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In this section, we will learn about: How direction fields and Euler’s method help us solve certain differential equations.

1.
9

DIFFERENTIAL EQUATIONS

DIFFERENTIAL EQUATIONS

2.
DIFFERENTIAL EQUATIONS

Unfortunately, it’s impossible to solve

most differential equations in the sense

of obtaining an explicit formula for the

Unfortunately, it’s impossible to solve

most differential equations in the sense

of obtaining an explicit formula for the

3.
DIFFERENTIAL EQUATIONS

Despite the absence of an explicit solution,

we can still learn a lot about the solution

through either:

A graphical approach (direction fields)

A numerical approach (Euler’s method)

Despite the absence of an explicit solution,

we can still learn a lot about the solution

through either:

A graphical approach (direction fields)

A numerical approach (Euler’s method)

4.
DIFFERENTIAL EQUATIONS

9.2

Direction Fields

and Euler’s Method

In this section, we will learn about:

How direction fields and Euler’s method help us

solve certain differential equations.

9.2

Direction Fields

and Euler’s Method

In this section, we will learn about:

How direction fields and Euler’s method help us

solve certain differential equations.

5.
Suppose we are asked to sketch the graph

of the solution of the initial-value problem

y’ = x + y y(0) = 1

We don’t know a formula for the solution.

So, how can we possibly sketch its graph?

of the solution of the initial-value problem

y’ = x + y y(0) = 1

We don’t know a formula for the solution.

So, how can we possibly sketch its graph?

6.
Let’s think about what

the differential equation

the differential equation

7.
The equation y’ = x + y tells us that the slope

at any point (x, y) on the graph (called

the solution curve) is equal to the sum of

the x- and y-coordinates of the point.

at any point (x, y) on the graph (called

the solution curve) is equal to the sum of

the x- and y-coordinates of the point.

8.
In particular, as the curve passes

through the point (0, 1), its slope there

must be 0 + 1 = 1.

through the point (0, 1), its slope there

must be 0 + 1 = 1.

9.
So, a small portion of the solution curve

near the point (0, 1) looks like a short line

segment through (0, 1) with slope 1.

near the point (0, 1) looks like a short line

segment through (0, 1) with slope 1.

10.
As a guide to sketching the rest of the

curve, let’s draw short line segments at a

number of points (x, y) with slope x + y.

curve, let’s draw short line segments at a

number of points (x, y) with slope x + y.

11.
DIRECTION FIELD

The result is called a direction field.

For instance,

the line segment

at the point (1, 2)

has slope 1 + 2 = 3.

The result is called a direction field.

For instance,

the line segment

at the point (1, 2)

has slope 1 + 2 = 3.

12.
DIRECTION FIELDS

The direction field allows us to visualize

the general shape of the solution curves by

indicating the direction in which the curves

proceed at each point.

The direction field allows us to visualize

the general shape of the solution curves by

indicating the direction in which the curves

proceed at each point.

13.
DIRECTION FIELDS

Now, we can sketch the solution curve

through the point (0, 1) by following

the direction field as in this figure.

Notice that we have

drawn the curve so that

it is parallel to nearby

line segments.

Now, we can sketch the solution curve

through the point (0, 1) by following

the direction field as in this figure.

Notice that we have

drawn the curve so that

it is parallel to nearby

line segments.

14.
DIRECTION FIELDS

In general, suppose we have a first-order

differential equation of the form

y’ = F(x, y)

where F(x, y) is some expression in x and y.

The differential equation says that the slope

of a solution curve at a point (x, y) on the curve

is F(x, y).

In general, suppose we have a first-order

differential equation of the form

y’ = F(x, y)

where F(x, y) is some expression in x and y.

The differential equation says that the slope

of a solution curve at a point (x, y) on the curve

is F(x, y).

15.
SLOPE FIELD

If we draw short line segments with slope

F(x, y) at several points (x, y), the result

is called a direction field (or slope field).

These line segments indicate the direction

in which a solution curve is heading.

So, the direction field helps us visualize

the general shape of these curves.

If we draw short line segments with slope

F(x, y) at several points (x, y), the result

is called a direction field (or slope field).

These line segments indicate the direction

in which a solution curve is heading.

So, the direction field helps us visualize

the general shape of these curves.

16.
DIRECTION FIELDS Example 1

a. Sketch the direction field for the differential

equation y’ = x2 + y2 – 1.

b. Use part (a) to sketch the solution curve

that passes through the origin.

a. Sketch the direction field for the differential

equation y’ = x2 + y2 – 1.

b. Use part (a) to sketch the solution curve

that passes through the origin.

