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An initial value problem and its exact solution \(y(x)\) are given. Apply Euler's method twice to approximate to this solution on the interval \(\left[0, \frac{1}{2}\right]\), first with step size \(h=0.25\), then with step size \(h=0.1 .\) Compare the threedecimal-place values of the two approximations at \(x=\frac{1}{2}\) with the value \(y\left(\frac{1}{2}\right)\) of the actual solution. $$ y^{\prime}=y+1, y(0)=1 ; y(x)=2 e^{x}-1 $$

Short Answer

Expert verified
At \(x = 0.5\), the approximations are 1.875 (step size 0.25) and 2.221 (step size 0.1); the exact value is 2.297. Euler's method with a smaller step size (0.1) gives a closer approximation.

Step by step solution

01

Understand Euler's Method Formula

Euler's method is used to approximate the solution of an initial value problem. The formula for Euler's method is given by:\[ y_{n+1} = y_n + h f(x_n, y_n) \]where \(h\) is the step size, \(f(x, y) = y' = y + 1\), \(y_n\) is the current value, and \(x_n\) is the current point.
02

Apply Euler's Method with Step Size 0.25

Given step size \(h = 0.25\), initial condition \(y(0) = 1\).Calculate successive approximations:- **At \(x = 0\):** \(y_0 = 1\)- **At \(x = 0.25\):** \[ y_1 = y_0 + 0.25(y_0 + 1) = 1 + 0.25(1 + 1) = 1.5 \]- **At \(x = 0.5\):** \[ y_2 = y_1 + 0.25(y_1 + 1) = 1.5 + 0.25(1.5 + 1) = 1.875 \]
03

Apply Euler's Method with Step Size 0.1

Given step size \(h = 0.1\), initial condition \(y(0) = 1\).Calculate successive approximations:- **At \(x = 0\):** \(y_0 = 1\)- **At \(x = 0.1\):** \[ y_1 = y_0 + 0.1(y_0 + 1) = 1 + 0.1(1 + 1) = 1.2 \]- **At \(x = 0.2\):** \[ y_2 = y_1 + 0.1(y_1 + 1) = 1.2 + 0.1(1.2 + 1) = 1.42 \]- **At \(x = 0.3\):** \[ y_3 = y_2 + 0.1(y_2 + 1) = 1.42 + 0.1(1.42 + 1) = 1.662 \]- **At \(x = 0.4\):** \[ y_4 = y_3 + 0.1(y_3 + 1) = 1.662 + 0.1(1.662 + 1) = 1.9282 \]- **At \(x = 0.5\):** \[ y_5 = y_4 + 0.1(y_4 + 1) = 1.9282 + 0.1(1.9282 + 1) = 2.22102 \]
04

Calculate Exact Solution at x = 0.5

The exact solution is given by:\[ y(x) = 2e^x - 1 \]Substitute \(x = 0.5\):\[ y\left(0.5\right) = 2e^{0.5} - 1 \approx 2(1.6487) - 1 = 2.2974 \approx 2.297 \text{ (rounded to three decimal places)} \]
05

Compare Approximations with the Exact Solution

The three-decimal-place approximations for \(x = 0.5\) using Euler's method are:- With \(h = 0.25\), \(y_2 = 1.875\)- With \(h = 0.1\), \(y_5 = 2.221\)The exact value is approximately 2.297. The approximation with \(h = 0.1\) is closer to the exact value than with \(h = 0.25\).

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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 (IVP) is a fundamental concept in differential equations. It involves a differential equation along with a specified initial condition. This condition tells us the value of the solution at a specific point. In our given exercise, the IVP consists of a differential equation \( y' = y + 1 \) and an initial condition \( y(0) = 1 \). This means that at \( x = 0 \), the value of \( y \) is 1.

The importance of the IVP lies in its ability to predict how a system will behave over time starting from this initial state. Solving the IVP allows us to understand the behavior of the function \( y(x) \) at any point in its domain. In this exercise, we also have the exact solution \( y(x) = 2e^x - 1 \), which gives us a means to compare the approximations from Euler's method to the actual solution.
Approximation
Approximation in mathematics helps us to find a value that is close to the exact solution, especially when the exact solution is difficult or impossible to determine analytically. Euler's method is a numerical technique used to approximate solutions of differential equations where the function is unknown or complex.

This method estimates the next value \( y_{n+1} \) by computing the slope \( f(x_n, y_n) \) at the current point \( (x_n, y_n) \) and stepping forward using this slope, scaled by a step size \( h \). The recursive formula is:
  • \( y_{n+1} = y_n + h f(x_n, y_n) \)

For example, when using a step size \( h = 0.25 \), starting at \( y(0) = 1 \), and applying the formula twice, we obtain approximations at \( x = 0.25 \) and \( x = 0.5 \). Euler's method provides us with a way to find approximate solutions that become more precise with smaller step sizes.
Step Size
One of the key parameters in Euler's method is the step size \( h \). The step size determines how far we move along the \( x \)-axis to estimate the next value \( y_{n+1} \) in the solution sequence. Smaller step sizes generally lead to more accurate approximations of the solution, though they also increase computational effort as more steps are needed.

In the given exercise, Euler's method was applied with two different step sizes: \( h = 0.25 \) and \( h = 0.1 \).
  • With \( h = 0.25 \), the approximation at \( x = 0.5 \) was \( y_2 = 1.875 \).
  • With \( h = 0.1 \), the approximation at \( x = 0.5 \) improved to \( y_5 = 2.221 \).

These results illustrate that a smaller step size provides an approximation closer to the exact solution \( y(0.5) = 2.297 \). Hence, the choice of step size is crucial for balancing between computational cost and the accuracy of the approximation.

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Most popular questions from this chapter

Use Euler's method with step sizes \(h=0.1,0.02,0.004\), and \(0.0008\) to approximate to four decimal places the values of the solution at ten 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 \(x\). $$ y^{\prime}=x+\sqrt[3]{y}, y(0)=-1 ; 0 \leqq x \leqq 2 $$

Separate variables and use partial fractions to solve the initial value problems in Problems \(1-8 .\) Use either the exact solution or a computer- generated slope field to sketch the graphs of several solutions of the given differential equation, and highlight the indicated particular solution. $$ \frac{d x}{d t}=9-4 x^{2}, x(0)=0 $$

Use the Runge-Kutta method with a computer system to find the desired solution values in Problems 27 and \(28 .\) Start with step size \(h=1\), and then use successively smaller step sizes until successive approximate solution values at \(x=2\) agree rounded off to five decimal places. \(y^{\prime}=x+\frac{1}{2} y^{2}, y(-2)=0 ; y(2)=?\)

In Problems, use a computer system or graphing calculator to plot a slope field and/or enough solution curves to indicate the stability or instability of each critical point of the given differential equation. (Some of these critical points may be semistable in the sense mentioned in Example 6.) $\$$ $$ \frac{d x}{d t}=x\left(x^{2}-4\right) $$

Consider the differential equation \(d x / d t=x+k x^{3}\) containing the parameter \(k .\) Analyze (as in Problem 21) the dependence of the number and nature of the critical points on the value of \(k\), and construct the corresponding bifurcation diagram.

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