MATH 2930 - Extra Credit 4
Monday, February 22, 2010
10:31 AM
3. Courtesy of Claire Paduano
4. Courtesy of Pam Snodgrass & George Lu
Notes Page 1
5. Courtesy of Lee Carlaw
Notes Page 2
Notes Page 3
HW12 solutions
10.6-12 (a) Mathematically, the problem is
2 uxx = ut , ux (0, t) = 0 = ux (L, t), u(x, 0) = sin( x).
L
Use the PDE method of separation of variables to write
2 X 00 T = XT 0
T0
X 00
= 2 = ,
X
T
which yields the two ordinary differential
HW13 solutions
10.7-12 (a) The 1D wave equation is given by
a2
2u
2u
=
.
x2
t2
Using the chain rule for partial derivatives, we change the variable x to s = x/L :
u
u ds
1 u
=
=
x
s dx
L s
It follows that
1 2u
2u
=
.
x2
L2 s2
Hence, in terms of the new va
HW11 solutions
10.5-3 Seek a solution of form u(x, t) = X(x)T (t) and the equation becomes
uxx + uxt + ut = X 00 T + X 0 T 0 + XT 0 = 0 X 00 T = T 0 (X 0 + X)
X 00
T0
=
.
X0 + X
T
For the last equality to hold, the two sides must equal to a constant, say
MATH 2930 : HW10 SOLUTIONS
Chapter 10.3 (page 612)
(10.3.1). (a) The function is assumed with period 2L = 2.
Z
Z 0
Z 1
1 L
f (x)dx =
(1)dx +
(1)dx = 0,
a0 =
L L
1
0
and for n > 0,
1
an =
L
bn =
1
L
Z
Z
L
nx
f (x) cos
dx =
L
L
L
f (x) sin
L
nx
dx =
L
Z
Z
MATH 2930 : HW8 SOLUTIONS
Chapter 4.2 (page 234)
(4.2.29). The characteristic equation is r3 + r = 0. We can factor out an r from this equation to
get r(r2 + 1) = 0. In this form, we see that the roots are r = 0, i
So the general solution is of the form y
HW2 solutions
t
2.1-16 y 0 + 2t = cos
is of the formR y 0 + p(t)y = g(t), so if we multiply the equation by the
t2
2
integrating factor (t) = e t dt = e2 ln |t| = t2 , we have t2 y 0 + 2ty = cos t = (t2 y)0 .
Integrate both sides and we obtain t2 y = sin
Diy Qs Section 5 Problems
Peter Tseng
March 1, 2010
9
We notice that the given equation x2 y 5xy + 9y = 0 (x > 0) ts the form
of the homogeneous second-order linear ODE y + p(x)y + q(x)y = 0 by rst
9
5
dividing by x2 , giving us p(x) = x , q(x) = x2 . As
HW 38 Solutions
May 3, 2013
14.(a) Find the solution of Laplaces equation in the rectangle 0 < x < a
and 0 < y < b that satises the boundary conditions
ux (0, y) = 0, ux (a, y) = 0, 0 < y < b
u(x, 0) = 0, u(x, b) = g(x), 0 x a.
Solution: The general form
MATH 2930 - Spring 2010
Worksheet 9
MATH 2930 - Dierential Equations - Spring 2010
Worksheet 9 - Select Solutions
1. (E&P 8.2, #24)
Use the power series method (i.e., assume solution y(x) =
order ODE
(x2 1)y + 2xy + 2xy = 0
cn xn ) to solve the 2nd-
Apart
MATH 2930 - Spring 2010
Worksheet 10
MATH 2930 - Dierential Equations - Spring 2010
Worksheet 10 - Select Solutions
For each of the following periodic functions f (t) whose value in one complete period is
given, nd its Fourier series.
1. f (t) = t for 0 <
MATH 2930 - Extra Credit 2
Wednesday, February 10, 2010
7:07 PM
3, 5: Courtesy of Anthony Basile
8: Courtesy of George Lu & Pam Snodgrass
Notes Page 1
9, 10, 11, 12: Courtesy of Caroline Chang & Mahina Wang
Notes Page 2
Notes Page 3
MATH 2930 - Extra Credit 3
Tuesday, February 16, 2010
10:49 PM
9, 10: Courtesy of David Smith
11: Courtesy of Jeffrey Hsiao
Notes Page 1
12: Courtesy of Claire Paduano
The bifurcation in this case looks like a pitchfork because when k>0, there are three c
MATH 2930 - Extra Credit 1
Tuesday, February 02, 2010
8:00 PM
3. y'=|y| (Solution provided by Caroline Chang & Mahina Wang, revised by Joe)
Plot of y'=|y|
Plot of y = y2
The slope fields of y = y2 slopes up quicker than the slope fields of y = |y|. As the
HW14 solutions
10.8-1 (a) The easiest way to solve this problem is to transform it to the standard Dirichlet problem
for a rectangle, the solution of which we already know (see Section 10.8, pp. 660661).
Lets write this standard problem in terms of new va