MATH 302 Differential Equations (Bueler)
16 April 2009
Selected Solutions to Assignment #8
These problems were graded at 3 points each for a total of 24 points.
5.2 #2.
Rewrite as
(i)
Dx 3y = 0
(ii)
2x + (D + 1)y = 0.
Combine by 2(i) + D(ii) to get: D(D +
MATH 302 Differential Equations (Bueler)
REVISED 24 March 2009
Selected Solutions to Assignment #5
(Revised 4.3 #4.)
These problems were graded at 3 points each for a total of 27 points.
4.1 #2.
(b)
(a)
m(cy)00 + b(cy)0 + k(cy) = c(my 00 + by 0 + ky) = c(
MATH 302 Differential Equations (Bueler)
16 February 2009
Selected Solutions to Assignment #3
These problems were graded at 3 points each for a total of 27 points.
2.4 #6.
2.4 #10.
separable, linear (and not exact)
It is exact because
1=
N
M
=
= 1.
y
x
Be
MATH 302 Differential Equations (Bueler)
28 February 2009
Solutions to Assignment #4
These problems were graded at 3 points each for a total of 21 points.
3.4 #2. This problem fits into the framework of Example 1. In particular, we are told b = 10 and g =
MATH 302 Differential Equations (Bueler)
24 March 2009
Selected Solutions to Assignment #6
These problems were graded at 3 points each for a total of 21 points.
(The Group Project on A#6 is treated as a separate 10 point assignment.)
4.4 #10.
The auxiliar
MATH 302 Differential Equations (Bueler)
2 April 2009
Selected Solutions to Assignment #7
These problems were graded at 3 points each, except for 4.9 #4 which was 6 points, for a total of 21 points.
4.7 #33. The answer Ctet is in the back of the book. To
MATH 302 Differential Equations (Bueler)
1 May 2009
A Solution, from Assignment #11
Recall Assignment #11 is NOT DUE!
8.2 #8.
(d) We have the series
cos x = 1
X
x2 x4
x2j
+
=
(1)j
2!
2!
(2j)!
j=0
2j
x
It is easiest to apply the original ratio test with
MATH 302 Differential Equations (Bueler)
1 May 2009
Selected Solutions to Assignment #10
These problems were graded at 5 points each for a total of 25 points.
7.5 #4.
The Laplace transform of the ODE, including the initial values, is
2
1
s Y (s) sy(0) y
MATH 302 Differential Equations (Bueler)
18 February 2009
Selected Solutions to Assignment #2
Revised. Corrections in 2.2 #12 and 2.3 #8.
1.4 #8.
The full table of my Eulers method calculations looks like this:
steps
x
0
1
2
4
8
y
y
y
y
0
0
0
0
8
4
3
8
2
MATH 302 Differential Equations (Bueler)
2 February 2009
Selected Solutions to Assignment #1
These problems were graded at 3 points each for a total of 27 points.
1.1 #4.
PDE, second order, dependent variable is u, independent are x and y
1.1 #14.
dx
= kx
MATH 302 Differential Equations (Bueler)
27 April 2009
Selected Solutions to Assignment #9
These problems were graded at 3 points each for a total of 27 points.
7.2 #4.
An integration-by-parts:
Z
Z
t= Z
(3 s)1 e(3s)t dt
L te3t (s) =
est te3t dt =
te(3s
Math F401: Introduction to Real Analysis
Fall 2016 Syllabus
Course Description
This course is a rigorous study of the ideas underlying calculus and an introduction to the
real numbers. Rather than the computational focus of your previous calculus classes,
Math 401: Homework 2
Due September 12, 2016
Solutions
Exercise 1.4.1: Recall that I stands for the set of irrational numbers.
(a) Show that if a, b Q then ab and a + b Q as well.
(b) Show that if a Q and t I then a + t I and if a , 0 then at I as well.
(c
Math F401: Homework 5 Solutions
October 6, 2016
1. Suppose cfw_n j
j=1 is a sequence of natural numbers such that n j < n j+1 for all j N. Show
that n j j for all j N
2. Show that a subsequence of a convergent sequence converges to the same limit. Be sur
Math F401: Homework 5
Due: October 12, 2016
1. Use the worksheet from class to write up a nice proof of the Alternating Series Test.
Solution:
Let (a n )n be a decreasing sequence of non-negative numbers such that lim a n = 0. We
wish to show that (1)n+1
Math F401: Homework 4
Due: September 29, 2016
1. Abbott 2.2.6
Show that limits, if they exist, must be unique. In other words, assume lim a n = l1 and
lim a n = l2 , and prove l1 = l2 .
Solution:
Let > 0. Since lim a n = l1 , there exists an N1 such that
Math 401: Homework 3
Due September 19, 2016
Solutions
Exercise 1.4.7: Finish the proof of Theorem 1.4.5 by showing that the assumption 2 > 2
contradicts the assumption that = sup A.
Solution:
We first observe that if x > 0 and x2 > 2 then x is an upper bo
Math F401: Homework 9
Due: November 9, 2016
1. Hand in your proof of the Intermediate Value Theorem.
Solution:
Suppose f [a, b] is continuous and f (a) < f (b). Suppose v ( f (a), f (b). Let Av = cfw_x
[a, b] f (x) < v. Observe that a Av and that Av is b
Math F401: Homework 8 Solutions
Due: November 2, 2016
1. Abbott 4.3.7
Solution, part a:
Let c R. Let (x n ) be a sequence of rational numbers coverging to c, and let (y n ) be a
sequence of irrational numbers coverging to c. The existence of these sequenc
Math 401: Homework 1
September 7, 2016
Solutions
Exercise 1.2.5: Use the triangle inequality to establish the following inequalities:
(a) |a b| |a| + |b|;
(b) |a| |b| |a b|.
Solution:
(a) Proof. By the triangle inequality and the fact that |x| = | x| for
1
, chhm o 3 Vambkg:
Fwd/0:601:19 mm'ables P.- If a K
.fk g H45 0% g) ) (gift)
(997) ,y) WW?) ) e y
W W W W
W
Er- 'm 4 MM 0F F/xy) = h/xry)
1:401) Jammy) elk: aqua
U>0
154/4 con/g5 \/
EX an/,2) = 75+ 22 A'WS/b/a/e/<W%S M"
W; of (4:04?) a c Cudi.
value,