Problem 8

Question

Determine the type of each differential equation: unlimited growth, limited growth, logistic growth, or none of these. (Do not solve, just identify the type.) $$ y^{\prime}=6 y $$

Step-by-Step Solution

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Answer
The equation represents unlimited growth.
1Step 1: Identifying Components of the Equation
The given differential equation is \( y' = 6y \). This is a first-order, linear differential equation where the derivative \( y' \) is proportional to \( y \). The coefficient 6 is a positive constant.
2Step 2: Recognizing the Pattern as Unlimited Growth
Notice that the equation \( y' = 6y \) is in the form \( y' = ky \), which is the standard form for an unlimited growth model. Here, \( k = 6 \) is a constant that determines the rate of growth. When \( k > 0 \), it indicates exponential or unlimited growth.

Key Concepts

Unlimited Growth ModelLogistic Growth ModelFirst-Order Linear Differential Equation
Unlimited Growth Model
In mathematics, the unlimited growth model describes a process where the rate of change of a quantity is directly proportional to the current quantity itself. This typically results in exponential growth over time. The equation that represents this model is usually in the form \( y' = ky \), where \( y' \) is the derivative of \( y \) with respect to time \( t \), and \( k \) is a positive constant.

This positive constant \( k \) indicates that as time progresses, the quantity \( y \) will grow exponentially, leading to what's often called unlimited or exponential growth. The typical real-world examples include populations in an idealized environment, money that grows with compound interest, or any self-replicating process not limited by resources.

The important characteristics to remember are:
  • The growth is continuous and accelerates over time.
  • The mathematical representation is simple: \( y' = ky \).
  • There are no limitations or constraints applied to the growth within the model.
In the context of the equation \( y' = 6y \), since \( k = 6 \) is greater than zero, it directly fits the unlimited growth model.
Logistic Growth Model
Unlike the unlimited growth model, the logistic growth model includes limits on growth. It is commonly used to describe populations where there is a carrying capacity or maximum population that the environment can sustainably support. The differential equation for this model takes a different form than the simple exponential growth model.

Typically, the logistic growth model is described by the equation:
\[ y' = ky(1 - \frac{y}{L}) \]
where:
  • \( y' \) is the rate of growth.
  • \( k \) is a constant representing the rate of growth.
  • \( L \) is the carrying capacity or maximum population limit.
This form indicates that the growth rate decreases as \( y \) approaches \( L \). Hence, in the early stages, when \( y \) is much smaller than \( L \), the growth resembles the unlimited model. But as \( y \) grows closer to \( L \), the growth slows down, eventually stopping as \( y \) reaches \( L \).

The logistic growth model is often applied to real-world scenarios where resources are limited, such as biological populations and resource-limited economic growth.
First-Order Linear Differential Equation
A first-order linear differential equation involves a derivative that is first-order, which means it contains only the first derivative of the unknown function. The equation is in the form:
\[ y' + P(x)y = Q(x) \]
where:
  • \( y' \) is the first derivative of \( y \).
  • \( P(x) \) and \( Q(x) \) are functions of \( x \).
These equations are called linear because the terms involving the unknown function and its derivative are linear. That is, they are not multiplied or divided by each other, nor are they raised to any powers other than one.

The general solution to a first-order linear differential equation usually involves finding an integrating factor that simplifies the equation into a form that can be easily integrated.

The given equation \( y' = 6y \) can also be categorized as a first-order linear differential equation. Here, the function \( P(x) \) equals \(-6\) and \( Q(x) \) equals \(0\). The simplicity and linearity of these equations make them foundational in calculus and essential tools in applied mathematics.