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Chapter 2: Problem 11

In each problem find the exact solution of the given initial value problem. Then apply Euler's method twice to approximate (to four decimal places) this solution on the given interval, first with step size \(h=0.01\), then with step size \(h=0.005 .\) Make a table showing the approximate values and the actual value, together with the percentage error in the more accurate approximation, for \(x\) an integral multiple of 0.2. Throughout, primes denote derivatives with respect to \(x .\) \(y^{\prime}=y-2, y(0)=1 ; 0 \leqq x \leqq 1\)

Short Answer

Expert verified
The exact solution is \( y = 2 - e^x \). Euler's method approximations vary by step size; smaller steps give higher accuracy. Use a table for detailed comparison.

Step by step solution

01

Solve the Differential Equation

The given differential equation is \( y' = y - 2 \). This is a first-order linear differential equation.To solve it, we first rewrite the equation as \( y' - y = -2 \).This has the standard form \( y' + P(x)y = Q(x) \), where \( P(x) = -1 \) and \( Q(x) = -2 \). The integrating factor is \( e^{\int -1 \, dx} = e^{-x} \).Multiplying the entire differential equation by the integrating factor, we have:\[ e^{-x} y' - e^{-x} y = -2e^{-x} \]This can be rewritten as:\[ \frac{d}{dx}(e^{-x}y) = -2e^{-x} \]Integrating both sides with respect to \( x \), we get:\[ e^{-x}y = \int -2e^{-x} \, dx = 2e^{-x} + C \]Thus, \( y = 2 + Ce^{x} \).Given the initial condition \( y(0) = 1 \), we substitute to find \( C \):\[ 1 = 2 + Ce^{0} = 2 + C \rightarrow C = -1 \]So, the exact solution is:\[ y = 2 - e^x \].
02

Apply Euler's Method with h=0.01

Euler's method iterates using the formula:\[ y_{n+1} = y_n + h f(x_n, y_n) \]Given \( y'(x) = y(x) - 2 \), the function \( f(x_n, y_n) = y_n - 2 \).**With Step Size \( h = 0.01 \):**- Start with \( (x_0, y_0) = (0, 1) \).- For each \( n \), calculate\( y_{n+1} \). For example: - \( y_0 = 1 \) - \( y_1 = 1 + 0.01(1 - 2) = 0.99 \) - Continue this process for \( x_n = 0.2, 0.4, 0.6, 0.8, 1.0 \).Repeat the calculations to approximate \( y \) values.
03

Apply Euler's Method with h=0.005

**With Step Size \( h = 0.005 \):**- Again, start with \( (x_0, y_0) = (0, 1) \).- Follow the calculation process, now using \( h = 0.005 \).- Compute \( y_{n+1} \) for each \( n \) similarly.- Evaluate \( y \) at \( x = 0.2, 0.4, 0.6, 0.8, 1.0 \) using this step size.
04

Compare Results

Calculate the exact solution \( y = 2 - e^x \) at each needed \( x \) value for comparison.For each integral multiple of 0.2, create a table with:1. The exact value from the solution \( y = 2 - e^x \).2. Approximations from both step sizes.3. Calculate the percentage error for \( h = 0.005 \). Use the formula:\[\text{Percentage Error} = \left(\frac{|\text{Exact} - \text{Approx.}|}{\text{Exact}}\right) \times 100\%\]
05

Construct the Table

Create a table listing:- The integral multiples of \( x = 0.2 \).- Exact \( y \) values using \( y = 2 - e^x \).- Approximate \( y \) values obtained with \( h = 0.01 \).- Approximate \( y \) values obtained with \( h = 0.005 \).- The percentage error for the approximation with \( h = 0.005 \).This will help visualize the accuracy of Euler's method with different step sizes.

