Barrons AP Calculus - David Bock

FIGURE N9–5

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 Use Euler’s method to approximate the y-values with four steps, starting at point P 0 (1,

0) and letting Δx = 0.5.

SOLUTION: The slope at P 0 = (x 0 , y 0 ) = (1,0) is To find the y-coordinate

of P 1 (x 1 , y 1 ), we add Δy to y 0. Since (^) ≈ we estimate
Δy = (slope at P 0 ) · Δx = 3 · (0.5) = 1.5.
Then
y 1 = y 0 + Δy = 0 + 1.5 = 1.5
and
P 1 = (1.5,1.5).
To find the y-coordinate of P 2 (x 2 , y 2 ) we add Δy to y 1 , where
Δy = (slope at P 1 ) ·
Then
y 2 = y 1 + Δy = 1.5 + 1.0 = 2.5
and
P 2 = (2.0,2.5).
To find the y-coordinate of P 3 (x 3 , y 3 ) we add Δy to y 2 , where
Δy = (slope at P 2 ) ·
Then