- Page 15:
-
(Spotted by Dr. Bradley)
In the text just after equation (2), the $y^{(n)}$
should be $y^{(n-1)}$, as in:
where $f$ is a function of $x, y, y', \ldots, y^{(n-1)}$.
- Page 23, last paragraph:
-
(Spotted by Dr. Peercy)
The expression
$\partial f/\partial y = \frac13 y^{2/3}$ should be
$\partial f/\partial y = \frac13 y^{-2/3}$.
- Page 36, Exercise #6:
-
The answer in the back of the book has a typo;
the correct answer is $y = ce^{-x^2} + 1/2$.
- Page 42, Exercise #20:
-
(Spotted by Christine Sweigart, reported by Dr. Lo)
The answer in the back of the book is garbled. It is meant to be
$c(1-y)e^{x^2}=1+y$. Better yet, $y$ may be computed explicitly as
$\ds y=\frac{1+ce^{-x^2}}{1-ce^{-x^2}}$.
- Page 42, Exercise #23:
-
(Spotted by Dr. Soane)
The solution in the back of the book is correct but incomplete—the
constant function $y(x)=1$ is also a solution.
The inadvertent omission is brought about by the necessity of dividing
by $y-y^2$ to separate the variables; the division is permissible only if
$y$ is other than 0 or 1.
- Page 43, Exercise #32:
-
(Spotted by Anthony Santini, reported by Dr. Kogan)
The answer in the back of the book is garbled.
It should be
\[
y = \Big( 1 - \int_1^x e^{t^2} \,dt \Big)^{-1}.
\]
- Page 43, Exercise #33:
-
(Spotted by Dr. Stegenga)
In the statement of the problem, the $1+y^2$ term is meant to be
outside the square root, as in
\[
\frac{dy}{dx} = \sqrt{1 + \sin x} \; (1+y^2).
\]
- Page 43, Exercise #38:
-
(Spotted by Christine Sweigart, reported by Dr. Lo)
The answer in the back of the book is wrong. It should be $y = cx^2 - x$.
- Page 45, Exercise #45:
-
(Spotted by Dr. Mayans)
The differential equation to be solved is $y'^2 + y^2 = 1$.
The answer $y=\sin(x+c)$ given in the back of the book is correct but
incomplete—the constant functions $y(x)=1$ and $y(x)=-1$ are
also solutions.
Aside: The graphs of the constant functions $y(x)=1$
and $y(x)=-1$ form the envelopes of the family of
functions $y=\sin(x+c)$. According to Exercise 44, the
envelope of a family of solutions to a differential equation
is also a solution of the differential equation.
- Page 45, Exercise #47:
-
(Spotted by Dr. Mayans)
The answer given in the back of the book is off by one hour.
The correct answer is 11:23 a.m.
Aside:
This is not truly a differential equations
problem as it amounts to solving a trivial equation of the form
$dy/dt=f(t)$.
- Page 50:
-
(Spotted by Dr. Soane)
In equation (5) the rightmost term is missing a negative sign in the exponent.
It should be $Q_0 e^{-0.00012378t}$.
- Page 56, Exercise #17:
-
(Spotted by Nick Selock with assistance from Samuel Khuvis and Ari Rabe,
reported by Dr. Peercy)
The problem's description leads to the model $dP/dt = -k$ (with $k = 0.015$).
Its placement in this section on exponential decay is confusing.
Nevertheless, the answer in the back of the book is correct.
- Page 56, Exercise #18: new
-
(Spotted by Momo Mizuguchi)
The answer $5e^{0.3\ln2}$ given in the back of the book is correct, however
its decimal approximation is 6.15572 which rounds to 6.16, not
6.15.
- Page 56, Exercise #20:
-
(Spotted by Michael Norris)
The problem's statement implies that two hours after
10:00 p.m. is
12:00 p.m.
This goes against the convention of referring to midnight as
12:00 a.m.
Aside: The Wikipedia article
Confusion at noon and midnight presents an informative discussion of this issue.
- Page 57, Exercise #24: new
-
(Spotted by Menachem Greenfeld)
The answer in the back of the book is missing the word “million”.
It should be $5e^{4\ln2}\text{ million} = 80 \text{ million}$.
- Page 57, Exercise #27:
-
(Spotted by Rachana Haliyur, reported by Dr. Peercy)
The answer to part (b) should be 2435 A.D.
