Differential Equations and Linear Algebra 2250
Midterm Exam 2
Exam Date: 17 April 2015 at 7:25am
Instructions: This in—class exam is designed to be completed in 80 minutes. No calculators, notes, tables or
I ookx. No answer check is expected. Details count 3/4, answers count I /l.
Chapter 4
Problem 1 (Linear Algebra) Do all parts.
1(a) [10%] For any matrix A, nnllity(A) equals the number of free variables for the problem AiJ
many pivot columns in a 1000 x 54 matrix A with nullity(A) = 21? Answer:
d.
How
7
tI1i4
1(b)
[40%] Consider the following matrix and its reduced row echelon form:
A=(
1
3
—2
—2
—6
4
—2
—5
3
0
1
—1
fi
1’
4 ,
—3)
rref(A)
) Show all linear algebra steps used to solve Ax
1
(b
—-
) Report a basis for the solution space S of Af
2
(b
) Report the dimension of S.
3
(b
/ /—2.—L
I
‘
-L q
I
j\
?-‘-
)
I
-2
i—z
C)
(_
2
0
I
0
I
I
q 3-I
COW)O(
I
-7.
t)2)—3)
2
I
=
(
f\o0
i.
—2
o
-•2.
—2
I
o
tc In
0
V4Ag
a)
3) i)
7,a’X + 2
I.
Lea—
5
X
1
t)
Ll 2-4 3L
LZLEDD\
trv
c LEE
*
Use this page to start your solution.
14 ‘v
—2.
ut
—1i —l
CDrn3( 1)3,
+
1)2..)
Pi--t
23
1 1
0 0
.
ç
I
C’
(?,
2
x
for
=
=
—20
0 1
0 0
a
)L1
)Lç
—
)
2251) Mid tenn Exan i 2 S20 15 [1 7AprilJ
Name.
Chapter 4
Problem 1 (Linear Algebra) Continued.
1 (c) [20%j Check lic inclcl)cfl(lence tests which apply to prove that vectors
1
o
o
‘
1
1
3
o
‘
1
1
3
0
ure independent in the vector space R.’
. Demerits are given for missing a box and also for checicing the box
1
for an inapplicable test.
j
Wronskian test
Rank test
LI
Determinant test
Atom test
Pivot test
LI
Sampling test
Wronskian of functions f, g, Ii. nonzero at x = xo implies indepen
dence of f, g, Ii.
Vectors
v, V3 are independent if their augmented matrix has
rank 3.
Vectors i1
, V2, if
1
3 are independent if their square augmented matrix
has nonzero determinant.
Any finite set of distinct Euler solution atoms is independent.
,
Vectors hi, v2, V3 are independent if their augmented matrix A has
3 pivot columns.
Let samples a, b, c be given and fr functions f, g, h ‘define
A=
( f(a)
f(b)
f(c)
Then det(A)
Use this page to start your solution.
g(a)
g(b)
g(c)
li(u)
h(b)
h(c)
0 implies independence of
f, g, h.
2250 MidI en n Ixi in 2 H20 15 [ [7Apnil]
Name.
Chapter 4
Problem I (Linear Algebra) Continued.
4 (leilote the rOWS of the matrix
, V, V
2
1 (d) I3o,1 Let vj, V
--2
2
1
1
()
0
0
0
-6 o\
5 1 I
3 0 I
2
, v, v, a largest set of independent vectors.
2
, v
1
Extract from the list v
C
—z
0
O
5’
0
L
N
/ —2
2I
\)
i
I
oiJ
\c2
Q/L,+
-‘
Use this page to start your solution.
—2
2
0
j
I
c29c)/
0
Do cP
\2o
*
p1
2250 Midterm Exam
Name.
2 S2015 [l7Aprilj
Chapter 5
Problem 2 (Linear Differential Equations of Order ri.)
Do all parts.
2(a) [20%j Mi.rk with X (lie ftuictiotis which cannot he a solution of a Linear homogeneous differential
(cluatiomi with com stamil cocflicwiit x. Before marking, apply all algebraic simphfications. Test your choices
against this theomeni
The qc’ni’ral solution of a linear homogeneous nth order differential equation with constant coeffici cuts is a lznear combination of Euler solution atoms.
1.
2(b) [20%} Determine a basis of solutions for a homogeneous constant-coefficient linear differential equa
tion, given the characteristic equation has roots 0,0,0; —1, —1; 1 + 2i, 1 2i, 1 + 2i, 1 2i.
—
)
Use this page to start your solution.
ZY
—
eX zx
7 x eP
_tY)
LD
c
E
c
z
-
U)
—
I
v
UF
4
—I
‘4
-
c-i
4
ii
t.J
I
U
tJ)r
Il
‘4
0
0
U)
0
0
U)
U)
ci)
U)
2250 MidI erm Exiun 2 S201 5 [1 7Aprilj
Name.
Chapter 5
Problem 2 (Linear Differential Equations of Order ri) Continued.
