From 5f0231fc9dec5610369398c16724c9eb3792fec3 Mon Sep 17 00:00:00 2001 From: Worker Agent Date: Thu, 4 Jun 2026 15:37:21 -0500 Subject: [PATCH] ch2: first-order solution methods with examples --- chapters/ch02_first_order.tex | 483 +++++++++++++++++++++++++++++++++- 1 file changed, 472 insertions(+), 11 deletions(-) diff --git a/chapters/ch02_first_order.tex b/chapters/ch02_first_order.tex index 3a46719..53fc804 100644 --- a/chapters/ch02_first_order.tex +++ b/chapters/ch02_first_order.tex @@ -1,44 +1,505 @@ \section{First-Order Methods} \label{ch:first_order} +\usetikzlibrary{positioning} + \subsection{Separable Equations} \label{sec:ch02_separable} -% Content goes here +A first-order ODE is \textbf{separable} when it can be written as +\begin{equation} + \frac{\diff y}{\diff x} = g(x)\,h(y) +\end{equation} +i.e.\ the right-hand side factors into a function of $x$ times a function of $y$. -\subsection{Linear First-Order (Integrating Factor)} +\begin{keyresult} + \textbf{Separable equations.} Rewrite as $\dfrac{\diff y}{h(y)} = g(x)\,\diff x$ and integrate both sides: + \[ + \int \frac{\diff y}{h(y)} = \int g(x)\,\diff x + C. + \] + Solve for $y$ explicitly when possible; otherwise leave the solution in implicit form. +\end{keyresult} + +\begin{hintbox} + If $h(y_0)=0$, the constant function $y(x)=y_0$ is always a solution (an \emph{equilibrium}). These may be lost when dividing by $h(y)$, so check separately. +\end{hintbox} + +\begin{workedexample} + Solve $\dfrac{\diff y}{\diff x} = x\,y$. + + \textbf{Solution.} Separate variables: + \[ + \frac{\diff y}{y} = x\,\diff x. + \] + Integrate: + \[ + \ln|y| = \frac{x^2}{2} + C. + \] + Exponentiate: + \[ + |y| = e^{C}\,e^{x^2/2} \quad\Longrightarrow\quad y = C_1\,e^{x^2/2}, + \] + where $C_1 = \pm e^{C}$ is an arbitrary nonzero constant. Including the equilibrium $y=0$ (lost when dividing by $y$), the general solution is + \[ + y(x) = C\,e^{x^2/2}, \qquad C \in \R. + \] +\end{workedexample} + +\begin{workedexample} + Solve $\dfrac{\diff y}{\diff x} = \dfrac{2x}{3y^2}$. + + \textbf{Solution.} Separate: + \[ + 3y^2\,\diff y = 2x\,\diff x. + \] + Integrate: + \[ + y^3 = x^2 + C. + \] + This implicit form is perfectly acceptable. Solving explicitly gives $y = \sqrt[3]{x^2 + C}$. +\end{workedexample} + +\subsection{Linear First-Order Equations} \label{sec:ch02_linear_first_order} -% Content goes here +A first-order ODE is \textbf{linear} if it can be written in the \emph{standard form} +\begin{equation} + \frac{\diff y}{\diff x} + p(x)\,y = g(x). + \label{eq:linear_std} +\end{equation} +The coefficient of $\diff y/\diff x$ \textbf{must be $1$}; if the equation arrives with a leading coefficient $a(x)$, divide through first. + +\begin{keyresult} + \textbf{Integrating factor.} For \cref{eq:linear_std}, the integrating factor is + \[ + \mu(x) = \exp\!