HW #9 Solutions1
7.6.2:
First recall the denition of the inner product
(, ) =
(x)(x) dx
where the bar denotes complex conjugation.2 Relative to this inner product the
operator x
(multiplication by x) and the momentum operator
p = i
d
dx
are self-adjoin
HW #10 Solutions1
8.6.1:
The PDE we wish to solve is
u
u
2u
+ (1 a) , < x < , t > 0
=
t
x2
x
(1)
with the initial condition u(x, 0) = f (x). This is the one-dimensional convectiondiffusion equation. From Wikipedia: The convection-diffusion equation is
a c
Math 22B, Section 001, CRN 30209
Winter 2017
Reading Assignments
All reading assignments refer to the textbook Lectures on Differential Equations. You will
get more out of the lectures if you do the reading assignments before attending the lectures.
Class
Math 22B, Section 001, CRN 30209
Winter 2017
Some Important Dates
Monday, January 9: First class meeting of Math 22B.
Monday, January 16: No class, MLK Day
Friday, February 17: Midterm Exam; Bring a new bluebook. Covers material of Chps. 15.
Friday, Febru
Homework 1: Math 22B - Tavernetti, Fall 2016
Due Wednesday, Sept 28 in class
(20 points)
Instructions : Solve all problems. Print out your solutions when computer results are asked for
(do not include the dfield8.m code), work neatly, label your plots, sh
Assignment 8
Math 22B
Due June 3. 2015
For each of the following systems of equations:
(i) Rewrite in matrix form.
(ii) Find the eigenvalues and eigenvectors of the matrix.
(iii) Find the general solution which is real valued.
(iv) Calculate the Wronskian
Assignment 7
Math 22B
Due May 22, 2015
1.Solve the following dierential equations using the Laplace Transform.
(a) y + 3y + 2y = 0, y(0) = 1, y (0) = 0
(b) y + 2 y = cos(2t), omega2 = 4, y(0) = 1, y (0) = 0
(c) y (4) 4y + 6y 4y + y = 0, y(0) = 0, y (0) =
Assignment 8
Math 22B
Due June 3, 2015
For each of the following systems of equations:
(i) Rewrite in matrix form.
(ii) Find the eigenvalues and eigenvectors of the matrix.
(iii) Find the general solution which is real valued.
(iv) Calculate the Wronskian
Assignment 5
Math 22B
Due May 1, 2015
1. (a) Show y1 (x) = cos(kx) and y2 (x) = sin(kx) are solutions to y + k 2 y = 0, where k = 0.
(b) Find the Wronskian for this pair of solutions. For what initial conditions can you nd a solution.
2. Find a pair of fu
Final Review
Math 22B
For each dierential equation, write its order, show it is linear or non-linear, and if it
is linear, write whether it is homogeneous or non-homogeneous. For review, you should
also review Fourier series, systems of dierential equatio
Practice Midterm
Math 22B
1. For the following dierential equations, write down the order, determine whether they are linear
or nonlinear, and homogeneous or non-homogeneous.
3
y
(a) d 3 + ( dy )2 = 0
dt
dt
(b) dy + 2y 3 = e4t
dt
(c) dy + t3 y = 0
dt
2. S
Assignment 6
Math 22B
Due May 15, 2015
1. Use the method of undetermined coecients to nd the general solutions to the following dierential equations.
(a) y + 2y = 3 + 4 sin(2t)
(b) 2y + 3y + y = t2 + 3 sin(t)
(c) y + y + 4y = 2 sinh(t)
2. Use the variatio
HW #9 Solutions1
7.5.2:
First recall the denition of the inner product
(, ) =
(x)(x) dx
where the bar denotes complex conjugation.2 Relative to this inner product the
operator x (multiplication by x) and the momentum operator
p = i
d
dx
are self-adjoint (
HW #8 Solutions1
6.9.7 We are asked to solve the two-dimensional Helmholtz equation
U + k 2 U = 0
in the rectangular domain 0 x a, 0 y b subject to the boundary
conditions that U (x, y) vanish on the boundary of the rectangle. We write
U (x, y) = X(x)Y (y
HW #7 Solutions1
6.9.3 We begin with (6.34) of the notes
1
2
2
(1 + k1 )1 + k1 2
2
k2 1 (2 + k2 )2 .
=
=
Dene
1
2
=
.
Then the above DEs are in matrix form
= A
where
A=
(1)
2
(1 + k1 )
k1
2
k2
(2 + k2 )
We look for a solution of (1) of the form
(t) = eit
HW #6 Solutions1
5.5.6:
We are again asked to analyze the DE
dx
= Ax
dt
where now A is a 2 2 matrix given by (5.16). The eigenvalues of A are the
roots of the characteristic polynomial of A. A simple calculation shows we must
nd the roots to
2 4 + 3 + 2 =
HW #7 Solutions1
6.9.1: Observe that in this case the masses at site 1 and site N are connected
with a spring. (Visualize the system of N particles to be on a circle.) Thus there
are no end eects as in the case of the lectures. The only change in the matr
HW #8 Solutions1
6.9.8:
Assuming a solution of the form
uj (t) = eit vj , vj independent of t,
one simply substitutes this into the equation
mj
d2 uj
+ k (uj 1 + 2uj uj +1 ) = 0 for j = 1, 2, 3, . . . , 2N,
dt2
(1)
to obtain the equation for the vj
mj 2 v
HW #6 Solutions1
5.5.6:
We are again asked to analyze the DE
dx
= Ax
dt
where now A is a 2 2 matrix given by (5.16). The eigenvalues of A are the
roots of the characteristic polynomial of A. A simple calculation shows we must
nd the roots to
2 4 + 3 + 2 =
HW #5 Solutions1
5.5.1:
The rst step is to write the harmonic oscillator equation
2
x + 0 x = 0
as a rst order system. Let
x
x
X=
Then the rst order matrix equation is
dX
= AX
dx
where
0
2
0
A=
1
0
.
Now we compute the eigenvalues and eigenvectors of A. T
HW #4 Solutions1
3.5.6:
The DE given is a 2nd order linear inhomogeneous equation. We rst
nd a particular solution. Based upon our experience with forced oscillations,
we guess a particular solution of the form
yp (t) = A cos(t)
Substituting this into the
HW #2 Solutions1
2.3.6:
We are given that the force exerted by the spring is
F = F (x) = kx x3 ,
where k and are positive constants and x is the displacement from equilibrium.
It is reasonable to assume the force is an odd function of x since the spring s
HW #3 Solutions1
3.5.1:
1. By Eulers formula
ei4 = cos 4 + i sin 4.
(1)
By the addition theorem and Eulers formula
4
ei4 = ei
= (cos + i sin )4 .
Expanding the last expression, and using the fact that i2 = 1 we obtain
ei4 = cos4 + 4i cos3 sin 6 cos2 sin2
Math 22B
Comments on January 14, 2013, lecture
In deriving the small 0 expansion of the period of the pendulum T , the
following expansion was used:
cos cos 0 =
12
14
2
4
( 0 )
0 +
2!
4!
Here is a better explanation of this than given in the lectures:
HW #1 Solutions1
1.3.2:
We are asked to linearize the DE
g
+ sin = 0
near the point = . Let = then = and
sin() = sin( + ) = sin() .
Thus the linearized DE is
g
= 0.
The dierence between this linearization and the one at = 0 is the sign in front
of . We
L1.3(#1)
1)
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