Calculus AB and Calculus BC

CHAPTER 9 Differential Equations

C. EULER’S METHOD

BC ONLY

In §B we found solution curves to first-order differential equations graphically, using slope fields. Here we will find solutions numerically, using Euler’s method to find points on solution curves.

When we use a slope field we start at an initial point, then move step by step so the slope segments are always tangent to the solution curve. With Euler’s method we again select a starting point; but now we calculate the slope at that point (from the given d.e.), use the initial point and that slope to locate a new point, use the new point and calculate the slope at it (again from the d.e.) to locate still another point, and so on. The method is illustrated in Example 6.

BC ONLY

EXAMPLE 6

Let Image Use Euler’s method to approximate the y-values with four steps, starting at point P0 (1, 0) and letting Δx = 0.5.

SOLUTION: The slope at P0 = (x0, y0) = (1,0) is Image To find the y-coordinate

of P1 (x1, y1), we add Δy to y0. Since ImageImage we estimate Image

Δy = (slope at P0) · Δx = 3 · (0.5) = 1.5.

Then

y1 = y0 + Δy = 0 + 1.5 = 1.5

and

P1 = (1.5,1.5).

To find the y-coordinate of P2 (x2, y2) we add Δy to y1, where

Δy = (slope at P1) · Image

Then

y2 = y1 + Δy = 1.5 + 1.0 = 2.5

and

P2 = (2.0,2.5).

To find the y-coordinate of P3 (x3, y3) we add Δy to y2, where

Δy = (slope at P2) · Image

Then

y3 = y2 + Δy = 2.5 + 0.75 = 3.25,

P3 = (2.5, 3.25),

and so on.

The table summarizes all the data, for the four steps specified, from x = 1 to x = 3:

TABLE FOR Image

Image

The table gives us the numerical solution of Image using Euler’s method. Figure N9–6a shows the graphical solution, which agrees with the data from the table, for x increasing from 1 to 3 by four steps with Δx equal to 0.5. Figure N9–6b shows this Euler graph and the particular solution of Image passing through the point (1,0), which is y = 3 ln x.

Image

FIGURE N9–6a

Image

FIGURE N9–6b

We observe that, since y for 3 ln x equals Image the true curve is concave down and below the Euler graph.

The last column in the table shows the true values (to three decimal places) of y. The Euler approximation for 3 ln 3 is 3.85; the true value is 3.296. The Euler approximation with four steps is not very good! However, see what happens as we increase the number n of steps:

n

EULER APPROXIMATION

ERROR

4

3.85

0.554

10

3.505

0.209

20

3.398

0.102

40

3.346

0.050

80

3.321

0.025

Doubling the number of steps cuts the error approximately in half.

EXAMPLE 7

Given the d.e. Image = x + y with initial condition y(0) = 0, use Euler’s method with Δx = 0.1 to estimate y when x = 0.5.

SOLUTION: Here are the relevant computations:

x

y

(SLOPE) · Δx = (x + y) · (0.1) = Δ y

P0

0

0

0(0.1) = 0

P1

0.1

0

(0.1)(0.1) = 0.01

P2

0.2

0.01

(0.21)(0.1) = 0.021

P3

0.3

0.031

(0.331)(0.1) = 0.033

P4

0.4

0.064

(0.464)(0.1) = 0.046

P5

0.5

0.110

A Caution: Euler’s method approximates the solution by substituting short line segments in place of the actual curve. It can be quite accurate when the step sizes are small, but only if the curve does not have discontinuities, cusps, or asymptotes.

For example, the reader may verify that the curve Image for the domain Image solves the differential equation Image with initial condition y = −1 when x = 2. The domain restriction is important. Recall that a particular solution must be differentiable on an interval containing the initial point. If we attempt to approximate this solution using Euler’s method with step size Δx = 1, the first step carries us to point (3, −3), beyond the discontinuity at Image and thus outside the domain of the solution. The accompanying graph (Figure N9–7) shows that this is nowhere near the solution curve with initial point y = 1 when x = 3 (and whose domain is Image). Here, Euler’s method fails because it leaps blindly across the vertical asymptote at Image

Always pay attention to the domain of any particular solution.

BC ONLY

Image

FIGURE N9–7