Be sure this exam has 15 pages including the cover
The University of British Columbia
MATH 215/255, Sections 101–105
Final Exam – December 2014
Family Name
Given Name
Student Number
Signature
Circle Section:
101 Henriot
102 Tsai
103 Shih
104 Dontsov
105 Zhao
No notes nor calculators.
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6. Candidates must follow any additional examination rules or
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Problem
Points
1
8
2
12
3
10
4
13
5
8
6
12
7
10
8
12
9
15
Total:
100
Score
December 2014
Math 215/255 Sections 101–105 Final Exam
Page 2 of 15
1. Solve the initial value problem
(3xy + y 2 ) + (x2 + xy)y 0 = 0,
(3 points)
y(1) = 2.
(a) Verify that µ(x) = x is an integrating factor, that is, x(3xy + y 2 )dx + x(x2 + xy)dy = 0 is
exact.
Answer.
Let M = x(3xy + y 2 ) = 3x2 y + xy 2 and N = x(x2 + xy) = x3 + x2 y, then
My = 3x2 + 2xy,
and Nx = 3x2 + 2xy.
Then My = Nx . So x(3xy + y 2 )dx + x(x2 + xy)dy = 0 is exact.
(5 points)
(b) Solve the initial value problem.
Answer.
Since M dx + N dy = 0 is exact. Then there exists some φ(x, y) such that
φx = M = 3x2 y + xy 2 ,
and φy = N = x3 + x2 y.
Since φx = 3x2 y + xy 2 , then
Z
1
φ(x, y) = (3x2 y + xy 2 ) dx + g(y) = x3 y + x2 y 2 + g(y).
2
Since φy = x3 + x2 y, then
x3 + x2 y = φy = x3 + x2 y + g 0 (y),
which implies that g 0 (y) = 0, so g(y) = constant. Hence
1
x3 y + x2 y 2 = C.
2
Since y(1) = 2, then 2 + 2 = C, that is, C = 4. Therefore, the solution is:
1
x3 y + x2 y 2 = 4.
2
December 2014
Math 215/255 Sections 101–105 Final Exam
Page 3 of 15
2. A second order chemical reaction can be modeled by the equation
dy
= α(y − p)(y − q),
dx
where α, p and q are constants
(4 points)
(a) Assume that α > 0 and p > q > 0. Find equilibrium points and classify stabilities of
these equilibrium points.
Answer.
Let’s solve α(y − p)(y − q) = 0, since α > 0 and p > q > 0, then y = p or
y = q. So we have two equilibrium points:
y = p,
and y = q.
The phase diagram is:
So y = p is unstable, y = q is stable.
(4 points)
(b) Assume that α = 1, p = 0, q = 1 and y(0) = −1, solve the initial value problem and
determine the limiting value of y(x) as x → ∞.
Answer.
Since α = 1, p = 0 and q = 1, then our differential equation becomes
dy
= y(y − 1).
dx
dy
1
1
So
= dx, that is,
−
dy = dx. Then
y(y − 1)
y−1 y
y − 1
= x + C.
ln y Since y(0) = −1 < 0, then y(x) < 0 for all x ≥ 0, then
y−1
ln
= x + C.
y
Then
y(x) =
1
.
1 − ex+C
Since y(0) = −1, then C = ln 2. So solution is y(x) =
lim y(x) = 0.
x→∞
1
, which implies that
1 − 2ex
December 2014
(4 points)
Math 215/255 Sections 101–105 Final Exam
Page 4 of 15
(c) Assume that α = 1, p = q = 0 and y(0) = 1. Use Euler’s method to approximate y(2)
with the step size h = 1.
Answer.
Since α = 1, p = q = 0 and y(0) = 1, then our differential equation becomes
dy
= y2,
dx
and y(0) = 1.
Let f (y) = y 2 , x0 = 1 and y0 = y(x0 ) = y(0) = 1, then
x1 = x0 + h
= 1
y1 = y0 + f (y0 )h
= 1 + 12
= 2
x2 = x0 + 2h
= 2
y2 = y1 + f (y1 )h
= 2 + 22 · 1
= 6.
December 2014
Math 215/255 Sections 101–105 Final Exam
(5 points) 3. (a) Solve y 00 + 4y 0 + 13y = 0,
Page 5 of 15
y 0 (0) = 5.
y(0) = 2,
The characteristic equation is r2 + 4r + 13 = (r + 2)2 + 9 = 0 with roots
Answer.
r = −2 ± 3i.