\left(\int p(x)\,\diff x\right). + \] + Multiplying the equation by $\mu(x)$ produces + \[ + \frac{\diff}{\diff x}\bigl[\mu(x)\,y\bigr] = \mu(x)\,g(x), + \] + so the general solution is + \[ + y(x) = \frac{1}{\mu(x)}\left(\int \mu(x)\,g(x)\,\diff x + C\right). + \] +\end{keyresult} + +\textbf{Derivation.} With $\mu(x) = \exp\!\bigl(\int p(x)\,\diff x\bigr)$, the chain rule gives +$\mu'(x) = \mu(x)\,p(x)$. Multiplying \cref{eq:linear_std} by $\mu(x)$: +\[ + \mu y' + \mu p\,y = \mu g \quad\Longrightarrow\quad \mu y' + \mu' y = \mu g. +\] +The left side is exactly the product-rule derivative $\dfrac{\diff}{\diff x}[\mu y]$, yielding the result. + +\begin{hintbox} + \textbf{Pitfall.} If the equation is $a(x)y' + b(x)y = f(x)$ with $a(x)\neq 1$, you \emph{must} divide by $a(x)$ to obtain standard form before computing $\mu(x)$. Forgetting this is the most common error. +\end{hintbox} + +\begin{workedexample} + Solve $y' + \dfrac{2}{x}\,y = x^3$. + + \textbf{Solution.} The equation is already in standard form with $p(x)=\dfrac{2}{x}$ and $g(x)=x^3$. + Compute the integrating factor: + \[ + \mu(x) = \exp\!\left(\int \frac{2}{x}\,\diff x\right) + = \exp\!\bigl(2\ln|x|\bigr) + = |x|^2 = x^2. + \] + (We drop the absolute value since $x^2 \ge 0$.) + + Multiply the entire equation by $x^2$: + \[ + x^2 y' + 2x\,y = x^5 + \quad\Longrightarrow\quad + \frac{\diff}{\diff x}\bigl[x^2 y\bigr] = x^5. + \] + Integrate: + \[ + x^2 y = \frac{x^6}{6} + C. + \] + Divide by $x^2$: + \[ + y(x) = \frac{x^4}{6} + \frac{C}{x^2}. + \] +\end{workedexample} + +\begin{workedexample} + Solve $2x\,y' + 3y = 6x$. + + \textbf{Solution.} The leading coefficient is $2x \neq 1$. Divide through: + \[ + y' + \frac{3}{2x}\,y = 3. + \] + Now $p(x) = \dfrac{3}{2x}$ and $g(x) = 3$. Integrating factor: + \[ + \mu(x) = \exp\!\left(\int \frac{3}{2x}\,\diff x\right) + = \exp\!\left(\tfrac{3}{2}\ln|x|\right) + = |x|^{3/2}. + \] + Work with $x>0$ for simplicity, so $\mu(x) = x^{3/2}$. Multiply: + \[ + x^{3/2}y' + \tfrac{3}{2}x^{1/2}y = 3x^{3/2} + \quad\Longrightarrow\quad + \frac{\diff}{\diff x}\!\left[x^{3/2}y\right] = 3x^{3/2}. + \] + Integrate: + \[ + x^{3/2}y = \int 3x^{3/2}\,\diff x = 3\cdot\frac{2}{5}x^{5/2} + C = \frac{6}{5}x^{5/2} + C. + \] + Solve for $y$: + \[ + y(x) = \frac{6}{5}x + \frac{C}{x^{3/2}}. + \] +\end{workedexample} \subsection{Exact Equations} \label{sec:ch02_exact} -% Content goes here +An ODE written in differential form +\begin{equation} + M(x,y)\,\diff x + N(x,y)\,\diff y = 0 +\end{equation} +is \textbf{exact} if there exists a scalar function $\psi(x,y)$ such that +$\pd{\psi}{x} = M$ and $\pd{\psi}{y} = N$. The solution is then given implicitly by +$\psi(x,y) = C$. + +\begin{keyresult} + \textbf{Exactness test.