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5.2 Euler’s Method 139

3 3

2 2

1 1

–4 –2 y(x)=0 2 4 –4 –2 y(x)=0 2 4

–1 –1

–2 –2

–3 –3
FIGURE 5.2 Integral curves in the direction field FIGURE 5.3 Direction field and some integral curves
2

for y = y . for y = sin(xy).

SECTION 5.1 PROBLEMS

In each of Problems 1 through 6, draw a direction field for 2. y = x cos(2x) − y; y(1) = 0
the differential equation and some solution curves. Use it 2

3. y = y sin(x) − 3x ; y(0) = 1
to sketch the integral curve of the solution of the initial
x

value problem. 4. y = e − y; y(−2) = 1
2
5. y − y cos(x) = 1 − x ; y(2) = 2

1. y = sin(y); y(1) = π/2 6. y = 2y + 3; y(0) = 1

5.2 Euler’s Method
In this section, we present Euler’s method for generating approximate numerical values of the
solution of an initial value problem

y = f (x, y); y(x 0 ) = y 0
at selected points x 0 , x 1 = x 0 + h, x 2 = x 0 + 2h,···, and x n = x 0 + nh.Here n is a positive integer
(the number of iterations to be performed); and h is a (small) positive number called the step
size. This number h is the distance between successive points at which approximate values of the
solution are computed.
The idea behind Euler’s method is conceptually simple. First choose n and h.Weare given
y(x 0 ) = y 0 . Calculate f (x 0 , y 0 ) and draw the line having this slope through (x 0 , y 0 ). This line is
tangent to the solution at (x 0 , y 0 ). Move along this tangent line to the point (x 1 , y 1 ), where x 1 =
x 0 + h. Use this number y 1 as the approximation to y(x 1 ) at x 1 . This is illustrated in Figure 5.4.
We have some hope that this is a “good” approximation for h “small” because a tangent line at a
point fits the curve closely near that point. Note that (x 1 , y 1 ) is probably not on the integral curve
through (x 0 , y 0 ) but is on the tangent to this curve at (x 0 , y 0 ).

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October 14, 2010 14:19 THM/NEIL Page-139 27410_05_ch05_p137-144