ch4: second-order homogeneous, Wronskian, Abel's identity

chapters/ch04_second_order_homogeneous.tex

@@ -1,49 +1,751 @@
+% =============================================================================
+% ch04_second_order_homogeneous.tex
+% Chapter 4: Second-Order Linear Homogeneous Equations
+% =============================================================================
+
 \section{Second-Order Homogeneous}
 \label{ch:second_order_homogeneous}
 
-\subsection{Characteristic Equation}
+\subsection{The Characteristic Equation}
 \label{sec:ch04_characteristic_equation}
 
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+We consider the \textbf{second-order linear homogeneous differential equation with constant coefficients}:
+\begin{equation}
+  \label{eq:second_order_homogeneous}
+  a\,y'' + b\,y' + c\,y = 0,
+  \qquad a, b, c \in \R,\;\; a \neq 0.
+\end{equation}
+This is the simplest class of second-order ODEs that is both tractable analytically and broadly applicable to physical systems (mechanical vibrations, electrical circuits, etc.).
 
-\subsection{Three Root Cases}
+\paragraph{Derivation.} The key insight is that the exponential function $y = e^{rx}$ has derivatives proportional to itself, making it a natural candidate for solutions of linear equations with constant coefficients. Substituting the \textbf{ansatz}
-\label{sec:ch04_three_root_cases}
+\[
+  y = e^{rx}, \qquad y' = r\,e^{rx}, \qquad y'' = r^2\,e^{rx}
+\]
+into \cref{eq:second_order_homogeneous} gives
+\[
+  a\,r^2 e^{rx} + b\,r\,e^{rx} + c\,e^{rx} = 0.
+\]
+Since $e^{rx} \neq 0$ for all $x$, we divide through and obtain the \textbf{characteristic equation} (also called the \textbf{auxiliary equation}):
+\begin{equation}
+  \label{eq:characteristic}
+  a\,r^2 + b\,r + c = 0.
+\end{equation}
+This is an ordinary quadratic equation. Its roots are given by the quadratic formula:
+\begin{equation}
+  \label{eq:quadratic_formula}
+  r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
+\end{equation}
 
