Problem 2
Question
Find the equilibrium points and for each determine whether or not it is stable. a. \(\quad y^{\prime}=y-1\) b. \(\quad y^{\prime}=-y+1\) c. \(y^{\prime}=y^{2}-1\) d. \(\quad y^{\prime}=1-y^{2}\) e. \(\quad y^{\prime}=e^{-y}-1\) f. \(\quad y^{\prime}=e^{y}-1\) g. \(\quad y^{\prime}=\sin y\) h. \(\quad y^{\prime}=-y+y^{2}\) i. \(\quad y^{\prime}=-y^{3}\) j. \(\quad y^{\prime}=y^{3}\) For parts i. and \(\mathrm{j}\). draw the phase plane with arrows to determine the question of stability.
Step-by-Step Solution
Verified Answer
a. Stable, b. Stable, c. \( y=1 \) stable, \( y=-1 \) unstable, d. Stable, e. Stable, f. Unstable, g. Depends on \( n \), h. \( y=1 \) stable, \( y=0 \) unstable, i. Stable, j. Unstable.
1Step 1: Understanding Equilibrium Points
An equilibrium point occurs when the derivative \( y' = 0 \). For each equation, set \( y' = 0 \) and solve for \( y \).
2Step 2: Stability Analysis Method
To determine stability, utilize phase line analysis for simple cases. For nonlinear cases, check the sign of the derivative change \( y' \) around equilibrium points.
3Step 3: Solve Part a: \( y' = y - 1 \)
Set \( y - 1 = 0 \). The equilibrium point is \( y = 1 \). Since \( y' = y - 1 \), the derivative is positive for \( y > 1 \) (unstable) and negative for \( y < 1 \) (stable). Therefore, \( y=1 \) is a stable equilibrium point.
4Step 4: Solve Part b: \( y' = -y + 1 \)
Set \( -y + 1 = 0 \). The equilibrium point is \( y = 1 \). \( y' < 0 \) for \( y > 1 \) and \( y' > 0 \) for \( y < 1 \). Thus, \( y=1 \) is a stable equilibrium point.
5Step 5: Solve Part c: \( y' = y^2 - 1 \)
Set \( y^2 - 1 = 0 \). The equilibrium points are \( y = 1 \) and \( y = -1 \). For \( y = -1 \), \( y' > 0 \) for \( y > -1 \), thus unstable. For \( y = 1 \), \( y' < 0 \) for \( y > 1 \), thus stable.
6Step 6: Solve Part d: \( y' = 1 - y^2 \)
Set \( 1 - y^2 = 0 \). The equilibrium points are \( y = 1 \) and \( y = -1 \). For both points, the phase line shows \( y = 1 \) and \( y = -1 \) are stable.
7Step 7: Solve Part e: \( y' = e^{-y} - 1 \)
Set \( e^{-y} - 1 = 0 \). The equilibrium point is \( y = 0 \). \( y' > 0 \) for \( y < 0 \) and \( y' < 0 \) for \( y > 0 \) making \( y = 0 \) a stable point.
8Step 8: Solve Part f: \( y' = e^y - 1 \)
Set \( e^y - 1 = 0 \). The equilibrium point is \( y = 0 \). \( y' < 0 \) for \( y < 0 \) and \( y' > 0 \) for \( y > 0 \), making \( y = 0 \) unstable.
9Step 9: Solve Part g: \( y' = \\sin y \)
Set \( \sin y = 0 \). The equilibrium points are \( y = n\pi \) where \( n \) is an integer. Analysis shows stable at \( y = 2n\pi \) and unstable at \( y = (2n + 1)\pi \).
10Step 10: Solve Part h: \( y' = -y + y^2 \)
Rearrange as \( y' = y(y - 1) \). Set \( y(y - 1) = 0 \), equilibrium at \( y = 0, 1 \). \( y = 0 \) is unstable, \( y = 1 \) is stable.
11Step 11: Phase Plane Analysis for Part i: \( y' = -y^3 \)
Draw phase line: \( y' < 0 \) for all \( y eq 0 \), \( y' = 0 \) at \( y = 0 \). The equilibrium at \( y = 0 \) is stable due to flow towards equilibrium from both sides.
12Step 12: Phase Plane Analysis for Part j: \( y' = y^3 \)
Draw phase line: \( y' > 0 \) for all \( y eq 0 \), \( y' = 0 \) at \( y = 0 \). The equilibrium at \( y = 0 \) is unstable due to flow away from equilibrium on both sides.
Key Concepts
Stability AnalysisPhase Line AnalysisDerivative ChangeNonlinear Systems
Stability Analysis
When dealing with differential equations, one critical aspect is analyzing the stability of equilibrium points. An equilibrium point occurs when the derivative, often denoted as \( y' \), is equal to zero. To determine if these points are stable or unstable, stability analysis is employed, which involves evaluating the behavior of the system around these points. If small perturbations or changes in \( y \) lead to the system returning to the equilibrium point, it is considered stable. Conversely, if perturbations cause the system to move away from the equilibrium, the point is unstable.
For example, if we take the equation \( y' = y - 1 \), setting \( y - 1 = 0 \) gives the equilibrium point of \( y = 1 \). By examining the derivative around this point, we see that it is negative when \( y < 1 \) and positive when \( y > 1 \), suggesting that perturbations will revert to the equilibrium, making it stable.
