Problem 77 - Calculus for Biology and Medicine

Question

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. $$ \frac{d y}{d t}=2-3 y $$

Step-by-Step Solution

Answer

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

Step 1: 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 \( y = \frac{2}{3} \).

Step 2: 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 \( y = \frac{2}{3} \): \[ f'(\frac{2}{3}) = -3 \]

Step 3: 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.

Key Concepts

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.