Sect9.3 - SECTIO 4 Q 3 2‘ C/eqr'lfl poliso met(ea We...

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Unformatted text preview: SECTIO 4/ Q. 3 2‘}- « C/eqr'lfl. poliso met/(ea We @ d9 ' Y+743xg rag/709 Hema/ Spa/Q, 0'9 QOt04 51??!on '64+XL3(9~ eaaa'fivn zero, Hence [0, 0) 6/6” f5 01 QFI’HCc-g/ FOI/I’ é. ghow Q/Masé /x/;éa/‘; Con$§c>f€l glow) \ I.‘ 2 PC059— rsw] (Q ' I a Q ‘ M > I rec M9 ~31” : r‘ n-w r‘ M“; 49 O \n/y, r~->o Cease/w ga 2) r X03? 9‘ m 1“ I . "‘“S‘G‘r‘w‘e ( L ‘ r I : I." ‘ 2. I ‘ Co 9- ~ ' Z : n-ao Pflo r‘ f‘ ‘5 L324 9) C5 H—enca, %e_ syfilem ,5 q/moéé [mean 17er (0/0) "T112. Calrisponolir'zg llheqr Exfng mof/‘rly 4: ( 1/ A BfgWWqIUfibb arr-=91 95¢ 4") 1.. r 1. I-., ,A‘: (4-)) +°I = I’LQAJ‘A +7: Aflgys =0 _“~2‘<2W" ' . /\' 'T l-Iiao q5amlo7’zfil‘fmcv apjr‘q) 14049, 6-) 32; = (9”) (fix) 0% - @ DeJermnyQ, %£ Carri“!qu penafb Lam CW): 0 anal 01*) [fiI-x) ; e Kt“? 0f g‘zx qnol xabl 0r yl‘X cram» poms 642,50 (L/JL/LzaoB ® FWD) \H‘Q Conacpcmollhg [thwf‘ Syslgm Ineor‘ inficw/ P01716- F-c CQHUQWA: 362 +3“ ~—’¢9~><~><z G: Uhflfgn): Lly';x +”/x~xq Ff g-wax I??? guy- G}: ~0¢4+°I-6l>< 03:4»; (,0 z _ e . '1‘) Q ‘ I ,9, ( a a» / L7 q‘xgrmm-» 3 LI 1,, ~ I I = —3 "I Maw/V5? >L~aA ~/6 >0 QiJVfiL/Cl) X a 6 ; 1:677 Pcifié Ll-‘x o C 7% ~ Q 6 - 9,, . / 06 Ora liq/149° 6A+x1+ )1 g ‘ ‘ 8’ ‘A X: 26.3” 65:76“ 2. _g: HA I poflaé VVVVV Section 9.3 Problem 5 part d with(DEtools): ' eqnl:=diff(x(t),t)=(2+x)*(y—x);eqn2;=diff(y(t),t)=(4-x)*(y+x) I a V ean .=a—tx(t) = (2 +x) (y -x) a eqn2==5;y(t)=(4-x)(y+x) ica:=[ol-2I I [OI-1! I [011,0]! [0'2] I [0'3]: icb:=[0,3,3],[0,5,5]: icc:=[oI-1I1] I [OI-3’3] I [DION-1]: initsé=ica,icb,icc: ' . . DEplot([eqn1,eqn2],[x,y],t=-10..10,{inits},arrows=thin,x=-3.. 5,y=-1..5,stepsize=0.02); ////,.>r.>,.u NMNN‘HNHNN \-‘ \\ \\\\_‘\‘ (S \ \\‘m\\'\@~\\\m A ‘ ,\N \\\\\\A\\ "Piaf-H f—A—An-AHNN ‘-.. NNN \\ \K‘N \\\\MHHHHHW \ \\\N\N\N\H-a~u \Nhr—a-(f! u~\‘\‘\ 1 ‘u\&\o\ hwx\\\'\ 1 f / ////‘/ Remember we found that: (0,0) is an unstable saddle point (4,4) is an asymptotically stable spiral point (—2,2) is an unstable improper node /O’2én.2/ X Cb—x—EL); 0 =3 X30} or [~xy;0 Cri’Hc’q‘ POIW‘E QR (OJ J C I] J Qnoi g9‘ "X 025441 3~¥~Qa3 O 3‘x-6H8kx : tzo :5 x;~/J a; 0/, 25 > -)g 3 [’2X'g “"X 6‘? = ‘3 6 > %~x~°/0cz hear (C’s/9.) “Va 0 QWUm/aeb Iii/53-x O 3635.))637O20 -3/8 _ /9 as”; ' >2? MVa Wmfim‘mdéf 3510/2, Mo prgpor‘ moo/Q hear CA0) ('I q} gypnuoyltaes "4 “I \: C—I—D (94): O 0 3L 0 9"} M O Li~>36a~A¥© 0 2r} /\ 3 I B IWFTOp—Ql‘ I’lC) o/Q ' POM/7 0‘ D 0“ pqu 9‘3 ,0 6 (L) :1’5 :F find“ "1:0 4:)‘111’75‘1‘4zo ‘7 Y;‘)’\ 4+ ,1 XI. 1. o ' _ i1 : x31 =6 ‘1 ‘ exotpk. M (51) M (—51) 4, a; DP _ _ wquuM 2* I?