Be sure this exam has 6 pages including the cover
The University of British Columbia
MATH 215/255
Midterm Exam II – November 2015
Signature
Name
Student Number
Circle Section:
Course Number
101 Zhao
102 Tsai
103 Kolokolnikov
104 Zhao
This exam consists of 4 questions worth 40 marks. No notes nor calculators.
Problem
1
2
3
4
Total
Points:
13
7
10
10
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 2015
Math 215/255 Midterm 2
Page 2 of 6
(6 points) 1. (a) Use the method of undetermined coefficients to find a particular solution to the
differential equation:
y 00 + 2y 0 + 2y = 5 sin(x).
Answer. The characteristic equation of y 00 + 2y 0 + 2y = 0 is r2 + 2r + 2 = 0, then
r1 = −1 + i and r2 = −1 − i. Let yp (x) = A cos(x) + B sin(x) be a particular solution to
y 00 + 2y 0 + 2y = 5 sin(x), then
yp0 (x) = −A sin(x) + B cos(x),
and yp00 (x) = −A cos(x) − B sin(x).
Then we have
yp00 + 2yp0 + 2yp = −A cos(x) − B sin(x) + 2(−A sin(x) + B cos(x)) + 2(A cos(x) + B sin(x))
= −A cos(x) − B sin(x) − 2A sin(x) + 2B cos(x) + 2A cos(x) + 2B sin(x)
= (A + 2B) cos(x) + (B − 2A) sin(x)
= 5 sin(x).
So we have
A + 2B = 0,
and B − 2A = 5.
Hence A = −2 and B = 1. Then we have
yp (x) = −2 cos(x) + sin(x).
(7 points)
(b) Use the method of variation of parameters to find a particular solution to problem:
y 00 − y = 2e−x .
Answer. The characteristic equation of y 00 − y = 0 is r2 − 1 = 0, then r1 = 1 and
r2 = −1. So the general solution to y 00 − y = 0 is yc (x) = C1 ex + C2 e−x . Now let
yp (x) = u1 (x)ex + u2 (x)e−x be a particular solution to y 00 − y = 2e−x for some functions.
By the method of variation of parameters, we have
u01 (x)ex + u02 (x)e−x = 0
u01 (x)ex − u02 (x)e−x = 2e−x .
So we get
u01 (x) = e−2x ,
and u02 (x) = −1.
So we can take
Z
u1 (x) =
−2x
e
1
dx = − e−2x ,
2
Z
and u2 (x) = −
1 dx = −x.
Hence we know that
1
1
yp (x) = u1 (x)ex + u2 (x)e−x = − e−2x · ex − xe−x = − e−x − xe−x .
2
2
November 2015
Math 215/255 Midterm 2
Page 3 of 6
(7 points) 2. Find a particular solution to the problem:
y 00 − 3y 0 = 2ex − 6x.
Hint: You can use either method of undetermined coefficients or variation of parameters.
Answer. Method I: The characteristic equation of y 00 − 3y 0 = 0 is r2 − 3r = 0, then r1 = 0
and r2 = 3. Let yp (x) = Aex + x(Bx + C) = Aex + Bx2 + Cx be a solution to
y 00 − 3y 0 = 2ex − 6x, then
yp0 (x) = Aex + 2Bx + C,
and yp00 (x) = Aex + 2B
So we get
yp00 − 3yp0 = Aex + 2B − 3(Aex + 2Bx + C)
= Aex + 2B − 3Aex − 6Bx − 3C
= −2Aex − 6Bx + 2B − 3C
= 2ex − 6x.
So we get
−2A = 2,
6B = 6,
2B − 3C = 0.
So A = −1, B = 1 and C = 23 . Hence the solution is:
2
yp (x) = −ex + x2 + x.
3
Method II: The characteristic equation of y 00 − 3y 0 = 0 is r2 − 3r = 0, then r1 = 0 and
r2 = 3. So the general solution to y 00 − 3y 0 = 0 is yc (x) = C1 + C2 e3x . Now let
yp (x) = u1 (x) + u2 (x)e3x be a particular solution to y 00 − 3y 0 = 2ex−6 for some functions. By
the method of variation of parameters, we have
u01 (x) + u02 (x)e3x = 0
3u02 (x)e3x = 2ex − 6x.
So we get
2
u01 (x) = − ex + 2x,
3
2
and u02 (x) = e−2x − 2xe−3x .
3
So we can take
2 x
u1 (x) =
− e + 2x
3
Z 2 −2x
1
u2 (x) =
e
− 2xe−3x dx = − e−2x +
3
3
Z 2
dx = − ex + x2
3
2 −3x 2 −3x
xe
+ e .
