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Chapter 9: Problem 21

Solve the exponential growth/decay initial value problem for \(y\) as a function of \(t\) by thinking of the differential equation as a first-order linear equation with \(P(x)=-k\) and \(Q(x)=0 :\) $$\frac{d y}{d t}=k y \quad(k\( constant \)), \quad y(0)=y_{0}$$

Short Answer

Expert verified
The solution is \( y(t) = y_0 e^{kt} \).

Step by step solution

01

Recognize the Form of Differential Equation

The equation \( \frac{d y}{d t} = k y \) is a first-order linear differential equation where \( k \) is a constant. The general form of a first-order linear differential equation is \( \frac{dy}{dt} + P(t)y = Q(t) \). Here, we identify \( P(t) = -k \) and \( Q(t) = 0 \).
02

Rearrange and Integrate the Equation

Rearrange the differential equation to separate variables: \[ \frac{1}{y} \frac{d y}{d t} = k \]. Integrating both sides with respect to \( t \), we get: \[ \int \frac{1}{y} dy = \int k \, dt \].
03

Calculate the Integrals

The integration results in:\[ \ln |y| = kt + C \]where \( C \) is the constant of integration. This follows from the fact that the integral of \( \frac{1}{y} \) is \( \ln |y| \) and the integral of \( k \) with respect to \( t \) is \( kt \).
04

Solve for \(y\) in Terms of \(t\)

Exponentiate both sides to solve for \( y \):\[ |y| = e^{kt + C} \]. Since \( e^{C} \) is a constant, it can be written as another constant \( C' \), so:\[ |y| = C' e^{kt} \]. We can express \( y \) as:\[ y = C' e^{kt} \], assuming \( C' > 0 \).
05

Apply the Initial Condition

Given the initial condition \( y(0) = y_0 \), substitute \( t = 0 \) into the equation: \[ y_0 = C' e^{k\cdot0} = C' \]. This means \( C' = y_0 \). The equation becomes:\[ y(t) = y_0 e^{kt} \].

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Key Concepts

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

First-Order Linear Differential Equation
A first-order linear differential equation is one where the highest derivative is first order. In our exercise, the equation \( \frac{dy}{dt} = ky \) represents such an equation. Generally, the form of a first-order linear differential equation is \( \frac{dy}{dt} + P(t)y = Q(t) \). In this problem:
  • \( P(t) \) is \(-k\), a constant coefficient.
  • \( Q(t) \) is 0, meaning there's no additional function altering the rate of change.
This structure is commonly useful in modeling exponential growth or decay in various scientific and real-world scenarios, such as population dynamics or radioactive decay. The presence of the coefficient \( k \) determines whether the process is growth (\( k > 0 \)) or decay (\( k < 0 \)).
Initial Value Problem
An initial value problem involves solving a differential equation with a given initial condition. The initial condition is provided as \( y(0) = y_0 \), where \( t = 0 \) is the starting point and \( y_0 \) is the initial amount or state of the system.In our specific problem, this initial value condition is essential:
  • It allows us to solve for the constant of integration \( C' \).
  • By applying the condition, we find that \( y(0) = y_0 \) such that \( C' = y_0 \).
  • This step tailors the generic solution of the differential equation to fit the situation described by the initial state.
The resolution of initial value problems is crucial as it ensures that the mathematical model aligns correctly with known data or conditions.
Separation of Variables
Separation of variables is a technique used to solve differential equations by isolating the dependent and independent variables on opposite sides. To apply this method to \( \frac{dy}{dt} = ky \), we rearrange the terms:
  • Move \( y \) to the left side: \( \frac{1}{y} \frac{dy}{dt} = k \).
  • The equation shows that the rate of change of \( y \) is proportional to \( y \) itself, a hallmark of exponential functions.
The next step involves integrating both sides, which brings us closer to finding a solution.This method is especially effective for exponential growth and decay problems because it simplifies the differential equation, making it easier to find an exact solution.
Integration of Differential Equations
Once the variables are separated in a differential equation, the next step is integration. By integrating \( \int \frac{1}{y} dy = \int k \, dt \), we find:
  • The left integral becomes \( \ln |y| \).
  • The right integral becomes \( kt + C \), where \( C \) is an integration constant.
To solve for \( y \), we exponentiate both sides to eliminate the logarithm:\[ |y| = e^{kt + C} \]As \( e^C \) is another constant (\( C' \)), we write the final expression as \( y = C'e^{kt} \). Adding the initial condition \( y(0) = y_0 \) determines that \( C' = y_0 \), leading to the complete solution:
  • \( y(t) = y_0 e^{kt} \)
This integral form is pivotal because it provides a function that describes how quantities evolve over time, capturing essential exponential behaviors.

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Most popular questions from this chapter

The spread of information Sociologists recognize a phenomenon called social diffusion, which is the spreading of a piece of information, technological innovation, or cultural fad among a population. The members of the population can be divided into two classes: those who have the information and those who do not. In a fixed population whose size is known, it is reasonable to assume that the rate of diffusion is proportional to the number who have the information times the number yet to receive it. If \(X\) denotes the number of individuals who have the information in a population of \(N\) people, then a mathematical model for social diffusion is given by $$\frac{d X}{d t}=k X(N-X)$$ where \(t\) represents time in days and \(k\) is a positive constant. a. Discuss the reasonableness of the model. b. Construct a phase line identifying the signs of \(X^{\prime}\) and \(X^{\prime \prime}\) . c. Sketch representative solution curves. d. Predict the value of \(X\) for which the information is spreading most rapidly. How many people eventually receive the information?

In Exercises 29 and \(30,\) obtain a slope field and graph the particular solution over the specified interval. Use your CAS DE solver to find the general solution of the differential equation. $$ \begin{array}{l}{\text { A logistic equation } y^{\prime}=y(2-y), y(0)=1 / 2 ; 0 \leq x \leq 4} \\ {0 \leq y \leq 3}\end{array} $$

Show that the solution of the initial value problem is \(y^{\prime}=x+y, \quad y\left(x_{0}\right)=y_{0}\) \(y=-1-x+\left(1+x_{0}+y_{0}\right) e^{x-x_{0}}\)

Solve the following initial value problem for \(u\) as a function of \(t :\) $$ \frac{d u}{d t}+\frac{k}{m} u=0 \quad(k \text { and } m \text { positive constants }), \quad u(0)=u_{0} $$ a. as a first-order linear equation. b. as a separable equation.

a. Identify the equilibrium values. Which are stable and which are unstable? b. Construct a phase line. Identify the signs of \(y^{\prime}\) and \(y^{\prime \prime}\) . c. Sketch several solution curves. \(\frac{d y}{d x}=y^{3}-y\)
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