Be sure this exam has 8 pages including the cover
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
Page 2 of 8
(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.
Answer.
The characteristic equation is
r2 + r + α = 0,
so that the roots are
√
1 − 4α
(1 pt.).
2
To have an oscillatory nature, solution should consist of functions sin(·) or cos(·). This is
only possible when the roots are complex. To make sure that the roots are complex, the
discriminant should be negative, i.e. 1 − 4α < 0, or α > 1/4 (1 pt.). Using the computed
roots, the general solution is
√
√
1
1
y = C1 e− 2 t cos 21 4α−1 t + C2 e− 2 t sin 21 4α−1 t , (1 pt.)
r1,2 =
(6 points)
−1 ±
(1 pt.)
where C1 and C2 are constants.
(b) Find a particular solution of the equation
y 00 + y 0 + y = 2te2t .
Hint: method of undetermined coefficients is less tedious.
Answer. We are going to use the hint and use method of undetermined coefficients.
First, the characteristic equation for the given equation is
r2 + r + 1 = 0,
so that the roots are
√
3
1
r1,2 =
=− ±i
. (1 pt.)
2
2
2
Applying the method of undetermined coefficients, the particular solution has the
following form yp = (A + Bt)e2t (2 pts.), where A and B are constants. The derivatives
of yp can be computed as
−1 ±
√
1−4
yp0 = Be2t + (2A + 2Bt)e2t ,
yp00 = 4Be2t + (4A + 4Bt)e2t .
Substituting yp into the original differential equation, we obtain
4Be2t + (4A + 4Bt)e2t + Be2t + (2A + 2Bt)e2t + (A + Bt)e2t = 2te2t ,
or, after some simplifications,
(5B + 7A)e2t + 7Bte2t = 2te2t .
Since e2t and te2t are linearly independent, then
5B + 7A = 0,
7B = 2,
so that A = −10/49 and B = 2/7 (2 pts.). Finally, the solution is
yp =
2
(−5 + 7t)e2t . (1 pt.)
49
November 2014
Math 215/255 Midterm Exam II
Page 3 of 8
(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.
Answer.
Since there is no damping in the system, the governing equation is
mx00 + kx = cos(t−π/4),
or x00 + ω02 x =
1
cos(t−π/4),
m
(1 pt.)
p
where ω0 = k/m is the natural frequency. Since the system is in the state of pure
resonance (so that the amplitude of the solution grows linearly), then the frequency of
the force is equal to the natural frequency, i.e. ω0 = 1 (1 pt.).
Method 1. We search for a particular solution in the form
xp = At sin(t−π/4) + Bt cos(t−π/4). Substituting it into the differential equation
1
x00 + x = m
cos(t−π/4) gives
− At sin(t−π/4) − Bt cos(t−π/4) + 2A cos(t−π/4) − 2B sin(t−π/4)
1
+ At sin(t−π/4) + Bt cos(t−π/4) =
cos(t−π/4).
m
Clearly, A = 1/(2m) and B = 0, so that the particular solution is
xp =
t
sin(t−π/4). (3 pt.)
2m
Method 2. We search for a particular solution in the form xp = At sin(t) + Bt cos(t).
1
Substituting it into the differential equation x00 + x = m
cos(t−π/4) gives
− At sin(t) − Bt cos(t) + 2A cos(t) − 2B sin(t)
√
√
1
2
2
+ At sin(t) + Bt cos(t) =
cos(t−π/4) =
cos(t) +
sin(t).
m
2m
2m
√
√
In this case, A = 2/(4m) and B = − 2/(4m), so that the particular solution
√
√
2
2
xp =
t sin(t) −
t cos(t), (3 pt.)
4m
4m
which is equivalent to the solution obtained by method
1.
√
The amplitude of the particular solutionp
is C = A2 + B 2 = t/(2m) = 2t. This implies
that m = 1/4. Using the fact that ω0 = k/m = 1, we obtain k = 1/4 (1 pt.).
November 2014
(4 points)
Math 215/255 Midterm Exam II
Page 4 of 8
(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.
Answer.
Using the fact that m = 1/4 and k = 1/4, the governing equation becomes
1 00
1
x + cx0 + x = cos(t−π/4),
4
4
or
x00 + 4cx0 + x = 4 cos(t−π/4). (1 pt.)
Since the student examined steady periodic solution, we need to find a particular
solution to the above equation.
Method 1. Using method of undetermined coefficients, we search for a particular solution
in the following form xp = C cos(t − γ) (1 pt.), where C = 5. Substituting it into the
equation gives
−5 cos(t − γ) − 20c sin(t − γ) + 5 cos(t − γ) = 4 cos(t−π/4).
This allows us to find c = 1/5 and γ = 3π/4 (2 pts.).
Method 2. Using method of undetermined coefficients, we search for a particular
solution
√
in the following form xp = A sin(t−π/4) + B cos(t−π/4) (1 pt.), where A2 + B 2 = 5.
Substituting it into the equation gives
− A sin(t−π/4) − B cos(t−π/4) − 4c A cos(t−π/4) − B sin(t−π/4)
+ A sin(t−π/4) + B cos(t−π/4) = 4 cos(t−π/4).
