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For Problems \(77-88\) find all equilibria, and, by calculating the eigenvalue of the differential equation, determine which equilibria are stable and which are unstable. $$ \frac{d y}{d t}=2-3 y $$

Short Answer

Expert verified
The equilibrium point \( y = \frac{2}{3} \) is stable.

Step by step solution

01

Find Equilibria

To find the equilibria of the differential equation \( \frac{dy}{dt} = 2 - 3y \), we need to set the derivative equal to zero and solve for \( y \).\[ 2 - 3y = 0 \] Solve this equation: \[ 3y = 2 \] \[ y = \frac{2}{3} \] So, the equilibrium is at \( y = \frac{2}{3} \).
02

Calculate Eigenvalue (Jacobian Evaluation)

The stability of an equilibrium point in a one-dimensional differential equation is determined by the sign of the derivative at the equilibrium. Thus, differentiate \( f(y) = 2 - 3y \) with respect to \( y \). \[ f'(y) = -3 \] Evaluate this at the equilibrium point \( y = \frac{2}{3} \): \[ f'(\frac{2}{3}) = -3 \]
03

Determine Stability

An equilibrium is stable if the derivative is negative at that point. Since \( f'(\frac{2}{3}) = -3 \) is negative, the equilibrium at \( y = \frac{2}{3} \) is stable.

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Key Concepts

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

Differential Equations
In the context of mathematics, a differential equation is a relationship that connects a function with its derivatives. These equations are crucial for modeling how certain quantities change over time or space.
In simpler terms, they describe how things like population sizes, temperatures, or speeds vary, often depending on the current state of the system, such as at a given moment in time.
Differential equations can be classified into different types. The one in the exercise is a first-order linear differential equation, which means it includes only the first derivative and no higher-order derivatives, and it has constant coefficients.
Understanding these equations is essential since they form the backbone of various scientific fields and engineering, describing phenomena in physics, biology, and economics, to name a few.
Stability Analysis
Stability analysis is a technique used to determine whether small changes in a system's initial conditions will die out over time, return to equilibrium, or cause the system to behave unpredictably.
It tells us if an equilibrium point, like the one found at \( y = \frac{2}{3} \) in the exercise, remains stable or becomes unstable when small disturbances occur.
There are a couple of key points to consider regarding stability:
  • Stable equilibrium: When small changes or disturbances to the system make it return to its equilibrium state over time.
  • Unstable equilibrium: When disturbances cause the system to move away further from its equilibrium point.
In our specific case, by finding that the derivative \( f'(y) = -3 \) is negative at the equilibrium point, it is determined that the equilibrium at \( y = \frac{2}{3} \) is stable. Negative derivatives typically suggest that the system will return to equilibrium after small perturbations, thus indicating stability.
Eigenvalues
Eigenvalues play a crucial role when analyzing the stability of equilibrium points in systems. They arise when calculating the Jacobian matrix, which in more complex systems involves matrices instead of just evaluating the derivative as in our one-dimensional example.
The analysis of eigenvalues helps us determine whether the small deviations from an equilibrium point will return to equilibrium, spiral out, or oscillate.
When dealing with linear differential equations as in our example, the eigenvalue simplifies to finding the derivative. If the eigenvalue (or in this case, the derivative) is negative, as \( f'(y) = -3 \), it indicates that any deviation will decay back to equilibrium, showcasing stability.
Understanding eigenvalues is essential because they provide insights into the dynamic behavior and responsiveness of a system.

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Most popular questions from this chapter

Find the equilibria of the following differential equations. $$ \frac{d y}{d t}=\frac{y-1}{y^{2}+1} $$

By breaking down each equation into two parts that you can sketch, determine how many equilib\mathrm{\\{} r i a ~ e a c h ~ d i f f e r e n t i a l ~ e q u a t i o n ~ h a s , ~ a n d ~ c l a s s i f y ~ t h e m ~ a s ~ s t a b l e ~ or unstable. You do not need to determine the location of the equilibria. $$ \frac{d N}{d t}=N^{2}-N+1 \quad N>0 $$

Determine whether the equilibrium at \(x=0\) is stable, unstable, or semi- stable. $$ \frac{d x}{d t}=x^{3}+x^{4} $$

A chemostat is a device that can be used to maintain a chemical at a particular concentration. Assume that the reaction \(A+B=C\) takes place in a chemostat that maintains \(A\) at a constant concentration \(a\). The chemical \(B\) has initial concentration \(b\) and is depleted by the reaction. (a) Explain why the concentration \(x(t)\) of \(C\) now obeys a differential equation: $$ d x / d t=k_{A B} a(b-x)-k_{C} x $$ (b) Find the equilibrium for \(x\) predicted by Equation (8.65). (c) Is the equilibrium that you found in part (b) stable or unstable?

In our compartment model we assumed that inflows and outflows are matched at \(q\) to keep the volume of water in the tank constant. It's often useful when modeling, for example, the flow of pollutant into a pristine environment, to consider what can occur if the inflows and outflows do not match. Let's assume that the tank initially contains a volume \(V_{0}\) of water. Water flows into the tank at rate \(q_{\mathrm{in}}\), and out of the tank at rate \(q_{\text {out. }}\) (You may assume \(q_{\text {in }}>q_{\text {out } .}\) ) Suppose that the water flowing into the tank contains a concentration \(C_{I}\) of solute. As usual we write \(C(t)\) for the concentration in the tank. (a) Show that the concentration in the tank can be modeled using a differential equation: $$ \frac{d}{d t}(C V)=q_{\text {in }} C_{I}-q_{\text {out }} C $$ (b) Previously we were able to treat \(V\) as a constant. Now \(V\) changes with time. Derive a formula for \(V(t)\). (c) By substituting your formula for \(V(t)\) into (a), derive a differential equation for \(C(t)\). (d) In general we cannot analyze the behavior of the solution \(C(t)\) using techniques from Section \(8.2 .\) Why not? (e) Let's assume \(C_{l}=0\). Then show that your equation from (c) can be written as: $$ \frac{d C}{d t}=\frac{-q_{\mathrm{in}} C}{V_{0}+\left(q_{\mathrm{in}}-q_{\mathrm{out}}\right) t} $$ (f) Assume some definite values for the constants in \((8.57):\) \(q_{\mathrm{in}}=2, q_{\text {out }}=1\), and \(V_{0}=20 .\) Assuming \(C(0)=1\), solve \((8.57)\) to find \(C(t) .\) Show that \(\lim _{t \rightarrow \infty} C(t)=0\).

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