- Page 65, Exercise #1: new
-
(Spotted by Momo Mizuguchi)
The answer to part (d) in the back of the book is given in
“lbs/gal”.
The proper expression is
“lb/gal”.
- Page 66, Exercise #4:
-
The answer in the back of the book is garbled. The correct answer is:
\[
\frac{dQ}{dt} + \Big(\frac{5}{100-2t}\Big) Q = 0,
\quad
Q(0) = 300.
\]
- Page 74, Exercise #6:
-
(Spotted by Rachana Haliyur, reported by Dr. Peercy)
Recorded temperatures are missing.
The answer of 30°F given in the back of the book is not obtainable.
The original 1994 hardcover edition—well, one of the hardcover
versions, since there are multiple versions of it around—gives
the pertinent data. It says:
“Suppose the temperature of the can
was 45°F after 0.5 hours and 55°F
after 1 hour.” With this data, the answer would be 28.33°F.
- Page 74, Exercise #8:
-
Based on the given data, the decay constant turns out to be
$k = (\ln 2)/2$ and the time of murder is $t = -2.561912628$.
This corresponds to 5:26 pm. In the answer given in the
back of the book, the value of $t$ is rounded to $-2.6$, consequently
the time of the murder is given as 5:24 pm, which is off by
2 minutes. The 5:24 pm value is repeated in Exercise #12 on
page 75.
- Page 74, Exercise #10:
-
(Spotted by Dr. Dovermann)
The correct answer depends on several unstated assumptions.
Is the cream's temperature the same or different from the
room temperature? Does
the cream's temperature remain constant as Mary waits,
(e.g., is it kept on ice or in a refrigerator,) or is it
brought to the table from the refrigerator and allowed to warm?
Analyzing the various cases would be an interesting and instructive
mini-project. Here is the answer to one of the many scenarios: If
Mary's cream remains at room temperature while she waits, then her
coffee will be exactly as hot as John's when they begin to drink.
- Page 75, Exercise #12:
-
The answer in the back of the book is entirely wrong. The
correct value for the decay constant is $k=0.2912836670$.
The murder occurred at $t=-3.498093089$, that is, approximately
at 4:30 pm.
Note:
The equation that determines $k$
has no solution in terms of elementary functions.
You need to compute $k$ numerically on a calculator.
- Page 77, Equation (4):
-
(Spotted by Marie Wagner)
The malformed fraction $\ds\frac{m^2g}{}$ should be $\ds\frac{m^2g}{k^2}$.
The hardcover edition has the correct expression; the error was introduced
in the paperback edition.
- Page 80:
-
(Spotted by Dr. Dovermann)
A square root sign is missing in equation (12). It should be:
\[
\frac{d\theta}{dt} = \sqrt{
\frac{v_d^2 - v_s^2}{v_s^2} } \; \frac{1}{t}.
\]
- Page 85, Exercise #20: new
-
(Spotted by Sean Shields)
In the equation given in part (b), the term
$\sqrt{2gt}$ should be $\sqrt{2g}\,t$, that is, the $t$ should
not be under the radical.
- Page 85, Exercise #22:
-
(Spotted by Dr. Dovermann)
In part (e) the condition $v(a)=0$ should be $v(1)=0$.
In part (g) the equation (24f) is not quite correct. It should be:
\[
y(x) = \frac{x}{2}\Big(
\frac{x^k}{1+k} - \frac{x^{-k}}{1 - k} \Big)
+ \frac{k}{1-k^2}.
\]
- Page 86, Exercise #23:
-
(Spotted by Dr. Dovermann)
The equation of motion, given in part (b),
has an extra minus sign. It should
be $\ds \frac{dv}{dt} = - \frac{gr}{R}$.
Additionally, the answer for part (b) in the back of the
book is off by a factor of two. The correct answer is 42.54 minutes.
- Page 89, Exercise #28: new
-
(Spotted by Sean Shields)
In part (b), “drive” should be “derive”. Furthermore,
the $h^3$ in the numerator should be $h^2$.
- Page 98, Exercise #20: new
-
(Spotted by Menachem Greenfeld)
In part (a) the $t=1$ should be $x=1$.
- Page 103, equation (9): new
-
(Spotted by Dr. Nathaniel Mays)
The equation is missing a $y_n$ term on the right hand side. It should be
\[
y_{n+1} = y_n + \frac{h}{6} \big( k_{n1} + 2k_{n2} + 2k_{n3} + k_{n4} \big).
