From b826747f3000348e5a1f1ba0dde3690fb03e07d6 Mon Sep 17 00:00:00 2001 From: Worker Agent Date: Thu, 4 Jun 2026 15:58:03 -0500 Subject: [PATCH] ch5: undetermined coefficients and variation of parameters --- chapters/ch05_second_order_nonhomogeneous.tex | 846 +++++++++++++++++- 1 file changed, 832 insertions(+), 14 deletions(-) diff --git a/chapters/ch05_second_order_nonhomogeneous.tex b/chapters/ch05_second_order_nonhomogeneous.tex index c4221dc..938559b 100644 --- a/chapters/ch05_second_order_nonhomogeneous.tex +++ b/chapters/ch05_second_order_nonhomogeneous.tex @@ -1,39 +1,857 @@ +% ============================================================================= +% ch05_second_order_nonhomogeneous.tex +% Chapter 5: Second-Order Linear Nonhomogeneous Equations +% ============================================================================= + \section{Second-Order Nonhomogeneous} \label{ch:second_order_nonhomogeneous} +We now extend the theory from \cref{ch:second_order_homogeneous} to the nonhomogeneous case. The general equation is +\begin{equation} + \label{eq:nonhomogeneous_general} + a\,y'' + b\,y' + c\,y = g(x), + \qquad a, b, c \in \R,\;\; a \neq 0, +\end{equation} +where $g(x)$ is a given \textbf{forcing function} (also called the \textbf{source term} or \textbf{nonhomogeneous term}). When $g(x) \equiv 0$, we recover the homogeneous equation \cref{eq:second_order_homogeneous}. The presence of $g(x)$ models external driving forces in physical systems --- a periodically driven pendulum, a forced electrical circuit, or a building shaken by an earthquake. + +The general strategy in all cases is the same: first solve the associated homogeneous equation, then find one particular solution of the nonhomogeneous equation. The two parts combine to give the complete solution. + \subsection{General Solution Structure} \label{sec:ch05_general_solution} -% Content goes here +\begin{theorem}[Structure of the General Solution] + \label{thm:nonhomogeneous_structure} + Let $L[y] = ay'' + by' + cy$ with constant $a, b, c$ and $a \neq 0$. If $y_h(x)$ is the general solution of the associated homogeneous equation $L[y] = 0$, and $y_p(x)$ is \emph{any} particular solution of the nonhomogeneous equation $L[y] = g(x)$, then the general solution of $L[y] = g(x)$ is + \[ + y(x) = y_h(x) + y_p(x). + \] +\end{theorem} -\subsection{Undetermined Coefficients} +\begin{proof} + Define the linear operator $L[y] = ay'' + by' + cy$. We need to show that $y = y_h + y_p$ satisfies $L[y] = g(x)$ for \emph{every} choice of the constants in $y_h$. + + By linearity of differentiation (the \textbf{superposition principle}, \cref{thm:superposition}): + \[ + L[y_h + y_p] = L[y_h] + L[y_p]. + \] + Since $y_h$ solves the homogeneous equation, $L[y_h] = 0$. Since $y_p$ is a particular solution, $L[y_p] = g(x)$. Therefore: + \[ + L[y_h + y_p] = 0 + g(x) = g(x). + \] + Thus $y_h + y_p$ is a solution of the nonhomogeneous equation. Conversely, if $y$ is any solution of $L[y] = g(x)$, then $L[y - y_p] = L[y] - L[y_p] = g(x) - g(x) = 0$, so $y - y_p = y_h$ for some homogeneous solution $y_h$, meaning $y = y_h + y_p$. +\end{proof} + +The homogeneous part $y_h(x)$ retains two arbitrary constants $c_1, c_2$ from the general solution found in \cref{ch:second_order_homogeneous}. The particular solution $y_p(x)$ contains no free parameters --- we only need \emph{one} specific function that works. + +\begin{hintbox} + \textbf{Two-step workflow for every nonhomogeneous problem.} + \begin{enumerate} + \item \textbf{Solve the homogeneous equation} $ay'' + by' + cy = 0$ using the characteristic equation method from \cref{ch:second_order_homogeneous}. Obtain $y_h(x) = c_1 y_1(x) + c_2 y_2(x)$. + \item \textbf{Find one particular solution} $y_p(x)$ using either the method of undetermined coefficients or variation of parameters. + \end{enumerate} + Combine: $y(x) = y_h(x) + y_p(x)$. +\end{hintbox} + +We now develop the two main methods for finding $y_p(x)$. + +\subsection{Method of Undetermined Coefficients} \label{sec:ch05_undetermined_coefficients} -% Content goes here +The method of \textbf{undetermined coefficients} (UC) is an algebraic technique that works when the forcing function $g(x)$ has a \textbf{simple form}. The idea is straightforward: guess the \emph{shape} of $y_p$ based on the form of $g(x)$, but leave the coefficients unknown. Then substitute the guess into the ODE and solve for those coefficients. + +\paragraph{When to use.