Chapter 8: Problem 4
What is the equilibrium solution of the equation \(y^{\prime}(t)=3 y-9\) Is it stable or unstable?
Short Answer
Expert verified
Answer: The equilibrium solution is \(y = 3\), and it is unstable.
Step by step solution
01
Find the equilibrium solution
Set the derivative equal to zero and solve for y:
$$
y'(t) = 3y - 9 = 0
$$
Rearrange the equation to solve for y:
$$
3y = 9
$$
Then, divide both sides by 3:
$$
y = \frac{9}{3} = 3
$$
The equilibrium solution is \(y = 3\).
02
Analyze the stability
We need to check whether perturbation around the equilibrium point grows or decays. Let's consider a small change in y, which we can denote as \(\Delta y\):
$$
y = 3 + \Delta y
$$
Now, compute the derivative for this new value of y:
$$
y'(t) = 3(3 + \Delta y) - 9
$$
Simplify the equation:
$$
y'(t) = 9 + 3\Delta y - 9 = 3\Delta y
$$
Since \(y'(t)\) is directly proportional to \(\Delta y\), any small positive perturbation around the equilibrium point (\(y=3\)) will cause an increase in y'(t), meaning the solution will grow. Similarly, a small negative perturbation will cause a decrease in y'(t), leading to a further decrease in y.
Therefore, the equilibrium solution \(y = 3\) is #tag_highlight#unstable#tag_highlight_end#.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differential Equations
Differential equations are mathematical expressions that relate a function with its derivatives. They can describe various phenomena such as growth, decay, or oscillation. In particular, they reflect how a quantity changes over time or space. For instance, consider the differential equation \(y^{\prime}(t)=3y-9\). Here, the rate of change of \(y(t)\) depends on the current value of \(y(t)\) itself. This kind of differential equation is often found in population dynamics and physics.
To solve a differential equation, we often look for an equilibrium solution, which is a constant solution where the rate of change \(y^{\prime}(t)\) is zero. This signifies that the system is in a state of balance.
To solve a differential equation, we often look for an equilibrium solution, which is a constant solution where the rate of change \(y^{\prime}(t)\) is zero. This signifies that the system is in a state of balance.
- To find it, set \(3y - 9 = 0\).
- Solving gives \(y = 3\), meaning when \(y = 3\), \(y'(t) = 0\).
Stability Analysis
Stability analysis is crucial in understanding how systems respond to small changes or perturbations. After finding the equilibrium point of a differential equation, the next step is to determine if it is stable or unstable. This analysis tells us if the system will return to equilibrium after a disturbance or if deviations will grow.
- An equilibrium point is stable if perturbations tend to decrease, bringing the system back to the equilibrium.
- Conversely, if they increase, leading the system away, the point is unstable.
Perturbation Method
The perturbation method is a technique used in stability analysis to study how small changes affect a system at equilibrium. By introducing a slight shift \(\Delta y\) from the equilibrium point, we can determine the response of the system.
This method is quite straightforward:
This method is quite straightforward:
- Start with the found equilibrium solution, which is \(y = 3\) in our case.
- Introduce a perturbation \(\Delta y\), leading to \(y = 3 + \Delta y\).
- Substitute this back into the system's equation.
- Simplify to find how the derivative depends on \(\Delta y\).