} The equation is exact \emph{if and only if} + \[ + \pd{M}{y} = \pd{N}{x}. + \] + When exact, the potential function is + \[ + \psi(x,y) = \int M\,\diff x \;+\; \int\!\left[N - \pd{}{y}\!\left(\int M\,\diff x\right)\right]\!\diff y. + \] + The solution is $\psi(x,y) = C$. +\end{keyresult} + +\begin{workedexample} + Solve $(2xy + y^3)\,\diff x + (x^2 + 3xy^2)\,\diff y = 0$. + + \textbf{Solution.} Here $M = 2xy + y^3$ and $N = x^2 + 3xy^2$. + Check exactness: + \[ + \pd{M}{y} = 2x + 3y^2, \qquad \pd{N}{x} = 2x + 3y^2. + \] + They match, so the equation is exact. + + Integrate $M$ with respect to $x$: + \[ + \int (2xy + y^3)\,\diff x = x^2y + xy^3 + h(y). + \] + Differentiate with respect to $y$ and equate to $N$: + \[ + \pd{}{y}\bigl[x^2y + xy^3 + h(y)\bigr] = x^2 + 3xy^2 + h'(y) \stackrel{!}{=} x^2 + 3xy^2. + \] + Thus $h'(y) = 0$, so $h(y)$ is constant. The potential function is + \[ + \psi(x,y) = x^2y + xy^3. + \] + The implicit solution is + \[ + x^2y + xy^3 = C. + \] +\end{workedexample} + +\paragraph{Integrating factors for non-exact equations.} +If $\pd{M}{y} \neq \pd{N}{x}$, the equation can sometimes be made exact by multiplying by an integrating factor $\mu$. + +\begin{keyresult} + \textbf{Integrating factors.} + \begin{itemize} + \item If $\displaystyle \frac{\pd{M}{y} - \pd{N}{x}}{N}$ depends \emph{only on $x$}, then + $\mu(x) = \exp\!\left(\int \frac{\pd{M}{y} - \pd{N}{x}}{N}\,\diff x\right)$. + \item If $\displaystyle \frac{\pd{N}{x} - \pd{M}{y}}{M}$ depends \emph{only on $y$}, then + $\mu(y) = \exp\!\left(\int \frac{\pd{N}{x} - \pd{M}{y}}{M}\,\diff y\right)$. + \end{itemize} +\end{keyresult} + +\begin{workedexample} + Solve $(3xy + y^2)\,\diff x + (x^2 + xy)\,\diff y = 0$. + + \textbf{Solution.} $M = 3xy + y^2$, $N = x^2 + xy$. Check: + \[ + \pd{M}{y} = 3x + 2y, \qquad \pd{N}{x} = 2x + y. + \] + Not exact. Test for an integrating factor depending on $x$ alone: + \[ + \frac{\pd{M}{y} - \pd{N}{x}}{N} + = \frac{(3x+2y) - (2x+y)}{x^2 + xy} + = \frac{x + y}{x(x+y)} + = \frac{1}{x}. + \] + This depends only on $x$, so + \[ + \mu(x) = \exp\!\left(\int \frac{1}{x}\,\diff x\right) = \exp(\ln|x|) = x. + \] + Multiply the original equation by $\mu = x$: + \[ + (3x^2y + xy^2)\,\diff x + (x^3 + x^2y)\,\diff y = 0. + \] + Verify: $\pd{\tilde{M}}{y} = 3x^2 + 2xy = \pd{\tilde{N}}{x}$. Exact. + + Integrate $\tilde{M}$ with respect to $x$: + \[ + \int (3x^2y + xy^2)\,\diff x = x^3y + \tfrac{1}{2}x^2y^2 + h(y). + \] + Differentiate with respect to $y$: + \[ + x^3 + x^2y + h'(y) \stackrel{!}{=} x^3 + x^2y \quad\Longrightarrow\quad h'(y)=0. + \] + The potential is $\psi(x,y) = x^3y + \dfrac{1}{2}x^2y^2$, and the solution is + \[ + x^3y + \tfrac{1}{2}x^2y^2 = C. + \] +\end{workedexample} \subsection{Bernoulli Equations} \label{sec:ch02_bernoulli} -% Content goes here +A \textbf{Bernoulli equation} has the form +\begin{equation} + \frac{\diff y}{\diff x} + p(x)\,y = g(x)\,y^n, \qquad n \neq 0,1. +\end{equation} +For $n=0$ the equation is linear; for $n=1$ it is also linear. The substitution $v = y^{1-n}$ transforms it into a linear equation. + +\begin{keyresult} + \textbf{Bernoulli substitution.