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Find all equilibria of the autonomous differential equation and construct the phase plot. (a) $$y^{\prime}=y+2 \ln y, \quad y>0$$ (b) $$y^{\prime}=y^{3}-a, \quad a \geqslant 0$$ (c) $$y^{\prime}=\frac{5}{2+y}, \quad y \geqslant 0$$

Short Answer

Expert verified
(a) Equilibrium at \( y \approx 0.27 \); (b) Equilibrium at \( y = a^{1/3} \); (c) No equilibria.

Step by step solution

01

(a): Set the Equation to Zero

For the equilibrium points, set \( y' = 0 \): \( y + 2 \ln y = 0 \). This gives us the equation \( y = -2 \ln y \).
02

(a): Solve for Equilibrium Points

Solving \( y = -2 \ln y \), try substituting special function values or use iterative methods. Numerical approximation finds \( y \approx 0.27 \) as the equilibrium.
03

(a): Construct Phase Plot

To construct the phase plot, analyze the sign of \( y' = y + 2 \ln y \). For \( y > 0.27 \), \( y' > 0 \); for \( y < 0.27 \), \( y' < 0 \). This indicates a repelling equilibrium point at \( y \approx 0.27 \).
04

(b): Set the Equation to Zero

Set \( y' = 0 \) for equilibrium: \( y^3 - a = 0 \), giving \( y^3 = a \).
05

(b): Solve for Equilibrium Points

The equilibrium point is \( y = a^{1/3} \). This provides one equilibrium point for each non-negative \( a \).
06

(b): Construct Phase Plot

For \( y > a^{1/3} \), \( y' > 0 \); for \( y < a^{1/3} \), \( y' < 0 \). Equilibria are stable when \( y \to a^{1/3} \).
07

(c): Set the Equation to Zero and Analyze

For \( y' = \frac{5}{2+y} \), there are no zeros since the fraction is always positive. No equilibria.
08

(c): Construct Phase Plot

Since \( y' > 0 \) for all \( y \geq 0 \), the plot shows that solutions are monotonically increasing for any \( y \geq 0 \).

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Key Concepts

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

Equilibrium Points
In the context of autonomous differential equations, equilibrium points, also known as critical or stationary points, are values of the dependent variable where the rate of change is zero. These points are crucial because they represent states of the system where there is no change over time.
For an equation like \( y' = y + 2 \ln y \), finding equilibrium involves setting \( y' = 0 \). This yields the equation \( y = -2 \ln y \). Solving this can be challenging; numerical methods often provide estimates such as \( y \approx 0.27 \).
For \( y' = y^3 - a \), equilibriums occur when \( y^3 = a \). The equilibrium point here is \( y = a^{1/3} \), indicating that for each non-negative \( a \), there is a corresponding equilibrium state.
In contrast, the function \( y' = \frac{5}{2+y} \) does not yield any equilibrium points since the derivative never equals zero; it is always positive, indicating constant growth without pause.
Phase Plot
The phase plot is a graphical representation that shows the behavior of solutions around the equilibrium points and helps us understand the dynamics of differential equations over time. It's an essential tool for visualizing how solutions change as the variable \( y \) changes.
For the equation \( y' = y + 2 \ln y \), we observe:
  • If \( y > 0.27 \), then \( y' > 0 \) indicating growth; solutions move away from this point.
  • If \( y < 0.27 \), then \( y' < 0 \), indicating decay; again, solutions diverge from this point.
This characterizes \( y \approx 0.27 \) as a repelling equilibrium.
In the case of \( y' = y^3 - a \), the phase plot shows:
  • Solutions for \( y > a^{1/3} \) indicate \( y' > 0 \), meaning growth toward \( y \to a^{1/3} \).
  • For \( y < a^{1/3} \), \( y' < 0 \), indicating the system moves toward the equilibrium.
Thus, \( y = a^{1/3} \) is attracting or stabilizing.
For \( y' = \frac{5}{2+y} \), since \( y' > 0 \) everywhere, the phase plot shows relentless growth; all trajectories are uprising, confirming no equilibrium exists.
Mathematical Analysis
Mathematical analysis of differential equations involves understanding the behavior of solutions, specifically how they evolve over time and which structures arise under certain conditions.
Let's delve into equation (a) - \( y' = y + 2 \ln y \). Analyzing it mathematically, we set \( y' = 0 \) to find when changes stop, yielding \( y = -2 \ln y \). Attempts at an analytical solution are difficult; numerical methods estimate \( y \approx 0.27 \). The system's analysis reveals this equilibrium is repelling, critical in predicting future states.
For \( y' = y^3 - a \), setting the derivative to zero provides \( y = a^{1/3} \), achieved directly by solving \( y^3 = a \). Mathematical rigor shows these as attracting points, crucial in forecasting steady states.
Finally, \( y' = \frac{5}{2+y} \) has no zeros, thus mathematical analysis tells us solutions escalate continuously, essential in anticipating constant growth dynamics. Understanding these aspects helps in modeling and predicting real-world systems governed by such equations.

