Euler’s Method BC Only Copyright © Cengage Learning. All rights reserved Day

2 6.1 day 2 Euler’s Method Leonhard Euler Leonhard Euler made a huge number of contributions to mathematics, almost half after he was totally blind. (When this portrait was made he had already lost most of the sight in his right eye.)

3 Leonhard Euler It was Euler who originated the following notations: (base of natural log) (function notation) (pi) (summation) (finite change)

4 There are many differential equations that can not be solved. We can still find an approximate solution. We will practice with an easy one that can be solved. Initial value:

5

6

7 1+0(.5) = 1 1+1(.5) = (.5) = (.5) = 4 4+4(.5) = 6

8 Exact Solution:

9 It is more accurate if a smaller value is used for dx. This is called Euler’s Method. It gets less accurate as you move away from the initial value.

10 Euler’s Method – BC Only Euler’s Method is a numerical approach to approximating the particular solution of the differential equation y' = F(x, y) that passes through the point (x 0, y 0 ). From the given information, you know that the graph of the solution passes through the point (x 0, y 0 ) and has a slope of F(x 0, y 0 ) at this point. This gives you a “starting point” for approximating the solution.

11 Euler’s Method From this starting point, you can proceed in the direction indicated by the slope. Using a small step h, move along the tangent line until you arrive at the point (x 1, y 1 ) where x 1 = x 0 + h and y 1 = y 0 + hF(x 0, y 0 ) as shown in Figure 6.6. Figure 6.6

12 Euler’s Method If you think of (x 1, y 1 ) as a new starting point, you can repeat the process to obtain a second point (x 2, y 2 ). The values of x i and y i are as follows.

13 Example 6 – Approximating a Solution Using Euler’s Method Use Euler’s Method to approximate the particular solution of the differential equation y' = x – y passing through the point (0, 1). Use a step of h = 0.1. Solution: Using h = 0.1, x 0 = 0, y 0 = 1, and F(x, y) = x – y, you have x 0 = 0, x 1 = 0.1, x 2 = 0.2, x 3 = 0.3,…, and y 1 = y 0 + hF(x 0, y 0 ) = 1 + (0 – 1)(0.1) = 0.9 y 2 = y 1 + hF(x 1, y 1 ) = (0.1 – 0.9)(0.1) = 0.82 y 3 = y 2 + hF(x 2, y 2 ) = (0.2 – 0.82)(0.1) =

14 Example 6 – Solution Figure 6.7 The first ten approximations are shown in the table. cont’d You can plot these values to see a graph of the approximate solution, as shown in Figure 6.7.

15 Homework  6.1 Day 2 (Euler’s Lesson) : Pg. 410: 69,71,73 and Slope Fields WS

16 The TI-89 has Euler’s Method built in. Example: We will do the slopefield first: 6: DIFF EQUATIONS Graph….. Y= We use: y1 for y t for x MODE

17 6: DIFF EQUATIONS Graph….. WINDOW t0=0 tmax=150 tstep=.2 tplot=0 xmin=0 xmax=300 xscl=10 ymin=0 ymax=150 yscl=10 ncurves=0 diftol=.001 fldres=14 not critical GRAPH We use: y1 for y t for x Y= MODE

18 t0=0 tmax=150 tstep=.2 tplot=0 xmin=0 xmax=300 xscl=10 ymin=0 ymax=150 yscl=10 ncurves=0 diftol=.001 fldres=14 WINDOW GRAPH

19 While the calculator is still displaying the graph: yi1=10 tstep =.2 If tstep is larger the graph is faster. If tstep is smaller the graph is more accurate. I Press and change Solution Method to EULER. WINDOW GRAPH Y=

20 To plot another curve with a different initial value: Either move the curser or enter the initial conditions when prompted.F8 You can also investigate the curve by using. F3 Trace

21 t0=0 tmax=10 tstep=.5 tplot=0 xmin=0 xmax=10 xscl=1 ymin=0 ymax=5 yscl=1 ncurves=0 Estep=1 fldres=14 GRAPH Now let’s use the calculator to reproduce our first graph: We use: y1 for y t for x I Change Fields to FLDOFF. WINDOW Y=

22 Use to confirm that the points are the same as the ones we found by hand. F3 Trace Table Press TblSet Pressand set:

23 This gives us a table of the points that we found in our first example. Table Press