Question

Differential Equations Class, please give general solution

1 -1 2 8. x'(t) 4 -1 ) x(t) 3 2 1 9. I'(t) = 3 4 -2 -1 î) (0)

Answer 1

3 2 2 17 Giver x'(x): 1 2 3 4) This is of th form X(t) = AX (4) where Lel an eigenvalue of A then del (A -AI) to 1 Abe 1-2 -1 + 3 2 1 -- 2-2 - 4 +2 -|-A 12-7- (->> | | | & (A) (1²-2-1)-3 (2-3)+2 ( 42 - 7) = 1 13 +2.72 +51 - 6 = - (1-1) (12- 7-6) 20 -(x-1)(x+2) (x-3) =0 > 1, -2,3 ^, = 1, A2 = -2 4 13 = 3 = 3 are the elgenvalul led to be eigenator Ах, Ах, A corresponding to 1,=1 (A-2, I)X = 0 7 (A-I)X,=0 -1 3 2 ng n2 n3

R2-R3 - 1 2 2 m3 - 1 -2 R3 R3 -2 R2 0 O -1 4 R, R, + R3 lio 0 42 -4 ng OI -4. 23 nit ng so 2, = -83 ng - 423=0 h2 = 423 ชา) u X, 22 -83 423 nz - 93 6) 23 Let 83= 1 then X2 is the eigenvector of A. corf. to x=1 let be eigen vector of À corresponding to X2 = -2 (A +2I) X2 4 3 3 2 -1 4 22 23 R2 A2 - 4 S -S O ) 2. m3

X(4): .84 1-c, etc, ezt 4 C, et + C282t + 2 Cg est Cze (ciet + czë where (,, (2463 are constant - 26 + Cze Answer, 3/4)=(-:) **). or, x'(&) X(t) = 1 hi(e) 3 sielt) 4 -1 22 30, -n2 maa! 42 -22 → Boy! = 30, -222 wazito 't we get a hal = 40,- 02 Differentiating get 2l = 30,1 - 2012 From ② using na 42, - ha 30,' -2(932,-012) = 30,1 - 89,+2, from o using 202 = -21'+3h, , we get. - 39,' -8 , -1, +381, 3 hi"-2211+52,

The characteristic equation of a 8? - 27 + 5 = 8= 2 + 4-20 2 I 41 2 2 > this is where * = 1121 of the form 8 &t ß q=1 4 B=2 rence, general solution of 0 is n. 4) = et (c, cos Bl + C₂ sin Bt) - mile) = f (c, cosal & Cg singf), مهار get 2 (c, cosa lt Ca sinat) & ex (-2C, sinad + 2 C2 cosat) using ③ f in ef (c, cosalt + Ca sinad) + ( + (-20, sirzt & 2 C2 (osad) = lote 30 (co cosat & Casinoat) ef (2 9,60524 + 2C281n2f) + - 2 22 - 202 11 et (21, sinad -2 C₂ cosat) ng е e, (cos 2 4+ 51024) +62 (sin2 4 -c0524)) where 6 4 6 are constant

Therefore X () = ( 27 (1) is the required half)) general solution where brill) f delt) is given by equation ③ to respectively