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+The nature of the roots depends on the \textbf{discriminant}
+\begin{equation}
+  \Delta = b^2 - 4ac.
+\end{equation}
+There are three cases.
+
+\begin{keyresult}
+  \textbf{Three cases for second-order homogeneous equations.}
+
+  \begin{center}
+  \begin{tabular}{l l l}
+    \toprule
+    \textbf{Discriminant} & \textbf{Roots} & \textbf{Solution form} \\
+    \midrule
+    $\Delta > 0$ & Two distinct real roots $r_1 \neq r_2$ & $y = c_1 e^{r_1 x} + c_2 e^{r_2 x}$ \\[6pt]
+    $\Delta = 0$ & One repeated real root $r = -b/(2a)$ & $y = c_1 e^{rx} + c_2 x\,e^{rx}$ \\[6pt]
+    $\Delta < 0$ & Complex conjugates $r = \alpha \pm i\beta$ & $y = e^{\alpha x}\bigl[c_1 \cos(\beta x) + c_2 \sin(\beta x)\bigr]$ \\
+    \bottomrule
+  \end{tabular}
+  \end{center}
+\end{keyresult}
+
+We treat each case in detail in the following subsections. First, two quick examples to illustrate the method.
+
+\begin{workedexample}
+  Solve $y'' - 5y' + 6y = 0$.
+
+  \textbf{Solution.} The characteristic equation is
+  \[
+    r^2 - 5r + 6 = 0 \quad\Longrightarrow\quad (r - 2)(r - 3) = 0.
+  \]
+  The roots are $r_1 = 2$ and $r_2 = 3$ (two distinct real roots, $\Delta = 25 - 24 = 1 > 0$). The general solution is
+  \[
+    y(x) = c_1 e^{2x} + c_2 e^{3x}.
+  \]
+\end{workedexample}
+
+\begin{workedexample}
+  Solve $y'' + 4y = 0$.
+
+  \textbf{Solution.} The characteristic equation is
+  \[
+    r^2 + 4 = 0 \quad\Longrightarrow\quad r^2 = -4 \quad\Longrightarrow\quad r = \pm 2i.
+  \]
+  Here $\alpha = 0$ and $\beta = 2$, so the general solution is
+  \[
+    y(x) = c_1 \cos(2x) + c_2 \sin(2x).
+  \]
+  This represents undamped oscillations (we will see the physical interpretation in \cref{sec:ch04_complex_roots}).
+\end{workedexample}
+
+\subsection{Case 1: Distinct Real Roots}
+\label{sec:ch04_distinct_real_roots}
+
+When $\Delta = b^2 - 4ac > 0$, the characteristic equation \cref{eq:characteristic} has two distinct real roots
+\[
+  r_1 = \frac{-b + \sqrt{\Delta}}{2a}, \qquad r_2 = \frac{-b - \sqrt{\Delta}}{2a}.
+\]
+
+\begin{keyresult}
+  \textbf{Distinct real roots.} If $r_1 \neq r_2$ are real, then $\{e^{r_1 x}, e^{r_2 x}\}$ is a fundamental set of solutions and the general solution is
+  \[
+    y(x) = c_1 e^{r_1 x} + c_2 e^{r_2 x}.
+  \]
+\end{keyresult}
+
+\paragraph{Verification.} Each exponential satisfies the ODE by construction (that is how the characteristic equation was derived). The general solution follows from the superposition principle (\cref{thm:superposition}): any linear combination of solutions is again a solution.
+
+\begin{workedexample}
+  Solve $y'' - y' - 2y = 0$ with initial conditions $y(0) = 3$ and $y'(0) = 1$.
+
+  \textbf{Solution.} The characteristic equation is
+  \[
+    r^2 - r - 2 = 0 \quad\Longrightarrow\quad (r - 2)(r + 1) = 0.
+  \]
+  Roots: $r_1 = 2$, $r_2 = -1$. The general solution is
+  \[
+    y(x) = c_1 e^{2x} + c_2 e^{-x}.
+  \]
+
+  Differentiate:
+  \[
+    y'(x) = 2c_1 e^{2x} - c_2 e^{-x}.
+  \]
+
+  Apply initial conditions:
+  \[
+    \begin{cases}
+      y(0) = c_1 + c_2 = 3, \\
+      y'(0) = 2c_1 - c_2 = 1.
+    \end{cases}
+  \]
+  Adding the two equations: $3c_1 = 4$, so $c_1 = \dfrac{4}{3}$. From the first equation: $c_2 = 3 - \dfrac{4}{3} = \dfrac{5}{3}$.
+
+  The solution is
+  \[
+    y(x) = \frac{4}{3}\,e^{2x} + \frac{5}{3}\,e^{-x}.
+  \]
+\end{workedexample}
+
+\begin{workedexample}
+  Solve $2y'' + 7y' + 3y = 0$ with $y(0) = 4$ and $y'(0) = -5$.
+
+  \textbf{Solution.} The characteristic equation is
+  \[
+    2r^2 + 7r + 3 = 0.
+  \]
+  Using the quadratic formula:
+  \[
+    r = \frac{-7 \pm \sqrt{49 - 24}}{4} = \frac{-7 \pm 5}{4}.
+  \]
+  So $r_1 = \dfrac{-7 + 5}{4} = -\dfrac{1}{2}$ and $r_2 = \dfrac{-7 - 5}{4} = -3$.
+
+  The general solution is
+  \[
+    y(x) = c_1 e^{-x/2} + c_2 e^{-3x}.
+  \]
+
+  Differentiate:
+  \[
+    y'(x) = -\tfrac{1}{2}\,c_1\,e^{-x/2} - 3c_2\,e^{-3x}.
+  \]
+
+  Apply initial conditions:
