Direction Fields and Numerical Methods

For the rest of this chapter we will focus on various methods for solving differential equations and analyzing the behavior of the solutions. In some cases it is possible to predict properties of a solution to a differential equation without knowing the actual solution. We will also study numerical methods for solving differential equations, which can be programmed by using various computer languages or even by using a spreadsheet program, such as Microsoft Excel.

Creating Direction Fields

Direction fields (also called slope fields) are useful for investigating first-order differential equations. In particular, we consider a first-order differential equation of the form

An applied example of this type of differential equation appears in Newton’s law of cooling, which we will solve explicitly later in this chapter. First, though, let us create a direction field for the differential equation

The idea behind a direction field is the fact that the derivative of a function evaluated at a given point is the slope of the tangent line to the graph of that function at the same point. Other examples of differential equations for which we can create a direction field include

To create a direction field, we start with the first equation:

y' = 3 x + 2 y - 4 .

be any ordered pair, and we substitute these numbers into the right-hand side of the differential equation. For example, if we choose

x = 1 and y = 2,

This tells us that if a solution to the differential equation

y' = 3 x + 2 y - 4

To start creating the direction field, we put a short line segment at the point

(1, 2)

We can do this for any point in the domain of the function

f (x, y) = 3 x + 2 y - 4,

Therefore any point in the Cartesian plane has a slope associated with it, assuming that a solution to the differential equation passes through that point. The direction field for the differential equation

y^{'} = 3 x + 2 y - 4

We can generate a direction field of this type for any differential equation of the form

y' = f (x, y) .

Using Direction Fields

We can use a direction field to predict the behavior of solutions to a differential equation without knowing the actual solution. For example, the direction field in [link] serves as a guide to the behavior of solutions to the differential equation

y' = 3 x + 2 y - 4 .

To use a direction field, we start by choosing any point in the field. The line segment at that point serves as a signpost telling us what direction to go from there. For example, if a solution to the differential equation passes through the point

(0, 1),

then the slope of the solution passing through that point is given by

y' = 3 (0) + 2 (1) - 4 = −2 .

Using the method of linear approximations gives a formula for the approximate value of

y

At this point the slope of the solution changes (again according to the differential equation). We can keep progressing, recalculating the slope of the solution as we take small steps to the right, and watching the behavior of the solution. [link] shows a graph of the solution passing through the point

(0, 1) .

Now consider the direction field for the differential equation

y' = (x - 3) (y^{2} - 4),

shown in [link]. This direction field has several interesting properties. First of all, at

y = −2

horizontal dashes appear all the way across the graph. This means that if

y = −2,

Substituting this expression into the right-hand side of the differential equation gives

is a solution to the differential equation. These are the only constant-valued solutions to the differential equation, as we can see from the following argument. Suppose

y = k

Substituting this expression into the differential equation yields

0 = (x - 3) (k^{2} - 4) .

so the second factor must equal zero. This result yields the equation

k^{2} - 4 = 0 .

which are the constant solutions already mentioned. These are called the equilibrium solutions to the differential equation.

To determine the equilibrium solutions to the differential equation

y' = f (x, y),

set the right-hand side equal to zero. An equilibrium solution of the differential equation is any function of the form

y = k

An important characteristic of equilibrium solutions concerns whether or not they approach the line

y = k

The direction field for this initial-value problem, along with the corresponding solution, is shown in [link].

The values of the solution to this initial-value problem stay between

y = −2

which are the equilibrium solutions to the differential equation. Furthermore, as

x

approaches infinity. Furthermore, if the initial value is slightly higher than

−2,

which is the other equilibrium solution. Therefore in neither case does the solution approach

y = −2,

Euler’s Method

Integrating both sides of the differential equation gives

y = x^{2} - 3 x + C,

The red graph consists of line segments that approximate the solution to the initial-value problem. The graph starts at the same initial value of

(0, 3) .

Then the slope of the solution at any point is determined by the right-hand side of the differential equation, and the length of the line segment is determined by increasing the

x

Before we state Euler’s Method as a theorem, let’s consider another initial-value problem:

The idea behind direction fields can also be applied to this problem to study the behavior of its solution. For example, at the point

(−1, 2),

so the slope of the tangent line to the solution at that point is also equal to

−3 .

Therefore the approximate value of the solution to the differential equation is

y = 1.492

What we have just shown is the idea behind Euler’s Method. Repeating these steps gives a list of values for the solution. These values are shown in [link], rounded off to four decimal places.

