- Page 113:
-
(Spotted by Sean Clements)
In the paragraph above the boxed display we have
“the general solution of the trivial
second order equation $y''=1$ is…”.
That equation is meant to be $y''=0$.
- Page 113:
-
(Spotted by Robert Weisser)
In boxed figure, the text on the graph's $y$ axis reads
$y(x_0)=y_0$,
$y'(x_0)=y'_0$.
The latter should be
$y'(x_0)=y_1$.
- Page 117, Exercise #30:
-
(Spotted by Dr. Nathaniel Mays)
This problem is wrongly listed under the heading of second order equations
with missing $y$. It should be
grouped with problems 32–37 under the heading of second order
equations with missing $x$.
- Page 117, Exercise #31:
-
(Spotted by Paul Rydeen)
The answer of $y= \frac{1}{k}\cosh kx$ given in the back of the book
is incomplete. The complete answer would involve the coordinates
$(x_1,y_1)$ and $(x_2,y_2)$ of the suspension points, as well as
the length $L$ of the cable. Here is a sketch of how to
go about obtaining the complete solution.
Solving equation (20) in the problem's statement we get
$y(x) = c_1 + \frac{1}{k} \cosh k(x-c_2)$ which involves three
parameters $c_1$, $c_2$, and $k$. The requirement that the cable goes through
the points $(x_1,y_1)$ and $(x_2,y_2)$ yields two equations,
$y_1 = c_1 + \frac{1}{k} \cosh k(x_1-c_2)$
and
$y_2 = c_1 + \frac{1}{k} \cosh k(x_2-c_2)$.
A third equation is obtained by setting the cable's length to $L$.
For that, we plug the function $y(x)$ obtained above into
the curve-length formula
$L = \int_{x_1}^{x_2} \sqrt{1 + y'(x)^2} \,dx$
from calculus, evaluate the integral, and arrive at
$
L = \frac{1}{k} \sinh k(x_2-c_2) - \frac{1}{k} \sinh k(x_1-c_2).
$
Then we solve (numerically)
the three equations simultaneously for the three unknowns $c_1$, $c_2$, and $k$.
For instance, if the suspension points are $(0,0)$ and $(10,2)$,
and the length of the cable is $15$, we get
$
c_1 = -7.1338,
c_2 = 4.5816,
k = 0.3205$, and therefore the
equation of the cable is
$y(x) = -7.1338 + 3.1196 \cosh(0.3205 x - 1.4686)$.
See the Wikipedia article
https://en.wikipedia.org/wiki/Catenary for a detailed discussion of the hanging cable/chain problem.
- Page 117, in the comments preceding Exercise #32:
-
(Spotted by Clayton Lively, reported by Dr. Lo)
The substitution $y=v'$ should be $y'=v$.
- Page 117, in the comments preceding Exercise #39:
-
(Spotted by Daniel Engbert)
The initial condition $y'(x_0)=y'$ is meant to be $y'(x_0)=y_1$.
- Page 122, Exercise #4:
-
(Spotted by Andrew Zuelsdorf)
The answer in the back of the book is
$W(f,g) = (m-n) e^{(m+n)x}$.
It should be
$W(f,g) = (n-m) e^{(m+n)x}$.
- Page 122, Exercise #9:
-
(Spotted by Faraz Vatan)
Each of the problems 1–10 consists of two parts.
First, you are asked to determine if the given pair of
functions $f$ and $g$ is linearly independent on the interval $(-1,1)$.
Second, you are asked to compute the Wronskian $W(f,g)$.
According to a note on page 118, a function pair is
linearly independent if one function is not a constant multiple of the other.
(You don't really need a Wronskian to determine that.)
In Exercise #9, we are given $f(x) = x$, $g(x) = |x|$.
Since $|x|$ is not equal to a constant times $x$ on the given interval, then
this pair is linearly independent, as
the answer in the back of the book correctly states.
As to the second part,
since the function $g(x) = |x|$ is not differentiable at $x=0$,
the Wronskian $W(f,g)$ is undefined at $x=0$.
Therefore the answer in the back of the book, which says $W(f,g)=0$, is wrong.
