Solutions Manualc to accompany System Dynamics, Third Edition by William J. Palm III University ...

85 Pages·2015·0.26 MB·English

System Dynamics 3rd Edition Palm Solutions Manual
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(cid:13)c
Solutions Manual
to accompany
System Dynamics, Third Edition
by
William J. Palm III
University of Rhode Island
Solutions to Problems in Chapter Two
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2.1 a) Nonlinear because of the yy¨ term. b) Nonlinear because of the sin y term. c)
√
Nonlinear because of the y term. d) Variable coefficient, but Linear. e) Nonlinear because
of the sin y term. f) Variable coefficient, but linear.
(cid:13)c
2014 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
2.2 a)
Z x Z t
4 dx = 3 tdt
2 0
3
x(t) = 2+ t2
8
b)
Z x Z t
5 dx = 2 e−4tdt
3 0
x(t) = 3.1−0.1e−4t
c) Let v = x˙.
Z v Z t
3 dv = 5 tdt
7 0
dx 5
v(t) = = 7+ t2
dt 6
Z x Z t(cid:18) 5 (cid:19)
dx = 7+ t2 dt
6
2 0
5
x(t) = 2+7t+ t3
18
d) Let v = x˙.
Z v Z t
4 dv = 7 e−2tdt
2 0
23 7
v(t) = − e−2t
8 8
Z x Z t(cid:18)23 7 (cid:19)
dx = − e−2t dt
8 8
4 0
57 23 7
x(t) = + t+ e−2t
16 8 16
e) x˙ = C , but x¨(0) = 5, so C = 5. x = 5t + C , but x(0) = 2, so C = 2. Thus
1 1 2 2
x = 5t+2.
(cid:13)c
2014 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
2.3 a)
Z x dx Z t
= dt = t
25−5x2
3 0
√ √ √
" ! !#
Z x dx 5 5x 3 5
= arctanh −arctanh = t
25−5x2 25 5 5
3
Let √
!
3 5
C = arctanh
5
Solve for x to obtain
√ √
x = 5tanh(5 5t+C)
b)
Z x dx Z t
= dt = t
36+4x2
10 0
1 x(cid:12)x
tan−1 (cid:12)(cid:12) = t
12 3(cid:12)
10
10
x(t) = 3 tan(12t+C) C = tan−1
3
c)
Z x xdx Z t
= dt
5x+25
4 0
x (cid:12)x x 4
(cid:12)
−ln(x+5)(cid:12) = −ln(x+5)− +ln 9 = t
5 (cid:12) 5 5
4
x−5 ln(x+5) = 5t+4−5 ln 9
So a closed form solution does not exist.
(continued on the next page)
(cid:13)c
2014 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
Problem 2.3 continued:
d)
Z x dx Z t
= −2 e−4tdt
x
5 0
1 (cid:16) (cid:17)
ln x|x = e−4t−1
5 2
x 1 (cid:16) (cid:17)
ln = e−4t−1
5 2
5
x(t) = √ e12e−4t
e
(cid:13)c
2014 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
2.4 From the transform definition, we have
" # " #
Z T Z T
L[mt] = lim mte−stdt = m lim te−stdt
T→∞ 0 T→∞ 0
The method of integration by parts states that
Z T Z T
udv = uv|T − vdu
0
0 0
Choosing u = t and dv = e−stdt, we have du = dt, v = −e−st/s, and
" #  (cid:12)T 
Z T e−st(cid:12) Z T e−st
L[mt] = mTl→im∞ 0 te−stdt = mTl→im∞t −s (cid:12)(cid:12)(cid:12) − 0 −s dt
0
 (cid:12)T (cid:12)T " #
e−st(cid:12) e−st (cid:12) Te−sT e−sT e0
= mTl→im∞t −s (cid:12)(cid:12)(cid:12) − (−s)2(cid:12)(cid:12)(cid:12)  = mTl→im∞ −s −0− (−s)2 + (−s)2
0 0
m
=
s2
because, if we choose the real part of s to be positive, then
lim Te−sT = 0
T→∞
(cid:13)c
2014 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
2.5 From the transform definition, we have
" #
Z T
L[t2] = lim t2e−stdt
T→∞ 0
The method of integration by parts states that
Z T Z T
udv = uv|T − vdu
0
0 0
Choosing u = t2 and dv = e−stdt, we have du = 2tdt, v = −e−st/s, and
" #  (cid:12)T 
Z T e−st(cid:12) Z T e−st
L[t2] = Tl→im∞ 0 t2e−stdt = Tl→im∞t2 −s (cid:12)(cid:12)(cid:12) − 0 −s 2tdt
0
" e−st 2 Z T # " e−st# 2 (cid:18) 1 (cid:19)
= lim −T2 + te−stdt = lim −T2 +
T→∞ s s 0 T→∞ s s s2
2
=
s3
because, if we choose the real part of s to be positive, then,
lim T2e−sT = 0
T→∞
(cid:13)c
2014 McGraw-Hill. This work is only for non-profit use by instructors in courses for which
the textbook has been adopted. Any other use without publisher’s consent is unlawful.
2.6 a)
10 2
X(s) = +
s s3
b)
6 1
X(s) = +
(s+5)2 s+3
c) From Property 8,
dY(s)
X(s) = −
ds
where y(t) = e−3tsin 5t. Thus
5 5
Y(s) = =
(s+3)2+52 s2+6s+34
dY(s) 10s+30
= −
ds (s2+6s+34)2
Thus
10s+30
X(s) =
(s2+6s+34)2
d) X(s) = e−5sG(s), where g(t) = t. Thus G(s) = 1/s2 and
e−5s
X(s) =
s2
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.
2.7
f(t) = 5u (t)−7u (t−6)+2u (t−14)
s s s
Thus
5 e−6s e−14s
F(s) = −7 +2
s s s
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2.8 a)
2 sin 3t
b)
5
4 cos 2t+ sin 2t
2
c)
2e−2tsin 3t
d)
5 5e−3t
−
3 3
e)
5e−3t 5e−7t
−
2 2
f)
e−3t 3e−7t
+
2 2
(cid:13)c
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the textbook has been adopted. Any other use without publisher’s consent is unlawful.