Minor spacing changes some minor wording changes

CHANGES

@@ -1,4 +1,6 @@
 <li><b>Change Exercise 3.1.106</b> to fix an erratum.  Rate of flow from tank 2 back into tank 1 should
 be \(r-s ,\) otherwise volumes do not stay constant and the problem doesn't quite all make sense.
 <li>Add a couple of minor clarifications to the end of Example 1.9.3.
-<li>A few minor clarifications and improvements in english and style.
+<li>Slightly tighter spacing in 3.1 leads to a slight change (for the better I
+think) in pagination in the section.
+<li>A few minor clarifications and improvements in English and style.

ch-higher-order-ode.tex

@@ -441,10 +441,12 @@ \subsection{Solving constant coefficient equations}
 We need to solve for $C_1$ and $C_2$.  To apply the initial conditions,
 we first find $y' = 2 C_1 e^{2x} + 4 C_2 e^{4x}$.  We plug $x=0$ into
 $y$ and $y'$ and solve.
-\begin{align*}
+\begin{equation*}
+\begin{aligned}
 -2 & = y(0) = C_1 + C_2 , \\
 6 & = y'(0) = 2 C_1 + 4 C_2 .
-\end{align*}
+\end{aligned}
+\end{equation*}
 Either apply some matrix algebra, or just solve these by high school
 math.  For example, divide the second equation by 2
 to obtain $3 = C_1 + 2 C_2$, and subtract the two equations to
@@ -603,7 +605,7 @@ \subsection{Complex numbers and Euler's formula}
 
 The numbers
 $i$ and $-i$ are the two roots of $r^2 + 1 = 0$.
-Engineers often use the letter $j$ instead of $i$ for the square
+Some engineers use the letter $j$ instead of $i$ for the square
 root of $-1$.  We use the mathematicians' convention and use $i$.
 
 \begin{exercise}
@@ -702,16 +704,20 @@ \subsection{Complex roots}
 y_1 = e^{(\alpha+i\beta)x} \qquad \text{and} \qquad y_2 = e^{(\alpha-i\beta)x} .
 \end{equation*}
 Then 
-\begin{align*}
+\begin{equation*}
+\begin{aligned}
 y_1 & = e^{\alpha x} \cos (\beta x) + i e^{\alpha x} \sin (\beta x) , \\
 y_2 & = e^{\alpha x} \cos (\beta x) - i e^{\alpha x} \sin (\beta x) .
-\end{align*}
+\end{aligned}
+\end{equation*}
 
 Linear combinations of solutions are also solutions.  Hence,
-\begin{align*}
+\begin{equation*}
+\begin{aligned}
 y_3 & = \frac{y_1 + y_2}{2} = e^{\alpha x} \cos (\beta x) , \\ 
 y_4 & = \frac{y_1 - y_2}{2i} = e^{\alpha x} \sin (\beta x) ,
-\end{align*}
+\end{aligned}
+\end{equation*}
 are also solutions.  Furthermore, they are real-valued.  It is not hard to
 see that they are linearly independent (not multiples of each other).
 Therefore, we have the following theorem.
@@ -1128,11 +1134,13 @@ \subsection{Constant coefficient higher order ODEs}
 