- Page 122, Exercise #10:
-
This problem asks you to determine if the pair of functions
$f(x) = \ln x$ and $g(x) = \ln x^2$ is linearly independent
on the interval $(-1,1)$. The answer in the back of the book
is wrong. The correct answer is that the question is meaningless
since the natural logarithm $\ln x$ is undefined when $x \le 0$.
- Page 123, Exercise #25:
-
(Spotted by Kaung San, reported by Dr. Peercy)
In equation (21) a minus sign is missing in the exponent. It should be:
\[
W(x) = c e^{-\int p(t) \, dt}.
\]
(Spotted by Paul Rydeen) In the equation above, $t$ should be $x$.
- Page 124, Exercise #30:
-
(Spotted by Paul Rydeen)
Problems 28–30 are expected
to be reducible to the form $y'' + \big[ g(x) y \big]' = 0$ but this one
does not fit into that category. Rather, it is of the form
$p(x) y'' + 2 p'(x) y' + p''(x) y = q(x)$, which reduces to
$\big[ p(x) y \big]'' = q(x)$, and which may be readily solved by
integrating $q(x)$ twice. In this instance,
$p(x) = x^2 - 2x$ and
$q(x) = e^{2x}$, which confirms the answer in the back of the book.
- Page 126, calculations leading to Theorem 3.5:
-
(Spotted by UHM)
Each of the integrals $\int p(s)\,ds$
in equations (6), (7), (9)
should have a minus sign in front.
Furthermore, the $r$ and $s$ variables introduced in the derivation
are not truly necessary since we are dealing with indefinite
integrals here. The following simpler form of equation (9)
should do:
\[
y_2(x) = y_1(x) \int \frac{e^{-\int p(x)\,dx}}{y_1^2(x)} \,dx .
\]
In any case, as noted at the bottom
of page 126, it is easier to carry out the steps
that lead to equation (9) than to use that equation directly.
- Page 128, Exercise #15:
-
(Spotted by Dr. Bradley)
The answer in the back of the book has a sign error. It should be
\[
y = -x \cos x + \sin x \ln|\sin x|.
\]
- Page 129, Exercise #18:
-
(Spotted by Justin Korn)
In the answer in the back of the book, the
$e^{-x^2}$
should be
$e^{x^2}$.
- Page 129, Exercise #20:
-
(Spotted by Paul Rydeen)
The answer in the back of the book is incorrect. I should be
\[
y = (x - 1) \int \frac{e^x}{x(x-1)^2} \,dx.
\]
- Page 129, Exercise #25(b):
-
(Spotted by Paul Rydeen)
In equation (18) the integration variable should be $x$, not $t$.
- Page 130 forward:
-
The choice of $m$ for the name of the variable in the
characteristic equation is unfortunate because it conflicts
with the symbol for mass in applications to dynamics.
This has been rectified later (page 166 onward)
by replacing it with $r$.
It would have been better to use the symbol $r$ instead of $m$
right from the start.
- Page 132, Example #2:
-
In equation (8), the initial condition should be $y'(0)=0$.
- Page 134, Exercise #23:
-
(Spotted by Dr. Dovermann)
There's a misprint in the hint. It should read
$x^2 d^2y/dx^2 = d^2y/dt^2 - dy/dt$.
- Page 137, line 2 from the top:
-
(Spotted by Dr. Sousedik)
The term $\frac{ix^3}{3!}$ should have a minus sign in front.
On the next line it has the correct (minus) sign.
- Page 138, Example #2:
-
(Spotted by UHM)
The characteristic equation is presented as $m^2+1=0$. It should
be $m^2+4=0$.
- Page 139, Exercise #11:
-
(Spotted by Corey Kegerreis, reported by Dr. Peercy)
The answer in the back of the book is missing a term. The correct
answer is $y = \cos 2x -\frac{1}{2} \sin 2x$.
- Page 140, Exercise #17:
-
The statement of this exercise and the corresponding answer are correct,
but the solution involves a fourth order differential equation which is
outside this chapter's scope.
To confine the solution to second order differential equations,
replace the equation $y''(x) = y(-x)$ by $y'(x) = y(-x)$ and follow the
given hint. The answer turns out to be $y(x) = c(\cos x + \sin x)$,
where $c$ is an arbitrary constant.
- Page 140, Exercise #19:
-
Equation (19) is meant be
\[
xy'' - y' - x^3 y = 0.
