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Find the solution \(y(t)\) by recognizing each differential equation as determining unlimited, limited, or logistic growth, and then finding the constants. $$ \begin{array}{l} y^{\prime}=\frac{2}{3}(1-y) \\ y(0)=0 \end{array} $$

Short Answer

Expert verified
The solution is \( y(t) = 1 - e^{-\frac{2}{3}t} \).

Step by step solution

01

Identify the Type of Differential Equation

The given differential equation is \( y' = \frac{2}{3}(1-y) \). It appears to follow the format of a \( y' = k(M-y) \) equation, indicating limited growth model behavior, where \( k \) is the growth rate and \( M \) is the carrying capacity.
02

Determine the Constants

From the equation \( y' = \frac{2}{3}(1-y) \), we can see that \( k = \frac{2}{3} \) and the carrying capacity \( M = 1 \). This is because the equation matches the limited growth form \( y' = k(M-y) \) where \( k \) is the coefficient of the term \( (1-y) \) and \( M \) is the constant being subtracted from \( y \).
03

Solve the Differential Equation

The standard solution for a limited growth differential equation \( y' = k(M-y) \) is given by: \[ y(t) = M - (M - y_0)e^{-kt} \] where \( y_0 \) is the initial condition, \( M = 1 \), \( y_0 = 0 \), and \( k = \frac{2}{3} \). Substituting these values in, we have: \[ y(t) = 1 - (1 - 0)e^{-\frac{2}{3}t} = 1 - e^{-\frac{2}{3}t} \].
04

Verify the Initial Condition

To ensure our solution satisfies the initial condition \( y(0) = 0 \), substitute \( t = 0 \) into the solution: \( y(0) = 1 - e^{-\frac{2}{3} \times 0} = 1 - 1 = 0 \). This confirms that our solution is correct.

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Key Concepts

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

Limited Growth Model
The limited growth model is a type of differential equation that describes systems where growth slows down as the solution approaches a maximum value. This is different from unlimited growth models, where no such cap is present. In the equation \( y' = \frac{2}{3}(1-y) \), the term \( (1-y) \) represents how the growth rate diminishes over time.
The formula for limited growth is typically written as \( y' = k(M-y) \), where:
  • \( k \) is the growth rate constant of the model.
  • \( M \) is the carrying capacity, representing the maximum value that \( y \) can approach as time goes on.
As \( y \) gets closer to \( M \), the term \( (M-y) \) becomes smaller, reducing the rate of change. This depicts a situation where resources or other factors inherently limit growth.
Carrying Capacity
Carrying capacity, denoted as \( M \) in the limited growth model, is the maximum value that a population size \( y \) can reach. This parameter plays a crucial role in modeling real-world scenarios like population dynamics or resource-limited growth.
In the equation \( y' = \frac{2}{3}(1-y) \), the carrying capacity \( M \) is 1. This means that the population, or whatever item being measured by \( y \), will stabilize at the value of 1 over time. If the current value of \( y \) is less than \( M \), the population can still grow, but this growth will slow as \( y \) nears 1.
Understanding carrying capacity is essential in predicting long-term behavior. It represents a balance point, where the rate of growth equals the rate of environmental resistance.
Initial Conditions
Initial conditions are important because they allow us to find a specific solution to a differential equation. In this exercise, the initial condition given is \( y(0) = 0 \). This information tells us the starting point of the population or quantity being modeled by the differential equation.
Using initial conditions helps determine the constant values in the exponential solution formula. For the limited growth model, the general solution is:
  • \( y(t) = M - (M - y_0)e^{-kt} \)
Where \( y_0 \) is the initial condition.
Plugging the initial condition \( y(0) = 0 \) into the equation ensures the function behaves correctly from the start. It also allows us to verify the solution by checking it against the given initial point.
Exponential Functions
Exponential functions play a role in modeling growth processes, like the limited growth model. They describe how rapidly things can increase or decrease. In the differential equation solution \( y(t) = 1 - e^{-\frac{2}{3}t} \), the exponential term \( e^{-\frac{2}{3}t} \) reveals how the growth rate changes over time.
This function shows decay due to the negative exponent, \( -\frac{2}{3}t \), causing rapid changes at first, which slow down as time passes.
Key properties of exponential functions include:
  • They start changing quickly but slow down as time increases if they involve exponential decay.
  • The initial rate of change depends on the coefficient of the exponent.
These functions are superb for describing processes where change itself decreases over time, making them invaluable in modeling scenarios like the limited growth model.

