Question

What is the equilibrium solution of the equation \(y^{\prime}(t)=3 y-9\) Is it stable or unstable?

Short answer

Answer: The equilibrium solution is \(y = 3\), and it is unstable.

Step-by-step solution

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 \(y = 3\).

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#.

Key concepts

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 find it, set \(3y - 9 = 0\).
• Solving gives \(y = 3\), meaning when \(y = 3\), \(y'(t) = 0\).

Understanding these basic forms helps interpret the stability and behavior of systems described by differential equations.

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.

In the example \(y^{\prime}(t)=3y-9\), we found that \(y = 3\) is the equilibrium point. By substituting \(y = 3 + \Delta y\) into the equation, we derive \(y'(t) = 3\Delta y\). Here, \(y'(t)\) is directly proportional to \(\Delta y\). Thus, small positive changes lead to growth away from \(y = 3\), confirming that this equilibrium 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:

• 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 \(\Delta y\).

For our differential equation, this gives \(y'(t) = 3\Delta y\). The sign and magnitude of \(y'(t)\) relative to \(\Delta y\) help conclude about stability: positive \(y'(t)\) for positive \(\Delta y\) indicates instability, as we'd see the system moving away from equilibrium. This method provides a simple but powerful insight into dynamic systems.