} Undetermined coefficients applies only when $g(x)$ is one of the following (or a sum/product of them): +\begin{itemize} + \item A polynomial $P_n(x)$ of degree $n$. + \item An exponential $e^{\alpha x}$. + \item A sine or cosine function $\sin(\beta x)$ or $\cos(\beta x)$. + \item Products of the above, such as $e^{\alpha x} P_n(x)$ or $e^{\alpha x}\sin(\beta x)$. +\end{itemize} +If $g(x) = \ln(x)$, $1/x$, $\tan(x)$, or any other function outside this family, undetermined coefficients \textbf{will not work}. Use variation of parameters instead (\cref{sec:ch05_variation_of_parameters}). + +\begin{hintbox} + \textbf{Limitation warning.} Undetermined coefficients is an \emph{educated guessing} method. It exploits the fact that polynomials, exponentials, and trigonometric functions are closed under differentiation. Functions like $\ln x$ and $\tan x$ do not share this property, so the method breaks down. +\end{hintbox} + +\subsubsection{Guess Table} +\label{sec:ch05_guess_table} + +The following table gives the initial guess for $y_p$ based on the form of $g(x)$. The constants $A_k, B_k$ are the ``undetermined coefficients'' to be found by substitution. + +\begin{keyresult} + \textbf{Guess table for the method of undetermined coefficients.} + + \begin{center} + \begin{tabular}{l l} + \toprule + \textbf{Form of $g(x)$} & \textbf{Initial guess for $y_p(x)$} \\ + \midrule + $P_n(x) = a_n x^n + \cdots + a_0$ & $A_n x^n + A_{n-1} x^{n-1} + \cdots + A_0$ \\[8pt] + $e^{\alpha x}$ & $A\,e^{\alpha x}$ \\[8pt] + $\sin(\beta x)$ \textbf{or} $\cos(\beta x)$ & $A\sin(\beta x) + B\cos(\beta x)$ \\[8pt] + $e^{\alpha x} P_n(x)$ & $e^{\alpha x}(A_n x^n + \cdots + A_0)$ \\[8pt] + $e^{\alpha x}\sin(\beta x)$ \textbf{or} $e^{\alpha x}\cos(\beta x)$ & + \begin{tabular}[t]{@{}l@{}} + $e^{\alpha x}\bigl[(A_n x^n + \cdots + A_0)\cos(\beta x)$ \\ + \quad $+ (B_n x^n + \cdots + B_0)\sin(\beta x)\bigr]$ + \end{tabular} \\[12pt] + Product of two forms & Multiply the individual guesses \\[8pt] + Sum of two forms $g_1 + g_2$ & Sum the individual guesses \\ + \bottomrule + \end{tabular} + \end{center} +\end{keyresult} + +\subsubsection{Modification Rule} +\label{sec:ch05_modification_rule} + +There is one important caveat: if any term of your initial guess \textbf{already appears in the homogeneous solution} $y_h$, the guess will fail (substitution leads to $0 = g(x)$, a contradiction). In this case, you must \textbf{modify} the guess. + +\begin{hintbox} + \textbf{Modification rule.} If a term in the initial guess for $y_p$ is also a solution of the homogeneous equation: + \begin{enumerate} + \item Multiply that term by $x$. + \item If the modified term \emph{still} appears in $y_h$, multiply by $x$ again (i.e.\ by $x^2$). + \end{enumerate} + For a second-order equation, $x^2$ is always sufficient, because the homogeneous solution space has dimension~2. +\end{hintbox} + +\paragraph{Why this works.} The modification multiplies the guess by $x$ (or $x^2$), which introduces new functional forms (like $x e^{rx}$ or $x^2 e^{rx}$) that are \emph{not} solutions of the homogeneous equation, while preserving the algebraic structure needed for substitution. + +\subsubsection{Worked Examples} +\label{sec:ch05_uc_examples} + +\paragraph{Example 1: Polynomial forcing.} + +\begin{workedexample} + Solve $y'' - 3y' + 2y = 4x + 1$. + + \textbf{Solution.} \textit{Step 1: Homogeneous solution.} The characteristic equation is + \[ + r^2 - 3r + 2 = 0 \quad\Longrightarrow\quad (r-1)(r-2) = 0. + \] + Roots: $r_1 = 1$, $r_2 = 2$. The homogeneous solution is + \[ + y_h(x) = c_1 e^x + c_2 e^{2x}. + \] + + \textit{Step 2: Particular solution.} Here $g(x) = 4x + 1$, a first-degree polynomial. From the guess table, we try + \[ + y_p(x) = Ax + B. + \] + No term overlaps with $y_h$ (exponentials vs.\ polynomials), so no modification is needed. Differentiate: + \[ + y_p' = A, \qquad y_p'' = 0. + \] + Substitute into the ODE: + \[ + 0 - 3A + 2(Ax + B) = 4x + 1. + \] + Collect terms: + \[ + 2A\,x + (2B - 3A) = 4x + 1. + \] + Equate coefficients of like powers of $x$: + \[ + \begin{cases} + 2A = 4 \quad\Longrightarrow\quad A = 2, \\ + 2B - 3A = 1 \quad\Longrightarrow\quad 2B - 6 = 1 \quad\Longrightarrow\quad B = \dfrac{7}{2}. + \end{cases} + \] + The particular solution is $y_p(x) = 2x + \dfrac{7}{2}$. + + \textit{Step 3: General solution.