2(d) [:30%j A liiiogeiwoiis linear diflerential equation with constant coefficients has characteristic equation
of order 6 with roots 0, 0, 0, [ I •f 2i, I
2, listed according to multiplicity. The corresponding non—
liomogeiwois eciiatioii for imkitown y(.r) has right side f(x) = 2
(.r
0
3e__t_
_
F5ze
4x
r cos 2x+ j
sin 2x. Determine
the undetermined (:ocfficicIlts shortest trial solution for y,,.
To save time, do not evaluate the undetermined coefficients and do not find y,
(x)!
2
—
b
22xj
_(
(e
—
..)
4 rç f ç
Use this page to start your solution.
-
Co a
C;
2.Y)
ex)
ax
714
4J2k
R
2250 lvi id ten ii lxn in 2 S 20 15 [1 7Apri 11
Nan ic.
Chapter 10
Problem 3. (Laplace rFl.icory) (uinplcte all parts.
It is assiiiiied t lmt you have inem()nsed the basic i—item Laplace integral table and know the 6 basic rules for
Iap1ace integrals. No other tables or theory are rectuireci to solve the problems below. If you don’t. know a
la]) Ic ei itry, I lien I cave II ic expression mu eval i ial,ecl h)r partial credit
3(a) (20%]
Display the s-fractions for £(r(t)) and £c(t)) according to Laplace’s Method.
(
:r’-3:r+4y,
y’= 2y,
x(0) = 1, y(O)
2.
To save time, don’t. solve for x(t) or
ICY-)
—c)
L’)
÷c-)
I
t
Lj)
z2
t))
=
2..
1)
2
(-3)(s4z)
Q
2j)
(+2)
Use this page to start your solution.
C
C
Cd)
I!
-I
iL
Th
S.—
‘1
+
(1
1•’
ç%)
+
‘v,
N
t%J
+
CCD
Z
I
2250 IVlidterni Exam 2 S2015 i7A1.)rilj
NJa1HC.
Chapter 10
Problem 3. (Laplace Theory)
3 c) [21))4j Ii II iii II i 1 )1U1lc Sp1CC’H
(
ContimLed.
iii
the Laplace tal )Ie:
Backward Table
Forward Table
f(t.)
f(j(t))
2
S
3
+9
s—2
s 4s + 5
2
sin3t
cc(-&)
e.
—
1
+s+1
4-e
s+1
(s_1)2
)
4j)
S
2.
-
(— )
—
Lc4ç
—
IL
Z
-f—----.
Use this page to start your solution.
-=
4---.-
S -t
cZ(CA,S-r’1
5-, y—I
2250 lVlidl;crm Exam 2 S2015 [17Apr11]
Name:
Chapter 10
Problem 3. (Laplace Theory)
3(d)
Continued.
[30%] Laplace’s method is applied to the probleni
1
finding the answer x(t)
=
sinh(t) +
“ + 2x’ -I- x
x(0) = 0, x’(O)
—
1,
=
te. Display the Laplace solution steps used to obtain the answer
n(t). Check the correctness of your answer with identity sinh(u)
1
—u
—
(e)
(Y)=
(f)
u
=
I
[_
S(x)
(
-
) )
=
2
-? 4-)
42£
i))
—
c—I
(i-i)(
2
2
r+))
—--—--
=
4
c
.s.
—)
5÷l
I)
*
LiA.
A
+‘)
c(—’)
Use this page to start your solution.
t)#f
=
X(+)
2250 MkOcrm fixai n 2 X20 15 [1 7Aprii]
Name.
Chapter 6
Problem 4. (Eigenanalysis) Conipicte all parts.
/6 4 4
The cigenpairs are reported to be
4(a) oX] Lt A
6 11
8
-
)
—
.
15
°
W)
Circle any incorrect elgenpairs.
(:Z)
t)(z)=(::)
H) (‘) (;‘)
Use this page to start your
solution.
()
(r)
/
2250 Midtcrni 1xain 2 S20 15 [1 7April]
Name.
Chapter 6
Problem
4(b)
4. (Eigenanalysis)
Cont imitd.
[40%] Find t11( (igdnpair of t 1w matrix A
=
DD
(
X:
-
)
q
1
(
—
oq
.
4
(
Z
2L2±L5 V
LzzEt’
I
o
,i’5
1
10
1/
/)
,1
3/
\
3/
L)
/)Q
\//‘,
f
‘
f—2)
I
k
[:,
3
x=
XDh
I
3
13
Lx
0
+3
çx
St
Use this page to start your solution.
=
Ho
I0
‘3
9
‘oN
1-,
Jo!
11
D
2
2250 Midtenn Fixain 2 S20 15 [l7ApriIj
Name.
Chapter 6
Problem 4. (Eigenanalysis)
(knit,inned.
rFIn, vector ge eri.I oIution i(/) of a linear (liflereiltial system
nivolvmg time (‘igeilpn.irs of i. X 71. cliagoiia.lu’ab1c matrix A.
4(c) 1:30%!
Assitmime A -=
0.9
o
•
0.3
0
“\
.
F nid the general solution u ol the systeni
.
.
0
Lk
1,U
C.
t
3
y
)
4-
L)
ArM244)7f
Use this page to start your solution.
=
d
—-U
Aml is given 1 y a lrnm1a
AU.