1.D. iiuibei.
Math 2250-4
FINAL EXAM
December 17, 2013
This exam is closed-book and closed-note. You may use a scientific calculator, but not one which is
capable of graphing or of solving differential or linear algebra equations. Laplace Transform Tables are
included with this exam. In order to receive full or partial credit on any problem, you must show all
of your work and justify your conclusions. This exam counts for 30% of your course grade. It has
been written so that there are 150 points possible, and the point values for each problem are indicated in the
right-hand margin. Good Luck!
problem
possible
score
1
40
2
15
3
15
4
15
5
25
6
10
7
30
total
150
1) Constant coefficient first order linear differential equations with constant right hand sides can be solved
with most of the techniques we studied in this course. Find the general solution to the differential equation
for x = x(t) below, using the specified techniques.
x’(t) + 2x(t)
=
-8.
j) Use the integrating factor technique for first order linear differential equations.
2c(f
i1
2t
/
2
(10 pointS)
—
-
-
e
-
jj) Use the technique for separable differential equations.
(10 points)
-)
1
-?
-
1!
-
-
—
I
—1 Le
/1
2
j, Use the technique based on superposition of particular and homogeneous solutions. For your
convenience, the diFferential equation is repeated here:
x’(t) +2x(!)=
-8.
(10 points)
/
—9
j.). Use Laplace transform. Note that in this case the free parameter in your solution will be the initial
value x
.
0
(10 points)
2
s-
4_L.
i”- S
j
-
I
c
S4
vl
A
/
‘‘
—
S
4
—
L
3
). Here is a matrix and its reduced row echelon form:
24
10
10-210
3 -1 -5 0
0 C)
0 1 -1 3 0 (2
-2
A
2
;. reduced row echelon form of A:
:=
0
1 -1 3 -3
3
2-89-2
(j
0 0
0 0 1 C)
00
0000
2.1 Find the solution space of vectors that solve the homogeneous matrix equation A= Q Write your
explicit solution (with free parameters), in linear combination form.
(10 points)
t
-2
-
2
r
\
=
0
It is easy to check that
24
1
3 -1 -5 0
0
2
-2
—1 3 -3
0
1
3
2-89-2
0
1
0
1
-2
—4
1
to write down the general solution to the inhornogeneous matrix
Use this fact and your work from
equation A . = where . = [1, -2, -4, 7]” is the vector on the right side of the matrix equation above.
Explain your reasoning.
(5 points)
-
S jh
t’c’
1
‘
0
4
Find eigenvalues and eigenspace bases for the matrix below. Be careful to check your work, because
this matrix shows up in later problems. Hint: the eigenvalues should be negative integers.
F-4
41
2
-6f
A:=I
I
(15 points)
H- 4
I
2
--\
((+C
-=
-
-
to).
(4
4
)(X)
7Q<,
1
_Q
cD
2
.
[n’
4
-l()
2\O
5
41 Consider the first order system of differential equations, that uses the matrix from the previous
exercise:
x’(t)
4).
-4
=
4
x
2 -6
y
y’(t)
Use your work from to write down the general solution to this system of differential equations
(4 points)
-
H
L’ i
4j Classify the equilbrium point at the origin. (Recall, the names we give these equilibrium points are
nodal sink, nodal source, saddle point, spiral sink, spiral source, and stable center.) Then sketch the phase
portrait for this linear system, using the eigendata of the matrix and the general solution from
(6 points)
4—L
(CG S
G
+1)
0
&0
/? +
3
c
1
o()
F
-2
2
3
6
4
Consider the second order system of differential equations that has the same matrix:
[xh’(t)
iF
j
-4
[
411x1
2 6
[y”(t)
This system could arise from a two mass, three spring system of the sort we have discussed in class.
What is the general solution in this case?
—)
(5 points)
v
j
c
4
r—
2,
-
t)
•
7/
/
[c]
vc
7
j. Consider a general input-output model with two compartments as indicated below. The compartments
1 (t), x,(t) respectively. The flow rates (volume per time) are
2 and solute amounts x
. V
1
contain volumes V
,c
1
.
5
1 i I .6. The two input concentrations (solute amount per volume) are c
indicated by r
,
.