} + \[ + y(x) = c_1 e^x + c_2 e^{2x} + 2x + \frac{7}{2}. + \] +\end{workedexample} + +\paragraph{Example 2: Exponential forcing.} + +\begin{workedexample} + Solve $y'' + y' - 6y = 12e^{3x}$. + + \textbf{Solution.} \textit{Step 1: Homogeneous solution.} The characteristic equation is + \[ + r^2 + r - 6 = 0 \quad\Longrightarrow\quad (r-2)(r+3) = 0. + \] + Roots: $r_1 = 2$, $r_2 = -3$. Thus + \[ + y_h(x) = c_1 e^{2x} + c_2 e^{-3x}. + \] + + \textit{Step 2: Particular solution.} Here $g(x) = 12e^{3x}$. From the guess table, we try + \[ + y_p(x) = A\,e^{3x}. + \] + Check for overlap: $e^{3x}$ is not in $y_h$ (which contains $e^{2x}$ and $e^{-3x}$), so no modification needed. Differentiate: + \[ + y_p' = 3A\,e^{3x}, \qquad y_p'' = 9A\,e^{3x}. + \] + Substitute: + \[ + 9A\,e^{3x} + 3A\,e^{3x} - 6A\,e^{3x} = 12e^{3x}. + \] + Factor out $e^{3x}$ (never zero): + \[ + (9 + 3 - 6)A = 12 \quad\Longrightarrow\quad 6A = 12 \quad\Longrightarrow\quad A = 2. + \] + So $y_p(x) = 2e^{3x}$. + + \textit{Step 3: General solution.} + \[ + y(x) = c_1 e^{2x} + c_2 e^{-3x} + 2e^{3x}. + \] +\end{workedexample} + +\paragraph{Example 3: Trigonometric forcing.} + +\begin{workedexample} + Solve $y'' + 4y = 8\cos(2x)$. + + \textbf{Solution.} \textit{Step 1: Homogeneous solution.} The characteristic equation is + \[ + r^2 + 4 = 0 \quad\Longrightarrow\quad r = \pm 2i. + \] + Thus + \[ + y_h(x) = c_1 \cos(2x) + c_2 \sin(2x). + \] + + \textit{Step 2: Particular solution.} Here $g(x) = 8\cos(2x)$. From the guess table, we try + \[ + y_p(x) = A\cos(2x) + B\sin(2x). + \] + \textbf{Overlap detected!} Both $\cos(2x)$ and $\sin(2x)$ appear in $y_h$. We must apply the modification rule: multiply the entire guess by $x$: + \[ + y_p(x) = x\bigl[A\cos(2x) + B\sin(2x)\bigr] = Ax\cos(2x) + Bx\sin(2x). + \] + Check again: $x\cos(2x)$ and $x\sin(2x)$ are \emph{not} in $y_h$, so we are good. Differentiate (product rule): + \[ + \begin{aligned} + y_p' &= A\cos(2x) - 2Ax\sin(2x) + B\sin(2x) + 2Bx\cos(2x), \\ + y_p'' &= -2A\sin(2x) - 2A\sin(2x) - 4Ax\cos(2x) + 2B\cos(2x) + 2B\cos(2x) - 4Bx\sin(2x) \\ + &= -4A\sin(2x) - 4Ax\cos(2x) + 4B\cos(2x) - 4Bx\sin(2x). + \end{aligned} + \] + Substitute into $y'' + 4y = 8\cos(2x)$: + \[ + \bigl[-4A\sin(2x) - 4Ax\cos(2x) + 4B\cos(2x) - 4Bx\sin(2x)\bigr] + + 4\bigl[Ax\cos(2x) + Bx\sin(2x)\bigr] = 8\cos(2x). + \] + The $x$-terms cancel: + \[ + -4A\sin(2x) + 4B\cos(2x) = 8\cos(2x). + \] + Equate coefficients: + \[ + \begin{cases} + -4A = 0 \quad\Longrightarrow\quad A = 0, \\ + 4B = 8 \quad\Longrightarrow\quad B = 2. + \end{cases} + \] + So $y_p(x) = 2x\sin(2x)$. + + \textit{Step 3: General solution.} + \[ + y(x) = c_1 \cos(2x) + c_2 \sin(2x) + 2x\sin(2x). + \] +\end{workedexample} + +\paragraph{Example 4: Product of exponential and trigonometric.} + +\begin{workedexample} + Solve $y'' - 2y' + 5y = 10e^x\cos(2x)$. + + \textbf{Solution.} \textit{Step 1: Homogeneous solution.} The characteristic equation is + \[ + r^2 - 2r + 5 = 0. + \] + Discriminant: $\Delta = 4 - 20 = -16 < 0$. Roots: + \[ + r = \frac{2 \pm \sqrt{-16}}{2} = 1 \pm 2i. + \] + Thus + \[ + y_h(x) = e^x\bigl[c_1 \cos(2x) + c_2 \sin(2x)\bigr]. + \] + + \textit{Step 2: Particular solution.} Here $g(x) = 10e^x\cos(2x)$. From the guess table for $e^{\alpha x}\cos(\beta x)$ with $\alpha = 1, \beta = 2$, we try + \[ + y_p(x) = e^x\bigl[A\cos(2x) + B\sin(2x)\bigr]. + \] + \textbf{Overlap detected!} Both $e^x\cos(2x)$ and $e^x\sin(2x)$ appear in $y_h$. Apply the modification rule: multiply by $x$: + \[ + y_p(x) = x\,e^x\bigl[A\cos(2x) + B\sin(2x)\bigr]. + \] + Now differentiate. Let $y_p = x\,e^x\,(A\cos(2x) + B\sin(2x))$. Set $u = x\,e^x$ and $v = A\cos(2x) + B\sin(2x)$ for clarity. Then $u' = e^x(1 + x)$ and $v' = -2A\sin(2x) + 2B\cos(2x)$. + \[ + \begin{aligned} + y_p' &= u'v + uv' = e^x(1+x)\bigl[A\cos(2x) + B\sin(2x)\bigr] + x\,e^x\bigl[-2A\sin(2x) + 2B\cos(2x)\bigr], \\ + y_p'' &= e^x(1+x)\bigl[-2A\sin(2x) + 2B\cos(2x)\bigr] + e^x\bigl[A\cos(2x) + B\sin(2x)\bigr] \\ + &\quad + e^x(1+x)\bigl[-2A\sin(2x) + 2B\cos(2x)\bigr] + x\,e^x\bigl[-4A\cos(2x) - 4B\sin(2x)\bigr] \\ + &= 2e^x(1+x)\bigl[-2A\sin(2x) + 2B\cos(2x)\bigr] + e^x\bigl[A\cos(2x) + B\sin(2x)\bigr] \\ + &\quad - 4x\,e^x\bigl[A\cos(2x) + B\sin(2x)\bigr]. + \end{aligned} + \] + Substitute $y_p$, $y_p'$, and $y_p''$ into the ODE $y'' - 2y' + 5y$. Collect the coefficient of $e^x\cos(2x)$ and $e^x\sin(2x)$ (the $x$-terms will cancel due to the modification rule): + + Coefficient of $e^x\cos(2x)$: $4B + A - 2A + 5A = 4B + 4A$. + + Coefficient of $e^x\sin(2x)$: $-4A + B - 2B + 5B = -4A + 4B$. + + Set equal to the RHS $10e^x\cos(2x)$: + \[ + \begin{cases} + 4A + 4B = 10, \\ + -4A + 4B = 0 \quad\Longrightarrow\quad A = B. + \end{cases} + \] + From the second equation, $A = B$. Substituting into the first: $8A = 10$, so $A = \dfrac{5}{4}$ and $B = \dfrac{5}{4}$. + + The particular solution is + \[ + y_p(x) = \frac{5}{4}\,x\,e^x\bigl[\cos(2x) + \sin(2x)\bigr]. + \] + + \textit{Step 3: General solution.