Question

Estimate the following solutions using Euler's method with \(n=5\) steps over the interval \(t=[0,1] .\) If you are able to solve the initialvalue problem exactly, compare your solution with the exact solution. If you are unable to solve the initial-value problem, the exact solution will be provided for you to compare with Euler's method. How accurate is Euler's method? $$ y^{\prime}=2^{x}, y(0)=0, \text { Exact solution is } y=\frac{2^{x}-1}{\ln (2)} $$

Short answer

Euler's method yields a value of 1.7449 at \( t = 1 \), with 20.93% error from the exact value of 1.4427.

Step-by-step solution

Identify the Differential Equation and Initial Condition

We are given the differential equation \( y' = 2^x \) with the initial condition \( y(0) = 0 \). Our goal is to approximate the solution using Euler's method with \( n = 5 \) steps over the interval \( t = [0, 1] \).

Determine Step Size

The interval \([0, 1]\) is divided into 5 equal steps. The step size \( h \) is calculated by dividing the interval length by the number of steps: \[ h = \frac{1 - 0}{5} = 0.2 \]

Apply Euler's Method Iteratively

Euler's Method updates the solution using the formula: \[ f(x,y) = 2^{x} \; \text{and} \; y_{n+1} = y_n + h \cdot f(x_n, y_n) \]- **Step 0**: \( x_0 = 0 \), \( y_0 = 0 \)- **Step 1**: \( y_1 = y_0 + 0.2 \cdot 2^{x_0} = 0 + 0.2 \cdot 1 = 0.2 \)- **Step 2**: \( x_1 = 0.2 \), \( y_2 = y_1 + 0.2 \cdot 2^{x_1} = 0.2 + 0.2 \cdot 1.1487 = 0.4297 \)- **Step 3**: \( x_2 = 0.4 \), \( y_3 = y_2 + 0.2 \cdot 2^{x_2} = 0.4297 + 0.2 \cdot 1.3195 = 0.6936 \)- **Step 4**: \( x_3 = 0.6 \), \( y_4 = y_3 + 0.2 \cdot 2^{x_3} = 0.6936 + 0.2 \cdot 1.5157 = 0.9967 \)- **Step 5**: \( x_4 = 0.8 \), \( y_5 = y_4 + 0.2 \cdot 2^{x_4} = 0.9967 + 0.2 \cdot 1.7411 = 1.3449 \)- **Step 6**: \( x_5 = 1.0 \), \( y_6 = y_5 + 0.2 \cdot 2^{x_5} = 1.3449 + 0.2 \cdot 2 = 1.7449 \)

Calculate the Exact Solution at t=1

The exact solution is given by \( y = \frac{2^x - 1}{\ln(2)} \). Calculate it at \( x = 1 \): \[ y = \frac{2^1 - 1}{\ln(2)} = \frac{1}{0.6931} \approx 1.4427 \]

Evaluate Accuracy of Euler's Method

Euler's method approximation at \( x = 1 \) is \( y_6 = 1.7449 \). The exact solution's value is approximately \( 1.4427 \). The error can be calculated as:\[ \text{Error} = |1.7449 - 1.4427| = 0.3022 \]. The relative error proportion is approximately \( \frac{0.3022}{1.4427} \approx 0.2093 \), or 20.93% error.

Key concepts

Differential Equations

Differential equations are equations that involve a function and its derivatives. In our exercise, the differential equation is given as \( y' = 2^x \). Essentially, it tells us how the function \( y \) changes with respect to \( x \). The goal of solving a differential equation is to find a function \( y(x) \) that satisfies this relationship for all values in a given range.

Understanding the behavior of differential equations is important, as they are widely used to model natural phenomena in fields such as physics, biology, and economics. For instance, they can describe how a population grows over time or how an object moves under the influence of forces.

Numerical Approximation

Numerical approximation is a method used to find approximate solutions for mathematical problems that cannot be solved exactly. Euler's method, applied in this exercise, is a straightforward numerical technique for approximating solutions to first-order differential equations.

It works by predicting the future value of a function based on its current value and slope. The slope is determined by the differential equation itself. Euler's method uses a series of small steps (in our case, 5 steps) to incrementally construct an approximate solution over a certain interval, such as \( [0, 1] \).

Although it can introduce errors, this approach makes it possible to solve complex differential equations that defy analytical solutions. It's a foundation for more sophisticated methods and serves as an entry point into numerical analysis.

Initial Value Problem

An initial value problem in the context of differential equations involves finding a function whose derivative equals a given function and is also consistent with a specified initial condition. For our problem, we are looking for a function \( y(x) \) where its derivative \( y' \) equals \( 2^x \) and must satisfy the initial condition \( y(0) = 0 \).

This initial condition is critical because it helps to uniquely define the function within the mathematical space of potential solutions. Without it, the problem could have infinitely many solutions.

The task in Euler's method is to use this initial condition combined with incremental calculation to predict the future values of the function, ensuring they align as closely as possible with the underlying differential equation.

Exact Solution Comparison

Comparing the numerical and exact solutions provides insights into the precision of numerical methods. The exact solution for the provided differential equation \( y' = 2^x \) is \( y = \frac{2^x - 1}{\ln(2)} \). This formula returns the function's exact value at any given point, such as \( x = 1 \).

Using Euler's method, we approximated the value at \( x = 1 \) to be 1.7449, while the exact solution delivers a more precise result of 1.4427.

Evaluating the difference or error between these two provides a measure of accuracy for Euler's method. Here, the error is found to be 0.3022, signifying a 20.93% difference.

• Large errors, as seen here, highlight the limitations of basic numerical methods.
• Nevertheless, numerical methods remain crucial for scenarios where exact solutions aren't feasible.

The comparison underscores why understanding both numerical and analytical methods is vital for effectively tackling real-world problems.