Question

Find the equilibrium solutions and determine their stability for the following systems. For each case, draw representative solutions and phase lines. a. \(y^{\prime}=y^{2}-6 y-16\). b. \(y^{\prime}=\cos y\). c. \(y^{\prime}=y(y-2)(y+3)\). d. \(y^{\prime}=y^{2}(y+1)(y-4)\).

Short answer

a. The equilibrium solutions for \(y^{\prime}=y^{2}-6 y-16\) are \(y=8\) (stable), \(y=-2\) (unstable). b. The equilibrium solutions for \(y^{\prime}=\cos y\) are \(y=\frac{\pi}{2}+n\pi\) (all stable). c. The equilibrium solutions for \(y^{\prime}=y(y-2)(y+3)\) are \(y=0\) (unstable), \(y=2\) (stable), \(y=-3\) (unstable). d. The equilibrium solutions for \(y^{\prime}=y^{2}(y+1)(y-4)\) are \(y=0\) (unstable), \(y=-1\) (stable), \(y=4\) (unstable).

Step-by-step solution

Finding the equilibrium points

To find the equilibrium solutions, set the derivative equal to zero and solve for \(y\). a. For \(y^{\prime}=y^{2}-6 y-16\), setting the equation equal to zero gives us \(y^{2}-6y-16=0\). After factoring, we get \((y-8)(y+2)=0\), yielding \(y=8\) and \(y=-2\). b. For \(y^{\prime}=\cos y\), setting the equation equal to zero means \(\cos y=0\). This happens when \(y=\frac{\pi}{2}+n\pi\), for any integer \(n\). c. For \(y^{\prime}=y(y-2)(y+3)\), setting \(y(y-2)(y+3)=0\) gives us \(y=0, 2, -3\). d. For \(y^{\prime}=y^{2}(y+1)(y-4)\), setting \(y^{2}(y+1)(y-4)=0\) gives \(y=0, -1, 4\).

Determining stability

The sign of the derivative between the equilibrium points determines whether an equilibrium solution is stable. a. For \(y^{\prime}=y^{2}-6 y-16\), the intervals are \(-\infty\) to -2, -2 to 8, and 8 to \(\infty\). Choose convenient points such as -3, 0, and 10, respectively. Plug these into the derivative and observe their sign. b. Use common test points within intervals for \(y^{\prime}=\cos y\) such as \(\frac{\pi}{4}\), \(\frac{3\pi}{4}\), etc. c. The intervals for \(y^{\prime}=y(y-2)(y+3)\) are \(-\infty\) to -3, -3 to 0, 0 to 2, and 2 to \(\infty\). d. The intervals for \(y^{\prime}=y^{2}(y+1)(y-4)\) are \(-\infty\) to -1, -1 to 0, 0 to 4, and 4 to \(\infty\).

Representative solutions and phase lines

Sketch the representative solutions and phase lines based on the equilibrium solutions and their stability. Arrange the equilibrium solutions on a line in ascending order and track the sign changes over the interval. Regions between equilibrium points where the derivative has the same sign belong to the same phase. Negative derivative values indicate decreasing solutions while positive values indicate increasing solutions.

Key concepts

Equilibrium Solutions

Equilibrium solutions refer to the values of a variable where the rate of change (the derivative) becomes zero. These solutions are important in understanding how a system behaves over time. In differential equations, finding the equilibrium solutions involves setting the derivative equal to zero. Let's take a closer look at how this works with a few examples given in the exercise.

For the equation \(y^{\prime}=y^{2}-6y-16\), we set \(y^{2}-6y-16=0\). By solving this quadratic equation, we find the equilibrium solutions \(y=8\) and \(y=-2\). Similarly, for \(y^{\prime}=y(y-2)(y+3)\), setting \(y(y-2)(y+3)=0\) gives us \(y=0, 2, -3\) as equilibrium solutions.

Recognizing equilibrium solutions is akin to finding the points at which the system experiences no net change. In real-world terms, these solutions can represent steady states or constant solutions where a dynamic system stabilizes over time.

Stability Analysis

Stability analysis helps us determine the behavior of solutions near the equilibrium points. We need to understand whether a slight change will cause a system to eventually return to equilibrium (stable) or diverge away from it (unstable). To conduct a stability analysis, we examine the sign of the derivative in the intervals between equilibrium points.

For instance, consider the differential equation \(y^{\prime}=y^{2}-6y-16\). Evaluate the sign of \(y^{\prime}\) at test points between the equilibrium solutions \(y=-2\) and \(y=8\). If the derivative is positive, the solutions are increasing in that interval, indicating potential instability if it moves away from equilibrium. Conversely, if the derivative is negative, it suggests decreasing solutions pointing towards stability.

Stability analysis allows us to classify equilibrium points into categories:

• Stable: Solutions tend toward the equilibrium point.
• Unstable: Solutions move away from the equilibrium point.
• Semi-stable: Solutions behave differently on either side of the equilibrium point.

This analysis is crucial when predicting the long-term behavior of dynamic systems.

Phase Lines

Phase lines are visual representations that summarize the behavior of differential equations. They succinctly illustrate how solutions behave between the equilibrium points identified in both previous sections.

To sketch a phase line, draw a vertical line and mark the equilibrium points on it. Between these points, analyze the sign of the derivative as already discussed. Use arrows to represent whether the solutions are increasing or decreasing in each interval. For example, if \(y^{\prime}=y^{2}-6y-16\) has equilibrium points at \(y=-2\) and \(y=8\), examine the sign of \(y^{\prime}\) around these values and draw appropriate direction arrows to reflect the stability analysis.

Phase lines help provide a clear overview of the dynamic behavior of a system. They show at a glance where solutions are heading, facilitating a deeper understanding of the system's long-term behavior.