Solving Romeo & Juliet's Love Dynamics: 1st Order Linear System of DEs

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Need some help figuring this out...

Juliets love for Romeo cools in proportion to his love for her. Romeo's love for Juliet grows in proportion to her love for him.
When they first meet Romeo is immediately attracted to Juliet, but she is as yet indifferent.
Time is measured in days (0-50), and their love is measured on a scale from -5 to 5.
-5 : hatred
-2.5 disgust
0 - indifference
2.5 - attraction
5 - love

1. Model the situation by a 1st order linear system of differential equations with initial condition x(0)=2 , y(0)=0

Im looking at this site as a guide:
http://www.aw-bc.com/ide/idefiles/media/pdf/Documents/PART4/LABS/LAB18.PDF

but on part 1.1 I don't understand how they got the equations
dx/dt= -0.2y and dy/dt= 0.8x

any help would be greatly appreciated. thanks!
 
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I think they've just picked arbitrary negative and positive numerical values; these are givens, rather than something you can derive.

Assuming x(t) is Juliet's love, and y(t) Romeo's, the statement "Juliets love for Romeo cools in proportion to his love for her. Romeo's love for Juliet grows in proportion to her love for him", tells us that dx/dt = (a negative number) * x, while dy/dt = (a positive number) * y. But we can't tell from this alone what the actual numbers will be.

If you look further down the page, you'll see that they modify the story, and change the equations to suit the new scenario.
 
So I can choose any values as long as dx/dt = (negative value)y and dy/dt= (positive value)x?
Also is there any other way I can model the situation using the love scale?
 
Oh, I just noticed, in their version, the scenario in 1.1 is the other way round from the way you stated it:

"Romeo's love for Juliet cools in proportion to her love for him. Juliet's love for Romeo grows in proportion to his love for her."

In that case, x must be Romeo's love, and y Juliet's.

But in answer to your question, yes, as long as the positive coefficient stays positive, and the negative one negative, and you don't introduce any other terms, I think your modified system would also be consistent with the verbal description here. And, unless I'm mistaken, any model with the same qualitative assumptions would take that form.
 
With dx/t= -ay and dy/dt= bx (a and b both positive), we can write this as the matrix equation
\begin{pmatrix}x' \\ y'\end{pmatrix}= \begin{pmatrix}0 & -a \\ b & 0\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix}

The characteristic equation for that matrix is \lambda^2+ ab= 0 so that \lambda^2= -ab which, since a and b are both positive, has roots \lambda= \pm i\sqrt{ab}. That is, this problem has imaginary eigenvalues which means that its solution trajectories are circles around the equilibrium solution, which is (0, 0).

Exactly how much Romeo and Juliet will love or hate each other depends upon those coefficients but, in any case, there will be times when Romeo and Juliet love each other, times when they hate each other, times when one loves but the other hates and this will keep happening, cyclicly.
 
:cry: it'll all end in tears! :cry:
 
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