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Unformatted text preview: Math 22B Solutions
Homework 8
Spring 2008 Section 7.5
2. Solution Let x = ζert . Substitute into the ODE to get:
1−r
−2
3
−4 − r ζ1
ζ2 0
0 = For a nonzero solution, we must have det(A − rI ) = r2 +3r +2 = 0, which has
roots r1 = −1 and r2 = −2. For r = −1, we get ζ1 = ζ2 . A corresponding
eigenvector is v1 = (1, 1)T . For r = −2, we get 3ζ1 = 2ζ2 . A corresponding
eigenvector is v2 = (2, 3)T . So x = c1 (1, 1)T e−t + c2 (2, 3)T e−2t . Also see ﬁgure
1 for the graph.
24. See ﬁgure 2
Section 7.6
13. Solution (a) det(A − rI ) = (α − r)2 + 1 so r2 − 2rα + α2 + 1 = 0 gives
us r = α ± i.
(b) The equilibrium is when α = 0.
(c) See ﬁgure 3.
28. Solution (a) x1 = u, x2 = u ⇒ x1 = x2 and x2 = u . So x2 =
−k
x.
m1
(b) det(A − rI ) = r2 + k
m = 0. So r = ±i −k
u
m k
.
m (c) Origin is the center. See ﬁgure 4.
(d) The general solution is
u(t) = c1 cos k
t + c2 sin
m k
t
m So the solution to the system is
x = c1 ( m
sin
k k
, cos
m kT
t) + c2 (
m
1 m
cos
k k
, − sin
m kT
t)
m = The natural frequency is Im (r1,2 ).
Section 7.7
3. Solution The eigenvalues and eigenvectors were found in Prob. 3 of
section 7.5. The general soln of the system is x = c1 (1, 1)T et + c2 (1, 3)T e−t .
Given the initial conditions x(0) = e, we solve the equations c1 + c2 = 1
3
and c1 + 3c2 = 0, to obtain c1 = 2 , c2 = −1 . The corresponding solution
2
3t
1 −t 3 t
3 −t T
is x = ( 2 e − 2 e , 2 e − 2 e ) . Given the initial conditions x(0) = e(2) ,
we solve the equations c1 + c2 = 0 and c1 + 3c2 = 1 to obtain c1 = −1 and
2
1
c2 = 1 . The corresponding equation is x = ( −1 et + 2 e−t , −1 et + 3 e−t )T . The
2
2
2
2
fundamental matrix easily follows. 2 ...
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This note was uploaded on 03/05/2009 for the course MATH 22B taught by Professor Hunter during the Spring '08 term at UC Davis.
 Spring '08
 Hunter
 Equations

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