Problem 10 - Calculus

Question

Draw a direction field of arrows for \(y^{\prime}=y-1,\) with solution curves \(y=e^{x}+1\) and \(y=1\).

Step-by-Step Solution

Answer

Draw arrows with slope \( y - 1 \) on the plane, plotting solution curves \( y = e^{x} + 1 \) and \( y = 1 \).

Step 1: Understand the Equation

The differential equation given is \( y' = y - 1 \). This means that the slope of the solution at any point \( (x, y) \) is \( y - 1 \). To draw the direction field, we need to evaluate the slope at various points on the \( xy \)-plane.

Step 2: Calculate Slopes for Several Points

Evaluate the slope \( y' = y - 1 \) at several points. For example, at \( y = 0 \), the slope is \( y' = 0 - 1 = -1 \), at \( y = 1 \), the slope is \( y' = 1 - 1 = 0 \), and at \( y = 2 \), the slope is \( y' = 2 - 1 = 1 \). Continue this evaluation for various points to get a sense of the direction of arrows.

Step 3: Draw the Direction Field

Plot points on the plane and draw small arrows representing the slopes calculated in the previous step. At each point \( (x, y) \), draw an arrow whose slope is \( y - 1 \). For example, at any point where \( y = 1 \), the arrows should be horizontal.

Step 4: Add Solution Curves

Superimpose the solution curves onto the direction field. The curves are \( y = e^{x} + 1 \) and the constant solution \( y = 1 \) (a horizontal line). These curves should align with the direction indicated by the arrows for any given value of \( x \).

Step 5: Analyze the Behavior

Notice that the solution curve \( y = e^{x} + 1 \) approaches the horizontal line \( y = 1 \) as \( x \to -\infty \). This is consistent with the direction field where arrows above the line \( y = 1 \) generally point downwards, moving solutions towards \( y = 1 \).

Key Concepts

Differential Equations

Differential equations are mathematical expressions that relate functions and their derivatives. They form the foundation for modeling a wide variety of real-world phenomena, from population growth to physics. In our exercise, the differential equation is \( y' = y - 1 \), indicating the rate of change of \( y \) concerning \( x \). In simple terms, it tells us how steep the curve at any given \( y \) value will be.

• The expression \( y' = y - 1 \) shows that if \( y \) is greater than 1, \( y' \) will be positive, indicating an increase.
• If \( y \) equals 1, \( y' \) is zero, indicating a constant function (no change in \( y \)).
• For \( y < 1 \), \( y' \) is negative, meaning \( y \) decreases as \( x \) increases.

Understanding these relationships gives us insights into how solution curves behave, which helps when sketching or analyzing graphs.

Solution Curves

Solution curves represent the paths traced by solutions of a differential equation given initial conditions. For the equation \( y' = y - 1 \), we analyze two solution curves: \( y = e^{x} + 1 \) and \( y = 1 \).

The curve \( y = 1 \) stands out because it's a constant solution. It means no matter what \( x \) value you choose, \( y \) will always be 1. This solution is evident in the direction field as horizontal.

• The curve \( y = e^{x} + 1 \) is dynamic. It starts above the line \( y = 1 \) and represents exponential growth added onto the constant solution. As \( x \) decreases, \( y \) approaches 1 from above, showing how any deviations from the equilibrium are corrected over time.

The analysis of these curves shows how changes in initial conditions or constants affect the outcomes in differential equations.

Slope Field

A slope field, or direction field, provides a visual map of a differential equation's solution without explicitly solving it. For \( y' = y - 1 \), the slope field illustrates the gradient of solutions at various points in the plane.

Here are some key steps to drawing a slope field:

• Evaluate the differential equation to find the slope at different \( y \)-values.
• Plot vectors or small line segments at each point that indicate the slope \( y - 1 \).
• These vectors should follow the slope; for instance, horizontal when \( y = 1 \).

Visualizing the slope field helps you observe patterns and the general direction solutions will take, especially when actual solution equations align with these slopes.

Phase Portrait

A phase portrait is a graphical representation that combines the behavior of all possible solutions of a differential equation. It essentially maps out the trajectories that solutions take as time progresses and parameters change.

In this problem, adding solution curves \( y = e^{x} + 1 \) and \( y = 1 \) into a slope field creates a phase portrait:

• This visualization helps identify equilibrium points, like \( y = 1 \) where all change ceases, laying out paths in steady-state conditions.
• The curves also highlight approaching behaviors, like \( y = e^{x} + 1 \) converging towards \( y = 1 \) as \( x \to -fty \).
• Phase portraits can reveal the stability of solutions, pointing out where solutions tend to stabilize or diverge.

Thus, these graphical displays are valuable tools for better understanding of differential equations beyond mere algebraic solutions.