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In Problems 39-48 you should treat \(h\) as a constant. For what values of \(h\) (if any) does each equation have equilibria? Use \(a\) graphical argument to show which of the equilibria (if any) are stable. $$ \frac{d y}{d t}=e^{y}-(1-y) $$

Short Answer

Expert verified
The equation has equilibria where \( e^y = 1-y \); stability is checked graphically; \( h \) does not affect equilibria.

Step by step solution

01

Set the equation to zero

To find the equilibria of the differential equation, set the derivative of the function equal to zero: \[ e^y - (1 - y) = 0 \] This represents the condition where the rate of change of \( y \) is zero, which means the system is at equilibrium.
02

Solve for the equilibrium points

Rearrange the equation to solve for \( y \): \[ e^y = 1 - y \] This is a transcendental equation, and the solutions for \( y \) are the equilibrium points.
03

Use graphical method to determine solutions

Graph the functions \( e^y \) and \( 1 - y \) to find points of intersection. These intersections are solutions to the equation \( e^y = 1 - y \) and represent the equilibrium points.
04

Analyze the stability of equilibria

To determine the stability of the equilibria, inspect the graph. Evaluate the behavior of the derivative sign changes around each equilibrium point: - If \( \frac{d y}{d t} > 0 \) for \( y \) just less than the equilibrium and \( \frac{d y}{d t} < 0 \) just more, the equilibrium is stable.- Conversely, if the derivative sign changes from negative to positive, the equilibrium is unstable.
05

Determine if \( h \) affects equilibrium

In this problem, \( h \) is constant and does not appear in the differential equation or affect the static equilibria directly, so it does not determine the presence of equilibria. Equilibria depend on solving the equation \( e^y = 1 - y \).

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Key Concepts

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

Equilibrium Points
In a differential equation, equilibrium points are where the rate of change of the variable equals zero. These points are crucial because they define states where the system doesn't change over time. To find equilibrium points, we set the derivative of the function to zero.

For our exercise, this means solving the equation \(e^y - (1 - y) = 0\). This equation shows when the change in \(y\) over time is zero. Essentially, these are the solutions where \(e^y = 1 - y\). Solving this particular equation might not be straightforward due to its transcendental nature. However, possible solutions can often be found by observing graphical intersections, trial and error, or numerical methods.

Each equilibrium point represents a spot where the differential dynamics pause. It's crucial to identify these points in any differential equation to understand how the system behaves at rest or in steady states.
Stability Analysis
Stability analysis tells us whether an equilibrium point is stable or unstable. Stability here means small disturbances or changes will or will not return to the equilibrium state. If slight shifts in \(y\) return to equilibrium, it's stable. Otherwise, it's unstable.

After identifying equilibrium points, analyze the behavior of the function around them. For example, if you have an equilibrium at \(y_0\), check the signs of \(\frac{dy}{dt}\) slightly above and below \(y_0\). If \(\frac{dy}{dt}\) is positive before \(y_0\) and negative after, this suggests a stable environment.

Conversely, if the derivative changes from negative to positive around an equilibrium, it indicates that small changes in \(y\) drive the system away from equilibrium—revealing instability. Hence, evaluating the behavior near these points allows us to predict the system's response to perturbations.
Graphical Method
The graphical method provides a visual approach to solving transcendental equations and is especially useful when solutions are not easily calculated. By plotting the components of an equation, such as \(e^y\) and \(1-y\) in our problem, we can visually identify intersections or equilibrium points.

These intersections on the graph represent where the rates of change equate to zero, corresponding to equilibrium points of the differential equation. The advantage of this method is its simplicity and intuitive approach to dealing with complex equations.

Also, examining the slope or derivative around these intersections helps determine stability. When looking at a graph, pay attention to how the slope behaves around equilibrium intersections—the curve's direction insightfully reveals stability or instability. The graphical method, therefore, is a practical tool in both finding and analyzing equilibrium in differential equations.

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Most popular questions from this chapter

You should treat \(h\) as a constant. For what values of \(h\) (if any) does each equation have equilibria? Use a graphical argument to show which of the equilibria (if any) are stable. $$ \frac{d x}{d t}=x^{2}-h x $$

In Problems you will need to solve differential equations by separation of variables. In these problems it will not always be possible to solve explicitly for \(y\) in terms of \(t ;\) instead your solution may take the form of an implicit function relating the two variables. $$ \frac{d y}{d t}=\frac{t^{2}+1}{\cos y+\sin y} \text { where } y(0)=0 $$

Determine whether the equilibrium at \(x=0\) is stable, unstable, or semi- stable. $$ \frac{d x}{d t}=x^{3} e^{-x} $$

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