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Chapter 7: Problem 8

In Problems 7-10, 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 \( y = 4e^{-x} \), and it passes through the point \( (0, 4) \).

Step by step solution

01

Identify the Type of Differential Equation

The given differential equation is \( y' = -y \). This is a first-order linear ordinary differential equation.
02

Solve the Differential Equation Using Separation of Variables

To solve \( y' = -y \), we can use the method of separation of variables. Rewrite the equation as \( \frac{dy}{dx} = -y \). Separating the variables gives us \( \frac{dy}{y} = -dx \).
03

Integrate Both Sides

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

Solve for \( y \)

To solve for \( y \), exponentiate both sides: \( |y| = e^{-x+C} = e^{-x} \cdot e^C \). Let \( e^C = C_1 \, > 0 \), thus \( y = C_1 e^{-x} \).
05

Apply the Initial Condition

Use the initial condition \( y(0) = 4 \) to find \( C_1 \). Substitute \( x = 0 \) and \( y = 4 \) into the equation: \( 4 = C_1 e^0 = C_1 \). So, \( C_1 = 4 \).
06

Write the Particular Solution

The particular solution to the differential equation is \( y = 4e^{-x} \).
07

Plot the Slope Field and Particular Solution

To plot the slope field, draw small line segments representing the slope \( -y \) at various points in the plane. For the particular solution \( y = 4e^{-x} \), plot this curve on the same graph, showing it passing through the point \((0, 4)\).

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Key Concepts

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

Slope Field
A slope field, also known as a direction field, is a visual representation of a differential equation. It consists of short line segments at various points in a plane, indicating the slope of the solution at those points based on the differential equation. In our given equation, \( y' = -y \), the slope at any point \((x, y)\) is \(-y\).
To create a slope field for this equation, follow these steps:
  • Choose points on a grid and calculate the slope \(-y\) at each point.
  • Draw short line segments at each grid point with the calculated slope.
  • Observe the pattern of these segments; they guide us to the general shape of the solutions.
Slope fields help us visualize how a solution behaves over a range of values without solving the differential equation explicitly. They offer an initial step to understanding the nature of solutions.
Separation of Variables
Separation of variables is a method for solving first-order ordinary differential equations that can be rewritten so that all terms involving one variable and its differential are on one side of the equation, and all terms involving the other variable are on the other side.
For the equation \( y' = -y \):
  • Rewrite as \( \frac{dy}{dx} = -y \).
  • Separate the variables: \( \frac{dy}{y} = -dx \).
  • This segregation allows us to integrate each side independently.
The integration of \( \frac{dy}{y} = -dx \) gives:
  • \( \ln|y| = -x + C \), where \( C \) is the integration constant.
This method simplifies the process of finding general solutions to differential equations by reducing them to simple integrations.
Initial Condition
In solving differential equations, an initial condition is a value that helps us find a specific solution, known as a particular solution, from the general solution. Initial conditions are usually given in the form of \( y(x_0) = y_0 \).
For our problem, the initial condition \( y(0) = 4 \) is used to determine the constant \( C_1 \) in the general solution \( y = C_1 e^{-x} \).
Applying this:
  • Substitute \( x = 0 \) and \( y = 4 \) into \( y = C_1 e^{-x} \).
  • Solve: \( 4 = C_1 e^0 = C_1 \).
  • Thus, \( C_1 = 4 \).
Initial conditions are essential as they provide the necessary specifics needed to pinpoint one distinct solution from infinitely many possibilities deriving from the general solution.
Particular Solution
A particular solution to a differential equation is a specific solution that satisfies the initial conditions provided. It contrasts with the general solution, which contains a constant \( C \) and represents a family of solutions.
For the differential equation \( y' = -y \) and the initial condition \( y(0) = 4 \), we derived the particular solution:
  • Start with the general form: \( y = C_1 e^{-x} \).
  • Utilize the initial condition: since \( C_1 = 4 \), the particular solution becomes \( y = 4e^{-x} \).
This particular solution uniquely solves our differential equation with the given initial condition, accurately describing the behavior of the system over time. Additionally, when plotted on the slope field, it confirms that the curve passes through the point \((0, 4)\), illustrating agreement with the initial condition.

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Most popular questions from this chapter

In Problems 1-4, a slope field is given for a differential equation of the form \(y^{\prime}=f(x, y)\). Use the slope field to sketch the solution that satisfies the given initial condition. In each case, find \(\lim _{x \rightarrow \infty} y(x)\) and approximate \(y(2)\). $$ y(0)=5 $$

Apply the Improved Euler Method to the equation \(y^{\prime}=y, y(0)=1\), with \(h=0.2,0.1,0.05,0.01,0.005\) to approximate the solution on the interval \([0,1]\). (Note that the exact solution is \(y=e^{x}\), so \(y(1)=e\).) Compute the error in approximating \(y(1)\) (see Example 3 and the subsequent discussion) and fill in the following table. For the Improved Euler Method, is the error proportional to \(h, h^{2}\), or some other power of \(h\) ? $$ \begin{array}{lcc} \hline h & \begin{array}{c} \text { Error from } \\ \text { Euler's Method } \end{array} & \begin{array}{c} \text { Error from } \\ \text { Improved Euler Method } \end{array} \\ \hline 0.2 & 0.229962 & 0.015574 \\ 0.1 & 0.124540 & \\ 0.05 & 0.064984 & 0.001091 \\ 0.01 & 0.013468 & 0.000045 \\ 0.005 & 0.006765 & \\ \hline \end{array} $$

In Problems 1-54, perform the indicated integrations. \(\int_{0}^{\pi / 6} 2^{\cos x} \sin x d x\)

In Problems 1-54, perform the indicated integrations. \(\int \frac{t^{2} \cos ^{2}\left(t^{3}-2\right)}{\sin ^{2}\left(t^{3}-2\right)} d t\)

In Problems 1-54, perform the indicated integrations. \(\int \frac{x}{x^{2}+4} d x\)
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