\]
Furthermore (spotted by Menachem Greenfeld) the answer to part (b)
in the back of the book needs to be adjusted to
$v' = x^2 + \frac{1}{x} v - v^2$.
- Page 140, Exercise #20:
-
(Spotted by Paul Rydeen)
The book's answers for parts (b) and (e)
violate the stipulation that $-\pi < \theta \le \pi$. The correct answers
are $e^{-\pi/2}$ and $\sqrt{2} e^{-\pi/4}$, respectively.
The book's solutions would have been correct if the the angle range
were $0 \le \theta < 2\pi$.
- Page 147, Example #2:
-
(Spotted by Momo Mizuguchi)
Just before equation (23) we have:
“Hence the general solution of the nonhomogeneous solution is”.
The second
“solution”
should be
“equation”.
- Page 148, Exercise #11:
-
(Spotted by Abigail Jackson)
In the last line, the right side of
the equation should be $c_1 f_1(x) + c_2 f_2(x)$. (That is,
the $y$'s should be changed to $f$'s.)
- Page 148, Exercise #13:
-
(Spotted by Christine Sweigart, reported by Dr. Lo)
The solution to part (b) in the back of the book is wrong.
It should be:
\[
y = c_1 x + \frac{c_2}{x} - 1.
\]
- Page 150, Example #4:
-
(Spotted by Dr. Gobbert)
The expressions for
$y_p$,
$y'_p$ and
$y''_p$ are correct, however in substituting them into the
differential equation, the positions of $y_p$ and $y''_p$
have been swapped. Had this been done correctly, the final result
in equation (13) would have been:
\[
y_p(x) = \frac{3}{2} \cos 2x - \frac{1}{2} \sin 2x.
\]
- Page 153, Table 3.2:
-
(Comment by UHM)
The entire Table 3.2 is equivalent to the following single rule:
Consider the differential equation
\[
ay'' + by' + cy
= e^{\alpha x} \big[ P_n(x) \cos \beta x + Q_n(x) \sin \beta x \,\big],
\]
where $a$, $b$, $c$, $\alpha$ and $\beta$
are constants and $P_n(x)$ and $Q_n(x)$ are polynomials
of up to $n\,$th degree for some $n \ge 0$.
Then a particular solution is:
\begin{align*}
y_p(x) = x^s e^{\alpha x} \big[
&(A_0 + A_1 x + \cdots + A_n x^n) \cos \beta x \\
+&(B_0 + B_1 x + \cdots + B_n x^n) \sin \beta x
\,\big].
\end{align*}
The exponent $s$ is the multiplicity of $\alpha + i\beta$ as a root of the
characteristic polynomial $am^2 + bm + c$.
Thus, if $\alpha + i\beta$ is not a root, then $s=0$,
if $\alpha + i\beta$ is a root of multiplicity one then $s=1$,
and
if $\alpha + i\beta$ is a root of multiplicity two then $s=2$.
I suggest using this single rule instead of the nine rules
in the textbook's Table 3.2.
See the related comments below under the headings of
Page 154, Example #8
and
Page 156, Exercises #31 and #35.
- Page 154, Example #8:
-
- (spotted by Celia Drew and reported by Dr. Dean)
In the solution to part (d), the expression
$f(x) = 2x\cos3x + \sin3x$ should be
$f(x) = 2x\sin3x + \cos3x$ but this does not affect the rest of
the solution.
- (spotted by Celia Drew and reported by Dr. Dean)
In the solution to part (e), the expression
$f(x) = xe^{-2x}\cos x$ should be
$f(x) = xe^{-2x}\sin x$. Also, where it says
“the terms of $f(x)$ are of type 8”
it should say
“the terms of $f(x)$ are of types 8 and 9”
since the types 8 and 9” form an inseparable pair.
-
(Spotted by Momo Mizuguchi)
In the solution to part (e), every $\sin3x$ and $\cos3x$ should
be $\sin x$ and $\cos x$.
In any case, the 9 rules of the textbook's Table 3.2
make things more complicated than necessary. Instead, I suggest that you
use the single rule formulated in the boxed comment above.
See the following item for usage samples.
- Page 156, Exercise #7:
-
(Spotted by Christine Sweigart, reported by Dr. Lo,
also by Kevin Stern)
The solution in the back of the book is not quite correct.
