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
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