1. Euler’s Method BC Only Copyright © Cengage Learning. All rights reserved. 6.1 6.1 Day 2 2014

2. 2 6.1 day 2 Euler’s Method Leonhard Euler 1707 - 1783 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. 3 Leonhard Euler 1707 - 1783 It was Euler who originated the following notations: (base of natural log) (function notation) (pi) (summation) (finite change)

4. 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:

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6. 6

7. 7 1+0(.5) = 1 1+1(.5) = 1.5 1.5+2(.5) = 2.5 2.5+3(.5) = 4 4+4(.5) = 6

8. 8 Exact Solution:

9. 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. 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. 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. 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. 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 ) = 0.9 + (0.1 – 0.9)(0.1) = 0.82 y 3 = y 2 + hF(x 2, y 2 ) = 0.82 + (0.2 – 0.82)(0.1) = 0.758.

14. 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. 15 Homework  6.1 Day 2 (Euler’s Lesson) : Pg. 410: 69,71,73 and Slope Fields WS

16. 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. 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. 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. 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. 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. 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. 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. 23 This gives us a table of the points that we found in our first example. Table Press