It should be:
\[
y = c_1 + c_2 e^{-4x} + \frac{1}{8} x^2 - \frac{1}{16} x.
\]
- Page 156, Exercises #31 and #35:
-
(Spotted by Anthony Simms, reported by Dr. Peercy)
The answers in the back of the book are wrong. The confusion is
due to the less than ideal explanation of the 9 entries
of the Table 3.2 on page 153. Here we solve
these exercises by following the single rule formulated
in the boxed comment above under the heading of
Page 153, Table 3.2.
-
Exercise #31: $y'' + 4y = (x^2 + 2x -1) e^x \sin 2x$
This corresponds to $\alpha=1$, $\beta=2$, $n=2$.
Since $\alpha + i\beta$, that is $1 + 2i$, is different from the roots
of the characteristic polynomial $m^2 + 4$ (which are $\pm 2i$)
then $s=0$ and the particular solution takes the form:
\[
y_p = e^x \big[
(A_0 + A_1 x + A_2 x^2) \cos 2x
+ (B_0 + B_1 x + B_2 x^2) \sin 2x \big].
\]
-
Exercise #35: $y'' + 2y' + y = x^2 e^{-x} \sin 2x$
This corresponds to $\alpha=-1$, $\beta=2$, $n=2$.
Since $\alpha + i\beta$, that is $-1 + 2i$, is different from the roots
of the characteristic polynomial $m^2 + 2m + 1$ (which are $-1$ and $-1$)
then $s=0$ and the particular solution takes the form:
\[
y_p = e^{-x} \big[ (A_0 + A_1 x + A_2 x^2) \cos 2x
+ (B_0 + B_1 x + B_2 x^2) \sin 2x \big].
\]
- Page 157, Exercise #43:
-
(Spotted by Paul Rydeen)
In equations (34) and (35), $F_0$ should be $E_0$.
Furthermore, the method of presenting an oscillatory solution
in terms of amplitude and phase angle comes later on page 167.
- Page 157, Exercise #45:
-
(Spotted by Paul Rydeen)
In the statement of the problem,
“Suppose that the dart has velocity $v$”,
should be
“Suppose that the dart's initial velocity is $v_0$”.
Furthermore, the answers in the back of the book need correcting.
Letting where $d = \sqrt{x_0^2 + y_9^2}$, the correct answer to part (a) is
\[
\text{(target)}~~y_1 = y_0 - \frac{1}{2} g t^2,
\quad
\text{(dart)}~~y_2 = \frac{y_0 v_0 t}{d} - \frac{1}{2} g t^2.
\]
while the correct answer to part (c) is
$\ds y = y_0 - \frac{1}{2} g \Bigl( \frac{d}{v_0} \Bigr)^2$,
Aside: Although this is an interesting problem, its solution
is unrelated to this section's techniques.
- Page 158:
-
The system of equations (9) and its solution in (10)
correspond to the differential equation (1), namely
\[
y'' + p(x) y' + q(x) y = f(x),
\]
where the coefficient of $y''$ is 1.
It would have been just as easy to work out the more general case
where the coefficient of $y''$ is other than 1, as in:
\[
a(x)y'' + b(x)y' + c(x)y = f(x).
\]
For this modified equation, the system of equations (9) would
have changed to
\begin{align*}
y_1 v_1' + y_2 v_2' &= 0 \\
a(x) \big(y_1' v_1' + y_2' v_2'\big) &= f(x),
\end{align*}
and the corresponding solution in (10) would have been
\[
v_1' = - \frac{y_2(x) f(x)}{a(x) W(y_1,y_2)}
\quad\text{and}\quad
v_2' = \frac{y_1(x) f(x)}{a(x) W(y_1,y_2)},
\]
where
$W(y_1,y_2)
= \det\big( \begin{smallmatrix} y_1 & y_2 \\ y'_1 & y'_2 \end{smallmatrix} \big)
= y_1 y_2' - y_1' y_2$, as usual.
I suggest using these instead of the textbook's equations (10).
- Page 159, the boxed Historical Note:
-
(Spotted by Daniel Conlon, reported by Dr. Lo)
It refers to Euler and Lagrange as seventeenth century mathematicians.
Actually they were eighteenth century mathematicians.
