mathlab 4.docx - Woo Jung Kim A10017873 Section A03 TA Mary...

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Woo Jung KimA10017873Section A03TA: Mary RadcliffeExercise 4.1(a)>> b=[1.2,2.5;4,0.7]b = 4.0 2.50004.0000 0.7000(b)>> [egivec,eigval]=eig(b)egivec = 0.6501 -0.58990.7599 0.8075eigval = 4.1221 00 -2.2221Exercise 4.2(a)(4) x' = x + 3y y' = -x - 8y(b)>> A=[1,3;-1,-8]A = 1 3-1 -8>> [egivec,eigval]=eig(A)egivec = 0.9934 -0.3276-0.1148 0.9448eigval = 0.6533 00 -7.6533(c)
(d)I expected the solution to approach zero as t gets larger, and my expectation matches with the graph above. Exercise 4.3(6) x’=2.7x –yY’=4.2x+3.5y
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(b)Write the general solution of (6)
(C) Because Eigenvalues’ real parts are positives, the solutions that positive exponents of exponentials dictate will approach the infinity as t gets large. Exercise 4.4(a) >> A = [1.25, -.97, 4.6; -2.6, -5.2, -.31; 1.18, -10.3, 1.12]A = 1.2500 -0.9700 4.6000-2.6000 -5.2000 -0.31001.1800 -10.3000 1.1200>> [egivec,eigval]=eig(A)egivec = 0.7351 -0.4490 - 0.2591i -0.4490 + 0.2591i-0.1961 0.3375 - 0.2242i 0.3375 + 0.2242i0.6490 0.7530 0.7530 eigval = 5.5698 0 0 -4.1999 + 2.6606i 0 0 -4.1999 - 2.6606i(b)With few examples, I found out that only a real part of complex solutionis used to decide the stability. In this case, the eigenvalues are one positive and two negatives. Since any system with one or more positive is always unstalbe, the graph will be unstable. Exercise 4.5Ais the 4 x 4 matrix
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