For example, if we take the equation \( y' = y - 1 \), setting \( y - 1 = 0 \) gives the equilibrium point of \( y = 1 \). By examining the derivative around this point, we see that it is negative when \( y < 1 \) and positive when \( y > 1 \), suggesting that perturbations will revert to the equilibrium, making it stable.
Phase Line Analysis
Phase line analysis is a visual method for studying the behavior of solutions to differential equations. It's especially useful for equations like \( y' = y^2 - 1 \), where equilibrium points are determined by setting \( y^2 - 1 = 0 \), yielding \( y = 1 \) and \( y = -1 \).
By drawing a simple line and labeling the equilibrium points, we can analyze the direction of \( y' \) between and beyond these points. This involves checking whether \( y' \) is positive or negative in different regions of the line. For instance, if \( y' > 0 \) before \( y = -1 \) and \( y' < 0 \) after \( y = 1 \), these points help visualize how solutions behave as \( y \) approaches these equilibrium states. This analysis can guide whether such points are attracting (stable) or repelling (unstable).
The convenience of phase line diagrams lies in their ability to convey the stability of equilibria efficiently, allowing us to depict the system's overall dynamics at a glance.
By drawing a simple line and labeling the equilibrium points, we can analyze the direction of \( y' \) between and beyond these points. This involves checking whether \( y' \) is positive or negative in different regions of the line. For instance, if \( y' > 0 \) before \( y = -1 \) and \( y' < 0 \) after \( y = 1 \), these points help visualize how solutions behave as \( y \) approaches these equilibrium states. This analysis can guide whether such points are attracting (stable) or repelling (unstable).
The convenience of phase line diagrams lies in their ability to convey the stability of equilibria efficiently, allowing us to depict the system's overall dynamics at a glance.
Derivative Change
Understanding how the derivative \( y' \) changes with respect to \( y \) provides insights into system stability. A key part of analyzing equilibrium points involves observing the sign of \( y' \) around these points.
For example, consider the function \( y' = e^{-y} - 1 \). Setting \( e^{-y} - 1 = 0 \) gives \( y = 0 \) as an equilibrium point. By observing the derivative, we notice that \( y' > 0 \) when \( y < 0 \), and \( y' < 0 \) when \( y > 0 \). This means that if \( y \) moves above or below 0, the derivative's sign ensures \( y \) returns to the equilibrium, confirming it as a stable equilibrium point.
Monitoring changes in the derivative helps predict how likely a system is to return to equilibrium after disturbances, a crucial aspect in fields requiring system stability such as engineering and physics.
For example, consider the function \( y' = e^{-y} - 1 \). Setting \( e^{-y} - 1 = 0 \) gives \( y = 0 \) as an equilibrium point. By observing the derivative, we notice that \( y' > 0 \) when \( y < 0 \), and \( y' < 0 \) when \( y > 0 \). This means that if \( y \) moves above or below 0, the derivative's sign ensures \( y \) returns to the equilibrium, confirming it as a stable equilibrium point.
Monitoring changes in the derivative helps predict how likely a system is to return to equilibrium after disturbances, a crucial aspect in fields requiring system stability such as engineering and physics.
Nonlinear Systems
Nonlinear systems, characterized by differential equations that are not simply a direct proportion, are often more complex and diverse in behavior than linear systems. Such systems can lead to multiple equilibrium points and unique stability properties.
Take \( y' = y^2 - 1 \) for instance. Unlike linear systems, here equilibrium points depend on the dynamics of \( y^2 \). With nonlinear behavior, the stability around equilibrium points can shift more dramatically, affecting how solutions evolve over time. Often, nonlinear systems require analytical methods like phase line analysis or numerical simulations to fully understand their characteristics.
These systems are common in many real-world scenarios, ranging from population dynamics to electronic circuits. Understanding nonlinear dynamics allows for better modeling and prediction of complex systems and their behavior. Recognizing how those systems stabilize can be crucial in designing models for practical applications.
Take \( y' = y^2 - 1 \) for instance. Unlike linear systems, here equilibrium points depend on the dynamics of \( y^2 \). With nonlinear behavior, the stability around equilibrium points can shift more dramatically, affecting how solutions evolve over time. Often, nonlinear systems require analytical methods like phase line analysis or numerical simulations to fully understand their characteristics.
These systems are common in many real-world scenarios, ranging from population dynamics to electronic circuits. Understanding nonlinear dynamics allows for better modeling and prediction of complex systems and their behavior. Recognizing how those systems stabilize can be crucial in designing models for practical applications.
Other exercises in this chapter
Problem 2
Find the unique solutions to a. \(\quad y(0)=5 \quad y^{\prime}+2 y=0\) b. \(\quad y(0)=0 \quad y^{\prime}+2 y=0\) c. \(\quad y(0)=4 \quad y^{\prime}+3 y=t\) d.
View solution Problem 2
Consider the pair of differential equations $$u(0)=1 \quad u^{\prime}(t)=0.5 \times u(t)-0.2 \times u(t) \times v(t)$$ \(v(0)=2 \quad v^{\prime}(t)=0.1 \times u
View solution Problem 2
Which of the following possible solutions satisfies the initial condition and the differential equation. Possible Solution \(\quad\) Initial Condition \(\quad\)
View solution Problem 2
Find the values of \(P(t)\) for which \(P^{\prime}(t)=0\) in each of the equations Verhulst \(\quad p^{\prime}(t)=r \times p(t) \times\left(1-\frac{P(t)}{M}\rig
View solution