Homework 12 MAP 2302/3305 Fall 2009
Solutions will be posted on Friday, 4 December
Problem 12.1
12.1.a
Textbook, p. 428, Problem 1
Solution:
To nd the eigenvalues:
A=
3 4
1 1
det(A I ) = (3 )(1 ) + 4 = 2 2 + 1
1 = 2 = 1
To nd the eigenvector:
(A 1 I )v =
Solutions for Homework 11 MAP 2302/3305 Fall 2009
Problem 11.1
Problems 1, 2, and 5 all use the same fact for their solution.
b
c
au + bu + cu = g (t) u = u u + g (t)
a
a
x1 = u
x1 = x2 = u
x2 = u
b
c
x2 = x2 x1 + g (t)
a
a
x1
x2
=
0
1
c
b
a a
x1
x2
0
g (
Solutions to Homework 10 MAP 2302/3305 Fall 2009
Problem 10.1
10.1.a
Textbook, p. 336, Problem 1
Solution:
y + y = f (t), y (0) = 0, y (0) = 1
a = 1, b = 0, c = 1
f (t) =
1 0 t < 3
0 3 t
We have
f (t) = 1 u3 (t)
1 e3s
F (s) =
s
s
F (s)
1
Y (s) = 2
+2
(s
Homework 9 MAP 2302/3305 Fall 2009
Solutions will be posted on Wednesday, 11 November
Problem 9.1
9.1.a
Textbook, p. 311, Problem 1
Solution:
f (t) is continuous within each open interval and approaches a nite limit at t = 0, t = 1
and t = 2. It is contin
Solutions for Homework 8 MAP 2302/3305 Fall 2009
Problem 8.1
8.1.a
Textbook, p. 215, Problem 1
Solution:
The relevant trigonometric identity is
cos A cos B = 2 sin
AB
A+B
sin
2
2
Taking A = 9t and B = 7t yields
cos 9t cos 7t = 2 sin(t) sin 8t
= 2 sin t si
Solutions for Homework 7 MAP 2302/3305 Fall 2009
Problem 7.1
7.1.a
Textbook, p. 202, Problem 1
Solution:
A > 0,
u = 3 cos 2t + 4 sin 2t
0 = 2, A = 3, B = 4
2
A + B 2 = 25 = R2 R = 5
4
tan =
3
B > 0, in rst quadrant, i.e., 0 /2
4
= arctan 0.927295
3
u = 4
Solutions for Homework 6 MAP 2302/3305 Fall 2009
Problem 6.1
6.1.a
Textbook, p. 183, Problem 1
Solution:
y 2y 3y = 3e2t
y 2y 3y = 0
2
r 2r 3 = 0 r1 = 1, r2 = 3
cfw_y1 , y2 = cfw_et , e3t is a fundamental set
g = 3e2t is not a solution to the homogeneous
Solutions for Homework 4 MAP 2302/3305 Fall 2009
Problem 4.1
4.1.a
Textbook, p. 88, Problem 2
Solution:
y = f (y ) = ay + by 2
a > 0 b > 0 < y0 <
f (y ) is a quadratic opening up with roots y = 0 and y = a/b. f = a + 2by has a root
at a/(2b) and is an in
Solutions for Homework 2 MAP 2302/3305 Fall 2009
Problem 2.1
Textbook, p. 40, Problem 30
Solution:
y y = 1 + 3(sin t), y (0) = y0
p(t) = 1, g (t) = 1 + 3(sin t)
(t) = et
y (t) = Cet + et
et 1 + 3(sin t) dt
= Cet 1 + 3et
et (sin t)dt
et
(sin t) (cos t)
2
3
Solutions for Homework 1 MAP 2302/3305 Fall 2009
Problem 1.1
1.1.a
Textbook, p. 24, Problem 1
Solution:
d2 y
dy
+ t + 2y = sin t
2
dt
dt
Highest derivative order is 2 and
t2
d2 y
dy
+ 1 (t) + 0 (t)y = g (t)
2
dt
dt
2 (t) = t2 , 1 (t) = t, 0 (t) = 2, g (t)
MAP 2302/MAP 3305 Final Exam
Friday, 11 December 2009
7:30 AM 9:30 AM
In-class, 102 Love Building
Closed Book. Two sheets of notes and calculators
allowed.
Show all work on these pages.
All answers must be justied for full credit.
Question
1.
2.
3.
4.
5.
MAP 2302/MAP 3305 Exam 2
102 Love Building
Monday, 30 November 2009
1:25 PM 2:15 PM
Closed Book. Two sheets of notes and calculators
allowed.
Show all work on these pages.
All answers must be justied for full credit.
Question
1.
2.
3.
4.
Total
Points
Poin
MAP 2302/MAP 3305 Exam 1
102 Love Building
Friday October 16, 2009
1:25 PM 2:15 PM
Closed Book. Two sheets of notes and calculators
allowed.
Show all work on these pages.
All answers must be justied for full credit.
Question Points
Points
Possible Awarded