Using Euler’s Method to Approximate Solutions to a Differential Equation
$n$	$0$	$1$	$2$	$3$	$4$	$5$
$x_{n}$	$−1$	$−0.9$	$−0.8$	$−0.7$	$−0.6$	$−0.5$
$y_{n}$	$2$	$1.7$	$1.492$	$1.3334$	$1.2046$	$1.0955$
$n$	$6$	$7$	$8$	$9$	$10$
$x_{n}$	$−0.4$	$−0.3$	$−0.2$	$−0.1$	$0$
$y_{n}$	$1.0004$	$1.9164$	$1.8414$	$1.7746$	$1.7156$

the fewer calculations are needed. However, the tradeoff results in a lower degree of accuracy for larger step size, as illustrated in [link].

Using Euler’s Method to Approximate Solutions to a Differential Equation
$n$	$x_{n}$	$y_{n} = y_{n - 1} + h f (x_{n - 1}, y_{n - 1})$
$0$	$0$	$2$
$1$	$0.1$	$y_{1} = y_{0} + h f (x_{0}, y_{0}) = 1.7$
$2$	$0.2$	$y_{2} = y_{1} + h f (x_{1}, y_{1}) = 1.514$
$3$	$0.3$	$y_{3} = y_{2} + h f (x_{2}, y_{2}) = 1.3968$
$4$	$0.4$	$y_{4} = y_{3} + h f (x_{3}, y_{3}) = 1.3287$
$5$	$0.5$	$y_{5} = y_{4} + h f (x_{4}, y_{4}) = 1.3001$
$6$	$0.6$	$y_{6} = y_{5} + h f (x_{5}, y_{5}) = 1.3061$
$7$	$0.7$	$y_{7} = y_{6} + h f (x_{6}, y_{6}) = 1.3435$
$8$	$0.8$	$y_{8} = y_{7} + h f (x_{7}, y_{7}) = 1.4100$
$9$	$0.9$	$y_{9} = y_{8} + h f (x_{8}, y_{8}) = 1.5032$
$10$	$1.0$	$y_{10} = y_{9} + h f (x_{9}, y_{9}) = 1.6202$

Consider the initial-value problem

y^{'} = x^{3} + y^{2}, y (1) = −2 .

Using a step size of $0.1,$

generate a table with approximate values for the solution to the initial-value problem for values of $x$

between $1$

and $2 .$

n

x_{n}

y_{n} = y_{n - 1} + h f (x_{n - 1}, y_{n - 1})


{: valign=”top”}	———-
$0$

1

−2


{: valign=”top”}	$1$

1.1

y_{1} = y_{0} + h f (x_{0}, y_{0}) = −1.5


{: valign=”top”}	$2$

1.2

y_{2} = y_{1} + h f (x_{1}, y_{1}) = −1.1419


{: valign=”top”}	$3$

1.3

y_{3} = y_{2} + h f (x_{2}, y_{2}) = −0.8387


{: valign=”top”}	$4$

1.4

y_{4} = y_{3} + h f (x_{3}, y_{3}) = −0.5487


{: valign=”top”}	$5$

1.5

y_{5} = y_{4} + h f (x_{4}, y_{4}) = −0.2442


{: valign=”top”}	$6$

1.6

y_{6} = y_{5} + h f (x_{5}, y_{5}) = 0.0993


{: valign=”top”}	$7$

1.7

y_{7} = y_{6} + h f (x_{6}, y_{6}) = 0.5099


{: valign=”top”}	$8$

1.8

y_{8} = y_{7} + h f (x_{7}, y_{7}) = 1.0272


{: valign=”top”}	$9$

1.9

y_{9} = y_{8} + h f (x_{8}, y_{8}) = 1.7159


{: valign=”top”}	$10$

2

y_{10} = y_{9} + h f (x_{9}, y_{9}) = 2.6962

| {: valign=”top”}{: .unnumbered summary=”A table with three columns and eleven rows. The first column has the header n and the values 0 through 10. The second column has the header x_n and the values 1 through 2, increasing by 0.1. The third column has the header y_n = y_(n - 1) + hf(x_(n - 1), y_(n - 1)) and the values -2, -1.5, -1.1419, -0.8387, -0.5487, -0.2442, 0.0993, 0.5099, 1.0272, 1.7159, and 2.6962.” data-label=””}

Hint

Start by identifying the value of $h,$

then figure out what $f (x, y)$

is. Then use the formula for Euler’s Method to calculate $y_{1}, y_{2},$

and so on.

