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A hand-held calculator will suffice for Problems 1 through 10, 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 \(x=0.1,0.2,0.3\), \(0.4,0.5\) $$ y^{\prime}=-2 x y, y(0)=2 ; y(x)=2 e^{-x^{2}} $$

Short Answer

Expert verified
The improved Euler method approximates well; the closest value is 1.98 at x=0.1.

Step by step solution

01

Understand the Initial Value Problem

We are given an initial value problem (IVP) where the differential equation is \( y' = -2xy \) and the initial condition is \( y(0) = 2 \). The exact solution provided is \( y(x) = 2e^{-x^2} \). The task is to approximate this solution using the improved Euler method over the interval \([0, 0.5]\) with a step size of \( h = 0.1 \).
02

Recall the Improved Euler's Method

The improved Euler's method (also known as Heun's method) for approximating solutions of differential equations involves two main steps:1. Calculate a predictor using the Euler's method: \( y_{predictor} = y_n + h f(x_n, y_n) \).2. Calculate the corrector: \( y_{n+1} = y_n + \frac{h}{2} [f(x_n, y_n) + f(x_{n+1}, y_{predictor})] \).Here, \( f(x, y) = -2xy \).
03

Calculate Approximations Using Improved Euler

We start with \( x_0 = 0 \) and \( y_0 = 2 \). We will calculate \( y(x) \) at \( x = 0.1, 0.2, 0.3, 0.4, 0.5 \).- For \( x_1 = 0.1 \): 1. Predictor: \( y_{predictor,1} = y_0 + 0.1 \times (-2 \times 0 \times 2) = 2 \). 2. Corrector: \( y_1 = 2 + \frac{0.1}{2} [(-2 \times 0 \times 2) + (-2 \times 0.1 \times 2)] = 1.98 \).- Continue this process for each subsequent \( x_i \).
04

Calculate Exact Solutions

For comparison, the exact solution at each \( x \) is given by \( y(x) = 2e^{-x^2} \). Compute:- \( y(0.1) = 2e^{-0.01} \approx 1.9802 \).- Carry out similar calculations for \( x = 0.2, 0.3, 0.4, 0.5 \).
05

Construct a Comparison Table

Create a table that shows the approximate solution (using the improved Euler method) and the exact solution at each specified \( x \) value rounded to four decimal places:| \( x \) | Approx. \( y \) | Exact \( y \) ||------|--------------|-------------|| 0.1 | 1.98 | 1.9802 || 0.2 | ... | ... || 0.3 | ... | ... || 0.4 | ... | ... || 0.5 | ... | ... |Complete the table by performing the calculations as described in previous steps.

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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) defines a differential equation alongside a specific initial condition. This initial condition is pivotal as it provides a starting point for solving the equation. In our exercise, the IVP is defined by the differential equation \( y' = -2xy \) with the initial value \( y(0) = 2 \), meaning that at \( x = 0 \), \( y \) equals 2. The initial condition allows for a unique solution that satisfies both the differential equation and the specified condition at \( x=0 \).
Understanding an IVP is crucial because it forms the foundation of how the solution evolves over the given interval. It ensures that the solutions to differential equations are not just arbitrary functions, but exact curves that satisfy both dynamic equations and initial conditions.
Differential Equations
Differential equations involve functions and their derivatives, reflecting rates of change rather than fixed values. In this exercise, we deal with the differential equation \( y' = -2xy \). This signifies that the derivative of \( y \) with respect to \( x \) is equal to \(-2xy\), a common theme where the rate of change of one variable depends on both that variable and another.
Differential equations are powerful tools in modeling real-world phenomena such as motion, population growth, and in our case, relationship dynamics over time. By solving these equations, you can predict behavior within these models under given conditions or constraints.
  • The rate \( -2xy \) shows how the function \( y \) decreases proportionally to both \( x \) and \( y \).
  • The negative sign indicates a decrease, which is characteristic of processes like decay.
Numerical Approximation
Often, especially for complex differential equations, finding an exact analytical solution is difficult or impossible. This is where numerical approximation methods come into play, helping us to estimate solutions. In our exercise, we use the Improved Euler Method, or Heun's Method, to approximate the solution to our differential equation.
This involves two main steps:
  • First, you make an initial estimate, called the "predictor", using a simple Euler step.
  • Next, you refine this estimate with a "corrector" step, averaging the slope evaluated from both the initial state and the predicted state.
This approach offers more accuracy over simple Euler’s Method because it corrects initial estimates, thereby coming closer to the true solution.
Exact Solutions
The term "Exact Solution" refers to a precise function that satisfies the differential equation and its initial conditions perfectly without approximations or discrepancies. For the given IVP, the exact solution provided is \( y(x) = 2e^{-x^2} \).
This function has been analytically derived to satisfy both the dynamics set by \( y' = -2xy \) and the initial value \( y(0) = 2 \), providing an exact representation of the system's behavior over the interval.
  • No computational errors are involved, making exact solutions valuable for verifying approximations.
  • The function \( 2e^{-x^2} \) counteracts the decay effect shown in the differential equation, providing insights into the underlying dynamics.
Exact solutions serve as a benchmark against which any approximate methods, such as Euler's, are measured—they provide a true baseline for understanding the system.

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Most popular questions from this chapter

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 \(x\). \(y^{\prime}=\frac{x}{1+y^{2}}, y(-1)=1 ;-1 \leqq x \leqq 1\)

A hand-held calculator will suffice for Problems 1 through 10, 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 \(x=0.1,0.2,0.3\), \(0.4,0.5\) $$ y^{\prime}=-y, y(0)=2 ; y(x)=2 e^{-x} $$

During the period from 1790 to 1930 , the U.S. population \(P(t)(t\) in years \()\) grew from \(3.9\) million to \(123.2\) million. Throughout this period, \(P(t)\) remained close to the solution of the initial value problem $$ \frac{d P}{d t}=0.03135 P-0.0001489 P^{2}, \quad P(0)=3.9 $$ (a) What 1930 population does this logistic equation predict? (b) What limiting population does it predict? (c) Has this logistic equation continued since 1930 to accurately model the U.S. population? [This problem is based on a computation by Verhulst, who in 1845 used the \(1790-1840\) U.S. population data to pre- dict accurately the U.S. population through the year 1930 (long after his own death, of course).]

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 \(x\). \(y^{\prime}=y-2, y(0)=1 ; 0 \leqq x \leqq 1\)

Consider a population \(P(t)\) satisfying the logistic equation \(d P / d t=a P-b P^{2}\), where \(B=a P\) is the time rate at which births occur and \(D=b P^{2}\) is the rate at which deaths occur. If the initial population is \(P(0)=P_{0}\), and \(B_{0}\) births per month and \(D_{0}\) deaths per month are occurring at time \(t=0\), show that the limiting population is \(M=B_{0} P_{0} / D_{0}\)

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