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832 Answers to Selected Problems

6
6
4
4
2 2
0 0
y
–6 –4 –2 0 2 4 6 –15 –10 –5 0 5 10 15
x
–2 –2

–4 –4

–6
–6
FIGURE A.11 Phase portrait for Problem
FIGURE A.12 Phase portrait for Problem
1, Section 10.6.
3, Section 10.6.
600
4
400

2
200
–10 0 10 20 30
y 0
–800 –400 0 400 800
x –2
–200

–4
–400
–6
–600
FIGURE A.13 Phase portrait for Problem FIGURE A.14 Phase portrait for Problem
5, Section 10.6. 7, Section 10.6.

3. Eigenvalues are ±2i; the origin is a center. Solution is

(c 1 − 2c 2 )sin(2t) + (2c 1 + c 2 )cos(2t)
X =
c 1 sin(2t) + c 2 cos(2t)
Figure A.12 is a phase portrait.
5. 4 ± 5i, and the origin is a spiral point. The solution is
4t 4t
(3c 1 − 5c 2 )e sin(5t) + (5c 1 + 3c 2 )e cos(5t)
X = 4t 4t
2c 1 e sin(5t) + 2c 2 e cos(5t)
Figure A.13 is a phase portrait.
7. 3,3, the origin is an improper node. The solution is
3t 3t
c 1 e + c 2 te
X = 3t 3t
(c 1 + c 2 )e + c 2 te
Figure A.14 is a phase portrait.

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October 14, 2010 17:50 THM/NEIL Page-832 27410_25_Ans_p801-866