Be sure this exam has 8 pages including the cover and Table of Laplace transforms
The University of British Columbia
MATH 215/255
Midterm Exam II – November 2015
Name
Signature
Student Number
Course Number
Circle Section:
101 Zhao
102 Tsai
103 Kolokolnikov
104 Zhao
This exam consists of 6 questions worth 40 marks. No notes nor calculators.
Problem
Points
1
12
2
6
3
6
4
5
5
6
6
5
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.
October 2015
Math 215/255 Midterm 2
Page 2 of 8
(12 points) 1. Use the method of undetermined coefficients to find a particular solution to: (a) y 00 − 4y = ex ;
(b) y 00 − 4y = sin(2x); (c) y 00 − 4y = e2x .
Answer. (a) Let y = Aex then A(1 − 4) = 1 so that y = − 13 ex . (b) Let y = A sin(2x) then
−4A − 4A = 1 so that y = − 18 sin(2x). (c) Since e2x is a solution to the homogeneous
problem, we try an anzatz y = Axe2x . We find 4A = 1 so that y = 14 xe2x .
October 2015
Math 215/255 Midterm 2
Page 3 of 8
(6 points) 2. It is known that the homogeneous ODE y 00 + x1 y 0 − x42 y = 0 admits two solutions, y1 = x2 and
y2 = x−2 . Find a particular solution to the inhomogeneous ODE y 00 + x1 y 0 − x42 y = 1.
u01 x2 + u02 x−2 = 0
so that
2u01 x − 2u02 x−3 = 1
1 4
u01 = 1/(4x); u02 = −x3 /4; u1 = 41 log x, u2 = − 16
x ; so that yp = 41 x2 log x −
1 2
yp = 4 x log x (by removing the homogeneous part).
Answer.
Let yp = u1 y1 + u2 y2 ; then we solve:
1 2
16 x
or
October 2015
(6 points) 3. Let G(s) =
Answer.
Math 215/255 Midterm 2
1
1
s−1 (s−1)2 +4 .
Page 4 of 8
Find L−1 {G(s)}.
Solution 1. Use partial fractions to write
1
A
1
B + C (s − 1)
=
;
+
2
s − 1 (s − 1) + 4
s−1
(s − 1)2 + 4
the constants A, B, C are determined below. Therefore
1
1
B
−1
L
= Aet + et sin(2t) + Cet cos(2t).
s − 1 (s − 1)2 + 4
2
The constants A, B, C satisfy A u2 + 4 + (B + Cu) u = 1 (where u = s − 1). Plugging u = 0
we get A = 1/4. Then the coefficient in front of u yields B = 0 and coefficient in front of u2
yields C = −A = −1/4, so that
L−1 {G(s)} =
et
(1 − cos(2t)) .
4
Solution 2. We have:
1
s−1
1
(s − 1)2 + 4
−1
L
L
−1
= et ;
1 t
e cos (2t) ;
2
=
so by convolution property,
1
1
−1
L
=
s−1
(s − 1)2 + 4
=
=
=
t
e
∗
1 t
e sin (2t)
2
Z
1 t t−τ τ
e e sin (2τ ) dτ
2 0
Z
1 t t
e
sin (2τ ) dτ
2
0
1 t
e (1 − cos (2t)) .
4
Solution 3. First, note that
−1
L
1 1
s s2 + 4
Z
L
=
0
=
=
t
−1
1
2
s +4
Z
1 t
sin (2t)
2 0
1
(1 − sin (2t))
4
Therefore by shifting property,
1
1
1
−1
t −1 1
L
= eL
s − 1 (s − 1)2 + 4
s s2 + 4
1 t
=
e (1 − cos (2t)) .
4
October 2015
Math 215/255 Midterm 2
Page 5 of 8
(5 points) 4. Let
f (t) =
et ,
0,
t≤2
.
t≥2
Find L{f (t)}.
Answer.
Write
f (t) = (1 − u(t − 2))et .
Next note that
e2 e−2s
L u(t − 2)et = L u(t − 2)et+2−2 = e−2s L et+2 = e2 e−2s L et =
.
s−1
So then
L{f (t)} =
1
1 − e2 e−2s .
s−1
October 2015
Math 215/255 Midterm 2
Page 6 of 8
(6 points) 5. Consider the ODE x00 − 2x0 + 6x = f (t) subject to initial conditions x(0) = 0 = x0 (0), where
f (t) is as given in question 4.
(a) Let X(s) = L{f (t)}. Find X(s).
Answer.
s2 X − 2sX + 6X =
so that
X=
or
X=
(s −
1 − e2 e−2s
s−1
1
1 − e2 e−2s
− 2s + 6)
1)(s2
1
1 − e2 e−2s
2
(s − 1)((s − 1) + 5)
(b) Determine x(t). HINT: You should find the answer to question 3 useful here.
n
o
1
1
Answer. First do L−1 (s−1)
2
(s−1) +5
Similarly to Q3, we have
1
1
1
=
2
(s − 1) (s − 1) + 5
5
so that
L−1
1
1
(s − 1) (s − 1)2 + 5
=
1
s−1
−
s − 1 (s − 1)2 + 5
1
5
√ 1
et − √ et cos
5t
5
and then by shifting property,
√
1
e−2s
1 t−2
1 t−2
−1
5 (t − 2)
u(t − 2)
L
=
e
− √ e cos
(s − 1) (s − 1)2 + 5
5
5
so that
x = (1) − e2 (2)
√ 1 √
1 t
1 t
1 t
t
=
5t
−
5 (t − 2)
u(t − 2).
e − √ e cos
e − √ e cos
5
5
5
5
(1)
(2)
October 2015
Math 215/255 Midterm 2
Page 7 of 8
(5 points) 6. Use Laplace’s transform and convolution to find the solution to the ODE x00 + x = f (t)
subject to initial conditions x(0) = 0, x0 (0) = 3. Here, f (t) is any function and your solution
should be in terms of certain integrals of f (t).
F (s)+3
1+s2
where X(s) = L{x(t)}, F (s) = L{f (t)}. Therefore
Rt
x(t) = sin(t) ∗ f (t) + 3 sin(t) = 0 sin(t − τ )f (τ )dτ + 3 sin(t).
Answer.
We get X =
October 2015
Math 215/255 Midterm 2
Page 8 of 8
Table of Laplace transforms
f (t) = L−1 {F (s)}
F (s) = L{f (t)}
1
, s>0
s
1
, s > −a
e−at
s+a
n!
tn , n positive integer
, s>0
n+1
s a
sin(at)
, s>0
s 2 + a2
s
cos(at)
, s>0
2
s + a2
a
sinh(at)
, s > |a|
s 2 − a2
s
, s > |a|
cosh(at)
s 2 − a2
e−as
, s>0
u(t − a)
s
u(t − a)f (t − a)
e−as F (s)
1. 1
2.
3.
4.
5.
6.
7.
8.
9.
10. e−at f (t)
Z t
11.
f (t − τ )g(τ )dτ
0
Z t
12.
f (τ )dτ
F (s + a)
F (s)G(s)
13. δ(t − a)
F (s)
s
e−as
14. f (n) (t)
sn F (s) − sn−1 f (0) − ... − f (n−1) (0)
0
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).