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Mathematics LibreTexts

3.4: Direction Fields for First Order Equations

  • Page ID
    15015
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

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    It's impossible to find explicit formulas for solutions of some differential equations. Even if there are such formulas, they may be so complicated that they're useless. In this case we may resort to graphical or numerical methods to get some idea of how the solutions of the given equation behave.

    In Section~3.4 we'll take up the question of existence of solutions of a first order equation

    \begin{equation} \label{eq:3.4.1} y'=f(x,y). \end{equation}

    In this section we'll simply assume that \eqref{eq:3.4.1} has solutions and discuss a graphical method for approximating them.

    In Chapter~3 we discuss numerical methods for obtaining approximate solutions of \eqref{eq:3.4.1}. Recall that a solution of \eqref{eq:3.4.1} is a function \(y=y(x)\) such that \(y'(x)=f(x,y(x))\) for all values of \(x\) in some interval, and an integral curve is either the graph of a solution or is made up of segments that are graphs of solutions.Therefore, not being able to solve \eqref{eq:3.4.1} is equivalent to not knowing the equations of integral curves of \eqref{eq:3.4.1}. However, it's easy to calculate the slopes of these curves. To be specific, the slope of an integral curve of \eqref{eq:3.4.1} through a given point \((x_0,y_0)\) is given by the number \(f(x_0,y_0)\). This is the basis of \( \textcolor{blue}{\mbox{the method of direction field}} \)

    If \(f\) is defined on a set \(R\), we can construct a \( \textcolor{blue}{\mbox{direction field}} \) for \eqref{eq:3.4.1} in \(R\) by drawing a short line segment through each point \((x,y)\) in \(R\) with slope \(f(x,y)\). Of course, as a practical matter, we can't actually draw line segments through \( \textcolor{blue}{\mbox{every}} \) point in \(R\); rather, we must select a finite set of points in \(R\).

    For example, suppose \(f\) is defined on the closed rectangular region \(R:\{a\le x\le b, c\le y\le d\}.\) Let \(a= x_0< x_1< \cdots< x_m=b\) be equally spaced points in \([a,b]\) and \(c=y_0<y_1<\cdots<y_n=d\) be equally spaced points in \([c,d]\). We say that the points \((x_i,y_j),\quad 0\le i\le m,\quad 0\le j\le n,\) form a \( \textcolor{blue}{\mbox{rectangular grid}} \)

    Figure~\ref{figure:3.4.1}).

    Through each point in the grid we draw a short line segment with slope \(f(x_i,y_j)\). The result is an approximation to a direction field for \eqref{eq:3.4.1} in \(R\). If the grid points are sufficiently numerous and close together, we can draw approximate integral curves of \eqref{eq:3.4.1} by drawing curves through points in the grid tangent to the line segments associated with the points in the grid.

    \begin{figure}[H] \centering \includegraphics[width = 0.7\textwidth]{fig010301} \color{blue} \caption{\quad A rectangular grid} \label{figure:3.4.1} \end{figure}

    Unfortunately, approximating a direction field and graphing integral curves in this way is too tedious to be done effectively by hand. However, there is software for doing this. As you'll see, the combination of direction fields and integral curves gives useful insights into the behavior of the solutions of the differential equation even if we can't obtain exact solutions.

    We'll study numerical methods for solving a single first order equation \eqref{eq:3.4.1} in Chapter~3. These methods can be used to plot solution curves of \eqref{eq:3.4.1} in a rectangular region \(R\) \( \textcolor{blue}{\mbox{if \(f\) is continuous on}} \) \(R\).

    Figures~\ref{figure:3.4.2}, \ref{figure:3.4.3}, and \ref{figure:3.4.4} show direction fields and solution curves for the differential equations \(y'=\frac{x^2-y^2}{1+x^2+y^2}, y'=1+xy^2, y'=\frac{x-y}{1+x^2}\) which are all of the form \eqref{eq:3.4.1} with \(f\) continuous for all \((x,y)\).

    \begin{figure}[htbp]

    \color{blue}

    \begin{minipage}[b]{0.5\linewidth}

    \centering

    \scalebox{.6}{

    \includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig010302}}

    \caption{\,A direction field and integral curves for

    $y =\dst{\frac{x^{2}-y^{2}}{1+x^{2}+y^{2}}}$}

    \label{figure:1.3.2}

    \end{minipage}

    \hspace{0.6cm}

    \begin{minipage}[b]{0.5\linewidth}

    \centering

    \scalebox{.7}{

    \includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig010303}}

    \caption{A direction field and integral curves

    for $y'=1+xy^2$}

    \label{figure:1.3.3}

    \end{minipage}

    \end{figure}

    \begin{figure}[tbp]