+  \[
+    \begin{cases}
+      y(0) = c_1 + c_2 = 4, \\
+      y'(0) = -\tfrac{1}{2}\,c_1 - 3c_2 = -5.
+    \end{cases}
+  \]
+  From the first equation: $c_1 = 4 - c_2$. Substitute into the second:
+  \[
+    -\tfrac{1}{2}(4 - c_2) - 3c_2 = -5 \quad\Longrightarrow\quad -2 + \tfrac{1}{2}c_2 - 3c_2 = -5.
+  \]
+  \[
+    -\tfrac{5}{2}c_2 = -3 \quad\Longrightarrow\quad c_2 = \frac{6}{5}, \qquad c_1 = 4 - \frac{6}{5} = \frac{14}{5}.
+  \]
+
+  The solution is
+  \[
+    y(x) = \frac{14}{5}\,e^{-x/2} + \frac{6}{5}\,e^{-3x}.
+  \]
+\end{workedexample}
+
+\begin{workedexample}
+  Solve $y'' - 4y = 0$ with $y(0) = 0$ and $y(1) = e^2 - e^{-2}$.
+
+  \textbf{Solution.} The characteristic equation is $r^2 - 4 = 0$, so $r = \pm 2$. The general solution is
+  \[
+    y(x) = c_1 e^{2x} + c_2 e^{-2x}.
+  \]
+
+  Apply the first condition:
+  \[
+    y(0) = c_1 + c_2 = 0 \quad\Longrightarrow\quad c_2 = -c_1.
+  \]
+  The solution simplifies to $y(x) = c_1(e^{2x} - e^{-2x})$.
+
+  Apply the second condition:
+  \[
+    y(1) = c_1(e^2 - e^{-2}) = e^2 - e^{-2} \quad\Longrightarrow\quad c_1 = 1.
+  \]
+
+  The solution is
+  \[
+    y(x) = e^{2x} - e^{-2x} = 2\sinh(2x).
+  \]
+\end{workedexample}
+
+\subsection{Case 2: Repeated Real Roots}
+\label{sec:ch04_repeated_roots}
+
+When $\Delta = 0$, the characteristic equation has a single (double) root
+\[
+  r = -\frac{b}{2a}.
+\]
+
+\begin{keyresult}
+  \textbf{Repeated real root.} If $r$ is a double root, the general solution is
+  \[
+    y(x) = c_1 e^{rx} + c_2\,x\,e^{rx}.
+  \]
+\end{keyresult}
+
+\paragraph{Why the $x$ factor?} With only one root $r$, the function $e^{rx}$ gives us just one solution. We need a second linearly independent solution to form the general solution of a second-order equation. The method of \textbf{reduction of order} provides this second solution.
+
+\subparagraph{Derivation via reduction of order.}
+Suppose $y_1(x) = e^{rx}$ is one solution of \cref{eq:second_order_homogeneous} (with $\Delta = 0$). We seek a second solution of the form
+\[
+  y_2(x) = v(x)\,y_1(x) = v(x)\,e^{rx},
+\]
+where $v(x)$ is an unknown function to be determined. Compute the derivatives:
+\[
+  y_2' = (v' + rv)\,e^{rx}, \qquad y_2'' = (v'' + 2rv' + r^2 v)\,e^{rx}.
+\]
+Substitute $y_2$ into the ODE $y'' + \frac{b}{a}y' + \frac{c}{a}y = 0$ (dividing \cref{eq:second_order_homogeneous} by $a$):
+\[
+  \bigl(v'' + 2rv' + r^2 v\bigr)e^{rx} + \frac{b}{a}\bigl(v' + rv\bigr)e^{rx} + \frac{c}{a}\,v\,e^{rx} = 0.
+\]
+Dividing by $e^{rx}$ and collecting terms:
+\[
+  v'' + \left(2r + \frac{b}{a}\right)v' + \left(r^2 + \frac{b}{a}r + \frac{c}{a}\right)v = 0.
+\]
+The $v$-coefficient is exactly the characteristic polynomial evaluated at $r$, which vanishes because $r$ is a root:
+\[
+  r^2 + \frac{b}{a}r + \frac{c}{a} = 0.
+\]
+For a repeated root, $r = -b/(2a)$, so the $v'$-coefficient is also zero:
+\[
+  2r + \frac{b}{a} = 2\!\left(-\frac{b}{2a}\right) + \frac{b}{a} = 0.
+\]
+We are left with the simple equation
+\[
+  v''(x) = 0 \quad\Longrightarrow\quad v(x) = Ax + B.
+\]
+Choosing $A = 1$ and $B = 0$ gives $y_2(x) = x\,e^{rx}$, a second linearly independent solution. The general solution is
+\[
+  y(x) = c_1 e^{rx} + c_2 x\,e^{rx}.
+\]
+
+\begin{hintbox}
+  \textbf{Quick check.} You can verify that $y = x\,e^{rx}$ satisfies $ay'' + by' + cy = 0$ when $r = -b/(2a)$ and $b^2 = 4ac$ by direct substitution. This is always a safe verification step.
+\end{hintbox}
+
+\begin{workedexample}
+  Solve $y'' - 6y' + 9y = 0$.
+
+  \textbf{Solution.} The characteristic equation is
+  \[
+    r^2 - 6r + 9 = 0 \quad\Longrightarrow\quad (r - 3)^2 = 0.
+  \]
+  Double root: $r = 3$. The general solution is
+  \[
+    y(x) = c_1 e^{3x} + c_2 x\,e^{3x}.
+  \]
+\end{workedexample}
+
+\begin{workedexample}
+  Solve $y'' + 4y' + 4y = 0$ with $y(0) = 2$ and $y'(0) = 0$.
+
+  \textbf{Solution.} The characteristic equation is
+  \[
+    r^2 + 4r + 4 = 0 \quad\Longrightarrow\quad (r + 2)^2 = 0.
+  \]
