Question

(1 point) xi(t) Let x(t) = be a solution to the system of differential equations: x2(t) xy(t) x'z(t) –6 x (1) 2 xi(t) x2(t) 3 x2(t) = If x(0) find x(t). 2 3 Put the eigenvalues in ascending order when you enter xi(t), x2(t) below. xi(t) = exp( t)+ expo t) x2(t) = exp( t)+ expl t)

Answer 1

SOLUTIONS The given system can be written as Z!) -6 - 1 2 2C+) X2 Cf) 23 () -3 X'(+) = AX(+) Eigen value equation for the coefficient mahix /A-APl=0 U 1*3416 -6-d | 2 -3-1 (6+1)(3+1) +2=0 12+9 4+ 20 =0 (x+4) (a+s=o x = -4 and = -5 Eigen vector corresponding t = -4 [A + 4D]x=0 76** 5][] [:] (2 7] [3]-[:] R₂ R2+R, [:][*]-[: -2% - x2=0

Here R2 is free variable (by pivot) Hence con be chosen X2 = 1 T -2x-1=0 xp =-Y Hence ] Now eigen vector for d2=-5 [A+51]x so -> 2 2][*]:10] [ ][:]=[] n - x - x = 0 similarily R2 = 1 - 2 = -1 Hence V2 = Hence [-] solution X (t) = ct ett + Cz Vz etzt *CH) = 0 [13] <*+62 [7e (11) Now given (r (0) = )

Hence C [3] [***]8+42[] 31-12 =) 29 -C2 ct C2 >> - 1 G - 2=2 at = 3 - 9=10, C2 = -7 (CM)=10 [*]***--[;)]* "x, (4) Hence X(t)= X₂ (4) In ascending order [X, (+) = 7est 5 xz (t) = -7 est -et + lo eet