(1pt)
y(t) = c1 e−2t cos 3t + c2 e−2t sin 3t.
(1pt)
The general solution is
Hence
y 0 (t) = (−2c1 + 3c2 )e−2t cos 3t + (−3c1 − 2c2 )e−2t sin 3t.
The initial conditions give
c1 + 0 = 2,
(−2c1 + 3c2 ) + 0 = 5.
(2pt)
Thus c1 = 2, c2 = 3, and
y(t) = 2e−2t cos 3t + 3e−2t sin 3t.
(5 points)
(1pt)
(b) Find a particular solution of
y 00 + 3y 0 = e−3t
using either the method of undetermined coefficients, or the method of variation of
parameters.
Answer. #1. (method of undetermined coefficients.) The characteristic equation is
r2 + 3r = 0, with roots −3, 0. Hence
yc (t) = c1 + c2 e−3t .
For f =
set
e−3t ,
the initial guess is Y (t) =
ae−3t .
However
(2pt)
ae−3t
is part of yc and hence we
Y (t) = ate−3t .
(1pt)
We have
Y 0 = (a − 3at)e−3t ,
Y 00 = (−3a − 3a + 9at)e−3t ,
Y 00 + 3Y 0 = −3ae−3t .
Thus a =
− 31
(1pt)
and
1
Y (t) = − te−3t .
3
Answer #2. (method of variation of parameters.) The characteristic equation is
r2 + 3r = 0, with roots −3, 0. Hence two independent solutions are
y1 (t) = 1,
y2 (t) = e−3t .
(1pt)
(2pt)
A particular solution is of the form Y (t) = y1 (t)u1 (t) + y2 (t)u2 (t) with
u01 + e−3t u02 = 0,
Thus
We can solve u1 = − 19 e−3t
0 u01 − 3e−3t u02 = e−3t .
1
1
u02 = − , u01 = e−3t .
3
3
1
and u2 = − 3 t, and hence
1
1
1
1
Y = 1(− e−3t ) + e−3t (− t) = − e−3t − te−3t .
9
3
9
3
(2pt)
(1pt)
December 2014
Math 215/255 Sections 101–105 Final Exam
Page 6 of 15
4. Consider a vibrating system described by the initial value problem
u00 + cu0 + 4u = cos 2t,
u(0) = 0,
u0 (0) = 2.
where c > 0 is the damping coefficient.
(5 points)
(a) Find the steady periodic part of the solution (the part of the solution which remains as
t → ∞) of this problem, and find its amplitude. Do not find the transient part.
Answer.
The steady state is the particular solution of the form
U (t) = a cos 2t + b sin 2t.
Thus U 0 = 2b cos 2t − 2a sin 2t and U 00 = −4a cos 2t − 4b sin 2t. Substitution into the
equation yields
c(2b cos 2t − 2a sin 2t) = cos 2t.
Thus a = 0, b =
1
2c ,
and
U (t) =
1
sin 2t.
2c
Its amplitude is
1
.
2c
(2 points)
(b) Let A(c) denote the maximum amplitude of the steady state solutions of the systems
u00 + cu0 + 4u = cos ωt,
u(0) = 0,
u0 (0) = 2
among all possible ω > 0. What happens to A(c) as c → 0+ ? Explain why.
Hint. You do not need to solve A(c) explicitly.
Answer.
A(c) goes to infinity, because A(c) ≥
1
2c
and limc→0+
1
2c
= ∞.
December 2014
(6 points)
Math 215/255 Sections 101–105 Final Exam
Page 7 of 15
(c) Find a particular solution of
y 00 + y =
Hint:
R
Answer.
tan t dt = − ln | cos t|,
R
1
,
cos t
−
π
π
<t< .
2
2
cot t dt = ln | sin t|
The homogeneous equation has two independent solutions
y1 (t) = sin t,
y2 (t) = cos t.
Using the method of variation of parameters, a particular solution is given by
Y (t) = y1 (t)u1 (t) + y2 (t)u2 (t), in which
y1 u01 + y2 u02 = 0,
y10 u01 + y20 u02 = f.
