MA | 262 | Linear Algebra and Differenti...
MA | 262 | Linear Algebra and Differential Equations | exam |MA262 — FINAL EXAM — SPRING 2016 — MAY 2, 2016 TEST NUMBER 01 INSTRUCTIONS: 1. Do not open the exam booklet until you are instructed to do so. 2. Before you open the booklet ﬁll in the information below and use a # 2 pencil to ﬁll in the required information on the scantron. 3. MARK YOUR TEST NUMBER ON YOUR SCANTRON 4. Once you are allowed to open the exam, make sure you h
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MA | 262 | Linear Algebra and Diffe...
MA | 262 | Linear Algebra and Differential Equations | exam |MA262 — FINAL EXAM — SPRING 2016 — MAY 2, 2016 TEST NUMBER 01 INSTRUCTIONS: 1. Do not open the exam booklet until you are instructed to do so. 2. Before you open the booklet ﬁll in the information below and use a # 2 pencil to ﬁll in the required information on the scantron. 3. MARK YOUR TEST NUMBER ON YOUR SCANTRON 4. Once you are allowed to open the exam, make sure you h
##### Page 9
15.
It is given that
y
p
=
-
1
2
x
cos
x
is a particular solution to the diﬀerential equation
y
′′
+
y
= sin
x
.
Let
y
be the solution to the initial value problem
y
′′
+
y
= sin
x, y
(0) = 0
,y
(0) = 1
.
Find the value of
y
(
π
2
).
A.
π
4
B. 3
π
C.
3
2
D.
1
2
E. 1
16.
Find all the values of
a
and
b
such that the system of equations
x
1
+2
x
3
=1
x
1
+3
x
2
+
x
3
=0
x
1
+
ax
3
=
b
has
inﬁnitely many solution.
A.
a
=1
,b
=2
B.
a
̸
=2
,b
̸
=1
C.
a
=2
,b
̸
=1
D.
a
̸
=1
,b
̸
=2
E.
a
=2
,b
=1

##### Page 10
17.
For the inverse of the matrix
0
3
0
2
1
1
4
4
1
, the entry in the third row and second column is
A. 2
B.
-
1
C. 3
D.
-
6
E. 4
18.
If
y
1
=
t
4
is a solution of the diﬀerential equation
t
2
y
′′
-
7
ty
+ 16
y
= 0 for
t>
0, and
y
2
=
y
1
v
is another solution of the same diﬀerential equation, then
v
satisﬁes which of the following
equation?
A.
t
2
v
′′
-
7
tv
+ 16
v
=0
B.
tv
′′
+
v
=0
C.
tv
+4
v
=0
D.
tv
′′
-
v
=0
E.
tv
-
4
v
=0

##### Page 11
19.
Let
A
=
[
3
4
4
-
3
]
.
The eigenvalues of
A
are
λ
1
= 5 and
λ
2
=
-
5
,
and corresponding eigenvectors
are
V
1
=
[
2
1
]
and
V
2
=
[
1
-
2
]
.
Let
x
(
t
) be the solution to the initial value problem
x
(
t
)=
A
x
(
t
)
,
x
(0) =
[
4
-
3
]
.
Then
x
(
1
5
ln 2) is equal to
A.
x
(
1
5
ln 2) =
[
1
3
]
B.
x
(
1
5
ln 2) =
[
5
0
]
C.
x
(
1
5
ln 2) =
[
2
7
]
D.
x
(
1
5
ln 2) =
[
4
9
]
E.
x
(
1
5
ln 2) =
[
1
2
]

##### Page 12
20.
Find a particular solution of the following system of nonhomogeneous diﬀerential equations
x
=
[
1
1
4
-
2
]
x
+
[
1
0
]
e
-
2
t
.
of the form
x
p
(
t
)=
e
-
2
t
[
a
b
]
.
A.
x
p
(
t
)=
e
-
2
t
[
0
-
1
]
B.
x
p
(
t
)=
e
-
2
t
[
1
1
]
C.
x
p
(
t
)=
e
-
2
t
[
2
3
]
D.
x
p
(
t
)=
e
-
2
t
[
1
2
]
E.
x
p
(
t
)=
e
-
2
t
[
3
4
]

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