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## Numerical Methods for Ece

by: Karolann Wiegand

19

0

2

# Numerical Methods for Ece ECE 2331

Karolann Wiegand
UH
GPA 3.51

Staff

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COURSE
PROF.
Staff
TYPE
Class Notes
PAGES
2
WORDS
KARMA
25 ?

## Popular in Electrical Engineering

This 2 page Class Notes was uploaded by Karolann Wiegand on Saturday September 19, 2015. The Class Notes belongs to ECE 2331 at University of Houston taught by Staff in Fall. Since its upload, it has received 19 views. For similar materials see /class/208310/ece-2331-university-of-houston in Electrical Engineering at University of Houston.

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Date Created: 09/19/15
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 25303oon8696 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 29o22903 8597 Using a 3rdorderRK yi1yih6K14K2K3 where K1 fXiyi K2 fXi5hyi5hK1 Page 1 of 2 Using quotthequot 4th0rderRKyi1yih6K12K22K3K4 where K1 fXiyi K2 fXi5hyi5hK1 K3 fXi5hyi5th K4 fXihyihK3 Page 2 of 2

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