    \centering

    \includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig010304}

    \color{blue}

    \caption{A direction and integral curves for

    $y'=\dst{\frac{x-y}{1+x^2}}$}

    \label{figure:1.3.4}

    \end{figure}

    The methods of Chapter~3 won't work for the equation \begin{equation} \label{eq:3.4.2} y'=-x/y \end{equation}

    if \(R\) contains part of the \(x\)-axis, since \(f(x,y)=-x/y\) is undefined when \(y=0\). Similarly, they won't work for the equation

    \begin{equation} \label{eq:3.4.3} y'={x^2\over1-x^2-y^2} \end{equation}

    if \(R\) contains any part of the unit circle \(x^2+y^2=1\), because the right side of \eqref{eq:3.4.3} is undefined if \(x^2+y^2=1\). However, \eqref{eq:3.4.2} and \eqref{eq:3.4.3} can written as \begin{equation} \label{eq:3.4.4} y'={A(x,y)\over B(x,y)} \end{equation}

    where \(A\) and \(B\) are continuous on any rectangle \(R\). Because of this, some differential equation software is based on numerically solving pairs of equations of the form \begin{equation} \label{eq:1.3.5} {dx\over dt}=B(x,y),\quad {dy\over dt}=A(x,y) \end{equation}

    where \(x\) and \(y\) are regarded as functions of a parameter \(t\). If \(x=x(t)\) and \(y=y(t)\) satisfy these equations, then \(y'={dy\over dx}={dy\over dt}\left/{dx\over dt}\right.={A(x,y)\over B(x,y)},\) so \(y=y(x)\) satisfies \eqref{eq:3.4.4}.

    Eqns.~\eqref{eq:3.4.2} and \eqref{eq:3.4.3} can be reformulated as in \eqref{eq:3.4.4} with \({dx\over dt}=-y,\quad {dy\over dt}=x\) and \({dx\over dt}=1-x^2-y^2,\quad {dy\over dt}=x^2\) respectively. Even if \(f\) is continuous and otherwise "nice'' throughout \(R\), your software may require you to reformulate the equation \(y'=f(x,y)\) as \({dx\over dt}=1,\quad {dy\over dt}=f(x,y)\) which is of the form \eqref{eq:3.4.5} with \(A(x,y)=f(x,y)\) and \(B(x,y)=1\).

    Figure~\ref{figure:3.4.5} shows a direction field and some integral curves for \eqref{eq:3.4.2}. As we saw earlier, the integral curves of \eqref{eq:3.4.2} are circles centered at the origin.

    \begin{figure}[H]

    \centering

    \includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig010305}

    \color{blue}

    \caption{A direction field and integral curves for

    $y'=-\dst{x\over y}$}

    \label{figure:1.3.5}

    \end{figure}

    Figure~\ref{figure:3.4.6} shows a direction field and some integral curves for \eqref{eq:3.4.3}. The integral curves near the top and bottom are solution curves. However, the integral curves near the middle are more complicated. For example, Figure~\ref{figure:3.4.7} shows the integral curve through the origin. The vertices of the dashed rectangle are on the circle \(x^2+y^2=1\) \((\approx.846, b\approx.533)\), where all integral curves of \eqref{eq:3.4.3} have infinite slope. There are three solution curves of \eqref{eq:3.4.3} on the integral curve in the figure: the segment above the level \(y=b\) is the graph of a solution on \((-\infty,a)\), the segment below the level \(y=-b\) is the graph of a solution on \((-a,\infty)\), and the segment between these two levels is the graph of a solution on \((-a,a)\).

    As you study from this book, you'll often be asked to use computer software and graphics. Exercises with this intent are marked as \Cex\, (computer or calculator required), \CGex\, (computer and/or graphics required), or \Lex\ (laboratory work requiring software and/or graphics). Often you may not completely understand how the software does what it does. This is similar to the situation most people are in when they drive automobiles or watch television, and it doesn't decrease the value of using modern technology as an aid to learning. Just be careful that you use the technology as a supplement to thought rather than a substitute for it.

    \begin{figure}[htbp]

    \color{blue}

    \begin{minipage}[b]{0.5\linewidth}

    \centering

    \scalebox{.7}{

    \includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig010306} }

    \caption{A direction field and integral curves for

    $y'=\dst{x^2\over1-x^2-y^2}$}

    \label{figure:1.3.6}

    \end{minipage}

    \hspace{0.6cm}

    \begin{minipage}[b]{0.5\linewidth}

    \centering

    \scalebox{.7}{

    \includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{fig010307} }

    \caption{}

    \label{figure:1.3.7}

    \end{minipage}

    \end{figure}

    Contributors

    Trench, William F., "Elementary Differential Equations" (2013). Faculty Authored and Edited Books & CDs. 8.
    https://digitalcommons.trinity.edu/mono/8