Solutions to end-of-chapter problems
Engineering Economy, 7th edition
Leland Blank and Anthony Tarquin
Chapter 14
Effects of Inflation
14.1 (a) There is no difference.
(b) Todays dollars are inflated compared to dollars of 2 years ago. Therefore, in order

Homework 15
1
May 16, 2011
7.2
2 Derive the Legendre Polynomials P3 , P4 , P5 .
P3 (x): We calculate 3 = 3 4 = 12, and have a5 = a7 = = 0. We are free to choose
a0 = a2 = = 0, so our polynomial solution to Legendres equation is
y(x) = a1 x + a3 x3
From th

Homework 11
1
October 30, 2010
6.4
1 We wish to solve
ut k 2 uxx
< x < +,
T1
T2
u(x, 0) =
t>0
x<0
x>0
(6.2)
We know that the solution to the Cauchy problem is
+
(xy)2
1
(y)e 4k2 t dy
u(x, t) =
2k t
We can write in terms of T1 and T2 by splitting at ze

Homework 12
November 10, 2010
1
Section 9.2
1
Problem 1
We have the wave equation in 3D,
utt c2 u = 0,
u = u(x, y, z, t)
Show that if we choose to separate only the time variable, by letting u(x, y, z, t) = (x, y, z)T (t),
we arrive at the Helmholtz equat

Homework 13
1
May 16, 2011
9.3
1. a. Calculate in cartesian and polar coordinates that xy = 0. In cartesian coordinates, we
have
2 xy 2 xy
+
xy =
x2
y 2
y x
=
+
x y
=0
In polar coordinates, we want to show that [r2 sin cos ] = 0. But note that we can us

Homework 10
1
October 29, 2010
6.3
1c Let f (x) = xec|x| ; nd F[f (x)]. Though this is an application of a formula you will prove
in problem 9, for now you can integrate by parts in the denition easily enough. We would
like to calculate
+
1
xec|x| eix dx

Homework 8
1
October 11, 2010
Denition 1.1. Laplace Transform The Laplace Transform of a function f (x, t), denoted by
F (x, s) = L[f (x, t)], is dened to be
+
F (x, s) = L[f (x, t)] =
est f (x, t)dt
(1.1)
0
Though we wont be calculating this integral, f

Homework 7
1
October 5, 2010
4.4
1b Given the problem
utt = uxx ,
0 < x < L,
u(x, 0) = f (x)
u(0, t) = T
(4.1)
0<x<L
(4.2)
t>0
ut (x, 0) = g(x),
t>0
(4.3)
ux (L, t) = a,
Introduce a new function v(x, t) which will have homogeneous boundary conditions in t

Homework 6
September 22, 2010
1
Recall from page 81, the denition of a periodic function. In particular, we have
Denition 1. We say that a function is periodic of period T = 0 if f (x + T ) = f (x) for all values
of x in the domain of f . The collection o

Homework 9
October 24, 2010
1
6.2
Before we do 2, lets go ahead and do 7. We can use the formula from there to solve 2 immediately.
7 We wish to nd
F s [xf (x)] =
2
+
xf (x) sin(x)dx
0
But if we take the derivative with respect to of the Fourier Cosine tr

Homework 3
1
September 7, 2010
See the end for a proof of something used in class (exercise 36)!
1.6
3,24 We wish to solve the PDE
y 2 ux + x2 uy = 0,
u = u(x, y)
so we assume that there is a product solution, u(x, y) = X(x)Y (y). Plugging in gives us
y 2

Homework 5
1
September 7, 2010
Note again that having the answer in any and all of these problems is not terribly impressive
from a grading standpoint, which is why we provide solutions: it is knowing and doing the work in
between the problem and answer w

Homework 2
September 7, 2010
1
Recall, the denition of linearity for an operator L[u]:
Denition 1 (Linearity of PDE operator). A PDE in operator form, i.e. expressed as L[u] = f
like in class, is linear if for any two solutions u1 and u2 and any real numb

Homework 14
December 1, 2010
7.3
8 We start with the Bessel equation,
x2 y + xy + (x2 2 )y = 0
and apply the substitution u =
xy, so y = u/ x. Then
1
1 3
y = x 2 u x 2 u
2
and
1
3
3 5
y = x 2 u x 2 u + x 2 u
4
So we substitute to nd
x2 y + xy + (x2 2 )y =