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? $$ y^{\prime}=y, y(0)=-1 . \text { Exact solution is } y=-e^{x} . $$

Short answer

Euler's method gives -2.48832, compared to the exact -2.71828, a small error of 0.22996.

Step-by-step solution

Understand Euler's Method

Euler's method approximates the solution of a differential equation by taking small steps along the curve, using the derivative to find the slope at each step. We have the equation \( y' = y \), and we'll estimate the value of \( y \) over the interval \( [0,1] \) with \( n=5 \) steps. The initial condition is \( y(0) = -1 \), and the exact solution is \( y = -e^t \).

Calculate Step Size

To use Euler's method, first calculate the step size \( h \). Since the interval is \([0, 1]\) and we have \( n = 5 \) steps, the step size is \[ h = \frac{1-0}{5} = 0.2. \]

Apply Euler's Method Iteratively

Start with \( t_0 = 0 \) and \( y_0 = -1 \). For each step, use Euler's formula: \( y_{n+1} = y_n + h f(t_n, y_n) \). Given \( f(t, y) = y \), apply this to each of the next steps. Calculate \( y_1, y_2, y_3, y_4, \) and \( y_5 \).

First Iteration (n=0 to n=1)

Calculate \( y_1 \): \[ y_1 = y_0 + h y_0 = -1 + 0.2(-1) = -1 - 0.2 = -1.2 \].

Second Iteration (n=1 to n=2)

Calculate \( y_2 \): \[ y_2 = y_1 + h y_1 = -1.2 + 0.2(-1.2) = -1.2 - 0.24 = -1.44 \].

Third Iteration (n=2 to n=3)

Calculate \( y_3 \): \[ y_3 = y_2 + h y_2 = -1.44 + 0.2(-1.44) = -1.44 - 0.288 = -1.728 \].

Fourth Iteration (n=3 to n=4)

Calculate \( y_4 \): \[ y_4 = y_3 + h y_3 = -1.728 + 0.2(-1.728) = -1.728 - 0.3456 = -2.0736 \].

Fifth Iteration (n=4 to n=5)

Calculate \( y_5 \): \[ y_5 = y_4 + h y_4 = -2.0736 + 0.2(-2.0736) = -2.0736 - 0.41472 = -2.48832 \].

Compute Exact Solution for Comparison

The exact solution at \( t = 1 \) is \( y = -e^1 = -e \approx -2.71828 \).

Compare Euler's Method Result with Exact Solution

Euler's method gives \( y_5 \approx -2.48832 \) at \( t = 1 \), while the exact solution is \( -2.71828 \). The approximation differs by \( -2.71828 - (-2.48832) = -0.22996 \), indicating a small error with Euler's method.

Key concepts

differential equations

Differential equations are mathematical equations that describe how a certain quantity changes over time. They involve derivatives, which represent rates of change. In this context, we're dealing with a simple differential equation:

• \( y' = y \) - This equation states that the rate of change of \( y \) with respect to \( t \) is equal to \( y \) itself.
• It is a first-order linear differential equation.

Understanding these equations is crucial because they model real-world phenomena such as the rate of growth of populations, physical motion, or chemical reactions. In our exercise, solving the differential equation gives us the behavior of \( y \) over the interval \( [0, 1] \).
The exact solution provided, \( y = -e^t \), helps validate that Euler's Method gives a reasonably close approximation.

numerical approximation

Numerical approximation is a way to find an approximate solution to a mathematical problem when an exact answer is difficult or impossible to obtain. Euler's Method, which is our method of choice here, is a straightforward algorithm for numerical approximation. Here’s what Euler's Method does:

• Start at the initial point provided by an initial value, such as \( y(0) = -1 \).
• Then repeat calculations iteratively over each small interval (or step), in this case using step size \( h = 0.2 \).
• At every step, adjust the approximation using the derivative to "predict" the direction of the curve.

By following this method over our interval \( t=[0,1] \), we achieve successive approximations \( y_1, y_2, \ldots, y_5 \). The beauty of numerical approximation is that it provides a practical way to tackle complex problems without exploring labor-intensive analytical methods.

initial value problem

An initial value problem is a type of differential equation along with an initial condition that specifies the value of the unknown function at a starting point. For our exercise:

• The differential equation is \( y' = y \).
• The initial value is \( y(0) = -1 \).

This initial value helps us to determine a specific solution trajectory from many others. Such problems are common in real-life applications where conditions at the start of an observation period are known.
Additionally, the initial value anchors our numerical approximation using Euler's Method and serves as the starting point for estimating future values.