“ )1 (f 1‘ \ '7 J1) 131‘ (MW a [356 ‘GISL. )‘ ' vl a‘ 7— \A‘fithwx; —‘( (’1‘TY) *7. t r +2f fl 7— 0 as M fe‘h'uall shbl - .. — ‘ _L."___.J___L 1“ W1tq+ 8/ : -‘tk -7 531114 Rofw'f 9. . ‘W/D) “WI 0 -I - ~ 1 hfjmvaflucs ~Y‘ (-I-Y\ —Z = Y ‘9 7-Y “I ‘3 O Y‘: ~1t\‘4‘f§ :-lt\f3‘ __7 smug {M— 2. LUNCH“; L943 ~ ’0' S" Section 9.3 Problem 6 part d > with(DEtools): > eqn1:=diff(x(t),t)=x-x*2-x*y;eqn2:=diff(y(t),t)=3*y-x*y-2*y“2 I a e 1:=- t =x- 2—x cm, atx() x y a «2an:=5;y(t)=3y-~xy-2y2 ica==[0,3/2,-1],[0,3/2,1],[0,3/2,2],[0,3/2,3]: iCb:=[oI_3/2l—1] I [OI—3/2I1] I [OI-3/2I2] l [OI-3/2l: icc:=[°I—1I—1]l[OI-1/2I-11Iloll/zl—1]I[ol1l-;1]: iCd:=[oI-1I I [OI-l/zl I [oil/2’ I [olllz ice:=[0,-1/2,1/2],[0,-1/2,1],[0,1/2,1/2]: inits:=ica,icb,icc,icd,ice: > DEplot([eqn1,eqn2],[x,y],t=-5..5,{inits},arrows=thin;x=—3/2.. 3/2,y=-1..3,stepsize=0.02); VVVVVV Remember we found that: (0,0) is an unstable improper node . (0,3/2) is an asymptotically stable improper node (1,0) is an unstable saddle point (—1,2) is an unstable saddle point C? 2-[1 6r > > VVVVVVV Section 9.3 Problem 7 part d with(DEtools): eqn1:=diff(x(t),t)=1-y;eqn2:=diff(y(t),t)=x“2—y“2; a 1:- =1- eqn atx(t) y em2F§fl0=fio9 a: [0,3/2,0],[0,3/2,1],[0,3/2,2]: 10b: [OI—3/2I I [OI—3/2I1] I [OI-3/2I2]: ice: [OI-1’0]! [olololl [0’1]: iod:=[oI-1l I [OIOI I loll]: ice:=[0,-1/2,2],[0,-1,0.9],[0,-1,1.1]: inits:=ica,icb,icc,icd,ice: ica: DEplot([eqn1,eqn21,[x,y],t=—5..5,{inits},arrows=thin,x=-3/2.. 3/2,y=0..2,stepsize=0.02); [HHWWHW/ l/lsl [Ill 1!!!! 1/ / ////ll 111 it!!! 11// xlllll [ll lllllllllx —/ llfilt 11 I!!! / mm f”? Remember we found that: > (1,1) is an asymptotically stable spiral point (-1,1) is an unstable saddle point 161.3 $7 7 3-: = >< (913) Save 6x730»;ka x; or make ymv‘mw be °’::/ avg C Mm.‘ Paws (0)03/ Co) I) J (.2’_23J (5W3) I @1890 001/1496 ' d «1-) 3 .>‘) I *7 Z'I'ABC’AB "'9' - X+A€¢Q g CXKQCA-l) >0 >2 l'~a V VVVVVV Section 9.3 Problem 9 part d~ with(DEtools): v eqnl :=diff (x (t) , t)=- (x-y) * (l-x-y) ;eqn2 :=diff (y (t) , t)=x* (2+y); a €an:=§x(t)=-(x-y)(1-x-y) a ean :=§y(t)=x(2+y) ica==[0,1.11,[0,2,2].