Be sure this exam has 12 pages including the cover
The University of British Columbia
MATH 215/255, Sections 101–105
Final Exam – December 2009
Name
Signature
Student Number
Circle Section:
101 Liu
102 Tsai
103 Seon
104 Dridi
105 Smith
No notes nor calculators.
1.
Each candidate should be prepared to produce his li-
problem
max
1.
12
2.
10
3.
13
4.
10
5.
20
6.
10
7.
10
8.
15
total
100
brary/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.
score
December 2009
Math 215/255 Sections 101–105 Final Exam
(12 points) 1. Find the general solution of each of the following differential equations.
(a) −y 0 + y = 4e3x .
Answer.
The integrating factor is
Z
µ(t) = exp( −dx) = e−x .
On multiplying the DE by µ(t) = e−x , we get
e−x y 0 − e−x y = −4e2x ,
Hence
e−x y =
Z
or
(e−x y)0 = −4e2x .
(−4e2x )dx = −2e2x + c.
y = −2e3x + c ex .
(b) y 0 = 4x(y − 1)1/2 ,
Answer.
y > 1.
The equation is separable and can be rewritten as
dy
= 4xdx.
(y − 1)1/2
Integrating, we get
2(y − 1)1/2 = 2x2 + 2c.
Thus
y = 1 + (x2 + c)2 .
Page 2 of 12
December 2009
Math 215/255 Sections 101–105 Final Exam
Page 3 of 12
(10 points) 2. Solve the initial value problem
(6xy − y 3 )dx + (4y + 3x2 + bxy 2 )dy = 0,
y(0) = 3,
where b is a constant such that the differential equation is exact.
Answer. Let M = 6xy − y 3 and N = 4y + 3x2 + bxy 2 . Since the DE is exact, we get
My = Nx , or 6x − 3y 2 = 6x + by 2 . Thus b = −3.
There is a ψ(x, y) such that
ψx = 6xy − y 3 ,
2
ψy = 4y + 3x − 3xy
(1)
2
(2)
By (1),
Z
ψ=
(6xy − y 3 )dx = 3x2 y − xy 3 + h(y).
By (2),
4y + 3x2 − 3xy 2 = ψy = 3x2 − 3xy 2 + h0 (y),
or h0 (y) = 4y. Hence h(y) = 2y 2 + c. Thus the general solution of the DE is given implicitly
by the equation
3x2 y − xy 3 + 2y 2 = k, k is a constant.
Using the initial condition y(0) = 3, we get k = 18. Hence the solution of the IVP is given
implicitly by the equation
3x2 y − xy 3 + 2y 2 = 18.
December 2009
Math 215/255 Sections 101–105 Final Exam
Page 4 of 12
(13 points) 3. Use the method of variation of parameters to find the general solution of the given differential
equation. Then check your answer by using the method of undetermined coefficients :
y 00 + 2y 0 + y = 3e−t
Answer.
The solution of the homogeneous equation is
yc (t) = C1 e−t + C2 te−t
The functions y1 = e−t and y2 = te−t form a fundamental set of solutions. The Wronskian of
these functions is W (y1 , y2 ) = e−2t . Using the method of variation of parameters, the
particular solution is given by
Yp (t) = u1 (t)y1 (t) + u2 (t)y2 (t)
in which
te−t (3e−t )
dt = −3t2 /2
W (t)
Z −t −t
e (3e )
u2 (t) =
dt = 3t
W (t)
Z
u1 (t) = −
Hence the particular solution found by using the method of variation of parameters is
3
Yp (t) = −3t2 e−t /2 + 3t2 e−t = t2 e−t
2
Now by using the method of undetermined coefficients, since the nonhomogenous term is e−t
the basic form of the particular solution Yp is Yp = Ae−t . Since Ae−t duplicates the term
C1 e−t in yc we should multiply by t2 (indeed, te−t duplicates C2 te−t in yc ).
Thus, we set for the particular solution of the non homogeneous problem,
Yp (t) = At2 e−t
Upon substitution, we obtain A = 3/2 which gives,
3
Yp (t) = t2 e−t
2
So we find the same solution than by using the method of variation of parameters.
