Direction Fields and Euler’s Method

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Sharp Tutor Wed, Jan 19, 2022 11:10 PM UTC

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

1. 9
DIFFERENTIAL EQUATIONS

2. DIFFERENTIAL EQUATIONS
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)

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.

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?

6. Let’s think about what
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.

8. In particular, as the curve passes
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.

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.

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.

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.

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.

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).

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.

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.

17. DIRECTION FIELDS Example 1 a
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.

19. DIRECTION FIELDS Example 1 a
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.

21. DIRECTION FIELDS Example 1 b
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.

23. DIRECTION FIELDS BY COMPUTERS
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.

25. DIRECTION FIELDS
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

27. The circuit also contains:
 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).

29. One of Kirchhoff’s laws says that
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.

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.

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.

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?

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

35. ELECTRICALS Example 2 a
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→ ∞

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.

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

39. ELECTRICALS Example 2 d
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.

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

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.

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.

44. DIRECTION FIELDS
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

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).

47. DIRECTION FIELDS
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.

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.

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.

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.

52. EULER’S METHOD
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.

54. EULER’S METHOD
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.

56. EULER’S METHOD
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.

58. EULER’S METHOD
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.

60. EULER’S METHOD
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)

62. EULER’S METHOD
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

64. EULER’S METHOD Example 3
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

66. EULER’S METHOD Example 3
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.

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.

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.

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).

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.

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.

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.

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

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

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