Name
GMay2Ol5
Math 2250-10 Final Exam for 7:15am oh
May 2015
Scores
CliI-2. ft,
c1:i.
Chl and Ch2. (First Order Differential Equations)
[20%] Chl-Ch2(a):
4
T\
Find the posit ion
/‘
0h4. jçjh
Ch!5.
7(t) from t
he velcicit y iimdcl
and the position niodel
t’(t),
[10%] Chl-Ch2(b):
Find all equilibrium solutions
for ij’
a(0)
=
0,2I,
(c’i(t))
((1)
0
100.
C1i6.
//
C1i7.
,‘
0
ChD. /
2
ri(2
3
r
+
2
(O5(i,I))(1J
3 —f 2).
:
ChIo. fO
[20%] Ch1-Ch2(c):
A
11h
Given y
/y 2y
2
2 +
2.r
=
—
I
—-—-----——
‘
‘ ‘\
I
find the non—equilibrium solution in implicit form.
To save time, do not solve for y explicitly.
[20%] Chl-Ch2(d):
Solve the linear homogeneous equation 2
=
y using the integrating factor shortcut.
[10%] Chl-Ch2(e):
Draw a phase line diagram for the differeiitial equation
A
=
(3 + 7.)(7.2
9)(4
--
7.2)3
Label the equilibrium points, display the signs of d.r/dt, and classify each equilibrium point as funnel,
spout or node. To save time, do not draw a phase portrait.
[20%] Chl-Ch2(f):
Solve the linear drag model 1000k
=
50
—
200m’ using superposition v
=
h
0
—I-
lJ.
fod
\/
c
()
, C5
6
5Ce ,v(°)O
()
()
b)
x(°)
/
5e
r/
—
fl
(
(y
I
—
,—
‘--
2
0
‘_()(r2)
O
‘
p
-I—
ç;l
9
0
F
0
cy
<
(1
(t3*c
no
II
+
—
‘+,
C
Th
L
0-
>c
‘3
-o
I)
(
-
—Th
-fr
C
-Th
>s
a’
N
I’
r
Name
6May2015
Math 2250-10 Final Exam for 7:15am on 6 May 2015
Ch3. (Linear Systems and Matrices)
[10%] C1i3(a): Consider a 3 x 5 matrix A and
B
rref(A)
=
its
reduced
row
echelon form:
(2 1) 2 4
0 0
1 1
0 0 0 0 0
(
=
(1) Explain in detail why Ai = 13 and B:i = 13 have exactly the same solutions.
(2) Show the linear algebra steps used to find the scalar general solution to the system Bi
(3) Report a basis for the solution space S of Ai = 13.
(4) Report the dimension of S.
[20%] Ch3(b): Define
matrix
13.
A and vector b by the equations
A
A=(1
For the
Ai
system
h, find r1, a by Cramer’s Rule, showing all details (details count. 75%).
=
[40%] Ch3(c):
Deteriniiie which values of k correspond to (1) a unique solution, (2) infinitely many solutions and (3)
no solution. for the systeiii Ax = b given by
(0
A=I1
D
(J,-J
k—2
4
4
L—3\
k
3)
(—i
1
b=I
‘(ThC
1.
c
cI
‘
7
‘(]
(ô
r
)
-t 1Lj
t
>
/
‘50
-
-
‘i/
1
1
\
7
7 7
0
\
y
7
)
—I
I
H-I
\o
j
((
/ \ ( /
L]
/
I
—
—
O
><
-h
-fJ
I’
U’
-
-U-
c)n
-
A
\]Ui
7ç-
-
tJ
‘I
>c
—
%
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fl
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ON)
I,
YD
I
Name
6May2015
Math 2250-10 Final Exam for 7:15am on 6 May 2015
Ch4. (Vector Spaces)
(
-
[3Yz] Ch4(a): Some independence tests below apply to pi’ovc that vectors x i.2 ,r(,.r arc independent
in the vector space of all continuous functions oil
oc < a K o. Mark one nwthod and display the
details of application (details count 75
‘Nronskian test
Wronskian of functions f, q, Ii nonzero at .r .i() nnphes inclepen
deuce of f, cj, h.
