1550078481-Ordinary_Differential_Equations__Roberts_

The Initial Value Problem y' = f(x, y); y(c) = d 85

Table 2.1 Euler's approximation to the IVP (7) y' = y + x; y(O) = 1

on [O, 1] with stepsize h = .2

f(xn, Yn) = Yn+1 =

n Xn Yn Yn +xn hj(xn, Yn) Yn + hj(xn, Yn)

0 .o 1.0 1.0 .2 1.2

1 .2 1.2 1.4. 28 1.48

2 .4 1.48 1.88 .376 1.856

3 .6 1.856 2.456 .4912 2.3472

4 .8 2.3472 3.1472 .62944 2.97664

5 1.0 2.97664

Table 2.2 Euler's approximation to the IVP (7) y' = y + x; y(O) = 1

on [O, 1] with stepsize h = .1

f(xn, Yn) = Yn+l =

n Xn Yn Yn +xn hj(xn, Yn) Yn + hj(xn, Yn)

0 .o 1.0 1.0 .1 1.1

1 .1 1.1 1.2. 12 1.22

2 .2 1.22 1.42 .142 1.362

3 .3 1.362 1.662 .1662 1.5282

4 .4 1.5282 1.9282 .19282 1.72102

5 .5 l. 72102 2.22102 .222102 1.94312

6 .6 1.94312 2.54312 .254312 2.19743

7 .7 2.19743 2.89743 .28974 3 2.48718

8 .8 2.48718 3.28718 .328718 2.81590

9 .9 2.81590 3.71590 .371590 3.18748

10 1.0 3.18748

Since the given differential equation is linear, we easily calculate the exact

solution of the IVP (7) to be ¢(x) = 2e x - x - l. So the value of the exact

solution at x = 1 is ¢(1) = 2e - 2 ~ 3.43656. Hence, the actual total error

due to discretization and round-off at x = 1 for h = .2 is 3.43656-2.97664 =

.45992, which is about one-third of the estimated error of 1.54645.