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(1 point) Solve the initial value problem dx 1.5 2. -1.5 1,5) X, x(0) = (-3)...


(1 point) Solve the initial value problem dx 1.5 2. -1.5 1,5) X, x(0) = (-3) dt -1 Give your solution in real form. 3e^(1/2)
(1 point) Solve the initial value problem dx 1.5 2. -1.5 1,5) X, x(0) = (-3) dt -1 Give your solution in real form. 3e^(1/2) x(t) = -2e^(-1/2t) Use the phase plotter pplane9.m in MATLAB to determine how the solution curves (trajectories)of the system x' Ax behave. O A. The solution curves converge to different points. OB. The solution curves race towards zero and then veer away towards infinity (Saddle) C. All of the solution curves run away from 0. (Unstable node) D. All of the solution curves converge towards 0. (Stable node)
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Solution- dx = 1.25 -5]x, XL0) = (13) 1.5 А 2 15 IA-AI1=0 5 (2-2)(-1-1)+152-0 -2-2a tata?+2.25=0 x²_, +.25=0 4,12_40+1=0 > (2-260.5t_ iste 0.52 PA 0:56 [is] X(+) = G, eest, ts (V, ++ k ] 2017 (3756) [:]-[a] 1) -9, -C2=3 Cit/3/2=2 = -3 Cite 36+(2 -b +

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