Page 851 - Advanced_Engineering_Mathematics o'neil

Answers to Selected Problems 831

7.
∞
1
Bt k k
e = I n + B t
k!
k=1
so
∞
1
k
Bt
−1 k
Pe P −1 = PI n P −1 + PB P t
k!
k=1
∞
1

−1 k k
= I n + (PBP ) t
k!
k=1
At
= e (PBP −1 )t = e .
Section 10.5 Applications and Illustrations of Techniques
1. The capacitor charge is maximum when the capacitor voltage is maximum. This voltage is
V C = 10(q 1 − q 2 ) = 5i 3 .
Setting dV c /dt = 0 yields t = (9/2)ln(10/9) ≈ 0.474 seconds. At this time, the capacitor voltage is 6.97 volts.
3.
x 1 (t) = 20 + 25e −t/10 − 5e −3t/50 pounds ,
x 2 (t) = 30 − 25e −t/10 − 5e −3t/50 pounds
The brine solution in tank 1 has minimum concentration at t = 25ln(25/3) minutes. At this time there is
√
20 − (6 3)/125 ≈ 19.9 pounds of salt in tank 1.
5. Designate y 1 (t) as the position of the upper weight relative to its equilibrium position and y 2 (t) the position of the
lower weight relative to its equilibrium position. Then
2 3 √
y 1 (t) = cos(2t) + cos(2 6t),
5 5
6 1 √
y 2 (t) = cos(2t) − cos(2 6t)
5 5
7. Let y 1 (t) be the displacement function of the left mass, and y 2 (t) the displacement function of the right mass (from
their equilibrium positions). The solution is
y 1 (t) = cos(3t), y 2 (t) =−cos(3t).
9. The currents are
1
i 1 (t) = − 2te −10t amp ,
10
1

i 2 (t) = − t e −10t amp
√ 10
3 26 −t −2t
11. y 1 (t) = 2e sin(2t) + e (3cos(t) + sin(t)) − 3cos(t) + sin(t) ,
40
3 −t −2t
y 2 (t) = e (8cos(2t) + 4sin(2t)) + e (7cos(t) − sin(t))
40
−15cos(t) + 15sin(t)]
Section 10.6 Phase Portraits
1. Eigenvalues are −2,−2 and the origin is an improper node. The solution is
−2t −2t
c 1 e + 5(c 1 − c 2 )te
X = −2t −2t
c 2 e + 5(c 1 − c 2 )te
A phase portrait is given in Figure A.11.

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October 14, 2010 17:50 THM/NEIL Page-831 27410_25_Ans_p801-866