Question

Use Euler's method to find a numerical solution of \(\frac{\mathrm{d} y}{\mathrm{~d} x}=-2 y, y(0)=1\), for \(0 \leq x \leq 0.5 .\) First take \(h=0.1\), then \(h=0.05\), and compare your answers at \(x=0.5\) with the exact solution obtained by separating the variables. Work throughout to six decimal places.

Short answer

Using Euler's method, we calculated the approximate value of \(y(0.5)\) for two different step sizes, \(h=0.1\) and \(h=0.05\). For \(h=0.1\), the approximate value of \(y(0.5)\) is 2.488, while for \(h=0.05\), it is 2.362. In comparison with the exact solution \(y(0.5) \approx 0.3679\), neither of the estimated values achieved good accuracy. The solution with \(h=0.05\) was slightly closer to the exact solution than the one with \(h=0.1\), but to obtain better accuracy, it is necessary to either increase the number of steps or use another numerical method.

Step-by-step solution

Review Euler's Method

Euler's method estimates the value of the unknown function as it moves along the independent variable for small intervals. The general formula to apply Euler's method is: \(y_{n+1} = y_n + h*f(x_n, y_n)\), where \(h\) is the step size, \(y_n\) is the current value of \(y\), \(x_n\) is the current value of \(x\), and \(f(x_n, y_n)\) is the function representing the derivative of \(y\) with respect to \(x\). In this problem, \(f(x, y) = -2y\).

Apply the Euler's method for \(h=0.1\)

Let's iterate the Euler's method with \(h = 0.1\), starting from \(x_0=0\) and \(y_0= 1\): 1. For \(n=0\): \(f(x_0, y_0) = -2(1) = -2\). Hence, \(y_1 = y_0 + h * f(x_0, y_0) = 1 -0.1*(-2) = 1+0.2=1.2\) 2. For \(n=1\): \(f(x_1, y_1) = -2(1.2) = -2.4\). Hence, \(y_2 = y_1 + h * f(x_1, y_1) = 1.2 -0.1*(-2.4) = 1.2+0.24=1.44\) 3. For \(n=2\): \(f(x_2, y_2) = -2(1.44) = -2.88\). Hence, \(y_3 = y_2 + h * f(x_2, y_2) = 1.44 -0.1*(-2.88) = 1.44+0.288=1.728\) 4. For \(n=3\): \(f(x_3, y_3) = -2(1.728) = -3.456\). Hence, \(y_4 = y_3 + h * f(x_3, y_3) = 1.728 -0.1*(-3.456) = 1.728+0.345=2.073\) 5. For \(n=4\): \(f(x_4, y_4) = -2(2.073) = -4.146\). Hence, \(y_5 = y_4 + h * f(x_4, y_4) = 2.073 -0.1*(-4.146) = 2.073+0.415 =2.488\) So, we got \(y_5 \approx 2.488\), which corresponds to the value of \(y(0.5)\).

Apply the Euler's method for \(h=0.05\)

Iterating the Euler's method with \(h = 0.05\), increasing the number of steps, and repeating the process: For \(n=0\): \(y_1 \approx 1.100\) For \(n=1\): \(y_2 \approx 1.210\) For \(n=2\): \(y_3 \approx 1.331\) For \(n=3\): \(y_4 \approx 1.464\) For \(n=4\): \(y_5 \approx 1.611\) For the rest of the steps, the results are as follows: For \(n=6\): \(y_7 \approx 1.774\) For \(n=7\): \(y_8 \approx 1.952\) For \(n=8\): \(y_9 \approx 2.147\) For \(n=9\): \(y_{10} \approx 2.362\) So, we got \(y_{10} \approx 2.362\), at the end.

Calculate exact solution and compare results

In order to obtain the exact solution, we need to first understand the differential equation by separating variables: \(\frac{dy}{dx} = -2y\) \(\frac{dy}{y} = -2 dx\) Integrating both sides and using the initial condition \(y(0)=1\), we get \(\ln|y| = -2x + C\) As \(y(0) = 1\), this implies that \(C = 0\). Therefore, the exact solution is \(y(x) = e^{-2x}\) Now, let's calculate \(y(0.5)\) using the exact solution: \(y(0.5) = e^{-2(0.5)} = e^{-1} \approx 0.3679\) Now, let's compare the estimated results for \(h=0.1\) and \(h=0.05\) with the exact solution: \(h=0.1\) -> \(y(0.5) \approx 2.488\) \(h=0.05\) -> \(y(0.5) \approx 2.362\) Exact solution -> \(y(0.5) \approx 0.3679\) From the comparison, we can see that the approximate value for \(y(0.5)\) using Euler's method is not very accurate, and the solution with \(h=0.05\) is closer to the exact solution than \(h=0.1\). It is necessary to either increase the step amount or use another numerical method to achieve better accuracy.

Key concepts

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. They arise in various scientific and engineering fields, modeling how things change. When you encounter a differential equation, it usually means that you are dealing with a system that evolves over time or space.

These equations come in two main types:

• Ordinary Differential Equations (ODEs): Involve functions of a single variable and their derivatives. They express the rate of change of a function with respect to one variable, usually time. An example is the equation \(\frac{dy}{dx} = -2y\).
• Partial Differential Equations (PDEs): Involve functions of multiple variables and their partial derivatives. They are more complex as they describe phenomena changing with multiple variables.

Understanding the solutions to these equations helps predict system behavior and make informed decisions based on these predictions. They could help in fields ranging from physics to economics, wherever dynamic changes need analysis.

Euler's method, used in the exercise, is a way to approach solving ODEs when exact solutions are difficult to find.

Numerical Analysis

Numerical analysis is the study of algorithms that use numerical approximation for solving mathematical problems. It's essential when an exact analytical solution is unavailable, or impractical to compute by hand.

Euler's method is a foundational technique in numerical analysis. It is a simple yet effective way to approximate solutions of ordinary differential equations (ODEs) over an interval. The central concept is to use a series of linear steps to trace the solution curve based on given derivative information. However, the method is not always accurate, especially over large intervals or with larger step sizes.

Why might we choose a numerical approach like Euler's method?

• Complexity: Some equations are too complex for direct solutions.
• Initial Conditions: Help solve initial value problems where you know the initial state but need future values.
• Flexibility: Numerical methods can handle equations that don't have a simple closed-form solution.

As with the exercise, where Euler's method approximates the solution to a differential equation, it often serves as the first step in exploring more sophisticated numerical techniques.

Initial Value Problem

An initial value problem in the context of differential equations refers to finding a function that satisfies a differential equation and also fulfills an initial condition at a particular point. It sees frequent application in physics and engineering where starting conditions influence a system's evolution over time.

Consider the differential equation: \(\frac{dy}{dx} = -2y\) with the initial condition \(y(0) = 1\). Here, solving the initial value problem means finding the function \(y(x)\) that meets both the differential equation and the condition that \(y(0)\) equals 1.

Solving such problems accurately typically involves:

• Identifying the initial condition.
• Using numerical or analytical methods to predict future values, such as Euler's method applied here.
• Comparing numerical solutions to exact solutions (if they exist), to understand accuracy.

The exercise used Euler's method, changing the step size, and compared estimates to the exact solution. This process is common in initial value problems, to refine results and understand the dynamics the model represents.