Question

Problem 1. Each of the following linear systems has one eigenvalue and one line of eigenvectors. For each system, (a) find the eigenvalue; (b) find an eigenvector; (c) sketch the direction field; (d) sketch the phase portrait, including the solution curve with initial condition Yo = (1,0); and (e) find the general solution;

Answer 1

0 :13 y = A4 A-141 GM-a-62 +9 = x= 30-34 +9. = (x-3) (x-3). G (2) = (2-3) 2 7= 3,3 Eigen value al 3,3.. .(A .1)4 = 20

lool . -% +X2 = 0. X = X x = -X2 (A-3) y = V, lo – a, t. = 1. Eigen vectors tors au v. =( 1), -1) are general soln. (ti) 4 (+) = (vi+ (Vt+1e na 904) = gebe [!] + 60 e?([!] ++81). Y(t) = Set + Ctest.

4it) $ (64(+) 23+ [(a+calt +1)) e 3 .(iv) Initial vondition y(0) = ) 400) = $(9 + c(0)) atle =0. G =1 Citle=0 – ce= -1. y(t) = . {. (!-t) est - t-te 960) S (1-4) ? - test same and positive (V) Eigen values are I so node s unstable.

CA (x) = x² + 2x + x² + x tatl. = (x+1) (+1) = (+1)? (1) Eigen values ase , (4 + )4 = 121-70% [ TD =13% a + H2 =0. Ny=-". 2.- HH1 (n+1) zval 14:47 %, f n = 1. i Figen vectors V, V, - [ ]

il general soln cho 4[) - LGV, F.CZ (vit +va) jent (1) = 4 + 6 (L]+/) 464) - S (9 + Colt +1)) et Taut) et { [ate lttt)) et atlet) é anitical condition 910) = 10l 960) = 3 Iata al 1 -4 = 0. G = 0; Gal. in ů (t) = { Ca (t het - Cote-t.

(1) Eigen so values node & lim are negative & same Aysmptotically stable. alt)=0, lim gt) = 0. 7-7700 ty to 1. 4.1-3 174 1-3 17 A 3 71 2 . (a) = x² + 4x = x (x+4)=o. are & 04. (1) Eigen values () - 01) - 13 F -19) -. - 3 2 + 2 =0. CS Scanned with CamScanner 3!

13 3 11 N (A +43) - (11-12 . 121-182 %,+ XEO: -) 91,--N2". V =H7 (1) General soln YEt = Gevf he e ورام و 414)= G & * [3] +62 4* ? Y(t) = a + p 4t 134 & Gent (iv) Initial condition 410) | . 410) - ? ata = 134 - C. no 4G & 1 c = 1/4

C1 = 2 = 3 47+) - + 38547 non-zlor, elements. sign: (1) 0 Eigen values are - 4 & opposite point & unstable. so saddle lim 400 glt). = 34 lim alt) = 4. so to