/*! 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 7 An initial value problem and its... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 three-decimal-place values of the two approximations at \(x=\frac{1}{2}\) with the value \(y\left(\frac{1}{2}\right)\) of the actual solution. \(y^{\prime}=-3 x^{2} y, y(0)=3 ; y(x)=3 e^{-x^{3}}\)

Short Answer

Expert verified
For \(x=0.5\), Euler's method gives 2.859 with \(h=0.25\), 2.860 with \(h=0.1\), both close to the actual 2.861.

Step by step solution

01

Understand the Problem Requirements

We need to approximate the solution of the differential equation \(y' = -3x^2y\) on the interval \([0, \frac{1}{2}]\) using Euler's method with two different step sizes: \(h = 0.25\) and \(h = 0.1\). We will then compare these approximations to the actual solution value at \(x=\frac{1}{2}\), which is given to be \(y(x) = 3e^{-x^3}\).
02

Euler's Method Overview

Euler’s method is used to approximate solutions of ordinary differential equations. From the equation \(y' = f(x, y)\), we use the formula \(y_{n+1} = y_n + hf(x_n, y_n)\) to find the next value, where \(h\) is the step size, \(y_n\) is the current approximation, and \(x_n\) is the current x value.
03

Apply Euler's Method with Step Size h=0.25

Start at \(x_0 = 0\) with \(y(0) = 3\). Calculate \(y_1\):- \(x_1 = 0 + 0.25 = 0.25\) - \(y_1 = 3 + 0.25(-3 \cdot 0^2 \cdot 3) = 3\).Calculate \(y_2\):- \(x_2 = 0.25 + 0.25 = 0.5\) - \(y_2 = 3 + 0.25(-3 \cdot 0.25^2 \cdot 3) = 3 - 0.140625 = 2.859\).
04

Apply Euler's Method with Step Size h=0.1

Start again at \(x_0 = 0\) with \(y(0) = 3\).Calculate incrementally:1. \(x_1 = 0.1\), \(y_1 = 3 + 0.1(-3 \cdot 0^2 \cdot 3) = 3\)2. \(x_2 = 0.2\), \(y_2 = 3 + 0.1(-3 \cdot 0.1^2 \cdot 3) = 3 - 0.009\)3. \(x_3 = 0.3\), \(y_3 = 2.991 + 0.1(-3 \cdot 0.2^2 \cdot 2.991) = 2.973\)4. \(x_4 = 0.4\), \(y_4 = 2.973 + 0.1(-3 \cdot 0.3^2 \cdot 2.973) = 2.937\)5. \(x_5 = 0.5\), \(y_5 = 2.937 + 0.1(-3 \cdot 0.4^2 \cdot 2.937) = 2.860\).
05

Calculate Exact Solution at x=0.5

The exact solution given is \(y(x) = 3e^{-x^3}\). Calculate \(y(0.5) = 3e^{-(0.5)^3} = 3e^{-0.125} \approx 2.861\).
06

Compare Results and Conclude

The approximation with \(h=0.25\) at \(x=0.5\) is \(2.859\), and with \(h=0.1\) it is \(2.860\). Both are close to the exact solution \(2.861\). The smaller step size \(h=0.1\) gives a more accurate approximation.

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 91Ó°ÊÓ!

Key Concepts

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

Understanding Initial Value Problems
An initial value problem in mathematics involves solving a differential equation along with specific conditions, usually provided at a single point. These conditions help in determining the unique solution to the problem. In our exercise, the initial value problem is presented with the differential equation \( y' = -3x^2y \) and the initial condition \( y(0) = 3 \). This means that at \( x = 0 \), the value of \( y \) is 3. The goal is to find the value of \( y \) for other points, such as \( x = \frac{1}{2} \), which is a common task in calculus and numerical analysis.
Understanding initial value problems is crucial because they provide a starting point for solving differential equations. This establishes the initial state of a system and guides us in predicting its evolution over time. In many real-world scenarios, such as physics and engineering, identifying the initial values is vital for accurate modeling.
Step Size Approximation with Euler's Method
When solving differential equations using numerical methods like Euler's Method, the step size \( h \) is critically important. It determines how far forward we jump at each iteration. In the given problem, we apply Euler's method twice—with different step sizes, \( h = 0.25 \) and \( h = 0.1 \).
Step size affects the accuracy of the approximation. A smaller step size, such as 0.1, allows more points to be calculated, leading to a smoother and more precise approximation of the solution curve. Conversely, a larger step size like 0.25 may result in a solution that deviates more from the exact curve.
  • With \( h = 0.25 \), fewer points are calculated, leading to an approximation of \( y(0.5) = 2.859 \).
  • With \( h = 0.1 \), a more detailed approximation is achieved, resulting in \( y(0.5) = 2.860 \).
Euler's method, though simple, can guide us toward understanding the behavior of solutions to differential equations using appropriate step sizes.
Exploring Ordinary Differential Equations
Ordinary differential equations (ODEs) are equations involving functions of one independent variable and their derivatives. An ODE describes the relationship between these derivatives and defines how a quantity changes over time or space. In the context of our problem, the ODE given is \( y' = -3x^2y \), which describes how \( y \) changes with respect to \( x \).
ODEs appear in various fields such as physics, biology, and economics, where they are used to model continuous systems. Understanding these equations allows us to predict future behavior based on initial conditions. When exact solutions to ODEs are challenging to find, numerical methods, like Euler’s method, become invaluable tools. They provide approximations that help in understanding possible behavior of complex systems, offering insights even when only partial information is available.

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

A hand-held calculator will suffice for, where an initial value problem and its exact solution are given. Apply the improved Euler method to approximate this solution on the interval \([0,0.5]\) with step size \(h=0.1 .\) Construct a table showing four-decimal-place values of the approximate solution and actual solution at the points \(x=0.1,0.2,0.3,0.4\), \(0.5\). $$ y^{\prime}=y-x-1, y(0)=1 ; y(x)=2+x-e^{x} $$

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 three-decimal-place values of the two approximations at \(x=\frac{1}{2}\) with the value \(y\left(\frac{1}{2}\right)\) of the actual solution. \(y^{\prime}=x-y, y(0)=1 ; y(x)=2 e^{-x}+x-1\)

A programmable calculator or a computer will be useful for Problems 11 through 16. In each problem find the exact solution of the given initial value problem. Then apply the RungeKutta method twice to approximate (to five decimal places) this solution on the given interval, first with step size \(h=0.2\), then with step size \(h=0.1 .\) 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 \(x\). $$ y^{2} y^{\prime}=2 x^{5}, y(2)=3 ; 2 \leqq x \leqq 3 $$

If \(4 h>k M^{2}\), show that \(x(t)=0\) after a finite period of time, so the lake is fished out (whatever the initial population). [Suggestion: Complete the square to rewrite the differential equation in the form \(d x / d t=-k\left[(x-a)^{2}+b^{2}\right]\). Then solve explicitly by separation of variables.] The results of this and the previous problem (together with Example 4) show that \(h=\frac{1}{4} k M^{2}\) is a critical harvesting rate for a logistic population. At any lesser harvesting rate the population approaches a limiting population \(N\) that is less than \(M\) (why?), whereas at any greater harvesting rate the population reaches extinction.

Use the improved Euler method with a computer system to find the desired solution values .Start with step size \(h=0.1\), and then use successively smaller step sizes until successive approximate solution values at \(x=2\) agree rounded off to four decimal places. $$ y^{\prime}=x^{2}+y^{2}-1, y(0)=0 ; y(2)=? $$

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.