- Page 159, in the sentence immediately after equation (7):
-
(Spotted by Moon Skillin)
In “only the first derivatives or $v_1$ and $v_2$”,
the “or” should be “of”.
- Page 160, Equation (12):
-
(Spotted by Dr. Soane)
Both integrals are indefinite integrals (that is, antiderivatives)
therefore naming the integration variables $t$ is misleading and not
quite correct. Change all occurrences of $t$ to $x$.
- Page 161, Example 2:
-
The solution is incorrect because the coefficient of $y''$ in
equation (19) is $x^2$, not 1, contrary to the assumption
that was built into the preceding analysis. In fact, the
differential equation, as given, cannot be solved in terms of elementary
functions.
To retain the solution given in the book, change equation (19) to
\[
x^2 y'' - 2xy' + 2y = x^3 \sin x,
\]
then follow the method suggested in the boxed formula above.
- Page 162, Exercise #9:
-
(Spotted by Dr. Bradley)
The answer in the back of the book has a sign error. The term
$-x \cos x$ should be
$+x \cos x$.
- Page 162, Exercise #10:
-
(Spotted by Dr. Bradley)
The answer in the back of the book has a sign error. The term
$+x \cos x$ should be
$-x \cos x$.
- Page 162, Exercise #16:
-
(Spotted by Dr. Bradley)
The answer in the back of the book is given as
$y_p = x^{1/2}$. It should be
$y_p = x^{-1/2}$.
- Page 162, Exercise #17:
-
(Spotted by Dr. Soane)
The answer in the back of the book requires integration limits.
It should be:
\[
y_p(x) = \int_0^x \sinh(x-s) \, e^{-s^2} \, ds.
\]
Note: See Exercise 24 on page 163 for a similar case.
- Page 165, line 3:
-
(Spotted by Andrew Waite)
The second $\dot{u}$ should be $\ddot{u}$.
- Page 166, Equation (3):
-
(Spotted by Haley Boyd)
The coefficient of $\dot{u}$ should be 0.20, not 0.02.
- Page 167, Figure 3.9:
-
Along the horizontal axis, the markings
\[
\delta, \qquad \delta + \pi/\omega_0, \qquad \delta + 2\pi/\omega_0
\]
should be
\[
\delta/\omega_0, \qquad (\delta + \pi)/\omega_0, \qquad
(\delta + 2\pi)/\omega_0.
\]
- Page 167, first footnote:
-
(Spotted by Paul Rydeen)
Where it says “attached to a spring constant with constant”,
it should say “attached to a spring with constant”.
- Page 167, second footnote:
-
(Spotted by Dr. Bradley)
Where it says Eq. (9), it should
say Eq. (10). Furthermore, the expression
$\tan\phi = c_2/c_1$ should be
$\tan\phi = c_1/c_2$.
- Page 168, near the bottom:
-
(Spotted by Dr. Bradley)
The equation $\tan\delta = \frac{c_1}{c_2} = -1$ should be
$\tan\delta = \frac{c_2}{c_1} = -1$.
- Page 168, near the bottom:
-
We are told
Since $c_1$ is positive and $c_2$ is negative, we see from Figure 3.10
that the phase angle $\delta$ is in the second quadrant.
Hence $\delta = 3\pi/4$.
This is not correct. Since $c_1$ is positive and $c_2$ is negative,
the phase angle $\delta$ is in the fourth quadrant.
Hence $\delta = -\pi/4$.
Consequently, the equation (14) and the graphs on page 169 are
incorrect.
A related issue, spotted by Dr. Gobbert, is that the equation
\[
\tan \delta = \frac{c_1}{c_2}
\]
near the bottom of page 168 is incorrect. The correct version is
\[
\tan \delta = \frac{c_2}{c_1}
\]
which is obtained by dividing the expressions for
$\sin\delta$ and $\cos\delta$ given in equation (9) on page 167.
- Page 170, Exercise #4:
-
To get the answer in the back of the book, change
the initial condition $\dot{u}(0) = 1$ to $\dot{u}(0) = 3$.
- Page 170, Exercise #7:
-
(Spotted by Christopher Gongora)
In the back of the book, the $\dot{u}$ in the answer to part (b)
should be $\ddot{u}$. Furthermore,
the two initial conditions there have run into each other.