That is,
sin t u01 + cos t u02 = 0,
cos t u01 − sin t u02 =
1
.
cos t
Thus
u01 = 1,
u02 = − tan t
Z
u1 (t) = 1dt = t,
Z
u2 (t) = −
tan t dt = ln | cos t| = ln(cos t)
Hence a particular solution is
Y (t) = t sin t + cos t ln(cos t)
(−
π
π
<t< )
2
2
December 2014
Math 215/255 Sections 101–105 Final Exam
Page 8 of 15
(8 points) 5. Use the Laplace transform to solve the system x00 (t) + 2x0 (t) + 2x(t) = 2 with initial
conditions x(0) = 0, x0 (0) = 0.
Answer.
Writing X(s) for the Laplace transform of x(t), we obtain
(s2 + 2s + 2)X(s) =
2
s
and therefore
X(s) =
s(s2
2
.
+ 2s + 2)
By partial fraction decomposition, we have
X(s) =
s2
−s − 2
1
+ .
+ 2s + 2 s
By completion of the square, this can be put in a convenient form to later apply the first
shifting formula:
X(s) =
−(s + 1)
1
1
−
+ .
2
2
(s + 1) + 1 (s + 1) + 1 s
Taking inverse Laplace transforms and applying the said shifting formula, we obtain
n 1 o
s o
−t −1
(t)
−
e
L
(t) + 1
s2 + 1
s2 + 1
= −e−t (cos t + sin t) + 1.
x(t) = −e−t L−1
n
December 2014
Math 215/255 Sections 101–105 Final Exam
Page 9 of 15
(6 points) 6. (a) Solve the system x00 (t) + 4x(t) = δ(t − 2) with initial conditions x(0) = 0 and x0 (0) = 0.
Answer.
Writing X(s) = L{x(t)} and taking Laplace transforms, we obtain
(s2 + 4)X(s) = e−2s
⇒
X(s) =
e−2s
.
s2 + 4
By the second shifting formula, we have therefore
1 o
(t − 2) · u(t − 2)
s2 + 4
= 21 sin(2(t − 2)) · u(t − 2).
x(t) = L−1
(6 points)
n
(b) Find the Laplace transform of


1 if t < 1
f (t) = t if 1 ≤ t < 2


0 if t ≥ 2.
Answer.
We can rewrite the piecewise definition of f using Heavyside functions:
f (t) = 1 · (1 − u(t − 1)) + t · u(t − 1) − u(t − 2) + 0 · u(t − 2)
= 1 + (t − 1)u(t − 1) − tu(t − 2)
= 1 + (t − 1)u(t − 1) − (t − 2)u(t − 2) − 2u(t − 2).
This last expression makes it easier to apply the second shifting formula:
2e−2s
1
+ e−s L{t}(s) − e−2s L{t}(s) −
s
s
1
1 e−s
2
= +
− e−2s 2 +
.
s
s
s
s
F (s) =
December 2014
Math 215/255 Sections 101–105 Final Exam
Page 10 of 15
(7 points) 7. (a) Find general solution of
dx
= Ax,
dt
A=
p 4
−1 p
,
assuming that p is real.
Answer. We first find the eigenvalues of A by solving det(A − λI) = (p − λ)2 + 4 = 0.
This gives two complex conjugated eigenvalues λ1,2 = p ± 2i. The eigenvector that
corresponds to λ1 = p + 2i is
−2i 4
0
2
v1 =
⇒ v1 =
.
−1 −2i
0
i
The general solution is then given by
2 2 (p+2i)t
e(p+2i)t ,
e
+ C2 Im
x = C1 Re
i
i
which can be simplified to
x = C1 e
(3 points)
pt
2 cos(2t)
− sin(2t)
+ C2 e
pt
2 sin(2t)
cos(2t)
.
(b) Describe the behaviour of the system (do not draw phase portrait) for all possible real
values of p.
Answer. Since the eigenvalues are λ1,2 = p ± 2i and Re(λ1,2 ) = p is real, there are only
three cases. If p > 0, then the system behaves like a spiral source. If p < 0, then the system
behaves like a spiral sink. If p = 0, then the system behaves like a center.
December 2014
Math 215/255 Sections 101–105 Final Exam
Page 11 of 15
(12 points) 8. Solve
dx
= Ax −
dt
0
3t + 2
,
x(0) =
0
1
,
A=
0 1
3 2
.
