Be sure this exam has 6 pages including the cover
The University of British Columbia
MATH 215/255
Midterm Exam II – November 2015
Name
Signature
Student Number
Course Number
Circle Section:
101 Zhao
102 Tsai
103 Kolokolnikov
104 Zhao
This exam consists of 4 questions worth 40 marks. No notes nor calculators.
Problem
Points
1
12
2
7
3
10
4
11
Total:
40
Score
1. Each candidate should be prepared to produce his library/AMS card upon request.
2. Read and observe the following rules:
No candidate shall be permitted to enter the examination room after the expiration of one half hour, or to leave
during the first half hour of the examination.
Candidates are not permitted to ask questions of the invigilators, except in cases of supposed errors or ambiguities
in examination questions.
CAUTION - Candidates guilty of any of the following or similar practices shall be immediately dismissed from the
examination and shall be liable to disciplinary action.
(a) Making use of any books, papers or memoranda, other than those authorized by the examiners.
(b) Speaking or communicating with other candidates.
(c) Purposely exposing written papers to the view of other candidates. The plea of accident or forgetfulness
shall not be received.
3. Smoking is not permitted during examinations.
November 2015
Math 215/255 Midterm 2
1. Use the method of undetermined coefficients to find a particular solution to:
(4 points)
(a) y 00 − 4y = ex .
(4 points)
(b) y 00 − 4y = sin(2x).
(4 points)
(c) y 00 − 4y = e2x .
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November 2015
Math 215/255 Midterm 2
Page 3 of 6
(7 points) 2. Use the method of variation of parameters to find a particular solution to problem:
y 00 − y = 2e−x .
November 2015
Math 215/255 Midterm 2
Page 4 of 6
(10 points) 3. Use the Laplace transform to solve the following initial value problem:
y 00 + 2y 0 + 5y = δ(t − 1),
y(0) = 1,
y 0 (0) = 2.
November 2015
(6 points) 4. (a) Let F (s) =
Math 215/255 Midterm 2
1
1
. Find L−1 [F (s)](t).
·
s + 1 (s + 1)2 + 4
(5 points)
(b) Find the Laplace transform of f (t) =
et , if 0 ≤ t < 2,
.
0, if t ≥ 2.
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November 2015
Math 215/255 Midterm 2
Page 6 of 6
Table of Laplace transforms
f (t) = L−1 {F (s)}
F (s) = L{f (t)}
1. 1
1
,
s
s>0
1
, s > −a
s+a
n!
tn , n positive integer n+1 , s > 0
s a
sin(at)
, s>0
s 2 + a2
s
, s>0
cos(at)
s 2 + a2
a
sinh(at)
, s > |a|
s 2 − a2
s
cosh(at)
, s > |a|
s 2 − a2
e−as
u(t − a)
, s>0
s
u(t − a)f (t − a)
e−as F (s)
2. e−at
3.
4.
5.
6.
7.
8.
9.
10. e−at f (t)
Z t
11.
f (t − τ )g(τ )dτ
0
Z t
12.
f (τ )dτ
F (s + a)
F (s)G(s)
13. δ(t − a)
F (s)
s
−as
e
14. f (n) (t)
sn F (s) − sn−1 f (0) − ... − f (n−1) (0)
0
Variation of parameters
If y1 (x) and y2 (x) are two solutions of Ly = 0, then the particular solution
of Ly = f (x) is
yp (x) = u1 (x)y1 (x) + u2 (x)y2 (x),
y1 u01 + y2 u02 = 0,
y10 u01 + y20 u02 = f (x).