17.
DIRECTION FIELDS Example 1 a

We start by computing

the slope at several points

in the following chart.

We start by computing

the slope at several points

in the following chart.

18.
DIRECTION FIELDS Example 1 a

Then, we draw short

line segments with these

slopes at these points.

Then, we draw short

line segments with these

slopes at these points.

19.
DIRECTION FIELDS Example 1 a

The result is

this direction field.

The result is

this direction field.

20.
DIRECTION FIELDS Example 1 b

We start at the origin and move to the right

in the direction of the line segment (which has

slope -1).

We continue to draw

the solution curve so

that it moves parallel to

the nearby line segments.

We start at the origin and move to the right

in the direction of the line segment (which has

slope -1).

We continue to draw

the solution curve so

that it moves parallel to

the nearby line segments.

21.
DIRECTION FIELDS Example 1 b

This is the resulting solution

Returning to the origin,

we draw the curve

to the left as well.

This is the resulting solution

Returning to the origin,

we draw the curve

to the left as well.

22.
DIRECTION FIELDS

The more line segments we draw in

a direction field, the clearer the picture

Of course, it’s tedious to compute slopes and draw

line segments for a huge number of points by hand.

However, computers are well suited for this task.

The more line segments we draw in

a direction field, the clearer the picture

Of course, it’s tedious to compute slopes and draw

line segments for a huge number of points by hand.

However, computers are well suited for this task.

23.
DIRECTION FIELDS BY COMPUTERS

This is a more detailed, computer-drawn

direction field for the differential equation

in Example 1.

This is a more detailed, computer-drawn

direction field for the differential equation

in Example 1.

24.
DIRECTION FIELDS BY COMPUTERS

It enables us to draw, with reasonable

accuracy, the solution curves shown here

with y-intercepts

-2, -1, 0, 1, and 2.

It enables us to draw, with reasonable

accuracy, the solution curves shown here

with y-intercepts

-2, -1, 0, 1, and 2.

25.
DIRECTION FIELDS

Now, let’s see how

direction fields give insight

into physical situations.

Now, let’s see how

direction fields give insight

into physical situations.

26.
This simple electric circuit contains:

An electromotive force (usually a battery or generator)

that produces a voltage of E(t) volts (V) and a current of

I(t) amperes (A) at time t

An electromotive force (usually a battery or generator)

that produces a voltage of E(t) volts (V) and a current of

I(t) amperes (A) at time t

27.
The circuit also contains:

A resistor with a resistance of R ohms (Ω)

An inductor with an inductance of L henries (H)

A resistor with a resistance of R ohms (Ω)

An inductor with an inductance of L henries (H)

28.
Ohm’s Law gives the drop in voltage

due to the resistor as RI.

The voltage drop due to the inductor

is L(dI/dt).

due to the resistor as RI.

The voltage drop due to the inductor

is L(dI/dt).

29.
One of Kirchhoff’s laws says that

the sum of the voltage drops is equal

to the supplied voltage E(t).

the sum of the voltage drops is equal

to the supplied voltage E(t).

30.
ELECTRICALS Equation 1

Thus, we have

dI

L + RI =E (t )

dt

which is a first-order differential equation

that models the current I at time t.

Thus, we have

dI

L + RI =E (t )

dt

which is a first-order differential equation

that models the current I at time t.

31.
ELECTRICALS Example 2

In this simple circuit, suppose that:

The resistance is 12 Ω.

The inductance is 4 H.

A battery gives a constant voltage of 60 V.

In this simple circuit, suppose that:

The resistance is 12 Ω.

The inductance is 4 H.

A battery gives a constant voltage of 60 V.

32.
ELECTRICALS Example 2

a. Identify any equilibrium solutions.

b. If the switch is closed when t = 0 so

the current starts with I(0) = 0, use

the direction field to sketch the solution

curve.

a. Identify any equilibrium solutions.

b. If the switch is closed when t = 0 so

the current starts with I(0) = 0, use

the direction field to sketch the solution

curve.

33.
ELECTRICALS Example 2

c. Draw a direction field for Equation 1

with those values.

d. What can you say about the limiting value

of the current?

c. Draw a direction field for Equation 1

with those values.

d. What can you say about the limiting value

of the current?

