# Name____________ Student I.D. Math 2250-1 Exam 2

```Name____________
Student I.D.
Math 2250-1
Exam 2
November 8 2012
Please show all work for full credit. 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. In order to receive full or partial credit on any problem, you must show all of your
work and justify your conclusions. There are 100 points possible. The point values for each problem are
indicated in the right-hand margin. There is a Laplace transform table at the end of this test. Good Luck!
Score
POSSIBLE
1
25
2
25
3
25
4
15
5
10
TOTAL_________________ 100
jj Here is a matrix and its reduced row echelon form:
F 1 -2 -2 0 1
A
:
3 -6 -5
L-2
1
4
9
F 1 -2 0 2 3 10
0
reduced row echelon form of A:
42-2-40
[0
j Find the general solution to the homogeneous matrix equation A=
combination form.
-
Zr
.
0 1
1
1
ooooj
zcv L
(10 points)
r2
2
[1
Li
IjH
0
01 I
LoJ Loj
I
1-3
1&deg;
L
4
!
-
LJ
I
I
Lj
jj). What is the dimension of the solution space to A= that you found in j? Explain.
(5 points)
..
Lit
.vw-
r
2
-z
I0
l&ocirc;
I
LOJ
Li
r3
lo I
0
i
J
&plusmn;
1r
. Find a basis for this
3
j.&ccedil;. The span of the five columns of the matrix A is a certain subspace of ER
subspace, and describe what the subspace is geometrically.
(10 points)
For your convenience, here is the matrix and its reduced form, from page 1:
F 1 -2 -2 I 0 I 1 1
F 1 1-2102 3 1
A
3 -6 -5
:=
1
4
reduced row echelon form of A:
[-2 4 2 2 -4]
0
I
[0/
0 1
/
1 )1
oo ooj
LT
VI
V
A
Le
re(A
Ccce
_J
1 -2V
V
-J
iV
1
2v
3
v
3v
-
+V;.
—,
-J
-J
—,
-
s+
s
I
Vft5,
(2 —‘) p(a
Consider the linear homogeneous differential equation fory(x)
y”(x) —4y”(x)
+4y’(x)0.
Find a basis for the solution space to this DE.
(10 points)
(r
r— rZ
(-
)
OX
2
(r-)
Lc
r
2
C
0,2
4-c x.
1-C\t
GU&amp;o
ZN.
1,
‘Zx
xe
1
Use superposition and the method of undetermined coefficients to find the general solutiony(x) to
y”(x)—4y”(x)+4y’(x)16x +e
x
(15 points)
L()
x
/
z
ji
4L4
Vc
(c
-
Ce
O
g
f (2A
9(f1
2
C
i
=)
)
)
ce
t’.e
)
(-)
f-Q
--=)fl4’2
-iO
,
‘
(
2
So
C
6,(eK
jg). Use Chapter 5 techniques to solve the initial value problem for this forced damped harmonic oscillator
differential equation:
x’’ (t) + 4x’(t) + 13x(t) 40 sin(3 t)
x(0) 0
x’ (0) 0
You may use the fact that this DE has a particular solution
x(t) —3 cos(3 1) + sin(3 t)
(You do not need to check this fact; you can just use it.)
—
(l5points)
r2t
-24
—J
s3
4Si’
Xo[)
=
2
I
3t
+C2g
1
3-c
—)
&divide;c
l
\
42
j
)
2
. Identify the “steady periodic” and “transient” parts of your solution to
periodic part in amplitude-phase form. (Express the phase angle o as an inverse trig function; you don’t
need a decimal value.)
(10 points)
+ s
.
t31
fl=
=
CA,So(rf=
C-
4-
1’
Sv’-c(
-I
ITi5
c’J7)
4.1 Re-solve the IVP from problem 3,
x’ (t) + 4x’ (t) + 13 x(t) 40 sin(3 t)
x(O) = 0
x’ (0) 0
using Laplace transform techniques (and the table at the end of the test). If you wish, you may use the
particular solution x(t) = -3 cos (3 t) + sin (3 t) given in problem to deduce two of the partial fraction
coefficients forX(s). This will save you time.
(15 points)
sX(c i
120
—
-
S,.
/
-
12-v
120
+
+
Sb
(-
—
I
3
(
2
.4-
-
P1—
3
Use Laplace transforms (and the table you’ve been provided) to solve the forced oscillator initial value
problem
0
F
2
x’’(t) +w
x(t) —cos(w
0
t)
0
x(O)x
0
x’ (0) v
0
—
(8 points)
5X()
X)(
-x 0 -v6
2
G3
0/
k)
2z
txbvo
-
-
‘2
2 )2
What word describes the sort of beha ior exhibited by solutions to the differential equation in 5
a?
(2 points)
/
Table of Laplace Transforms
transforms of particular functions
This table summarizes the general properties of Laplace transforms and the Laplace
derived in Chapter 10,
Function
Transform
f(t)
F(s)
aF(s) + hG(S)
bg(c)
af (1
f’(r)
sF(s)—f)O)
f”(t)
(s)
s
F
2
)
1
fa)(
s&deg;F(s)
/
franform
Punctiwa
—
—
sf(O)
(s
coskr
—
s_ +
.-
sin kt
f’(O)
sf(O)
—
—
f ‘(O)
coshkt
—
k
sinhkt
f(r)dr
F(s
e”’f(1)
a)
e’ coskr
F(s)
e”’ sinki
—
2
(s —a)
+k
2
k
u(t
f
—
a)f(r
f(r)g(r
—
—
a)
C’
r)dr
F(s)G(s)
(sinkt
2 +k
(s—a)
2
—
ktcoskt)
tf(t)
—F’(s)
sinkt
t” f(t)
(—1 )F (s)
(sin ki + kt cos kt)
fF(u)dcJ
f(r).
period p
‘,
fe’fQ)dl
)
2
‘+k
U(t—a)
(r
e’
(square wave)
5
—
a!
F(a + 1)
IF i]] (staircase)
U
Il_all
H
—
2)2
1
k
2
-
—a)
(_ULt/a1
2+
(s
tanh
2
c
s(1
—
e’)
```