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