Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Plot a slope field for each differential equation. Use the method of separation of variables (Section 4.9) or an integrating factor (Section 7.7) to find a particular solution of the differential equation that satisfies the given initial condition, and plot the particular solution. $$ y^{\prime}=-y ; y(0)=4 $$

Short Answer

Expert verified
The particular solution is \( y = 4e^{-x} \).

Step by step solution

01

Understanding the Differential Equation

The given differential equation is \( y' = -y \). This is a first-order linear differential equation that can be solved using separation of variables.
02

Separation of Variables

Rewrite the equation as \( \frac{dy}{dx} = -y \). Separate the variables to get \( \frac{dy}{y} = -dx \).
03

Integrating Both Sides

Integrate both sides: \( \int \frac{dy}{y} = \int -dx \). This results in \( \ln|y| = -x + C \), where \( C \) is the constant of integration.
04

Solving for \( y \)

Solve for \( y \) by exponentiating both sides: \( |y| = e^{-x+C} \), which simplifies to \( y = Ce^{-x} \). We apply absolute value when integrating, but since \( y \) must be positive from the initial condition, we use \( y = Ce^{-x} \).
05

Apply the Initial Condition

Use the initial condition \( y(0) = 4 \): \( 4 = Ce^{0} \), which implies \( C = 4 \). Thus, the particular solution is \( y = 4e^{-x} \).
06

Plotting the Slope Field

To plot the slope field for \( y' = -y \), at each point \((x, y)\), calculate the slope \(-y\). This results in lines with slopes dependent only on the \( y \)-value.
07

Plotting the Particular Solution

On the same graph where you plotted the slope field, graph the particular solution \( y = 4e^{-x} \). It should show how the initial condition modifies the general behavior depicted by the slope field.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Slope Field
A slope field is a visual tool that helps us understand differential equations. It is a grid of small line segments or arrows that indicate the slope of the solution at various points. In the context of the equation \( y' = -y \), the slope at any point \( (x, y) \) is equal to \(-y\).
This means all slopes in the field depend solely on the \( y \)-value, not \( x \). For instance, when \( y = 2 \), the slope is \(-2\) regardless of the \( x \)-coordinate. Slope fields help us visualize the direction that solutions to the differential equation will follow.
By plotting a slope field, you can see the general shape and behavior of solutions, and how they might change with different initial conditions. It's like a road map for finding solutions without actually solving the equation.
Separation of Variables
Separation of variables is a simple method used to solve first-order differential equations. It involves rearranging an equation so that all terms involving one variable are on one side, and terms involving another variable are on the other. For \( y' = -y \), we rewrite it as \( \frac{dy}{y} = -dx \).
This step is crucial because it allows us to integrate both sides independently: the left side with respect to \( y \) and the right side with respect to \( x \).
Once the equation is separated, the integration process reveals the natural logarithm on the left side: \( \ln|y| \) and a linear term on the right: \(-x + C\) (where \( C \) is the integration constant).
This method is particularly useful because it simplifies the process of solving differential equations by reducing it to straightforward integration.
Initial Condition
An initial condition is a specific value a solution to a differential equation must satisfy, which is usually given in the problem statement. In this case, we have \( y(0) = 4 \). This condition tells us the value of \( y \) when \( x = 0 \).
Initial conditions are essential because they allow us to pinpoint the exact solution from a family of solutions.
Without an initial condition, a differential equation can often have many solutions, differing by a constant. Applying \( y(0) = 4 \) to the general solution \( y = Ce^{-x} \) helps us find the particular solution by determining the constant \( C \). It turns out \( C \) equals 4 in our specific case, so our particular solution becomes \( y = 4e^{-x} \). This solution is plotted on the slope field to show how it follows the directions indicated by the slopes, starting from the initial point.
Particular Solution
The particular solution of a differential equation is the specific solution that satisfies the given initial condition. After performing separation of variables and integrating, we obtained the general solution \( y = Ce^{-x} \).
We found that, using the initial condition \( y(0) = 4 \), \( C = 4 \). Thus, the particular solution is \( y = 4e^{-x} \).
This solution not only satisfies the differential equation but also adheres to the initial condition provided. The particular solution is critical because it represents the actual scenario from the range of potential behaviors indicated by the slope field.
When plotted on the same graph as the slope field, the particular solution gives a path that accurately tracks from the initial value through each point, reflecting the flow of the slope field.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use a CAS to evaluate the definite integrals. If the CAS does not give an exact answer in terms of elementary functions, then give a numerical approximation. $$ \int_{1}^{4} \frac{\sqrt{t}}{1+t^{8}} d t $$

. Suppose that \(f(t)\) has the property that \(f^{\prime}(a)=f^{\prime}(b)=0\) and that \(f(t)\) has two continuous derivatives. Use integration by parts to prove that \(\int_{a}^{b} f^{\prime \prime}(t) f(t) d t \leq 0 .\) Hint \(:\) Use integration by parts by differentiating \(f(t)\) and integrating \(f^{\prime \prime}(t) .\) This result has many applications in the field of applied mathematics.

In Problems 55-61, derive the given reduction formula using integration by parts. $$ \int x^{\alpha} e^{\beta x} d x=\frac{x^{\alpha} e^{\beta x}}{\beta}-\frac{\alpha}{\beta} \int x^{\alpha-1} e^{\beta x} d x $$

Find the error in the following "proof" that \(0=1 .\) In \(\int(1 / t) d t\), set \(u=1 / t\) and \(d v=d t .\) Then \(d u=-t^{-2} d t\) and \(u v=1\). Integration by parts gives $$ \int(1 / t) d t=1-\int(-1 / t) d t $$

The Law of Mass Action in chemistry results in the differential equation $$ \frac{d x}{d t}=k(a-x)(b-x), \quad k>0, \quad a>0, \quad b>0 $$ where \(x\) is the amount of a substance at time \(t\) resulting from the reaction of two others. Assume that \(x=0\) when \(t=0\). (a) Solve this differential equation in the case \(b>a\). (b) Show that \(x \rightarrow a\) as \(t \rightarrow \infty\) (if \(b>a\) ). (c) Suppose that \(a=2\) and \(b=4\), and that 1 gram of the substance is formed in 20 minutes. How much will be present in 1 hour? (d) Solve the differential equation if \(a=b\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free