[0,3,3]: icb:=[0,-2,2],[0,—3,3]: icc:=[0,-1,-1],[0,-3,—3],[0,—1,—3]: icd:=[0,1,-1]r[0,1,-3],[0,1,-2]}[0,3,-3]: inits:=ica,icb,icc,icd: DEplot([eqn1,eqn2],[x,y],t=-10..10,{inits},arrows=thin,x=-5.. 5,y=-5..5,stepsize=0.025); ' ' Remember we found that: (0,0) is an unstable saddle point {-2,-2) is an asymptotically stable improper node (0,1) is an asymptotically stable spiral point {3,-2) is an unstable improper node IQ irat)’; Section 9.3, Problem 25 Problem. In this problem we show how small changes in the coefficients of a system of linear equations can affect a critical point that is a center. Consider 0 1 the system a?” = :5. Show that the eigenvalues are ii so that (0,0) is a ' —1 0 l 6 1 —’ I I 0 center. Now cons1der the system a?” = x, Where |e| 1s arbltrarlly small. , —1 6 Show that the eigenvalues are e :i: 2‘. Thus no matter how small {6| 7:4 0 is, the cneter becomes a spiral point. If 6 < 0, the spiral point is asymptotically stable; if e > 0, the spiral point is unstable. \ 0 1 Solution, For the first matrix, , one finds the eigenvalues by —10 ' —A 1 subtracting A from the diagonal elements — giving one A , and then -1 __ taking the determinant. This gives one the expression A2 + 1, Which, to equal zero, requires that A = 2' or A = —2'. Therefore the equilibrium at (0,0) must be a center. 6 1 For the second matrix, ( —1 e ), one again subtracts A from the diagonal, 6 — A . giving the matrix A . The determinant of this matrix is (6— A)2 +1 :2 __1 E._ A2 — 26A + (1 + 62). Using the quadratic formula to find the eigenvalues for which the determinant equals zero, gives one A = e+i and A = e — i. This will be a spiral; it will spiral in if 6 < 0 (and so be asymptotically stable) and spiral out (and so be unstable) if e > 0. These are drastic changes from the behavior of a center. /23r.10 ...
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This homework help was uploaded on 04/09/2008 for the course MATH 2400 taught by Professor Yoon during the Spring '04 term at Rensselaer Polytechnic Institute.

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Sect9.3 - SECTIO 4 Q 3 2‘ C/eqr'lfl poliso met(ea We...

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