Problem 30 - Calculus for Biology and Medicine

Question

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. $$ \frac{d y}{d t}=\ln y-e^{-y} \quad y>0 $$

Step-by-Step Solution

Answer

There is one equilibrium, and it is unstable.

Step 1: Understand the Differential Equation

We have the differential equation \( \frac{dy}{dt} = \ln y - e^{-y} \), valid for \( y > 0 \). An equilibrium point is where the rate of change \( \frac{dy}{dt} \) is zero, meaning \( \ln y - e^{-y} = 0 \).

Step 2: Rewrite the Equation to Find Equilibria

Equilibrium occurs when \( \ln y = e^{-y} \). This will be our equation to find equilibria values of \( y \).

Step 3: Analyze Functions Separately

Consider two functions: \( f(y) = \ln y \) and \( g(y) = e^{-y} \). We need to compare these two functions to find where they intersect.

Step 4: Sketch Both Functions' Behaviors

Sketch the graph of \( f(y) = \ln y \), which is increasing and undefined for \( y \leq 0 \), and \( g(y) = e^{-y} \), which is decreasing, always positive, and tends to zero as \( y \to \infty \).

Step 5: Determine Intersection Points

Since \( \ln y \) increases while \( e^{-y} \) decreases, they will intersect at precisely one positive value of \( y \). Thus, there is one equilibrium point.

Step 6: Classify Stability Using Derivatives

To determine stability, consider the derivative \( \frac{d}{dy} (\ln y - e^{-y}) = \frac{1}{y} + e^{-y} \). At equilibrium, \( \frac{dy}{dt} \) switches signs, thus the slope \( 1/y + e^{-y} \) will be positive, indicating the equilibrium is unstable.

Key Concepts

Equilibrium Points

Understanding equilibrium points is essential when analyzing differential equations like \(\frac{dy}{dt} = \ln y - e^{-y}\). Equilibrium points occur where the rate of change is zero. This means the differential equation becomes a balance point. Mathematically, this is expressed as \(\ln y - e^{-y} = 0\). To find these equilibrium points, we set the left-hand side equal to zero and solve for \(y\). In our case, it simplifies to \(\ln y = e^{-y}\). Moving forward, finding intersections of the two components, \(f(y) = \ln y\) and \(g(y) = e^{-y}\), helps. When these functions intersect, they equal each other, revealing an equilibrium point. Here, sketching the graph aids in visual understanding.Remember that for \(y > 0\), \(\ln y\) is an increasing function, while \(e^{-y}\) is decreasing. Eventually, they cross precisely at one positive value of \(y\). This marks our single equilibrium point within the constraints provided.

Stability Analysis

Once an equilibrium point is identified, understanding its stability is the next crucial step. Stability analysis involves determining whether small perturbations around the equilibrium will return to the point or cause deviation. In this context, we use derivatives to assess stability. Specifically, we examine the change in the function's slope at the equilibrium point. For our equation, the derivative is \(\frac{d}{dy} (\ln y - e^{-y}) = \frac{1}{y} + e^{-y}\). Analyzing this derivative at the equilibrium can inform us about the surrounding dynamics. If the derivative is positive, a small increase in \(y\) would tend to further increase \(y\), suggesting instability. In this case, after computing the derivative at the equilibrium, we find it is positive. Consequently, the equilibrium is deemed unstable as small variations cause the system to move away from the equilibrium point. Stability analysis, therefore, is essential to predict system behaviors without directly solving the differential equation.

Derivative Applications

Derivatives offer valuable insights into the behavior of functions within differential equations. Here, derivatives help in analyzing the stability of the system by assessing changes at equilibrium points. Let’s break down their applications.

• Rate of Change: Derivatives provide the rate at which a system's state is changing over time. This is fundamental in determining equilibrium points where the rate of change equates to zero.


• Slope Analysis: They are critical in understanding the local behavior of a function. A positive slope around an equilibrium suggests that any small movement will lead away from the equilibrium, whereas a negative slope implies stability.


• Function Behavior: Derivatives clarify how a function behaves as inputs change, which is pivotal for sketching curves and understanding intersections that determine equilibrium points.

Using derivatives, we determine stability and visualize how slight variations in starting conditions can affect overall system behavior. Their application transcends beyond simple calculus and helps predict and analyze complex systems.