Errata in Farlow's Differential Equations — Chapter 6

Page 316:

(Spotted by Dr. Sousedik) Within the shaded box, all 9 occurrences of $x_0$ should be $t_0$,

Page 319, equation immediately preceding equation (17):

(Spotted by Menachem Greenfeld) On the right-hand side, the term $x_2 c_1 e^{3t}$ should be $c_1 e^{3t}$. Equation (17), however, is correct.

Page 323, Exercise #24:

(Spotted by Ethan Piercy, reported by Dr. Peercy) The answer for $r(t)$ in the back of the book is correct, but $j(t)$ has its sign reversed. It should be $$j(t) = -c_1 \sqrt{\frac{b}{a}} \cos \bigl( \sqrt{ab} t \bigr) + c_2 \sqrt{\frac{b}{a}} \sin \bigl( \sqrt{ab} t \bigr)$$

Page 330, equation (20):

(Spotted by Noah Zambidis) The $\ds\frac{d}{dx}$ should be $\ds\frac{d}{dt}$.

Page 338, line 2:

(Spotted by Ryan Olson) $x_{ij}(x)$ should be $x_{ij}(t)$.

Page 344, Figure 6.3:

(Spotted by Momo Mizuguchi) The point labeled $(-1,2)$ should be labeled $(1,-2)$.

Pages 349–350, Example 5:

(Spotted by Paul Rydeen) In equation (28) the $2$ and $-2$ have been swapped. The correct version is $$\boldsymbol{\xi}^{(1)} = c \begin{bmatrix} 1 \\ -2 \end{bmatrix} \qquad \boldsymbol{\xi}^{(2)} = c \begin{bmatrix} 1 \\ 2 \end{bmatrix}$$ The $2$ and $-2$ should be swapped in the same way in equations (29a), (29b), and (30).

Page 352, Exercise #18:

(Spotted by Paul Rydeen) In the answer for $x_2(t)$ in the back of the book, $e^{2t}$ should be $e^{3t}$.

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 353, Exercise #30:

(Spotted by Paul Rydeen) The answer to part (b) in the back of the book in incorrect. It should be $$\begin{bmatrix} I(t) \\ X(t) \end{bmatrix} = c_1 e^{-k_1 t} \begin{bmatrix} k_2 - k_1 \\ k_1 \end{bmatrix} + c_2 e^{-k_2 t} \begin{bmatrix} 0 \\ 1 \end{bmatrix}$$

Page 357, Equation (14):

The expression $(1-\lambda)(\lambda^2+2)$ should be $(1-\lambda)(\lambda^2+4)$.

Page 358–359, Exercises #13, 14, 18:

(Spotted by Paul Rydeen) The solution to each of these exercises should be of the form given in equation (18).The answers in the back of the book are missing the $c$ multiplier.

Page 358, Exercise #15:

(Spotted by Paul Rydeen) In the comments preceding Exercise 15, the vectors $x^{(1)}(t)$ and $x^{(2)}(t)$, and the matrix $X(t)$ should be $$x^{(1)}(t) = \begin{bmatrix} e^{-t} \\ -2 e^{-t} \end{bmatrix}, \quad x^{(2)}(t) = \begin{bmatrix} e^{3t} \\ 2 e^{3t} \end{bmatrix}, \quad X(t) = \begin{bmatrix} e^{-t} & e^{3t} \\ -2 e^{-t} & 2e^{3t} \end{bmatrix}.$$

Page 361, Example 1:

(Spotted by Matthew Lastner) In line 1 of equation (7), the matrix marked $A^4$ should be the identity matrix, that is: $$A^4 \rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}.$$ The rest of the calculation is correct as is.

Page 364, in the boxed statement of Theorem 6.9:

(Spotted by Paul Rydeen) The general solution is given as $\dot{x}$. If should be $x$ without a dot.

Page 366, Exercise #5:

(Spotted by Paul Rydeen) The solution in the back of the book needs to be fixed. The last term of the solution which is given as
$-\left[\begin{array}{r} -10 \\ 1 \end{array}\right]$,
should be
$-\left[\begin{array}{r} \frac{14}{3} \\ -\frac{11}{3} \end{array}\right]$.

Page 366, Exercise #7:

(Spotted by Paul Rydeen) The solution in the back of the book needs to be fixed. The last term of the solution which is given as $\frac{1}{2}\left[\begin{array}{r} e^{5t} \\ e^{5t} \end{array}\right]$, should be $\left[\begin{array}{r} 0 \\ -e^{5t} \end{array}\right]$, or equivalently, $e^{5t} \! \left[\begin{array}{r} 0 \\ -1 \end{array}\right]$.

Page 366, Exercise #11:

(Spotted by Paul Rydeen) The right-hand side of the equation should have $e^{-At}$ rather than $e^{At}$.

Page 368, last line before the footnote:

It says: “The following example illustrates that nonlinear equations…”. The “nonlinear” should be “nonhomogeneous”.

Page 369, Example 2, Equation (8):

(Spotted by Paul Rydeen) $X_1(x)$ should be $X_1(s)$.

Page 371, Example 3, Equation (14):

(Spotted by Paul Rydeen) In the second equation of the system (14), the $-m_1$ should be $m_1$.