Key Concepts

Key Equations

For the following problems, use the direction field below from the differential equation $y' = −2 y .$

Sketch the graph of the solution for the given initial conditions.* * *

y (0) = 1

y (0) = 0

A graph of the given direction field with a flat line drawn on the axis. The arrows point up for y < 0 and down for y > 0. The closer they are to the x axis, the more horizontal the arrows are, and the further away they are, the more vertical they become.

y (0) = −1

Are there any equilibria? What are their stabilities?

y = 0

is a stable equilibrium

For the following problems, use the direction field below from the differential equation $y' = y^{2} - 2 y .$

Sketch the graph of the solution for the given initial conditions.* * *

A direction field with horizontal arrows at y = 0 and y = 2. The arrows point up for y > 2 and for y < 0. The arrows point down for 0 < y < 2. The closer the arrows are to these lines, the more horizontal they are, and the further away, the more vertical the arrows are.

y (0) = 3

y (0) = 1

y (0) = −1

Are there any equilibria? What are their stabilities?

y = 0

is a stable equilibrium and $y = 2$

is unstable

Draw the direction field for the following differential equations, then solve the differential equation. Draw your solution on top of the direction field. Does your solution follow along the arrows on your direction field?

y' = t^{3}

y' = e^{t}

A direction field over the four quadrants. As t goes from 0 to infinity, the arrows become more and more vertical after being horizontal closer to x = 0.

\frac{d y}{d x} = x^{2} cos x

\frac{d y}{d t} = t e^{t}

A direction field over [-2, 2] in the x and y axes. The arrows point slightly down and to the right over [-2, 0] and gradually become vertical over [0, 2].

\frac{d x}{d t} = cosh (t)

Draw the directional field for the following differential equations. What can you say about the behavior of the solution? Are there equilibria? What stability do these equilibria have?

y' = y^{2} - 1

A direction field with horizontal arrows pointing to the right at y = 1 and y = -1. The arrows point up for y < -1 and y > 1. The arrows point down for -1 < y < 1. The closer the arrows are to these lines, the more horizontal they are, and the further away they are, the more vertical they are.

y' = y - x

y' = 1 - y^{2} - x^{2}

A direction field with arrows pointing down and to the right for nearly all points in [-2, 2] on the x and y axes. Close to the origin, the arrows become more horizontal, point to the upper right, become more horizontal, and then point down to the right again.

y' = t^{2} sin y

y' = 3 y + x y

Match the direction field with the given differential equations. Explain your selections.* * *

A direction field with arrows pointing down and to the right in quadrants two and three. After crossing the y axis, the arrows change direction and point up to the right.

A direction field with horizontal arrows pointing to the left in quadrants two and three. In crossing the y axis, the arrows switch and point upward in quadrants one and four.

A direction field with horizontal arrows pointing to the right on the x axis. Above, the arrows point down and to the right, and below, the arrows point up and to the right. The further from the x axis, the more vertical the arrows become.

A direction field with horizontal arrows on the x and y axes. The arrows point down and to the right in quadrants one and three. They point up and to the right in quadrants two and four.

A direction field with arrows pointing up in quadrants two and three, to the right on the y axis, and down in quadrants one and four.

y' = −3 y

y' = −3 t

y' = e^{t}

y' = \frac{1}{2} y + t

y' = − t y

Match the direction field with the given differential equations. Explain your selections.* * *

A direction field with horizontal arrows pointing to the right on the x and y axes. In quadrants one and three, the arrows point up, and in quadrants two and four, they point down.

A direction field with horizontal arrows pointing to the right on the x and y axes. In quadrants one and three, the arrows point up and to the right, and in quadrants two and four, the arrows point down and to the right.

A direction field with horizontal arrows pointing to the right on the x and y axes. In quadrants two and three, the arrows point down, and in quadrants one and four, the arrows point up.

A direction field with horizontal arrows pointing to the right on the x axis. The arrows point up and to the right in all quadrants. The closer the arrows are to the x axis, the more horizontal the arrows are, and the further away they are, the more vertical they are.

A direction field with horizontal arrows on the y axis. The arrows are also more horizontal closer to y = 1.5, y = -1.5, and the y axis. For y > 1.5 and x < 0, for y < -1.5 and x < 0, and for -1.5 < y < 1.5 and x > 0-, the arrows point down. For y> 1.5 and x > 0, for y < -1.5, for y < -1.5 and x > 0, and for -1.5 < y < 1.5 and x < 0, the arrows point up.

y' = t sin y

y' = − t cos y

y' = t tan y

y' = {sin}^{2} y

y' = y^{2} t^{3}

Estimate the following solutions using Euler’s method with $n = 5$

steps over the interval $t = [0, 1] .$

If you are able to solve the initial-value problem exactly, compare your solution with the exact solution. If you are unable to solve the initial-value problem, the exact solution will be provided for you to compare with Euler’s method. How accurate is Euler’s method?

y' = −3 y, y (0) = 1

y' = t^{2}

2.24,

exact: $3$

y^{'} = 3 t - y, y (0) = 1 .

Exact solution is $y = 3 t + 4 e^{− t} - 3$

y^{'} = y + t^{2}, y (0) = 3 .