U
k
1
4
,
1
What-qia14ties-l3etwe-en the flow rates r
2 remain constant?
volumes V
. V
1
).
1
V
/
rrY—Y
vf
I7
...,
6 are necessary and sufficient to guarantee that the
r
(4 points)
o
ç
c+v—r
—
3
a
Assuming the equalities in hold, what first order system of differential equations governs the rates of
(t) ?
7
1 (t),x
change forx
(6 points)
v’— r ±r
-r
•
1
V
2.
-
x/
_
2
c)cS
c
—
(r x-
2
v
/
//
0.15
0,c
=
1
;c
=
5
ho
2 = 100 gal. ify that the constant volumes are consistent with the rate balancing required in
1 = V
V
Then show that the genràl system in hreduces to the following system of DEs for the given parameter
values:
1
X
(t)
t
1
x
0
_4 4
+
=
2
x
60
2 -6
(t)
1
7
x
r
er
,r
0
0,r
0O400
pos
0,
20
2O
400,r2
6
Sup
=
1
=
7
=
3
=
.,
(4 points)
rr
‘°
--o-o
4o(.i)
—‘49x
—
—
lw)
8
-
&
r
rt
ft
0
1
-.
x
c,—
o
“-f
c
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1
—
ni
n
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t_
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CD
CD
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CD
0
0
.
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—
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+
F,
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a
(ID
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c)
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(—
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We have studied and returned to the rigid rod pendulum several times in this course. This is the freelyrotating configuration indicated in the diagram below, in which the rod is assumed to be massless of length
L, with a mass in at the end, and with angle displacement 0(f) measured counterclockwise from the
vertical.
—
Using conservation of energy in the no-drag case, we have derived the autonomous second order
differential equation that describes the angle 9(t), arriving at
O”(t) + ffsin(o(r)) =0.
Derive the differential equation above, using conservation of energy. Hint: Express the total energy in
terms of 0(t) and 0’ (t)and set its time derivative equal to zero since the total energy function is constant
once the pendulum is set in motion.
(5 points)
rE
v
c1 LL’
L-Lse
Csru—N
2
•LO
o
(LLs)
a
11
2i1 The second order DE in this problem,
O’’(t) + fsin(O(t)) =0
is equivalent to the autonomous first order system of two differential equations
x’ (t) =y
y’
(t)
fsin(x)
Explain this equivalence.
(4 points)
(
(
e/
I
c (‘e1
t5
1
L
Sy-.
L
Find all equilibrium (i. nstant solutions to the first order system of DE’ s abovhv
these equilibrium solutioeto the constant solutions and corresponding configurations of the
second order differential uation for rigid-rod pendulum.
(6 points)
o
0
r
Oj
c- Sc
V
1
i
LJ
SiL
s
---
A’r’ 5
o- 4ç.
c-Q
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LU
Table of Laplace Transforms
particular functions
ihs table summanzes the general properties nt Laplace transforms and the Laplace transforms of
10.
derived in Chapter
Fswct1an
Iranaform
f(!)
F(s)
a.ftr) ± bg(!)
oF(s) + bG(s)
f’Q)
SF(S)
f”(i)
(x)
,c
F
2
>(t)
f
”
t
s”F(s)
Punchon
.5—a
—
—
sf(0)
—
s’ .1(0)
k
sinki
f’(Oi
-
±k
2
s
-
—
•f” (0)
±
cosh kt
—
F(s)
ff(r)dr
e°’f(t)
f(r)g(t—r)d
([U)
k
sinhkr
S
—
F(s—a)
em coskt
F(s)
e° sink!
u(1 —a)f(t —a)
S
a
k
+k
2
(s—a)
—(sink: —ktcoskt)
3
2k
2 +k
(s
)
2
—F’(s)
—sink!
2k
(s2
)
2
-f k
(2
)
2
+k
sink; -I-kicosk!)
.1(1)
u(t
period p
—
2
(s—a)
F(s)G(s)
t”f(f)
f(t).
S
cos kt
f(O)
—
—
(S
—
a)
eJ(t —a)
J—eP
o
ti/al
(.
)
1
(square wave)
S
(staircase)
5-
I
as
tanh
s
2
—
so
—
—
a!
sn+I
T(a + 1)
14