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The University of British Columbia
MATH 215/255-103/104
Midterm Exam II – November 2014
Name
Signature
Student Number
Section Number
This exam consists of 4 questions worth 10 marks each. No notes nor calculators.
Question
Points
1
10
2
10
3
10
4
10
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 2014
Math 215/255 Midterm Exam II
(4 points) 1. (a) Find all values of α for which the solution of the equation
y 00 + y 0 + αy = 0
has oscillatory nature. Write down the general solution.
(6 points)
(b) Find a particular solution of the equation
y 00 + y 0 + y = 2te2t .
Hint: method of undetermined coefficients is less tedious.
Page 2 of 6
November 2014
Math 215/255 Midterm Exam II
Page 3 of 6
(6 points) 2. (a) A student made a frictionless (i.e. damping is c = 0) oscillator by connecting a mass to a
wall by a spring. Unfortunately, he forgot to measure the mass, m, and the spring
stiffness constant k. He decided to use the material he learned at the ODE course about
resonance to find these parameters. He applied the periodic force F = cos(ωt−π/4) to
the system and found that ω = 1 corresponds to pure resonance. In particular, the
amplitude of the oscillations grows linearly with time for some initial conditions, i.e.
C = αt, where α = 2. Use this information to help the student to determine m and k.
(4 points)
(b) Then, the student decided to add damping and tried to measure it (mass m and spring
constant k are the same as in part (a)). He applied the same force F = cos(ωt−π/4) to
the system and examined the steady periodic solution. The measurements for ω = 1
revealed that the amplitude of the response no longer grows linearly, but instead is
constant, C = 5. Use this information to find damping coefficient c.
November 2014
Math 215/255 Midterm Exam II
(10 points) 3. Solve the following initial value problem for ODEs with the Laplace transform:
x000 (t) + 2x0 (t) = 0, x(0) = 0, x0 (0) = 3, x00 (0) = 2.
Page 4 of 6
November 2014
Math 215/255 Midterm Exam II
Page 5 of 6
(3 points) 4. (a) Find the impulse response for
L(x) = x00 (t) + 2x0 (t) − 2x, x(0) = 0, x0 (0) = 0.
(2 points)
(b) Find the solution of
x00 (t) + 2x0 (t) − 2x = f (t), x(0) = 0, x0 (0) = 0,
for a general function f (t). Express the solution by a definite integral. Hint: can make
use of the impulse response in (a).
(5 points)
(c) Find the solution in (b) if f is the Heaviside function f (t) = u(t − 2), i.e., f (t) = 1, if
t ≥ 2, and zero otherwise. Hint: you can use any method you know.
November 2014
Math 215/255 Midterm Exam II
Page 6 of 6
Table of Laplace transforms
f (t) = L−1 {F (s)}
F (s) = L{f (t)}
1. 1
2. e−at
3. tn , n positive integer
4. sin(at)
5. cos(at)
6. sinh(at)
7. cosh(at)
8. u(t − a)
1
, s>0
s
1
, s > −a
s+a
n!
, s>0
n+1
s
a
, s>0
2
s + a2
s
, s>0
2
s + a2
a
, s > |a|
s2 − a2
s
, s > |a|
2
s − a2
e−as
, s>0
s
9. u(t − a)f (t − a)
e−as F (s)
10. e−at f (t)
Z t
11.
f (t − τ )g(τ )dτ
F (s + a)
F (s)G(s)
0
Z
t
0
F (s)
s
13. δ(t − a)
e−as
14. f (n) (t)
sn F (s) − sn−1 f (0) − ... − f (n−1) (0)
12.
f (τ )dτ
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).