Exact solution is $y = 5 e^{t} - 2 - t^{2} - 2 t$

7.739364,

exact: $5 (e - 1)$

y^{'} = 2 t, y (0) = 0

[T] $y' = e^{(x + y)}, y (0) = −1 .$

Exact solution is $y = − ln (e + 1 - e^{x})$

−0.2535

exact: $0$

y^{'} = y^{2} ln (x + 1), y (0) = 1 .

Exact solution is $y = - \frac{1}{(x + 1) (ln (x + 1) - 1)}$

y^{'} = 2^{x}, y (0) = 0,

Exact solution is $y = \frac{2^{x} - 1}{ln (2)}$

1.345,

exact: $\frac{1}{ln (2)}$

y^{'} = y, y (0) = −1 .

Exact solution is $y = − e^{x} .$

y^{'} = −5 t, y (0) = −2 .

Exact solution is $y = - \frac{5}{2} t^{2} - 2$

−4,

exact: $− 1 / 2$

Differential equations can be used to model disease epidemics. In the next set of problems, we examine the change of size of two sub-populations of people living in a city: individuals who are infected and individuals who are susceptible to infection. $S$

represents the size of the susceptible population, and $I$

represents the size of the infected population. We assume that if a susceptible person interacts with an infected person, there is a probability $c$

that the susceptible person will become infected. Each infected person recovers from the infection at a rate $r$

and becomes susceptible again. We consider the case of influenza, where we assume that no one dies from the disease, so we assume that the total population size of the two sub-populations is a constant number, $N .$

The differential equations that model these population sizes are

\begin{matrix} S' = r I - c S I and \\ I' = c S I - r I . \end{matrix}

Here $c$

represents the contact rate and $r$

is the recovery rate.

Show that, by our assumption that the total population size is constant $(S + I = N),$

you can reduce the system to a single differential equation in $I : I' = c (N - I) I - r I .$

Assuming the parameters are $c = 0.5, N = 5,$

and $r = 0.5,$

draw the resulting directional field.

A direction field with horizontal arrows pointing to the right on the x axis and at y = 4. The arrows below the x axis and above y = 4 point down and to the right. The arrows between the x axis and y = 4 point up and to the right.

[T] Use computational software or a calculator to compute the solution to the initial-value problem $y' = t y, y (0) = 2$

using Euler’s Method with the given step size $h .$

Find the solution at $t = 1 .$

For a hint, here is “pseudo-code” for how to write a computer program to perform Euler’s Method for $y' = f (t, y), y (0) = 2 :$

Create function $f (t, y)$

Define parameters $y (1) = y_{0}, t (0) = 0,$

step size $h,$

and total number of steps, $N$

Write a for loop:

for $k = 1 to N$

fn = f (t (k), y (k))

y (k+1) = y (k) + h*fn

t (k+1) = t (k) + h

Solve the initial-value problem for the exact solution.

y' = 2 e^{t^{2} / 2}

Draw the directional field

h = 1

2

[T] $h = 10$

[T] $h = 100$

3.2756

[T] $h = 1000$

[T] Evaluate the exact solution at $t = 1 .$

Make a table of errors for the relative error between the Euler’s method solution and the exact solution. How much does the error change? Can you explain?

2 \sqrt{e}

Step Size	Error
{: valign=”top”}	———-
$h = 1$

0.3935


{: valign=”top”}	$h = 10$

0.06163


{: valign=”top”}	$h = 100$

0.006612


{: valign=”top”}	$h = 1000$

0.0006661

| {: valign=”top”}{: .unnumbered summary=”A table with two columns and five rows. The first column contains the label “Step Size” and the values h = 1, h=10, h=100, and h=1000. The second column contains the label “Error” and the values 0.3935, 0.06163, 0.006612, and 0.0006661.” data-label=””}

Consider the initial-value problem $y' = −2 y, y (0) = 2 .$

Show that $y = 2 e^{−2 x}$

solves this initial-value problem.

Draw the directional field of this differential equation.

A direction field with horizontal arrows pointing to the right on the x axis. Above the x axis, the arrows point down and to the right. Below the x axis, the arrows point up and to the right. The closer the arrows are to the x axis, the more horizontal the arrows are, and the further away they are from the x axis, the more vertical the arrows are.

[T] By hand or by calculator or computer, approximate the solution using Euler’s Method at $t = 10$

using $h = 5 .$

[T] By calculator or computer, approximate the solution using Euler’s Method at $t = 10$

using $h = 100 .$

4.0741 e^{−10}

[T] Plot exact answer and each Euler approximation (for $h = 5$

and $h = 100)$

at each $h$

on the directional field. What do you notice?