Any finite set of distinct Euler solnt ion atoms is iiiclepenclcnt
Let samples a, b, c be given and for inactions f, q, Ii define
——
Atom test
r_i
Sampling test
([(ci)
f(b)
f(c)
A
\
Then det (A)
y(a)
q(b)
cj(c)
/u(a)
Ii(b)
Ii(c)
0 implies incteiwnclenee of
f, g, Ii.
[20j Ch4(b): Give an exampie of three vectors v
, v
1
, v
2
3 for which the nullity of their augmented
A is two.
matrix
A
207o1 Ch4(c): Find a 4 x 4 system of linear equations
the fraction
for
the constants a, b, c, il
in
the J)artial fraction
decomposition of
3i.2
(i
—
14.1’ +3
(:r
2
+ 1)
—
2)2
To save time, do not solve for a, b, c, d.
[30%} Ch4(d):
The 5 x 6 matrix A below has some independent columns. Report a largest set of independent
columns
of A, according to the Pivot Theorem.
A=
/
o)
00
—3 0
—1 0
60
20
000
—2 0 1
0 0 1
600
200
0
—1
0
3
1
(( C(cc
vn3
/
-
C
4
(r
e
t,
’
c1
OO
?)q)
c)
22
2
-
L
--3
—2
j
rn
0
3 —2
I
2
c
z
ccc-’
‘
-
( (
F
( 3
f)
L
I
oO
-
—
7
c
(PVz
I
2
-
r’Ero
I
()
C> C)
c
Q(Th()
C
O--/
1
j
I
L
DOO
-
(‘)
ri/
°
2
0 C) m)
o
J
i)-?-H
j
C
I
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3
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0
e
C_
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6May2015
Name
Math 2250-10 Final Exam for 7:15am on 6 IVIay 2015
Ch5. (Linear Equations of Higher Order)
[10%] Ch5(a) Find a basis for t he solot ion space of y”
f
[20%] Ch5(b):
Solve h)) t lie general solot ion y of the
—
4y’ + 5y
=—
0.
d’
I 4
16T
6
d
(quat ion
=
0.
[2OtX ] Ch5(c): Find a basis for t lie solution space of a linear coust ant coefficient homogeneous differ—
1)2(7.2 + 2r 4 5)2 = 0.
e(liiatioli is ccc + I )(r
A
(‘relitial equlitOfl, giveil tin’ (‘haracteristic
[20] Ch5(d):
system wit Ii
True
True
A
Given
6, r
in
(ir”(f) -l 2.r’(t) + 2.r(t) =Sc os(wt), which represents a clanipecl forced spring—mass
2, k = 2, answer the following questions.
or False
r
)< [ Practical mechanical resonance is at input frequency w
j or False
.
The homogeneous problem is over-damped.
[30%] Ch5(e): Determine for
(
=
+ 4
= 2’
22
+
+
CI
4 xcos(2r) the shortest trial solution for Yp
according to the met hod of uiidetermmed cocfhcients. Do not evaluate the undetermined coefficients!
2
2
-
mc
c’ co ()
) (I
“
()
-“
4
71/\
‘
L
‘r
/
C-’•
-
4
‘
ic
(r I)
/
(,, X
C
‘j
S I r\
(o
/
c
4
k
o()
2
(rt2,
r
—
C
/
/
X
1
CoS(2
X
hr5
1J1(
()OU
((
-
jLC
-
c
Z
+
L
(?)
i--’
—7
d
(2x
y
(Z)
N
>c
Q)
0
\j
x
0
Th
-
Q)
>‘.