 Suppose we were given some initial conditions $y(0) = 1$, $y'(0) = 2$,
 and $y''(0) = 3$.  Then
-\begin{align*}
+\begin{equation*}
+\begin{aligned}
 1 = y(0) & = C_1 + C_2 + C_3 , \\
 2 = y'(0) & = -C_1 + C_2 + 3C_3 , \\
 3 = y''(0) & = C_1 + C_2 + 9C_3 .
-\end{align*}
+\end{aligned}
+\end{equation*}
 It is possible to find the solution by high school algebra, but it would be a
 pain.
 The sensible way to solve a system of equations such as this is to use
@@ -2327,31 +2335,39 @@ \subsection{Variation of parameters}
 formula for the solution we could just plug into, but instead of memorizing
 that, it is better, and easier, to
 just repeat what we do below.  In our case the two equations are
-\begin{align*}
+\begin{equation*}
+\begin{aligned}
 u_1' \cos (x) + u_2' \sin (x) &= 0 ,\\
 -u_1' \sin (x) + u_2' \cos (x) &= \tan (x) .
-\end{align*}
+\end{aligned}
+\end{equation*}
 Hence
-\begin{align*}
+\begin{equation*}
+\begin{aligned}
 u_1' \cos (x) \sin (x) + u_2' \sin^2 (x) & = 0 ,\\
 -u_1' \sin (x) \cos (x) + u_2' \cos^2 (x) & = \tan (x) \cos (x) = \sin (x) .
-\end{align*}
+\end{aligned}
+\end{equation*}
 And thus
-\begin{align*}
+\begin{equation*}
+\begin{aligned}
 & u_2' \bigl(\sin^2 (x) + \cos^2 (x)\bigr) = \sin (x) , \\
 & u_2' = \sin (x) , \\
 & u_1' = \frac{- \sin^2 (x)}{\cos (x)} = - \tan (x) \sin (x) .
-\end{align*}
+\end{aligned}
+\end{equation*}
 We integrate $u_1'$ and $u_2'$ to get $u_1$ and $u_2$.
-\begin{align*}
+\begin{equation*}
+\begin{aligned}
 & u_1 = \int u_1'\,dx 
 = \int - \tan (x) \sin (x)\,dx
 = \frac{1}{2}
 \ln \left\lvert \frac{\sin (x)-1}{\sin (x) + 1} \right\rvert
 + \sin (x) , \\
 & u_2 = \int u_2'\,dx 
 = \int \sin (x)\,dx = -\cos (x) .
-\end{align*}
+\end{aligned}
+\end{equation*}
 So our particular solution is
 \begin{multline*}
 y_p = u_1 y_1 + u_2 y_2 =
@@ -2756,12 +2772,14 @@ \subsection{Damped forced motion and practical resonance}
 \end{equation*}
 
 We solve for $A$ and $B$:
-\begin{align*}
+\begin{equation*}
+\begin{aligned}
 & A=\frac{(\omega_0^2-\omega^2) F_0}
 {m{(2\omega p)}^2+m{(\omega_0^2-\omega^2)}^2} , \\
 & B=\frac{2 \omega p F_0}
 {m{(2\omega p)}^2+m{(\omega_0^2-\omega^2)}^2} .
-\end{align*}
+\end{aligned}
+\end{equation*}
 We also compute $C = \sqrt{A^2+B^2}$
 to be
 \begin{equation*}

ch-systems.tex

@@ -22,21 +22,25 @@ \subsection{Systems}
 For example, $y_1'' = f(y_1',y_2',y_1,y_2,x)$.
 Usually, when we have two dependent variables we have two equations
 such as
-\begin{align*}
+\begin{equation*}
+\begin{aligned}
 y_1'' & = f_1(y_1',y_2',y_1,y_2,x) , \\
 y_2'' & = f_2(y_1',y_2',y_1,y_2,x) ,
-\end{align*}
+\end{aligned}
+\end{equation*}
 for some functions $f_1$ and $f_2$.  We call the above a
 \emph{\myindex{system of differential equations}}.
 More precisely, the above is a \emph{\myindex{second order system}}
 of ODEs as second
 order derivatives appear.
 The system
-\begin{align*}
+\begin{equation*}
+\begin{aligned}
 x_1' & = g_1(x_1,x_2,x_3,t) , \\
 x_2' & = g_2(x_1,x_2,x_3,t) , \\
 x_3' & = g_3(x_1,x_2,x_3,t) ,
-\end{align*}
+\end{aligned}
+\end{equation*}
 is a \emph{\myindex{first order system}}, where $x_1,x_2,x_3$ are
 the dependent variables,
 and $t$ is the independent variable.
@@ -46,11 +50,13 @@ \subsection{Systems}
 For the system above, a
 \emph{solution}\index{solution to a system}
 is a set of three functions $x_1(t)$, $x_2(t)$, $x_3(t)$, such that
-\begin{align*}
+\begin{equation*}
+\begin{aligned}
 x_1'(t) &= g_1\bigl(x_1(t),x_2(t),x_3(t),t\bigr) , \\
 x_2'(t) &= g_2\bigl(x_1(t),x_2(t),x_3(t),t\bigr) , \\
 x_3'(t) &= g_3\bigl(x_1(t),x_2(t),x_3(t),t\bigr) .
-\end{align*}
+\end{aligned}
+\end{equation*}
 
