Question

At least one of the answers above is NOT correct. (1 point) Consider the differential equation = AX(t) dt where A is a given matrix. Determine whether the equilibrium (0,0) is a stable spiral, an unstable spiral, or a neutral spiral (center). Also determine whether the trajectories travel clockwise or counterclockwise. | 3 91 A = | -16 -5] The equilibrium (0,0) is a stable spiral Trajectories travel counterclockwise The equilibrium (0,0) is a stable spiral Trajectories travel counterclockwise, The equilibrium (0,0) is an unstable spiral Trajectories travel counterclockwise

Answer 1

1. The given system Renz Where A=(-5) 32 This system is equivalent to i = 32+gy} ġ = -164-57) The characteristie egun of A is 1A-211-0 1-6-5-21-0 => 3-1)(-5-1) +144 =0 + -15 +51-31+22+144 =0 12+ 22 +129=0 .227 : -21 122-4-1-129 27 -21 -125 = 2² 4.1.129 2.1 - = -24515i 2 :-2, = -1+55 i & 72=-1-55 ::. The equilibrium point (0.03 is and stable spiral. we put, x=1, y=0 then, from © ž=3 x = -16 10 so, the trajectory travels clockwise direction.

2. The given system Ax where A= (3) PT = (5) The system is equivalent to á = 3x+6y ? = -92-3y S2 The characteriste equr of A is 1 A-»I|=0 - 1 3-2 6 10 1 9 -3-4 | = (3-1) (-3-1) + 54 =0 =) -9 +34-32 + 12 +54 = 0 - 72 + 45 - 0 → a= + 355 i -2, = 375 i 12 = -315 i :. The equilibriuna point (0,0) is noutral spiral (center). we put, x=1, y=0 then form © ; = 3 =-9<o So, the trajectory travel clockwise direction

3. The given system å = Ax where A= (1-8). 62 = ( 9 ) 125 9 The system is equivalent to å = -x-84 } -③ 20 - ý = 25x+9y The characteristic equn of A is 1A-111=0 - 1-1-2 -8 1 25 9-1 = (-1-1)(9-x) + 200 =0 = -9-92+2+32 +200=0 =) 72-87 + 19130 az 87 8- 4.1.191 81 V-700 2:1 8£ 1057i. 4517 i & 72 = 4-5571; The equilibrium point (0,0) is unstable - spiral. We put X=1, y=0 i from *; Â =-1 3 = 25 20 so, the trajectory travels anticlockwise direction, ) = 8 1057i :: 7= 4+557

Ansi 1. Stable spiral clockwise dirn neutral spiral (center) clockwise direction 3, unstable spiral antieboek wise direction.

NOTE Note: The system of ode: i where, I is given matrix let. Roots of the characteristic equn. Still Roots of characteristic I eques Type of equilibrium | ft. stability spiral unstable 21, 42 = ptice (070) 71,72 = 8tim (oco) 22,12= till spiral stable centre stable To determine direction of trajectory: We put X=(1.0) then, ý 30 then anticlockwise direction To then clockwise direction