} Let $v = y^{1-n}$. Then + \[ + v' = (1-n)\,y^{-n}\,y', + \] + and substituting into the Bernoulli equation yields the \emph{linear} equation in $v$: + \[ + v' + (1-n)\,p(x)\,v = (1-n)\,g(x). + \] +\end{keyresult} + +\begin{workedexample} + Solve $y' + \dfrac{1}{x}\,y = x\,y^3$. + + \textbf{Solution.} This is Bernoulli with $p(x)=\dfrac{1}{x}$, $g(x)=x$, and $n=3$. + Substitute $v = y^{1-3} = y^{-2}$. Then $v' = -2y^{-3}y'$. + + Multiply the original equation by $-2y^{-3}$: + \[ + -2y^{-3}y' - 2y^{-3}\cdot\frac{1}{x}\,y = -2y^{-3}\cdot x\,y^3. + \] + Recognizing $v = y^{-2}$ and $v' = -2y^{-3}y'$: + \[ + v' - \frac{2}{x}\,v = -2x. + \] + This is a linear equation in $v$. Integrating factor: + \[ + \mu(x) = \exp\!\left(\int -\frac{2}{x}\,\diff x\right) = \exp(-2\ln|x|) = \frac{1}{x^2}. + \] + Multiply by $\mu$: + \[ + \frac{1}{x^2}v' - \frac{2}{x^3}v = -\frac{2}{x} + \quad\Longrightarrow\quad + \frac{\diff}{\diff x}\!\left[\frac{v}{x^2}\right] = -\frac{2}{x}. + \] + Integrate: + \[ + \frac{v}{x^2} = -2\ln|x| + C + \quad\Longrightarrow\quad + v = x^2\bigl(C - 2\ln|x|\bigr). + \] + Substitute back $v = y^{-2}$: + \[ + \frac{1}{y^2} = x^2\bigl(C - 2\ln|x|\bigr) + \quad\Longrightarrow\quad + y(x) = \pm\frac{1}{x\sqrt{\,C - 2\ln|x|\,}}. + \] +\end{workedexample} + +\begin{workedexample} + Solve $y' - \dfrac{2}{x}\,y = x^2\,y^2$. + + \textbf{Solution.} Bernoulli with $n=2$, $p(x)=-\dfrac{2}{x}$, $g(x)=x^2$. + Substitute $v = y^{1-2} = y^{-1} = \dfrac{1}{y}$. Then $v' = -y^{-2}y'$. + + Multiply the original equation by $-y^{-2}$: + \[ + -y^{-2}y' + \frac{2}{x}\,y^{-1} = -x^2. + \] + In terms of $v$: + \[ + v' + \frac{2}{x}\,v = -x^2. + \] + Integrating factor: + \[ + \mu(x) = \exp\!\left(\int \frac{2}{x}\,\diff x\right) = x^2. + \] + Multiply: + \[ + x^2v' + 2xv = -x^4 + \quad\Longrightarrow\quad + \frac{\diff}{\diff x}\!\left[x^2v\right] = -x^4. + \] + Integrate: + \[ + x^2v = -\frac{x^5}{5} + C + \quad\Longrightarrow\quad + v = -\frac{x^3}{5} + \frac{C}{x^2}. + \] + Since $v = 1/y$: + \[ + y(x) = \frac{1}{-\dfrac{x^3}{5} + \dfrac{C}{x^2}} + = \frac{5x^2}{C - x^5}. + \] +\end{workedexample} \subsection{Homogeneous Substitutions} \label{sec:ch02_homogeneous_substitutions} -% Content goes here +A first-order ODE is \textbf{homogeneous} (of degree zero) if +\begin{equation} + \frac{\diff y}{\diff x} = F\!\left(\frac{y}{x}\right). +\end{equation} +That is, the right-hand side is a function of the ratio $y/x$ alone. + +\begin{keyresult} + \textbf{Homogeneous substitution.