- Page 35, Figure 2.2
-
The figure indicates that
the line x=π/4 is the vertical asymptote to the various solutions.
That's not correct.
The vertical asymptote is the line x=0 since
the solution goes to infinity as x approaches zero due to the
x3 in the denominator.
- Page 36, Exercise #6:
-
The answer in the back of the book has a typo;
the correct answer is y=ce−x2+1/2.
- Page 36, Exercise #8:
-
(Spotted by Trey Mace)
The answer in the back of the book assumes, unnecessarily, that x>0.
The proper form of the solution is y=cx+ln|x|x.
The domain of the solution can be either −∞<x<0 or 0<x<∞.
- Page 36, Exercises #9 and #15:
-
(Spotted by Trey Mace)
The answers in the back of the book assume, unnecessarily, that x>0.
The domain of the solutions can be either −∞<x<0 or 0<x<∞.
- 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)ex2=1+y. Better yet, y may be computed explicitly as
y=1+ce−x21−ce−x2.
(Spotted by Dr. Jean Bragard)
Oops! The “correction” above is incorrect!
The correct solution is c(1−y)e1/x2=1+y, or equivalently,
y=ce1/x2−1ce1/x2+1.
- 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−y2 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=(1−∫x1et2dt)−1.
- Page 43, Exercise #33:
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(Spotted by Dr. Stegenga)
In the statement of the problem, the 1+y2 term is meant to be
outside the square root, as in
dydx=√1+sinx(1+y2).
- 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=cx2−x.
- Page 45, Exercise #45:
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(Spotted by Dr. Mayans)
The differential equation to be solved is y′2+y2=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 #46:
-
Not an error, however this is not truly a differential equations
problem as it amounts to solving a trivial equation of the form
dy/dt=f(t).
- 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.
Furthermore, the statement
“where ts is the time that it started snowing”
is incorrect. It should be
“where ts is the length of time that it had been snowing
before the snowplow began to move”.
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 46, Equation (27c):
-
(Spotted by Paul Rydeen)
The ds should be outside of the radical sign.
- Page 49, Example 1:
-
It says that the half-life of C–14 is approximately 5600 years.
That number is outdated.
According to the latest data, C–14's half-life is approximately
5730 years. However for the sake of consistency with the calculations
of the rest of this chapter, it is good advice
to adhere to that 5600 number.
- Page 50:
-
(Spotted by Dr. Soane)
In equation (5) the rightmost term is missing a negative sign in the exponent.
It should be Q0e−0.00012378t.
- Page 50:
-
(Spotted by Daniel Engbert)
The 15,336 in equation (7) should be 15,327.
- Page 56, Exercise #15:
-
(Spotted by Paul Rydeen)
The answer in the back of the book lacks sufficient precision.
A better value for the coefficient 0.000004 is 4.3481×10−6 which is larger by more than 8 percent.
- 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:
-
(Spotted by Momo Mizuguchi)
The answer 5e0.3ln2 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:
-
(Spotted by Menachem Greenfeld)
The answer in the back of the book is missing the word “million”.
It should be 5e4ln2 million=80 million.
- Page 57, Exercise #27:
-
(Spotted by Rachana Haliyur, reported by Dr. Peercy)
The answer to part (b) should be 2435 A.D.
- Pages 58–59, Exercise #34:
-
(Spotted by Celia Drew and James Loy, and reported by Dr. Dean)
The numerical answer in the back of the book
is correct but the exponential
form of the answer is missing a factor of 24. It was meant to say
24e(0.08)(366)≐$124,839,000,000,000.
However…
James Loy points out that
24e(0.08)(366)≐$124,839,974,659,582.22,
therefore the properly rounded answer is $124,840,000,000,000.
Moreover…
Max Bobbin points out that most of the numbers on the
associated diagram in Figure 2.14 on page 59 are incorrect.
For instance, the value corresponding to the year 1990 is shown as
2.79×1016 while it should be 1.0638×1014.
The first four numbers in the figure's gray inset correspond
to an interest rate of 10% rather than 8%.
It's not clear where the fifth number comes from.
- Page 62, between equations (4) and (5):
-
(Spotted by Celia Drew, reported by Dr. Dean)
It says μ(x)=e3t/50. That μ(x) should be μ(t).
- Page 64:
-
(Spotted by Jessica Alignay, reported by Dr. Hoffman)
In the equation that comes just after equation (10), the
(100−t2) on the left-hand side should be (100−t)2.
