Question

(1 point) xi(t) Let x(t) be a solution to the system of differential equations: x2(t) = xl () x"(t) -15 xi(t) 20x1(1) 4 x2(t) + 3 x2(t) = If x(0) = find x(t). -5 Put the eigenvalues in ascending order when you enter xi(t), x2(t) below. xi(t) = expo t)+ exp( t) x2(t) = exp( t)+ exp( 1 t)

Answer 1

Consider the given IVP x'=AX with X(o)= (0=[5] The characteristic equation is 1A-XIl=0 20 -15-7 - 4 =0(15-))(3-) - (20)(-4) -0 3-71 → +12 +35-O > >+7)(+5) = 0 The eigen values of A are =-5, -7 when &=-5 we have (A+SI)x=0 10 -4 E] [60] from 1st row: 40%, -48220 22 = -52, eigen vector v when d=-5 is via 82 20 8 -- [] 69:16 The for a = 2 when &=-7 we have (A+TI)x=0 -8 -4 20 10 ra U2 from ist row: -80,- 492=0 » 82=-28, eigen vector v₂ when 72=-7 is V2 =) for al. The general solution is -5t X(t) = Cie -5 + C₂ e

use xco) - [S "[:]*[:]» (, 1.2 (26, +C₂ → XCO) = G -56,- 26₂ By equating 20₂ + C2 = 5 and 50,- 2 C₂ =-5 By solving these equations, we get c, =-5, C2215 Thus, the solution of the given Ivp is x, ctr= -5t2 --5e 22Ct) -5 "Tt +15 e ► 2,(t)= -10 exp(-5t) + 15 exp(-7t) az(t) = 25 eap(-5+)-30 eXpG7t).