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Most popular questions from this chapter

BIOMEDICAL: Heart Function In the reservoir model, the heart is viewed as a balloon that swells as it fills with blood (during a period called the systole), and then at time \(t_{0}\) it shuts a valve and contracts to force the blood out (the diastole). Let \(p(t)\) represent the pressure in the heart at time \(t\) a. During the diastole, which lasts from \(t_{0}\) to time \(T\), \(p(t)\) satisfies the differential equation $$\frac{d p}{d t}=-\frac{K}{R} p$$ Find the general solution \(p(t)\) of this differential equation. (K and \(R\) are positive constants determined, respectively, by the strength of the heart and the resistance of the arteries. The differential equation states that as the heart contracts, the pressure decreases \((d p / d t\) is negative) in proportion to itself. \()\) b. Find the particular solution that satisfies the condition \(p\left(t_{0}\right)=p_{0} . \quad\left(p_{0}\right.\) is a constant representing the pressure at the transition time \(t_{0}\).) c. During the systole, which lasts from time 0 to time \(t_{0}\), the pressure \(p(t)\) satisfies the differential equation $$ \frac{d p}{d t}=K I_{0}-\frac{K}{R} p $$ Find the general solution of this differential equation. (I \(_{0}\) is a positive constant representing the constant rate of blood flow into the heart while it is expanding.) [Hint: Use the same \(u\) -substitution technique that was used in Example \(7 .\) d. Find the particular solution that satisfies the condi- $$ \text { tion } p\left(t_{0}\right)=p_{0} $$ e. In parts (b) and (d) you found the formulas for the pressure \(p(t)\) during the diastole \(\left(t_{0} \leq t \leq T\right)\) and the systole \(\left(0 \leq t \leq t_{0}\right) .\) Since the heart behaves in a cyclic fashion, these functions must satisfy \(p(T)=p(0) .\) Equate the solutions at these times (use the correct formula for each time) to derive the important relationship $$ R=\frac{p_{0}}{I_{0}} \frac{1-e^{-K T / R}}{1-e^{-K t_{0} / R}} $$.

The following exercises require the use of a slope field program. For each differential equation: a. Use a graphing calculator slope field program to graph the slope field for the differential equation on the window [-5,5] by [-5,5]. b. Sketch the slope field on a piece of paper and draw a solution curve that follows the slopes and that passes through the given point. $$ \begin{array}{l} \frac{d y}{d x}=x \ln \left(y^{2}+1\right) \\ \text { point: }(0,-2) \end{array} $$

Think of the slope field for the differential equation \(\frac{d y}{d x}=x y .\) What is the sign of the slope in quadrant \(I\) (where \(x\) and \(y\) are both positive)? What is the sign of the slope in each of the other three quadrants? Check your answers by looking at the slope field on page 596.

For each initial value problem, calculate the Euler approximation for the solution on the interval [0,1] using \(n=4\) segments. Draw the graph of your approximation. (Carry out the calculations "by hand" with the aid of a calculator, rounding to two decimal places. Answers may differ slightly, depending on when you do the rounding.) $$ \begin{array}{l} y^{\prime}=8 x^{2}-y \\ y(0)=2 \end{array} $$

For each initial value problem, calculate the Euler approximation for the solution on the interval [0,1] using \(n=4\) segments. Draw the graph of your approximation. (Carry out the calculations "by hand" with the aid of a calculator, rounding to two decimal places. Answers may differ slightly, depending on when you do the rounding.) $$ \begin{array}{l} y^{\prime}+2 y=e^{4 x} \\ y(0)=2 \end{array} $$

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