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The University of British Columbia
MATH 215/255
Midterm Exam II – November 2014
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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
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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 2014
Math 215/255 Midterm Exam II
Page 2 of 6
(5 points) 1. (a) Find a particular solution to y 00 + 3y = 9x2 + 3x.
Answer.
Let’s try yp = Ax2 + Bx + C, so that we have to solve
2A + 3(Ax2 + Bx + C) = 9x2 + 3x.
Then we have
3A = 9,
3B = 3,
and
2A + 3C = 0.
Therefore A = 3, B = 1 and C = −2. So a particular solution can be
yp = 3x2 + x − 2 .
(5 points)
(b) Find the inverse Laplace transform f (t) of F (s) =
Answer.
4
.
+ 4)
s(s2
By the partial fractions, we have
4
A Bs + C
A(s2 + 4) + s(Bs + C)
(A + B)s2 + Cs + 4A + C
=
+
=
=
.
s(s2 + 4)
s
s2 + 4
s(s2 + 4)
s(s2 + 4)
Then
A + B = 0,
C = 0,
and
4A + C = 4.
So we have
A = 1,
That is we have
and B = −1,
and C = 0.
4
1
s
= − 2
.
s(s2 + 4)
s s +4
By looking up the table, we know that
s
−1 1
−1
L
= 1, and L
= cos(2t).
s
s2 + 4
Then we know that
−1
f (t) = L
−1
[F (s)] = L
1
s
−1
−L
= 1 − cos(2t).
s
s2 + 4
November 2014
Math 215/255 Midterm Exam II
Page 3 of 6
(5 points) 2. (a) Find a particular solution to y 00 + 2y 0 + y = 2e−x using the method of undetermined
coefficients.
Answer.
Trying y = Ae−x or y = Axe−x fails (one obtains y 00 + 2y 0 + y = 0), so we try
2
−x
y = Ax e . We have
y = Ax2 e−x ,
y 0 = A(2x − x2 )e−x ,
y 00 = A(2 − 4x + x2 )e−x .
Therefore
y 00 + 2y 0 + y = Ae−x (2 − 4x + x2 + 4x − 2x2 + x2 ) = 2Ae−x .
We set A = 1 to make the RHS equal to 2, and obtain the particular solution
y = x2 e−x .
(5 points)
(b) Find a particular solution to the same equation y 00 + 2y 0 + y = 2e−x , this time using the
method of variation of parameters.
Answer.
The associated homogeneous equation y 00 + 2y + 1 = 0 has a characteristic
equation (r + 1)2 = 0, with unique root r = 1. Therefore it admits two linearly
independent solutions
y1 = e−x ,
y2 = xe−x .
Following the method of variation of parameters, we try y = y1 u1 + y2 u2 , where u1 and
u2 are two functions satistfying the system of equations
y1 u01 + y2 u02 = 0,
y10 u01 + y20 u02 = 2e−x .
Computing the derivatives of y1 , y2 , this becomes
e−x u01 + xe−x u02 = 0
−e−x u01 + (1 − x)e−x u02 = 2e−x .
The first line gives u01 = −xu02 , which can be inserted in the second line to obtain
e−x u02 = 2e−x . Therefore u02 = 2 and u2 = 2x, and consequently u01 = −2x and u1 = −x2 .
Finally, we obtain the particular solution
y = −x2 e−x + 2x · xe−x = x2 e−x .
November 2014
Math 215/255 Midterm Exam II
Page 4 of 6
3. Suppose the displacement x(t) of a damped mass-spring system subject to sinusoidal forcing
of amplitude F0 is modelled by:
x00 + 4x0 + 13x = F0 sin(t),
(5 points)
x(0) = 0,
x0 (0) = 1.
(a) Find the solution x(t) when F0 = 0 (no forcing).
Answer.
The characteristic equation is r2 + 4r + 13 = 0. Its roots are
r = −2 ± 3i.
The general solution to the homogeneous equation is
x(t) = e−2t (c1 cos 3t + c2 sin 3t).
(1)
Thus
x0 (t) = e−2t [(−2c1 + 3c2 ) cos 3t + (−3c1 − 2c2 ) sin 3t].
Invoking the initial conditions,
0 = c1 ,
1 = −2c1 + 3c2 .
Thus c1 = 0, c2 = 1/3, and
1
x(t) = e−2t sin 3t.
3
(5 points)
(2)
(b) Now take F0 = 40, and find the steady periodic solution (the part of the solution x(t)
which remains as t → ∞). Do not find the transient part.
Answer.
The steady periodic solution is a particular solution of the form
U (t) = A cos t + B sin t.
In view of (1), there is no resonance and it needs no modification.
We have U 0 = B cos t − A sin t, U 00 = −U , and
U 00 + 4U 0 + 13U = (12A + 4B) cos t + (12B − 4A) sin t = 40 sin t.
Thus
12A + 4B = 0,
12B − 4A = 40.
We get
B = −3A,
A = −1,
B = 3.
Thus
U (t) = − cos t + 3 sin t.
(3)
November 2014
Math 215/255 Midterm Exam II
Page 5 of 6
(5 points) 4. (a) Find the Laplace transform of the solution to the following initial value problem:
x00 − x0 + x = f (t),
where
f (t) =
and x0 (0) = 1,
x(0) = 0,
1, if 0 ≤ t < 1,
t, if t ≥ 1.
Answer. Let X(s) = L[x(t)], apply the Laplace transform on the both sides of
x00 − x0 = f (t), then L[x00 ] − L[x0 ] + L[x] = L[f (t)]. By the transforms of derivatives (No.
14 in the table), we have
L[x0 ] = sL[x] − x(0) = sX(s),
and L[x00 ] = s2 L[x] − sx(0) − x0 (0) = s2 X(s) − 1.
Then we have s2 X(s) − 1 − sX(s) + X(s) = L[f (t)], tht is,
X(s) =
1 + L[f (t)]
.
s2 − s + 1
By the definition of f (t), we know that
f (t) = 1 − u(t − 1) + tu(t − 1) = 1 + (t − 1)u(t − 1).
By looking up the table, we have
L[f (t)] = L[1] + L[(t − 1)u(t − 1)] =
1
1 e−s
+ e−s L[t] = + 2 .
s
s
s
So we know that
−s
1 + 1s + es2
1
1
e−s
X(s) = 2
= 2
+
+
.
s −s+1
s − s + 1 s(s2 − s + 1) s2 (s2 − s + 1)
(5 points)
(b) Solve the initial value problem:
x00 + x0 = δ(t − 1),
x(0) = x0 (0) = 0.
Answer. Let X(s) = L[x(t)], apply the Laplace transform on the both sides of
x00 − x0 = δ(t − 1), then L[x00 ] + L[x0 ] = L[δ(t − 1)]. By the transforms of derivatives (No.
14 in the table), we have
L[x0 ] = sL[x] − x(0) = sX(s),
and L[x00 ] = s2 L[x] − sx(0) − x0 (0) = s2 X(s).
Then we have
s2 X(s) + sX(s) = L[δ(t − 1)].
By looking up the table, we have L[δ(t − 1)] = e−s , which implies that
X(s) =
e−s
.
s2 + s
By the partial fractions, we have
s2
1
1
1
= −
.
+s
s s+1
By looking up the table, we have
−s −s e
1
−1 e
−1
−1
x(t) = L
−L
= u(t−1)−u(t−1)L
(t−1) = u(t−1)−u(t−1)e−(t−1) .
s
s+1
s+1
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).