Question

Direction field analysis Consider the first-order initial value problem \(y^{\prime}(t)=a y+b, y(0)=A,\) for \(t \geq 0,\) where \(a, b,\) and \(A\) are real numbers. a. Explain why \(y=-b / a\) is an equilibrium solution and corresponds to a horizontal line in the direction field. b. Draw a representative direction field in the case that \(a>0\) Show that if \(A>-b / a\), then the solution increases for \(t \geq 0,\) and that if \(A<-b / a,\) then the solution decreases for \(t \geq 0\). c. Draw a representative direction field in the case that \(a<0\) Show that if \(A>-b / a\), then the solution decreases for \(t \geq 0,\) and that if \(A<-b / a,\) then the solution increases for \(t \geq 0\).

Short answer

Answer: The given solution is an equilibrium solution because when substituting it into the first derivative, the slope of the tangent lines is 0, resulting in horizontal lines in the direction field. For the case when a > 0, the solution increases if A > -b/a and decreases if A < -b/a. For the case when a < 0, the solution decreases if A > -b/a and increases if A < -b/a.

Step-by-step solution

Part a: Equilibrium Solution

Firstly, let's find the equilibrium solution. An equilibrium solution is a constant solution, which means that \(y'(t) = 0\). From the given equation, we can write: \(0 = ay + b\). Solving for y, we get the equilibrium solution: \(y_{eq} = -\frac{b}{a}\) Now, to explain why this corresponds to a horizontal line in the direction field, remember that the slope of the tangent lines in a direction field is given by the first derivative, \(y'(t)\). In this case, \(y'(t) = ay + b\). When we plug the equilibrium solution into the first derivative, we get: \(y'(t) = a\left(-\frac{b}{a}\right) + b = 0\) Since the slope of the tangent lines is 0, the corresponding lines in the direction field are horizontal.

Part b: Direction Field and Solution Behavior for a > 0

For the case when a > 0, we can analyze the behavior of the solution for different values of A by checking the sign of the first derivative, \(y'(t)\): 1. If \(A > -\frac{b}{a}\), then \(y(0) > y_{eq}\). By substituting into \(y'(t) = ay + b\), we can see that \(y'(t) = a(A-y_{eq}) > 0\). This means the solution increases for \(t \geq 0\). 2. If \(A < -\frac{b}{a}\), then \(y(0) < y_{eq}\). By substituting into \(y'(t) = ay + b\), we can see that \(y'(t) = a(y_{eq}-A) > 0\). This means the solution decreases for \(t \geq 0\). A representative direction field for this case would show tangent lines with positive slopes above the horizontal line \(y = y_{eq}\), and tangent lines with negative slopes below this horizontal line.

Part c: Direction Field and Solution Behavior for a < 0

For the case when a < 0, we will again analyze the behavior of the solution for different values of A by checking the sign of the first derivative, \(y'(t)\): 1. If \(A > -\frac{b}{a}\), then \(y(0) > y_{eq}\). By substituting into \(y'(t) = ay + b\), we can see that \(y'(t) = a(y_{eq}-A) < 0\). This means the solution decreases for \(t \geq 0\). 2. If \(A < -\frac{b}{a}\), then \(y(0) < y_{eq}\). By substituting into \(y'(t) = ay + b\), we can see that \(y'(t) = a(A-y_{eq}) < 0\). This means the solution increases for \(t \geq 0\). A representative direction field for this case would show tangent lines with negative slopes above the horizontal line \(y = y_{eq}\), and tangent lines with positive slopes below this horizontal line.

Key concepts

First-order Initial Value Problem

Understanding the concept of a first-order initial value problem (IVP) is vital when dealing with differential equations. It consists of a differential equation along with a condition that specifies the value of the unknown function at a particular point, typically the starting point.

In our example, the IVP is given by the equation \(y'(t) = ay + b\), where \(y(0) = A\) specifies the condition at \(t = 0\). Essentially, this can be viewed as a recipe that tells you the rate at which a quantity, represented by \(y\), is changing at any time \(t\). The constants \(a\), \(b\), and \(A\) are parameters that influence the behavior of the solution.

This type of problem is crucial since it lays down the foundation for predicting how the system evolves over time from a given starting point. The solution to the IVP will be a function that describes the behavior of \(y\) over time, adhering to both the differential equation and the initial condition. This approach is widely used in fields like physics, economics, and biology to model real-world situations.

Equilibrium Solution

An equilibrium solution represents a state of balance in a dynamical system, where the value of the function does not change over time. To identify such a solution, the derivative of the function, which indicates the rate of change, is set to zero.

In the context of our differential equation \(y'(t) = ay + b\), the equilibrium solution is found by setting \(y'(t) = 0\). By solving for \(y\), we discover that the equilibrium solution is \(y_{eq} = -\frac{b}{a}\). This solution is significant because it tells us the condition under which the system remains stable and unchanging.

The equilibrium solution corresponds to a horizontal line in the direction field, as the slope of the tangent at any point along this line is zero. Therefore, the system neither grows nor declines but remains constant. Recognizing equilibrium solutions helps us understand long-term behavior and stability for various systems.

Slope of Tangent Lines

The slope of tangent lines in direction fields gives us a visual representation of the solution's trajectory over time. It is a direct reflection of the rate of change of the function at any given point.

For our differential equation \(y'(t) = ay + b\), the slope of the tangent line at any point \((t, y)\) in the direction field is determined by substituting the value of \(y\) into the equation. For example, points above the equilibrium solution will have a different slope compared to points below it, depending on the sign of \(a\).

Visualizing the Slope

In a direction field, slopes of the tangent lines are represented by little line segments called 'arrows' or 'field elements.' These arrows suggest the behavior of the solution: whether it increases or decreases. For instance, arrows pointing upwards indicate an increasing solution, while those pointing downwards indicate a decreasing one. Drawing these slopes accurately gives us a better understanding of how solutions behave, especially with varying initial values.