- Page 65, Exercise #1:
-
(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:
dQdt+(5100−2t)Q=0,Q(0)=300.
- Page 66, Exercise #11:
-
(Spotted by Paul Rydeen)
The answer in the back of the book is correct, however the
problem's original variable x has been renamed Q.
- Page 71, in the middle of the page:
-
(Spotted by Dr. Robert Kouba)
We have: “To give meaning to 1/k,
temperature T(k) of an object…”.
Here the T(k) should be T(t).
- 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.
(Spotted by Paul Rydeen)
If the data above is correct, then the can must have been taken out of the
refrigerator's freezer section!
- Page 74, Exercise #8:
-
Based on the given data, the decay constant turns out to be
k=(ln2)/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 75, Exercise #13:
-
The solution in the back of the is correct except for the numerical values
of the temperature in parts (c) and (d), which should be
61.07°F and 68.93°F, respectively.
Moreover,
(Spotted by Paul Rydeen)
The answer to part (b) in the back of the book is expressed
as Tss(s). That should be Tss(t).
- Page 77, Equation (4):
-
(Spotted by Marie Wagner)
The malformed fraction m2g should be m2gk2.
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:
dθdt=√v2d−v2sv2s1t.
- Page 80
-
(Spotted by Sean Clements)
In the middle of the page,
in the line immediately preceding the boxed equations (14),
reference is made to the
“general spiral equation (1)”. That should be
“general spiral equation (7)”.
- Page 83: Exercise #9:
-
(Spotted by Sophie Peet)
The goal of this exercise is to find a function g(x,y) so that
dg=xdx+ydyx2+y2. The answer ln(x2+y2)
in the back of the book is missing a factor of 1/2. It should be
g=12ln(x2+y2).
- Page 84, Exercise #12:
-
The answer of 28.8 sec in the back of the book is correct.
But the comment “He would have fallen
6280 ft” is not; it should be 7013 ft.
- Page 84, Exercise #17:
-
(Spotted by Hailey Lynch, reported by Dr. Dean)
The answer in the back of the book is missing a negative
sign in the exponent and also suffers from improper rounding.
The correct expressions are:
v(t)=1207(1−e−t/375)≈17.14285714(1−e−0.002666666667t)≈17.1(1−e−0.0027t).
In particular, the ship's limiting velocity is approximately 17.1 ft/sec
or 11.7 mph.
- Page 85, Exercise #20:
-
(Spotted by Sean Shields)
In the equation given in part (b), the term
√2gt should be √2gt, that is, the t should
not be under the radical.
- Page 85, Exercise #20:
-
(Spotted by Paul Rydeen)
Part (e) refers to a “a 1–inch hole”.
The “1–inch”
is the hole's radius, not the diameter.
- 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)=x2(xk1+k−x−k1−k)+k1−k2.
- Page 86, Exercise #23:
-
(Spotted by Dr. Dovermann)
The equation of motion, given in part (b),
has an extra minus sign. It should
be dvdt=−grR.
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:
-
(Spotted by Sean Shields)
In part (b), “drive” should be “derive”. Furthermore,
the h3 in the numerator should be h2.
- Page 94, Equation (6):
-
(Spotted by Mayur Darji, reported by Dr. Dean)
The v′ on left-hand side of the equation should be mv′.
- Page 97:
-
(Spotted by Andrew Zuelsdorf)
In the line just before equation (17) we have:
“If Newton's method with step size h is used to compute N
approximations …”.
“Newton” in that sentence is supposed to be “Euler”.
- Page 98, Exercise #20:
-
(Spotted by Menachem Greenfeld)
In part (a) the t=1 should be x=1.
- Page 99, Exercise #25:
-
The equation y=x2−y2 is meant to be y′=x2−y2.
- Page 103, equation (9):
-
(Spotted by Dr. Nathaniel Mays)
The equation is missing a yn term on the right-hand side. It should be
yn+1=yn+h6(kn1+2kn2+2kn3+kn4).
- Page 103, equation (10b):
-
The formula for kn2 is incorrect. It should be
kn2=f(xn+12h,yn+12hkn1).
The formula is stated correctly in the boxed summary on page 104
and the BASIC program on page 105.
- Page 105, in the boxed program:
-
(Spotted by Trey Mace)
The line immediately preceding the calculation of k4 reads
yy = y + h * k3/2.
That should be
yy = y + h * k3.
- Page 108, Exercise #18:
-
(Spotted by Trey Mace)
The initial value problem is stated as
y′=1yy(1)=0.
That is meant to be
y′=1xy(1)=0.