3
9
Hence we know that
yp (x) = u1 (x) + u2 (x)e3x
2 x
1 −2x 2 −3x 2 −3x
2
3x
− e
+ xe
+ e
= − e +x +e
3
3
3
9
2
2
= −ex + x2 + x + .
3
9
November 2015
Math 215/255 Midterm 2
Page 4 of 6
(10 points) 3. Use the Laplace transform to solve the following initial value problem:
y 00 + 2y 0 + 2y = δ(t − 1),
y(0) = 1,
y 0 (0) = 2.
Answer. Let Y (s) = L[y(t)](s), apply the Laplace transform on the both sides of
y 00 + 2y 0 + 2y = δ(t − 1), then L[y 00 ] + 2L[y 0 ] + 2L[y] = L[δ(t − 1)]. By the transform of
derivatives, since y(0) = 1 and y 0 (0) = 0, then
L[y 00 ] = s2 L[y] − sy(0) − y 0 (0) = s2 Y (s) − s − 2,
and L[y 0 ] = sL[y] − y(0) = sY (s) − 1.
By looking up the table, we know that L[δ(t − 1)] = e−s . So we get
s2 Y (s) − s − 2 + 2sY (s) − 2 + 2Y (s) = e−s .
So we get
Y (s) =
s + 4 + e−s
s+4
e−s
=
+
.
s2 + 2s + 2
s2 + 2s + 2 s2 + 2s + 2
Notice that s2 + 2s + 2 = (s + 1)2 + 1, by the first and second shifting properties, we have
y(t) = L−1 [Y (s)](t)
s+1+3
e−s
−1
−1
= L
(t) + L
(t)
(s + 1)2 + 1
(s + 1)2 + 1
1
−t −1 s + 3
−1
= e L
(t) + u(t − 1)L
(t − 1)
s2 + 1
(s + 1)2 + 1
s
3
−t
−1
−1
= e
L
(t) + L
(t) + u(t − 1)e−(t−1) sin(t − 1)
s2 + 1
s2 + 1
= e−t [cos(t) + 3 sin(t)] + u(t − 1)e−(t−1) sin(t − 1).
November 2015
Math 215/255 Midterm 2
Page 5 of 6
(5 points) 4. (a) Use the Laplace transform and convolution to find the solution to x00 + x = f (t) subject
to initial value conditions x(0) = 0 and x0 (0) = 3. Here f (t) is any function and your
solution should be in terms of certain integrals of f (t).
Answer. Let X(s) = L[x(t)](s) and F (s) = L[f (t)](s), apply the Laplace transform on
the both sides of x00 + x = f (t), then L[x00 ] + L[x] = L[f (t)](s) = F (s). By the transform
of derivative, then
L[x00 ] = s2 L[x] − sx(0) − x0 (0) = s2 X(s) − 3,
and L[x0 ] = sL[x] − x(0) = sX(s).
Hence we get s2 X(s) − 3 + X(s) = F (s), which implies that
F (s) + 3
F (s)
3
= 2
+
.
s2 + 1
s + 1 s2 + 1
1
−1
By looking up the table, we know that L
(t) = sin(t). By the transform of
s2 + 1
convolution, we know that
1
3
−1
−1
x(t) = L [X(s)](t) = L
F (s) · 2
+
(t)
s + 1 s2 + 1
1
−1
−1
= L [F (s)] ∗ L
(t) + 3 sin(t)
s2 + 1
Z t
=
f (τ ) sin(t − τ ) dτ + 3 sin(t).
X(s) =
0
(5 points)
(b) Find the Laplace transform of f (t) =
Answer.
t,
if 0 ≤ t < 1,
.
2 − t, if t ≥ 1.
It’s easy to see that
f (t) = t[u(t − 0) − u(t − 1)] + (2 − t)[u(t − 1) − u(t − ∞)]
= t − tu(t − 1) + (2 − t)u(t − 1)
= t + (2 − 2t)u(t − 1)
= t − 2(t − 1)u(t − 1).
By the second shifting property, we have
L[f (t)](s) = L[t] − 2L[(t − 1)u(t − 1)]
1
= 2 − 2e−s L[t](s)
s
1
2e−s
= 2− 2 .
s
s
November 2015
Math 215/255 Midterm 2
Page 6 of 6
Table of Laplace transforms
f (t) = L−1 {F (s)}
F (s) = L{f (t)}
1
, s>0
s
1
e−at
, s > −a
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|
2
s − a2
s
cosh(at)
, s > |a|
2
s − a2
e−as
u(t − a)
, s>0
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
f (t − τ )g(τ )dτ
11.
Z0 t
f (τ )dτ
12.
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).