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Answers to Selected Problems 833
10
2
8
–1.5 –1 –0.5 0 0.5
6
0
y
–2 4
2
–4
0
–6 8 12 16 20 24
x
FIGURE A.15 Phase portrait for Problem
FIGURE A.16 Phase portrait for Problem
9, Section 10.6.
13(a), Section 10.6.
10
8
6
y
4
2
0
0 10 20 30 40 50 60 70
x
FIGURE A.17 Phase portrait for Problem
13(c), Section 10.6.
√
9. −2 ± 3i, spiral point. The solution is
√ √
−2t −2t
c 1 e cos( 3t) − c 2 e sin( 3t)
X = √ √
c 1 e −2t sin( 3t) + 3c 2 e −2t cos( 3t)
Figure A.15 is a phase portrait.
11.
x = ax 1 − bx 1 x 2 − Hx 1 , x =−kx 2 + cx 1 x 2 − Hx 2
1 2
13. Figure A.16 is a phase portrait for (a), and Figure A.17 for (c).
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October 14, 2010 17:50 THM/NEIL Page-833 27410_25_Ans_p801-866

