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The University of British Columbia
MATH 215/255, Section 102
Midterm Exam II – November 2009
Name
Signature
Student Number
This exam consists of 4 questions worth 10 marks each. No notes nor calculators.
Problem
1.
max score
10
2.
10
3.
10
4.
10
total
40
score
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during the first half hour of the examination.
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in examination questions.
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examination and shall be liable to disciplinary action.
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shall not be received.
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November 2009
Math 215/255 Section 102 Midterm Exam II
Page 2 of 6
(5 points) 1. (a) Find the general solution of the equation
y 00 + 2y 0 + y = 0.
Answer.
The characteristic equation is
r2 + 2r + 1 = 0,
which has repeated roots r = −1, −1. Thus a fundamental set of solutions is
y1 (t) = e−t ,
y2 (t) = te−t .
The general solution is
y(t) = c1 e−t + c2 te−t .
(5 points)
(b) Find a particular solution of the equation
y 00 + 2y 0 + y = 2e2t .
Answer.
Since e2t is not a solution of the homogeneous equation, we may assume
Y (t) = Ae2t .
Thus
Y 0 (t) = 2Ae2t ,
Y 00 (t) = 4Ae2t ,
and
Y 00 + 2Y 0 + Y = 9Ae2t .
Thus 9A = 2 and A = 29 ,
Y (t) =
2 2t
e .
9
Alternative solution by variation of parameters: Let Y (t) = u1 (t)e−t + u2 (t)te−t
and solve
u01 e−t + u02 te−t = 0, −u01 e−t + u02 (1 − t)e−t = 2e2t .
Their sum gives u02 e−t = 2e2t , thus u02 = 2e3t and u01 = −2te3t . Integration gives
u2 = 23 e3t and u1 = (− 23 t + 29 )e3t (we don’t need the constants). Thus
2
2
2
2
Y (t) = (− t + )e3t e−t + e3t te−t = e2t .
3
9
3
9
Alternative solution by reduction of order: Let Y (t) = v(t)e−t . Then
Y 0 = (v 0 − v)e−t , Y 0 = (v 00 − 2v 0 + v)e−t and
Y 00 + 2Y 0 + Y = [(v 00 − 2v 0 + v) + 2(v 0 − v) + v]e−t = 2e2t .
We get v 00 = 2e3t . Thus v 0 = 32 e3t + c1 , v = 92 e3t + c1 t + c2 . Taking c1 = c2 = 0,
Y = 92 e3t e−t = 29 e2t .
November 2009
Math 215/255 Section 102 Midterm Exam II
Page 3 of 6
.) A mass of 1 kg is hung
(7 points) 2. (a) A spring is stretched 0.2 m by a force of 5 N. (1 N = 1 kg·m
s2
from the spring and is also attached to a viscous damper that exerts a force of 18 N
when the velocity of the mass is 3 m/s. If the mass is set in motion from its equilibrium
position with a downward velocity of 0.4 m/s, determine its position u(t) at any time t.
Here u is the displacement from the equilibrium position measured positive downward.
(The mass spring system has equation mu00 + γu0 + ku = 0 where m is the mass attached
to the spring, γ is the damping coefficient so that the damping force is given by −γu0 ,
and k is the spring constant so that the force when the spring is stretched by u is −ku.)
Answer.
The spring constant is k = 5/0.2 = 25 (N/m).
The damping coefficient is γ = 18/3 = 6 (N-s/m).
Hence the equation of motion is
u00 + 6u0 + 25u = 0.
The initial conditions are
u(0) = 0,
u0 (0) = 0.4.
The characteristic equation is r2 + 6r + 25 = 0. Its roots are
r = −3 ± 4i.
The general solution is
u(t) = Ae−3t cos 4t + Be−3t sin 4t.
Invoking the initial conditions, we have A = 0,
u0 (t) = 4Be−3t cos 4t − 3Be−3t sin 4t,
0.4 = 4B,
and B = 0.1. Thus
1 −3t
e sin 4t.
10
Alternative solution: After deriving the equation and the initial conditions, Laplace
transform gives (s2 + 6s + 25)U (s) = 0.4. Thus
1 −1
4
1
= e−3t sin 4t.
u(t) = L
10
(s + 3)2 + 42
10
u(t) =
(3 points)
(b) Find the quasi frequency µ and the ratio of µ to the natural frequency ω0 of the
corresponding undamped motion. (When γ 2 < 4km, the quasi frequency µ is given by
1 p
µ=
4km − γ 2 . The natural frequency is the same quantity when γ = 0.)
