Chapter 9: Problem 6
In Exercises \(1-8\) \begin{equation}\begin{array}{l}{\text { a. Identify the equilibrium values. Which are stable and which }} \\ {\text { are unstable? }} \\ {\text { b. Construct a phase line. Identify the signs of } y^{\prime} \text { and } y^{\prime \prime} \text { . }} \\ {\text { c. Sketch several solution curves. }}\end{array}\end{equation} $$y^{\prime}=y-\sqrt{y}, \quad y>0$$
Short Answer
Step by step solution
Find Equilibrium Values
Determine Stability of the Equilibrium
Construct the Phase Line
Identify the Signs of Derivatives
Sketch Solution Curves
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Phase Line
Consider when \( y < 1 \): the derivative \( y' < 0 \), indicating that any initial value below 1 will move toward \( y = 1 \). Mark this by drawing arrows pointing toward one from the left. When \( y > 1 \), \( y' > 0 \); values above 1 will also move back toward the equilibrium at \( y = 1 \). Here, you would draw arrows pointing toward one from the right. This balanced movement from both directions shows that \( y = 1 \) is a stable equilibrium point.
Derivative Signs
To find the stability around the equilibrium \( y = 1 \), check the derivative signs for values just below and just above. For \( y = 0.5 \), calculate \( y' = 0.5 - \sqrt{0.5} \approx -0.207 \), so the graph decreases towards zero. For \( y = 1.5 \), \( y' = 1.5 - \sqrt{1.5} \approx 0.776 \); the graph increases. Therefore, we see stability since trajectories move back to \( y = 1 \) regardless of starting position.
The second derivative \( y'' \) tells us about concavity. For \( y = 1 \), \( y'' = 1 - \frac{1}{2\sqrt{1}} = 0.5 \) implies it is concave up around this point, confirming \( y = 1 \) is a minimum.
Concavity
In this example, we derived \( y'' = 1 - \frac{1}{2\sqrt{y}} \). For values close to the stable equilibrium \( y = 1 \), substitute to get \( y'' = 1 - 0.5 = 0.5 \), indicating the graph curves upwards, or is concave up.
This tells us more context for sketching solution curves: when approaching the stability point at \( y = 1 \), the graph turns upward, creating a U-shape. This clue complements the information from the first derivative and phase line analysis.
Solution Curves
Since \( y' \) shows stability around \( y = 1 \), the solutions are attracted to this point from both sides. Points where \( y < 1 \) fall towards one, creating increasingly shallower angles as the graph slides up to \( y = 1 \). Points starting at \( y > 1 \) head down towards the line as well.
The second derivative hints solutions will keep a concave paradise, bending upward as they near \( y = 1 \). This analysis outlines the overall shape as a soothing curve gently bending upward approaching the equilibrium, which matches our expectations for stability and concavity.