+  Double root: $r = -2$. The general solution is
+  \[
+    y(x) = c_1 e^{-2x} + c_2 x\,e^{-2x}.
+  \]
+
+  Differentiate:
+  \[
+    y'(x) = -2c_1 e^{-2x} + c_2 e^{-2x} - 2c_2 x\,e^{-2x}
+           = \bigl(-2c_1 + c_2 - 2c_2 x\bigr)e^{-2x}.
+  \]
+
+  Apply initial conditions:
+  \[
+    \begin{cases}
+      y(0) = c_1 = 2, \\
+      y'(0) = -2c_1 + c_2 = 0 \quad\Longrightarrow\quad c_2 = 2c_1 = 4.
+    \end{cases}
+  \]
+
+  The solution is
+  \[
+    y(x) = 2\,e^{-2x} + 4x\,e^{-2x} = 2(1 + 2x)\,e^{-2x}.
+  \]
+\end{workedexample}
+
+\subsection{Case 3: Complex Conjugate Roots}
+\label{sec:ch04_complex_roots}
+
+When $\Delta = b^2 - 4ac < 0$, the characteristic equation has two complex conjugate roots:
+\[
+  r = \alpha \pm i\beta, \qquad
+  \alpha = -\frac{b}{2a}, \quad
+  \beta = \frac{\sqrt{4ac - b^2}}{2a}.
+\]
+
+\begin{keyresult}
+  \textbf{Complex conjugate roots.} If $r = \alpha \pm i\beta$ with $\beta > 0$, the general real-valued solution is
+  \[
+    y(x) = e^{\alpha x}\Bigl[c_1 \cos(\beta x) + c_2 \sin(\beta x)\Bigr].
+  \]
+  Equivalently, using amplitude–phase form:
+  \[
+    y(x) = A\,e^{\alpha x}\cos(\beta x - \phi),
+  \]
+  where $A = \sqrt{c_1^2 + c_2^2}$ and $\phi = \arctan(c_2/c_1)$.
+\end{keyresult}
+
+\paragraph{Derivation from Euler's formula.}
+The complex-valued solutions are $e^{(\alpha + i\beta)x}$ and $e^{(\alpha - i\beta)x}$. By Euler's formula,
+\[
+  e^{(\alpha + i\beta)x} = e^{\alpha x}\,e^{i\beta x}
+  = e^{\alpha x}\bigl[\cos(\beta x) + i\,\sin(\beta x)\bigr].
+\]
+Because the ODE has real coefficients, both the real and imaginary parts are themselves real solutions:
+\begin{align*}
+  \Re\!\left[e^{(\alpha+i\beta)x}\right] &= e^{\alpha x}\cos(\beta x), \\
+  \Im\!\left[e^{(\alpha+i\beta)x}\right] &= e^{\alpha x}\sin(\beta x).
+\end{align*}
+These two functions are linearly independent (one is not a constant multiple of the other), so they form a fundamental set. The general solution is their linear combination, giving the formula above.
+
+\paragraph{Physical interpretation: damped oscillations.}
+The parameter $\alpha$ governs the exponential growth ($\alpha > 0$) or decay ($\alpha < 0$) of the amplitude, while $\beta$ determines the angular frequency of the oscillation. In mechanical systems, this corresponds to \textbf{underdamped motion}: the mass oscillates about equilibrium while the amplitude decays exponentially (if $\alpha < 0$). When $\alpha = 0$ (i.e.\ $b = 0$), we have pure undamped oscillations with constant amplitude.
+
+\begin{workedexample}
+  Solve $y'' + 2y' + 5y = 0$.
+
+  \textbf{Solution.} The characteristic equation is
+  \[
+    r^2 + 2r + 5 = 0.
+  \]
+  Discriminant: $\Delta = 4 - 20 = -16 < 0$. The roots are
+  \[
+    r = \frac{-2 \pm \sqrt{-16}}{2} = -1 \pm 2i.
+  \]
+  Here $\alpha = -1$ and $\beta = 2$. The general solution is
+  \[
+    y(x) = e^{-x}\Bigl[c_1 \cos(2x) + c_2 \sin(2x)\Bigr].
+  \]
+  This represents damped oscillations with amplitude decaying like $e^{-x}$.
+\end{workedexample}
+
+\begin{workedexample}
+  Solve $y'' + 4y' + 13y = 0$ with $y(0) = 2$ and $y'(0) = 3$.
+
+  \textbf{Solution.} The characteristic equation is
+  \[
+    r^2 + 4r + 13 = 0.
+  \]
+  Discriminant: $\Delta = 16 - 52 = -36 < 0$. The roots are
+  \[
+    r = \frac{-4 \pm \sqrt{-36}}{2} = -2 \pm 3i.
+  \]
+  So $\alpha = -2$ and $\beta = 3$. The general solution is
+  \[
+    y(x) = e^{-2x}\Bigl[c_1 \cos(3x) + c_2 \sin(3x)\Bigr].
+  \]
+
+  Differentiate (product rule):
+  \[
+    y'(x) = -2\,e^{-2x}\bigl[c_1 \cos(3x) + c_2 \sin(3x)\bigr]
+            + e^{-2x}\bigl[-3c_1 \sin(3x) + 3c_2 \cos(3x)\bigr].
+  \]
+
+  Apply initial conditions:
+  \[
+    \begin{cases}
+      y(0) = c_1 = 2, \\[6pt]
+      y'(0) = -2c_1 + 3c_2 = 3 \quad\Longrightarrow\quad -4 + 3c_2 = 3 \quad\Longrightarrow\quad c_2 = \dfrac{7}{3}.
+    \end{cases}
+  \]
+
+  The solution is
+  \[
+    y(x) = e^{-2x}\left(2\cos(3x) + \frac{7}{3}\sin(3x)\right).
+  \]
+\end{workedexample}
 