Answer. We start by finding the complimentary solution. The eigenvalues of matrix A can
be found from −λ(2 − λ) − 3 = 0, which gives λ1 = 3 and λ2 = −1. The corresponding
eigenvectors are
−3 1
0
1
v1 =
⇒ v1 =
,
3 −1
0
3
1 1
0
1
v2 =
⇒ v2 =
.
3 3
0
−1
The complimentary solution is then
xc = C1
1
3
3t
e + C2
1
−1
e−t .
Method 1. By undetermined coefficients, we seek for the particular solution in the form
xp = at + b, where a and b are constant vectors with unknown components. Substituting xp
into the equation gives
0
0
a = Aat + Ab −
−
t,
2
3
which implies that
a = Ab −
0
2
,
Aa =
0
3
.
Computing the the inverse of A gives
1
=
3
−1
A
−2 1
3 0
,
which can be used to find
a=A
−1
0
3
=
1
0
−1
,
b=A
1
2
=
0
1
.
So, the particular solution is
xp =
t
1
.
Method 2. To use variation of parameters, we first need to find fundamental matrix solution.
Using the obtained complimentary solution, one has
3t
1 e−3t e−3t
e
e−t
−1
X=
,
X =
.
3e3t −e−t
3et −et
4
Particular solution can be found using
3t
R
Z
1
0
e
e−t
(−3t
− 2)e−3t dt
−1
R
xp = X X
dt =
.
−3t − 2
(3t + 2)et dt
4 3e3t −e−t
December 2014
Math 215/255 Sections 101–105 Final Exam
Page 12 of 15
The integrals can be calculated as
R
(−3t
− 2)e−3t dt
(t + 1)e−3t
R
=
.
(3t + 2)et dt
(3t − 1)et
Finally, the particular solution is
3t
1
e
e−t
(t + 1)e−3t
t
xp =
=
.
3t
−t
t
−e
(3t − 1)e
1
4 3e
The general solution is
x = C1
1
3
1
3
3t
e + C2
1
−1
e
−t
+
0
1
t
1
0
1
.
Using the initial condition gives
x(0) = C1
+ C2
1
−1
+
which implies that C1 = C2 = 0. Finally, the solution is
t
.
x=
1
=
,
December 2014
Math 215/255 Sections 101–105 Final Exam
Page 13 of 15
9. Consider the nonlinear system
y2
dx
= x2 −
+ 1,
dt
2
(5 points)
dy
= −4x − 2y.
dt
(1)
(a) Find the equilibria (critical points) for the system (1).
2
Answer.
The equilibria are solutions of the equations x2 − y2 + 1 = 0 and
−4x − 2y = 0. From the second equation, we obtain y = −2x. Substituting this into the
first equation, we have −x2 + 1 = 0. Hence the equilibria are (1, −2) and (−1, 2).
(6 points)
(b) Find the Jacobian (partial derivative) matrix for the system (1), compute the linearized
system at each equilibrium, and compute the eigenvalues for each of the coefficient
matrices.
Answer.
The Jacobian matrix is
2x −y
−4 −2
.
u
2
2
u0
.
=
v
−4 −2
v0
The characteristic equation is λ2 + 4 = 0 and the eigenvalues are ±2i.
0 u
−2 −2
u
.
=
The linearized system at (−1, 2) is
v
−4 −2
v0
The characteristic equation is λ2 + 4λ − 4 = 0, and the eigenvalues are
√
√
−4 ± 32
= −2 ± 2 2.
2
The linearized system at (1, −2) is
December 2014
(4 points)
Math 215/255 Sections 101–105 Final Exam
Page 14 of 15
(c) Identify the equilibrium of (1) at which the linearized system (not system (1)) has a
center. Classify the other equilibrium and indicate whether it is (asymptotically) stable
or unstable in system (1).
Answer.
The eigenvalues of the coefficient matrix for the linearized system at
equilibrium (1, −2) are are purely imaginary. Therefore the linearized system at (1, −2)
has a center.
The coefficient matrix for the linearized system at equilibrium (−1, 2) has a positive
eigenvalue and a negative eigenvalue. Hence, the equilibrium is a saddle, which is
unstable, in system (1).
December 2014
Math 215/255 Sections 101–105 Final Exam
Page 15 of 15
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).