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Euler's

To approximate the solution of the initial-value problem

y'=f(t,y), a ::::t ::::b, yea) = ex,

at (N + 1) equally spaced numbers in the interval [a, b]:

INPUT endpoints a, b; integer N; initial condition ex.

OUTPUT approximation w to y at the (N + 1) values of t.

Step 1 Seth = (b - a)jN;t =a;w = ex;

OUTPUT (t, w).

Step 2 For i = 1, 2, . .. , N do Steps 3, 4.

Step3 Setw=w+h!(t,w); (ComputeWi.)t =a + ih. (Compute ti.)

Step 4 OUTPUT(t, w).

Step 5 STOP.

To interpret Euler's method geometrically,note that when Wi is a close approximationto y(ti), the assumption that the problemis well-posed implies that

f(ti, Wi) ~ y'(ti) = f(ti, y(ti».

The graph of the function highlighting y(ti) is shown in Figure 5.2(a). One step inEuler's method appears in Figure 5.2(b), and a series of steps appears in Figure 5.3.

Figure 5.2

..

y

y(tN) = y(b)

~

,.

{~,K..

Figure5.3

y

y' = f(t, y).y(a) = ex

y' = f(t, y),y(a) = ex

y(t 2)

y(tl)y(to) = a

WIa

II I I I

to=a tl t2!

tN = b

.-tto = a tIt 2 ... tN =b

(a) (b)

i

".',}

, y

y(b)WN

W2

WIa

to=a t1 t2 ... t,v=b

~

..

E;<. ~YY\l!e-Suppose Euler's method is used to approximate the solution to the initial-value problem

, ')

y =y-r+l, 0 < t < 2, yea) = 0.5,

with N = 10.Then h = 0.2, tj = 0.2i, Wo= 0.5, and

WHI = Wj + h(wj - t? + 1)= Wj + O.2[wj - 0.04i2+ 1]= 1.2wj - 0.008i2+ 0.2,

for i = 0, I, . .. ,9. The exact solution is yet) = (t + 1)2- 0.5et. Table 5.1 shows thecomparison between the approximate values at tj and the actual values. 8

Table 5.1 ti Wi Yi = yeti) Iv- - w-!. I I

0.0 0.5000000 0.5000000 0.00000000.2 0.8000000 0.8292986 0.02929860.4 1.1520000 1.2140877 0.06208770.6 1.5504000 1.6489406 0.09854060.8 1.9884800 2.1272295 0.13874951.0 2.4581760 2.6408591 0.18268311.2 2.9498112 3.1799415 0.23013031.4 3.4517734 3.7324000 0.28062661.6 3.9501281 4.2834838 0.33335571.8 4.4281538 4.8151763 0.38702252.0 4.8657845 5.3054720 0.4396874

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