##y(0) = 0##. What is the solution to this initial value problem?
Relevant Equations
Characteristic Equation
lambda^2 - (trA) lambda + det A
From solving the characteristic equations, I got that ##\lambda = 0.5 \pm 1.5i##. Since using either value yields the same answer, let ##\lambda = 0.5 - 1.5i##. Then from solving the system for the eigenvector, I get that the eigenvector is ##{i}\choose{1.5}##. Hence the complex solution is ##{i}\choose{1.5}## ##e^{(0.5 - 1.5i)t}##.

Using the Euler's formula ##e^{iwt} = \cos(\omega t) + i\sin(\omega t)##, I get the real parts of ##x## and ##y## is given by

##{x}\choose{y}## = ##e^{0.5t}## ##{0}\choose{1.5}## ##\cos(1.5t)## + ##e^{0.5t}## ##{1}\choose{0}## ##\sin(1.5t)##

And given that $x(0) = 1$ and $y(0) = 0$, I arrived at:
##x(t) = \sin(1.5t) e^{0.5t} + e^{0.5t}##
##y(t) = 1.5\cos(1.5t) e^{0.5t} - 1.5e^{0.5t}##

However, these equations turned out to be the wrong model.
Where might I have gone wrong? Any help would really be appreciated!

## Answers and Replies

Mentor
Homework Statement:: The population of foxes and rabbits on Nantucket Island has been studied by biologists. They measure the populations relative to a baseline, in hundreds of animals. (So ##x(2)=5## means that there are 500 more foxes than the baseline value, and ##y(2)=−5## means that there are 500 fewer rabbits than the baseline value.)

The biologists have established the following relationship between ##x(t)## (foxes' population) and ##y(t)## (rabbits' population): ##x' = 0.5x + y## ##y' = -2.25x + 0.5y##

Suppose that at ##t=0## there are ##100## more foxes than the baseline: ##x(0) = 1##; the rabbit population is at the baseline value, ##y(0) = 0##. What is the solution to this initial value problem?
Relevant Equations:: Characteristic Equation
lambda^2 - (trA) lambda + det A
JessicaHelena said:
From solving the characteristic equations, I got that ##\lambda = 0.5 \pm 1.5i##. Since using either value yields the same answer, let ##\lambda = 0.5 - 1.5i##. Then from solving the system for the eigenvector, I get that the eigenvector is ##{i}\choose{1.5}##. Hence the complex solution is ##{i}\choose{1.5}## ##e^{(0.5 - 1.5i)t}##.

Using the Euler's formula ##e^{iwt} = \cos(\omega t) + i\sin(\omega t)##, I get the real parts of ##x## and ##y## is given by

##{x}\choose{y}## = ##e^{0.5t}## ##{0}\choose{1.5}## ##\cos(1.5t)## + ##e^{0.5t}## ##{1}\choose{0}## ##\sin(1.5t)##

And given that $x(0) = 1$ and $y(0) = 0$, I arrived at:
##x(t) = \sin(1.5t) e^{0.5t} + e^{0.5t}##
##y(t) = 1.5\cos(1.5t) e^{0.5t} - 1.5e^{0.5t}##

However, these equations turned out to be the wrong model.
Where might I have gone wrong? Any help would really be appreciated!
I don't see anything wrong in your work, except that your solution uses only one of the eigenvalue/eigenvector pairs, so you're not getting the full general solution. Before substituting x(0) and y(0), expand your equations for ##{x(t)}\choose{y(t)}## with the other eigenvalue/eigenvector pair.

JessicaHelena
you're not getting the full general solution

@Mark44 I had thought that just using one eigenvalue would yield the full general solution.
When you say
expand your equations for (x(t)y(t))(x(t)y(t)){x(t)}\choose{y(t)} with the other eigenvalue/eigenvector pair
does that mean I need to work out the x(t) and y(t) for with the other eigenvalue/eigenvector pair, and add the two x(t) s for the general x(t) solution and add the two y(t) s for the general y(t) solution?

Mentor
does that mean I need to work out the x(t) and y(t) for with the other eigenvalue/eigenvector pair, and add the two x(t) s for the general x(t) solution and add the two y(t) s for the general y(t) solution?
Yes, and I believe this is why you aren't getting the right solution.

You already have this:
Hence the complex solution is ##{i}\choose{1.5}## ##e^{(0.5 - 1.5i)t}##.
This is the basic solution for one eigenvalue/eigenvector. You know the other eigenvalue (1/2 + 3i/2), so it shouldn't take much work to find its associated eigenvalue to get the other basic solution.
The general solution will be a linear combination of these two basic solutions. Use that solution with your initial conditions.

JessicaHelena
@Mark44

Hmm okay, so the first complex solution I get is ##{i}\choose{1.5}## ##e^{(0.5-1.5i)t}## and the other complex solution I get is ##{i}\choose{-1.5}## ##e^{(0.5+1.5i)t}##.

The general solution to this system can be given by
##{x}\choose{y}## = ##c_1 e^{\lambda_1} v_1 + c_2 e^{\lambda_2} v_2##
where lambdas are eigenvalues and v's are eigenvectors.

Since we're modelling populations, we cannot have imaginary #'s. Hence I just take the real parts only, and get that

##{x}\choose{y}## = ##e^{0.5t} (c_1-c_2)## ##{\sin(1.5t)}\choose{1.5\cos(1.5t)}##

And I'm also given that x(0) = 1 and y(0) = 0.
However, I'm not really sure what to do with the c_1 and c_2, or how to use the information above (initial conditions) in general to arrive at my solution. Somehow this part is harder.

Mentor
Hence I just take the real parts only
This might be where your problem lies.
I get this, based on your work:
##\begin{bmatrix} x(t) \\ y(t) \end{bmatrix} = e^{0.5t} \begin{bmatrix}c_1 i e^{-1.5it} + c_2ie^{1.5it} \\ c_1 1.5 e^{-1.5it} -c_2 1.5 e^{1.5it}\end{bmatrix}##
Now substitute your initial conditions. For what it's worth, I get ##c_1 = c_2 = \frac{-i}2##.

JessicaHelena
@Mark44 Oh so x(t) and y(t) can be imaginary as well?

JessicaHelena
@Mark44
Thank you—I did get the answers right! Could you tell me how you got the c_1,c_2 values?

Last edited:
Mentor
Oh so x(t) and y(t) can be imaginary as well?
Turns out they aren't.

Here are the equations:
I get this, based on your work:
##\begin{bmatrix} x(t) \\ y(t) \end{bmatrix} = e^{0.5t} \begin{bmatrix}c_1 i e^{-1.5it} + c_2ie^{1.5it} \\ c_1 1.5 e^{-1.5it} -c_2 1.5 e^{1.5it}\end{bmatrix}##
Using x(0) = 1 and y(0) = 0, the system above becomes
##x(0) = c_1 \cdot 1 + c_2 \cdot 1 = 1##
##y(0) = 3/2 c_1 \cdot 1 - 3/2 c_2 = 0##
Solve this system for ##c_1## and ##c_2##.

I used the values for these constants and got ##x(t) = e^{t/2}\cos(3t/2)##. I stopped there and didn't work out y(t).