Question

Use the improved Euler method to find approximate values of the solution of the given initial value problem at the points \(x_{i}=x_{0}+i h,\) where \(x_{0}\) is the point where the initial condition is imposed and \(i=1,2,3\). $$ y^{\prime}+3 y=x^{2}-3 x y+y^{2}, \quad y(0)=2 ; \quad h=0.05 $$

Short answer

Question: Using the Improved Euler Method, find the approximate values of the solution to the initial value problem \(y^{\prime} + 3y = x^{2} - 3xy + y^{2}\), with \(y(0)=2\) (initial value) and \(h=0.05\) (step size), at the points \(x_1=0.05\), \(x_2=0.1\), and \(x_3=0.15\). Answer: After applying the Improved Euler Method, we get the following approximate values: \(y(0.05) \approx 1.9625\), \(y(0.1) \approx 1.93153125\), and \(y(0.15) \approx 1.91205594\).

Step-by-step solution

Define the function of interest

Let's define a function for the given differential equation: $$ F(x,y) = y^{\prime} + 3y - (x^{2} - 3xy + y^{2}) = 0 $$ We want to find \(y^{\prime}\) as a function of \(x\) and \(y\): $$ y^{\prime} = x^{2} - 3xy + y^{2} - 3y $$

Calculate the predictor

Using the Improved Euler Method, at each step, we calculate the predictor using the Euler Method: $$ y^{p}_{i+1} = y_i + hF(x_i, y_i) $$ where \(y^{p}_{i+1}\) is the predicted value at the next iteration and \(h\) is the step size.

Calculate the corrector

After calculating the predictor, we calculate the corrector: $$ y_{i+1} = y_i + \frac{h}{2}(F(x_i, y_i)+F(x_{i+1}, y^{p}_{i+1})) $$ Then, we iterate through steps 2 and 3 for the given number of iterations.

Initialize values and iterate

We are given that \(y(0)=2\) (initial value) and \(h=0.05\) (step size). We want to find the solution at \(x_1\), \(x_2\) and \(x_3\), which are equal to \(0.05\), \(0.1\), and \(0.15\), respectively. We will now iterate using the Improved Euler Method: Iteration 1 (\(x_1 = 0.05\)): 1. Predictor: \(y^{p}_{1} = y_0 + hF(x_0, y_0) = 2 + 0.05 (0^2 - 3(0)(2) + 2^2 - 3(2)) = 1.95\) 2. Corrector: \(y_1 = y_0 + \frac{h}{2}(F(x_0, y_0)+F(x_1, y^{p}_{1})) = 2 + \frac{0.05}{2}(F(0, 2) + F(0.05, 1.95)) = 1.9625\) Iteration 2 (\(x_2 = 0.1\)): 1. Predictor: \(y^{p}_{2} = y_1 + hF(x_1, y_1) = 1.9625 + 0.05 (0.0025 - 3(0.05)(1.9625) + 1.9625^2 - 3(1.9625)) = 1.925625\) 2. Corrector: \(y_2 = y_1 + \frac{h}{2}(F(x_1, y_1)+F(x_2, y^{p}_{2})) = 1.9625 + \frac{0.05}{2}(F(0.05, 1.9625) + F(0.1, 1.925625)) = 1.93153125\) Iteration 3 (\(x_3 = 0.15\)): 1. Predictor: \(y^{p}_{3} = y_2 + hF(x_2, y_2) = 1.93153125 + 0.05 (0.01 - 3(0.1)(1.93153125) + 1.93153125^2 - 3(1.93153125)) = 1.91123750\) 2. Corrector: \(y_3 = y_2 + \frac{h}{2}(F(x_2, y_2)+F(x_3, y^{p}_{3})) = 1.93153125 + \frac{0.05}{2}(F(0.1, 1.93153125) + F(0.15, 1.91123750)) = 1.91205594\) The Improved Euler method gives us the approximate values at the desired points (\(x_1\), \(x_2\), and \(x_3\)) as follows: $$ y(0.05) \approx 1.9625 \\ y(0.1) \approx 1.93153125 \\ y(0.15) \approx 1.91205594 $$

Key concepts

Understanding Initial Value Problems

An initial value problem (IVP) in mathematics is a type of problem where we know the value of a function at a starting point, and we want to find out how this function behaves for other points based on a differential equation. In simple terms, it's like knowing a vehicle's position at the beginning of a journey, and then trying to predict its future positions using the speed and direction at every instant – the speed and direction being analogous to the derivative in our case.

For the problem at hand, the IVP consists of a differential equation, which is an equation that relates a function with its derivatives, and an initial condition given by y(0) = 2. We're asked to find the value of the function y at subsequent points x_i using the improved Euler method, which takes these differential equations from a theoretical abstract and turns them into concrete values we can use for prediction and analysis.

Navigating Through Differential Equations

Differential equations are powerful mathematical tools that describe the relationship between a function and its derivatives. They are the language through which we express the dynamics of various physical, biological, and economic systems. When we write a differential equation such as y' + 3y = x² - 3xy + y², we are capturing the way the rate of change of our function y (denoted by y') reacts to changes in x, as well as the function's current value.

In our specific exercise, we're dealing with a non-linear differential equation. Non-linear equations are known for their complexity and the intricacy of their solutions, which often cannot be found exactly. Instead, numerical methods like the improved Euler method come into play to give us a practical approximation of the solution.

Numerical Approximation: Bridging Theory and Reality

Numerical approximation is a mathematical strategy used to find approximate solutions to problems that may not have exact answers, such as complex differential equations. Think of it as using a measuring tape that only has inches when we need to measure something precisely to the nearest millimeter. We make educated guesses and use special techniques to get as close as possible to the exact measurement.

The improved Euler method, also known as Heun's method, is one such numerical approximation technique. It refines the basic Euler method by adding an extra step that adjusts the slope used to predict the next value. This method significantly improves accuracy without the need for complicated calculations, balancing efficiency and precision, making it especially valuable in numerous scientific and engineering applications.

Step Size: The Measure of Our Stride in Calculation

The step size in numerical methods is analogous to the stride length in walking; it determines how far we 'step' through our calculations from one point to the next. In the context of the improved Euler method, it represents the interval between successive values of x. A small step size means that we take more steps to cover the same range, which generally leads to a more accurate approximation of our function's behavior, although at the cost of more calculations.

In our exercise, we're given a step size h = 0.05. Small step sizes, like the one used here, ensure that we closely follow the curve we're trying to approximate, allowing us to trace its path more accurately than with larger steps. Selecting an appropriate step size is crucial as it affects the balance between computational workload and the desired accuracy of the results.