Problem 5 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The equilibrium point is \( y = 2 \).

Step 1: Identify Equilibrium Condition

Equilibrium points occur when the derivative \( \frac{dy}{dt} \) is equal to zero. This means we need to solve the equation: \[ \frac{y-2}{y+1} = 0 \]

Step 2: Solve for \( y \)

To solve \( \frac{y-2}{y+1} = 0 \), we set the numerator of the fraction to zero, as a fraction is zero when its numerator is zero and its denominator is not zero: \[ y - 2 = 0 \]

Step 3: Express Solution for \( y \)

Solving the equation \( y - 2 = 0 \), we find \[ y = 2 \] as the solution, provided the denominator is not zero, which is true when \( y + 1 eq 0 \). Therefore, \( y = 2 \) is indeed an equilibrium point.

Key Concepts

Understanding Equilibrium Points

Equilibrium points in differential equations are key to analyzing the behavior of dynamic systems. When we talk about equilibrium, we're looking at a state where the system doesn't change; it's stable, not moving anywhere else. In mathematical terms, this occurs when the rate of change, given by the derivative \( \frac{dy}{dt} \), is zero. This means the function is not increasing or decreasing at that point. For the differential equation \( \frac{dy}{dt} = \frac{y-2}{y+1} \), equilibrium occurs when this fraction equals zero. This simplifies our task to finding when:

• The numerator (\( y - 2 \)) is zero.
• The denominator (\( y + 1 \)) is not zero.

Solving \( y - 2 = 0 \) gives us the equilibrium point \( y = 2 \). Ensuring \( y + 1 eq 0 \) means \( y eq -1 \), preserving our equilibrium condition.

Understanding the Derivative

The derivative \( \frac{dy}{dt} \) is a fundamental concept in calculus and differential equations. It measures how a function changes as its input changes. In the context of our differential equation, it describes how the variable \( y \) changes with respect to the variable \( t \). When you have a derivative set to zero, it indicates that there's no change at that point, signifying an equilibrium state. For the given function, \( \frac{dy}{dt} = \frac{y-2}{y+1} \), the derivative's value depends on both the current state \( y \) and its relationship to other parts of the function.

• If the derivative is positive, \( y \) is increasing.
• If the derivative is negative, \( y \) is decreasing.
• If the derivative is zero, \( y \) remains unchanged.

This mathematical tool helps predict the behavior of dynamical systems, making the derivative pivotal in locating and confirming equilibrium points.

Solution Method for Finding Equilibria

The process of finding equilibrium points doesn't have to be complex once the steps are clear. Here's how you solve it:1. **Identify Equilibrium Condition**: Start by setting the derivative to zero, which for our problem is \( \frac{dy}{dt} = \frac{y-2}{y+1} = 0 \). This step alone sets you on the path to finding equilibrium.2. **Solve the Equation**: Concentrate on simplifying and solving \( \frac{y-2}{y+1} = 0 \). Remember, a fraction equals zero when its numerator is zero, so you solve \( y - 2 = 0 \) which easily gives \( y = 2 \).3. **Check Conditions for Validity**: Ensure that the denominator, \( y + 1 \), is not zero to prevent division by zero conditions. Here, \( y eq -1 \), ensuring our solution is valid.These steps make the process manageable and provide a clear method to find equilibrium points for any differential equation setup. Remembering these steps aids in tackling similar problems with confidence and precision.