Question

Compute tables for the Euler Method and Modified Euler Method by hand, for the IVP x' = t - x, x(0) = 1. To make these a reasonable length, you are going to find values of x(1), instead of x(2) (as in the Examples). The exact solution is x(t) = t - 1 + 2e-?, which gives x(1) = 0.735759. For the IVP x' = t - x, x(0) = 1, do the following. (Round your answers to six decimal places.) (a) Do 4 steps of the Modified Euler Method, with At = 0.25, to find an approximation to x(1). x(1) = Do 8 steps of the Modified Euler Method, with At = 0.125, to find an approximation to x(1). x(1) = (b) Complete the following table (values are from Example 2.6.2). At x(1) error in (1) 0.5 0.781250 0.045491 0.25 Enter a number. 0.125 (c) Are the errors in x(1) what you would expect from a second-order method? Explain. This is what you should expect with a second-order method. The errors are multiplied by approximately 174 4 as At is multiplied by

Answer 1

Co) 4 the given t-n IVP is Euler Muttod 4(t, )= t-x ht(igti) At = 0.25 t Ki 0 25 075 1 O 25 O625 0125 St O O-59375 0-5 0 0312$ O625 O632812S O039062 S O75 O59375 0 72460935 00917968 75 0-6328125 ( correet to 6 diuimal plaus) xCi) O 724609 Modified Euler Method For mula K2f(ti whure h&(tini) K = K= Ki+ k2 2 Heu h=At 0.125 Ist iterationi- K2=-0.09375 K 0125 to =0 010.93 F5 x Co125)0890625

2nd t 0.125 0-890625 KIE -0-04527283 0-011962 K2 007858276 KO-25) O-8453 52 1 8 11 3Td 0-25 20843 5218 00744 1902 K = 0 04949164, K=- 0 06195533 K2 = x (0-35)= 0 8 33 9685 17 4th t3 0 375 0-783396 & 17 k= - 0 04 004651 00 2904 3 4 KI 0 05 104961 К> C0-5) O7433 50 34

5th 4 O7433 50 34 0 030418 79 11 =001 099144 K -0.020705115 x(o625) O722645225 0.625 0 226422 2 KT 0 012205G5 K, = O004945OS k=-oo036303 Co O71901493 th =0750 O719014 93 K 0003873 13 k2 001901399 9 K 0.011 4437 Co-75) O-7304584 0.73045849 k,=0.03143423 k =0.0247r096 0-01806769 0-75520945 (corTeet to 6 deeimal laes) x Ci) O755209