\]
- Page 117, Exercise #30: new
-
(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, in the comments preceding Exercise #32:
-
(Spotted by Clayton Lively, reported by Dr. Lo)
The substitution $y=v'$ should be $y'=v$.
- 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:
\[
W(x) = c e^{-\int p(t) \, dt}.
\]
- 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 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 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 #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 147, Example #2: new
-
(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 instead of the textbook's Table 3.2.
See the item under the heading
“Page 156, Exercises #31 and #35”
further down this page for usage samples.
- Page 154, Example #8: new
-
(Spotted by Momo Mizuguchi)
In the solution to part (e), every $\sin3x$ and $\cos3x$ should
be $\sin x$ and $\cos x$.
- 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 entries
of the Table 3.2 on page 153. Here we solve
these exercises by following the general rule formulated
in the boxed comment above.
-
Exercise #31: $y'' + 4y = (x^2 + 2x -1) e^x \sin 2x$
This corresponds to $\alpha=1$, $\beta=2$.
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 $\alpha=-1$, $\beta=2$.
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 156, Exercise #7:
-
(Spotted by Christine Sweigart, reported by Dr. Lo)
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 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 160, Equation (12): new
-
(Spotted by Dr. Soane)
Both integrals are indefinitie 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: new
-
(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:
-
(Spotted by Dr. Bradley)
In the second footnote, 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 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 #14:
-
The numerical data as given leads to quite an unpleasant calculation.
The answer in the back of the book is certainly wrong.
Here is a restatement of the problem with data that leads to a tamer
calculation.
An 8-lb weight stretches a spring 8 inches, thereby reaching its equilibrium
position. Assuming a damping constant for the system of 4lb/(ft/sec), the
weight is pulled down by 3 inches below its equilibrium position and given a
downward velocity of 3ft/sec. When will the mass attain its maximum
displacement below equilibrium? What is maximum displacement? Sketch the
graph of the motion.
Answer: $ t = \frac{1}{8} \ln 2$.
- Page 180, Exercise #19:
-
This exercise, involving forced vibrations, belongs to the next section.
- 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 187, Exercise #5: new
-
(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.
- 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 246, the boxed Historical Note:
-
It refers to Laplace's “seminal treatise on probability,
Mécanique Céleste”. The correct citation is
Laplace's treatise on probability,
Théorie analytique des probabilités, Paris, 1812.
Aside:
It is interesting to note that the Laplace transform had its origins in
the theory of probability, not differential equations.
An extensive historical account of the development
of the Laplace transform is given in:
- Michael A. B. Deakin,
The Development of the Laplace Transform, 1737–1937.
I. Euler to Spitzer, 1737–1880.
Archive for history of exact sciences 25 (1981), no. 4,
pp. 343–390.
- Michael A. B. Deakin,
The Development of the Laplace Transform, 1737–1937.
II. Poincaré to Doetsch, 1880–1937.
Archive for history of exact sciences 26 (1982), no. 4,
pp. 351–381.
- Page 253, Exercise #46:
-
(Spotted by UHM)
In part (a), the denominator
$1 - (a + ik)$
should be
$s - (a + ik)$.
- Page 254: new
-
(Spotted by Momo Mizuguchi)
Under the Purpose paragraph, on the “derivative property”
line, the expression $f\{0\}$ should be $f(0)$.
- Page 255, Theorem 5.4 (boxed): new
-
(Spotted by Momo Mizuguchi)
The term $s^{n-2} f''(0)$ should be $s^{n-3} f''(0)$.
- Page 259, Exercise #7:
-
(Spotted by UHM)
The answer in the back of the book is off by a factor of 2.
The correct answer is $\ds \frac{4(3s^2 - 4)}{(s^2+4)^3}$.
- Page 259, Exercise #12: new
-
(Spotted by Dr. Bell)
The answer in the back of the book has a typo. The ‘4’
in the numerator should be ‘6’.
- Page 259, Exercise #26: new
-
(Spotted by Momo Mizuguchi)
The answer in the back of the book is off by a factor of 2.
The correct answer is $\ds \frac{4(3s^2 - 4)}{(s^2+4)^3}$.
- Page 266, Table 5.4: new
-
(Spotted by Momo Mizuguchi)
In item 5, the large parentheses are meant to be large curly braces, as in
$\mathscr{L}^{-1}\Big\{F(s-a)\Big\}$.
