Problem 8 - Calculus

Question

Find all equilibrium points. Give answers as ordered pairs \((x, y).\) $$\begin{aligned} &\frac{d x}{d t}=x^{2}-x y\\\ &\frac{d y}{d t}=15 y-3 y^{2} \end{aligned}$$

Step-by-Step Solution

Answer

The equilibrium points are (0, 0), (0, 5), and (5, 5).

Step 1: Set the first equation to zero

To find equilibrium points, we first set the derivatives to zero because at equilibrium, there is no change with respect to time. For the equation \( \frac{dx}{dt} = x^2 - xy \), set it to zero: \[ x^2 - xy = 0 \]

Step 2: Factor the first equation

Notice that the equation from Step 1 can be factored: \[ x(x - y) = 0 \]. This gives two possibilities: \( x = 0 \) or \( x = y \).

Step 3: Set the second equation to zero

Now, consider the second equation \( \frac{dy}{dt} = 15y - 3y^2 \), and set it to zero: \[ 15y - 3y^2 = 0 \].

Step 4: Factor the second equation

Factor the equation from Step 3: \[ 3y(5 - y) = 0 \]. This gives two possibilities: \( y = 0 \) or \( y = 5 \).

Step 5: Find combinations for equilibrium points where \(x = 0\)

Substitute \( x = 0 \) in the possibilities for \( y \): 1. If \( y = 0 \), then an equilibrium point is \( (0, 0) \).2. If \( y = 5 \), then another equilibrium point is \( (0, 5) \).

Step 6: Find combinations for equilibrium points where \(x = y\)

Substitute \( x = y \) in the possibilities for \( y \): 1. \( x = 0 \) was already considered, so now use \( y = 5 \). If \( x = y \), then the equilibrium point is \( (5, 5) \).

Step 7: List all equilibrium points

Based on the solutions from the previous steps, the equilibrium points found are \( (0, 0) \), \( (0, 5) \), and \( (5, 5) \).

Key Concepts

Differential equations

Differential equations are mathematical tools used to describe how things change. They involve derivatives, which represent rates of change. In the problem presented, there are two differential equations. These equations capture how two variables, typically representing quantities like populations or concentrations, evolve over time.

The set of equations given are:

• \( \frac{dx}{dt} = x^2 - xy \)
• \( \frac{dy}{dt} = 15y - 3y^2 \)

Each differential equation depends on the variables involved: \( x \) and \( y \). Here, by calculating derivatives, we learn about the speed and direction of change for each variable. This analysis can preview complex dynamic behaviors and help find solutions that describe balance or steady states in a system.

System of equations

A system of equations involves finding solutions that satisfy multiple equations at the same time. In this problem, we have a pair of differential equations which makes it a system of equations. To solve, we seek pairs of \( (x, y) \) values that fulfill both equations simultaneously when their derivatives are zero.

When we make the derivatives zero, we focus on finding points where the system does not change over time—these are called equilibrium points. By setting each differential equation to zero:

• \( x^2 - xy = 0 \)
• \( 15y - 3y^2 = 0 \)

This reduces a dynamic system to an algebraic problem, allowing us to solve for equilibrium points where time-driven changes cease.

Factorization

Factorization is a key method to simplify equations and find solutions more easily. It involves expressing a complex expression as a product of simpler ones. In this exercise, we apply factorization to identify equilibrium points.

Consider the first equation:

• \( x^2 - xy = 0 \)

To solve, we factor it as:

• \( x(x - y) = 0 \)

Factorization reveals two potential solutions: \( x = 0 \) or \( x = y \). Similarly, for the second equation:

• \( 15y - 3y^2 = 0 \)

Factored as:

• \( 3y(5 - y) = 0 \)

This unveils \( y = 0 \) or \( y = 5 \), giving possible values for \( y \). By combining these results, we find equilibrium points straightforwardly.

Equilibrium analysis

Equilibrium analysis is the assessment of conditions where variables in a system show no net change. This is the goal when solving for equilibrium points in differential equations. The system can reach a state where, although dynamic factors are still present, the balance between them results in zero change.

To determine equilibrium points, set the time derivatives to zero, as change halts at equilibrium:

• \( x^2 - xy = 0 \)
• \( 15y - 3y^2 = 0 \)

The key outcomes from this analysis include the equilibrium points: \( (0, 0) \), \( (0, 5) \), and \( (5, 5) \). These represent the scenarios where the system's rate of change in terms of \( x \) and \( y \) has stabilized, often revealing insights about the system's long-term behavior.