Calculus AB and Calculus BC
CHAPTER 9 Differential Equations
C. EULER’S METHOD
BC ONLY
In §B we found solution curves to firstorder 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 yvalues 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 ycoordinate
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 ycoordinate 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 ycoordinate of P_{3} (x_{3}, y_{3}) we add Δy to y_{2}, where
Δy = (slope at P_{2}) ·
Then
y_{3} = y_{2} + Δy = 2.5 + 0.75 = 3.25,
P_{3} = (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
The table gives us the numerical solution of 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 passing through the point (1,0), which is y = 3 ln x.
FIGURE N9–6a
FIGURE N9–6b
We observe that, since y ″ for 3 ln x equals 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. = 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 

P_{0} 
0 
0 
0(0.1) = 0 
P_{1} 
0.1 
0 
(0.1)(0.1) = 0.01 
P_{2} 
0.2 
0.01 
(0.21)(0.1) = 0.021 
P_{3} 
0.3 
0.031 
(0.331)(0.1) = 0.033 
P_{4} 
0.4 
0.064 
(0.464)(0.1) = 0.046 
P_{5} 
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 for the domain solves the differential equation 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 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 ). Here, Euler’s method fails because it leaps blindly across the vertical asymptote at
Always pay attention to the domain of any particular solution.
BC ONLY
FIGURE N9–7