 \subsection{Superposition Principle}
 \label{sec:ch04_superposition}
 
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+The entire theory rests on the linearity of the differential operator.
+
+\begin{theorem}[Superposition Principle]
+  \label{thm:superposition}
+  Let $L[y] = ay'' + by' + cy$ with constant $a, b, c$ and $a \neq 0$. If $y_1(x)$ and $y_2(x)$ are solutions of $L[y] = 0$, then any linear combination
+  \[
+    y(x) = c_1 y_1(x) + c_2 y_2(x)
+  \]
+  is also a solution, for arbitrary constants $c_1, c_2 \in \R$.
+\end{theorem}
+
+\begin{proof}
+  Compute:
+  \[
+    L[c_1 y_1 + c_2 y_2]
+    = a(c_1 y_1 + c_2 y_2)'' + b(c_1 y_1 + c_2 y_2)' + c(c_1 y_1 + c_2 y_2).
+  \]
+  By linearity of differentiation:
+  \[
+    = c_1\bigl(ay_1'' + by_1' + cy_1\bigr) + c_2\bigl(ay_2'' + by_2' + cy_2\bigr)
+    = c_1 \cdot 0 + c_2 \cdot 0 = 0.
+  \]
+  Therefore $c_1 y_1 + c_2 y_2$ satisfies the ODE.
+\end{proof}
+
+\begin{definition}[Fundamental Set of Solutions]
+  \label{def:fundamental_set}
+  Two solutions $\{y_1, y_2\}$ of \cref{eq:second_order_homogeneous} form a \textbf{fundamental set of solutions} on an interval $I$ if they are linearly independent on $I$. The general solution is then
+  \[
+    y(x) = c_1 y_1(x) + c_2 y_2(x),
+  \]
+  where $c_1, c_2$ are arbitrary constants.
+\end{definition}
+
+\begin{workedexample}
+  Verify that $y_1(x) = e^{3x}$ and $y_2(x) = e^{-2x}$ form a fundamental set for $y'' - y' - 6y = 0$, and write the general solution.
+
+  \textbf{Solution.} Check each function:
+  \[
+    y_1 = e^{3x} \;\Rightarrow\; y_1' = 3e^{3x}, \; y_1'' = 9e^{3x}.
+  \]
+  \[
+    L[y_1] = 9e^{3x} - 3e^{3x} - 6e^{3x} = 0. \quad \checkmark
+  \]
+
+  \[
+    y_2 = e^{-2x} \;\Rightarrow\; y_2' = -2e^{-2x}, \; y_2'' = 4e^{-2x}.
+  \]
+  \[
+    L[y_2] = 4e^{-2x} + 2e^{-2x} - 6e^{-2x} = 0. \quad \checkmark
+  \]
+
+  Linear independence: the ratio $y_1/y_2 = e^{5x}$ is not constant, so $y_1$ and $y_2$ are linearly independent. They form a fundamental set.
+
+  The general solution is
+  \[
+    y(x) = c_1 e^{3x} + c_2 e^{-2x}.
+  \]
+\end{workedexample}
+
+\begin{workedexample}
+  Show that $\{y_1, y_2\} = \{e^{2x}, 2e^{2x}\}$ is \emph{not} a fundamental set for $y'' - 4y = 0$, even though both are solutions.
+
+  \textbf{Solution.} Clearly $y_2 = 2y_1$, so the two functions are linearly dependent. Any linear combination gives
+  \[
+    c_1 e^{2x} + c_2 (2e^{2x}) = (c_1 + 2c_2)e^{2x},
+  \]
+  which spans only a one-dimensional solution space. We cannot represent all solutions (we are missing the $e^{-2x}$ component). A proper fundamental set is $\{e^{2x}, e^{-2x}\}$.
+\end{workedexample}
 