0
c
:-.1
v
N
>c
/
x
,-
—
“
-c)
—
tO
Li’
—
C
6May20 15
flfl 1 C
Math 2250-10 Final Exam for 7:15am on 6 May 2015
Cho. (Eigoiivalues nd Eigenvectors)
f207 J ChG(F): Let A
).
(
Circle possible
eigenpairs of
A.
(2
[20{J Ch6(b): Find the 2 x 2 matrix A which has
eigellpairs
()). (.()).
A
[20] C1i6(c): Find the eigenvectors correspoiidmg to complex eigenvalues —1 ± 31 hr the 2 x 2 matrix
3
4(’
3
A
—
1 0
0 0
[30] Ch6(d): Let A
-
c
‘
))
(‘
.)
1
--5
Display the details for finding all cigenpairs of A.
2
L
(
-
:
) V (,
H)
,
(-2’\
/
-
(()
A
0’.
(
:.
A:
A
•‘
)
‘)
2)
110
)
(t
(t)
I
C’
/
2-
(
)
-
Ii.
-
-
-
-
-
-
(I
‘)
)
)
0
—I
-
)
-‘
1
: ()
C-.’
I
\J) c
7
-
—I
I
/
/
7N
t__
Li
it
1.
,—
---,
C,-’
I
.,-\
—
—
-T
I
(‘I
cc
1.1-
-J
0
—
—
><,“
—©
.—
-q ..
—
—
__)
—
‘5-
-Q
—
c-I,
-r
—
.-
—
—
Li)
—
0
‘;<
A.
—c
—---——-—
‘?
_f\;__
—
—--- ----
—
30
C
(I
-
:
1
,
‘
—
- ©
C)
(j—
C
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—
,
‘5
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-
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2
_,1
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r
6May20 15
N anie
Math 2250-10 Final Exam for 7:15am on 6 May 2015
Ch7. (Linear Systems of Differential
ib
Equat.ioiis)
The methods studied are (1) Linear Cascade method, (2) Cayley-Hamilton-Ziebur method and the
2 x 2 shortcut, (3) Eigenanalysis method, (4) Laplace’s method, including the Laplace resolvent
shortcut.
j\_
/
:30V J ChY(a): Solve fur lIe general solu t ion .r(t ) .ij(t) in the systeni below. l.Jse
apphe, [mm the lectures or any (hapter of t lit’ textlu)ok.
(ic
(It
(i.e
v.
2
[30J Ch’T(b): Consider the scalar system
I
r’
=
3.1’
LI’
=
.1’,
z’
=
‘
A
niet hod t lint
c-f y,
—
(It
A
ally
Then choose one niethod
Report two possible methods that apply to solve for .1, !J,
solution details and the answer (details count 75(,4 ). (
and
chsplay the
[40j Ch7(c): Define
/
4
2
—4
—3
A=I
0
—.
10
0
12
)
The eigenvalues of A are 7, 2, 2. Apply the eigenanalysis nwthod, which requires eigenvalues and
eigenvectors, to solve the differential system u’ = Art.
(3
‘IA
(Dx
(x
€ 0
0
-?\7
7)
r
1.
.-l.
-‘-
I.-.’--
f
‘-
L
::...
(
i
.-2_
çL\
if\
\,\G
ç
4
(‘
—‘(,
U
XJ I!
(‘
\
)
7
:
\j
‘ft’
(_)
(.)
‘
L.,
1
[
A
L
(L
7’
\.
r
\
1f\
C)
/
\
J\
(g
;
C)
tç
L)
1°
C)
/
0
(I’D
-
‘
C)
—
0
to
c
1
C)
/0 L-
O
C)
C---3
(\\
9
\_
\,
‘I
?-- C-)
C:3O
(0
U
Q
0
Ci
C)
C)
\. \
7
I
\_
-
/
\--
a
fl\:-
L:
:,
-—
97
)z
(
‘
\f)
7;•\<
4
flO
c
)
f\
‘D
•c)
4-,
cç
c)
(7
-.