} + \[ + y(x) = e^x\Bigl[c_1\cos(2x) + c_2\sin(2x)\Bigr] + \frac{5}{4}\,x\,e^x\bigl[\cos(2x) + \sin(2x)\bigr]. + \] +\end{workedexample} + +\paragraph{Example 5: Modification rule (repeated-root overlap).} + +\begin{workedexample} + Solve $y'' - 4y' + 4y = 6e^{2x}$. + + \textbf{Solution.} \textit{Step 1: Homogeneous solution.} The characteristic equation is + \[ + r^2 - 4r + 4 = 0 \quad\Longrightarrow\quad (r-2)^2 = 0. + \] + Double root: $r = 2$. Thus + \[ + y_h(x) = c_1 e^{2x} + c_2 x\,e^{2x}. + \] + + \textit{Step 2: Particular solution.} Here $g(x) = 6e^{2x}$. The initial guess is $y_p = A\,e^{2x}$. + + \textbf{Overlap check:} $e^{2x}$ is in $y_h$. Modify: multiply by $x \Rightarrow y_p = Ax\,e^{2x}$. + + \textbf{Overlap check again:} $x\,e^{2x}$ is \emph{also} in $y_h$. Modify again: multiply by $x$ once more $\Rightarrow y_p = Ax^2 e^{2x}$. + + Now $x^2 e^{2x}$ is \emph{not} in $y_h$, so this is our final guess. Differentiate (product rule): + \[ + \begin{aligned} + y_p &= A x^2 e^{2x}, \\ + y_p' &= A(2x e^{2x} + 2x^2 e^{2x}) = 2A x\,e^{2x} + 2A x^2 e^{2x}, \\ + y_p'' &= 2A e^{2x} + 4A x\,e^{2x} + 4A x\,e^{2x} + 4A x^2 e^{2x} \\ + &= 2A e^{2x} + 8A x\,e^{2x} + 4A x^2 e^{2x}. + \end{aligned} + \] + Substitute into $y'' - 4y' + 4y = 6e^{2x}$: + \[ + \begin{aligned} + &\bigl[2A e^{2x} + 8A x\,e^{2x} + 4A x^2 e^{2x}\bigr] + - 4\bigl[2A x\,e^{2x} + 2A x^2 e^{2x}\bigr] + + 4\bigl[A x^2 e^{2x}\bigr] \\ + &= 2A e^{2x} + (8A - 8A)x\,e^{2x} + (4A - 8A + 4A)x^2 e^{2x} \\ + &= 2A e^{2x}. + \end{aligned} + \] + Set equal to $6e^{2x}$: + \[ + 2A = 6 \quad\Longrightarrow\quad A = 3. + \] + The particular solution is $y_p(x) = 3x^2 e^{2x}$. + + \textit{Step 3: General solution.} + \[ + y(x) = c_1 e^{2x} + c_2 x\,e^{2x} + 3x^2 e^{2x}. + \] +\end{workedexample} + +\paragraph{Example 6: Polynomial times exponential.} + +\begin{workedexample} + Solve $y'' + 2y' + y = 3x\,e^{-x}$. + + \textbf{Solution.} \textit{Step 1: Homogeneous solution.} The characteristic equation is + \[ + r^2 + 2r + 1 = 0 \quad\Longrightarrow\quad (r+1)^2 = 0. + \] + Double root: $r = -1$. Thus + \[ + y_h(x) = c_1 e^{-x} + c_2 x\,e^{-x}. + \] + + \textit{Step 2: Particular solution.} Here $g(x) = 3x\,e^{-x} = e^{-x} P_1(x)$ where $P_1(x) = 3x$ is a first-degree polynomial. From the guess table, the initial guess would be + \[ + y_p(x) = e^{-x}(Ax + B). + \] + \textbf{Overlap check:} $e^{-x}$ and $x\,e^{-x}$ are both in $y_h$. We must multiply the entire guess by $x^2$: + \[ + y_p(x) = x^2 e^{-x}(Ax + B) = A x^3 e^{-x} + B x^2 e^{-x}. + \] + Differentiate: + \[ + \begin{aligned} + y_p &= (Ax^3 + Bx^2)e^{-x}, \\ + y_p' &= (3Ax^2 + 2Bx)e^{-x} - (Ax^3 + Bx^2)e^{-x} + = (-Ax^3 + (3A-B)x^2 + 2Bx)\,e^{-x}, \\ + y_p'' &= (-3Ax^2 + 2(3A-B)x + 2B)e^{-x} - (-Ax^3 + (3A-B)x^2 + 2Bx)e^{-x} \\ + &= \bigl[Ax^3 + (-3A + 2(3A-B) - (3A-B))x^2 + \dots\bigr]e^{-x}. + \end{aligned} + \] + Let me compute more carefully. Using the product rule systematically: + \[ + \begin{aligned} + y_p &= (Ax^3 + Bx^2)e^{-x}, \\ + y_p' &= (3Ax^2 + 2Bx - Ax^3 - Bx^2)e^{-x}, \\ + y_p'' &= (6Ax + 2B - 3Ax^2 - 2Bx - (3Ax^2 + 2Bx - Ax^3 - Bx^2))e^{-x} \\ + &= (Ax^3 - 5Ax^2 + (6A-4B)x + 2B)\,e^{-x}. + \end{aligned} + \] + Now substitute into $y'' + 2y' + y$: + \[ + \begin{aligned} + y_p'' + 2y_p' + y_p &= e^{-x}\Bigl[ + (Ax^3 - 5Ax^2 + (6A-4B)x + 2B) \\ + &\quad + 2(-Ax^3 + (3A-B)x^2 + 2Bx) + + (Ax^3 + Bx^2) + \Bigr] \\ + &= e^{-x}\Bigl[ + (A - 2A + A)x^3 + (-5A + 6A - 2B + B)x^2 + (6A - 4B + 4B)x + 2B + \Bigr] \\ + &= e^{-x}\Bigl[ (A-B)x^2 + 6Ax + 2B \Bigr]. + \end{aligned} + \] + Set equal to $g(x) = 3x\,e^{-x}$: + \[ + (A - B)x^2 + 6Ax + 2B = 3x. + \] + Equate coefficients: + \[ + \begin{cases} + A - B = 0 \quad\Longrightarrow\quad A = B, \\ + 6A = 3 \quad\Longrightarrow\quad A = \dfrac{1}{2}, \\ + 2B = 0 \quad\Longrightarrow\quad B = 