 \subsection{Wronskian and Linear Independence}
 \label{sec:ch04_wronskian}
 
-% Content goes here
+To systematically determine whether two solutions form a fundamental set, we use the \textbf{Wronskian}.
+
+\begin{definition}[Wronskian]
+  \label{def:wronskian}
+  For two differentiable functions $y_1(x)$ and $y_2(x)$, the \textbf{Wronskian} is
+  \[
+    W(y_1, y_2)(x) = \det\begin{pmatrix} y_1(x) & y_2(x) \\[4pt] y_1'(x) & y_2'(x) \end{pmatrix}
+    = y_1(x)\,y_2'(x) - y_1'(x)\,y_2(x).
+  \]
+\end{definition}
+
+\begin{theorem}[Wronskian Test]
+  \label{thm:wronskian_test}
+  Let $y_1$ and $y_2$ be two solutions of \cref{eq:second_order_homogeneous} on an interval $I$. Then:
+  \begin{enumerate}
+    \item If $W(y_1, y_2)(x_0) \neq 0$ for some $x_0 \in I$, then $\{y_1, y_2\}$ is a fundamental set on $I$.
+    \item If $W(y_1, y_2)(x) = 0$ for all $x \in I$, then $y_1$ and $y_2$ are linearly dependent on $I$.
+  \end{enumerate}
+  In particular, $W(y_1, y_2)(x)$ is either identically zero on $I$ or never zero on $I$.
+\end{theorem}
+
+The Wronskian gives a practical computational test: evaluate $W$ at any single point; if it is nonzero, the solutions are linearly independent.
+
+\begin{workedexample}
+  Compute the Wronskian of $y_1(x) = e^{3x}$ and $y_2(x) = e^{-2x}$.
+
+  \textbf{Solution.} We have $y_1' = 3e^{3x}$ and $y_2' = -2e^{-2x}$.
+  \[
+    W(y_1, y_2)(x) = e^{3x} \cdot (-2e^{-2x}) - 3e^{3x} \cdot e^{-2x}
+    = -2e^{x} - 3e^{x} = -5e^{x}.
+  \]
+  Since $W(x) = -5e^x \neq 0$ for all $x$, the two functions are linearly independent. They form a fundamental set.
+\end{workedexample}
+
+\begin{workedexample}
+  Compute the Wronskian of $y_1(x) = e^{rx}$ and $y_2(x) = x\,e^{rx}$ (the repeated-root case).
+
+  \textbf{Solution.} We have $y_1' = r\,e^{rx}$ and $y_2' = (1 + rx)\,e^{rx}$.
+  \[
+    W(y_1, y_2)(x) = e^{rx} \cdot (1 + rx)\,e^{rx} - r\,e^{rx} \cdot x\,e^{rx}
+    = (1 + rx)\,e^{2rx} - rx\,e^{2rx} = e^{2rx}.
+  \]
+  Since $W(x) = e^{2rx} \neq 0$ for all $x$, these two functions are linearly independent. This confirms that $\{e^{rx}, x\,e^{rx}\}$ is a valid fundamental set for the repeated-root case.
+\end{workedexample}
+
+\begin{remark}
+  The connection between the Wronskian and fundamental sets is fundamental (pardon the pun): a pair of solutions is a fundamental set \emph{if and only if} their Wronskian is nonzero on the interval. This is the standard criterion used in practice.
+\end{remark}
 
 \subsection{Abel's Identity}
 \label{sec:ch04_abels_identity}
 
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+Abel's identity provides a formula for the Wronskian without computing derivatives, by relating it directly to the coefficients of the ODE.
+
+\begin{theorem}[Abel's Identity]
+  \label{thm:abels_identity}
+  Let $y_1$ and $y_2$ be two solutions of the standard-form equation
+  \begin{equation}
+    \label{eq:standard_form}
+    y'' + p(x)\,y' + q(x)\,y = 0
+  \end{equation}
+  on an interval $I$, where $p(x)$ and $q(x)$ are continuous. Then the Wronskian satisfies
+  \[
+    W(x) = W(x_0)\,\exp\!\left(-\int_{x_0}^{x} p(t)\,\diff t\right)
+  \]
+  for any $x_0 \in I$.
+\end{theorem}
+
+\begin{proof}
+  Let $W(x) = y_1 y_2' - y_1' y_2$. Differentiate:
+  \[
+    W' = y_1' y_2' + y_1 y_2'' - y_1'' y_2 - y_1' y_2'
+       = y_1 y_2'' - y_1'' y_2.
+  \]
+  Since $y_1$ and $y_2$ both satisfy $y'' + p(x)y' + q(x)y = 0$, we have
+  \[
+    y_1'' = -p(x)y_1' - q(x)y_1, \qquad
+    y_2'' = -p(x)y_2' - q(x)y_2.
+  \]
+  Substitute:
+  \begin{align*}
+    W' &= y_1\bigl[-p y_2' - q y_2\bigr] - \bigl[-p y_1' - q y_1\bigr]y_2 \\
+       &= -p\,y_1 y_2' - q\,y_1 y_2 + p\,y_1' y_2 + q\,y_1 y_2 \\
+       &= -p\,(y_1 y_2' - y_1' y_2) = -p(x)\,W(x).
+  \end{align*}
+  Thus $W' + p(x)W = 0$, a first-order separable equation. The solution is
+  \[
+    W(x) = W(x_0)\,\exp\!\left(-\int_{x_0}^{x} p(t)\,\diff t\right).
+  \]
+\end{proof}
+
+\begin{remark}
+  For the constant-coefficient equation \cref{eq:second_order_homogeneous}, dividing by $a$ puts it in standard form with $p(x) = b/a$ (constant). Abel's identity then gives
+  \[
+    W(x) = W(x_0)\,\exp\!\left(-\frac{b}{a}(x - x_0)\right).
+  \]
+  Since $W(x_0) \neq 0$ for a fundamental set, this confirms that $W(x)$ is never zero (exponential of a real number is always positive).
+\end{remark}
+
+\begin{workedexample}
+  Use Abel's identity to find the Wronskian of two solutions of $y'' + 3y' + 2y = 0$, given that $W(0) = 5$.
+
+  \textbf{Solution.} In standard form, $p(x) = 3$. By Abel's identity:
+  \[
+    W(x) = W(0)\,\exp\!\left(-\int_0^x 3\,\diff t\right)
+          = 5\,e^{-3x}.
+  \]
+  The Wronskian is $W(x) = 5e^{-3x}$, which is always positive (confirming the solutions are linearly independent).
+\end{workedexample}
+
+\begin{workedexample}
+  Verify Abel's identity for $y'' - 4y = 0$ using the fundamental set $\{e^{2x}, e^{-2x}\}$.
+
+  \textbf{Solution.} In standard form, $p(x) = 0$. Abel's identity predicts
+  \[
+    W(x) = W(x_0)\,\exp\!\left(-\int_{x_0}^{x} 0\,\diff t\right) = W(x_0).
+  \]
+  The Wronskian should be constant. Compute it directly:
+  \[
+    W(e^{2x}, e^{-2x}) = e^{2x}(-2e^{-2x}) - (2e^{2x})e^{-2x} = -2 - 2 = -4.
+  \]
+  Indeed, $W(x) = -4$ is constant, in agreement with Abel's identity.
+\end{workedexample}
+
+\begin{hintbox}
+  \textbf{Practical tip.} Abel's identity is especially useful when you know one solution and need to find another. It tells you the Wronskian \emph{without} computing derivatives, and combined with the formula $v' = W/y_1^2$ from reduction of order (\cref{sec:ch04_reduction_of_order}), it directly yields the second solution.
+\end{hintbox}
 
