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exam3_sol

# exam3_sol - t ab-——" | 3X 3 ‘2 X5“" x6“ 3...

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Unformatted text preview: t ab -——-"' .. | __ 3X 3 ‘2 X5“ "' x6“ 3 F a” a; a; 4, ﬂ 1" A- ’x ,J’ ‘ 5L7. j y l. {‘20 pm) U539 a generalized Newt()ll—Ri-lphsml method 1.0 solve the ibllowing: y2+2ry=5 ‘F' 1': 314' 21‘;- r 1‘2 — y t 3 4:“- x244 Use i111 initial guess of .rr—-y-—1. Perform ONE iteration. “I 32:. Z q aizqzx 3?: z-f J” (2 “l {I = -€f#l-7) -‘/Y F" 1-- warm-w) -‘t LN“): —3.Lle 2- l2l} 1H5) Tilt? following expression has 1109.11 proposed for the HPI'Ul'lt'l ('lr.‘.1‘i\'21tivn 01' 21 function. _ :J‘Lr — :sh.) - “fu- + 2h) + Lam- + h} + 2113-) Fm hi, 1 ONE-'2] Find the values of (1, b that. make this uxpression correct. Use your cxpl'msicm m ﬁnd the Si‘(i()Il(l'TlCl‘lvalTlVE' of the following at x—.1 llﬁillg [17032: = (I'J‘p(—.1t} — 21-2 E me a u q“ i<X+3k)- SL5) +35'tx)k 4' 2 cu“) In" 4- 2‘2 chxnﬂ + 2“ £00“ .- " ‘- I a mu 'Rr-H-‘d's Vii) 4' 1V“) la 1- 1- r (10% 4- .E;" (“L + git; can.“ I 1 I'\ Run: an ammu- H‘UM t 7}: mm 5:33;"qu LOC— w~w¥ _|+a+b+2 =0 ﬁrj- rd all £cx) l‘I (21*«31-‘5‘ ° 7%“ "‘4 '4‘ Pout: TL‘B U ¢“°"0\'~ 3" C0519)! Q+b+l=° 1a. 1-9-3 1° “2‘1-25—239 2K '1‘!) “'3 =9 -5 -5'=-~‘D 8D Cl;ch Fu‘ ken-x.) -27+32-5' = [-L;+q(.g)—s(%ﬂ=l 6 ° chm-u {u *3 h"+ 3 “7' [(ll.) 7- Q¥ﬂ(""’)’1cl'5)z s-‘UNW 1' 5- 1. I/ Hm) =ewt—w)-w-~0*= “1‘7” ’33" " (:(l.2.) = ewtai-n-um)» :. £21732 1;“) 1 OWL-4) -1. 544.32.: In «v -(-‘t.ﬁazh)+ “IL-3,6739) -51—25’76’3’) 4101.432.) :W (9.1)" Fug-£6713” cf > 0.53;! ‘/ the!" 3. (‘20 pts) Consider the following numerical values for a function, ﬁx): Estimate the integral: [:0 rte) dx (a) Using the single segment trapezoid rule. (1.)) Using the two segment trapezoid rule. (6) Using the 1/3 Simpson rule. C“) I «=— 53; [4309+ 3W3 2;. if z o 4- 0.353 [Ire-5*) J“)? 2 4. loelﬁf) ’7 (fl; Marisa?) _-’.-“ Lint.)de '3: 0’ J m R 2- ": I A. .70 +033 =£§Yo+ 9.7194- BT: 1§"[.F(nt-‘IJUT)+£ (C7 )1 3 [Elk (33):} 4. (20 pts) USe Heunis method to solve for the following set of ODE’s. %=—2y+4cxp(“tl Z dz yz2 a??? =a(:7,2-) Initial conditions: y(0):2, z(0)=-’1. Assume At=0.1. Solve for the FIRST time step. He... :1) F.'u+ EXPLIcIT ‘EuzLE-g %'e:2°+\$(‘9°to)A+-:2+[—"l+q30_| :2 - (q 1 :7 ‘-l+l 1—39 Jo! 2 {—— AMT-£1 D. '= 29o * i Huey) + “9.903” 6 2t :2 2,, + g; [3(3°J%a)-r§l‘9. 33.6)] M ‘3 .. 21.4 { 0+[-4+‘lexf7€O-l )}o.[ =- .‘?8 m )2 + QM”); 2 =4“??? -—3—-§°-' {tic .-: .29 -§ gluing-3) Q-é : 5. (‘20 pm) Given ﬂ 2 {1982‘ — 1998:; - art" (1’? I I W ‘- I d_y (H _ moo;- —- 20001; , 2 LXI: ) with HO) —1 and MU) — 2. {a} U50 tlii‘ implicit Euler method to obtain a solution using; a. step size of U.|. Solve for the FIRST time step. {bl Use explivil. Eulm‘ and obtain a solution using a similar time Hi'op for flu) FIRST time step. [5: your answer stable? What would be a reasonath time step to take for stability"? I’thcdr xi = X, + -F (xv-3.34% 3| =. 30 + ﬁtx.p,3¢t X: = Xe +(‘173x, 47783.) o.| =9 73,3y'-'?7-33' 3 _’ 9. =30 +(1000x’-1ooogl)o.l =9 Manx, —Zo|_0§’ ._—_ -1 ' (18.3 )t, “\$3.37. = H 3' = 4.8057237 W-S’x, - rims; 7-4974» x E -2 Hummus”) l MIOO "'_' ' 1.212;, .—.- — 0.771 ‘ ——i______\ x = x0 + (Wan—mung!) I :1! + (mt!) -—1‘t1?(n.))(o.n :: ‘3 -- 9:. +Q°“°"~ ‘ 3°” New I 1:: 1 + (,1000 -' wom)[0.\) 7 '293-3’ (-293- )‘\Q I4-—---------' S" L I° I‘ I‘ N 3 IC. CO wanfL ’ U, o y .. ) ‘(coﬂﬂEcT\oN“ +0 (X6539) 3}!th be" 8” | At, :: 0.00M or JMIW! ...
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exam3_sol - t ab-——" | 3X 3 ‘2 X5“" x6“ 3...

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