Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

Question

Use Euler's Method with \(h=0.2\) to approximate the solution over the indicated interval. $$ y^{\prime}=-2 x y, y(1)=2,[1,2] $$

Step-by-Step Solution

Answer

At \( x = 2 \), \( y \approx 0.02786 \).

Step 1: Identify the differential equation and initial condition

We are working with the differential equation \( y' = -2xy \) and the initial condition \( y(1) = 2 \). The task is to approximate values of \( y \) over the interval \([1, 2]\) using Euler's Method with a step size \( h = 0.2 \).

Step 2: Set up Euler's Method formula

Euler's Method formula is given by \( y_{n+1} = y_n + h \, f(x_n, y_n) \). Here, \( f(x, y) = -2xy \), and \( h = 0.2 \). We will use the iterative process \( y_{n+1} = y_n - 0.2 \cdot 2x_ny_n \).

Step 3: Calculate the first iteration

Start with \( x_0 = 1 \) and \( y_0 = 2 \). Calculate \( y_1 \): \[ y_1 = y_0 - 0.2 \times 2 \times x_0 \times y_0 = 2 - 0.2 \times 2 \times 1 \times 2 = 2 - 0.8 = 1.2 .\]

Step 4: Calculate the second iteration

Set \( x_1 = 1.2 \) and \( y_1 = 1.2 \). Calculate \( y_2 \): \[ y_2 = y_1 - 0.2 \times 2 \times x_1 \times y_1 = 1.2 - 0.2 \times 2 \times 1.2 \times 1.2 \approx 1.2 - 0.576 = 0.624\].

Step 5: Calculate the third iteration

Set \( x_2 = 1.4 \) and \( y_2 = 0.624 \). Calculate \( y_3 \): \[ y_3 = y_2 - 0.2 \times 2 \times x_2 \times y_2 = 0.624 - 0.2 \times 2 \times 1.4 \times 0.624 \approx 0.624 - 0.34944 = 0.27456\].

Step 6: Calculate the fourth iteration

Set \( x_3 = 1.6 \) and \( y_3 = 0.27456 \). Calculate \( y_4 \): \[ y_4 = y_3 - 0.2 \times 2 \times x_3 \times y_3 = 0.27456 - 0.2 \times 2 \times 1.6 \times 0.27456 \approx 0.27456 - 0.1750656 = 0.0994944\].

Step 7: Calculate the fifth iteration

Set \( x_4 = 1.8 \) and \( y_4 = 0.0994944 \). Calculate \( y_5 \): \[ y_5 = y_4 - 0.2 \times 2 \times x_4 \times y_4 = 0.0994944 - 0.2 \times 2 \times 1.8 \times 0.0994944 \approx 0.0994944 - 0.071634528 = 0.027859872\].

Key Concepts

Differential Equations

Differential equations are mathematical equations that describe how a particular quantity changes over time. They are essential in modeling various real-world phenomena in engineering, physics, biology, and other fields. In this article, we are focusing on ordinary differential equations (ODEs), which involve functions of a single variable and their derivatives.
• Our equation here is a first-order differential equation because it involves the first derivative of the function. • The equation is structured as: \( y' = -2xy \), where \( y' \) denotes the derivative of \( y \) with respect to \( x \). Understanding how to solve these equations is crucial, as they can model everything from the growth rate of a population to the motion of a pendulum. Sometimes, we don't have a simple formula for the solution, which brings us to numerical methods like Euler's Method.

Numerical Methods

Numerical methods provide a way to solve complex mathematical problems using approximations. They are particularly useful for differential equations that cannot be solved analytically (with a simple formula). Euler's Method, one of the simplest numerical methods, provides an iterative approach to approximate solutions. It's especially useful in scenarios where the initial value is known, but an explicit solution is not available.
The basic idea is to use a small step size \( h \) to incrementally build an approximate solution across an interval. For our problem, \( h = 0.2 \) is chosen. Each step follows the formula:- \( y_{n+1} = y_n + h \cdot f(x_n, y_n) \), where \( f(x, y) \) represents the function derived from our differential equation. Using this process repeatedly provides increasingly refined approximations to our initial problem.

Initial Value Problems

Initial value problems (IVPs) specify the value of the unknown function at a given point, often simplifying the process of finding an approximate solution. Our initial condition here is \( y(1) = 2 \), which serves as the starting point for applying Euler's Method.
Here's how it works:- Starting with the initial condition, we use Euler's iterative formula to calculate successive values.- Each step estimates the function's next value based on its current value and the derivative's behavior over a small interval.By relying on the known starting point, initial value problems help us trace the trajectory of the solution through a range, providing insights into the behavior of dynamic systems. Thus, initial value problems are foundational in fields like physics and engineering, where starting conditions dictate future behavior.