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Math 215/255 Final Exam (Dec 2005)
Last Name
:
First name
:
Student #
:
Signature
:
Circle your section #
:
Burggraf=101, Peterson=102, Khadra=103, Burghelea=104, Li=105
I have read and understood the instructions below:
Please sign:
Instructions:
1. No notes or books are to be used in this exam.
2. You are allowed to bring a lettersized formula sheet and a smallscreen, nongraphic, non
programmable calculator.
3. Justify every answer, and show your work. Unsupported answers will receive no credit.
4. You will be given 2.5 hrs to write this exam. Read over the exam before you begin. You are
asked to stay in your seat during the last 5 minutes of the exam, until all exams are collected.
5. At the end of the hour you will be given the instruction “Put away all writing implements and
remain seated.”
Continuing to write after this instruction will be considered as cheating
.
6.
Academic dishonesty:
Exposing your paper to another student, copying material from another
student, or representing your work as that of another student constitutes academic dishonesty.
Cases of academic dishonesty may lead to a zero grade in the exam, a zero grade in the course,
and other measures, such as suspension from this university.
Question
grade
value
1
12
2
12
3
14
4
14
5
12
6
16
7
20
Total
100
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View Full Document Question 1:
[12 marks]
Solve each one of the following Frstorder initial value problems for a realvalued solution
y
(
t
) in explicit
form. Also, determine the domain of deFnition for each solution.
(a)
y
′
=
y
sin
t
+ 2
te
−
cos
t
,
y
(0) = 1.
(b) 1
−
(1
−
t
)
yy
′
= 0,
y
(0) = 1.
Solution:
(a) Rewrite the eqn in the normal form
y
′
+ (
−
sin
t
)
y
= 2
te
−
cos
t
. Thus the integrating factor is
μ
(
t
) =
e
R
(
−
sin
t
)
dt
=
e
cos
t
. Thus
(
e
cos
t
y
)
′
=
e
cos
t
(
2
te
−
cos
t
)
= 2
t
⇒
e
cos
t
y
=
i
(2
t
)
dt
=
t
2
+
C
⇒
y
(
t
) =
e
−
cos
t
(
t
2
+
C
)
.
y
(0) = 1
⇒
C
=
e
⇒
y
(
t
) =
e
−
cos
t
(
t
2
+
e
)
.
It is deFne on (
−∞
,
∞
).
(b) This eqn is not exact but is separable.
yy
′
=
1
1
−
t
⇒
1
2
y
2
(
t
) =
−
ln

t
−
1

+
C
⇒
y
2
(
t
) =
−
ln(
t
−
1)
2
+
C
⇒
y
(
t
) =
±
r
C
−
ln(
t
−
1)
2
.
y
(0) = 1
⇒
C
= 1
⇒
y
(
t
) =
r
1
−
ln(
t
−
1)
2
. It is deFne on the interval (1
−
√
e,
1).
Page 2 of 14
Question 2:
[12 marks]
Answer “True” or “False” to the statements below.
Put your answers in the boxes.
(20 points)
(a) Suppose the Wronskian of two functions
f
(
t
) and
g
(
t
) is
W
(
f, g
)(
t
) =
t
(
t
−
1) which is zero at
t
= 0
,
1. Then,
f
(
t
) and
g
(
t
) must be linearly dependent functions.
False.
(b) The Laplace transform of the initial value problem
y
′′′
+
y
′′
+
y
′
= 0
, y
(0) = 1
, y
′
(0) = 2
, y
′′
(0) = 3
yields
Y
(
s
) = (
s
2
+ 3
s
+ 6)
/
(
s
3
+
s
2
+
s
).
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This test prep was uploaded on 03/28/2012 for the course MATH 255 taught by Professor Akosmagyar during the Spring '12 term at The University of British Columbia.
 Spring '12
 AkosMagyar

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