Higher Engineering Mathematics

NUMERICAL METHODS FOR FIRST ORDER DIFFERENTIAL EQUATIONS 465

I

Table 49.7

xy

2.0 1

2.2 1.2

2.4 1.421818

2.6 1.664849

2.8 1.928718

3.0 2.213187

49.4 An improved Euler method

In Euler’s method of Section 49.2, the gradient (y′) 0
atP(x 0 ,y 0 )in Fig. 49.9 across the whole intervalhis
used to obtain an approximate value ofy 1 at point
Q.QRin Fig. 49.9 is the resulting error in the result.

0 x 0 x 1 x

h

P

R

Q

y

y 0

Figure 49.9

In an improved Euler method, called theEuler-
Cauchy method, the gradient atP(x 0 ,y 0 )across half
the interval is used and then continues with a line
whose gradient approximates to the gradient of the
curve atx 1 , shown in Fig. 49.10.
LetyP 1 be the predicted value at pointRusing
Euler’s method, i.e. lengthRZ, where

yP 1 =y 0 +h(y′) 0 (3)

The error shown asQTin Fig. 49.10 is now less than
the errorQRused in the basic Euler method and the
calculated results will be of greater accuracy. The
corrected value,yC 1 in the improved Euler method
is given by:

yC 1 =y 0 +^12 h[(y′) 0 +f(x 1 ,yP 1 )] (4)

0 x 0 x 0 + 12 hx 1 x

P S

R

T

Q

y

h

z

Figure 49.10

The following worked problems demonstrate how

equations (3) and (4) are used in the Euler-Cauchy

method.

Problem 4. Apply the Euler-Cauchy method to

solve the differential equation

dy

dx

=y−x

in the range 0(0.1)0.5, given the initial condi-

tions that atx=0,y=2.

dy

dx

=y′=y−x

Since the initial conditions arex 0 =0 andy 0 = 2

then (y′) 0 = 2 − 0 =2. Interval h= 0 .1, hence

x 1 =x 0 +h= 0 + 0. 1 = 0 .1.

From equation (3),

yP 1 =y 0 +h(y′) 0 = 2 +(0.1)(2)= 2. 2

From equation (4),

yC 1 =y 0 +^12 h[(y′) 0 +f(x 1 ,yP 1 )]

=y 0 +^12 h[(y′) 0 +(yP 1 −x 1 )],

in this case

= 2 +^12 (0.1)[2+(2. 2 − 0 .1)]=2.205

(y′) 1 =yC 1 −x 1 = 2. 205 − 0. 1 = 2. 105

If we produce a table of values, as in Euler’s method,

we have so far determined lines 1 and 2 of Table 49.8.

The results in line 2 are now taken asx 0 ,y 0 and

(y′) 0 for the next interval and the process is repeated.