Hence the general solution is
3
y(t) = C1 e−t + C2 te−t + t2 e−t
2
December 2009
Math 215/255 Sections 101–105 Final Exam
Page 5 of 12
(10 points) 4. A mass of 1 g stretches a spring 5 cm. Suppose that the mass is also attached to a viscous
damper with a damping constant of 20 dyn.s/cm. If the mass is pulled down an additional 2
cm and then released, find its position u at any time t.
(In CGS, the force unit is 1 dyn = 1 g·cm/s2 , and the standard gravity is g = 980 cm/s2 . Use
384 = 6*64).
Answer.
is
The spring constant is k = (1)(980)/5 = 196 dyne/cm. The I.V.P. for the motion
u00 + 20u0 + 196u = 0
with
u(0) = 2
u0 (0) = 0
Here u is measured in cm and t in sec. The general solution of the D.E. is
√
√
u = Ae−10t cos 4 6t + Be−10t sin 4 6t
√
Invoking the initial conditions, we require that A = 2 and B = 5/ 6.
The solution is
√
√
5
u = e−10t (2 cos 4 6t + √ sin 4 6t)
6
December 2009
Math 215/255 Sections 101–105 Final Exam
Page 6 of 12
5. Solve the following initial value problems.
(10 points)
(a)
(
y 00 + y
= uπ/2 + δ(t − π/6)
y(0) = y 0 (0) = 0
Answer.
Taking the Laplace transform of both sides of the ODE and then applying
the initial conditions, we obtain
π
e− 2 s
+
s
s2 Y (s) + Y (s) =
Z
∞
e−st δ(t − π/6)dt
0
π
π
e− 2 s
+ e− 6 s .
s
=
This yields
π
π
e− 2 s
e− 6 s
Y (s) =
+
.
s(s2 + 1) (s2 + 1)
Now, using partial fractions we have
π
π
π
e− 2 s
e− 2 s
s
se− 2 s
− π2 s 1
=
=e
−
− 2
.
s(s2 + 1)
s (s2 + 1)
s
(s + 1)
It follows
(
−1
y(t) = L
π
e− 2 s
s
)
(
−1
−L
π
se− 2 s
(s2 + 1)
(
)
−1
+L
π
e− 6 s
(s2 + 1)
Hence, according to the Laplace transform table, we have
y(t) = uπ/2 (t) − uπ/2 (t) sin(t) + uπ/6 sin(t − π/6)
)
.
December 2009
(10 points)
Math 215/255 Sections 101–105 Final Exam
Page 7 of 12
(b)
(
y 00 + 4y
1
k
=
[(t − 5)u5 (t) − (t − 5 − k)u5+k (t)]
y(0) = y 0 (0) = 0
Answer.
Taking the Laplace transform of both sides of the ODE and then applying the initial
conditions, we obtain
!
1 e−5s e−(5+k)s
2
s Y (s) + 4Y (s) =
−
.
k
s2
s2
Solving for Y (s), we obtain
1
Y (s) =
k
e−(5+k)s
e−5s
−
s2 (s2 + 4) s2 (s2 + 4)
!
.
We have (partial fractions)
1
1
=
2
2
s (s + 4)
4
It follows that
−1
L
1
1
− 2
2
s
s +4
1
s2 (s2
+ 4)
=
.
t
1
− sin 2t.
4 8
Hence, according to the Laplace transform table, we have
y(t) =
where h(t) =
t
1
− sin 2t.
4 8
1
[h(t − 5)u5 (t) − h(t − 5 − k)u5+k (t)]
k
December 2009
Math 215/255 Sections 101–105 Final Exam
Page 8 of 12
(10 points) 6. For the linear system
d
X=
dt
0 4
1 0
X,
find the general solution and plot a few trajectories.
Answer.
Denote the matrix by A. Its eigenvalues satisfy
0 = det(A − rI) = r2 − 4,
so that the eigenvalues are
r = ±2.
−2 4
0
The eigenvector ξ corresponding to r = 2 satisfies
ξ=
. One can take
1 −2
0
2
1
ξ=
.