2m
Answer.
We already have µ = 4.
The undamped equation is
The natural frequency ω0 =
The ratio is
√
u00 + 25u = 0.
25 = 5.
µ/ω0 = 4/5.
November 2009
Math 215/255 Section 102 Midterm Exam II
(7 points) 3. (a) Use Laplace transform to find the solution of the initial value problem
(
4 if 0 ≤ t < π
y 00 + 4y =
0 if t ≥ π
y(0) = 1,
y 0 (0) = 6.
Answer.
L{y 00 + 4y} = s2 Y (s) − s − 6 + 4Y (s).
Since RHS = 4 − 4uπ (t), its Laplace transform is
4
4
− e−πs .
s
s
Thus (s2 + 4)Y (s) = s + 6 +
Y (s) =
s2
4
s
− e−πs 4s , and
s
2
4
4
+3· 2
+
− e−πs 2
.
2
+4
s + 4 s(s + 4)
s(s + 4)
By partial fraction
1
s
4
= − 2
.
+ 4)
s s +4
s(s2
Thus
y(t) = cos 2t + 3 sin 2t + 1 − cos 2t − uπ (t)[1 − cos 2(t − π)]
= 1 + 3 sin 2t − uπ (t)[1 − cos 2t].
Z
(3 points)
(b) Find the Laplace transform of the function f (t) =
t
(t − τ )2 eτ dτ .
0
Answer.
f (t) = t2 ∗ et .
L[f ] = L[t2 ] · L[et ] =
2
1
·
.
3
s s−1
Page 4 of 6
November 2009
Math 215/255 Section 102 Midterm Exam II
Page 5 of 6
(6 points) 4. (a) Use Laplace transform to find the solution of the initial value problem
y 00 + 2y 0 + y = δ(t − 3),
Answer.
y(0) = 1, y 0 (0) = 0.
Let L{y} = Y (s). Then L{y 0 } = sY − 1 and L{y 00 } = s(sY − 1). Thus
(s2 + 2s + 1)Y (s) − s − 2 = L{δ(t − 3)} = e−3s .
Y (s) =
s + 2 + e−3s
1
1
e−3s
=
+
+
.
(s + 1)2
s + 1 (s + 1)2 (s + 1)2
1
−t ,
Since L−1 { (s+1)
2 } = te
y(t) = e−t + te−t + u3 (t)(t − 3)e−(t−3) .
(4 points)
(b) Find the inverse Laplace transform of
F (s) = e−3s
s
.
s2 + 4s + 5
Answer.
s2
s
s+2
2
s
=
=
−
2
2
+ 4s + 5
(s + 2) + 1
(s + 2) + 1 (s + 2)2 + 1
Its inverse Laplace transform is
e−2t cos t − 2e−2t sin t
Thus
L−1 {F }(t) = u3 (t)e−2(t−3) [cos(t − 3) − 2 sin(t − 3)]
November 2009
Math 215/255 Section 102 Midterm Exam II
Table of Laplace transforms
f (t) = L−1 {F (s)}
1. 1
2. eat
3. tn , n positive integer
4. tp , p > −1
5. sin(at)
6. cos(at)
7. sinh(at)
8. cosh(at)
9. eat sin(bt)
F (s) = L{f (t)}
1
, s>0
s
1
, s>a
s−a
n!
, s>0
n+1
s
Γ(p + 1)
, s>0
sp+1
a
, s>0
2
s + a2
s
, s>0
s2 + a2
a
, s > |a|
2
s − a2
s
, s > |a|
2
s − a2
b
, s>a
(s − a)2 + b2
10. eat cos(bt)
s−a
, s>a
(s − a)2 + b2
11. tn eat , n positive integer
n!
,
(s − a)n+1
12. uc (t)
e−cs
, s>0
s
13. uc (t)f (t − c)
e−cs F (s)
14. ect f (t)
F (s − c)
15. f (ct)
Z t
16.
f (t − τ )g(τ )dτ
1 s
F
,
c
c
s>a
c>0
F (s)G(s)
0
17. δ(t − c)
e−cs
18. f (n) (t)
sn F (s) − sn−1 f (0) − ... − f (n−1) (0)
19. (−t)n f (t)
F (n) (s)
Page 6 of 6