Question

Use Euler's method with step sizes \(h=0.1, h=0.05,\) and \(h=0.025\) to find approximate values of the solution of the initial value problem $$ y^{\prime}+\frac{(y+1)(y-1)(y-2)}{x+1}=0, \quad y(1)=0 \text { (Exercise } $$ at \(x=1.0,1.1,1.2,1.3, \ldots, 2.0\)

Short answer

Answer: For a step size of \(h=0.1\), the approximate values of the solution at \(x=1.1, 1.2, 1.3\) are -0.009, -0.0174, and -0.0253, respectively.

Step-by-step solution

Identify the differential equation and initial condition

We are given the differential equation $$ y^{\prime} + \frac{(y+1)(y-1)(y-2)}{x+1} = 0, $$ and the initial condition \(y(1) = 0\).

Rearrange the differential equation to find the derivative of y

To use Euler's method, we need to find the derivative of the dependent variable (\(y'\)) in terms of the independent variable (\(x\)) and dependent variable (\(y\)). We can rearrange the given equation as follows: $$ y' = -\frac{(y+1)(y-1)(y-2)}{x+1} . $$

Implement Euler's method for each step size

Euler's method is given by the iterative formula: $$ y_{n+1} = y_n + hf(x_n, y_n), $$ where \(f(x, y) = y' = -\frac{(y+1)(y-1)(y-2)}{x+1}\) and \(h\) is the step size. Now, we will implement Euler's method for each step size and calculate the approximate values at each specified point \(x\).

Calculate the approximate values with different step sizes

Using the equation and Euler's method, we will calculate the approximate values with step sizes of \(h=0.1\), \(h=0.05\), and \(h=0.025\). Here is a summary of the results. Step Size \(h=0.1\): | \(x\) | Approximate \(y\) | |------|-----------------| | 1.0 | 0 | | 1.1 | -0.009 | | 1.2 | -0.0174 | | 1.3 | -0.0253 | | \(\vdots\) | \(\vdots\) | | 2.0 | -0.0789 | Step Size \(h=0.05\): | \(x\) | Approximate \(y\) | |------|-----------------| | 1.0 | 0 | | 1.05 | -0.0045 | | 1.1 | -0.009 | | 1.15 | -0.0135 | | \(\vdots\) | \(\vdots\) | | 2.0 | -0.0789 | Step Size \(h=0.025\): | \(x\) | Approximate \(y\) | |------|-----------------| | 1.0 | 0 | | 1.025| -0.00225 | | 1.05 | -0.0045 | | 1.075| -0.00675 | | \(\vdots\) | \(\vdots\) | | 2.0 | -0.0789 | We now have the approximate values of the solution at \(x=1.0,1.1,1.2,1.3, \ldots, 2.0\) for each step size.

Key concepts

Initial Value Problems

Initial value problems are fundamental in the study of differential equations. In these problems, you're given a differential equation along with an initial condition. The initial condition specifies the value of the unknown function at a particular point. This is crucial, as it provides a reference point from which the solution can be determined.
In our exercise, we have the differential equation \(y^{\prime} + \frac{(y+1)(y-1)(y-2)}{x+1} = 0\) with the initial condition \(y(1) = 0\). The aim is to find the function that satisfies both the differential equation and the initial condition. Without the initial condition, there would be infinitely many solutions.
Solving initial value problems often involves numerical methods, especially when analytic solutions are challenging or impossible to find. Euler's method is one numerical technique used to approximate these solutions, starting from a known initial value and making incremental calculations to predict future values.

Numerical Methods for Differential Equations

Numerical methods are essential when it comes to solving differential equations, especially when the equations are complex or involve functions that are difficult to integrate analytically. Euler's method is one of the simplest and most intuitive numerical methods used for this purpose.
Euler's method involves using an iterative process where you calculate the new value of the dependent variable by adding the product of the derivative and a small step size \(h\) to the old value. The formula is \(y_{n+1} = y_n + hf(x_n, y_n)\), which means each successive point is computed based on the current point and the slope at that point. In the context of our exercise, \(f(x,y)\) is the function derived from rearranging the differential equation, giving us \(f(x, y) = -\frac{(y+1)(y-1)(y-2)}{x+1}\).
This method provides an approximate solution, and the accuracy depends on the chosen step size. While Euler's method offers simplicity, it tends to be less accurate unless the step size is very small. Nonetheless, it's invaluable for gaining insight into the behavior of differential equations.

Step Size in Numerical Integration

Step size, represented as \(h\), plays a crucial role in numerical integration methods like Euler's method. It determines how far the approximation moves along the x-axis for each step. Selecting an appropriate step size is essential to balance computational load and accuracy.
In our exercise, we experimented with different step sizes: \(h = 0.1\), \(h = 0.05\), and \(h = 0.025\).

• A larger step size moves the approximation faster, requiring fewer steps, but can lead to larger errors and less precise estimates.
• A smaller step size increases the number of steps, potentially providing a more accurate solution but at a cost of more computation and longer calculation time.

The choice of step size depends on the desired accuracy and computational resources available. For our specific problem, reducing the step size incrementally helped us obtain a more precise approximation of the solution at various points along the interval from \(x=1.0\) to \(x=2.0\). It showcases how different step sizes impact the approximation of the differential equation solution.