/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 Obtain a slope field and add to ... [FREE SOLUTION] | 91Ó°ÊÓ

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Obtain a slope field and add to it graphs of the solution curves passing through the given points.\(y^{\prime}=y\) with a.\( (0,1)\) b.\( (0,2)\) c. \((0,-1)\)

Short Answer

Expert verified
The solution curves are \( y=e^x \), \( y=2e^x \), \( y=-e^x \).

Step by step solution

01

Understand the Differential Equation

The given differential equation is \( y' = y \). This is a first-order linear differential equation where the rate of change of the function \( y \) is equal to the function itself.
02

Analyze the Slope Field

The slope field for a differential equation is a graphical representation of the slopes of the solution curves at given points in the plane. For the equation \( y' = y \), the slope at any point \((x, y)\) is equal to the y-coordinate at that point, which means that at every point \( (x, y) \), the slope is \( y \).
03

Draw the Slope Field

To draw the slope field, plot short line segments at various points \((x, y)\) in the plane. Each segment should have a slope equal to the y-coordinate of that point. For example, at \( (0, 1) \), the slope is 1; at \( (0, 2) \), the slope is 2; at \( (0, -1) \), the slope is -1.
04

Solve the Differential Equation Generally

We need to find the general solution of \( y' = y \). This is a separable differential equation and can be solved by rearranging as \( \frac{dy}{y} = dx \), then integrating both sides to get: \[ \ln |y| = x + C \] where \( C \) is the integration constant. Solving for \( y \) gives \( y = Ce^x \).
05

Find Particular Solutions

Use the initial conditions given to find specific solutions. - For \((0,1)\), \( y = Ce^0 = 1 \) gives \( C = 1 \). So the solution is \( y = e^x \).- For \((0,2)\), \( y = Ce^0 = 2 \) gives \( C = 2 \). So the solution is \( y = 2e^x \).- For \((0,-1)\), \( y = Ce^0 = -1 \) gives \( C = -1 \). So the solution is \( y = -e^x \).
06

Add the Solution Curves to the Slope Field

Add the graphs of the solution curves to the slope field. These curves are exponential functions based on the particular solutions derived in Step 5. Draw the curves: - \( y = e^x \) that passes through \( (0, 1) \).- \( y = 2e^x \) that passes through \( (0, 2) \).- \( y = -e^x \) that passes through \( (0, -1) \).

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Key Concepts

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

Differential Equations
Differential equations are equations that relate a function with its derivatives. They describe how a quantity changes in relation to another. In the given exercise, we have the differential equation \( y' = y \), which is a simple first-order linear differential equation. This equation tells us that the rate of change of \( y \) with respect to \( x \) is equal to the current value of \( y \).
Understanding such equations is important as they appear in various scientific fields to model real-world phenomena, such as population growth, heat conduction, or motion. To solve differential equations like this one, we find a function \( y(x) \) that satisfies the given relationship throughout its domain. This often involves mathematical techniques like integration or special transformations.
Solution Curves
A solution curve of a differential equation is a curve in the coordinate plane that represents a set of solutions to the differential equation for certain initial conditions. For \( y' = y \), a general solution is found as \( y = Ce^x \), where \( C \) is a constant.
The special points given in the exercise - \((0,1)\), \((0,2)\), and \((0,-1)\) - dictate the particular solutions for each case. These specific requirements help us determine the exact value of \( C \) for each initial condition:
  • Through \((0,1)\), the curve is \( y = e^x \)
  • Through \((0,2)\), it's \( y = 2e^x \)
  • Through \((0,-1)\), it becomes \( y = -e^x \)
These curves represent exponential growth or decay and have specific characteristics based on their respective initial conditions.
Slope Analysis
Slope analysis is a method used to understand how the solution of a differential equation behaves at various points in the plane. It involves examining the slopes given by the differential equation at those points. For \( y' = y \), the slope at any point \((x,y)\) is exactly the value of \( y \) at that point.
This means that whenever \( y \) is positive, the slope will be positive, indicating upward growth. Conversely, if \( y \) is negative, the slope will be downward, suggesting exponential decay. With a slope of zero, there's neither growth nor decay. These slopes can be visualized through a slope field, where short line segments represent different slopes at various points, assisting us in piecing together a general view of solution behaviors.
Graphical Representation
A slope field is a graphical tool used to represent the behavior of a differential equation visually. It's a grid of lines overlaid on a plane, where each line indicates the slope of the solution curve at that point. By observing these tiny line segments, we can infer the shape and direction of potential solution curves even before solving the equation analytically.
For the equation \( y' = y \), creating a slope field involves drawing lines with slopes equal to the \( y \)-values at selected points. Then, you can add the solution curves that align with initial given points like \((0,1)\), \((0,2)\), and \((0,-1)\).
  • At \((0,1)\), draw the curve \( y = e^x \)
  • At \((0,2)\), illustrate \( y = 2e^x \)
  • At \((0,-1)\), show \( y = -e^x \)
These curves depict specific trajectories on the field, showing how the solutions change across the plane, encapsulating both exponential growth and decay.

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Most popular questions from this chapter

The autonomous differential equations represent models for population growth. For each exercise, use a phase line analysis to sketch solution curves for \(P(t),\) selecting different starting values \(P(0) .\) Which equilibria are stable, and which are unstable? $$\frac{d P}{d t}=1-2 P$$

An 800-L tank is half full of distilled water. At time \(t=0,\) a solution containing 50 grams / \(L\) of concentrate enters the tank at the rate of \(20 \mathrm{L} / \mathrm{min}\), and the well-stirred mixture is withdrawn at the rate of \(12 \mathrm{L} / \mathrm{min}\). a. At what time will the tank be full? b. At the time the tank is full, how many kilograms of concentrate will it contain?

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. $$y^{\prime}=y-\sqrt{y}, \quad y > 0$$

Suppose that a healthy population of some species is growing in a limited environment and that the current population \(P_{0}\) is fairly close to the carrying capacity \(M_{0} .\) You might imagine a population of fish living in a freshwater lake in a wilderness area. Suddenly a catastrophe such as the Mount St. Helens volcanic eruption contaminates the lake and destroys a significant part of the food and oxygen on which the fish depend. The result is a new environment with a carrying capacity \(M_{1}\) considerably less than \(M_{0}\) and, in fact, less than the current population \(P_{0} .\) Starting at some time before the catastrophe, sketch a "before-and-after" curve that shows how the fish population responds to the change in environment.

Solve the Bernoulli equations. \(x y^{\prime}+y=y^{-2}\)

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