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# Class Note for ECE 2331 at UH

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COURSE
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TYPE
Class Notes
PAGES
2
WORDS
KARMA
25 ?

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This 2 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Houston taught by a professor in Fall. Since its upload, it has received 37 views.

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Date Created: 02/06/15
First Order Initial Value Problem Example Given xy39 X24y y12 Solve on interval 1g x g 2 with h25 Solution Solve for y as a function of X and y y X4yX fXy Using Crude Euler yi1 yi K1h where K1 fXiyi First Step K1 f121 4421 7 so y125 y1725 g Second Step K1 f12525 125 425125 45 so y15y1254525 Third Step K1 f15362515 4362515 5333 so y175a15533325 Fourth Step K1 f1754958175 449581756167 so y2 y175616725 Using Heun yi1 yi h2K1K2 where K1 fXiyi K2 fxihyihK1 First Step K1 f12 7 K2 f125225 7 f12525125 425125 45 so y125 y1 252 725 1181 Second Step K1 f1251181 253 K2 f12525118125325f15548803667 so yl 5a1 25252 2530366n8696 Third Step K1 f158696 8189 K2 f175869625 8189f17566492303 so yl 75a1 5252818923037960 Fourth Step K1 f1757960 069461lt2 f2796025 06946f277864427 so y2y1 75252069464427 8427 Using Improved PolygonMidpoint yi1yihK2 where K1 fXiyi K2 fXi5hyi5hK1 First Step K1 f12 7 K2 f15252525 7 f11251125 2875 so y125 y1 25 2875 1281 Second Step K1 f1251281 285 K2 f1255251281 285525f1375925 1316 so yl 5a1 2525 1 3169523 Third Step K1 f159523 1039 K2 f151259523125 1039f16258223 3992 so yl 75a1 525 39928525 Fourth Step1lt1 f1758525 19851lt2 f1751258525125 1985f187582771093 so y2yl752510938798 Using Ralston39s 23 yi1yih3K12K2 where K1 fXiyi K2 fxi75hyi75hK1 First Step K1 f12 7 K2 f1752527525 7 f118756875 1128 so y125 y1 253 72 1128 1229 Second Step K1 f1251229 26821lt2 f12575251229 26827525f14387258 5822 so yl 5a1 2525326822 58229081 Third Step K1 f159081 9217 K2 f157525908175259ZIDf1688735305546 so yl 75a1 525392172 055468221 Fourth Step1lt1 f1758221 12901lt2 f17512582217525 1290f193879792903 so y2y1 75253 1 29022903 8597 Using a 3rdorderRK yi1yih6K14K2K3 where K1 fXiyi K2 fXi5hyi5hK1 K3 fXihyihK12th klfxy 1u2 hk12 k2 h hk12hk2 k3 hk14k2k36 1 2 7 1125 1125 2875 125 23125 615 0972917 125 0972917 186333 1375 074 077773 15 104988 12997 0711503 15 0711503 039734 1625 0661835 000413 175 0808771 009862 0690149 175 0690149 0172517 1875 0711713 0356678 2 0825359 0349283 0771337 2 0771337 Page 1 of 2 Using quotthequot 4th0rderRKyi1yih6K12K22K3K4 where K1 fXiyi K2 fXi5hyi5hK1 K3 fXi5hyi5th K4 fXihyihK3 Page 2 of 2 k1fxy 1M2 hk12 k2 1 2 hk22 k3 h hk3 k4 hk12k22k3k4 1 2 7 1125 1125 2875 11251640625 470833 1250822917 138333 101875 125 101875 201 1375 07675 085773 13750911534127674 150699566 036551 0741899 150741899 04784 16250682099 005401 16250735147018459 175 0695750159713 070873 175070873 0130031 1875 0724990328354187507497810275468 207776030444793 078300 2078300 Recap summary the exact solution is y 116XA416 Xquot2 exact euler heun oly ralston 3rk 4rk 1 2 2 2 2 2 2 2 125 101135 025 118125 128125 1228618 0972917 101875 15 073714 03625 0869583 0952273 0908124 0711503 0741899 175 0705891 0495833 0796012 0852462 0822075 0690149 0708736 2 078125 065 0842668 0879798 0859699 0771337 0783006

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