- 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 252, Exercise #27:
-
(Spotted by Erica Ford)
The
$1 \le t < 1$
should be
$0 \le t < 1$.
- Page 253, Exercise #46:
-
(Spotted by UHM)
In part (a), the denominator
$1 - (a + ik)$
should be
$s - (a + ik)$.
- Page 254:
-
(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):
-
(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:
-
(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 #17:
-
(Spotted by Matthew Bleakney, reported by Dr. Sousedik)
The answer in the back of the book has a typo—the sign between
the two fractions should be a minus, not a plus.
- Page 259, Exercise #18:
-
(Spotted by Paul Rydeen)
In the book's answer, the second fraction should have a "2"
in the denominator just like the first fraction does.
- Page 259, Exercise #26:
-
(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 259, Exercise #42:
-
(Spotted by Salen Nhean, reported by Dr. Sousedik)
The upper bound in the inner integral should be $\tau$, not $u$.
- Page 266, Table 5.4:
-
(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 #12:
-
(Spotted by Marina Congedo, reported by Dr. Hoffman)
The answer in the back of the book is slightly off. It should be
$\ds\frac{7}{\sqrt{3}} e^{-2t} \sin \sqrt{3} t$.
- Page 267, Exercise #28:
-
(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.
(Spotted by Paul Rydeen)
Every $e^{-s}$ in this exercise should be $e^{-s}/s$.
- Page 267, Exercise #29:
-
(Spotted by Paul Rydeen)
The expression given for $A_k$ is sloppy and does not stand
scrutiny. The proper expression is
$A_k = \lim_{s\to r_k} (s-r_k) P(s)/Q(s)$
which, through the application of l'Hôpital's rule, simplifies to
$A_k = P(r_k)/Q'(r_k)$, where $Q'$ is the derivative of $Q$.
Consequently, in equation (24) the denominator $Q(r_k)$ should be $Q'(r_k)$,
as in
\[
\mathscr{L}^{-1}\bigl(F(s)\bigr)
= \sum_{k=1}^n \frac{P(r_k)}{Q'(r_k)} e^{r_k t}.
\]
- 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
\[
y(t) = \frac{1}{3} \sin t + \frac{1}{3} \sin 2t.
\]
(Additional comment from Dr. Sousedik) Consequently,
the graph in Figure 5.7 does not represent the problem's
correct solution.
- 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 274, Exercise #21:
-
(Spotted by Pi Fisher)
Equation (19) should be
\[
\mathscr{L}\big\{ t^2 y'(t) \big\} = s Y''(s) + 2 Y'(s).
\]
Moreover, the wisdom of suggesting to apply Property 3
from Table 5.4 of page 266 is questionable. Applying
Property 7 from Table 5.2 of page 258 makes better sense.
- Page 275, Exercise #25:
-
(Spotted by Paul Rydeen)
The answer in the back of the book reads:
“The solution of $y' = ky$ grows faster ($e^t > \cosh kt$)”.
That is meant to be $e^{kt} > \cosh kt$.
But that's immaterial since the conclusion is fundamentally incorrect.
The fact is, $k > \sqrt{k}$ if $k >1$, but $k < \sqrt{k}$ if $k <1$.
Therefore, the solution of $y' = ky$ grows faster than that of the
other equation if $k>1$, and it grows
slower if $k<1$.
- 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 282, Exercise #11:
-
(Spotted by Tanner Saslow)
[This pertains only to the
original 1994 hardcover edition—well,
one of the hardcover versions, since there are multiple versions of it around.]
The problem asks for the Laplace transform of $u(t-1)e^{-t}$,
while the answer in the back of the book corresponds to the Laplace transform
of $u(t)$. In Dover's softcover edition the problem's statement
has been changed to $u(t)$ and it matches the book's answer.
- Page 282, Exercise #27:
-
(Spotted by Paul Rydeen)
The answer in the back of the book is not correct. It should be
\[
\frac{s+1}{s^2} e^{-s} - \frac{3s+1}{s^2} e^{-3s}.
\]
- Page 283, Exercise #31:
-
(Spotted by Paul Rydeen)
The exercise, and the accompanying figure, refer to
“square cosine wave”.
The correct description would be
“square sine wave”.
- Page 284, sixth line from the bottom:
-
(Spotted by Celia Drew, reported by Dr. Dean)
The $\mathscr{L}\big\{ u(t-1) \big\}$ should be
$\mathscr{L}\big\{ u(t-\pi) \big\}$.
- Page 286, equation (7):
-
(Spotted by Dr. Sousedik)
There are two typos here:
- In the equation's top line, the $e^t$ should be $e^{-t}$.
- In the equation's bottom line, there should be a minus
sign in front of the $\frac{1}{2}$.
- 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:
-
(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 291, Figures 5.26 and 5.27:
-
(Spotted by Paul Rydeen)
The labels on the graphs' horizontal axes are
1, 2, 3, ….
They should be
π, 2π, 3π, ….
- Page 291, Exercise #17:
-
(Spotted by Paul Rydeen)
The two occurrences of 2/π in the problem's statement
should be π/2.
- 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.
This misuse of the word “energy” occurs again near
the bottoms of pages 295 and 297.
- Page 294:
-
(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 296, equation (4):
-
(Spotted by Celia Drew, reported by Dr. Dean)
In the final term the $e^{hs}$ should be $e^{-hs}$.
- Page 297, Equation (5):
-
(Spotted by Philip Ruijten)
A factor of $s$ is missing in the last fraction.
The entire formula should be:
\[
\mathscr{L}\big\{\delta(t)\big\}
= \lim_{h\to0} \Big( \frac{1-e^{-hs}}{hs} \Big)
= \lim_{h\to0} \Big( \frac{se^{-hs}}{s} \Big)
= 1.
\]
- 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:
-
(Spotted by Momo Mizuguchi)
In the answer in the back of the book, the term
$2\sin t$
should be just
$\sin t$.
- Page 302, Exercise #25:
-
(Spotted by Paul Rydeen)
In part (c), the expression given for $y(N)$ has an $e^{-aN}$ in
the numerator. That should be $e^{-kN}$.
- Page 304, Figure 5.37:
-
(Spotted by Paul Rydeen)
Both the figure and the corresponding text require several adjustments.
In particular, the function given as $g(\tau)=e^{-\tau}$ is meant to have a
multiplicative factor of $>1$.
For the illustration below, which is drawn to-scale,
we take $g(\tau)=1.4\,e^{-\tau}$, and $t_1=0.5$, $t_2=0.7$, $t_3=1.5$.
The resulting convolution, plotted in the rightmost diagram, is
\[
(f \ast g)(t) =
\begin{cases}
1.4\,(1 - e^{-t}) & \text{if } t \le 1,
\\
1.4\,(e - 1) e^{-t} & \text{if } t > 1.
\end{cases}
\]
- 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 311, Exercise #28:
-
(Spotted by Paul Rydeen)
In equation (21) the $I_{-1/2}$ should be $I_{1/2}$.
Aside: In the definition (21) of the fractional derivative,
we integrate by a half order first, then differentiate by one order,
and thus arrive at the derivative of $-1/2$ order. The result is
the Riemann-Liouville (R–E) fractional
derivative of order $-1/2$.
As seen in part (a) of this exercise, the
R–E derivative of a constant is nonzero! That's not terribly useful.
In the newer Davison–Essex definition of the fractional derivative
we reverse the order of the operations—we differentiate by one order first,
and then integrate by $1/2$ order to obtain the derivative of $-1/2$ order.
That way the fractional derivative of a constant comes out zero.
![[graph]](farlow-page304.png)