Problem 60 - Calculus Single Variable

Question

An initial value problem is given, along with its exact solution. (Read the instructions for Exercises \(47-50\) for terminology.) Verify that the given solution is correct by substituting it into the given differential equation and the initial value condition. Calculate the Euler's Method approximation \(y_{1}=y_{0}+F\left(x_{0}, y_{0}\right) \Delta x\) of \(y\left(x_{1}\right)\) where \(\Delta x=x_{1}-x_{0} .\) Let \(m_{1}=\left(F\left(x_{0}, y_{0}\right)+F\left(x_{1}, y_{1}\right)\right) / 2\) and \(z_{1}=y_{0}+\) \(m_{1} \Delta x .\) This is the Improved Euler Method approximation of \(y\left(x_{1}\right) .\) Calculate \(z_{1} .\) By evaluating \(y\left(x_{1}\right),\) determine which of the two approximations, \(y_{1}\) or \(z_{1},\) is more accurate. $$ \begin{array}{l} d y / d x=x-y, y(-2)=-1, x_{1}=-2.15 ; \text { Exact solution: } \\ y(x)=2 \exp (-2-x)+x-1 \end{array} $$

Step-by-Step Solution

Answer

The Improved Euler Method approximation, \( z_1 = -0.8275 \), is more accurate than the Euler's Method approximation, \( y_1 = -0.85 \).

Step 1: Verify the Exact Solution

First, substitute the exact solution \( y(x) = 2e^{-2-x} + x - 1 \) into the differential equation and the initial condition. Compute \( \frac{dy}{dx} \) of the given solution, which is \( \frac{d}{dx}(2e^{-2-x} + x - 1) = -2e^{-2-x} + 1 \). Substitute \( x = -2 \) and \( y = -1 \) into the differential equation \( \frac{dy}{dx} = x - y \) to check consistency: \( -2e^{-2-(-2)} + 1 = -1 - (-1) = 0 \). The given solution satisfies the differential equation at the initial condition, confirming it's correct.

Step 2: Calculate Euler's Method Approximation

Using Euler's Method, find the approximation \( y_1 \) at \( x_1 = -2.15 \). The formula is \( y_1 = y_0 + F(x_0, y_0) \Delta x \), where \( F(x, y) = x - y \). Given \( \Delta x = x_1 - x_0 = -2.15 + 2 = -0.15 \). \( F(-2, -1) = -2 - (-1) = -1 \), so \( y_1 = -1 + (-1)(-0.15) = -1 + 0.15 = -0.85 \).

Step 3: Calculate Improved Euler Method Approximation

For the Improved Euler Method, use the formula \( m_1 = \frac{F(x_0, y_0) + F(x_1, y_1)}{2} \). First, calculate \( F(x_1, y_1) = -2.15 - (-0.85) = -1.3 \). Then, \( m_1 = \frac{-1 + (-1.3)}{2} = \frac{-2.3}{2} = -1.15 \). Using \( z_1 = y_0 + m_1 \Delta x \), we get \( z_1 = -1 + (-1.15)(-0.15) = -1 + 0.1725 = -0.8275 \).

Step 4: Compare Approximations with Exact Value

Evaluate the exact solution at \( x_1 = -2.15 \): \( y(-2.15) = 2e^{-2-(-2.15)} + (-2.15) - 1 \). Calculating gives \( 2e^{-0.15} - 3.15 \). Approximating \( e^{-0.15} \approx 0.86071 \), the result is \( y(-2.15) \approx 1.72142 - 3.15 = -1.42858 \). Compare \( y_1 = -0.85 \) and \( z_1 = -0.8275 \) with the exact value. \( z_1 \approx -0.8275 \) is closer to \(-1.42858\) than \( y_1 \).

Key Concepts

Euler's Method

Euler's Method is a straightforward numerical technique for solving ordinary differential equations (ODEs) with a given initial value. It is especially useful when an exact solution is difficult to obtain. The core idea of this method is to use the slope at a known point to find the value of the solution after a small step, effectively creating a tangent line approximation.
To apply Euler's Method, we follow these steps:

• Identify the initial point \(x_0, y_0\).
• Calculate the slope \(F(x_0, y_0)\) from the differential equation \(dy/dx = F(x, y)\).
• Determine the step size \(\Delta x\).
• Approximate the next value as \(y_1 = y_0 + F(x_0, y_0) \Delta x\).

This method offers a simple way to approximate solutions, although its accuracy depends on the step size. A smaller step size typically yields more precise results, but requires more computations.

Improved Euler Method

The Improved Euler Method, also known as the Heun's Method, refines the basic Euler's Method by incorporating an additional step to enhance accuracy. It is considered a more precise method for approximating solutions to differential equations.
Here's how it works:

• Start with the standard Euler's step to find a preliminary slope at \(x_0\): \(F(x_0, y_0)\).
• Calculate an improved slope by averaging the slope at \(x_0\) and the slope at the estimated next point \(x_1 \): \( F(x_1, y_1)\).
• Compute the mean slope as \(m_1 = \frac{F(x_0, y_0) + F(x_1, y_1)}{2}\).
• Use this average slope in the updated formula: \(z_1 = y_0 + m_1 \Delta x\).

The Improved Euler Method takes an intermediate step which leads to greater precision than the basic Euler's Method, and is more reliable for problems requiring higher accuracy.

Initial Value Problem

An Initial Value Problem (IVP) is a type of differential equation problem where the solution is required to satisfy a specific condition at the initial point. This initial condition is what sets the starting point for solutions of the differential equation.
For IVPs, you are usually given:

• A differential equation in the form \(dy/dx = F(x, y)\).
• An initial condition expressed as \(y(x_0) = y_0\).

The goal is to find a function \(y(x)\) that satisfies both the differential equation throughout its domain and the initial condition at \(x_0\).
IVPs are commonly encountered in real-world applications such as physics and engineering, where the outcome of a system's behavior is predicted from its initial state. Solving an IVP often requires numerical methods like Euler's Method or Improved Euler Method when an analytical solution is challenging to derive.

Exact Solution Verification

Exact Solution Verification involves confirming that a proposed solution to a differential equation truly satisfies both the equation and any initial conditions. This process is crucial for validating solutions before using them in practical applications.
Here's the process for verifying an exact solution:

• Substitute the proposed solution into the differential equation.
• Differentiate the solution to find its derivative, \(\frac{dy}{dx}\).
• Check that the computed derivative and equation match the given differential equation.
• Additionally, ensure that the solution satisfies the initial conditions provided, \(y(x_0) = y_0\).

Verifying an exact solution is essential to confirm its validity. This ensures any predictions or insights drawn from the solution are rooted in accurate mathematical analysis, making it a foundational step in solving differential equations effectively.