0. + \end{cases} + \] + Wait --- $A = B$ from the first equation, but $A = 1/2$ and $B = 0$ from the others. Let me recheck the $x^2$ coefficient in $y_p''$. + + Let me redo this more carefully with a different approach. Set $y_p = e^{-x}(Ax^3 + Bx^2)$. + + \[ + \begin{aligned} + y_p' &= -e^{-x}(Ax^3 + Bx^2) + e^{-x}(3Ax^2 + 2Bx) \\ + &= e^{-x}\bigl(-Ax^3 + (3A-B)x^2 + 2Bx\bigr), \\ + y_p'' &= -e^{-x}\bigl(-Ax^3 + (3A-B)x^2 + 2Bx\bigr) + + e^{-x}\bigl(-3Ax^2 + 2(3A-B)x + 2B\bigr) \\ + &= e^{-x}\Bigl(Ax^3 - (3A-B+3A)x^2 + (2(3A-B)-2B)x + 2B\Bigr) \\ + &= e^{-x}\Bigl(Ax^3 - (6A-B)x^2 + (6A-4B)x + 2B\Bigr). + \end{aligned} + \] + Now $y'' + 2y' + y$: + \[ + \begin{aligned} + &e^{-x}\Bigl[ + \bigl(Ax^3 - (6A-B)x^2 + (6A-4B)x + 2B\bigr) \\ + &\quad + 2\bigl(-Ax^3 + (3A-B)x^2 + 2Bx\bigr) + + (Ax^3 + Bx^2) + \Bigr] \\ + &= e^{-x}\Bigl[ + (A - 2A + A)x^3 \\ + &\quad + \bigl(-(6A-B) + 2(3A-B) + B\bigr)x^2 \\ + &\quad + \bigl((6A-4B) + 4B\bigr)x + 2B + \Bigr] \\ + &= e^{-x}\Bigl[ 0\cdot x^3 + (A - B)x^2 + 6Ax + 2B \Bigr]. + \end{aligned} + \] + Set equal to $3x\,e^{-x}$: + \[ + \begin{cases} + A - B = 0 \quad\Longrightarrow\quad A = B, \\ + 6A = 3 \quad\Longrightarrow\quad A = \dfrac{1}{2}, \\ + 2B = 0 \quad\Longrightarrow\quad B = 0. + \end{cases} + \] + These are inconsistent. The issue is that the guess $x^2 e^{-x}(Ax + B)$ introduces a $Bx^2 e^{-x}$ term whose second derivative + $2y' + y$ does not vanish. Let me try the correct guess: since $e^{-x}$ and $xe^{-x}$ are in $y_h$, the unmodified guess for $xe^{-x}$ is $e^{-x}(Ax + B)$. Overlap on both terms $\Rightarrow$ multiply by $x^2$: + \[ + y_p = x^2 e^{-x}(Ax + B). + \] + This is what I computed. The inconsistency means I made an arithmetic error. Let me re-examine the $x^2$ coefficient: + $-(6A-B) + 2(3A-B) + B = -6A + B + 6A - 2B + B = 0$. So the $x^2$ coefficient is actually $0$, not $A - B$. Let me redo: + \[ + \text{Coefficient of } x^2: \quad -6A + B + 6A - 2B + B = 0. + \] + Good, that cancels. So the remaining equations are: + \[ + \begin{cases} + 6A = 3 \quad\Longrightarrow\quad A = \dfrac{1}{2}, \\ + 2B = 0 \quad\Longrightarrow\quad B = 0. + \end{cases} + \] + So $y_p(x) = \dfrac{1}{2}x^3 e^{-x}$. + + \textit{Step 3: General solution.} + \[ + y(x) = c_1 e^{-x} + c_2 x\,e^{-x} + \frac{1}{2}x^3 e^{-x}. + \] +\end{workedexample} \subsection{Variation of Parameters} \label{sec:ch05_variation_of_parameters} -% Content goes here +Variation of parameters (VoP) is a \textbf{universal} method for finding a particular solution. Unlike undetermined coefficients, it works for \emph{any} forcing function $g(x)$, provided we can evaluate the resulting integrals. The trade-off is that VoP typically involves more computation. + +\subsubsection{Full Derivation} +\label{sec:ch05_vop_derivation} + +\textbf{Setup.} Consider the nonhomogeneous equation in \textbf{normalized form}: +\begin{equation} + \label{eq:normalized_nonhomogeneous} + y'' + p(x)\,y' + q(x)\,y = g(x). +\end{equation} +Let $\{y_1(x), y_2(x)\}$ be a fundamental set of solutions for the associated homogeneous equation $y'' + p(x)y' + q(x)y = 0$. Their Wronskian is +\[ + W(x) = W(y_1, y_2)(x) = y_1(x)\,y_2'(x) - y_1'(x)\,y_2(x) \neq 0. +\] + +\textbf{Ansatz.} We look for a particular solution of the form +\begin{equation} + \label{eq:vop_ansatz} + y_p(x) = u_1(x)\,y_1(x) + u_2(x)\,y_2(x), +\end{equation} +where $u_1(x)$ and $u_2(x)$ are unknown functions to be determined. This is called ``variation of parameters'' because we have replaced the constant coefficients $c_1, c_2$ of the homogeneous solution with variable functions $u_1(x), u_2(x)$. + +Differentiate $y_p$: +\begin{equation} + \label{eq:vop_first_derivative} + y_p' = u_1' y_1 + u_1 y_1' + u_2' y_2 + u_2 y_2'. +\end{equation} +We have two unknown functions but only one equation (the ODE). To resolve this underdetermination, we impose an \textbf{auxiliary condition}: +\begin{equation} + \label{eq:vop_constraint} + u_1' y_1 + u_2' y_2 = 0. +\end{equation} +This is not a restriction --- it is a convenient choice that simplifies the algebra. With this constraint, the first derivative simplifies to +\[ + y_p' = u_1 y_1' + u_2 y_2'. +\] +Differentiate once more: +\begin{equation} + \label{eq:vop_second_derivative} + y_p'' = u_1' y_1' + u_1 y_1'' + u_2' y_2' + u_2 y_2''. +\end{equation} + +\textbf{Substitution.