Question

Each exercise lists the general solution of a linear system of the form $$ \begin{aligned} &x^{\prime}=a_{11} x+a_{12} y \\ &y^{\prime}=a_{21} x+a_{22} y \end{aligned} $$ where \(a_{11} a_{22}-a_{12} a_{21} \neq 0\). Determine whether the equilibrium point \(\mathbf{y}_{e}=\mathbf{0}\) is asymptotically stable, stable but not asymptotically stable, or unstable. $$ \begin{aligned} &x=c_{1} e^{-2 t}+c_{2} e^{-3 t} \\ &y=c_{1} e^{-2 t}-c_{2} e^{-3 t} \end{aligned} $$

Short answer

Question: Based on the provided general solution, classify the stability of the equilibrium point \(\mathbf{y}_{e}=\mathbf{0}\) as asymptotically stable, stable but not asymptotically stable, or unstable. Answer: The equilibrium point \(\mathbf{y}_{e}=\mathbf{0}\) is asymptotically stable.

Step-by-step solution

Behavior of \(e^{-2t}\)

Taking the limit of the function \(e^{-2t}\) as \(t \rightarrow \infty\), we have: $$\lim_{t \rightarrow \infty} e^{-2t} = 0$$

Behavior of \(e^{-3t}\)

Taking the limit of the function \(e^{-3t}\) as \(t \rightarrow \infty\), we have: $$\lim_{t \rightarrow \infty} e^{-3t} = 0$$ #Phase 2: Concluding the stability of the equilibrium point# As \(t \rightarrow \infty\), both factors \(e^{-2t}\) and \(e^{-3t}\) go to zero, regardless of the constants \(c_1\) and \(c_2\). Thus, both \(x\) and \(y\) will approach zero. Now, we can classify the stability of the equilibrium point \(\mathbf{y}_{e}=\mathbf{0}\) based on the observed behavior.

Classification of the stability

Since both \(x\) and \(y\) approach zero as \(t \rightarrow \infty\), the equilibrium point \(\mathbf{y}_{e}=\mathbf{0}\) is considered asymptotically stable.

Key concepts

Equilibrium Point

An equilibrium point in a system of equations refers to a state where all the variables remain constant over time. For the system given in the exercise, the equilibrium point is represented by \( \mathbf{y}_{e} = \mathbf{0} \). This means that when the system reaches this state, both variables \(x\) and \(y\) don't change unless disturbed by external forces.

When analyzing the stability of an equilibrium point, we are interested in seeing how the system behaves when it's slightly perturbed. If it returns to the equilibrium point over time, the point is said to be stable or asymptotically stable. However, if the system diverges away from it, the point is considered unstable.

Key features of an equilibrium point include:

• Constant state: Once reached, the variables do not change unless interrupted.
• Indicator of system stability: By analyzing it, we can determine how the system reacts to changes.

Understanding the behavior around the equilibrium point helps us predict long-term trends in the system.

Linear System of Differential Equations

A linear system of differential equations is a collection of equations that describe the rate at which variables change over time. In the exercise, we have the system:
\[ \begin{aligned} x^{\prime}=a_{11} x+a_{12} y \ y^{\prime}=a_{21} x+a_{22} y \end{aligned} \]
Such systems are foundational because they model a wide range of real-world phenomena, from mechanical systems to population dynamics.

The coefficients \(a_{11}, a_{12}, a_{21}, a_{22}\) dictate the interaction between the variables. Importantly, when the determinant of the coefficient matrix \(a_{11}a_{22} - a_{12}a_{21}\) is not equal to zero, the system has unique solutions.

Advantages of linear systems include:

• Simplicity: They are easier to solve compared to nonlinear equations.
• Predictability: Solutions can often be decomposed into simpler components, like exponential functions.
• Broad application: From physics to economics, these systems are widely used to describe change.

Understanding the dynamics within these systems enables us to predict future states and ensure system stability.

General Solution

The general solution of a differential system provides a formula that describes all possible behaviors of the system, determined by its initial conditions. In this exercise, the general solution is given by:
\[ \begin{aligned} x = c_{1} e^{-2t} + c_{2} e^{-3t} \ y = c_{1} e^{-2t} - c_{2} e^{-3t} \end{aligned} \]
This solution includes arbitrary constants \(c_1\) and \(c_2\), which are determined by the specific initial conditions. By solving for these constants based on initial states, we can predict the evolution of the system over time.

Important aspects of the general solution:

• Comprehensiveness: It encapsulates all possible trajectories of the system.
• Dependence on initial conditions: Determined by where the system starts.
• Insightful: Evaluating the general solution at different times gives insights into system behavior.

While the solution seems abstract, it serves as a powerful tool to understand the full scope of what the system can do.

Exponential Behavior

Exponential behavior in differential equations describes a type of growth or decay that is incredibly smooth, characterized by curves that either rise or fall dramatically. In this exercise, the functions \(e^{-2t}\) and \(e^{-3t}\) showcase exponential decay.

Exponential functions are crucial in the study of dynamics because they describe processes that change at rates proportional to their current value, like compound interest or radioactive decay.

The key characteristics include:

• Rapid change: Growth or decay can be very quick depending on the exponent's sign and magnitude.
• Predictability: The mathematical simplicity of exponentials makes them easy to calculate and predict.
• Relevance: They're seen in various applications, from heat dissipation to population growth models.

In our system, both \(e^{-2t}\) and \(e^{-3t}\) tend toward zero as \(t\) approaches infinity, resulting in the system reaching its equilibrium state smoothly over time.