} Let $y = vx$. Then + \[ + \frac{\diff y}{\diff x} = v + x\frac{\diff v}{\diff x}. + \] + The equation becomes + \[ + v + x\frac{\diff v}{\diff x} = F(v) + \quad\Longrightarrow\quad + \frac{\diff v}{\diff x} = \frac{F(v) - v}{x}, + \] + which is separable: + \[ + \int \frac{\diff v}{F(v) - v} = \int \frac{\diff x}{x} = \ln|x| + C. + \] +\end{keyresult} + +\begin{workedexample} + Solve $y' = \dfrac{x+y}{x-y}$. + + \textbf{Solution.} Rewrite the right side: + \[ + \frac{x+y}{x-y} = \frac{1 + y/x}{1 - y/x} = F\!\left(\frac{y}{x}\right). + \] + The equation is homogeneous. Set $y = vx$, so $y' = v + xv'$: + \[ + v + x\frac{\diff v}{\diff x} = \frac{1+v}{1-v}. + \] + Isolate the derivative: + \[ + x\frac{\diff v}{\diff x} = \frac{1+v}{1-v} - v = \frac{1+v - v(1-v)}{1-v} = \frac{1+v^2}{1-v}. + \] + Separate: + \[ + \frac{1-v}{1+v^2}\,\diff v = \frac{\diff x}{x}. + \] + Split the left integrand: + \[ + \int \frac{1}{1+v^2}\,\diff v - \int \frac{v}{1+v^2}\,\diff v = \int \frac{\diff x}{x}. + \] + These are standard integrals: + \[ + \arctan(v) - \tfrac{1}{2}\ln(1+v^2) = \ln|x| + C. + \] + Substitute back $v = y/x$: + \[ + \arctan\!\left(\frac{y}{x}\right) - \tfrac{1}{2}\ln\!\left(1 + \frac{y^2}{x^2}\right) = \ln|x| + C. + \] + Using $\ln(1 + y^2/x^2) = \ln\!\bigl((x^2+y^2)/x^2\bigr) = \ln(x^2+y^2) - 2\ln|x|$, the solution simplifies to + \[ + \arctan\!\left(\frac{y}{x}\right) - \tfrac{1}{2}\ln(x^2+y^2) = C. + \] +\end{workedexample} + +\begin{workedexample} + Solve $y' = \dfrac{xy + y^2}{x^2}$. + + \textbf{Solution.} Rewrite: + \[ + y' = \frac{y}{x} + \frac{y^2}{x^2} = F\!\left(\frac{y}{x}\right). + \] + Homogeneous with $F(v) = v + v^2$. Set $y = vx$: + \[ + v + x\frac{\diff v}{\diff x} = v + v^2 + \quad\Longrightarrow\quad + x\frac{\diff v}{\diff x} = v^2. + \] + Separate: + \[ + \frac{\diff v}{v^2} = \frac{\diff x}{x}. + \] + Integrate: + \[ + -\frac{1}{v} = \ln|x| + C + \quad\Longrightarrow\quad + v = -\frac{1}{\ln|x| + C}. + \] + Substitute back $v = y/x$: + \[ + y(x) = -\frac{x}{\ln|x| + C}. + \] + The equilibrium solution $y=0$ (corresponding to $v=0$) is also valid. +\end{workedexample} \subsection{Summary} \label{sec:ch02_summary} -% Summary table at end of chapter +The following flowchart guides method selection. Test methods in the order shown; the first applicable method solves the equation. + +\begin{center} +\begin{tikzpicture}[node distance=1.3cm and 1.2cm, + >=Stealth, + box/.style={draw, rounded corners=3pt, fill=LightSteelBlue!20, minimum width=3cm, align=center, font=\small, text width=3.2cm}, + sol/.style={draw, rounded corners=3pt, fill=ForestGreen!15, minimum width=3cm, align=center, font=\small, text width=3.0cm}, + arrow/.style={->, >=Stealth, thick, shorten <=1pt, shorten >=1pt}] + + \node[box, fill=LemonChiffon!40] (start) {\textbf{Given:} $y' = f(x,y)$}; + \node[box, below=of start] (t1) {\textbf{Can $f$ be written}\\as $g(x)\,h(y)$?