The proper form would be:
\[
\ddot{u} + 64u = 0, \quad u(0)=-\frac{1}{6}, \quad \dot{u}(0) = 1.
\]
- Page 170, Exercise #9:
-
(Spotted by Dr. Dean)
The answer in the back of the book is incorrect. It should be
\[
u(t) = -\frac{1}{3} \cos 8t - \frac{1}{4} \sin 8t
=
\frac{5}{12} \cos(8t - \delta),
\]
where $\delta = \pi + \tan^{-1} \frac{3}{4} \approx 3.79$.
- Page 170, Exercise #11:
-
(Spotted by Paul Rydeen)
In the answer in the back of the book,
all three occurrences of $\delta$ should be $\phi$.
- Page 170, Exercise #13:
-
(Spotted by Paul Rydeen)
It says that the kinetic energy is $1/2m\ddot{u}^2$.
That should be $1/2m\dot{u}^2$.
- Page 171, Exercise #17:
-
(Spotted by Paul Rydeen)
The equation $\ddot{x} + \omega_0 x = 0$
should be $\ddot{x} + \omega_0^2 x = 0$.
- Page 172, Exercise #19:
-
(Spotted by Paul Rydeen)
The equation $\ddot{u} = kx$
should be $\ddot{u} = ku$.
- Page 172, Exercise #21:
-
(Spotted by Paul Rydeen)
The statement of the problem, the accompanying figure, and the
answer in the back of the book are too badly garbled to afford
a quick fix. See
Gravity train
in Wikipedia for a proper statement of the problem and its solution.
- Page 174, Figure 3.16:
-
Along the horizontal axis, the markings
\[
\delta,
\qquad \delta + \pi/\mu,
\qquad \delta + 2\pi/\mu,
\qquad \delta + 3\pi/\mu
\]
should be
\[
\delta/\mu,
\qquad (\delta + \pi)/\mu,
\qquad (\delta + 2\pi)/\mu,
\qquad (\delta + 3\pi)/\mu.
\]
- Page 179, Exercise #6:
-
(Spotted by Dr. Dean)
The equation has a typo; it's meant to be
$\ddot{u} + 2\dot{u} + u = 0$.
- Page 179, Exercise #10:
-
(Spotted by Paul Rydeen)
The condition $k<0$ is meant to be $k > 0$.
- Page 179, Exercise #14:
-
In the problen's statement, change the amount of stretch from
$6\,$ft to $6\,$in in order to get the book's answer.
- Page 179, Exercise #15:
-
(Spotted by Paul Rydeen)
The book's answer is too vague. Here we provide some additional
details. From equations (5) and (6) on page 174 we know that
the amplitude of the motion is $R e^{-(d/(2m))t}$ and the
damped period is $4\pi m / \sqrt{4mk - d^2}$. Letting
$D = d/m$ and $K = k/m$, the expressions for the amplitude
and the damped period become $R e^{-t\,D/2}$ and
$4\pi/\sqrt{4K - D^2}$. Equate these to the given data and
solve for the unknowns $D$ and $K$.
Conclude that the differential equation (1) on page 173 takes the form
$m \ddot{u} + \bigl( 2/5 \ln(2) \bigr) \dot{u}
+ \big(\pi^2 + (\ln2)^2 / 25\big) \, u = 0$.
- Page 180, Exercise #19:
-
This exercise, involving forced vibrations, belongs to the next section.
Moreover (Spotted by Paul Rydeen) the damping constant $c$ should be changed to $d$ in order to
match the supplied figure.
- Page 181, Exercise #23:
-
(Spotted by Paul Rydeen)
This problem mixes up the $x$ and $y$ symbols. Make them either all $x$
or all $y$.
- Page 181, Exercise #25:
-
(Spotted by Paul Rydeen)
The damping coefficient $d$ is meant to be 1.
- Page 183, Equation (8):
-
The term $\ddot{u}$ should be $m\ddot{u}$.
- Pages 184 and 185, Example 1:
-
Equations (10) and (11) on page 185, and the unnumbered equation
that comes between them, need fixing.
Suppose the road's surface is given by a function
$h(x)$, where $x$ is the coordinate along the road.
