Question

Differential Equations and Matrix Algebra problem:

Problem 1 Find the general solution 2"(t) = ( 2) -(0) 0 1 2. r't) 0 1 0 3. x'(t) = (* (1)«ce)

Can you please show how to do numbers 2 and 3? Could you show once you find the eigenvalues, the steps you take for the Gaussian Elimination and row reducing to get the eigenvectors? I'm having trouble with the Gaussian Elimination portion of the problem, trying to get the bottom row of the matrix to be all zeros. For problem 3, I found the eigenvectors when lambda is equal to 0, but I'm stuck with lambda equal to 2. Thanks.

Answer 1

Q" Given that the 2017) 0 De'lt) - - 3 2 2-2 x!(t) = A 210) characteristic Squation 1A-Il fo -3-1 r2 SO JI - 189] 22 -2-X . (-3-4) (-2-7) - 2 =0 6+38 +24+72-230 &75x+4=0 tuxtaty=0 d (874) + 1 (2+4)=0 Coty) (71)co => (8+4)=0 ; 8+co d=-4 d=-4 & dal a=1 d=-1,-4 Eigen Vector Corresponding to do-1,-4: ifd, :-1 then (A-AI) X(t)=0 (A+I) 2,4)=0 1-3+1 vey)=(:) (2 2+1 2 6,76) =10) ป

-22 A-NI) 2) 2-1 R₂ 7 2 R₂ +VER, - (2) (23) au 10 ) .-28,+EX2=0 let take *2=k +22;= f 82% 21=22x 2 so Hitam); WEK (A) = x( V ; %%[+) = 1

d=-4 (A+UN) X244) = 0 Atux.= F3+4 & r. -2+4 ) Atuy R2 누 ( a 60%]K"]=6 R₂ - V2RI dit (222=0 42 =k =) di = - VZH₂ x=-Vek - so; alt) (ok) = *(-2) Hence general solution is 16 X17) - 1 2 alt) = alt) 0 x'(t) A x(+) 11 1 0 OWO TE here A= Ol Characteristic Equation 1A-XI) 1 0-1 0-0 10-0

1-) - (-1) (02-1) -1 (0-1)+1(1+0=0 -03+titltltd.co 3 do +30+2=0 03-31-220 of deu tien 7+8=2=0 day 2 1o-3 -2 O-II | -1 -2 (2²-1-27=0 d² 2x+8 -2=0 d (0-2) +!(1-2)=0 (d-2) 16+1220 d=21-1 d=-1,2 Eigen vector correspondling d = - thien (A-12) -11) - 63 R2 Ro Ri Rz -) R3-R, o

ar 212 23 (8) Orbitary Constant (f(n) {n} 3-1=2 KithfX₂=0 X2 = ki = kg X2= 2. = KitK2) = fkitka Xilt ci Kq d2=-2 then (A-II) 1 11 I 1 21 12 R2 72R₂ - Ri R372h-R1 2 1 0 31 13 R₂ > 3R₃ - R2 2 22 3 0 0 8 oog 2014 224 23=0 3x2 + x320 8 43=0 X=0

21,30 X250 alt)= ( ; 있 23=0 heuce the general solution ip Xit) X lt) = & +60) et pkite tge 248 ki K2 x It) = (0,+ c t )e-t(K,+82 ki K2 a & are constants ki & kz are Babitary from Rank of the matrix 2 the Constants and are Thee matrix n {(A)=3-1=2 2 (08bitasey Consecuits K, $ky). goullal Saleition is Alt) (Citate = (atat e + (6 +6+) e XL Plt) = ((1+(t)e (Citate 364) (C+(at) et 3 * O -t * [!] و کو

1 © xlt) = 6; ;) 11 d'It) = A alt) IA= A=(-1) O Sigoe Vector Congesponding to 1-0,23 here (A-VI) & co, 4,50 ; (A-AI) = (-1) tinc E.HA? 24-1=0 8-2x = 0 d(0-2) = 0 2 R₂ > RtiR 0 0 Oo X2 ditixa X2 =k X13-ik

k Wilt) = x=2 then (A-XI) = X/16) -- *11) (4-3) =<:: 603) 3] =6) R₂ > R2 i Ri -xit inco X =k ide x= X = ik hence thee xus] =>() x H = Geot (H) + Ges* (i] a lift ez eatli general Solution is XH)