34.
ELECTRICALS Example 2 a

If we put L = 4, R = 12, and E(t) = 60

in Equation 1, we get:

dI dI

4 + 12 I =60 or =15 −3I

dt dt

If we put L = 4, R = 12, and E(t) = 60

in Equation 1, we get:

dI dI

4 + 12 I =60 or =15 −3I

dt dt

35.
ELECTRICALS Example 2 a

The direction field for this differential

equation is shown here.

The direction field for this differential

equation is shown here.

36.
ELECTRICALS Example 2 b

It appears from the direction field that

all solutions approach the value 5 A.

That is,

lim I (t ) =5

t→ ∞

It appears from the direction field that

all solutions approach the value 5 A.

That is,

lim I (t ) =5

t→ ∞

37.
ELECTRICALS Example 2 c

It appears that the constant function

I(t) = 5 is an equilibrium solution.

Indeed, we can verify this directly from

the differential equation dI/dt = 15 – 3I.

It appears that the constant function

I(t) = 5 is an equilibrium solution.

Indeed, we can verify this directly from

the differential equation dI/dt = 15 – 3I.

38.
ELECTRICALS Example 2 c

If I(t) = 5, then

The left side is: dI/dt = 0

The right side is: 15 – 3(5) = 0

If I(t) = 5, then

The left side is: dI/dt = 0

The right side is: 15 – 3(5) = 0

39.
ELECTRICALS Example 2 d

We use the direction field to sketch

the solution curve that passes through (0, 0),

as shown in red.

We use the direction field to sketch

the solution curve that passes through (0, 0),

as shown in red.

40.
Notice that the line segments along any

horizontal line are parallel.

This is because

the independent

variable t does

not occur on

the right side

of the equation

I’ = 15 – 3I.

horizontal line are parallel.

This is because

the independent

variable t does

not occur on

the right side

of the equation

I’ = 15 – 3I.

41.
AUTONOMOUS DIFFERENTIAL EQUATION

In general, a differential equation

of the form

y’ = f (y)

in which the independent variable

is missing from the right side, is called

In general, a differential equation

of the form

y’ = f (y)

in which the independent variable

is missing from the right side, is called

42.
AUTONOMOUS DIFFERENTIAL EQUATIONS

For such an equation, the slopes

corresponding to two different points with

the same y-coordinate must be equal.

This means that, if we know one solution, we can

obtain infinitely many others—just by shifting the graph

of the known solution to the right or left.

For such an equation, the slopes

corresponding to two different points with

the same y-coordinate must be equal.

This means that, if we know one solution, we can

obtain infinitely many others—just by shifting the graph

of the known solution to the right or left.

43.
AUTONOMOUS DIFFERENTIAL EQUATIONS

These solutions result from shifting

the solution curve of Example 2 one and two

time units (namely, seconds) to the right.

They

correspond

to closing

the switch

when

t = 1 or t = 2.

These solutions result from shifting

the solution curve of Example 2 one and two

time units (namely, seconds) to the right.

They

correspond

to closing

the switch

when

t = 1 or t = 2.

44.
DIRECTION FIELDS

The basic idea behind direction

fields can be used to find numerical

approximations to solutions of differential

The basic idea behind direction

fields can be used to find numerical

approximations to solutions of differential

45.
DIRECTION FIELDS

We illustrate the method on the initial-value

problem that we used to introduce direction

y’ = x + y y(0) = 1

We illustrate the method on the initial-value

problem that we used to introduce direction

y’ = x + y y(0) = 1

46.
DIRECTION FIELDS

The differential equation tells us that:

y’(0) = 0 + 1 = 1

Hence, the solution curve has slope 1

at the point (0, 1).

The differential equation tells us that:

y’(0) = 0 + 1 = 1

Hence, the solution curve has slope 1

at the point (0, 1).

47.
DIRECTION FIELDS

As a first approximation to the solution,

we could use the linear approximation

L(x) = x + 1

As a first approximation to the solution,

we could use the linear approximation

L(x) = x + 1

48.
DIRECTION FIELDS

In other words, we could use the tangent line

at (0, 1) as a rough approximation to the

solution curve.

In other words, we could use the tangent line

at (0, 1) as a rough approximation to the

solution curve.

49.
EULER’S METHOD

Euler’s idea was to improve on this

approximation, by:

First, proceeding only a short distance along

this tangent line.

Then, making a midcourse correction by changing

direction as indicated by the direction field.

Euler’s idea was to improve on this

approximation, by:

First, proceeding only a short distance along

this tangent line.