 \subsection{Reduction of Order}
 \label{sec:ch04_reduction_of_order}
 
-% Content goes here
+Reduction of order is a systematic method for finding a second linearly independent solution when one solution $y_1(x)$ is already known. We saw this technique in \cref{sec:ch04_repeated_roots} to derive the $x\,e^{rx}$ factor; here we state it in full generality.
+
+\paragraph{Derivation.}
+Consider the standard-form equation $y'' + p(x)y' + q(x)y = 0$. Suppose $y_1(x)$ is a known solution. We seek a second solution of the form
+\begin{equation}
+  \label{eq:reduction_ansatz}
+  y_2(x) = v(x)\,y_1(x),
+\end{equation}
+where $v(x)$ is to be determined. Differentiate:
+\begin{align*}
+  y_2' &= v'y_1 + vy_1', \\
+  y_2'' &= v''y_1 + 2v'y_1' + vy_1''.
+\end{align*}
+Substitute into the ODE:
+\[
+  \bigl(v''y_1 + 2v'y_1' + vy_1''\bigr) + p(x)\bigl(v'y_1 + vy_1'\bigr) + q(x)\,v\,y_1 = 0.
+\]
+Collect terms:
+\[
+  v''y_1 + v'\bigl(2y_1' + p(x)y_1\bigr) + v\bigl(y_1'' + p(x)y_1' + q(x)y_1\bigr) = 0.
+\]
+The $v$-term vanishes because $y_1$ is a solution. We obtain a first-order equation in $u = v'$:
+\[
+  y_1\,u' + \bigl(2y_1' + p(x)y_1\bigr)u = 0.
+\]
+This is separable:
+\[
+  \frac{u'}{u} = -\frac{2y_1'}{y_1} - p(x) = -2\frac{d}{dx}\bigl(\ln|y_1|\bigr) - p(x).
+\]
+Integrating:
+\[
+  \ln|u| = -2\ln|y_1| - \int p(x)\,\diff x + C,
+\]
+so
+\[
+  u = \frac{v'}{y_1^2} \cdot y_1^2 = \frac{C}{y_1^2}\,\exp\!\left(-\int p(x)\,\diff x\right).
+\]
+Noting that $W(x) = W(x_0)\exp\!\left(-\int p\,\diff x\right)$ by Abel's identity, we can write this as
+\begin{equation}
+  \label{eq:reduction_formula}
+  v'(x) = \frac{W(x)}{y_1(x)^2}.
+\end{equation}
+
+\begin{keyresult}
+  \textbf{Reduction of order formula.}
+  Given one solution $y_1(x)$ of $y'' + p(x)y' + q(x)y = 0$, a second linearly independent solution is
+  \[
+    y_2(x) = y_1(x)\int \frac{1}{y_1(x)^2}\,\exp\!\left(-\int p(x)\,\diff x\right)\diff x.
+  \]
+\end{keyresult}
+
+\begin{workedexample}
+  Given that $y_1(x) = e^{3x}$ is a solution of $y'' - 6y' + 9y = 0$, find a second linearly independent solution using reduction of order.
+
+  \textbf{Solution.} The equation in standard form has $p(x) = -6$. By Abel's identity (or direct computation):
+  \[
+    \exp\!\left(-\int p(x)\,\diff x\right) = \exp\!\left(-\int (-6)\,\diff x\right) = e^{6x}.
+  \]
+  Using \cref{eq:reduction_formula}:
+  \[
+    v'(x) = \frac{e^{6x}}{(e^{3x})^2} = \frac{e^{6x}}{e^{6x}} = 1.
+  \]
+  Integrate: $v(x) = x$ (choosing the simplest antiderivative with $C=0$).
+  Therefore
+  \[
+    y_2(x) = v(x)\,y_1(x) = x\,e^{3x}.
+  \]
+  This recovers the repeated-root solution we derived earlier. The general solution is
+  \[
+    y(x) = c_1 e^{3x} + c_2 x\,e^{3x}.
+  \]
+\end{workedexample}
+
+\begin{workedexample}
+  Given that $y_1(x) = x$ is a solution of $x^2 y'' - x y' + y = 0$, find a second solution.
+
+  \textbf{Solution.} Write the equation in standard form by dividing by $x^2$:
+  \[
+    y'' - \frac{1}{x}y' + \frac{1}{x^2}y = 0.
+  \]
+  Here $p(x) = -\dfrac{1}{x}$. Compute the integrating factor:
+  \[
+    \exp\!\left(-\int p(x)\,\diff x\right) = \exp\!\left(-\int \!\left(-\frac{1}{x}\right)\!\diff x\right)
+    = \exp\!\bigl(\ln|x|\bigr) = |x|.
+  \]
+  For $x > 0$, this is simply $x$. Now apply the reduction of order formula:
+  \[
+    v'(x) = \frac{x}{y_1(x)^2} = \frac{x}{x^2} = \frac{1}{x}.
+  \]
+  Integrate: $v(x) = \ln|x|$. The second solution is
+  \[
+    y_2(x) = y_1(x)\,v(x) = x\,\ln|x|.
+  \]
+  The general solution (for $x > 0$) is
+  \[
+    y(x) = c_1\,x + c_2\,x\,\ln x.
+  \]
+\end{workedexample}
 