- Page 267, Exercise #28: new
-
(Spotted by Menachem Greenfeld)
In parts (a) and (b) the functions $LC$ and $RC$ are referred
to as “Laplace transforms”. However they should be referred to
as inverse Laplace transforms.
- Page 270, lines 1 and 3 from below:
-
(Spotted by Dr. Gobbert)
The coefficients of $\sin 2t$ on both of these lines should
be $\frac13$ rather than $\frac23$. That is,
\begin{align*}
Y(s)
&=
\frac{1}{3(s^2+1)} + \frac{2}{3(s^2+4)} \\
&=
\frac{1}{3} \mathscr{L}\{\sin t\}
+
\frac{1}{3} \mathscr{L}\{\sin 2t\}
\end{align*}
and
\[
y(t) = \frac{1}{3} \sin t + \frac{1}{3} \sin 2t.
\]
- Page 271, Equation (9):
-
(Spotted by UHM)
The term $3\big(\mathscr{L}\{y\} - y(0)\big)$
in the middle of the left hand side is missing an $s$.
It should be $3\big(s\mathscr{L}\{y\} - y(0)\big)$.
- Page 273, Equation (17):
-
There should be no minus sign in the exponent. The correct equation is:
\[
Y(s) = \frac{1}{s^2} + \frac{C}{s^2} e^{s^2/2}.
\]
- Page 275, Exercise #26:
-
(Spotted by UHM)
Equation (26b) should be $(D+2)y = v$.
- Page 282, Exercise #3:
-
(Spotted by UHM)
The answer in the back of the book has a typo—the first
minus should be a plus, as in:
\[u(t) + (4t - t^2 - 1)u(t-1) + (1 - 4t + t^2)u(t-2).\]
Better yet, this may be expressed more clearly as
\[u(t) + (4t - t^2 - 1)u(t-1) - (4t - t^2 - 1)u(t-2).\]
- Page 286, Example #3:
-
(Spotted by Dr. Gobbert)
The ‘1’ on the right hand side of the equation should be
‘$\pi$’, as in:
\[
y'' + 3y' + 2y =
\begin{cases}
t & (0 \le t < \pi) \\
\pi & (\pi \le t)
\end{cases}
\]
- Page 290, Exercise #4:
-
(Spotted by Donna Young)
The answer in the back of the book is not quite correct. It should be
$
y(t) = t - \sin t - u(t-1)\big( t - \cos(t-1) - \sin(t-1) \big).
$
- Page 290, Exercise #6:
-
The solution in the back of the book is malformed. The correct solution is
$y(t) = \sin t + u(t-3)\big[1 - \cos(t-3)\big]$.
- Page 290, Exercise #8:
-
(Spotted by Dr. Bell)
The answer in the back of the book has an extra $-u(t-\pi/2)$.
It should have been
\[
y(t) = \frac14 t - \frac18 \sin2t
-u\big(t-\frac{\pi}{2}\big)\Big(
\frac14\big(t-\frac{\pi}{2}\big)
-\frac18\sin2\big(t-\frac{\pi}{2}\big) \Big).
\]
- Page 290, Exercise #10:
-
(Spotted by Christine Sweigart, reported by Dr. Lo)
The answer in the back of the book is wrong. The correct solution is
\[
y(t)
= \frac{1}{3} \sin t - \frac{1}{6} \sin 2t
- u(t-2\pi) \big[
\frac{1}{3} \sin (t-2\pi) - \frac{1}{6} \sin 2(t-2\pi) \big].
\]
- Page 290, Exercise #11: new
-
(Spotted by Jeffrey Barden)
The answer in the back of the book has an extra factor of 2 in one of its
terms. Specifically, the $2\sqrt{3}$ in the middle of the
answer should be $\sqrt{3}$.
- Page 290, Exercise #12:
-
(Spotted by Dr. Bell)
The answer in the back of the book is incorrect.
It should have been
\[
y(t)
= 2 - 2e^{-t}\cos2t - e^{-t}\sin2t
- u(t-\pi) \Big(
2 - 2e^{-(t-\pi)}\cos2t - e^{-(t-\pi)} \sin2t \Big).