} Substitute $y_p$, $y_p'$, and $y_p''$ into the ODE \cref{eq:normalized_nonhomogeneous}: +\[ + \bigl[u_1' y_1' + u_1 y_1'' + u_2' y_2' + u_2 y_2''\bigr] + + p(x)\bigl[u_1 y_1' + u_2 y_2'\bigr] + + q(x)\bigl[u_1 y_1 + u_2 y_2\bigr] = g(x). +\] +Group terms by $u_1$ and $u_2$: +\[ + u_1'\,y_1' + u_2'\,y_2' + + u_1\bigl[y_1'' + p(x)y_1' + q(x)y_1\bigr] + + u_2\bigl[y_2'' + p(x)y_2' + q(x)y_2\bigr] = g(x). +\] +Since $y_1$ and $y_2$ are solutions of the homogeneous equation, the bracketed terms vanish. We are left with +\[ + u_1'\,y_1' + u_2'\,y_2' = g(x). +\] + +\textbf{System of equations.} Combining this with the constraint \cref{eq:vop_constraint}, we have a $2 \times 2$ linear system for $u_1'$ and $u_2'$: +\begin{equation} + \label{eq:vop_system} + \begin{cases} + u_1' y_1 + u_2' y_2 = 0, \\[4pt] + u_1' y_1' + u_2' y_2' = g(x). + \end{cases} +\end{equation} + +\textbf{Solution via Cramer's rule.} Write the system in matrix form: +\[ + \begin{pmatrix} y_1 & y_2 \\ y_1' & y_2' \end{pmatrix} + \begin{pmatrix} u_1' \\ u_2' \end{pmatrix} + = + \begin{pmatrix} 0 \\ g(x) \end{pmatrix}. +\] +The determinant of the coefficient matrix is the Wronskian: +\[ + \det = y_1 y_2' - y_1' y_2 = W(x). +\] +Since $\{y_1, y_2\}$ is a fundamental set, $W(x) \neq 0$ and the system is solvable. By Cramer's rule: +\begin{align} + u_1'(x) &= \frac{\det\begin{pmatrix} 0 & y_2 \\ g(x) & y_2' \end{pmatrix}}{W(x)} + = \frac{-y_2(x)\,g(x)}{W(x)}, \label{eq:vop_u1prime} \\ + u_2'(x) &= \frac{\det\begin{pmatrix} y_1 & 0 \\ y_1' & g(x) \end{pmatrix}}{W(x)} + = \frac{y_1(x)\,g(x)}{W(x)}. \label{eq:vop_u2prime} +\end{align} + +Integrate to find $u_1$ and $u_2$ (we may choose any antiderivatives; we set the constants of integration to zero since we need only \emph{one} particular solution): +\begin{align} + u_1(x) &= -\int \frac{y_2(x)\,g(x)}{W(x)}\,\diff x, \\ + u_2(x) &= \int \frac{y_1(x)\,g(x)}{W(x)}\,\diff x. +\end{align} + +\begin{keyresult} + \textbf{Variation of parameters formula.} + + Given a fundamental set $\{y_1, y_2\}$ of the homogeneous equation associated with \cref{eq:normalized_nonhomogeneous}, a particular solution is + \begin{equation} + \label{eq:vop_formula} + y_p(x) = -y_1(x)\int \frac{y_2(x)\,g(x)}{W(x)}\,\diff x + + y_2(x)\int \frac{y_1(x)\,g(x)}{W(x)}\,\diff x, + \end{equation} + where $W(x) = y_1 y_2' - y_1' y_2$. + + \textbf{CRITICAL:} The equation \emph{must} be in normalized form (coefficient of $y''$ equals~1) before applying this formula. For $a\,y'' + b\,y' + c\,y = \tilde{g}(x)$, first divide by $a$ so that the right-hand side becomes $g(x) = \tilde{g}(x)/a$. +\end{keyresult} + +\begin{hintbox} + \textbf{Normalized form warning.} A very common mistake is to apply the VoP formula to $a\,y'' + b\,y' + c\,y = \tilde{g}(x)$ directly, using $\tilde{g}(x)$ as $g(x)$. This is \textbf{incorrect}. The derivation assumed the coefficient of $y''$ is~1, so you \emph{must} divide the entire equation by $a$ first, making $g(x) = \tilde{g}(x)/a$. +\end{hintbox} + +\subsubsection{Worked Examples} +\label{sec:ch05_vop_examples} + +\begin{workedexample} + Solve $y'' + y = \sec(x)$ on the interval $(-\pi/2, \pi/2)$. + + \textbf{Solution.} \textit{Step 1: Homogeneous solution.} The characteristic equation is $r^2 + 1 = 0$, so $r = \pm i$. Thus + \[ + y_h(x) = c_1 \cos x + c_2 \sin x. + \] + We take $y_1(x) = \cos x$ and $y_2(x) = \sin x$. + + \textit{Step 2: Wronskian.} + \[ + W(x) = \cos x \cdot \cos x - (-\sin x) \cdot \sin x = \cos^2 x + \sin^2 x = 1. + \] + The Wronskian is constant, which simplifies the integrals. + + \textit{Step 3: Apply VoP.} The equation is already in normalized form ($g(x) = \sec x$). Compute: + \[ + \begin{aligned} + u_1(x) &= -\int \frac{y_2(x)\,g(x)}{W(x)}\,\diff x + = -\int \frac{\sin x \cdot \sec x}{1}\,\diff x + = -\int \tan x\,\diff x + = \ln|\cos x|, \\[8pt] + u_2(x) &= \int \frac{y_1(x)\,g(x)}{W(x)}\,\diff x + = \int \frac{\cos x \cdot \sec x}{1}\,\diff x + = \int 1\,\diff x + = x. + \end{aligned} + \] + + \textit{Step 4: Particular solution.} + \[ + y_p(x) = u_1(x)\,y_1(x) + u_2(x)\,y_2(x) + = \ln|\cos x|\,\cos x + x\,\sin x. + \] + + \textit{Step 5: General solution.} + \[ + y(x) = c_1 \cos x + c_2 \sin x + \cos x\,\ln|\cos x| + x\,\sin x. + \] +\end{workedexample} + +\begin{workedexample} + Solve $y'' - 4y' + 3y = \dfrac{e^{2x}}{1 + e^x}$ using variation of parameters. + + \textbf{Solution.} \textit{Step 1: Homogeneous solution.