Ulsan National Institute of Science and Technology
statics
MATH 101

Fall 2016
090901
coscos sinsin cos
sin sin
cos cos
cos
(p.816) 13.2 #21
cos sin cos sin
cos sin
(p.824) 13.3 #7
(p.830) 13.4 #1
cos sin
cos , sin
cos , cos
sin ,
Ulsan National Institute of Science and Technology
differential
MATH 101

Spring 2015
2015 Spring Complex Analysis 1 Homework 10 Solution
1. Evaluate the following integrals around the circle z = 3 in the positive direction:
(1)
ez
z2
Proof. Since ez =
n=0
zn
, we have
n!
ez
1
= 2
2
z
z
Hence, Resz=0
(1)n
n=0
ez
= 1. Thus,
z2
z=3
zn
1
Ulsan National Institute of Science and Technology
differential
MATH 101

Spring 2015
2015 Spring Complex Analysis 1 Homework 1 (Due 3/17)
Show all your work. Jumping to answer without minimum reasoning deserves no credit.
1. Prove that
(1) z is real if and only if z = z
(2) z is either real or pure imaginary if and only if z 2 = z 2
2. Le
Ulsan National Institute of Science and Technology
differential
MATH 101

Spring 2015
CHAPTER
7
Systems of First Order Linear
Equations
7.1
1. Introduce the variables x1 = u and x2 = u . It follows that x1 = x2 and
x2 = u = 2u 3 u .
In terms of the new variables, we obtain the system of two rst order ODEs
x1 = x2
x2 = 2x1 3 x2 .
3. First d
Ulsan National Institute of Science and Technology
differential
MATH 101

Spring 2015
CHAPTER
2
First Order Dierential Equations
2.1
5.(a)
(b) If y(0) > 2, solutions eventually have positive slopes, and hence increase without bound. If y(0) 2, solutions have negative slopes and decrease without
bound.
(c) The integrating factor is (t) = e
Ulsan National Institute of Science and Technology
differential
MATH 101

Spring 2015
CHAPTER
6
The Laplace Transform
6.1
3.
The function f (t) is continuous.
189
190
Chapter 6. The Laplace Transform
4.
The function f (t) has a jump discontinuity at t = 1, and is thus piecewise continuous.
7. Integration is a linear operation. It follows t
Ulsan National Institute of Science and Technology
differential
MATH 101

Spring 2015
CHAPTER
3
Second Order Linear Equations
3.1
1. Let y = ert , so that y = r ert and y = r2 ert . Direct substitution into the
dierential equation yields (r2 + 3r 4)ert = 0 . Canceling the exponential, the
characteristic equation is r2 + 3r 4 = 0 . The root
Ulsan National Institute of Science and Technology
differential
MATH 101

Spring 2015
CHAPTER
8
Numerical Methods
8.1
2. The Euler formula for this problem is yn+1 = yn + h(5 tn 2 yn ), in which
tn = t0 + nh . Since t0 = 0 , we can also write yn+1 = yn + 5nh2 2h yn with
y0 = 2 .
(a) Euler method with h = 0.05 :
tn
yn
n=2
0.1
1.73475
n=4
0.
Ulsan National Institute of Science and Technology
differential
MATH 101

Spring 2015
CHAPTER
5
Series Solutions of Second Order
Linear Equations
5.1
1. Apply the ratio test:
lim
n
(x 2)n+1
=
(x 2)n 
lim x 2 = x 2 .
n
Hence the series converges absolutely for x 2 < 1 . The radius of convergence
is = 1 . The series diverges for x =
Ulsan National Institute of Science and Technology
differential
MATH 101

Spring 2015
CHAPTER
4
Higher Order Linear Equations
4.1
1. The dierential equation is in standard form. Its coecients, as well as the
function g(t) = t , are continuous everywhere. Hence solutions are valid on the
entire real line.
3. Writing the equation in standard
Ulsan National Institute of Science and Technology
differential
MATH 101

Spring 2015
CHAPTER
1
Introduction
1.1
1.
For y > 2, the slopes are negative, therefore the solutions are decreasing. For y < 2,
the slopes are positive, hence the solutions are increasing. The equilibrium solution
appears to be y(t) = 2, to which all other solutions