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The University of British Columbia
MATH 215
Midterm Exam II – March 2010
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Name
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
1. Each candidate should be prepared to produce his library/AMS card upon request.
2. Read and observe the following rules:
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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.
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shall not be received.
3. Smoking is not permitted during examinations.
March 2010
Math 215 Midterm Exam II
Page 2 of 6
(5 points) 1. (a) Use the method of reduction of order to find a second solution y2 (t) of the equation
t2 y 00 − 4ty 0 + 4y = 0,
t > 0;
y1 (t) = t
which is independent of y1 .
Answer.
Let y2 = tv. Then y20 = tv 0 + v and y200 = tv 00 + 2v 0 . Thus
0 = t2 y200 − 4ty20 + 4y2 = (t3 v 00 + 2t2 v 0 ) − 4t2 v 0 − 4vtv + 4tv = t3 v 00 − 2t2 v 0 .
Let w = v 0 . We have
2
w0 − w = 0.
t
dt
2
Thus dw
w = 2 t and ln w = 2 ln t + c. We conclude w = At and v =
3
We may take A = 3 and B = 0 to get y2 (t) = tv(t) = t · t = t4 .
R
w = At3 /3 + B.
Remark. Many of you tried variation of parameter and wrote y = u1 y1 + u2 y2 . This
method is used to solve nonhomogeneous equations when both y1 and y2 are known, such
as part (b) below. It is not applicable to this problem.
(5 points)
(b) Find a particular solution of the equation
y 00 − 2y 0 = 4t.
Answer. The homogeneous part has characteristic equation r2 − 2r = 0 with roots 0
and 2. Thus its general solution is c1 + c2 e2t . The force 4t is a polynomial of degree one
and our first guess of a particular solution is Y (t) = A + Bt. However A is a solution of
the homogeneous equation and we multiply our initial guess by t to get
Y (t) = At + Bt2 .
We have Y 0 = A + 2Bt and Y 00 = 2B. Thus Y 00 − 2Y 0 = 2B − 2A − 4Bt = 4t. Thus
B = A = −1. Thus Y (t) = −t − t2 is a particular solution.
Alternative solution. Characteristic equation is r2 − 2r = 0. Homogeneous part has
general solution
y(t)
= c1 y1 (t) + c2 y2 (t) where y1 (t) = 1, y2 (t) = e2t , with Wronskian
1 e2t
W = det
= 2e2t . By the method of variation of parameter, we have
0 2e2t
yp = u1 y1 + u2 y2 where
Z 2t
Z
y2 g
e 4t
u1 (t) = −
=−
dt = − 2tdt = −t2 + c1 .
W
2e2t
Z
Z
Z
y1 g
4t
1
u2 (t) =
=
dt = 2te−2t dt = (−t − )e−2t + c2 .
2t
W
2e
2
Z
Thus, taking c1 = c2 = 0, yp (t) = t2 + e2t (−t − 12 )e−2t = −t2 − t − 12 .
March 2010
Math 215 Midterm Exam II
Page 3 of 6
2. Suppose the motion of a certain mass-spring system satisfies the differential equation
u00 + 2u0 + 5u = 10 cos t,
u(0) = 3,
u0 (0) = −4,
with units m, kg, and s. Here u(t) is the displacement from the equilibrium position.
(7 points)
(a) Determine the position u(t) at any time t.
Answer.
The characteristic equation is r2 + 2r + 5 = 0. Its roots are
r = −1 ± 2i.
The general solution to the homogeneous equation is
uc (t) = e−t [c1 cos 2t + c2 sin 2t].
A particular solution is of them
U (t) = A cos t + B sin t.
We have U 0 = B cos t − A sin t, U 00 = −U , and
U 00 + 2U 0 + 5U = (4A + 2B) cos t + (4B − 2A) sin t = 10 cos t.
Thus 4A + 2B = 10 and 4B − 2A = 0. We get A = 2B, B = 1 and A = 2. Thus
U (t) = 2 cos t + sin t and the general solution is u(t) = uc (t) + U (t), and thus
u0 (t) = e−t [(−c1 + 2c2 ) cos 2t + (−2c1 − c2 ) sin 2t] + cos t − 2 sin t.
Invoking the initial conditions,
3 = c1 + 2,
−4 = −c1 + 2c2 + 1.
Thus c1 = 1 and c2 = −2. Thus
u(t) = e−t [cos 2t − 2 sin 2t] + 2 cos t + sin t.
(1)
Common mistake: The initial conditions are for u(t) = uc (t) + U (t), not for uc (t)!
Alternative solution: Let L{u} = U (s). Then L{u0 } = sU − 3 and
L{u00 } = s(sU − 3) + 4. Laplace transform of the equation gives
10s
(s2 + 2s + 5)U (s) − 3s + 4 − 6 = 2
.
s +1
Thus
3s + 2
10s
U (s) = 2
+ 2
.
s + 2s + 5 (s + 2s + 5)(s2 + 1)
Partial fraction gives
10s
2s + 5
2s + 1
=− 2
+ 2
.
2
2
(s + 2s + 5)(s + 1)
s + 2s + 5 s + 1
Thus
U (s) =
(3 points)
(s + 1) − 4
2s + 1
+
,
s2 + 2s + 5 s2 + 1
and u(t) is as in (1).
(b) Identify the transient and steady state parts of the solution. Find the amplitude of the
steady state part.
Answer.
The transient part is e−t [cos 2t − 2 sin 2t]. √
√
The steady state part is 2 cos t + sin t, with amplitude 22 + 12 = 5.
March 2010
Math 215 Midterm Exam II
(10 points) 3. Use Laplace transform to find the solution of the initial value problem
(
1 if 0 ≤ t < 3
y 00 + 2y 0 + y =
0 if t ≥ 3
y 0 (0) = 0.
y(0) = 0,
Answer.
L{y 00 + 2y 0 + y} = (s + 1)2 Y (s).
Since RHS = 1 − u3 (t), its Laplace transform is
1
1
− e−3s .
s
s
Thus (s + 1)2 Y (s) =
1
s
− e−3s 1s , and
Y (s) =
1
(1 − e−3s ).
s(s + 1)2
By partial fraction
1
1
1
1
= −
−
.
2
s(s + 1)
s s + 1 (s + 1)2
Thus
y(t) = h(t) − u3 (t)h(t − 3),
h(t) = 1 − e−t − te−t .
Page 4 of 6
March 2010
Math 215 Midterm Exam II
Page 5 of 6
(7 points) 4. (a) Use Laplace transform to find the solution of the initial value problem
y 00 + 4y 0 + 5y = 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 + 4s + 5)Y (s) − s − 4 = L{δ(t − 3)} = e−3s .
Y (s) =
Y (s) =
1
s+4
+
e−3s .
s2 + 4s + 5 s2 + 4s + 5
(s + 2) + 2
1
+
e−3s .
2
(s + 2) + 1 (s + 2)2 + 1
Thus
y(t) = e−2t cos t + 2e−2t sin t + u3 (t)e−2(t−3) sin(t − 3).
Z
(3 points)
t
(b) Find the Laplace transform of the function f (t) =
sin(t − τ )e−2τ dτ .
0
Answer.
f (t) = sin t ∗ e−2t .
L[f ] = L[sin t] · L[e−2t ] =
s2
1
1
·
.
+1 s+2
March 2010
Math 215 Midterm Exam II
Page 6 of 6
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)2 + b2
11. tn eat , n positive integer
n!
,
(s − a)n+1
12. uc (t)
e−cs
,
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
s>a
s>a
s>0
1 s
F
,
c
c
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)