|Module 20 - Differential Equations and Euler's Method|
|Introduction | Lesson 1 | Lesson 2 | Lesson 3 | Self-Test|
|Lesson 20.2: Euler's Method|
In this lesson Euler's method is used to approximate a numerical solution to an initial-value problem. The method is based on linear approximations and uses a variation of the point-slope form of a linear equation: y1 = y0 + m(x1 x0).
Getting Linear Approximations
Suppose we want to solve a differential equation of the form dy / dx = m(x, y). In general, this means that the slope of the solution y = f(x) at any point (x, y) may depend on both x and y. We are going to approximate the values of the solution at a sequence of x-values given by x1, x2, x3, ..., etc. Given an initial value y0 = f(x0), the value of y1 = f(x1) can be estimated by the linear approximation.
f(x1) f(x0) + m(x0, y0) · (x1 x0)
Then the value of y2 = f(x2) can be approximated from the value of y1 = f(x1).
f(x2) f(x1) + m(x1, y1) · (x2 x1)
After approximating f(x2) the same method can be used to approximate f(x3).
f(x3) f(x2) + m(x2, y2) · (x3 x2)
Euler's method states that f(xn+1) can be approximated by using previously found values.
f(xn+1) f(xn) + m(xn, yn) · (xn+1 xn)
Using Euler's Method
In Lesson 20.1 you solved the logistic initial-value problem
y' = 0.037 * y * (25 y) and y(1) = 1
Use Euler's method to approximate the values of the solution to this differential equation. Let x0 = 1, y0 = 1, and x1 = 1.1. We compute m(x0, y0) = m(1, 1) using the differential equation with y replaced by y0 = 1.
m(x0, y0) = m(1, 1) = 0.037 * 1 * (25-1)
Note that x1 - x0 = 0.1, so
y1 = f(x1) = f(1.1)
f(x0) + m(x0, y0) · (x1 - x0)
Let x2 = 1.3, x3 = 1.2, etc. Next we approximate y2 = f(1.2) using the approximate value found for y1 as the value of m(x1, y1).
y(1.2) 1.0888 + 0.037 · 1.0888 · (25 1.0888) · 0.1
Then continue the process to find y3, y4, etc.
Using a Short Cut
There is a short cut that will speed up the process for subsequent approximations. The result of the most recent calculation is stored in a variable called "ans(1)". This can be entered into the Edit Line by pressing . is above . This approach will also improve the accuracy somewhat, since it uses all 14 places, rather than the 6 we entered by hand.
Because the most recent result represents y(1.2), the approximation for y(1.3) can be found by evaluating
y(1.3) ans(1) + 0.037 · ans(1) · (25 ans(1))· 0.1
Because ans(1) has now been updated to 1.28956, you can simply press
20.2.1 Approximate y(2) by pressing
five more times.
Using Euler's Method in Differential Equation Graphing Mode
Euler's method can be used in Differential Equation graphing mode.
The differential equation y' = 0.037.y.(25 y) with y(1) = 1 should be there.
A small "1" should appear in the upper right corner of the screen.
You should see the same values for y that you calculated with Euler's method.
The same value found for y(2) using Euler's method in 20.2.1 is shown as the y-coordinate.
Displaying a Table of Values
You can see the Euler method values in a table.
The values shown in the y1 column are found by using Euler's method, which should give you a better understanding of how the TI-89 Differential Equation graphing mode approximates the graph of the solution to an initial-value problem. The calculator computes approximate solution values for very small t-value iterations and displays short line segments that approximate the curve of the solution.
|< Back | Next >|
2007 All rights reserved. |