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Key Concepts

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

Initial Value Problem
An initial value problem involves solving a differential equation with a given starting condition. This type of problem needs both the differential equation and the specific initial condition set, which is the value of the unknown function at a certain point.
In the context of our example, the problem provides the first-order differential equation \( y' = y - 2 \) and the initial condition \( y(0) = 1 \).
This means we are looking for a function \( y(x) \) such that when we substitute \( x = 0 \) into the function, the output will be \( 1 \). The initial value forms the anchor point from which we solve the differential equation.
First-Order Linear Differential Equation
A first-order linear differential equation has the general form \( y' + P(x)y = Q(x) \), where \( P(x) \) and \( Q(x) \) are functions of \( x \).
The equation \( y' = y - 2 \) can be rewritten as \( y' - y = -2 \), matching the standard form.
These types of equations are fundamental in mathematical modeling of real-world processes, such as physics and engineering problems.
To solve them, we often use techniques like the integrating factor method, which provides a way to simplify and integrate the equation efficiently.
Percentage Error Calculation
Percentage error is a useful method to determine the accuracy of numerical approximations.
The percentage error compares the approximate solution to the exact solution to quantify the accuracy.
This is calculated using the formula:
  • \( \text{Percentage Error} = \left(\frac{|\text{Exact} - \text{Approx.}|}{\text{Exact}}\right) \times 100\% \)
In our exercise, using smaller step sizes in Euler's method usually results in a smaller percentage error, indicating a more accurate approximation.
By comparing the results of different step sizes, we can see how the accuracy of Euler's method changes, giving insights into how precise our numerical solution could be.
Integrating Factor Method
The integrating factor method is a key technique to solve first-order linear differential equations.
An integrating factor is chosen to transform the differential equation into an easily integrable form:
  • Here, the equation \( y' - y = -2 \) uses \( e^{-x} \) as the integrating factor.
  • This factor is calculated as \( e^{\int P(x) \, dx} \), which simplifies our equation.
By multiplying the entire differential equation by the integrating factor, the left side becomes conveniently integrable.
In our example, it transforms into \( \frac{d}{dx}(e^{-x} y) = -2e^{-x} \), allowing for straightforward integration.
This method results in the solution \( y = 2 - e^x \), which satisfies both the differential equation and the initial value condition.

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Most popular questions from this chapter

As in Problem 25 of Section \(2.4\), you bail out of a helicopter and immediately open your parachute, so your downward velocity satisfies the initial value problem $$ \frac{d v}{d t}=32-1.6 v, \quad v(0)=0 $$ (with \(t\) in seconds and \(v\) in \(\mathrm{ft} / \mathrm{s}\) ). Use the improved Euler method with a programmable calculator or computer to approximate the solution for \(0 \leqq t \leqq 2\), first with step size \(h=0.01\) and then with \(h=0.005\), rounding off approximate \(v\) -values to three decimal places. What percentage of the limiting velocity \(20 \mathrm{ft} / \mathrm{s}\) has been attained after 1 second? After 2 seconds?

The time rate of change of an alligator population \(P\) in a swamp is proportional to the square of \(P .\) The swamp contained a dozen alligators in 1988 , two dozen in 1998 . When will there be four dozen alligators in the swamp? What happens thereafter?

A hand-held calculator will suffice for Problems 1 through 10, where an initial value problem and its exact solution are given. Apply the Runge-Kutta method to approximate this solution on the interval \([0,0.5]\) with step size \(h=0.25 .\) Construct a table showing five-decimal-place values of the approximate solution and actual solution at the points \(x=0.25\) and \(0.5\). $$ y^{\prime}=-y, y(0)=2 ; y(x)=2 e^{-x} $$

In each problem find the exact solution of the given initial value problem. Then apply Euler's method twice to approximate (to four decimal places) this solution on the given interval, first with step size \(h=0.01\), then with step size \(h=0.005 .\) Make a table showing the approximate values and the actual value, together with the percentage error in the more accurate approximation, for \(x\) an integral multiple of 0.2. Throughout, primes denote derivatives with respect to \(x .\) \(y^{2} y^{\prime}=2 x^{5}, y(2)=3 ; 2 \leqq x \leqq 3\)

A computer with a printer is required for Problems 17 through 24. In these initial value problems, use the Runge-Kutta method with step sizes \(h=0.2,0.1,0.05\), and \(0.025\) to approximate to six decimal places the values of the solution at five equally spaced points of the given interval. Print the results in tabular form with appropriate headings to make it easy to gauge the effect of varying the step size \(h .\) Throughout, primes denote derivatives with respect to \(x\). $$ y^{\prime}=x^{2}-y^{2}, y(0)=1 ; 0 \leqq x \leqq 2 $$
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