\]
This may be expressed more succinctly as
\[
y(t) = g(t) - u(t-\pi) g(t-\pi),
\quad\text{where } g(t) = 2 - 2e^{-t}\cos2t - e^{-t}\sin2t.
\]
- Page 293, in the footnote:
-
It says “The area under the curve has units of
(force $\times$ time) which has units of energy.”
This is not correct. Energy would have units of
(force $\times$ distance). The integral of force with
respect to time is called impulse in physics, not energy.
- Page 294: new
-
(Spotted by Menachem Greenfeld)
The equation immediately above equation (2) is missing a factor of $m$
on the left hand side.
- Page 295, last paragraph:
-
It says
“Also, since $\delta(t)$ represents an impulse of
unit energy acting at $t=0$”. That should say
“Also, since $\delta(t)$ represents a unit impulse
at $t=0$”. Energy and impulse are not the same things;
see the previous remark.
- Page 296, Figure 5.32:
-
(Spotted by Dr. Gobbert)
In the diagram on the left, the height of the graph
is shown as $\frac{1}{2}h$. It should be $\frac{1}{2h}$.
- Page 296:
-
(Spotted by Dr. Gobbert)
Two lines below the boxed Historical Note we have
$\delta_h(t) = 1 - u(t-h)$.
This should be
$\delta_h(t) = \frac{1}{h}\big[1 - u(t-h)\big]$.
- Page 297, Equation (7):
-
(Spotted by Dr. Gobbert)
We have:
\[
\delta_h(t) = \frac{1}{h}\big[1 - u(1-h)\big]
\]
This should be
\[
\delta_h(t) = \frac{1}{h}\big[1 - u(t-h)\big]
\]
- Page 297, last line:
-
It says
“it is struck by a hammer in the downward direction, exerting
unit energy on the mass”. That
“unit energy”
should be
“unit impulse”.
Energy and impulse are not the same things; see the preceding
remarks on this issue.
- Page 301, Exercise #11: new
-
(Spotted by Momo Mizuguchi)
In the answer in the back of the book, the term
$2\sin t$
should be just
$\sin t$.
- Page 311, Exercise #17:
-
(Spotted by UHM)
The answer in the back of the book is not quite correct.
It should be $y = \frac12 \big[ \sin 2t + \sin 2t \ast f(t) \big]$.
- Page 311, Exercise #19:
-
(Spotted by Dr. Bell)
The answer in the back of the book has a typo. The term
$4\sin t$ in it should be $2\sin t$.
- Page 311, Exercise #23:
-
(Spotted by Bradley Scott)
The answer in the back of the book has a typo.
It should be $4t \ast e^{2t} = e^{2t} - 2t - 1$.
- Page 344, Figure 6.3: new
-
(Spotted by Momo Mizuguchi)
The point labeled $(-1,2)$ should be labeled $(1,-2)$.
- Page 353, Exercise #27:
-
(Spotted by Dr. Bell)
In part (b), the condition $(a-d)^2=0$ should be $(a-d)^2 + 4bc=0$.
- Page 357, Equation (14):
-
The expression $(1-\lambda)(\lambda^2+2)$ should be $(1-\lambda)(\lambda^2+4)$.
- Page 368, last line before the footnote:
-
It says:
“The following example illustrates that nonlinear equations…”.
The “nonlinear” should be
“nonhomogeneous”.
- Page 397: new
-
(Spotted by Menachem Greenfeld)
Top of the page, equation (b): The first $y_{n+1}$ should be $y_{n+2}$.
- Page 493, the boxed Historical Note:
-
It refers to “The Soviet mathematician Aleksandr Mikhailovich Liapunov
(1857–1918)…”. However Liapunov was not a Soviet
mathematician—he died before there was a Soviet Union. As the footnote
on page 488 correctly states, Liapunov was a Russian mathematician.
- Page 569: new
-
(Spotted by Menachem Greenfeld)
In the next to last step in equation (5), the denominator is
$c^2 - cdi + cdi - d^2i$
but it should be
$c^2 - cdi + cdi - d^2i^2$.
- Inside back cover: new
-
(Spotted by Menachem Greenfeld)
There is an extraneous parenthesis in formula 17. The correct expression is:
\[
\int \sqrt{u^2 \pm a^2} \, du
=
\frac{1}{2} \Big(
u \sqrt{u^2 \pm a^2}
\pm
a^2 \ln \big( u + \sqrt{u^2 \pm a^2} \, \big) \Big)
\]