Page 373, Exercise #8:

(Spotted by Paul Rydeen) The answer in the back of the book says $x(t) = \cdots$. It should say $\begin{bmatrix} x(t) \\ y(t) \end{bmatrix} = \cdots$.

Page 374, Exercise #16:

(Spotted by Paul Rydeen) In the first equation, the term $k_1m$ should be $-k_1m$.

Pages 376–378, Example 2:

(Spotted by Thomas Hsu) A numerical error has crept in Equation (4) on page 377—the eigenvectors are incorrect as shown. The corrected version of the equation is: $$x_h(t) = \begin{bmatrix} x_1(t)\\x_2(t) \end{bmatrix} = c_1 e^{-0.025 t} \begin{bmatrix} 1 \\ 1.732 \end{bmatrix} + c_2 e^{-0.095 t} \begin{bmatrix} 1 \\ -1.732 \end{bmatrix}.$$ Equation (5) on page 378 needs to be corrected accordingly. Upon applying the initial conditions we find that $c_1 = -26.93$ and $c_2 = 1.93$. Finally, we redo the graphs in Figure 6.13 on page 378 and we get:

Aside: If instead of decimal points we use true fractions in the computations, we see that the eigenvalues and eigenvectors, and the coefficients $c_1$ and $c_2$ are: $$\lambda = \frac{1}{50} \big[ -3 \pm \sqrt{3} \big], \quad \mathbf{v} = \begin{bmatrix} 1 \\ \pm \sqrt{3} \end{bmatrix}, \quad c_1 = -25\Big(\frac{1}{2} + \frac{1}{3} \sqrt{3}\Big), \quad c_2 = -25\Big(\frac{1}{2} - \frac{1}{3} \sqrt{3}\Big).$$

Page 379, Equation (8):

(Spotted by Paul Rydeen) The $m_1$ in the second line should be $m_2$.

Page 382, Equation (12):

(Spotted by Paul Rydeen) That equation is incorrect since it accounts for material entering each compartment but does not account for material leaving that compartment. The correct equation is $$\begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \\ \dot{x}_3 \end{bmatrix} = \begin{bmatrix} 0 &\quad a_{12} - a_{21} &\quad a_{13} - a_{31} \\ a_{21} - a_{12} &\quad 0 &\quad a_{23} - a_{32} \\ a_{31} - a_{13} &\quad a_{32} - a_{23} &\quad 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}.$$

Page 383, Exercises #6:

(Spotted by Paul Rydeen) In view of the comment above, the correct answer to part (a) is $$\begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \\ \dot{x}_3 \end{bmatrix} = \begin{bmatrix} 0 & -0.02 & -0.06 \\ 0.02 & 0 & 0.04 \\ 0.06 & -0.04 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}.$$

Page 383, Exercises #7:

(Spotted by Paul Rydeen) As in the previous comment, the correct system of differential equations is $$\begin{bmatrix} \dot{x}_1 \\ \dot{x}_2 \\ \dot{x}_3 \end{bmatrix} = \begin{bmatrix} 0 & -0.25 & 0.15 \\ 0.25 & 0 & -0.2 \\ -0.15 & 0.2 & 0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}.$$

Pages 386–387, Example 1:

(Spotted by Paul Rydeen) The first two lines of equation (9b) are correct but in the third line the final parenthesized 1 should be 2. On the next line, the answer $-0.1$ should be changed to $-0.099666\ldots$. That value is shown correctly in Table 6.1.

Page 389, Example 2:

(Spotted by Paul Rydeen) In the second line of the Solution, the $x_0=1$ should be $x_0=2$. The rest of the calculations uses the correct value of $x_0=2$.

Page 391, Exercises 4–8:

(Spotted by Paul Rydeen) Following the notation of Example 1, we introduce $x_1=y$, $x_2=y'$ (and in the case of Exercise 8, $x_3 = y''$), so the solutions would be things like $x_1(0.1) = \text{something}$, $x_2(0.1) = \text{something}$, etc. The answers given in the back of the book have $x(0.1) = \text{something}$, $y(0.1) = \text{something}$ which are inconsistent with that notation.

Page 391, Exercise 8:

(Spotted by Paul Rydeen) The book's answer is correct but incomplete. To covert this third order equation into a system, we let $x_1 = y$, $x_2 = y'$, $x_3 = y''$. Then we get $$\begin{align*} x_1' &= x_2, \\ x_2' &= x_3, \\ x_3' &= - x_1^2, \end{align*}$$ along with the initial conditions $x_1(0)=1, x_2(0) = 0, x_3(0) = 0$. Applying the Runge--Kutta algorithm with $h=1/10$ results in $$x_1\bigl(\frac{1}{10}\bigr) = \frac{5999}{6000}, \quad x_2\bigl(\frac{1}{10}\bigr) = -\frac{1}{200}, \quad x_3\bigl(\frac{1}{10}\bigr) = -\frac{95992001}{960000000},$$ or expressed in decimal fractions $$x_1(0.1) \approx 0.9998333333, \quad x_2(0.1) \approx -0.005, \quad x_3(0.1) \approx -0.0999916677.$$