} The characteristic equation is + \[ + r^2 - 4r + 3 = 0 \quad\Longrightarrow\quad (r-1)(r-3) = 0. + \] + Roots: $r_1 = 1$, $r_2 = 3$. Thus + \[ + y_h(x) = c_1 e^x + c_2 e^{3x}. + \] + Take $y_1(x) = e^x$ and $y_2(x) = e^{3x}$. + + \textit{Step 2: Wronskian.} + \[ + W(x) = e^x \cdot 3e^{3x} - e^x \cdot e^{3x} = 3e^{4x} - e^{4x} = 2e^{4x}. + \] + + \textit{Step 3: Apply VoP.} The equation is in normalized form with $g(x) = \dfrac{e^{2x}}{1 + e^x}$. Compute: + \[ + \begin{aligned} + u_1(x) &= -\int \frac{y_2(x)\,g(x)}{W(x)}\,\diff x + = -\int \frac{e^{3x} \cdot \dfrac{e^{2x}}{1 + e^x}}{2e^{4x}}\,\diff x \\ + &= -\frac{1}{2}\int \frac{e^{5x}}{e^{4x}(1 + e^x)}\,\diff x + = -\frac{1}{2}\int \frac{e^x}{1 + e^x}\,\diff x. + \end{aligned} + \] + Substituting $u = 1 + e^x$, $\diff u = e^x\,\diff x$: + \[ + u_1(x) = -\frac{1}{2}\int \frac{1}{u}\,\diff u = -\frac{1}{2}\ln|u| + = -\frac{1}{2}\ln(1 + e^x). + \] + + Next: + \[ + \begin{aligned} + u_2(x) &= \int \frac{y_1(x)\,g(x)}{W(x)}\,\diff x + = \int \frac{e^x \cdot \dfrac{e^{2x}}{1 + e^x}}{2e^{4x}}\,\diff x \\ + &= \frac{1}{2}\int \frac{e^{3x}}{e^{4x}(1 + e^x)}\,\diff x + = \frac{1}{2}\int \frac{1}{e^x(1 + e^x)}\,\diff x. + \end{aligned} + \] + Use partial fractions: $\dfrac{1}{e^x(1+e^x)} = \dfrac{1}{e^x} - \dfrac{1}{1+e^x} = e^{-x} - \dfrac{1}{1+e^x}$. + \[ + \begin{aligned} + u_2(x) &= \frac{1}{2}\int \left(e^{-x} - \frac{1}{1+e^x}\right)\diff x \\ + &= \frac{1}{2}\Bigl(-e^{-x} - \ln(1+e^x) + x\Bigr). + \end{aligned} + \] + (The second integral: $\int \frac{1}{1+e^x}\diff x = x - \ln(1+e^x)$, obtained by writing $\frac{1}{1+e^x} = 1 - \frac{e^x}{1+e^x}$.) + + \textit{Step 4: Particular solution.} + \[ + \begin{aligned} + y_p(x) &= u_1(x)\,y_1(x) + u_2(x)\,y_2(x) \\ + &= -\frac{1}{2}\ln(1+e^x)\,e^x + \frac{1}{2}\Bigl(-e^{-x} - \ln(1+e^x) + x\Bigr)e^{3x} \\ + &= -\frac{1}{2}e^x\ln(1+e^x) - \frac{1}{2}e^{2x} - \frac{1}{2}e^{3x}\ln(1+e^x) + \frac{1}{2}x\,e^{3x}. + \end{aligned} + \] + The term $-\frac{1}{2}e^{3x}\ln(1+e^x)$ can be absorbed into the homogeneous part for large $x$, but it is part of the particular solution as computed. The term $-\frac{1}{2}e^{2x}$ is not a homogeneous solution, so it stays. + + \textit{Step 5: General solution.} + \[ + y(x) = c_1 e^x + c_2 e^{3x} - \frac{1}{2}e^x\ln(1+e^x) + - \frac{1}{2}e^{3x}\ln(1+e^x) - \frac{1}{2}e^{2x} + \frac{1}{2}x\,e^{3x}. + \] +\end{workedexample} + +\begin{workedexample} + Solve $2y'' + 4y' + 2y = e^{-x}$ using variation of parameters. Note that this equation is \textbf{not} in normalized form. + + \textbf{Solution.} \textit{Step 0: Normalize the equation.} Divide the entire equation by $2$: + \[ + y'' + 2y' + y = \frac{1}{2}\,e^{-x}. + \] + Now $g(x) = \dfrac{1}{2}e^{-x}$. + + \textit{Step 1: Homogeneous solution.} The characteristic equation is + \[ + r^2 + 2r + 1 = 0 \quad\Longrightarrow\quad (r+1)^2 = 0. + \] + Double root $r = -1$. Take $y_1(x) = e^{-x}$ and $y_2(x) = x\,e^{-x}$. + + \textit{Step 2: Wronskian.} + \[ + W(x) = e^{-x}\bigl(e^{-x} - x\,e^{-x}\bigr) - (-e^{-x})(x\,e^{-x}) + = e^{-2x}(1 - x) + x\,e^{-2x} = e^{-2x}. + \] + + \textit{Step 3: Apply VoP.} + \[ + \begin{aligned} + u_1(x) &= -\int \frac{y_2(x)\,g(x)}{W(x)}\,\diff x + = -\int \frac{x\,e^{-x} \cdot \tfrac{1}{2}e^{-x}}{e^{-2x}}\,\diff x + = -\frac{1}{2}\int x\,\diff x + = -\frac{1}{4}x^2, \\[8pt] + u_2(x) &= \int \frac{y_1(x)\,g(x)}{W(x)}\,\diff x + = \int \frac{e^{-x} \cdot \tfrac{1}{2}e^{-x}}{e^{-2x}}\,\diff x + = \frac{1}{2}\int 1\,\diff x + = \frac{1}{2}x. + \end{aligned} + \] + + \textit{Step 4: Particular solution.} + \[ + y_p(x) = u_1(x)\,y_1(x) + u_2(x)\,y_2(x) + = -\frac{1}{4}x^2\,e^{-x} + \frac{1}{2}x^2\,e^{-x} + = \frac{1}{4}x^2\,e^{-x}. + \] + This is consistent with the undetermined coefficients result from Example~5 (the modification rule also gave $x^2 e^{-x}$ for this case). + + \textit{Step 5: General solution.} + \[ + y(x) = c_1 e^{-x} + c_2 x\,e^{-x} + \frac{1}{4}x^2\,e^{-x}. + \] +\end{workedexample} \subsection{Method Comparison} \label{sec:ch05_method_comparison} -% Content goes here +Both undetermined coefficients (UC) and variation of parameters (VoP) find particular solutions, but they have different strengths and limitations. + +\begin{table}[htbp] +\centering +\caption{Undetermined coefficients vs.