# Equilibrium and Stability of a set of equations

• xicor
Look at your book, or notes, or ask your instructor.In summary, the conversation discusses finding the equilibrium and determining its stability for the model in Section 1.4, assuming x and y are both strictly positive. The homework equations are the differential equations in Section 1.4, and the attempt at a solution involves solving the equations and determining what happens when t approaches infinity. The concept of equilibrium and its relation to constant solutions is also discussed. It is determined that the equilibrium points are (x,y)=(a2/b2,a1/b1) and the next step is to understand how to find the stability of the solutions near this point.
xicor

## Homework Statement

Find the equilibrium and determine its stability for the model of Sec 1.4. Assume x and y are both
strictly positive (nonzero).

## Homework Equations

Differential equations of section 1.4

(1/x)(dx/dt) = a1 - b1y
(1/y)(dy/dt) = -a2 + b2x

## The Attempt at a Solution

I solved the two differential equations and got x(t) = e^t(a1 -b1y) and y(t) = e^t(-a2 +b2x) so once I had these I though of what happens to the equations when t approaches infinity but that's where I'm unsure because x and y are apart of each other eqation. Either both of them approach infinity as t approaches infinity and there is no equilibrium or there is something more complex that I can't seem to figure out yet.

Thanks to anybody who helps.

xicor said:

## Homework Statement

Find the equilibrium and determine its stability for the model of Sec 1.4. Assume x and y are both
strictly positive (nonzero).

## Homework Equations

Differential equations of section 1.4

(1/x)(dx/dt) = a1 - b1y
(1/y)(dy/dt) = -a2 + b2x

## The Attempt at a Solution

I solved the two differential equations and got x(t) = e^t(a1 -b1y) and y(t) = e^t(-a2 +b2x) so once I had these I though of what happens to the equations when t approaches infinity but that's where I'm unsure because x and y are apart of each other eqation. Either both of them approach infinity as t approaches infinity and there is no equilibrium or there is something more complex that I can't seem to figure out yet.

Thanks to anybody who helps.

Those are NOT solutions to your differential equations. You are doing something very wrong. And you don't necessarily have to solve them. What are equilibrium points?

Well the concept I have so far is that equilibrium is where a state will persist forever but what I think I'm seeing now is that doesn't relate to the solutions but the differential equation where either (dx/dt) or dy/dt) are equal to zero. The equilibrium state would be where the differential equations are constant but I'm not sure how to use that definition in the forms the differential equations are currenly in.

xicor said:
Well the concept I have so far is that equilibrium is where a state will persist forever but what I think I'm seeing now is that doesn't relate to the solutions but the differential equation where either (dx/dt) or dy/dt) are equal to zero. The equilibrium state would be where the differential equations are constant but I'm not sure how to use that definition in the forms the differential equations are currenly in.

Yes, the equilibrium points are points where the solutions are constant functions. So dx/dt AND dy/dt are zero. Use that.

Well if I set the differential equations to zero I get the equilibrium points y=a1/b2 and x = a2/b2 but the question I now have is if these are the equilibrium points I don't see the usefulness in being able to figure out the stability since each differential equation is in respect of the other variable so the graph to find the stability would either always be zero or linear based on it's variable but not the equilibrium point. What is it I need to understand here in order to find the stability?

xicor said:
Well if I set the differential equations to zero I get the equilibrium points y=a1/b2 and x = a2/b2 but the question I now have is if these are the equilibrium points I don't see the usefulness in being able to figure out the stability since each differential equation is in respect of the other variable so the graph to find the stability would either always be zero or linear based on it's variable but not the equilibrium point. What is it I need to understand here in order to find the stability?

You only have one equilibrium point. (x,y)=(a2/b2,a1/b1). You need to figure out what the solutions look like near that point. You must have seen examples of how this is done in your course.

## 1. What is equilibrium in the context of a set of equations?

Equilibrium in a set of equations refers to a state where all variables are constant and there is no net change in the system. This means that the system is balanced and there is no external force or influence causing any changes.

## 2. How is stability defined in a set of equations?

Stability in a set of equations is the property of the system to return to equilibrium after being disturbed. It is a measure of how resistant the system is to changes and how quickly it can return to its original state.

## 3. What factors affect the stability of a set of equations?

The stability of a set of equations is affected by the eigenvalues of the system's Jacobian matrix, which represent the rate of change of the system's variables. The sign and magnitude of these eigenvalues determine the stability of the system.

## 4. How can the stability of a set of equations be analyzed?

The stability of a set of equations can be analyzed through several methods such as linear stability analysis, phase plane analysis, and Lyapunov stability analysis. These methods involve solving for the eigenvalues of the system and analyzing their properties to determine stability.

## 5. Why is it important to understand the equilibrium and stability of a set of equations?

Understanding the equilibrium and stability of a set of equations is crucial in various scientific fields such as physics, chemistry, and biology. It allows us to predict the behavior of systems and make informed decisions based on their stability. It also helps us identify critical points in a system and understand the effects of external factors on its stability.

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