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Chapter 9: Problem 1

a. Identify the equilibrium values. Which are stable and which are unstable? b. Construct a phase line. Identify the signs of \(y^{\prime}\) and \(y^{\prime \prime}\). c. Sketch several solution curves. $$\frac{d y}{d x}=(y+2)(y-3)$$

Short Answer

Expert verified
Equilibrium values are \( y = -2 \) (unstable) and \( y = 3 \) (stable).

Step by step solution

01

Find the Equilibrium Points

Equilibrium points are found by setting the derivative equal to zero. We need to solve \( \frac{d y}{d x} = (y+2)(y-3) = 0 \). This gives us two solutions: \( y = -2 \) and \( y = 3 \). These are the equilibrium points.
02

Determine Stability of Equilibrium Points

To determine stability, analyze the sign changes of \( \frac{d y}{d x} \) around the equilibrium points. For \( y < -2 \), \( \frac{d y}{d x} > 0 \); for \( -2 < y < 3 \), \( \frac{d y}{d x} < 0 \); and for \( y > 3 \), \( \frac{d y}{d x} > 0 \). Thus, \( y = -2 \) is unstable and \( y = 3 \) is stable.
03

Construct the Phase Line

On the phase line, plot \( y = -2 \) and \( y = 3 \). Use arrows to indicate the direction of \( y \). From \( y < -2 \) to \( -2 < y < 3 \), the direction is to the left (unstable), and from \( -2 < y < 3 \) to \( y > 3 \), the direction is to the right (stable). Thus, arrows on the phase line show stability and instability zones.
04

Identify Signs of Derivatives

For each region divided by equilibrium points, identify the signs of \( y' \) and \( y'' \). From Step 2, we know where \( y' \) is positive or negative. The second derivative \( y'' \) can be found by differentiating \( y' = (y+2)(y-3) \): \( y'' = \frac{d^2 y}{d x^2} = 2y -1 \). Investigate the sign of \( y'' \) in each interval.
05

Sketch Solution Curves

Using all the information from previous steps, draw solution curves. Start with horizontal lines at \( y = -2 \) and \( y = 3 \). For \( y < -2 \) and \( y > 3 \), draw curves moving away from \( y = -2 \) and towards \( y = 3 \) respectively, showing stability at \( y = 3 \) and instability at \( y = -2 \). Indicate changes in concavity based on \( y'' \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Points
In differential equations, equilibrium points represent the values for which the rate of change is zero. Hence, the system at these points is in a state of rest or balance. To find them, we solve the expression for the derivative set to zero. For our equation, \(\frac{d y}{d x}=(y+2)(y-3)\), this means:\[(y+2)(y-3) = 0\] This equation factors to find two solutions: \(y = -2\) and \(y = 3\). These are the equilibrium points where the slope of the function is zero, indicating no net change at those points. These points are crucial for understanding the behavior of the system as a whole.
Stability Analysis
Once we have identified equilibrium points, we need to determine their stability. Stability analysis helps us understand whether small deviations from these points will die out or grow over time.
For this, we analyze the sign of \(\frac{d y}{d x}\) just before and after each equilibrium point. This means investigating the intervals \(y < -2\), \(-2 < y < 3\), and \(y > 3\):
  • For \(y < -2\): \(\frac{d y}{d x} > 0\), meaning the function increases.
  • For \(-2 < y < 3\): \(\frac{d y}{d x} < 0\), indicating a decrease, stabilizing the behavior.
  • For \(y > 3\): \(\frac{d y}{d x} > 0\), where the function rises again.
Thus, \(y = -2\) is unstable because deviations away from \(y = -2\) will wander. Meanwhile, \(y = 3\) is stable as deviations return towards \(y = 3\), keeping the system calm and steady.
Phase Line
The phase line is a visual representation that allows us to summarize the stability and dynamics around the equilibrium points.
To construct it, place the equilibrium points \(y = -2\) and \(y = 3\) on a vertical line. Next, depict the behavior between these points with arrows:
  • From \(y < -2\): Arrow pointing upward, representing increasing \(y\).
  • From \(-2 < y < 3\): Arrow pointing downward, indicating a decrease in \(y\).
  • From \(y > 3\): Arrow pointing upward, symbolizing an increase in \(y\) again.
These arrows help identify zones of stability and instability, clearly showing where the system restores balance or diverges from it. The phase line is instrumental in visualizing these dynamics and aiding in sketching accurate solution curves.

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Most popular questions from this chapter

Suppose that a healthy population of some species is growing in a limited environment and that the current population \(P_{0}\) is fairly close to the carrying capacity \(M_{0} .\) You might imagine a population of fish living in a freshwater lake in a wilderness area. Suddenly a catastrophe such as the Mount St. Helens volcanic eruption contaminates the lake and destroys a significant part of the food and oxygen on which the fish depend. The result is a new environment with a carrying capacity \(M_{1}\) considerably less than \(M_{0}\) and, in fact, less than the current population \(P_{0} .\) Starting at some time before the catastrophe, sketch a "before-and-after" curve that shows how the fish population responds to the change in environment.

Is either of the following equations correct? Give reasons for your answers. a. \(\quad \frac{1}{\cos x} \int \cos x d x=\tan x+C\) b. \(\frac{1}{\cos x} \int \cos x d x=\tan x+\frac{C}{\cos x}\)

Is either of the following equations correct? Give reasons for your answers. a. \(x \int \frac{1}{x} d x=x \ln |x|+C \quad\) b. \(x \int \frac{1}{x} d x=x \ln |x|+C x\)

Find the orthogonal trajectories of the family of curves. Sketch several members of each family. $$y=m x$$

Have no explicit solution in terms of elementary functions. Use a CAS to explore graphically each of the differential equations.a. Use a CAS to plot the slope field of the differential equation $$y^{\prime}=\frac{3 x^{2}+4 x+2}{2(y-1)}.$$ over the region \(-3 \leq x \leq 3\) and \(-3 \leq y \leq 3\) b. Separate the variables and use a CAS integrator to find the general solution in implicit form. c. Using a CAS implicit function grapher, plot solution curves for the arbitrary constant values \(C=-6,-4,-2,0,2,4,6\) d. Find and graph the solution that satisfies the initial condition \(y(0)=-1.\)
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