Then, making a midcourse correction by changing

direction as indicated by the direction field.

50.
STEP SIZE

The figure shows what happens if we start out

along the tangent line but stop when x = 0.5

This horizontal

distance traveled

is called the

step size.

The figure shows what happens if we start out

along the tangent line but stop when x = 0.5

This horizontal

distance traveled

is called the

step size.

51.
EULER’S METHOD

Since L(0.5) = 1.5, we have y(0.5) ≈ 1.5

and we take (0.5, 1.5) as the starting point

for a new line segment.

Since L(0.5) = 1.5, we have y(0.5) ≈ 1.5

and we take (0.5, 1.5) as the starting point

for a new line segment.

52.
EULER’S METHOD

The differential equation tells us that:

y’(0.5) = 0.5 + 1.5 = 2

The differential equation tells us that:

y’(0.5) = 0.5 + 1.5 = 2

53.
EULER’S METHOD

So, we use the linear function

y = 1.5 + 2(x – 0.5) = 2x + 0.5

as an approximation to the solution for x > 0.5

This is the

orange segment.

So, we use the linear function

y = 1.5 + 2(x – 0.5) = 2x + 0.5

as an approximation to the solution for x > 0.5

This is the

orange segment.

54.
EULER’S METHOD

If we decrease the step size from 0.5 to

0.25, we get the better Euler approximation

shown here.

If we decrease the step size from 0.5 to

0.25, we get the better Euler approximation

shown here.

55.
EULER’S METHOD

In general, Euler’s method says:

Start at the point given by the initial value and proceed

in the direction indicated by the direction field.

Stop after a short time, look at the slope at

the new location, and proceed in that direction.

Keep stopping and changing direction according

to the direction field.

In general, Euler’s method says:

Start at the point given by the initial value and proceed

in the direction indicated by the direction field.

Stop after a short time, look at the slope at

the new location, and proceed in that direction.

Keep stopping and changing direction according

to the direction field.

56.
EULER’S METHOD

Euler’s method does not produce the exact

solution to an initial-value problem.

It gives approximations.

Euler’s method does not produce the exact

solution to an initial-value problem.

It gives approximations.

57.
EULER’S METHOD

However, by decreasing the step size

(thus increasing the number of midcourse

corrections), we obtain successively better

approximations to the exact solution.

However, by decreasing the step size

(thus increasing the number of midcourse

corrections), we obtain successively better

approximations to the exact solution.

58.
EULER’S METHOD

We understand this

when we compare

the three figures.

We understand this

when we compare

the three figures.

59.
EULER’S METHOD

For the general first-order initial-value problem

y’ = F(x, y), y(x0) = y0

our aim is to find approximate values for

the solution at equally spaced numbers

x0, x1 = x0 + h, x2 = x1 + h, ...,

where h is the step size.

For the general first-order initial-value problem

y’ = F(x, y), y(x0) = y0

our aim is to find approximate values for

the solution at equally spaced numbers

x0, x1 = x0 + h, x2 = x1 + h, ...,

where h is the step size.

60.
EULER’S METHOD

The differential equation tells us that

the slope at (x0, y0) is:

y’ = F(x0, y0)

The differential equation tells us that

the slope at (x0, y0) is:

y’ = F(x0, y0)

61.
EULER’S METHOD

So, the figure shows that the approximate

value of the solution when x = x1 is:

y1 = y0 + hF(x0, y0)

Similarly,

y2 = y1 + hF(x1, y1)

So, the figure shows that the approximate

value of the solution when x = x1 is:

y1 = y0 + hF(x0, y0)

Similarly,

y2 = y1 + hF(x1, y1)

62.
EULER’S METHOD

In general,

yn = yn-1 + hF(xn-1, yn-1)

In general,

yn = yn-1 + hF(xn-1, yn-1)

