
# 3.4: Direction Fields for First Order Equations

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$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$

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~2.3 we'll take up

the question of existence of solutions of a first order equation

\label{eq:1.3.1}

y'=f(x,y).

In this section we'll simply assume that \eqref{eq:1.3.1} has

solutions and discuss a graphical method for approximating them.

In Chapter~3 we discuss numerical methods for obtaining

approximate solutions of \eqref{eq:1.3.1}.

Recall that a solution of \eqref{eq:1.3.1} 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:1.3.1} is equivalent to not knowing the equations of

integral curves of \eqref{eq:1.3.1}. However, it's easy to calculate the

slopes of these curves. To be specific, the slope of an integral curve

of \eqref{eq:1.3.1} through a given point $(x_0,y_0)$ is given by the

number $f(x_0,y_0)$. This is the basis of {\color{blue}\it the method of direction

fields\/}.

If $f$ is defined on a set $R$, we can construct a {\color{blue}\it direction

field \/} for \eqref{eq:1.3.1} 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 {\color{blue}\it 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 {\color{blue}\it rectangular grid\/} (Figure~\ref{figure:1.3.1}). 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:1.3.1} in $R$. If the grid points are sufficiently numerous and

close together, we can draw approximate integral curves of \eqref{eq:1.3.1}

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}

\label{figure:1.3.1}

\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:1.3.1} in Chapter~3. These methods can be

used

to plot solution curves of \eqref{eq:1.3.1} in a rectangular region $R$

{\color{blue}\it if $f$ is continuous on\/} $R$. Figures~\ref{figure:1.3.2},

\ref{figure:1.3.3}, and \ref{figure:1.3.4} show direction fields and solution

curves for the differential equations

$$y'=\frac{x^2-y^2}{1+x^2+y^2}, \quad y'=1+xy^2,\text{\quad and \quad} y'=\frac{x-y}{1+x^2},$$

which are all of the form \eqref{eq:1.3.1} 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

\label{eq:1.3.2}

y'=-x/y

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

\label{eq:1.3.3}

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

if $R$ contains any part of the unit circle $x^2+y^2=1$, because the

right side of \eqref{eq:1.3.3} is undefined if $x^2+y^2=1$. However,

\eqref{eq:1.3.2} and \eqref{eq:1.3.3} can written as

\label{eq:1.3.4}

y'={A(x,y)\over B(x,y)}

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

\label{eq:1.3.5}

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:1.3.4}.

Eqns.~\eqref{eq:1.3.2} and \eqref{eq:1.3.3} can be reformulated as in

\eqref{eq:1.3.4} 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:1.3.5} with $A(x,y)=f(x,y)$ and

$B(x,y)=1$.

Figure~\ref{figure:1.3.5} shows a direction field and some integral curves

for \eqref{eq:1.3.2}. As we saw in

Example~\ref{example:1.2.1} and will verify

again in Section~2.2, the integral curves of \eqref{eq:1.3.2}

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:1.3.6} shows a direction field and some integral curves

for \eqref{eq:1.3.3}. 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:1.3.7} shows the integral

curve through the origin. The vertices of the dashed rectangle are on

the circle $x^2+y^2=1$ ($a\approx.846$, $b\approx.533$), where all

integral curves of \eqref{eq:1.3.3} have infinite slope. There are

three solution curves of \eqref{eq:1.3.3} 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 $(-a,a)$.

\technology

\medskip

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}

\enlargethispage{1in}

\exercises

\noindent

\emph{In Exercises~{\color{red}\rm\bf 1--11} a direction

field is drawn for the given equation. Sketch some integral curves.}

\begin{figure}[H]

\color{blue}

\centering

\includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{exer010301}

\caption*{{\color{red}\bf 1}\; A direction field for

$y'=\dst{\frac{x}{y}}$}

\end{figure}

\newpage

\begin{figure}[H]

\color{blue}

\centering

\includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{exer010302}

\caption*{{\color{red}\bf 2}\; A direction field for

$\dst{y'=\dst{2xy^2\over1+x^2}}$\quad}

\end{figure}

\begin{figure}[H]

\color{blue}

\centering

\includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{exer010303}

\caption*{{\color{red}\bf 3}\; A direction field for

$y'=x^2(1+y^2)$}

\end{figure}

\begin{figure}[H]

\color{blue}

\centering

\includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{exer010304}

\caption*{{\color{red}\bf 4}\; A direction field for

$y'=\dst{1\over1+x^2+y^2}$}

\end{figure}

\begin{figure}[H]

\color{blue}

\centering

\includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{exer010305}

\caption*{{\color{red}\bf 5}\; A direction field for

$y'=-(2xy^2+y^3)$}

\end{figure}

\begin{figure}[H]

\color{blue}

\centering

\includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{exer010306}

\caption*{{\color{red}\bf 6}\; A direction field for

$y'=(x^2+y^2)^{1/2}$}

\end{figure}

\begin{figure}[H]

\color{blue}

\centering

\includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{exer010307}

\caption*{{\color{red}\bf 7}\; A direction field for

$y'=\sin xy$}

\end{figure}

\begin{figure}[H]

\color{blue}

\centering

\includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{exer010308}

\caption*{{\color{red}\bf 8}\; A direction field for

$y'=e^{xy}$}

\end{figure}

\begin{figure}[H]

\color{blue}

\centering

\includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{exer010309}

\caption*{{\color{red}\bf 9}\; A direction field for

$y'=(x-y^2)(x^2-y)$}

\end{figure}

\begin{figure}[H]