 \subsection{Summary}
 \label{sec:ch04_summary}
 
-% Summary table at end of chapter
 \begin{table}[htbp]
 \centering
-\caption{Chapter Summary}
+\caption{Second-order linear homogeneous equations with constant coefficients}
-\label{tab:ch04_summary}
+\label{tab:ch04_cases}
-\begin{tabular}{l l}
+\begin{tabular}{l l l p{5cm}}
 \toprule
-\textbf{Concept} & \textbf{Key Formula/Method} \\
+\textbf{Discriminant} & \textbf{Roots} & \textbf{Solution} & \textbf{Key idea} \\
 \midrule
-TBD & TBD \\
+$\Delta > 0$ & $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ & $y = c_1 e^{r_1 x} + c_2 e^{r_2 x}$ & Two distinct real exponentials \\[8pt]
+$\Delta = 0$ & $r = -\dfrac{b}{2a}$ (double) & $y = c_1 e^{rx} + c_2 x e^{rx}$ & Extra $x$ from reduction of order \\[8pt]
+$\Delta < 0$ & $\alpha \pm i\beta$ & $e^{\alpha x}[c_1 \cos(\beta x) + c_2 \sin(\beta x)]$ & Damped oscillation via Euler's formula \\
 \bottomrule
 \end{tabular}
 \end{table}
+
+\begin{table}[htbp]
+\centering
+\caption{Theory results: Wronskian, Abel's identity, reduction of order}
+\label{tab:ch04_theory}
+\begin{tabular}{l p{6.5cm}}
+\toprule
+\textbf{Concept} & \textbf{Key formula} \\
+\midrule
+Wronskian & $W(y_1, y_2)(x) = y_1 y_2' - y_1' y_2$ \\
+Wronskian test & $W(x_0) \neq 0 \;\Rightarrow\;$ fundamental set on $I$ \\
+Abel's identity & $W(x) = W(x_0)\,\exp\!\left(-\displaystyle\int_{x_0}^{x} p(t)\,\diff t\right)$ \\
+Abel (constant coeff.) & $W(x) = W(x_0)\,\exp\!\left(-\dfrac{b}{a}(x - x_0)\right)$ \\
+Reduction of order & $y_2(x) = y_1(x)\displaystyle\int \frac{e^{-\int p(x)\,\diff x}}{y_1(x)^2}\,\diff x$ \\
+Superposition & $L[c_1 y_1 + c_2 y_2] = c_1 L[y_1] + c_2 L[y_2] = 0$ \\
+\bottomrule
+\end{tabular}
+\end{table}
+
+\begin{hintbox}
+  \textbf{Problem-solving workflow.}
+  \begin{enumerate}
+    \item Write the ODE in standard form (divide by $a$ if needed).
+    \item Form the characteristic equation $ar^2 + br + c = 0$.
+    \item Compute the discriminant $\Delta = b^2 - 4ac$.
+    \item Apply the appropriate case from \cref{tab:ch04_cases}.
+    \item If initial/boundary conditions are given, determine $c_1$ and $c_2$.
+  \end{enumerate}
+\end{hintbox}