The eigenvector ξ corresponding to r = −2 satisfies
ξ=
2
−1
2 4
1 2
2
−1
ξ=
.
The general solution is
X(t) = c1
Phase plane: omitted.
2
1
2t
e + c2
e−2t
0
. One can take
0
December 2009
Math 215/255 Sections 101–105 Final Exam
Page 9 of 12
(10 points) 7. Find the general solution of the following system.
d
1 1
2
X=
X+
et .
4 1
−1
dt
Answer.
Denote the matrix by A. Its eigenvalues satisfy
0 = det(A − rI) = (1 − r)(1 − r) − 4 = r2 − 2r − 3,
so that the eigenvalues are
r = 3, −1.
−2 1
0
The eigenvector ξ corresponding to r = 3 satisfies
ξ=
. One can take
4 −2
0
1
2
ξ=
.
The eigenvector ξ corresponding to r = −1 satisfies
ξ=
1
−2
2 1
4 2
ξ=
0
. One can take
0
.
The general solution to the homogeneous equation is
1
1
3t
e−t .
e + c2
Xc (t) = c1
−2
2
Based on the simple form of theright hand side, we use the method of undetermined
a1
coefficients, and set Xp =
et . Substitution into the ODE yields
a2
a1
a2
et =
1 1
4 1
a1
a2
et +
2
−1
et .
In scalar form, after canceling the exponential, we have
a1 = a1 + a2 + 2,
a2 = 4a1 + a2 − 1.
Thus a1 = 1/4 and a2 = −2. So the general solution is
1
1
1/4
3t
−t
e +
et .
X(t) = c1
e + c2
2
−2
−2
December 2009
Math 215/255 Sections 101–105 Final Exam
Page 10 of 12
8. Consider the nonlinear system
dx
= x(y − 1),
dt
(3 points)
(1)
(a) Find the Jacobian (partial derivative) matrix for the system (1).
Answer.
(4 points)
dy
= 2 + x − y.
dt
Defining F (x, y) = x(y − 1) and G(x, y) = x − y + 2, we have
Fx (x, y) Fy (x, y)
y−1 x
.
J(x, y) =
=
Gx (x, y) Gy (x, y)
1
−1
(b) Find the critical points for the system (1).
dx
will be zero only if x = 0 or y = 1. If x = 0, then we
dt
dy
= 2 − y, which will be zero when y = 2. Meanwhile, if y = 1, we will have
will have
dt
dy
= 1 + x, and this will be zero when x = −1. It follows that the system (1) has two
dt
critical points at (0, 2) and (−1, 1).
Answer.
We first note that
December 2009
(8 points)
Math 215/255 Sections 101–105 Final Exam
Page 11 of 12
(c) Use the linear systems near each critical point to classify them, specifying both the type
of critical point and whether it is (asymptotically) stable or unstable. If a critical point
cannot be unambiguously classified using the linear systems, please indicate this in your
answer.
Answer.
The linear system near (0, 2) is given by
0 u
1 0
u
,
=
0
v
1 −1
v
where u = x and v = y − 2. The eigenvalues of the coefficient matrix are given by the
roots of the quadratic polynomial r2 − 1 = 0, which are r = ±1. Since the eigenvalues
are real and of opposite sign, we may unambiguously classify the critical point of the
system (1) at (0, 2) as an unstable saddle point.
Meanwhile, the linear system near (−1, 1) is given by
0 u
0 −1
u
=
,
v0
1 −1
v
where now u = x + 1 and v = y − 1. The eigenvalues of the coefficient matrix in this case
2
are given by
√ the roots of the quadratic polynomial r + r + 1 = 0, which are
−1 ± i 3
. As the eigenvalues are complex conjugates with negative real part, we
r=
2
may unambiguously classify the critical point of the system (1) at (−1, 1) as an
asymptotically stable spiral point. December 2009
Math 215/255 Sections 101–105 Final Exam
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
(s − a)2 + b2
11. tn eat , n positive integer
n!
,
(s − a)n+1
12. uc (t)
e−cs
, s>0
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
1 s
F
,
c
c
s>a
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)
Page 12 of 12