Z5
>ç
-rn
H
c)::
“C :7.
=
C)
CD
a
Nanic
6May2015
Math 2250-10 Final Exam for 7:15am on 6 May 2015
C1i9. (Nonlinear Systems)
f
[10V Ch9(a):
I
ac?
Explniii
\Vliich ot the foiii types
your answer.
.spiinI, node, saddle can he asyniptotically stable at
C(1?t(;,
In parts (b), (c), (d), (e) below, consider the nonlinear systeiri
/
14.r
2.r 2
.ry,
y
/
•
lhtj
—
2y
9
—
;ry.
(1)
124J C1i9(h) : I)isplay details fr finding the equilibrium poiiits for the nonlinear system (1). There
are four answers, one of which is (4, 6).
i\
/4A
/
XJ Ch9(c) Consider again system (1). ( oinpiite t lie Jacobian matrix at (r.
1
[20
Jacobian niatrix at eqinhhrnim point (4. 6).
{3()X] Ch9(e): Consider again systelil (1). W hat classification can be deduced for eciuilibrium (4,6) of
(1). accordiiig to the Pasting Theorem? Explain fully (details count 75w).
nonlinear system
(?
k
ic ((kJ-e)
n
o
(k2-
‘I
C
l
i
l
“
C)
)
fi•rr
ç I
)
Then compute the
[20J Ch9(d): Classify the linear system for ecinilibrilnhl (4, 6) as a uncle, spiral, center. saddle.
‘o
b
q).
0)
(D:( -2\(o)
•)‘)(@‘,)
h\)
1
(t\
)
—s
7
t
C
r
I
—
•
,
I
0
_
cc
j
-
0
0
C
—
I
-
‘K
Cx’
C
c
Il
‘1
I
cx) Q)
Qi
)
—
II
-
A
;
I-_l
Q_ jQj
e: +i’
‘
C—-—
—
y
\_\
-
-
U
-
<
--i-.
7:,
(•
-
(_
s
-ç
-
--
I-.
—
_1
;
-,
:‘
-Q
9
—
-<5
7,
Q
10/
:
C”
cJ:,
C,-
0
7
r;-’
5
n
r-
-
c1
1
Cr:•.
-c
C’
(_j
--
&
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C
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u-.
in
—
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4Th
(-t
s:;- G
(1
(
.
(-p
0
6May20 15
Name
Math 2250-10 Final Exam for 7:15am on 6 May 2015
ChlO. (Laplace Transfornl Methods)
It is assumed that you know the miniinuni Forward Laplace integral table and the 8 basic rules For
Laplace integrals. No other tables or theory are recitnred to solve the problems below. If you clout know
a table entry, then leave the expression unevaluated for partial credit.
0
-
ChlO(a): Fill in the blank spaces in the Laplace tables. Each wrong answer subtracts 3 points
g
from the total of 40.
•N
1
,?
-V-V
J
[401j
—
-
-t
2c
f(t)
9
2s
(f(1))
-
2
1 -b
,2
1
3.s
2
1
.5—
5
D
2 +
s
4
(s
1
1)2 + 1
V
/
V,’r
•
V
7
-
/
2
( H
1210(b): Let
he the unit step.
30,4
)le
111
1 1 I(’ nw( ii anica 1 prol
.r” (t) + 4.i(t)
u(t)
5( u (t
—
=
1)
0. u(t)
I for t
—
u(t
-.
2)),
0 for / < 0. C’onipiitc’ C(r(t) given
.
.i’(O)
.r(0)
=
0.
To save time, do not solve for .1(t).
\
[3U/ ChlO(c): Solve for g(t) in the equation CL(t))
-(9’;
-
-
+
-VD/
V
( +h
/
/(
C
e
//
—‘
-
-
-z.
-
.
/
-D
t—
c..
(L(
r
—Z
q
/(V
C
-z)
I”
..