\color{blue}

\centering

\includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{exer010310}

\caption*{{\color{red}\bf 10}\; A direction field for

$y'=x^3y^2+xy^3$}

\end{figure}

\begin{figure}[H]

\color{blue}

\centering

\includegraphics[bb=-78 148 689 643,width=5.67in,height=3.66in,keepaspectratio]{exer010311}

\caption*{{\color{red}\bf 11}\; A direction field for

$y'=\sin(x-2y)$}

\end{figure}

\newpage

\emph{In Exercises~\ref{exer:1.3.12}-\ref{exer:1.3.22} construct a

direction

field and plot some integral curves in the indicated rectangular region.}

\begin{exerciselist}

\setcounter{exercise}{11}

\item\label{exer:1.3.12}\CGex

$y'=y(y-1); \quad \{-1\le x\le 2,\ -2\le y\le2\}$

\item\label{exer:1.3.13} \CGex

$y'=2-3xy; \quad \{-1\le x\le 4,\ -4\le y\le4\}$

\item\label{exer:1.3.14} \CGex

$y'=xy(y-1); \quad \{-2\le x\le2,\ -4\le y\le 4\}$

\item\label{exer:1.3.15}\CGex

$y'=3x+y; \quad \{-2\le x\le2,\ 0\le y\le 4\}$

\item\label{exer:1.3.16} \CGex

$y'=y-x^3; \quad \{-2\le x\le2,\ -2\le y\le 2\}$

\item\label{exer:1.3.17}\CGex

$y'=1-x^2-y^2; \quad \{-2\le x\le2,\ -2\le y\le 2\}$

\item\label{exer:1.3.18}\CGex

$y'=x(y^2-1); \quad \{-3\le x\le3,\ -3\le y\le 2\}$

\item\label{exer:1.3.19}\CGex

$y'=\dst{x\over y(y^2-1)}; \quad \{-2\le x\le2,\ -2\le y\le 2\}$

\item\label{exer:1.3.20}\CGex

$y'=\dst{xy^2\over y-1}; \quad \{-2\le x\le2,\ -1\le y\le 4\}$

\item\label{exer:1.3.21}\CGex

$y'=\dst{x(y^2-1)\over y}; \quad \{-1\le x\le1,\ -2\le y\le 2\}$

\item\label{exer:1.3.22}\CGex

$y'=-\dst{x^2+y^2\over1-x^2-y^2}; \quad \{-2\le x\le2,\ -2\le y\le 2\}$

\item\label{exer:1.3.23} \Lex

By suitably renaming the constants and dependent variables

in the equations

$$T' = -k(T-T_m) \eqno{\rm(A)}$$

and

$$G'=-\lambda G+r \eqno{\rm(B)}$$

discussed in Section~1.2 in connection with Newton's

law of cooling and absorption of glucose in the body, we can

write both as

$$y'=- ay+b, \eqno{\rm(C)}$$

where $a$ is a positive constant and $b$ is an arbitrary

constant. Thus, (A) is of the form (C) with $y=T$, $a=k$, and

$b=kT_m$, and (B) is of the form (C) with $y=G$,

$a=\lambda$, and $b=r$. We'll encounter equations of the form

(C) in many other applications in Chapter~2.

Choose a positive $a$ and an

arbitrary $b$.

Construct a direction field and plot some integral curves for (C) in

a rectangular

region of the form

$$\{0\le t\le T,\ c\le y\le d\}$$

of the $ty$-plane.

Vary $T$, $c$, and $d$ until you discover a common property

of all the solutions of (C). Repeat this experiment with various

choices of $a$ and $b$ until you can state

this property precisely in terms of $a$ and $b$.

\item\label{exer:1.3.24} \Lex

By suitably renaming the constants and dependent variables

in the equations

$$P'=aP(1-\alpha P) \eqno{\rm(A)}$$

and

$$I'=rI(S-I) \eqno{\rm(B)}$$

discussed in Section~1.1 in connection with Verhulst's

population

model and the spread of an epidemic, we can write both in the form

$$y'=ay-by^2, \eqno{\rm(C)}$$

where $a$ and $b$ are positive constants. Thus, (A) is of the form (C)

with $y=P$, $a=a$, and $b=a\alpha$, and (B) is of the form (C) with

$y=I$, $a=rS$, and $b=r$. In Chapter~2 we'll encounter

equations of the form

(C) in

other applications..

\begin{alist}

\item % (a)

Choose positive numbers $a$ and $b$. Construct a direction field and

plot some integral curves for (C) in a rectangular region of the form

$$\{0\le t\le T,\ 0\le y\le d\}$$

of the $ty$-plane. Vary $T$ and $d$ until you discover a common

property of all solutions of (C) with $y(0)>0$. Repeat this experiment

with various choices of $a$ and $b$ until you can state this property

precisely in terms of $a$ and $b$.

\item % (a)

Choose positive numbers $a$ and $b$. Construct a direction field and

plot some integral curves for (C) in a rectangular region of the form

$$\{0\le t\le T,\ c\le y\le 0\}$$

of the $ty$-plane. Vary $a$, $b$, $T$ and $c$ until you discover a

common property of all solutions of (C) with $y(0)<0$.

You can verify your results later by doing Exercise~2.2.~\hspace*{-3pt}\ref{exer:2.2.27}.

\end{alist}

\end{exerciselist}

### Contributors

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