Question

7.6(3)

(1 point) Consider the Initial Value Problem -L* 4)*, x0=[!] (a) Find the eigenvalues and eigenvectors for the coefficient matrix. 1 ,01 = , and 12 = (b) Find the solution to the initial value problem. Give your solution in real form. x(t) = Use the phase plotter pplane9.m in MATLAB to help you describe the trajectory An ellipse with clockwise orientation 1. Describe the trajectory.

Answer 1

- -5 -47x 210) c) | 4 -5 for eigen volver A- Arco 1-5rd Ylo T Y 5.11=0 (5-4)2 + 1600 25+ A 2 + 10x41650 ² +10 4+415o AE -10 = 100-4X41 A = -10 t Bi 2 di= -5+41 d2 = -5-41 for eigen space for t= -5-4, [A-AI] Y = 0 s-ui 474, 1 son 4 -41 1742.] 10 R - 'Rat Ri 1-41 -4 704, 7 ron lo olyaJalol 41 Y = 4Y2 Y = 142 1 Y2-1 = Y, so eigen vector for 1 now for to astui TA-AI) Y=o rui -4 14 1. roz -5-41' is ry Y2 Ro - iR2-R, ~ yil, = 442 = 72 = 14

No eigen veckor in ) - :) - To so solution it x17) - eest{ [ Jae- *** + [1] 2 elit? given xlo) :) - 13). ©{ [4] + [:16} MB] - iG & C2 = 3 at 1 C₂ = 3 + 20 = 3-31 + C = 3-31 2 93-16 3-13643) - 6-34-3.33 No 204) ** d 3-34) e 19 + 1931) **[:]] 20 () Answer

X1H) - / 3-31 )e-st fil6944-i Siny!) + ( Gs 41+ isin 4) 1 1 (Csut - isinut) til Gyt tisina) / +182) [ astf(i+1 (Gs 4t + Singh) 7 litid caht- & singh) | lais est s ces 4rt sinut 7 [ Gyt-singh X(t) = [ 3e St ( Ces 4f + sinyt). 3 e-St (es 48-sinut) Answer

X1H) - / 3-31 )e-st fil6944-i Siny!) + ( Gs 41+ isin 4) 1 1 (Csut - isinut) til Gyt tisina) / +182) [ astf(i+1 (Gs 4t + Singh) 7 litid caht- & singh) | lais est s ces 4rt sinut 7 [ Gyt-singh X(t) = [ 3e St ( Ces 4f + sinyt). 3 e-St (es 48-sinut) Answer