If the car is at $x=0$ when $t=0$, and travels at the constant speed $v$,
then it is not too hard to see that the differential equation of motion is
\(
m \ddot{u} = -k \big[u(t) - h(vt)\big],
\)
or equivalently,
\[
m \ddot{u} + k u = k h(vt).
\]
In particular, since the problem's statement postulates that
$h(x) = A \cos\big( \frac{2\pi x}{L} \big)$, then
the differential equation of motion is
\[
m \ddot{u} + k u = k A \cos\Big( \frac{2\pi vt}{L} \Big).
\]
This should replace the unnumbered equation that comes between
equations (10) and (11).
In view of this, equation (10) should be
\[
f(t) = k A \cos\Big( \frac{2\pi vt}{L} \Big),
\]
and equation (11) should be
\[
75 \ddot{u} + 7200 u = 7200 \cos\Big( \frac{2\pi vt}{25} \Big).
\]
Finally, near the lower right corner of Figure 3.24,
$f(t)$ should be $h(x)$.
The rest of the solution is correct as written.
- Page 185, equation (13):
-
(Spotted by Paul Rydeen)
The transient part of the equation is given as
$c_1e^{r_1t} + c_2e^{r_2t}$. That should be
$e^{-\alpha t} ( c_1 \cos \beta t + c_2 \sin \beta t )$
since the system is underdamped.
- Page 187, Exercise #5:
-
(Spotted by Sam Phipps)
The $t$ in the answer in the back of the book should be $3t$, as in:
$\ds u_{\text{st}} = \frac{4}{\sqrt{73}} \cos (3t - \delta)$.
- Page 187, Exercise #6:
-
(Spotted by Dr. Bell)
The solution to part (b) in the back of the book is wrong. Assuming
zero initial conditions, it should be:
\[
u = \frac{4\sqrt{3}}{3} t \sin\big(2\sqrt{3}t\big).
\]
- Page 187, Exercise #8:
-
(Spotted by Dr. Bell)
The solution in the back of the book is wrong. The vertical motion of the buoy,
measured from its water line (that's where the water surface contacts the buoy
when the buoy is at rest) is given by
\[
x(t) = A \cos \frac{2\pi}{5}t + B \sin \frac{2\pi}{5}t
+ \frac{25}{8} \sin \frac{2\pi}{7}t,
\]
where the arbitrary constants $A$ and $B$ depend on the initial conditions.
The terms with $A$ and $B$ correspond to the buoy's natural
oscillations. In reality they will go away after a while
due to the water's damping effect (although this is not
included in the model) and what will remain is the last term
which corresponds to the forced motion of the buoy due to the
action of the waves. The amplitude of the motion, 25/8, is slightly
larger than 3 ft, therefore the buoy will be completely
submerged in its lowest point.
- Page 187, Exercise #12:
-
(Spotted by Paul Rydeen)
What is listed as answer to Exercise 11 in the back of the book,
is answer to this exercise.
- Page 190:
-
(Spotted by Ryan Olson)
In the equation at the bottom of the page, the
term $c_2 e^x$ should be $c_3 e^x$.
- Pages 192–193:
-
(Spotted by Paul Rydeen)
The calculated constant $A=-1/95$ should be $A=-1/85$.
- Pate 194, Figure 3.27:
-
(An omission, not an error, spotted by UHM)
The flowchart is incomplete—in the lower left
corner, under “Roots $r_1,\ldots,r_n$”, the
case of repeated complex roots is not accounted for.
For completeness, the box dealing with complex roots should be
replaced with:
For each pair of complex roots $p+iq, p-iq$ of multiplicity $k$,
we obtain $2k$ linearly independent solutions
\begin{align*}
e^{px}\cos qx , \;
x e^{px}\cos qx , \;
\dots,\;
x^{k-1} e^{px}\cos qx , \\
e^{px}\sin qx , \;
x e^{px}\sin qx , \;
\dots,\;
x^{k-1} e^{px}\sin qx .
\end{align*}
-
- Page 196, Exercise #2:
-
(Spotted by Paul Rydeen)
The answer
$W(1,x,x^2,\ldots,x^n) = n!$
in the back of the book is incorrect.
It should be
$W(1,x,x^2,\ldots,x^n) = 1! \times 2! \times \cdots \times n!$.
This expression is sometimes called the superfactorial of $n$.
-