It is clear that B = 0, and thus A = 5. Moreover, 20c = 4, which gives c = 1/5 (2 pts.).
Method 3. Using method of undetermined coefficients, we search
√ for a particular solution
in the following form xp = A sin(t) + B cos(t) (1 pt.), where A2 + B 2 = 5. Substituting
it into the equation gives
− A sin(t) − B cos(t) − 4c A cos(t) − B sin(t)
√
+ A sin(t) + B cos(t) = 4 cos(t−π/4) = 2 2 cos(t) + sin(t) .
√
√
As a result, we have −4Ac = 2 2 and 4Bc = 2 2. Since
s √
2 √ 2 2 1
p
2
5 = C = A2 + B 2 =
−
+
= ,
2c
2c
c
we obtain c = 1/5 (2 pts.).
November 2014
Math 215/255 Midterm Exam II
Page 5 of 8
(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.
Answer.
Applying the Laplace transform to the equation and use the initial condition, we obtain
s3 X(s) − 3s − 2 + 2sX(s) = 0. (2points)
Thus,
X(s) =
=
=
3s + 2
s3 + 2s
1 −s + 3
+ 2
s
s +2
1
s
3
−
+
. (4points)
s s2 + 2 s2 + 2
x(t) = L−1 {X(s)}
1
s
1
= L−1 { } − L−1 { 2
} + 3L−1 { 2
}
s
s +2
s +2
√
√
3
= 1 − cos( 2t) + √ sin( 2t). (4points)
2
November 2014
Math 215/255 Midterm Exam II
Page 6 of 8
(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.
Answer.
a) Consider
x00 (t) + 2x0 (t) − 2x = δ(t), x(0) = 0, x0 (0) = 0.
Taking the Laplace transform to the equation, we obtain
s2 X(s) + 2sX(s) − 2X(s) = 1. (1 point)
Thus,
X(s) =
=
1
+ 2s − 2
1
. (1 point)
(s + 1)2 − 3
s2
Taking the inverse Laplace transform
x(t) = L−1 {X(s)}
√
e−t
= √ sinh( 3t). (1 point)
3
b) The solution is given by (x ∗ f )(t) and
√
e−τ
√ sinh( 3τ )f (t − τ )dτ
3
0
Z t
√
1
√
f (τ )e−(t−τ ) sinh( 3(t − τ ))dτ. (2 points)
3 0
Z
(x ∗ f )(t) =
=
t
c) Method 1: Using the formula in (b),
(x ∗ f )(t) =
=
Z t
√
1
√
u(τ − 2)e−(t−τ ) sinh( 3(t − τ ))dτ
3 0
Z t
√
1
√ u(t − 2)
e−(t−τ ) sinh( 3(t − τ ))dτ, (1 point)
3
2
November 2014
Math 215/255 Midterm Exam II
Page 7 of 8
where the Heaviside function outside of the integral appears since the integral is equal to
zero for t < 2. Then,
Z t
√
1
√ u(t − 2)
e−(t−τ ) sinh( 3(t − τ ))dτ
3
2
Z t
√
1 √
1
e−(t−τ ) [e 3(t−τ ) − e− 3(t−τ ) ]dτ (1 point)
= √ u(t − 2)
2
3
2
Z t √
√
1
√ u(t − 2) {e( 3−1)(t−τ ) − e−(1+ 3)(t−τ ) }dτ
=
2 3
2
√
√
1
1
1
√ u(t − 2){− √
e( 3−1)(t−τ ) − √
e−(1+ 3)(t−τ ) }|t2
=
2 3
3−1
3+1
√
√
1
1
1
√ u(t − 2){ √
[e( 3−1)(t−2) − 1] + √
[e−(1+ 3)(t−2) − 1]}
=
2 3
3−1
3+1
√
√
√
1
3 + 1 3(t−2)
3 − 1 −√3(t−2)
1
√ u(t − 2)e−(t−2) {
=
e
+
e
} − u(t − 2)
2
2
2
2 3
√
√
√
1
1
√ u(t − 2)e−(t−2) { 3 cosh( 3(t − 2)) + sinh( 3(t − 2))} − u(t − 2) (3 points)
=
2
2 3
Method 2: Taking the Laplace transform to the equation, we obtain
s2 X(s) + 2sX(s) − 2X(s) =
e−2s
. (1 point)
s
Thus,
1
e−2s
+ 2s − 2)
1
1
s+1
1
1
= e−2s [− +
+
]. (2 points)
2s 2 s2 + 2s − 2 2 s2 + 2s − 2
X(s) =
s(s2
Note that
L−1 {
and
L−1 {
s2
√
s+1
s+1
} = L−1 {
} = e−t cosh( 3t),
2
+ 2s − 2
(s + 1) − 3
√
1
1
−1
−t 1
√
}
=
L
{
}
=
e
sinh(
3t).
s2 + 2s − 2
(s + 1)2 − 3
3
Taking the inverse Laplace transform,
x(t) = L−1 {X(s)}
√
√
1 1
1
= u(t − 2)[− + e−(t−2) cosh( 3(t − 2)) + √ e−(t−2) sinh( 3(t − 2))]. (2 points)
2 2
2 3
November 2014
Math 215/255 Midterm Exam II
Page 8 of 8
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).