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Most popular questions from this chapter

Harvesting of renewable resources Suppose a population grows according to the logistic equation but is subject to a constant total harvest rate of \(H .\) If \(N(t)\) is the population size at time \(t,\) the population dynamics are \(\frac{d N}{d t}=r\left(1-\frac{N}{K}\right) N-H\) Different values of \(H\) result in different equilibrium population sizas; if \(H\) is lerge enough, we might expect extinction. (a) Suppose \(r=2, K=1000,\) and \(H=100 .\) Find all equilibria. (b) Determine whether each of the equilibria in part (a) is locally stable or unstable. Is the population predicted to go extinct?

Suppose that the population dynamics of a species obeys a modified version of the logistic differential equation having the following form: \(\frac{d N}{d t}=r\left(1-\frac{N}{K}\right)^{2} N\) where \(r \neq 0\) and \(K>0\) (a) Show that \(\hat{N}=0\) and \(\hat{N}=K\) are equilibria. (b) For which values of \(r\) is the equilibrium \(\hat{N}=0\) unstable? (b) For which values of \(r\) is the equilibrium \(\hat{N}=0\) unstable? (c) Apply the local stability criterion to the equilibrium \(\hat{N}=K .\) What do you think your answer means about the stability of this equilibrium? (Note: This is an example in which the local stability criterion is inconclusive.) (d) Construct two phase plots, one for the case where \(r>0\) and the other for \(r<0,\) and determine the stability of \(\hat{N}=K\) in each case. Does the answer match your reasoning in part \((c) ?\)

\(9-12\) Describe the motion of a particle with position \((x, y)\) as \(t\) varies in the given interval. \(x=3+2 \cos t, \quad y=1+2 \sin t, \quad \pi / 2 \leqslant t \leqslant 3 \pi / 2\)

(a) Sketch the curve by using the parametric equations to plot points. Indicate with an arrow the direction in which the curve is traced as \(t\) increases. (b) Eliminate the parameter to find a Cartesian equation of the curve. \(x=\sqrt{t}, \quad y=1-t\)

Predator-prey equations For each predator-prey system, determine which of the variables, \(x\) or \(y\) , represents the prey population and which represents the predator population. Is the growth of the prey restricted just by the predators or by other factors as well? Do the predators feed only on the prey or do they have additional food sources? Explain. $$\begin{aligned} \text { (a) } \frac{d x}{d t} &=-0.05 x+0.0001 x y \\\ \frac{d y}{d t} &=0.1 y-0.005 x y \\ \text { (b) } \frac{d x}{d t} &=0.2 x-0.0002 x^{2}-0.006 x y \\ \frac{d y}{d t} &=-0.015 y+0.00008 x y \end{aligned}$$

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