 We usually also have an
 \emph{initial condition}\index{initial condition for a system}.  Just like
@@ -68,10 +74,12 @@ \subsection{Systems}
 by solving for one variable and then for the second variable.
 Take 
 the first order system
-\begin{align*}
+\begin{equation*}
+\begin{aligned}
 y_1' & = y_1 , \\
 y_2' & = y_1 - y_2 ,
-\end{align*}
+\end{aligned}
+\end{equation*}
 with $y_1$, $y_2$ as the dependent variables and $x$ as the independent
 variable.  And consider initial conditions
 $y_1(0) = 1$, $y_2(0) = 2$.
@@ -113,10 +121,12 @@ \subsection{Systems}
 higher than one ($x$, $y$, $x'$ and $y'$, constants, and functions of $t$
 can appear, but not $xy$ or ${(y')}^2$ or $x^3$).  Another, more
 complicated, example of a linear system is
-\begin{align*}
+\begin{equation*}
+\begin{aligned}
 y_1'' &= e^t y_1' + t^2 y_1 + 5 y_2 + \sin(t), \\
 y_2'' &= t y_1'-y_2' + 2 y_1 + \cos(t).
-\end{align*}
+\end{aligned}
+\end{equation*}
 
 \subsection{Applications}
 
@@ -154,10 +164,12 @@ \subsection{Applications}
 \end{equation*}
 Similarly we find the rate $x_2'$, where the roles of $x_1$ and $x_2$
 are reversed.  All in all, the system of ODEs for this problem is
-\begin{align*}
+\begin{equation*}
+\begin{aligned}
 x_1' & = \frac{r}{V} (x_2-x_1), \\
 x_2' & = \frac{r}{V} (x_1-x_2).
-\end{align*}
+\end{aligned}
+\end{equation*}
 In this system we cannot solve for $x_1$ or $x_2$ separately.  We must
 solve for both $x_1$ and $x_2$ at once, which is intuitively clear since
 the amount of salt in one tank affects the amount in the other.
@@ -210,10 +222,12 @@ \subsection{Applications}
 thus the same thing with a negative sign.
 \myindex{Newton's second law} states that
 force equals mass times acceleration.  So the system of equations is
-\begin{align*}
+\begin{equation*}
+\begin{aligned}
 m_1 x_1'' & = k(x_2-x_1) , \\
 m_2 x_2'' & = - k(x_2-x_1) .
-\end{align*}
+\end{aligned}
+\end{equation*}
 
 Again, we cannot solve for the $x_1$ or $x_2$ variable separately.
 That we must solve for both $x_1$ and $x_2$ at once
@@ -1502,13 +1516,15 @@ \section{Linear systems of ODEs}
 functions.  Let $A(t)$ and $B(t)$ be matrix-valued
 functions.  Let $c$ a scalar and let $C$ be a constant matrix.
 Then
-\begin{align*}
+\begin{equation*}
+\begin{aligned}
 \bigl(A(t)+B(t)\bigr)' & = A'(t) + B'(t), \\
 \bigl(A(t)B(t)\bigr)' & = A'(t)B(t) + A(t)B'(t), \\
 \bigl(cA(t)\bigr)' & = cA'(t), \\
 \bigl(CA(t)\bigr)' & = CA'(t), \\
 \bigl(A(t)\,C\bigr)' & = A'(t)\,C .
-\end{align*}
+\end{aligned}
+\end{equation*}
 Note the order of the multiplication in the last two expressions.
 
 A \emph{\myindex{first order linear system of ODEs}} is a system that can be
@@ -1524,10 +1540,12 @@ \section{Linear systems of ODEs}
 $\vec{x}$ satisfying the vector equation.
 
 For example, the equations
-\begin{align*}
+\begin{equation*}
+\begin{aligned}
 x_1' &= 2t x_1 + e^t x_2 + t^2 , \\
 x_2' &= \frac{x_1}{t} -x_2 + e^t ,
-\end{align*}
+\end{aligned}
+\end{equation*}
 can be written as
 \begin{equation*}
 {\vec{x}}' = 
@@ -2420,7 +2438,7 @@ \subsection{Complex eigenvalues}
 In other words, if $\lambda = a+ib$ is an eigenvalue, then so is $\bar{\lambda} = a-ib$.
 And if $\vec{v}$ is an eigenvector corresponding to the eigenvalue
 $\lambda$, then $\overline{\vec{v}}$ is an eigenvector corresponding
-to the eigenvalue $\bar{\lambda}$.  
+to the eigenvalue~$\bar{\lambda}$.  
 
 Suppose $a + ib$ is a complex eigenvalue of $P$, and $\vec{v}$
 is a corresponding eigenvector.  Then