63.
EULER’S METHOD Example 3

Use Euler’s method with step size 0.1

to construct a table of approximate values

for the solution of the initial-value problem

y’ = x + y y(0) = 1

Use Euler’s method with step size 0.1

to construct a table of approximate values

for the solution of the initial-value problem

y’ = x + y y(0) = 1

64.
EULER’S METHOD Example 3

We are given that:

h = 0.1

x0 = 0

y0 = 1

F(x, y) = x + y

We are given that:

h = 0.1

x0 = 0

y0 = 1

F(x, y) = x + y

65.
EULER’S METHOD Example 3

Thus, we have:

y1 = y0 + h F(x0, y0) = 1 + 0.1(0 + 1) = 1.1

y2 = y1 + h F(x1, y1) = 1.1 + 0.1(0.1 + 1.1) = 1.22

y3 = y2 + h F(x2, y2) = 1.22 + 0.1(0.2 + 1.22) = 1.362

Thus, we have:

y1 = y0 + h F(x0, y0) = 1 + 0.1(0 + 1) = 1.1

y2 = y1 + h F(x1, y1) = 1.1 + 0.1(0.1 + 1.1) = 1.22

y3 = y2 + h F(x2, y2) = 1.22 + 0.1(0.2 + 1.22) = 1.362

66.
EULER’S METHOD Example 3

This means that,

if y(x) is the exact solution,

then y(0.3) ≈ 1.362

This means that,

if y(x) is the exact solution,

then y(0.3) ≈ 1.362

67.
EULER’S METHOD Example 3

Proceeding with similar calculations,

we get the values in the table.

Proceeding with similar calculations,

we get the values in the table.

68.
EULER’S METHOD

For a more accurate table of values

in Example 3, we could decrease

the step size.

However, for a large number of small steps,

the amount of computation is considerable.

So, we need to program a calculator or

computer to carry out these calculations.

For a more accurate table of values

in Example 3, we could decrease

the step size.

However, for a large number of small steps,

the amount of computation is considerable.

So, we need to program a calculator or

computer to carry out these calculations.

69.
EULER’S METHOD

This table shows the results of applying

Euler’s method with decreasing step size to

the initial-value problem of Example 3.

This table shows the results of applying

Euler’s method with decreasing step size to

the initial-value problem of Example 3.

70.
EULER’S METHOD

Notice that the Euler estimates seem to be

approaching limits—namely, the true values

of y(0.5) and y(1).

Notice that the Euler estimates seem to be

approaching limits—namely, the true values

of y(0.5) and y(1).

71.
EULER’S METHOD

These are graphs of the Euler approximations

with step sizes

0.5, 0.25, 0.1, 0.05, 0.02, 0.01, 0.005

They are

approaching

the exact

solution curve

as the step size

h approaches 0.

These are graphs of the Euler approximations

with step sizes

0.5, 0.25, 0.1, 0.05, 0.02, 0.01, 0.005

They are

approaching

the exact

solution curve

as the step size

h approaches 0.

72.
EULER’S METHOD Example 4

In Example 2, we discussed a simple electric

circuit with resistance 12 Ω, inductance 4 H,

and a battery with voltage 60 V.

In Example 2, we discussed a simple electric

circuit with resistance 12 Ω, inductance 4 H,

and a battery with voltage 60 V.

73.
EULER’S METHOD Example 4

If the switch is closed when t = 0, we modeled

the current I at time t by the initial-value

dI/dt = 15 – 3I I(0) = 0

Estimate the current in the circuit

half a second after the switch is closed.

If the switch is closed when t = 0, we modeled

the current I at time t by the initial-value

dI/dt = 15 – 3I I(0) = 0

Estimate the current in the circuit

half a second after the switch is closed.

74.
EULER’S METHOD Example 4

We use Euler’s method with:

F(t, I) = 15 – 3I

t0 = 0

I0 = 0

Step size h = 0.1 second

We use Euler’s method with:

F(t, I) = 15 – 3I

t0 = 0

I0 = 0

Step size h = 0.1 second

75.
EULER’S METHOD Example 4

I1 = 0 + 0.1(15 – 3 · 0) = 1.5

I2 = 1.5 + 0.1(15 – 3 · 1.5) = 2.55

I3 = 2.55 + 0.1(15 – 3 · 2.55) = 3.285

I4 = 3.285 + 0.1(15 – 3 · 3.285) = 3.7995

I5 = 3.7995 + 0.1(15 – 3 · 3.7995) = 4.15365

The current after 0.5 seconds is: I(0.5) ≈ 4.16 A

I1 = 0 + 0.1(15 – 3 · 0) = 1.5

I2 = 1.5 + 0.1(15 – 3 · 1.5) = 2.55

I3 = 2.55 + 0.1(15 – 3 · 2.55) = 3.285

I4 = 3.285 + 0.1(15 – 3 · 3.285) = 3.7995

I5 = 3.7995 + 0.1(15 – 3 · 3.7995) = 4.15365

The current after 0.5 seconds is: I(0.5) ≈ 4.16 A