Lecture Notes for MA 221
Douglas Bauer
1
Basic Denitions and Terminology
Dierential Equations (D. E.s)
Here are some examples.
1. y + y = 0.
Sol: y = c1 sin x, y = c2 cos x.
2. y = 2 Sol: y = 2x + c.
3. y + 2xy = ex .
2
2f 2f 2f
4.
+
+
= g (x, y, z ).
x2
17
Homogeneous Boundary Value Problems
Example: y 4y = 0, y (0) = 0, y (1) = 0.
Clearly y 0 is a solution. The general solution is
yh = c1 e2x + c2 e2x.
We now solve for c1 and c2.
y (0) = 0 0 = c1 + c2
y (1) = 0 0 = c1 e2 + c2 e2.
Thus we have 2 equation
15
Introduction to Laplace Transforms
Suppose f (x) is dened on [0, ). Then the Laplace Transform of f (x) is
L(f (x) = F (s) =
verges.
Note:
sx
e f (x)dx,
0
sx
e f (x)dx
0
= limR
provided the integral con-
R sx
e f (x)dx.
0
Note: If the integral conver
13
Review of Power Series
Most D. E.s do not have nice closed form solutions, i.e.,
solutions given by functions that we can easily recognize. Usually the solution is given as a power series.
un converges if limncfw_sn
n=1
cfw_sn is the sequence of partia
7
Linear Second Order D.E.s
This lecture, and maybe the next one, will be a little on the
theoretical side. We need to understand something about
the properties of solutions to 2nd order d.e.s before we start
learning how to solve them.
We will consider t
Lecture Notes for Discrete Math
Douglas Bauer
1
Symbolic Logic
Denition: A proposition is a sentence that is true or false
but not both.
10 is a prime number
4x2=8
George Washington was mayor of N.Y.C.
This class has 35 students
10100 can be written a
4
Functions
Before studying functions we will rst quickly dene a more
general idea, namely the notion of a relation. A function
turns out to be a special type of relation.
Denition: Let S and T be sets. A binary relation on SxT
is any subset of SxT . A bi
9
Tough Graphs and Hamilton Cycles
We would like a theorem that says A graph G is hamiltonian
if and only if G has property Q, where Q can be checked in
polynomial time.
However in the early 1970s it was discovered that it is NPcomplete to determine if a
8
Graphs and O-Notation
Knigsberg Bridge Problem
o
Problem: Cross every bridge exactly once and return to your
starting point.
Solution: Solved by Euler (1736).
Key Idea: Every time you enter a point you must be able to
leave. Hence the number of lines at
7
Mathematical Induction
Let Na = cfw_n Z | n a.
The following are equivalent:
1. Ordinary axiom of induction: If S Na such that
(a) a S
(b) n Na (n S n + 1 S )
then S = Na.
2. Strong axiom of induction: If S Na such that
(a) a S
(b) n Na (a, a + 1, a + 2
6
Relations
Let R be a relation on a set A, i.e., a subset of AxA.
Notation: xRy i (x, y ) R AxA.
Recall: A relation need not be a function.
Example: The relation R1 = cfw_(x, y ) RxR | x2 + y 2 = 1 is
not a function.
Some denitions
1. R is reexive i xRx