18.305 Exam 1
Oct. 7, 2011
(closed book)
1. Consider the solution of the dierential equation
d2 y
+ 2 x4 y = 0, > 1
dx2
satisfying the initial conditions y (0) = 0, y (0) = 1.
(a) Transform the dieren
Advanced Analytic Methods in Science and Engineering
MATH 18.305

Fall 2004
Singular Points of Ordinary Differential Equations
Lecture 7 Regular Singular Points of Ordinary
Differential Equations
Let us consider the Bessel equation
dy
d2y
+x
? p 2 y : ?x 2 y.
(6.22)
dx
dx 2
L
Advanced Analytic Methods in Science and Engineering
MATH 18.305

Fall 2004
The WKB Approximation
Lectures Nine and Ten The WKB
Approximation
The WKB method is a powerful tool to obtain solutions for many physical problems. It is
generally applicable to problems of wave propa
Advanced Analytic Methods in Science and Engineering
MATH 18.305

Fall 2004
Boundary Layers and Singular Perturbation
Lectures 16 and 17
Boundary Layers and Singular Perturbation
A Regular Perturbation
In some physical problems, the solution is dependent on a parameter K. Whe
Advanced Analytic Methods in Science and Engineering
MATH 18.305

Fall 2004
Asymptotic Expansions of Integrals
Lectures Fourteen and Fifteen
In the last lecture, we discuss the method of stationary phase which is applicable to the integral
(8.36).
If the integral (8.36) has n
Advanced Analytic Methods in Science and Engineering
MATH 18.305

Fall 2004
The WKB Approximation
Lecture Eleven
Turning Point
As we have mentioned, the WKB approximation is useful in problems of wave propagation. In
this section we demonstrate this by applying it to the wave
Advanced Analytic Methods in Science and Engineering
MATH 18.305

Fall 2004
Asymptotic Expansions of Integrals
Lecture Thirteen: Method of Stationary Phase
In 1911, E. Rutherford performed an experiment in which Fparticles were made to pass through
a thin metal foil. He meas
Advanced Analytic Methods in Science and Engineering
MATH 18.305

Fall 2004
Singular Points of Ordinary Differential Equations
Lecture Eight
Irregular Singular Points of Ordinary
Differential Equations
Solutions expanded around an irregular singular point are distinctive in o
Advanced Analytic Methods in Science and Engineering
MATH 18.305

Fall 2004
Asymptotic Expansions of Integrals
Lecture 12
The Laplace Method
We begin with integrals of the form
IR :
X a e ?Rvx hxdx,
b
(8.8)
where vx and hx are independent of the parameter R. The variable of i
Advanced Analytic Methods in Science and Engineering
MATH 18.305

Fall 2004
Boundary Layers and Singular Perturbation
Lecture 18
Interior Turning Points
In this lecture well examine some examples in which there is a turning point inside the interval
where (9.13) holds. This i
Advanced Analytic Methods in Science and Engineering
MATH 18.305

Fall 2004
Singular Points of Ordinary Differential Equations
Lecture 6
Singular Points of Ordinary Differential
Equations
In the last chapter, we applied the method of separation of variables to various PDEs an
18.305 Fall 2011, Solutions to HW 10
Rosalie BlangerRioux
e
Department of Mathematics, MIT
December 6, 2011
1
Problem 1
1.1
(a)
L
2
= x = (0 x + x(x2 + X 2 )
x
L
2
= X = (W0 X + X (x2 + X 2 )
X
2
2
w
18.305 Fall 2011, Solutions to HW 9
Rosalie BlangerRioux
e
Department of Mathematics, MIT
December 5, 2011
1
Problem 1
Jp (x) =
=
=
=
2
2
p
p
xp
(p + 1/2)
xp
(p + 1/2)
xp
2p (p + 1/2)
2
p
2
p
1
1
1
18.305 Exam 3 Dec. 12, 2011
(closed book)
1. Solve approximately
1
y + x(2 + x)y + y = 0, 0 < x < 2,
3
with the boundary conditions y (0) = 1, y (2) = 3.
a. Locate the boundary layer and determine its
18.305 Fall 2011, Solutions to HW 1
Rosalie BlangerRioux
e
Department of Mathematics, MIT
September 29, 2011
1
Chapter 1, Problem 1(a)
First note that when x = 0, 1 the equation is not rst order anym
18.305 Fall 2011, Solutions to HW 2
Rosalie BlangerRioux
e
Department of Mathematics, MIT
September 29, 2011
1
(a) We have (D2 x2 )y = 0. The dimension of D2 is 2, the dimension of x2 is 2. (b) We wr
18.305 Fall 2011, Solutions to HW 3
Rosalie BlangerRioux
e
Department of Mathematics, MIT
November 3, 2011
1
1.1
Problem 1
a.
We have
y + 2 x4 y = 0, y (1) = 0, y (1) = 1
so that p(x) = x2 , and P (x
18.305 Fall 2011, Solutions to HW 4
Rosalie BlangerRioux
e
Department of Mathematics, MIT
October 11, 2011
1
1.1
Chapter 8, Problem 3
Solutions to part (a)
1
3
et dt
I () =
1
We have
v (t) = t3 ,
and
18.305 Fall 2011, Solutions to HW 5
Rosalie BlangerRioux
e
Department of Mathematics, MIT
October 17, 2011
1
1.1
Chapter 8, Problem 8
Solutions to part (a)
1
3
eit dt
I () =
1
0
1
3
eit dt +
=
1
3
ei
18.305 Fall 2011, Solutions to HW 6
Rosalie BlangerRioux
e
Department of Mathematics, MIT
December 5, 2011
1
Problem 1
We are asked to solve y + xy + (n + 1)y = 0 for n a nonnegative integer.
(a) Us
18.305 Fall 2011, Solutions to HW 7
Rosalie BlangerRioux
e
Department of Mathematics, MIT
November 7, 2011
1
Problem 1
4
eix eix dx = 1/3
I () =
1/3 x
4
ei(t+t ) dt
4/3 .
using t =
and =
So f (t) = t
18.305 Fall 2011, Solutions to HW 8
Rosalie BlangerRioux
e
Department of Mathematics, MIT
December 5, 2011
1
Problem 1
We want to solve
x + x = 0, > 0, x(0) = 1, x (0) = 0.
(1)
1+t
We notice rst that
Chapter 1
d2 y
y = 2 + x2 + ex .
1.Find the general solution of
dx2
Solution:
m2 = 1 m = 1. Thus the complementary solution is
Y = Aex + Bex .
We also have
1
1
2 = 2 ,
x2 = (1 + D2 + )x2 = x2 2.
21
1