}; + \node[box, below=of t1] (t2) {\textbf{Is it linear}\\$y'+p(x)y=g(x)$?}; + \node[box, below=of t2] (t3) {\textbf{Is it Bernoulli}\\$y'+py=gy^n$?}; + \node[box, below=of t3] (t4) {\textbf{Is $f(x,y)=F(y/x)$?}\\(Homogeneous)}; + \node[box, below=of t4] (t5) {\textbf{Write as $M\,dx+N\,dy=0$.}\\Exact?}; + \node[box, below=of t5, fill=Salmon!20] (gen) {\textbf{Non-exact:}\\Try integrating\\factor or other methods}; + + % Solution boxes + \node[sol, right=2cm of t1] (s1) {Separable:\\Integrate}; + \node[sol, right=2cm of t2] (s2) {Integrating factor:\\$\mu=e^{\int p\,dx}$}; + \node[sol, right=2cm of t3] (s3) {Substitute\\$v=y^{1-n}$}; + \node[sol, right=2cm of t4] (s4) {Substitute\\$y=vx$}; + \node[sol, right=2cm of t5] (s5) {Potential\\$\psi(x,y)=C$}; + + % Decision arrows + \draw[arrow] (start) -- (t1); + \draw[arrow] (t1) -- node[above, font=\footnotesize, fill=white, inner sep=1pt] {No} (t2); + \draw[arrow] (t2) -- node[above, font=\footnotesize, fill=white, inner sep=1pt] {No} (t3); + \draw[arrow] (t3) -- node[above, font=\footnotesize, fill=white, inner sep=1pt] {No} (t4); + \draw[arrow] (t4) -- node[above, font=\footnotesize, fill=white, inner sep=1pt] {No} (t5); + \draw[arrow] (t5) -- node[above, font=\footnotesize, fill=white, inner sep=1pt] {No} (gen); + + % Yes arrows + \draw[arrow] (t1) -- node[above, font=\footnotesize, fill=white, inner sep=1pt] {Yes} (s1); + \draw[arrow] (t2) -- node[above, font=\footnotesize, fill=white, inner sep=1pt] {Yes} (s2); + \draw[arrow] (t3) -- node[above, font=\footnotesize, fill=white, inner sep=1pt] {Yes} (s3); + \draw[arrow] (t4) -- node[above, font=\footnotesize, fill=white, inner sep=1pt] {Yes} (s4); + \draw[arrow] (t5) -- node[above, font=\footnotesize, fill=white, inner sep=1pt] {Yes} (s5); +\end{tikzpicture} +\end{center} + \begin{table}[htbp] \centering -\caption{Chapter Summary} +\caption{First-order solution methods} \label{tab:ch02_summary} -\begin{tabular}{l l} +\begin{tabular}{l l p{6cm}} \toprule -\textbf{Concept} & \textbf{Key Formula/Method} \\ +\textbf{Type} & \textbf{Form} & \textbf{Method} \\ \midrule -TBD & TBD \\ +Separable & $y' = g(x)h(y)$ & Rewrite as $\dfrac{\diff y}{h(y)} = g(x)\,\diff x$; integrate both sides \\ +Linear & $y' + p(x)y = g(x)$ & Integrating factor $\mu(x) = \exp\!\left(\int p(x)\,\diff x\right)$ \\ +Exact & $M\,\diff x + N\,\diff y = 0,\;\pd{M}{y}=\pd{N}{x}$ & Find potential $\psi$ with $\pd{\psi}{x}=M,\;\pd{\psi}{y}=N$; $\psi=C$ \\ +Non-exact & $M\,\diff x + N\,\diff y = 0,\;\pd{M}{y}\neq\pd{N}{x}$ & Multiply by integrating factor $\mu(x)$ or $\mu(y)$ \\ +Bernoulli & $y' + p(x)y = g(x)y^n$ & Substitute $v=y^{1-n}$; solve resulting linear equation \\ +Homogeneous & $y' = F(y/x)$ & Substitute $y=vx$; solve resulting separable equation \\ \bottomrule \end{tabular} \end{table}