\ variation of parameters} +\label{tab:ch05_method_comparison} +\begin{tabular}{l l l} + \toprule + \textbf{Aspect} & \textbf{Undetermined Coefficients} & \textbf{Variation of Parameters} \\ + \midrule + Applicable $g(x)$ & Polynomials, $e^{\alpha x}$, $\sin(\beta x)$, $\cos(\beta x)$, and their sums/products & \emph{Any} continuous $g(x)$ (in principle) \\ + Requires $y_h$? & Only for overlap check (modification rule) & Yes --- need a full fundamental set \\ + Requires $W(x)$? & No & Yes \\ + Normalized form? & Not required & \textbf{Required} (coefficient of $y''$ must be~1) \\ + Computation & Algebra (solve linear system) & Integration (may be difficult or non-elementary) \\ + Speed (simple $g(x)$) & Fast --- algebraic, no integrals & Slower --- integrals needed \\ + Speed (complex $g(x)$) & \textbf{Not applicable} --- method fails & Always applicable (if integrals exist) \\ + \bottomrule +\end{tabular} +\end{table} + +\begin{hintbox} + \textbf{Decision guide.} + \begin{enumerate} + \item Is $g(x)$ a polynomial, exponential, sine/cosine, or a product of these? \textbf{Yes} $\Rightarrow$ use undetermined coefficients (simpler algebra). + \item Is $g(x)$ anything else ($\ln x$, $\tan x$, rational functions, etc.)? \textbf{Yes} $\Rightarrow$ use variation of parameters. + \item If you are unsure, variation of parameters is always a safe fallback (provided you can compute the integrals). + \end{enumerate} +\end{hintbox} + +\paragraph{Example comparison.} Consider $y'' + y' - 2y = 4x$. + +Using \textbf{undetermined coefficients}: guess $y_p = Ax + B$, substitute, solve the linear system. Takes about 5 lines of algebra. + +Using \textbf{variation of parameters}: find $y_h$, compute $W(x)$, evaluate two integrals involving $\int x e^x\,\diff x$ and $\int x e^{-2x}\,\diff x$ (by parts). Takes about 15 lines of computation. + +Both give the same answer, but UC is clearly faster here. However, if $g(x) = \ln x$, only VoP can be applied. \subsection{Summary} \label{sec:ch05_summary} -% Summary table at end of chapter \begin{table}[htbp] \centering -\caption{Chapter Summary} -\label{tab:ch05_summary} -\begin{tabular}{l l} -\toprule -\textbf{Concept} & \textbf{Key Formula/Method} \\ -\midrule -TBD & TBD \\ -\bottomrule +\caption{Nonhomogeneous second-order equations: complete reference} +\label{tab:ch05_summary_table} +\begin{tabular}{l p{8cm}} + \toprule + \textbf{Concept} & \textbf{Key formula / rule} \\ + \midrule + General solution & $y(x) = y_h(x) + y_p(x)$ (\cref{thm:nonhomogeneous_structure}) \\ + Homogeneous part & From \cref{ch:second_order_homogeneous}: $y_h = c_1 y_1 + c_2 y_2$ \\ + \midrule + \textbf{UC: polynomial $P_n(x)$} & $y_p = A_n x^n + A_{n-1} x^{n-1} + \cdots + A_0$ \\ + \textbf{UC: $e^{\alpha x}$} & $y_p = A\,e^{\alpha x}$ \\ + \textbf{UC: $\sin(\beta x)$ or $\cos(\beta x)$} & $y_p = A\sin(\beta x) + B\cos(\beta x)$ \\ + \textbf{UC: $e^{\alpha x} P_n(x)$} & $y_p = e^{\alpha x}(A_n x^n + \cdots + A_0)$ \\ + \textbf{UC: $e^{\alpha x}\sin(\beta x)$ or $e^{\alpha x}\cos(\beta x)$} & $y_p = e^{\alpha x}\bigl[(A_n x^n + \cdots + A_0)\cos(\beta x) + (B_n x^n + \cdots + B_0)\sin(\beta x)\bigr]$ \\ + \midrule + \textbf{UC modification rule} & If guess term $\in y_h$, multiply by $x$; if still $\in y_h$, multiply by $x^2$ \\ + \midrule + \textbf{VoP formula} & $y_p = -y_1\displaystyle\int\frac{y_2 g}{W}\,\diff x + y_2\displaystyle\int\frac{y_1 g}{W}\,\diff x$ \\ + \textbf{VoP constraint} & Equation must be normalized: $y'' + py' + qy = g$ (coeff.\ of $y''$ = 1) \\ + \textbf{VoP system} & $u_1' y_1 + u_2' y_2 = 0$, $\; u_1' y_1' + u_2' y_2' = g(x)$ \\ + \textbf{VoP Cramer's rule} & $u_1' = -\dfrac{y_2 g}{W}$, $\; u_2' = \dfrac{y_1 g}{W}$ \\ + \bottomrule \end{tabular} \end{table} + +\begin{hintbox} + \textbf{Problem-solving workflow for nonhomogeneous equations.} + \begin{enumerate} + \item \textbf{Identify the form of $g(x)$.} If it is a polynomial, exponential, trig function, or their products/sums, prefer undetermined coefficients. Otherwise, use variation of parameters. + \item \textbf{Solve the homogeneous equation} $ay'' + by' + cy = 0$ and obtain $y_h(x)$. + \item \textbf{If using UC:} Write the guess based on the table, check for overlap with $y_h$, modify if needed, substitute, and solve for coefficients. + \item \textbf{If using VoP:} \emph{First} normalize the equation (divide by $a$ if needed). Compute $W(x)$, then evaluate the two integrals for $u_1$ and $u_2$. + \item \textbf{Combine:} $y(x) = y_h(x) + y_p(x)$. Apply initial conditions if given. + \end{enumerate} +\end{hintbox}